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Calculus Volume 1










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Table of ContentsPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 1: Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1 Review of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Basic Classes of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341.3 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 601.4 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761.5 Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94Chapter 2: Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1232.1 A Preview of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1242.2 The Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1352.3 The Limit Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1602.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1792.5 The Precise Definition of a Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194Chapter 3: Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2133.1 Defining the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2143.2 The Derivative as a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2323.3 Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2473.4 Derivatives as Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2663.5 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2773.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2873.7 Derivatives of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2993.8 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3093.9 Derivatives of Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . 319Chapter 4: Applications of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3414.1 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3424.2 Linear Approximations and Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3544.3 Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3664.4 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3794.5 Derivatives and the Shape of a Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3904.6 Limits at Infinity and Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4074.7 Applied Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4394.8 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4544.9 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4724.10 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485Chapter 5: Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5075.1 Approximating Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5085.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5295.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5495.4 Integration Formulas and the Net Change Theorem . . . . . . . . . . . . . . . . . . . . . . 5665.5 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5845.6 Integrals Involving Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . 5955.7 Integrals Resulting in Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . 608Chapter 6: Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6236.1 Areas between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6246.2 Determining Volumes by Slicing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6366.3 Volumes of Revolution: Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . 6566.4 Arc Length of a Curve and Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 6716.5 Physical Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6856.6 Moments and Centers of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7036.7 Integrals, Exponential Functions, and Logarithms . . . . . . . . . . . . . . . . . . . . . . . 7216.8 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7346.9 Calculus of the Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745Appendix A: Table of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763Appendix B: Table of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 769Appendix C: Review of Pre-Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 861




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PREFACE
Welcome to Calculus Volume 1, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
About OpenStax
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About OpenStax ResourcesCustomization
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Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus, in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book.
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Errata
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Format
You can access this textbook for free in web view or PDF through openstax.org, and for a low cost in print.
About Calculus Volume 1
Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 1 covers functions, limits, derivatives, and integration.
Coverage and Scope
Our Calculus Volume 1 textbook adheres to the scope and sequence of most general calculus courses nationwide. We have worked to make calculus interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of the three volumes of Calculus have been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from mathematics educators dedicated to the project.
Volume 1Chapter 1: Functions and Graphs


Preface 1




Chapter 2: Limits
Chapter 3: Derivatives
Chapter 4: Applications of Derivatives
Chapter 5: Integration
Chapter 6: Applications of Integration


Volume 2Chapter 1: Integration
Chapter 2: Applications of Integration
Chapter 3: Techniques of Integration
Chapter 4: Introduction to Differential Equations
Chapter 5: Sequences and Series
Chapter 6: Power Series
Chapter 7: Parametric Equations and Polar Coordinates


Volume 3Chapter 1: Parametric Equations and Polar Coordinates
Chapter 2: Vectors in Space
Chapter 3: Vector-Valued Functions
Chapter 4: Differentiation of Functions of Several Variables
Chapter 5: Multiple Integration
Chapter 6: Vector Calculus
Chapter 7: Second-Order Differential Equations


Pedagogical Foundation
Throughout Calculus Volume 1 you will find examples and exercises that present classical ideas and techniques as well asmodern applications and methods. Derivations and explanations are based on years of classroom experience on the partof long-time calculus professors, striving for a balance of clarity and rigor that has proven successful with their students.Motivational applications cover important topics in probability, biology, ecology, business, and economics, as well as areasof physics, chemistry, engineering, and computer science. Student Projects in each chapter give students opportunities toexplore interesting sidelights in pure and applied mathematics, from determining a safe distance between the grandstand andthe track at a Formula One racetrack, to calculating the center of mass of the Grand Canyon Skywalk or the terminal speedof a skydiver. Chapter Opening Applications pose problems that are solved later in the chapter, using the ideas covered inthat chapter. Problems include the hydraulic force against the Hoover Dam, and the comparison of relative intensity of twoearthquakes. Definitions, Rules, and Theorems are highlighted throughout the text, including over 60 Proofs of theorems.
Assessments That Reinforce Key Concepts
In-chapter Exampleswalk students through problems by posing a question, stepping out a solution, and then asking studentsto practice the skill with a “Checkpoint” question. The book also includes assessments at the end of each chapter sostudents can apply what they’ve learned through practice problems. Many exercises are marked with a [T] to indicate theyare suitable for solution by technology, including calculators or Computer Algebra Systems (CAS). Answers for selectedexercises are available in the Answer Key at the back of the book. The book also includes assessments at the end of eachchapter so students can apply what they’ve learned through practice problems.
Early or Late Transcendentals
Calculus Volume 1 is designed to accommodate both Early and Late Transcendental approaches to calculus. Exponentialand logarithmic functions are introduced informally in Chapter 1 and presented in more rigorous terms in Chapter 6.Differentiation and integration of these functions is covered in Chapters 3–5 for instructors who want to include them withother types of functions. These discussions, however, are in separate sections that can be skipped for instructors who preferto wait until the integral definitions are given before teaching the calculus derivations of exponentials and logarithms.
Comprehensive Art Program
Our art program is designed to enhance students’ understanding of concepts through clear and effective illustrations,diagrams, and photographs.


2 Preface


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Additional ResourcesStudent and Instructor Resources
We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructorsolution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which can be requestedon your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.
Partner Resources
OpenStax Partners are our allies in the mission to make high-quality learning materials affordable and accessible to studentsand instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partnerresources for your text, visit your book page on openstax.org.
About the AuthorsSenior Contributing Authors
Gilbert Strang, Massachusetts Institute of TechnologyDr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus onlinetextbook is one of eleven that he has published and is the basis from which our final product has been derived and updatedfor today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University.
Edwin “Jed” Herman, University of Wisconsin-Stevens PointDr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University ofWisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student researchmentor, is experienced in course development/design, and is also an avid board game designer and player.
Contributing Authors
Catherine Abbott, Keuka College


Preface 3




Nicoleta Virginia Bila, Fayetteville State UniversitySheri J. Boyd, Rollins CollegeJoyati Debnath, Winona State UniversityValeree Falduto, Palm Beach State CollegeJoseph Lakey, New Mexico State UniversityJulie Levandosky, Framingham State UniversityDavid McCune, William Jewell CollegeMichelle Merriweather, Bronxville High SchoolKirsten R. Messer, Colorado State University - PuebloAlfred K. Mulzet, Florida State College at JacksonvilleWilliam Radulovich (retired), Florida State College at JacksonvilleErica M. Rutter, Arizona State UniversityDavid Smith, University of the Virgin IslandsElaine A. Terry, Saint Joseph’s UniversityDavid Torain, Hampton University
Reviewers
Marwan A. Abu-Sawwa, Florida State College at JacksonvilleKenneth J. Bernard, Virginia State UniversityJohn Beyers, University of MarylandCharles Buehrle, Franklin & Marshall CollegeMatthew Cathey, Wofford CollegeMichael Cohen, Hofstra UniversityWilliam DeSalazar, Broward County School SystemMurray Eisenberg, University of Massachusetts AmherstKristyanna Erickson, Cecil CollegeTiernan Fogarty, Oregon Institute of TechnologyDavid French, Tidewater Community CollegeMarilyn Gloyer, Virginia Commonwealth UniversityShawna Haider, Salt Lake Community CollegeLance Hemlow, Raritan Valley Community CollegeJerry Jared, The Blue Ridge SchoolPeter Jipsen, Chapman UniversityDavid Johnson, Lehigh UniversityM.R. Khadivi, Jackson State UniversityRobert J. Krueger, Concordia UniversityTor A. Kwembe, Jackson State UniversityJean-Marie Magnier, Springfield Technical Community CollegeCheryl Chute Miller, SUNY PotsdamBagisa Mukherjee, Penn State University, Worthington Scranton CampusKasso Okoudjou, University of Maryland College ParkPeter Olszewski, Penn State Erie, The Behrend CollegeSteven Purtee, Valencia CollegeAlice Ramos, Bethel CollegeDoug Shaw, University of Northern IowaHussain Elalaoui-Talibi, Tuskegee UniversityJeffrey Taub, Maine Maritime AcademyWilliam Thistleton, SUNY Polytechnic InstituteA. David Trubatch, Montclair State UniversityCarmen Wright, Jackson State UniversityZhenbu Zhang, Jackson State University


4 Preface


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1 | FUNCTIONS ANDGRAPHS


Figure 1.1 A portion of the San Andreas Fault in California. Major faults like this are the sites of most of the strongestearthquakes ever recorded. (credit: modification of work by Robb Hannawacker, NPS)


Chapter Outline
1.1 Review of Functions
1.2 Basic Classes of Functions
1.3 Trigonometric Functions
1.4 Inverse Functions
1.5 Exponential and Logarithmic Functions


Introduction
In the past few years, major earthquakes have occurred in several countries around the world. In January 2010, anearthquake of magnitude 7.3 hit Haiti. A magnitude 9 earthquake shook northeastern Japan in March 2011. In April 2014,an 8.2-magnitude earthquake struck off the coast of northern Chile. What do these numbers mean? In particular, howdoes a magnitude 9 earthquake compare with an earthquake of magnitude 8.2? Or 7.3? Later in this chapter, we showhow logarithmic functions are used to compare the relative intensity of two earthquakes based on the magnitude of eachearthquake (see Example 1.39).
Calculus is the mathematics that describes changes in functions. In this chapter, we review all the functions necessaryto study calculus. We define polynomial, rational, trigonometric, exponential, and logarithmic functions. We review howto evaluate these functions, and we show the properties of their graphs. We provide examples of equations with termsinvolving these functions and illustrate the algebraic techniques necessary to solve them. In short, this chapter provides thefoundation for the material to come. It is essential to be familiar and comfortable with these ideas before proceeding to theformal introduction of calculus in the next chapter.


Chapter 1 | Functions and Graphs 5




1.1 | Review of Functions
Learning Objectives


1.1.1 Use functional notation to evaluate a function.
1.1.2 Determine the domain and range of a function.
1.1.3 Draw the graph of a function.
1.1.4 Find the zeros of a function.
1.1.5 Recognize a function from a table of values.
1.1.6 Make new functions from two or more given functions.
1.1.7 Describe the symmetry properties of a function.


In this section, we provide a formal definition of a function and examine several ways in which functions arerepresented—namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. Wealso define composition of functions and symmetry properties. Most of this material will be a review for you, but it servesas a handy reference to remind you of some of the algebraic techniques useful for working with functions.
Functions
Given two sets A and B, a set with elements that are ordered pairs (x, y), where x is an element of A and y is an
element of B, is a relation from A to B. A relation from A to B defines a relationship between those two sets. A
function is a special type of relation in which each element of the first set is related to exactly one element of the secondset. The element of the first set is called the input; the element of the second set is called the output. Functions are used allthe time in mathematics to describe relationships between two sets. For any function, when we know the input, the output isdetermined, so we say that the output is a function of the input. For example, the area of a square is determined by its sidelength, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in theair can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function ofthe weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology tostudy them.
Definition
A function f consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output. The
set of inputs is called the domain of the function. The set of outputs is called the range of the function.


For example, consider the function f , where the domain is the set of all real numbers and the rule is to square the input.
Then, the input x = 3 is assigned to the output 32 = 9. Since every nonnegative real number has a real-value square root,
every nonnegative number is an element of the range of this function. Since there is no real number with a square that isnegative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative realnumbers.
For a general function f with domain D, we often use x to denote the input and y to denote the output associated with
x. When doing so, we refer to x as the independent variable and y as the dependent variable, because it depends on x.
Using function notation, we write y = f (x), and we read this equation as “y equals f of x.” For the squaring function
described earlier, we write f (x) = x2.
The concept of a function can be visualized using Figure 1.2, Figure 1.3, and Figure 1.4.


6 Chapter 1 | Functions and Graphs


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Figure 1.2 A function can be visualized as an input/outputdevice.


Figure 1.3 A function maps every element in the domain toexactly one element in the range. Although each input can besent to only one output, two different inputs can be sent to thesame output.


Figure 1.4 In this case, a graph of a function f has a domain
of {1, 2, 3} and a range of {1, 2}. The independent variable
is x and the dependent variable is y.


Visit this applet link (http://www.openstaxcollege.org/l/grapherrors) to see more about graphs offunctions.


We can also visualize a function by plotting points (x, y) in the coordinate plane where y = f (x). The graph of a function
is the set of all these points. For example, consider the function f , where the domain is the set D = {1, 2, 3} and the
rule is f (x) = 3 − x. In Figure 1.5, we plot a graph of this function.


Chapter 1 | Functions and Graphs 7




Figure 1.5 Here we see a graph of the function f with
domain {1, 2, 3} and rule f (x) = 3 − x. The graph consists
of the points (x, f (x)) for all x in the domain.


Every function has a domain. However, sometimes a function is described by an equation, as in f (x) = x2, with no
specific domain given. In this case, the domain is taken to be the set of all real numbers x for which f (x) is a real number.
For example, since any real number can be squared, if no other domain is specified, we consider the domain of f (x) = x2
to be the set of all real numbers. On the other hand, the square root function f (x) = x only gives a real output if x is
nonnegative. Therefore, the domain of the function f (x) = x is the set of nonnegative real numbers, sometimes called the
natural domain.
For the functions f (x) = x2 and f (x) = x, the domains are sets with an infinite number of elements. Clearly we cannot
list all these elements. When describing a set with an infinite number of elements, it is often helpful to use set-builder orinterval notation. When using set-builder notation to describe a subset of all real numbers, denoted ℝ, we write





⎨x|x has some property⎫⎭⎬.


We read this as the set of real numbers x such that x has some property. For example, if we were interested in the set of
real numbers that are greater than one but less than five, we could denote this set using set-builder notation by writing


{x|1 < x < 5}.


A set such as this, which contains all numbers greater than a and less than b, can also be denoted using the interval
notation (a, b). Therefore,


(1, 5) = ⎧⎩⎨x|1 < x < 5⎫⎭⎬.


The numbers 1 and 5 are called the endpoints of this set. If we want to consider the set that includes the endpoints, we
would denote this set by writing


[1, 5] = {x|1 ≤ x ≤ 5}.


We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegativereal numbers, we would use the set-builder notation
{x|0 ≤ x}.


The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would usethe symbol ∞, which refers to positive infinity, and we would write the set as
[0, ∞) = {x|0 ≤ x}.


It is important to note that ∞ is not a real number. It is used symbolically here to indicate that this set includes all real
numbers greater than or equal to zero. Similarly, if we wanted to describe the set of all nonpositive numbers, we could write


(−∞, 0] = {x|x ≤ 0}.


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1.1


Here, the notation −∞ refers to negative infinity, and it indicates that we are including all numbers less than or equal to
zero, no matter how small. The set


(−∞, ∞) = ⎧⎩⎨x|x is any real number⎫⎭⎬


refers to the set of all real numbers.
Some functions are defined using different equations for different parts of their domain. These types of functions are knownas piecewise-defined functions. For example, suppose we want to define a function f with a domain that is the set of all
real numbers such that f (x) = 3x + 1 for x ≥ 2 and f (x) = x2 for x < 2. We denote this function by writing


f (x) =




3x + 1 x ≥ 2


x2 x < 2
.


When evaluating this function for an input x, the equation to use depends on whether x ≥ 2 or x < 2. For example,
since 5 > 2, we use the fact that f (x) = 3x + 1 for x ≥ 2 and see that f (5) = 3(5) + 1 = 16. On the other hand, for
x = −1, we use the fact that f (x) = x2 for x < 2 and see that f (−1) = 1.
Example 1.1
Evaluating Functions
For the function f (x) = 3x2 + 2x − 1, evaluate


a. f (−2)
b. f ( 2)
c. f (a + h)


Solution
Substitute the given value for x in the formula for f (x).


a. f (−2) = 3(−2)2 + 2(−2) − 1 = 12 − 4 − 1 = 7
b. f ( 2) = 3( 2)2 + 2 2 − 1 = 6 + 2 2 − 1 = 5 + 2 2
c. f (a + h) = 3(a + h)2 + 2(a + h) − 1 = 3⎛⎝a2 + 2ah + h2⎞⎠+ 2a + 2h − 1


= 3a2 + 6ah + 3h2 + 2a + 2h − 1


For f (x) = x2 − 3x + 5, evaluate f (1) and f (a + h).


Example 1.2
Finding Domain and Range
For each of the following functions, determine the i. domain and ii. range.


Chapter 1 | Functions and Graphs 9




a. f (x) = (x − 4)2 + 5
b. f (x) = 3x + 2 − 1
c. f (x) = 3


x − 2


Solution
a. Consider f (x) = (x − 4)2 + 5.


i. Since f (x) = (x − 4)2 + 5 is a real number for any real number x, the domain of f is the
interval (−∞, ∞).


ii. Since (x − 4)2 ≥ 0, we know f (x) = (x − 4)2 + 5 ≥ 5. Therefore, the range must be a subset
of ⎧⎩⎨y|y ≥ 5⎫⎭⎬. To show that every element in this set is in the range, we need to show that for a
given y in that set, there is a real number x such that f (x) = (x − 4)2 + 5 = y. Solving this
equation for x, we see that we need x such that


(x − 4)2 = y − 5.


This equation is satisfied as long as there exists a real number x such that
x − 4 = ± y − 5.


Since y ≥ 5, the square root is well-defined. We conclude that for x = 4 ± y − 5, f (x) = y,
and therefore the range is ⎧⎩⎨y|y ≥ 5⎫⎭⎬.


b. Consider f (x) = 3x + 2 − 1.
i. To find the domain of f , we need the expression 3x + 2 ≥ 0. Solving this inequality, we


conclude that the domain is {x|x ≥ −2/3}.
ii. To find the range of f , we note that since 3x + 2 ≥ 0, f (x) = 3x + 2 − 1 ≥ −1. Therefore,


the range of f must be a subset of the set ⎧⎩⎨y|y ≥ −1⎫⎭⎬. To show that every element in this set is
in the range of f , we need to show that for all y in this set, there exists a real number x in the
domain such that f (x) = y. Let y ≥ −1. Then, f (x) = y if and only if


3x + 2 − 1 = y.


Solving this equation for x, we see that x must solve the equation
3x + 2 = y + 1.


Since y ≥ −1, such an x could exist. Squaring both sides of this equation, we have
3x + 2 = (y + 1)2.


Therefore, we need
3x = (y + 1)2 − 2,


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1.2


which implies
x = 1


3

⎝y + 1⎞⎠2 − 23


.


We just need to verify that x is in the domain of f . Since the domain of f consists of all real
numbers greater than or equal to −2/3, and


1
3

⎝y + 1⎞⎠2 − 23


≥ − 2
3
,


there does exist an x in the domain of f . We conclude that the range of f is ⎧⎩⎨y|y ≥ −1⎫⎭⎬.
c. Consider f (x) = 3/(x − 2).


i. Since 3/(x − 2) is defined when the denominator is nonzero, the domain is {x|x ≠ 2}.
ii. To find the range of f , we need to find the values of y such that there exists a real number x


in the domain with the property that
3


x − 2
= y.


Solving this equation for x, we find that
x = 3y + 2.


Therefore, as long as y ≠ 0, there exists a real number x in the domain such that f (x) = y.
Thus, the range is ⎧⎩⎨y|y ≠ 0⎫⎭⎬.


Find the domain and range for f (x) = 4 − 2x + 5.


Representing Functions
Typically, a function is represented using one or more of the following tools:


• A table
• A graph
• A formula


We can identify a function in each form, but we can also use them together. For instance, we can plot on a graph the valuesfrom a table or create a table from a formula.
Tables
Functions described using a table of values arise frequently in real-world applications. Consider the following simpleexample. We can describe temperature on a given day as a function of time of day. Suppose we record the temperature everyhour for a 24-hour period starting at midnight. We let our input variable x be the time after midnight, measured in hours,
and the output variable y be the temperature x hours after midnight, measured in degrees Fahrenheit. We record our data
in Table 1.1.


Chapter 1 | Functions and Graphs 11




Hours after Midnight Temperature (°F) Hours after Midnight Temperature (°F)
0 58 12 84
1 54 13 85
2 53 14 85
3 52 15 83
4 52 16 82
5 55 17 80
6 60 18 77
7 64 19 74
8 72 20 69
9 75 21 65
10 78 22 60
11 80 23 58


Table 1.1 Temperature as a Function of Time of Day
We can see from the table that temperature is a function of time, and the temperature decreases, then increases, and thendecreases again. However, we cannot get a clear picture of the behavior of the function without graphing it.
Graphs
Given a function f described by a table, we can provide a visual picture of the function in the form of a graph. Graphing
the temperatures listed in Table 1.1 can give us a better idea of their fluctuation throughout the day. Figure 1.6 shows theplot of the temperature function.


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Figure 1.6 The graph of the data from Table 1.1 showstemperature as a function of time.


From the points plotted on the graph in Figure 1.6, we can visualize the general shape of the graph. It is often usefulto connect the dots in the graph, which represent the data from the table. In this example, although we cannot make anydefinitive conclusion regarding what the temperature was at any time for which the temperature was not recorded, giventhe number of data points collected and the pattern in these points, it is reasonable to suspect that the temperatures at othertimes followed a similar pattern, as we can see in Figure 1.7.


Figure 1.7 Connecting the dots in Figure 1.6 shows thegeneral pattern of the data.
Algebraic Formulas
Sometimes we are not given the values of a function in table form, rather we are given the values in an explicit formula.
Formulas arise in many applications. For example, the area of a circle of radius r is given by the formula A(r) = πr2.
When an object is thrown upward from the ground with an initial velocity v0 ft/s, its height above the ground from the
time it is thrown until it hits the ground is given by the formula s(t) = −16t2 + v0 t. When P dollars are invested in an
account at an annual interest rate r compounded continuously, the amount of money after t years is given by the formula
A(t) = Pert. Algebraic formulas are important tools to calculate function values. Often we also represent these functions
visually in graph form.


Chapter 1 | Functions and Graphs 13




Given an algebraic formula for a function f , the graph of f is the set of points ⎛⎝x, f (x)⎞⎠, where x is in the domain of
f and f (x) is in the range. To graph a function given by a formula, it is helpful to begin by using the formula to create
a table of inputs and outputs. If the domain of f consists of an infinite number of values, we cannot list all of them, but
because listing some of the inputs and outputs can be very useful, it is often a good way to begin.
When creating a table of inputs and outputs, we typically check to determine whether zero is an output. Those values of
x where f (x) = 0 are called the zeros of a function. For example, the zeros of f (x) = x2 − 4 are x = ± 2. The zeros
determine where the graph of f intersects the x -axis, which gives us more information about the shape of the graph of
the function. The graph of a function may never intersect the x-axis, or it may intersect multiple (or even infinitely many)times.
Another point of interest is the y -intercept, if it exists. The y -intercept is given by ⎛⎝0, f (0)⎞⎠.
Since a function has exactly one output for each input, the graph of a function can have, at most, one y -intercept. If x = 0
is in the domain of a function f , then f has exactly one y -intercept. If x = 0 is not in the domain of f , then f has
no y -intercept. Similarly, for any real number c, if c is in the domain of f , there is exactly one output f (c), and the
line x = c intersects the graph of f exactly once. On the other hand, if c is not in the domain of f , f (c) is not defined
and the line x = c does not intersect the graph of f . This property is summarized in the vertical line test.


Rule: Vertical Line Test
Given a function f , every vertical line that may be drawn intersects the graph of f no more than once. If any vertical
line intersects a set of points more than once, the set of points does not represent a function.


We can use this test to determine whether a set of plotted points represents the graph of a function (Figure 1.8).


Figure 1.8 (a) The set of plotted points represents the graph ofa function because every vertical line intersects the set of points,at most, once. (b) The set of plotted points does not represent thegraph of a function because some vertical lines intersect the setof points more than once.


Example 1.3


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Finding Zeros and y -Intercepts of a Function
Consider the function f (x) = −4x + 2.


a. Find all zeros of f .
b. Find the y -intercept (if any).
c. Sketch a graph of f .


Solution
a. To find the zeros, solve f (x) = −4x + 2 = 0. We discover that f has one zero at x = 1/2.
b. The y -intercept is given by ⎛⎝0, f (0)⎞⎠ = (0, 2).
c. Given that f is a linear function of the form f (x) = mx + b that passes through the points (1/2, 0) and


(0, 2), we can sketch the graph of f (Figure 1.9).


Figure 1.9 The function f (x) = −4x + 2 is a line with
x -intercept (1/2, 0) and y -intercept (0, 2).


Example 1.4
Using Zeros and y -Intercepts to Sketch a Graph
Consider the function f (x) = x + 3 + 1.


a. Find all zeros of f .
b. Find the y -intercept (if any).
c. Sketch a graph of f .


Solution
a. To find the zeros, solve x + 3 + 1 = 0. This equation implies x + 3 = −1. Since x + 3 ≥ 0 for all


Chapter 1 | Functions and Graphs 15




1.3


x, this equation has no solutions, and therefore f has no zeros.
b. The y -intercept is given by ⎛⎝0, f (0)⎞⎠ = (0, 3 + 1).
c. To graph this function, we make a table of values. Since we need x + 3 ≥ 0, we need to choose values


of x ≥ −3. We choose values that make the square-root function easy to evaluate.
x −3 −2 1


f (x) 1 2 3


Table 1.2
Making use of the table and knowing that, since the function is a square root, the graph of f should be similar to
the graph of y = x, we sketch the graph (Figure 1.10).


Figure 1.10 The graph of f (x) = x + 3 + 1 has a
y -intercept but no x -intercepts.


Find the zeros of f (x) = x3 − 5x2 + 6x.


Example 1.5
Finding the Height of a Free-Falling Object
If a ball is dropped from a height of 100 ft, its height s at time t is given by the function s(t) = −16t2 + 100,
where s is measured in feet and t is measured in seconds. The domain is restricted to the interval [0, c], where
t = 0 is the time when the ball is dropped and t = c is the time when the ball hits the ground.


a. Create a table showing the height s(t) when t = 0, 0.5, 1, 1.5, 2, and 2.5. Using the data from the
table, determine the domain for this function. That is, find the time c when the ball hits the ground.


b. Sketch a graph of s.


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Solution
a.


t 0 0.5 1 1.5 2 2.5


s(t) 100 96 84 64 36 0


Table 1.3Height s as a Function of Time t
Since the ball hits the ground when t = 2.5, the domain of this function is the interval [0, 2.5].


b.


Note that for this function and the function f (x) = −4x + 2 graphed in Figure 1.9, the values of f (x) are getting
smaller as x is getting larger. A function with this property is said to be decreasing. On the other hand, for the function
f (x) = x + 3 + 1 graphed in Figure 1.10, the values of f (x) are getting larger as the values of x are getting larger.
A function with this property is said to be increasing. It is important to note, however, that a function can be increasing onsome interval or intervals and decreasing over a different interval or intervals. For example, using our temperature functionin Figure 1.6, we can see that the function is decreasing on the interval (0, 4), increasing on the interval (4, 14), and
then decreasing on the interval (14, 23). We make the idea of a function increasing or decreasing over a particular interval
more precise in the next definition.
Definition
We say that a function f is increasing on the interval I if for all x1, x2 ∈ I,


f (x1) ≤ f (x2) when x1 < x2.


We say f is strictly increasing on the interval I if for all x1, x2 ∈ I,


Chapter 1 | Functions and Graphs 17




f (x1) < f (x2) when x1 < x2.


We say that a function f is decreasing on the interval I if for all x1, x2 ∈ I,
f (x1) ≥ f (x2) if x1 < x2.


We say that a function f is strictly decreasing on the interval I if for all x1, x2 ∈ I,
f (x1) > f (x2) if x1 < x2.


For example, the function f (x) = 3x is increasing on the interval (−∞, ∞) because 3x1 < 3x2 whenever x1 < x2.
On the other hand, the function f (x) = −x3 is decreasing on the interval (−∞, ∞) because −x13 > − x23 whenever
x1 < x2 (Figure 1.11).


Figure 1.11 (a) The function f (x) = 3x is increasing on the interval (−∞, ∞). (b) The
function f (x) = −x3 is decreasing on the interval (−∞, ∞).


Combining Functions
Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when wecombine functions in different ways, using basic mathematical operations to create new functions. For example, if the costfor a company to manufacture x items is described by the function C(x) and the revenue created by the sale of x items is
described by the function R(x), then the profit on the manufacture and sale of x items is defined as P(x) = R(x) − C(x).
Using the difference between two functions, we created a new function.
Alternatively, we can create a new function by composing two functions. For example, given the functions f (x) = x2 and
g(x) = 3x + 1, the composite function f ∘g is defined such that



⎝ f ∘g⎞⎠(x) = f ⎛⎝g(x)⎞⎠ = ⎛⎝g(x)⎞⎠2 = (3x + 1)2.


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1.4


The composite function g ∘ f is defined such that

⎝g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠ = 3 f (x) + 1 = 3x2 + 1.


Note that these two new functions are different from each other.
Combining Functions with Mathematical Operators
To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Giventwo functions f and g, we can define four new functions:



⎝ f + g⎞⎠(x) = f (x) + g(x) Sum

⎝ f − g⎞⎠(x) = f (x) − g(x) Diffe ence

⎝ f · g⎞⎠(x) = f (x)g(x) Product




f
g

⎠(x) =


f (x)
g(x)


for g(x) ≠ 0 Quotient


Example 1.6
Combining Functions Using Mathematical Operations
Given the functions f (x) = 2x − 3 and g(x) = x2 − 1, find each of the following functions and state its
domain.


a. ( f + g)(x)
b. ( f − g)(x)
c. ( f · g)(x)
d. ⎛⎝ fg⎞⎠(x)


Solution
a. ⎛⎝ f + g⎞⎠(x) = (2x − 3) + (x2 − 1) = x2 + 2x − 4. The domain of this function is the interval (−∞, ∞).
b. ⎛⎝ f − g⎞⎠(x) = (2x − 3) − (x2 − 1) = −x2 + 2x − 2. The domain of this function is the interval


(−∞, ∞).


c. ⎛⎝ f · g⎞⎠(x) = (2x − 3)(x2 − 1) = 2x3 − 3x2 − 2x + 3. The domain of this function is the interval
(−∞, ∞).


d. ⎛⎝ fg⎞⎠(x) = 2x − 3x2 − 1. The domain of this function is {x|x ≠ ±1}.


For f (x) = x2 + 3 and g(x) = 2x − 5, find ⎛⎝ f /g⎞⎠(x) and state its domain.


Function Composition
When we compose functions, we take a function of a function. For example, suppose the temperature T on a given day is
described as a function of time t (measured in hours after midnight) as in Table 1.1. Suppose the cost C, to heat or cool
a building for 1 hour, can be described as a function of the temperature T . Combining these two functions, we can describe


Chapter 1 | Functions and Graphs 19




the cost of heating or cooling a building as a function of time by evaluating C⎛⎝T(t)⎞⎠. We have defined a new function,
denoted C ∘T , which is defined such that (C ∘T)(t) = C(T(t)) for all t in the domain of T . This new function is called
a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makessense to define this new function (C ∘T)(t). It does not make sense to consider (T ∘C)(t), because temperature is not a
function of cost.
Definition
Consider the function f with domain A and range B, and the function g with domain D and range E. If B is a
subset of D, then the composite function (g ∘ f )(x) is the function with domain A such that


(1.1)⎛⎝g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠.
A composite function g ∘ f can be viewed in two steps. First, the function f maps each input x in the domain of f to
its output f (x) in the range of f . Second, since the range of f is a subset of the domain of g, the output f (x) is an
element in the domain of g, and therefore it is mapped to an output g⎛⎝ f (x)⎞⎠ in the range of g. In Figure 1.12, we see a
visual image of a composite function.


Figure 1.12 For the composite function g ∘ f , we have

⎝g ∘ f ⎞⎠(1) = 4, ⎛⎝g ∘ f ⎞⎠(2) = 5, and ⎛⎝g ∘ f ⎞⎠(3) = 4.


Example 1.7
Compositions of Functions Defined by Formulas
Consider the functions f (x) = x2 + 1 and g(x) = 1/x.


a. Find (g ∘ f )(x) and state its domain and range.
b. Evaluate (g ∘ f )(4), (g ∘ f )(−1/2).
c. Find ( f ∘g)(x) and state its domain and range.
d. Evaluate ( f ∘g)(4), ( f ∘g)(−1/2).


Solution
a. We can find the formula for (g ∘ f )(x) in two different ways. We could write


(g ∘ f )(x) = g( f (x)) = g(x2 + 1) = 1
x2 + 1


.


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Alternatively, we could write
(g ∘ f )(x) = g⎛⎝ f (x)⎞⎠ = 1f (x)


= 1
x2 + 1


.


Since x2 + 1 ≠ 0 for all real numbers x, the domain of (g ∘ f )(x) is the set of all real numbers. Since
0 < 1/(x2 + 1) ≤ 1, the range is, at most, the interval (0, 1]. To show that the range is this entire
interval, we let y = 1/(x2 + 1) and solve this equation for x to show that for all y in the interval
(0, 1], there exists a real number x such that y = 1/(x2 + 1). Solving this equation for x, we see
that x2 + 1 = 1/y, which implies that


x = ± 1y − 1.


If y is in the interval (0, 1], the expression under the radical is nonnegative, and therefore there exists
a real number x such that 1/(x2 + 1) = y. We conclude that the range of g ∘ f is the interval (0, 1].


b. (g ∘ f )(4) = g( f (4)) = g(42 + 1) = g(17) = 1
17


(g ∘ f )⎛⎝−
1
2

⎠ = g

⎝ f

⎝−


1
2



⎠ = g



⎜⎛⎝−


1
2



2
+ 1



⎟ = g⎛⎝


5
4

⎠ =


4
5


c. We can find a formula for ( f ∘g)(x) in two ways. First, we could write
( f ∘g)(x) = f (g(x)) = f ⎛⎝


1
x

⎠ =


1
x



2
+ 1.


Alternatively, we could write
( f ∘g)(x) = f (g(x)) = (g(x))2 + 1 = ⎛⎝


1
x



2
+ 1.


The domain of f ∘g is the set of all real numbers x such that x ≠ 0. To find the range of f , we need
to find all values y for which there exists a real number x ≠ 0 such that




1
x



2
+ 1 = y.


Solving this equation for x, we see that we need x to satisfy


1
x



2
= y − 1,


which simplifies to
1
x = ± y − 1.


Finally, we obtain


Chapter 1 | Functions and Graphs 21




1.5


x = ± 1
y − 1


.


Since 1/ y − 1 is a real number if and only if y > 1, the range of f is the set ⎧⎩⎨y|y ≥ 1⎫⎭⎬.
d. ( f ∘g)(4) = f (g(4)) = f ⎛⎝14⎞⎠ = ⎛⎝14⎞⎠


2
+ 1 = 17


16


( f ∘g)⎛⎝−
1
2

⎠ = f

⎝g

⎝−


1
2



⎠ = f (−2) = (−2)


2 + 1 = 5


In Example 1.7, we can see that ⎛⎝ f ∘g⎞⎠(x) ≠ ⎛⎝g ∘ f ⎞⎠(x). This tells us, in general terms, that the order in which we compose
functions matters.


Let f (x) = 2 − 5x. Let g(x) = x. Find ⎛⎝ f ∘g⎞⎠(x).


Example 1.8
Composition of Functions Defined by Tables
Consider the functions f and g described by Table 1.4 and Table 1.5.


x −3 −2 −1 0 1 2 3 4


f (x) 0 4 2 4 −2 0 −2 4
Table 1.4


x −4 −2 0 2 4


g(x) 1 0 3 0 5
Table 1.5


a. Evaluate (g ∘ f )(3), ⎛⎝g ∘ f ⎞⎠(0).
b. State the domain and range of ⎛⎝g ∘ f ⎞⎠(x).
c. Evaluate ( f ∘ f )(3), ⎛⎝ f ∘ f ⎞⎠(1).
d. State the domain and range of ⎛⎝ f ∘ f ⎞⎠(x).


Solution


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1.6


a. ⎛⎝g ∘ f ⎞⎠(3) = g⎛⎝ f (3)⎞⎠ = g(−2) = 0
(g ∘ f )(0) = g(4) = 5


b. The domain of g ∘ f is the set {−3, −2, −1, 0, 1, 2, 3, 4}. Since the range of f is the set
{−2, 0, 2, 4}, the range of g ∘ f is the set {0, 3, 5}.


c. ⎛⎝ f ∘ f ⎞⎠(3) = f ⎛⎝ f (3)⎞⎠ = f (−2) = 4
( f ∘ f )(1) = f ( f (1)) = f (−2) = 4


d. The domain of f ∘ f is the set {−3, −2, −1, 0, 1, 2, 3, 4}. Since the range of f is the set
{−2, 0, 2, 4}, the range of f ∘ f is the set {0, 4}.


Example 1.9
Application Involving a Composite Function
A store is advertising a sale of 20% off all merchandise. Caroline has a coupon that entitles her to an additional
15% off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of
x dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by
using a composite function.
Solution
Since the sale price is 20% off the original price, if an item is x dollars, its sale price is given by f (x) = 0.80x.
Since the coupon entitles an individual to 15% off the price of any item, if an item is y dollars, the price, after
applying the coupon, is given by g(y) = 0.85y. Therefore, if the price is originally x dollars, its sale price will
be f (x) = 0.80x and then its final price after the coupon will be g( f (x)) = 0.85(0.80x) = 0.68x.


If items are on sale for 10% off their original price, and a customer has a coupon for an additional 30%
off, what will be the final price for an item that is originally x dollars, after applying the coupon to the sale
price?


Symmetry of Functions
The graphs of certain functions have symmetry properties that help us understand the function and the shape of its graph.
For example, consider the function f (x) = x4 − 2x2 − 3 shown in Figure 1.13(a). If we take the part of the curve that
lies to the right of the y-axis and flip it over the y-axis, it lays exactly on top of the curve to the left of the y-axis. In this
case, we say the function has symmetry about the y-axis. On the other hand, consider the function f (x) = x3 − 4x shown
in Figure 1.13(b). If we take the graph and rotate it 180° about the origin, the new graph will look exactly the same. In
this case, we say the function has symmetry about the origin.


Chapter 1 | Functions and Graphs 23




Figure 1.13 (a) A graph that is symmetric about the y -axis. (b) A graph that is symmetric
about the origin.


If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But withouta graph, how can we determine algebraically whether a function f has symmetry? Looking at Figure 1.14 again, we see
that since f is symmetric about the y -axis, if the point (x, y) is on the graph, the point (−x, y) is on the graph. In other
words, f (−x) = f (x). If a function f has this property, we say f is an even function, which has symmetry about the
y-axis. For example, f (x) = x2 is even because


f (−x) = (−x)2 = x2 = f (x).


In contrast, looking at Figure 1.14 again, if a function f is symmetric about the origin, then whenever the point (x, y) is
on the graph, the point (−x, −y) is also on the graph. In other words, f (−x) = − f (x). If f has this property, we say f
is an odd function, which has symmetry about the origin. For example, f (x) = x3 is odd because


f (−x) = (−x)3 = −x3 = − f (x).


Definition
If f (x) = f (−x) for all x in the domain of f , then f is an even function. An even function is symmetric about the
y-axis.
If f (−x) = − f (x) for all x in the domain of f , then f is an odd function. An odd function is symmetric about the
origin.


Example 1.10
Even and Odd Functions
Determine whether each of the following functions is even, odd, or neither.


a. f (x) = −5x4 + 7x2 − 2


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1.7


b. f (x) = 2x5 − 4x + 5
c. f (x) = 3x


x2 + 1


Solution
To determine whether a function is even or odd, we evaluate f (−x) and compare it to f(x) and − f (x).


a. f (−x) = −5(−x)4 + 7(−x)2 − 2 = −5x4 + 7x2 − 2 = f (x). Therefore, f is even.
b. f (−x) = 2(−x)5 − 4(−x) + 5 = −2x5 + 4x + 5. Now, f (−x) ≠ f (x). Furthermore, noting that


− f (x) = −2x5 + 4x − 5, we see that f (−x) ≠ − f (x). Therefore, f is neither even nor odd.
c. f (−x) = 3(−x)/((−x)2 + 1} = −3x/(x2 + 1) = −[3x/(x2 + 1)] = − f (x). Therefore, f is odd.


Determine whether f (x) = 4x3 − 5x is even, odd, or neither.


One symmetric function that arises frequently is the absolute value function, written as |x|. The absolute value function is
defined as


(1.2)
f (x) =






−x, x < 0
x, x ≥ 0


.


Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute valuefunction, we see that if x < 0, then |x| = −x > 0, and if x > 0, then |x| = x > 0. However, for x = 0, |x| = 0.
Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if x = 0, the output |x| = 0. We
conclude that the range of the absolute value function is ⎧⎩⎨y|y ≥ 0⎫⎭⎬. In Figure 1.14, we see that the absolute value function
is symmetric about the y-axis and is therefore an even function.


Figure 1.14 The graph of f (x) = |x| is symmetric about the
y -axis.


Chapter 1 | Functions and Graphs 25




1.8


Example 1.11
Working with the Absolute Value Function
Find the domain and range of the function f (x) = 2|x − 3| + 4.
Solution
Since the absolute value function is defined for all real numbers, the domain of this function is (−∞, ∞). Since
|x − 3| ≥ 0 for all x, the function f (x) = 2|x − 3| + 4 ≥ 4. Therefore, the range is, at most, the set ⎧⎩⎨y|y ≥ 4⎫⎭⎬.
To see that the range is, in fact, this whole set, we need to show that for y ≥ 4 there exists a real number x such
that


2|x − 3| + 4 = y.


A real number x satisfies this equation as long as
|x − 3| = 1


2
(y − 4).


Since y ≥ 4, we know y − 4 ≥ 0, and thus the right-hand side of the equation is nonnegative, so it is possible
that there is a solution. Furthermore,


|x − 3| =




−(x − 3) if x < 3
x − 3 if x ≥ 3


.


Therefore, we see there are two solutions:
x = ± 1


2
(y − 4) + 3.


The range of this function is ⎧⎩⎨y|y ≥ 4⎫⎭⎬.


For the function f (x) = |x + 2| − 4, find the domain and range.


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1.1 EXERCISES
For the following exercises, (a) determine the domain andthe range of each relation, and (b) state whether the relationis a function.
1.


x y x y


−3 9 1 1
−2 4 2 4
−1 1 3 9
0 0


2.
x y x y


−3 −2 1 1
−2 −8 2 8
−1 −1 3 −2
0 0


3.
x y x y


1 −3 1 1
2 −2 2 2
3 −1 3 3
0 0


4.
x y x y


1 1 5 1
2 1 6 1
3 1 7 1
4 1


5.
x y x y


3 3 15 1
5 2 21 2
8 1 33 3
10 0


6.
x y x y


−7 11 1 −2
−2 5 3 4
−2 1 6 11
0 −1


For the following exercises, find the values for eachfunction, if they exist, then simplify.
a. f (0) b. f (1) c. f (3) d. f (−x) e. f (a) f. f (a + h)
7. f (x) = 5x − 2


Chapter 1 | Functions and Graphs 27




8. f (x) = 4x2 − 3x + 1
9. f (x) = 2x
10. f (x) = |x − 7| + 8
11. f (x) = 6x + 5
12. f (x) = x − 2


3x + 7


13. f (x) = 9
For the following exercises, find the domain, range, and allzeros/intercepts, if any, of the functions.
14. f (x) = x


x2 − 16


15. g(x) = 8x − 1
16. h(x) = 3


x2 + 4


17. f (x) = −1 + x + 2
18. f (x) = 1


x − 9


19. g(x) = 3
x − 4


20. f (x) = 4|x + 5|
21. g(x) = 7


x − 5


For the following exercises, set up a table to sketch thegraph of each function using the following values:
x = −3, −2, −1, 0, 1, 2, 3.


22. f (x) = x2 + 1
x y x y


−3 10 1 2
−2 5 2 5
−1 2 3 10
0 1


23. f (x) = 3x − 6
x y x y


−3 −15 1 −3
−2 −12 2 0
−1 −9 3 3
0 −6


24. f (x) = 1
2
x + 1


x y x y


−3 −12 1 32


−2 0 2 2


−1 12 3 52


0 1


25. f (x) = 2|x|
x y x y


−3 6 1 2
−2 4 2 4
−1 2 3 6
0 0


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26. f (x) = −x2
x y x y


−3 −9 1 −1
−2 −4 2 −4
−1 −1 3 −9
0 0


27. f (x) = x3
x y x y


−3 −27 1 1
−2 −8 2 8
−1 −1 3 27
0 0


For the following exercises, use the vertical line test todetermine whether each of the given graphs represents afunction. Assume that a graph continues at both ends ifit extends beyond the given grid. If the graph represents afunction, then determine the following for each graph:
a. Domain and range
b. x -intercept, if any (estimate where necessary)
c. y -Intercept, if any (estimate where necessary)
d. The intervals for which the function is increasing
e. The intervals for which the function is decreasing
f. The intervals for which the function is constant
g. Symmetry about any axis and/or the origin
h. Whether the function is even, odd, or neither


28.


29.


30.


Chapter 1 | Functions and Graphs 29




31.


32.


33.


34.


35.


For the following exercises, for each pair of functions, finda. f + g b. f − g c. f · g d. f /g. Determine the domain
of each of these new functions.
36. f (x) = 3x + 4, g(x) = x − 2
37. f (x) = x − 8, g(x) = 5x2
38. f (x) = 3x2 + 4x + 1, g(x) = x + 1
39. f (x) = 9 − x2, g(x) = x2 − 2x − 3
40. f (x) = x, g(x) = x − 2
41. f (x) = 6 + 1x , g(x) = 1x
For the following exercises, for each pair of functions, find


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a. ⎛⎝ f ∘g⎞⎠(x) and b. ⎛⎝g ∘ f ⎞⎠(x) Simplify the results. Find the
domain of each of the results.
42. f (x) = 3x, g(x) = x + 5
43. f (x) = x + 4, g(x) = 4x − 1
44. f (x) = 2x + 4, g(x) = x2 − 2
45. f (x) = x2 + 7, g(x) = x2 − 3
46. f (x) = x, g(x) = x + 9
47. f (x) = 3


2x + 1
, g(x) = 2x


48. f (x) = |x + 1|, g(x) = x2 + x − 4


49. The table below lists the NBA championship winnersfor the years 2001 to 2012.
Year Winner
2001 LA Lakers
2002 LA Lakers
2003 San Antonio Spurs
2004 Detroit Pistons
2005 San Antonio Spurs
2006 Miami Heat
2007 San Antonio Spurs
2008 Boston Celtics
2009 LA Lakers
2010 LA Lakers
2011 Dallas Mavericks
2012 Miami Heat


a. Consider the relation in which the domain valuesare the years 2001 to 2012 and the range is thecorresponding winner. Is this relation a function?Explain why or why not.b. Consider the relation where the domain values arethe winners and the range is the correspondingyears. Is this relation a function? Explain why orwhy not.
50. [T] The area A of a square depends on the length of
the side s.


a. Write a function A(s) for the area of a square.
b. Find and interpret A(6.5).
c. Find the exact and the two-significant-digitapproximation to the length of the sides of a squarewith area 56 square units.


Chapter 1 | Functions and Graphs 31




51. [T] The volume of a cube depends on the length of thesides s.
a. Write a function V(s) for the area of a square.
b. Find and interpret V(11.8).


52. [T] A rental car company rents cars for a flat fee of$20 and an hourly charge of $10.25. Therefore, the totalcost C to rent a car is a function of the hours t the car is
rented plus the flat fee.a. Write the formula for the function that models thissituation.b. Find the total cost to rent a car for 2 days and 7hours.c. Determine how long the car was rented if the bill is$432.73.
53. [T] A vehicle has a 20-gal tank and gets 15 mpg.The number of miles N that can be driven depends on theamount of gas x in the tank.a. Write a formula that models this situation.b. Determine the number of miles the vehicle cantravel on (i) a full tank of gas and (ii) 3/4 of a tankof gas.c. Determine the domain and range of the function.d. Determine how many times the driver had to stopfor gas if she has driven a total of 578 mi.
54. [T] The volume V of a sphere depends on the length of
its radius as V = (4/3)πr3. Because Earth is not a perfect
sphere, we can use the mean radius when measuring fromthe center to its surface. The mean radius is the averagedistance from the physical center to the surface, based ona large number of samples. Find the volume of Earth with
mean radius 6.371 × 106 m.
55. [T] A certain bacterium grows in culture in a circularregion. The radius of the circle, measured in centimeters,
is given by r(t) = 6 − ⎡⎣5/⎛⎝t2 + 1⎞⎠⎤⎦, where t is time
measured in hours since a circle of a 1-cm radius of thebacterium was put into the culture.a. Express the area of the bacteria as a function oftime.b. Find the exact and approximate area of the bacterialculture in 3 hours.c. Express the circumference of the bacteria as afunction of time.d. Find the exact and approximate circumference ofthe bacteria in 3 hours.


56. [T] An American tourist visits Paris and must convertU.S. dollars to Euros, which can be done using the function
E(x) = 0.79x, where x is the number of U.S. dollars and
E(x) is the equivalent number of Euros. Since conversion
rates fluctuate, when the tourist returns to the United States2 weeks later, the conversion from Euros to U.S. dollarsis D(x) = 1.245x, where x is the number of Euros and
D(x) is the equivalent number of U.S. dollars.


a. Find the composite function that converts directlyfrom U.S. dollars to U.S. dollars via Euros. Did thistourist lose value in the conversion process?b. Use (a) to determine how many U.S. dollars thetourist would get back at the end of her trip if sheconverted an extra $200 when she arrived in Paris.
57. [T] The manager at a skateboard shop pays hisworkers a monthly salary S of $750 plus a commission of$8.50 for each skateboard they sell.a. Write a function y = S(x) that models a worker’s


monthly salary based on the number of skateboardsx he or she sells.b. Find the approximate monthly salary when aworker sells 25, 40, or 55 skateboards.c. Use the INTERSECT feature on a graphingcalculator to determine the number of skateboardsthat must be sold for a worker to earn a monthlyincome of $1400. (Hint: Find the intersection of thefunction and the line y = 1400.)


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58. [T] Use a graphing calculator to graph the half-circle
y = 25 − (x − 4)2. Then, use the INTERCEPT feature
to find the value of both the x - and y -intercepts.


Chapter 1 | Functions and Graphs 33




1.2 | Basic Classes of Functions
Learning Objectives


1.2.1 Calculate the slope of a linear function and interpret its meaning.
1.2.2 Recognize the degree of a polynomial.
1.2.3 Find the roots of a quadratic polynomial.
1.2.4 Describe the graphs of basic odd and even polynomial functions.
1.2.5 Identify a rational function.
1.2.6 Describe the graphs of power and root functions.
1.2.7 Explain the difference between algebraic and transcendental functions.
1.2.8 Graph a piecewise-defined function.
1.2.9 Sketch the graph of a function that has been shifted, stretched, or reflected from its initialgraph position.


We have studied the general characteristics of functions, so now let’s examine some specific classes of functions. Webegin by reviewing the basic properties of linear and quadratic functions, and then generalize to include higher-degreepolynomials. By combining root functions with polynomials, we can define general algebraic functions and distinguishthem from the transcendental functions we examine later in this chapter. We finish the section with examples of piecewise-defined functions and take a look at how to sketch the graph of a function that has been shifted, stretched, or reflected fromits initial form.
Linear Functions and Slope
The easiest type of function to consider is a linear function. Linear functions have the form f (x) = ax + b, where a and
b are constants. In Figure 1.15, we see examples of linear functions when a is positive, negative, and zero. Note that if
a > 0, the graph of the line rises as x increases. In other words, f (x) = ax + b is increasing on (−∞, ∞). If a < 0,
the graph of the line falls as x increases. In this case, f (x) = ax + b is decreasing on (−∞, ∞). If a = 0, the line is
horizontal.


Figure 1.15 These linear functions are increasing ordecreasing on (∞, ∞) and one function is a horizontal line.


As suggested by Figure 1.15, the graph of any linear function is a line. One of the distinguishing features of a line is itsslope. The slope is the change in y for each unit change in x. The slope measures both the steepness and the direction of
a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line pointsdownward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, weneed to determine the ratio of the change in y versus the change in x. To do so, we choose any two points (x1, y1) and
(x2, y2) on the line and calculate y2 − y1x2 − x1. In Figure 1.16, we see this ratio is independent of the points chosen.


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Figure 1.16 For any linear function, the slope
(y2 − y1)/(x2 − x1) is independent of the choice of points
(x1, y1) and (x2, y2) on the line.


Definition
Consider line L passing through points (x1, y1) and (x2, y2). Let Δy = y2 − y1 and Δx = x2 − x1 denote the
changes in y and x, respectively. The slope of the line is


(1.3)m = y2 − y1x2 − x1 = ΔyΔx.


We now examine the relationship between slope and the formula for a linear function. Consider the linear function givenby the formula f (x) = ax + b. As discussed earlier, we know the graph of a linear function is given by a line. We
can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating
(y2 − y1)/(x2 − x1) for any points (x1, y1) and (x2, y2) on the line. Evaluating the function f at x = 0, we see
that (0, b) is a point on this line. Evaluating this function at x = 1, we see that (1, a + b) is also a point on this line.
Therefore, the slope of this line is


(a + b) − b
1 − 0


= a.


We have shown that the coefficient a is the slope of the line. We can conclude that the formula f (x) = ax + b describes
a line with slope a. Furthermore, because this line intersects the y -axis at the point (0, b), we see that the y -intercept
for this linear function is (0, b). We conclude that the formula f (x) = ax + b tells us the slope, a, and the y -intercept,
(0, b), for this line. Since we often use the symbol m to denote the slope of a line, we can write


f (x) = mx + b


to denote the slope-intercept form of a linear function.
Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear functionpasses through the point (x1, y1) and the slope of the line is m. Since any other point (x, f (x)) on the graph of f must
satisfy the equation


m =
f (x) − y1
x − x1


,


Chapter 1 | Functions and Graphs 35




this linear function can be expressed by writing
f (x) − y1 = m(x − x1).


We call this equation the point-slope equation for that linear function.
Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using theslope-intercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot beexpressed in either of these forms. Instead, a vertical line is described by the equation x = k for some constant k. Since
neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation


ax + by = c,


where a, b are both not zero, to denote the standard form of a line.


Definition
Consider a line passing through the point (x1, y1) with slope m. The equation


(1.4)y − y1 = m(x − x1)
is the point-slope equation for that line.
Consider a line with slope m and y -intercept (0, b). The equation


(1.5)y = mx + b
is an equation for that line in slope-intercept form.
The standard form of a line is given by the equation


(1.6)ax + by = c,
where a and b are both not zero. This form is more general because it allows for a vertical line, x = k.


Example 1.12
Finding the Slope and Equations of Lines
Consider the line passing through the points (11, −4) and (−4, 5), as shown in Figure 1.17.


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1.9


Figure 1.17 Finding the equation of a linear function with a graph that is a line betweentwo given points.
a. Find the slope of the line.
b. Find an equation for this linear function in point-slope form.
c. Find an equation for this linear function in slope-intercept form.


Solution
a. The slope of the line is


m =
y2 − y1
x2 − x1


= 5 − (−4)
−4 − 11


= − 9
15


= − 3
5
.


b. To find an equation for the linear function in point-slope form, use the slope m = −3/5 and choose any
point on the line. If we choose the point (11, −4), we get the equation


f (x) + 4 = − 3
5
(x − 11).


c. To find an equation for the linear function in slope-intercept form, solve the equation in part b. for f (x).
When we do this, we get the equation


f (x) = − 3
5
x + 13


5
.


Consider the line passing through points (−3, 2) and (1, 4). Find the slope of the line.
Find an equation of that line in point-slope form. Find an equation of that line in slope-intercept form.


Example 1.13
A Linear Distance Function
Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer thefollowing questions, assuming Jessica runs at a constant pace.


Chapter 1 | Functions and Graphs 37




a. Describe the distance D (in miles) Jessica runs as a linear function of her run time t (in minutes).
b. Sketch a graph of D.
c. Interpret the meaning of the slope.


Solution
a. At time t = 0, Jessica is at her house, so D(0) = 0. At time t = 78 minutes, Jessica has finished


running 9 mi, so D(78) = 9. The slope of the linear function is
m = 9 − 0


78 − 0
= 3


26
.


The y -intercept is (0, 0), so the equation for this linear function is
D(t) = 3


26
t.


b. To graph D, use the fact that the graph passes through the origin and has slope m = 3/26.


c. The slope m = 3/26 ≈ 0.115 describes the distance (in miles) Jessica runs per minute, or her average
velocity.


Polynomials
A linear function is a special type of a more general class of functions: polynomials. A polynomial function is any functionthat can be written in the form


(1.7)f (x) = an xn + an − 1 xn − 1 +… + a1 x + a0
for some integer n ≥ 0 and constants an, an − 1 ,…,a0, where an ≠ 0. In the case when n = 0, we allow for a0 = 0;
if a0 = 0, the function f (x) = 0 is called the zero function. The value n is called the degree of the polynomial; the
constant an is called the leading coefficient. A linear function of the form f (x) = mx + b is a polynomial of degree 1
if m ≠ 0 and degree 0 if m = 0. A polynomial of degree 0 is also called a constant function. A polynomial function
of degree 2 is called a quadratic function. In particular, a quadratic function has the form f (x) = ax2 + bx + c, where
a ≠ 0. A polynomial function of degree 3 is called a cubic function.
Power Functions
Some polynomial functions are power functions. A power function is any function of the form f (x) = axb, where a and
b are any real numbers. The exponent in a power function can be any real number, but here we consider the case when the
exponent is a positive integer. (We consider other cases later.) If the exponent is a positive integer, then f (x) = axn is a
polynomial. If n is even, then f (x) = axn is an even function because f (−x) = a(−x)n = axn if n is even. If n is odd,
then f (x) = axn is an odd function because f (−x) = a(−x)n = −axn if n is odd (Figure 1.18).


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Figure 1.18 (a) For any even integer n, f (x) = axn is an even function. (b) For any odd
integer n, f (x) = axn is an odd function.


Behavior at Infinity
To determine the behavior of a function f as the inputs approach infinity, we look at the values f (x) as the inputs,
x, become larger. For some functions, the values of f (x) approach a finite number. For example, for the function
f (x) = 2 + 1/x, the values 1/x become closer and closer to zero for all values of x as they get larger and larger. For this
function, we say “ f (x) approaches two as x goes to infinity,” and we write f (x) → 2 as x → ∞. The line y = 2 is a
horizontal asymptote for the function f (x) = 2 + 1/x because the graph of the function gets closer to the line as x gets
larger.
For other functions, the values f (x) may not approach a finite number but instead may become larger for all values of x
as they get larger. In that case, we say “ f (x) approaches infinity as x approaches infinity,” and we write f (x) → ∞ as
x → ∞. For example, for the function f (x) = 3x2, the outputs f (x) become larger as the inputs x get larger. We can
conclude that the function f (x) = 3x2 approaches infinity as x approaches infinity, and we write 3x2 → ∞ as x → ∞.
The behavior as x → −∞ and the meaning of f (x) → −∞ as x → ∞ or x → −∞ can be defined similarly. We can
describe what happens to the values of f (x) as x → ∞ and as x → −∞ as the end behavior of the function.
To understand the end behavior for polynomial functions, we can focus on quadratic and cubic functions. The behavior for
higher-degree polynomials can be analyzed similarly. Consider a quadratic function f (x) = ax2 + bx + c. If a > 0, the
values f (x) → ∞ as x → ±∞. If a < 0, the values f (x) → −∞ as x → ±∞. Since the graph of a quadratic function
is a parabola, the parabola opens upward if a > 0; the parabola opens downward if a < 0. (See Figure 1.19(a).)
Now consider a cubic function f (x) = ax3 + bx2 + cx + d. If a > 0, then f (x) → ∞ as x → ∞ and f (x) → −∞
as x → −∞. If a < 0, then f (x) → −∞ as x → ∞ and f (x) → ∞ as x → −∞. As we can see from both of these
graphs, the leading term of the polynomial determines the end behavior. (See Figure 1.19(b).)


Chapter 1 | Functions and Graphs 39




Figure 1.19 (a) For a quadratic function, if the leading coefficient a > 0, the parabola opens
upward. If a < 0, the parabola opens downward. (b) For a cubic function f , if the leading
coefficient a > 0, the values f (x) → ∞ as x → ∞ and the values f (x) → −∞ as
x → −∞. If the leading coefficient a < 0, the opposite is true.


Zeros of Polynomial Functions
Another characteristic of the graph of a polynomial function is where it intersects the x -axis. To determine where a function
f intersects the x -axis, we need to solve the equation f (x) = 0 for .n the case of the linear function f (x) = mx + b,
the x -intercept is given by solving the equation mx + b = 0. In this case, we see that the x -intercept is given by
(−b/m, 0). In the case of a quadratic function, finding the x -intercept(s) requires finding the zeros of a quadratic equation:
ax2 + bx + c = 0. In some cases, it is easy to factor the polynomial ax2 + bx + c to find the zeros. If not, we make use
of the quadratic formula.
Rule: The Quadratic Formula
Consider the quadratic equation


ax2 + bx + c = 0,


where a ≠ 0. The solutions of this equation are given by the quadratic formula
(1.8)


x = −b ± b
2 − 4ac


2a
.


If the discriminant b2 − 4ac > 0, this formula tells us there are two real numbers that satisfy the quadratic equation.
If b2 − 4ac = 0, this formula tells us there is only one solution, and it is a real number. If b2 − 4ac < 0, no real
numbers satisfy the quadratic equation.


In the case of higher-degree polynomials, it may be more complicated to determine where the graph intersects the x -axis.
In some instances, it is possible to find the x -intercepts by factoring the polynomial to find its zeros. In other cases, it is
impossible to calculate the exact values of the x -intercepts. However, as we see later in the text, in cases such as this, we
can use analytical tools to approximate (to a very high degree) where the x -intercepts are located. Here we focus on the
graphs of polynomials for which we can calculate their zeros explicitly.


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Example 1.14
Graphing Polynomial Functions
For the following functions a. and b., i. describe the behavior of f (x) as x → ±∞, ii. find all zeros of f , and
iii. sketch a graph of f .


a. f (x) = −2x2 + 4x − 1
b. f (x) = x3 − 3x2 − 4x


Solution
a. The function f (x) = −2x2 + 4x − 1 is a quadratic function.


i. Because a = −2 < 0, as x → ±∞, f (x) → −∞.
ii. To find the zeros of f , use the quadratic formula. The zeros are


x = −4 ± 4
2 − 4(−2)(−1)
2(−2)


= −4 ± 8
−4


= −4 ± 2 2
−4


= 2 ± 2
2


.


iii. To sketch the graph of f , use the information from your previous answers and combine it with
the fact that the graph is a parabola opening downward.


b. The function f (x) = x3 − 3x2 − 4x is a cubic function.
i. Because a = 1 > 0, as x → ∞, f (x) → ∞. As x → −∞, f (x) → −∞.
ii. To find the zeros of f , we need to factor the polynomial. First, when we factor x out of all the


terms, we find
f (x) = x(x2 − 3x − 4).


Then, when we factor the quadratic function x2 − 3x − 4, we find
f (x) = x(x − 4)(x + 1).


Chapter 1 | Functions and Graphs 41




1.10


Therefore, the zeros of f are x = 0, 4, −1.
iii. Combining the results from parts i. and ii., draw a rough sketch of f .


Consider the quadratic function f (x) = 3x2 − 6x + 2. Find the zeros of f . Does the parabola open
upward or downward?


Mathematical Models
A large variety of real-world situations can be described usingmathematical models. A mathematical model is a method ofsimulating real-life situations with mathematical equations. Physicists, engineers, economists, and other researchers developmodels by combining observation with quantitative data to develop equations, functions, graphs, and other mathematicaltools to describe the behavior of various systems accurately. Models are useful because they help predict future outcomes.Examples of mathematical models include the study of population dynamics, investigations of weather patterns, andpredictions of product sales.
As an example, let’s consider a mathematical model that a company could use to describe its revenue for the sale of aparticular item. The amount of revenue R a company receives for the sale of n items sold at a price of p dollars per item
is described by the equation R = p · n. The company is interested in how the sales change as the price of the item changes.
Suppose the data in Table 1.6 show the number of units a company sells as a function of the price per item.


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p 6 8 10 12 14


n 19.4 18.5 16.2 13.8 12.2


Table 1.6 Number of Units Sold n (in Thousands) as a
Function of Price per Unit p (in Dollars)


In Figure 1.20, we see the graph the number of units sold (in thousands) as a function of price (in dollars). We note fromthe shape of the graph that the number of units sold is likely a linear function of price per item, and the data can be closelyapproximated by the linear function n = −1.04p + 26 for 0 ≤ p ≤ 25, where n predicts the number of units sold in
thousands. Using this linear function, the revenue (in thousands of dollars) can be estimated by the quadratic function


R(p) = p · ⎛⎝−1.04p + 26⎞⎠ = −1.04p2 + 26p


for 0 ≤ p ≤ 25. In Example 1.15, we use this quadratic function to predict the amount of revenue the company receives
depending on the price the company charges per item. Note that we cannot conclude definitively the actual number of unitssold for values of p, for which no data are collected. However, given the other data values and the graph shown, it seems
reasonable that the number of units sold (in thousands) if the price charged is p dollars may be close to the values predicted
by the linear function n = −1.04p + 26.


Figure 1.20 The data collected for the number of items sold as a function ofprice is roughly linear. We use the linear function n = −1.04p + 26 to estimate
this function.


Example 1.15


Chapter 1 | Functions and Graphs 43




Maximizing Revenue
A company is interested in predicting the amount of revenue it will receive depending on the price it charges for aparticular item. Using the data from Table 1.6, the company arrives at the following quadratic function to modelrevenue R as a function of price per item p:


R(p) = p · ⎛⎝−1.04p + 26⎞⎠ = −1.04p2 + 26p


for 0 ≤ p ≤ 25.
a. Predict the revenue if the company sells the item at a price of p = $5 and p = $17.
b. Find the zeros of this function and interpret the meaning of the zeros.
c. Sketch a graph of R.
d. Use the graph to determine the value of p that maximizes revenue. Find the maximum revenue.


Solution
a. Evaluating the revenue function at p = 5 and p = 17, we can conclude that


R(5) = −1.04(5)2 + 26(5) = 104, so revenue = $104,000;


R(17) = −1.04(17)2 + 26(17) = 141.44, so revenue = $144,440.


b. The zeros of this function can be found by solving the equation −1.04p2 + 26p = 0. When we factor
the quadratic expression, we get p⎛⎝−1.04p + 26⎞⎠ = 0. The solutions to this equation are given by
p = 0, 25. For these values of p, the revenue is zero. When p = $0, the revenue is zero because the
company is giving away its merchandise for free. When p = $25, the revenue is zero because the price
is too high, and no one will buy any items.


c. Knowing the fact that the function is quadratic, we also know the graph is a parabola. Since theleading coefficient is negative, the parabola opens downward. One property of parabolas is that they aresymmetric about the axis, so since the zeros are at p = 0 and p = 25, the parabola must be symmetric
about the line halfway between them, or p = 12.5.


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d. The function is a parabola with zeros at p = 0 and p = 25, and it is symmetric about the line
p = 12.5, so the maximum revenue occurs at a price of p = $12.50 per item. At that price, the revenue
is R(p) = −1.04(12.5)2 + 26(12.5) = $162, 500.


Algebraic Functions
By allowing for quotients and fractional powers in polynomial functions, we create a larger class of functions. An algebraicfunction is one that involves addition, subtraction, multiplication, division, rational powers, and roots. Two types ofalgebraic functions are rational functions and root functions.
Just as rational numbers are quotients of integers, rational functions are quotients of polynomials. In particular, a rationalfunction is any function of the form f (x) = p(x)/q(x), where p(x) and q(x) are polynomials. For example,


f (x) = 3x − 1
5x + 2


and g(x) = 4
x2 + 1


are rational functions. A root function is a power function of the form f (x) = x1/n, where n is a positive integer greater
than one. For example, f (x) = x1/2 = x is the square-root function and g(x) = x1/3 = x3 is the cube-root function. By
allowing for compositions of root functions and rational functions, we can create other algebraic functions. For example,
f (x) = 4 − x2 is an algebraic function.


Chapter 1 | Functions and Graphs 45




1.11


Example 1.16
Finding Domain and Range for Algebraic Functions
For each of the following functions, find the domain and range.


a. f (x) = 3x − 1
5x + 2


b. f (x) = 4 − x2


Solution
a. It is not possible to divide by zero, so the domain is the set of real numbers x such that x ≠ −2/5. To


find the range, we need to find the values y for which there exists a real number x such that
y = 3x − 1


5x + 2
.


When we multiply both sides of this equation by 5x + 2, we see that x must satisfy the equation
5xy + 2y = 3x − 1.


From this equation, we can see that x must satisfy
2y + 1 = x(3 − 5y).


If y = 3/5, this equation has no solution. On the other hand, as long as y ≠ 3/5,
x =


2y + 1
3 − 5y


satisfies this equation. We can conclude that the range of f is ⎧⎩⎨y|y ≠ 3/5⎫⎭⎬.
b. To find the domain of f , we need 4 − x2 ≥ 0. When we factor, we write


4 − x2 = (2 − x)(2 + x) ≥ 0. This inequality holds if and only if both terms are positive or both terms
are negative. For both terms to be positive, we need to find x such that


2 − x ≥ 0 and 2 + x ≥ 0.


These two inequalities reduce to 2 ≥ x and x ≥ −2. Therefore, the set {x| − 2 ≤ x ≤ 2} must be part
of the domain. For both terms to be negative, we need


2 − x ≤ 0 and 2 + x ≥ 0.


These two inequalities also reduce to 2 ≤ x and x ≥ −2. There are no values of x that satisfy both of
these inequalities. Thus, we can conclude the domain of this function is {x| − 2 ≤ x ≤ 2}.
If −2 ≤ x ≤ 2, then 0 ≤ 4 − x2 ≤ 4. Therefore, 0 ≤ 4 − x2 ≤ 2, and the range of f is



⎨y|0 ≤ y ≤ 2⎫⎭⎬.


Find the domain and range for the function f (x) = (5x + 2)/(2x − 1).


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The root functions f (x) = x1/n have defining characteristics depending on whether n is odd or even. For all even integers
n ≥ 2, the domain of f (x) = x1/n is the interval [0, ∞). For all odd integers n ≥ 1, the domain of f (x) = x1/n is
the set of all real numbers. Since x1/n = (−x)1/n for odd integers n, f (x) = x1/n is an odd function if n is odd. See the
graphs of root functions for different values of n in Figure 1.21.


Figure 1.21 (a) If n is even, the domain of f (x) = xn is [0, ∞). (b) If n is odd, the domain of f (x) = xn is
(−∞, ∞) and the function f (x) = xn is an odd function.


Example 1.17
Finding Domains for Algebraic Functions
For each of the following functions, determine the domain of the function.


a. f (x) = 3
x2 − 1


b. f (x) = 2x + 5
3x2 + 4


c. f (x) = 4 − 3x
d. f (x) = 2x − 13


Solution
a. You cannot divide by zero, so the domain is the set of values x such that x2 − 1 ≠ 0. Therefore, the


domain is {x|x ≠ ±1}.
b. You need to determine the values of x for which the denominator is zero. Since 3x2 + 4 ≥ 4 for all real


numbers x, the denominator is never zero. Therefore, the domain is (−∞, ∞).
c. Since the square root of a negative number is not a real number, the domain is the set of values x for


which 4 − 3x ≥ 0. Therefore, the domain is {x|x ≤ 4/3}.
d. The cube root is defined for all real numbers, so the domain is the interval (−∞, ∞).


Chapter 1 | Functions and Graphs 47




1.12


1.13


Find the domain for each of the following functions: f (x) = (5 − 2x)/(x2 + 2) and g(x) = 5x − 1.


Transcendental Functions
Thus far, we have discussed algebraic functions. Some functions, however, cannot be described by basic algebraicoperations. These functions are known as transcendental functions because they are said to “transcend,” or go beyond,algebra. The most common transcendental functions are trigonometric, exponential, and logarithmic functions. Atrigonometric function relates the ratios of two sides of a right triangle. They are sinx, cosx, tanx, cotx, secx, and cscx.
(We discuss trigonometric functions later in the chapter.) An exponential function is a function of the form f (x) = bx,
where the base b > 0, b ≠ 1. A logarithmic function is a function of the form f (x) = logb(x) for some constant
b > 0, b ≠ 1, where logb (x) = y if and only if by = x. (We also discuss exponential and logarithmic functions later in
the chapter.)
Example 1.18
Classifying Algebraic and Transcendental Functions
Classify each of the following functions, a. through c., as algebraic or transcendental.


a. f (x) = x3 + 1
4x + 2


b. f (x) = 2x2
c. f (x) = sin(2x)


Solution
a. Since this function involves basic algebraic operations only, it is an algebraic function.
b. This function cannot be written as a formula that involves only basic algebraic operations, so it istranscendental. (Note that algebraic functions can only have powers that are rational numbers.)
c. As in part b., this function cannot be written using a formula involving basic algebraic operations only;therefore, this function is transcendental.


Is f (x) = x/2 an algebraic or a transcendental function?


Piecewise-Defined Functions
Sometimes a function is defined by different formulas on different parts of its domain. A function with this property isknown as a piecewise-defined function. The absolute value function is an example of a piecewise-defined function becausethe formula changes with the sign of x:


f (x) =




−x, x < 0
x, x ≥ 0


.


Other piecewise-defined functions may be represented by completely different formulas, depending on the part of thedomain in which a point falls. To graph a piecewise-defined function, we graph each part of the function in its respectivedomain, on the same coordinate system. If the formula for a function is different for x < a and x > a, we need to pay
special attention to what happens at x = a when we graph the function. Sometimes the graph needs to include an open or


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1.14


closed circle to indicate the value of the function at x = a. We examine this in the next example.
Example 1.19
Graphing a Piecewise-Defined Function
Sketch a graph of the following piecewise-defined function:


f (x) =




x + 3, x < 1


(x − 2)2, x ≥ 1
.


Solution
Graph the linear function y = x + 3 on the interval (−∞, 1) and graph the quadratic function y = (x − 2)2
on the interval [1, ∞). Since the value of the function at x = 1 is given by the formula f (x) = (x − 2)2, we
see that f (1) = 1. To indicate this on the graph, we draw a closed circle at the point (1, 1). The value of the
function is given by f (x) = x + 2 for all x < 1, but not at x = 1. To indicate this on the graph, we draw an
open circle at (1, 4).


Figure 1.22 This piecewise-defined function is linear for
x < 1 and quadratic for x ≥ 1.


Sketch a graph of the function
f (x) =






2 − x, x ≤ 2
x + 2, x > 2


.


Example 1.20
Parking Fees Described by a Piecewise-Defined Function
In a big city, drivers are charged variable rates for parking in a parking garage. They are charged $10 for the firsthour or any part of the first hour and an additional $2 for each hour or part thereof up to a maximum of $30 for


Chapter 1 | Functions and Graphs 49




1.15


the day. The parking garage is open from 6 a.m. to 12 midnight.
a. Write a piecewise-defined function that describes the cost C to park in the parking garage as a function


of hours parked x.
b. Sketch a graph of this function C(x).


Solution
a. Since the parking garage is open 18 hours each day, the domain for this function is {x|0 < x ≤ 18}. The


cost to park a car at this parking garage can be described piecewise by the function


C(x) =













10, 0 < x ≤ 1
12, 1 < x ≤ 2
14, 2 < x ≤ 3
16, 3 < x ≤ 4



30, 10 < x ≤ 18


.


b. The graph of the function consists of several horizontal line segments.


The cost of mailing a letter is a function of the weight of the letter. Suppose the cost of mailing a letter is
49¢ for the first ounce and 21¢ for each additional ounce. Write a piecewise-defined function describing the
cost C as a function of the weight x for 0 < x ≤ 3, where C is measured in cents and x is measured in
ounces.


Transformations of Functions
We have seen several cases in which we have added, subtracted, or multiplied constants to form variations of simple
functions. In the previous example, for instance, we subtracted 2 from the argument of the function y = x2 to get the
function f (x) = (x − 2)2. This subtraction represents a shift of the function y = x2 two units to the right. A shift,
horizontally or vertically, is a type of transformation of a function. Other transformations include horizontal and verticalscalings, and reflections about the axes.
A vertical shift of a function occurs if we add or subtract the same constant to each output y. For c > 0, the graph of


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f (x) + c is a shift of the graph of f (x) up c units, whereas the graph of f (x) − c is a shift of the graph of f (x) down
c units. For example, the graph of the function f (x) = x3 + 4 is the graph of y = x3 shifted up 4 units; the graph of the
function f (x) = x3 − 4 is the graph of y = x3 shifted down 4 units (Figure 1.23).


Figure 1.23 (a) For c > 0, the graph of y = f (x) + c is a vertical shift up c units of
the graph of y = f (x). (b) For c > 0, the graph of y = f (x) − c is a vertical shift down
c units of the graph of y = f (x).


A horizontal shift of a function occurs if we add or subtract the same constant to each input x. For c > 0, the graph of
f (x + c) is a shift of the graph of f (x) to the left c units; the graph of f (x − c) is a shift of the graph of f (x) to the
right c units. Why does the graph shift left when adding a constant and shift right when subtracting a constant? To answer
this question, let’s look at an example.
Consider the function f (x) = |x + 3| and evaluate this function at x − 3. Since f (x − 3) = |x| and x − 3 < x, the graph
of f (x) = |x + 3| is the graph of y = |x| shifted left 3 units. Similarly, the graph of f (x) = |x − 3| is the graph of y = |x|
shifted right 3 units (Figure 1.24).


Chapter 1 | Functions and Graphs 51




Figure 1.24 (a) For c > 0, the graph of y = f (x + c) is a horizontal shift left c units of the graph of y = f (x). (b) For
c > 0, the graph of y = f (x − c) is a horizontal shift right c units of the graph of y = f (x).


A vertical scaling of a graph occurs if we multiply all outputs y of a function by the same positive constant. For c > 0,
the graph of the function c f (x) is the graph of f (x) scaled vertically by a factor of c. If c > 1, the values of the
outputs for the function c f (x) are larger than the values of the outputs for the function f (x); therefore, the graph has been
stretched vertically. If 0 < c < 1, then the outputs of the function c f (x) are smaller, so the graph has been compressed.
For example, the graph of the function f (x) = 3x2 is the graph of y = x2 stretched vertically by a factor of 3, whereas the
graph of f (x) = x2 /3 is the graph of y = x2 compressed vertically by a factor of 3 (Figure 1.25).


Figure 1.25 (a) If c > 1, the graph of y = c f (x) is a vertical stretch of the graph
of y = f (x). (b) If 0 < c < 1, the graph of y = c f (x) is a vertical compression of
the graph of y = f (x).


The horizontal scaling of a function occurs if we multiply the inputs x by the same positive constant. For c > 0, the
graph of the function f (cx) is the graph of f (x) scaled horizontally by a factor of c. If c > 1, the graph of f (cx) is the
graph of f (x) compressed horizontally. If 0 < c < 1, the graph of f (cx) is the graph of f (x) stretched horizontally. For


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example, consider the function f (x) = 2x and evaluate f at x/2. Since f (x/2) = x, the graph of f (x) = 2x is the
graph of y = x compressed horizontally. The graph of y = x/2 is a horizontal stretch of the graph of y = x (Figure
1.26).


Figure 1.26 (a) If c > 1, the graph of y = f (cx) is a horizontal compression of the graph
of y = f (x). (b) If 0 < c < 1, the graph of y = f (cx) is a horizontal stretch of the graph of
y = f (x).


We have explored what happens to the graph of a function f when we multiply f by a constant c > 0 to get a new
function c f (x). We have also discussed what happens to the graph of a function f when we multiply the independent
variable x by c > 0 to get a new function f (cx). However, we have not addressed what happens to the graph of the
function if the constant c is negative. If we have a constant c < 0, we can write c as a positive number multiplied by
−1; but, what kind of transformation do we get when we multiply the function or its argument by −1? When we multiply
all the outputs by −1, we get a reflection about the x -axis. When we multiply all inputs by −1, we get a reflection
about the y -axis. For example, the graph of f (x) = −(x3 + 1) is the graph of y = (x3 + 1) reflected about the x -axis.
The graph of f (x) = (−x)3 + 1 is the graph of y = x3 + 1 reflected about the y -axis (Figure 1.27).


Chapter 1 | Functions and Graphs 53




Figure 1.27 (a) The graph of y = − f (x) is the graph of
y = f (x) reflected about the x -axis. (b) The graph of
y = f (−x) is the graph of y = f (x) reflected about the
y -axis.


If the graph of a function consists of more than one transformation of another graph, it is important to transform the graphin the correct order. Given a function f (x), the graph of the related function y = c f ⎛⎝a(x + b)⎞⎠+ d can be obtained from
the graph of y = f (x) by performing the transformations in the following order.


1. Horizontal shift of the graph of y = f (x). If b > 0, shift left. If b < 0, shift right.
2. Horizontal scaling of the graph of y = f (x + b) by a factor of |a|. If a < 0, reflect the graph about the y -axis.
3. Vertical scaling of the graph of y = f (a(x + b)) by a factor of |c|. If c < 0, reflect the graph about the x -axis.
4. Vertical shift of the graph of y = c f (a(x + b)). If d > 0, shift up. If d < 0, shift down.


We can summarize the different transformations and their related effects on the graph of a function in the following table.


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Transformation of f(c > 0) Effect on the graph of f
f (x) + c Vertical shift up c units


f (x) − c Vertical shift down c units


f (x + c) Shift left by c units


f (x − c) Shift right by c units


c f (x)
Vertical stretch if c > 1;
vertical compression if 0 < c < 1


f (cx) Horizontal stretch if 0 < c < 1; horizontal compression if c > 1


− f (x) Reflection about the x -axis


f (−x) Reflection about the y -axis
Table 1.7 Transformations of Functions


Example 1.21
Transforming a Function
For each of the following functions, a. and b., sketch a graph by using a sequence of transformations of a well-known function.


a. f (x) = −|x + 2| − 3
b. f (x) = 3 −x + 1


Solution
a. Starting with the graph of y = |x|, shift 2 units to the left, reflect about the x -axis, and then shift down


3 units.


Chapter 1 | Functions and Graphs 55




1.16


Figure 1.28 The function f (x) = −|x + 2| − 3 can be
viewed as a sequence of three transformations of the function
y = |x|.


b. Starting with the graph of y = x, reflect about the y -axis, stretch the graph vertically by a factor of 3,
and move up 1 unit.


Figure 1.29 The function f (x) = 3 −x + 1 can be viewed
as a sequence of three transformations of the function y = x.


Describe how the function f (x) = −(x + 1)2 − 4 can be graphed using the graph of y = x2 and a
sequence of transformations.


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1.2 EXERCISES
For the following exercises, for each pair of points, a.find the slope of the line passing through the points andb. indicate whether the line is increasing, decreasing,horizontal, or vertical.
59. (−2, 4) and (1, 1)
60. (−1, 4) and (3, −1)
61. (3, 5) and (−1, 2)
62. (6, 4) and (4, −3)
63. (2, 3) and (5, 7)
64. (1, 9) and (−8, 5)
65. (2, 4) and (1, 4)
66. (1, 4) and (1, 0)
For the following exercises, write the equation of the linesatisfying the given conditions in slope-intercept form.
67. Slope = −6, passes through (1, 3)
68. Slope = 3, passes through (−3, 2)
69. Slope = 1


3
, passes through (0, 4)


70. Slope = 2
5
, x -intercept = 8


71. Passing through (2, 1) and (−2, −1)
72. Passing through (−3, 7) and (1, 2)
73. x -intercept = 5 and y -intercept = −3
74. x -Intercept = −6 and y -intercept = 9
For the following exercises, for each linear equation, a. givethe slope m and y -intercept b, if any, and b. graph the line.
75. y = 2x − 3
76. y = − 1


7
x + 1


77. f (x) = −6x
78. f (x) = −5x + 4


79. 4y + 24 = 0
80. 8x − 4 = 0
81. 2x + 3y = 6
82. 6x − 5y + 15 = 0
For the following exercises, for each polynomial, a. find thedegree; b. find the zeros, if any; c. find the y -intercept(s),
if any; d. use the leading coefficient to determine thegraph’s end behavior; and e. determine algebraicallywhether the polynomial is even, odd, or neither.
83. f (x) = 2x2 − 3x − 5
84. f (x) = −3x2 + 6x
85. f (x) = 1


2
x2 − 1


86. f (x) = x3 + 3x2 − x − 3
87. f (x) = 3x − x3
For the following exercises, use the graph of f (x) = x2 to
graph each transformed function g.
88. g(x) = x2 − 1
89. g(x) = (x + 3)2 + 1
For the following exercises, use the graph of f (x) = x to
graph each transformed function g.
90. g(x) = x + 2
91. g(x) = − x − 1
For the following exercises, use the graph of y = f (x) to
graph each transformed function g.


Chapter 1 | Functions and Graphs 57




92. g(x) = f (x) + 1
93. g(x) = f (x − 1) + 2
For the following exercises, for each of the piecewise-defined functions, a. evaluate at the given values of theindependent variable and b. sketch the graph.
94. f (x) = ⎧




4x + 3, x ≤ 0
−x + 1, x > 0


; f (−3); f (0); f (2)


95. f (x) = ⎧

⎨x


2 − 3, x < 0
4x − 3, x ≥ 0


; f (−4); f (0); f (2)


96. h(x) = ⎧


x + 1, x ≤ 5
4, x > 5


; h(0); h(π); h(5)


97. g(x) = ⎧



3
x − 2


, x ≠ 2


4, x = 2
; g(0); g(−4); g(2)


For the following exercises, determine whether thestatement is true or false. Explain why.
98. f (x) = (4x + 1)/(7x − 2) is a transcendental
function.
99. g(x) = x3 is an odd root function
100. A logarithmic function is an algebraic function.
101. A function of the form f (x) = xb, where b is a
real valued constant, is an exponential function.
102. The domain of an even root function is all realnumbers.


103. [T] A company purchases some computer equipmentfor $20,500. At the end of a 3-year period, the value of theequipment has decreased linearly to $12,300.a. Find a function y = V(t) that determines the value
V of the equipment at the end of t years.b. Find and interpret the meaning of the x - and y
-intercepts for this situation.c. What is the value of the equipment at the end of 5years?d. When will the value of the equipment be $3000?


104. [T] Total online shopping during the Christmasholidays has increased dramatically during the past 5 years.In 2012 (t = 0), total online holiday sales were $42.3
billion, whereas in 2013 they were $48.1 billion.a. Find a linear function S that estimates the totalonline holiday sales in the year t.b. Interpret the slope of the graph of S.c. Use part a. to predict the year when online shoppingduring Christmas will reach $60 billion.
105. [T] A family bakery makes cupcakes and sells themat local outdoor festivals. For a music festival, there is afixed cost of $125 to set up a cupcake stand. The ownerestimates that it costs $0.75 to make each cupcake. Theowner is interested in determining the total cost C as a
function of number of cupcakes made.a. Find a linear function that relates cost C to x, thenumber of cupcakes made.b. Find the cost to bake 160 cupcakes.c. If the owner sells the cupcakes for $1.50 apiece,how many cupcakes does she need to sell to startmaking profit? (Hint: Use the INTERSECTIONfunction on a calculator to find this number.)
106. [T] A house purchased for $250,000 is expected tobe worth twice its purchase price in 18 years.a. Find a linear function that models the price P ofthe house versus the number of years t since theoriginal purchase.b. Interpret the slope of the graph of P.c. Find the price of the house 15 years from when itwas originally purchased.
107. [T] A car was purchased for $26,000. The value ofthe car depreciates by $1500 per year.a. Find a linear function that models the value V of thecar after t years.b. Find and interpret V(4).
108. [T] A condominium in an upscale part of the city waspurchased for $432,000. In 35 years it is worth $60,500.Find the rate of depreciation.
109. [T] The total cost C (in thousands of dollars) toproduce a certain item is modeled by the function
C(x) = 10.50x + 28,500, where x is the number of items
produced. Determine the cost to produce 175 items.


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110. [T] A professor asks her class to report the amountof time t they spent writing two assignments. Most studentsreport that it takes them about 45 minutes to type a four-page assignment and about 1.5 hours to type a nine-pageassignment.a. Find the linear function y = N(t) that models this
situation, where N is the number of pages typed
and t is the time in minutes.b. Use part a. to determine how many pages can betyped in 2 hours.c. Use part a. to determine how long it takes to type a20-page assignment.


111. [T] The output (as a percent of total capacity) ofnuclear power plants in the United States can be modeledby the function P(t) = 1.8576t + 68.052, where t is time
in years and t = 0 corresponds to the beginning of 2000.
Use the model to predict the percentage output in 2015.
112. [T] The admissions office at a public universityestimates that 65% of the students offered admission to theclass of 2019 will actually enroll.a. Find the linear function y = N(x), where N is


the number of students that actually enroll and x is
the number of all students offered admission to theclass of 2019.b. If the university wants the 2019 freshman class sizeto be 1350, determine how many students should beadmitted.


Chapter 1 | Functions and Graphs 59




1.3 | Trigonometric Functions
Learning Objectives


1.3.1 Convert angle measures between degrees and radians.
1.3.2 Recognize the triangular and circular definitions of the basic trigonometric functions.
1.3.3 Write the basic trigonometric identities.
1.3.4 Identify the graphs and periods of the trigonometric functions.
1.3.5 Describe the shift of a sine or cosine graph from the equation of the function.


Trigonometric functions are used to model many phenomena, including sound waves, vibrations of strings, alternatingelectrical current, and the motion of pendulums. In fact, almost any repetitive, or cyclical, motion can be modeled by somecombination of trigonometric functions. In this section, we define the six basic trigonometric functions and look at some ofthe main identities involving these functions.
Radian Measure
To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians anddegrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1.The radian measure of an angle is defined as follows. Given an angle θ, let s be the length of the corresponding arc on
the unit circle (Figure 1.30). We say the angle corresponding to the arc of length 1 has radian measure 1.


Figure 1.30 The radian measure of an angle θ is the arc
length s of the associated arc on the unit circle.


Since an angle of 360° corresponds to the circumference of a circle, or an arc of length 2π, we conclude that an angle
with a degree measure of 360° has a radian measure of 2π. Similarly, we see that 180° is equivalent to π radians. Table
1.8 shows the relationship between common degree and radian values.


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1.17


Degrees Radians Degrees Radians
0 0 120 2π/3


30 π/6 135 3π/4


45 π/4 150 5π/6


60 π/3 180 π


90 π/2
Table 1.8 Common Angles Expressed in Degrees andRadians


Example 1.22
Converting between Radians and Degrees


a. Express 225° using radians.
b. Express 5π/3 rad using degrees.


Solution
Use the fact that 180° is equivalent to π radians as a conversion factor: 1 = π rad


180°
= 180°


π rad
.


a. 225° = 225° · π
180°


= 5π
4


rad
b. 5π


3
rad = 5π


3
· 180°π = 300°


Express 210° using radians. Express 11π/6 rad using degrees.


The Six Basic Trigonometric Functions
Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on anycircle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among thesides and angles of a triangle.
To define the trigonometric functions, first consider the unit circle centered at the origin and a point P = (x, y) on the unit
circle. Let θ be an angle with an initial side that lies along the positive x -axis and with a terminal side that is the line
segment OP. An angle in this position is said to be in standard position (Figure 1.31). We can then define the values of
the six trigonometric functions for θ in terms of the coordinates x and y.


Chapter 1 | Functions and Graphs 61




Figure 1.31 The angle θ is in standard position. The values
of the trigonometric functions for θ are defined in terms of the
coordinates x and y.


Definition
Let P = (x, y) be a point on the unit circle centered at the origin O. Let θ be an angle with an initial side along the
positive x -axis and a terminal side given by the line segment OP. The trigonometric functions are then defined as


(1.9)sinθ = y cscθ = 1y
cosθ = x secθ = 1x


tanθ =
y
x cotθ =


x
y


If x = 0, secθ and tanθ are undefined. If y = 0, then cotθ and cscθ are undefined.


We can see that for a point P = (x, y) on a circle of radius r with a corresponding angle θ, the coordinates x and y
satisfy


cosθ = xr
x = rcosθ


sinθ =
y
r


y = r sinθ.


The values of the other trigonometric functions can be expressed in terms of x, y, and r (Figure 1.32).


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Figure 1.32 For a point P = (x, y) on a circle of radius r,
the coordinates x and y satisfy x = rcosθ and y = r sinθ.


Table 1.9 shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determinethe values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometricfunctions are calculated easily from the values of sinθ and cosθ.
θ sinθ cosθ


0 0 1


π
6


1
2


3
2


π
4


2
2


2
2


π
3


3
2


1
2


π
2 1 0


Table 1.9 Values of sinθ
and cosθ at Major Angles
θ in the First Quadrant


Example 1.23
Evaluating Trigonometric Functions
Evaluate each of the following expressions.


a. sin⎛⎝2π3 ⎞⎠
b. cos⎛⎝−5π6 ⎞⎠


Chapter 1 | Functions and Graphs 63




c. tan⎛⎝15π4 ⎞⎠
Solution


a. On the unit circle, the angle θ = 2π
3
corresponds to the point ⎛⎝−12, 32 ⎞⎠. Therefore, sin⎛⎝2π3 ⎞⎠ = y = 32 .


b. An angle θ = − 5π
6


corresponds to a revolution in the negative direction, as shown. Therefore,
cos⎛⎝−



6

⎠ = x = −


3
2
.


c. An angle θ = 15π
4


= 2π + 7π
4
. Therefore, this angle corresponds to more than one revolution, as shown.


Knowing the fact that an angle of 7π
4


corresponds to the point ⎛⎝ 22 , − 22 ⎞⎠, we can conclude that
tan⎛⎝


15π
4

⎠ =


y
x = −1.


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1.18 Evaluate cos(3π/4) and sin(−π/6).


As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functionsevaluated at either of the acute angles of the triangle. Let θ be one of the acute angles. Let A be the length of the adjacent
leg, O be the length of the opposite leg, and H be the length of the hypotenuse. By inscribing the triangle into a circle of
radius H, as shown in Figure 1.33, we see that A, H, and O satisfy the following relationships with θ:


sinθ = O
H


cscθ = H
O


cosθ = A
H


secθ = H
A


tanθ = O
A


cotθ = A
O


Figure 1.33 By inscribing a right triangle in a circle, we canexpress the ratios of the side lengths in terms of thetrigonometric functions evaluated at θ.


Example 1.24
Constructing a Wooden Ramp
A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If thetop of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be 10°, how


Chapter 1 | Functions and Graphs 65




1.19


long does the ramp need to be?
Solution
Let x denote the length of the ramp. In the following image, we see that x needs to satisfy the equation
sin(10°) = 4/x. Solving this equation for x, we see that x = 4/sin(10°) ≈ 23.035 ft.


A house painter wants to lean a 20 -ft ladder against a house. If the angle between the base of the ladder
and the ground is to be 60°, how far from the house should she place the base of the ladder?


Trigonometric Identities
A trigonometric identity is an equation involving trigonometric functions that is true for all angles θ for which the
functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities arelisted next.
Rule: Trigonometric Identities
Reciprocal identities


tanθ = sinθ
cosθ


cotθ = cosθ
sinθ


cscθ = 1
sinθ


secθ = 1
cosθ


Pythagorean identities
sin2 θ + cos2 θ = 1 1 + tan2 θ = sec2 θ 1 + cot2 θ = csc2 θ


Addition and subtraction formulas
sin⎛⎝α ± β⎞⎠ = sinαcosβ ± cosαsinβ


cos(α ± β) = cosαcosβ ∓ sinαsinβ


Double-angle formulas
sin(2θ) = 2sinθcosθ


cos(2θ) = 2cos2 θ − 1 = 1 − 2sin2 θ = cos2 θ − sin2 θ


Example 1.25
Solving Trigonometric Equations
For each of the following equations, use a trigonometric identity to find all solutions.


a. 1 + cos(2θ) = cosθ
b. sin(2θ) = tanθ


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Solution
a. Using the double-angle formula for cos(2θ), we see that θ is a solution of


1 + cos(2θ) = cosθ


if and only if
1 + 2cos2 θ − 1 = cosθ,


which is true if and only if
2cos2 θ − cosθ = 0.


To solve this equation, it is important to note that we need to factor the left-hand side and not divide bothsides of the equation by cosθ. The problem with dividing by cosθ is that it is possible that cosθ is
zero. In fact, if we did divide both sides of the equation by cosθ, we would miss some of the solutions
of the original equation. Factoring the left-hand side of the equation, we see that θ is a solution of this
equation if and only if


cosθ(2cosθ − 1) = 0.


Since cosθ = 0 when
θ = π


2
, π
2
± π, π


2
± 2π,…,


and cosθ = 1/2 when
θ = π


3
, π
3
± 2π,… or θ = − π


3
, − π


3
± 2π,…,


we conclude that the set of solutions to this equation is
θ = π


2
+ nπ, θ = π


3
+ 2nπ, and θ = − π


3
+ 2nπ, n = 0, ± 1, ± 2,….


b. Using the double-angle formula for sin(2θ) and the reciprocal identity for tan(θ), the equation can be
written as


2sinθcosθ = sinθ
cosθ


.


To solve this equation, we multiply both sides by cosθ to eliminate the denominator, and say that if θ
satisfies this equation, then θ satisfies the equation


2sinθcos2 θ − sinθ = 0.


However, we need to be a little careful here. Even if θ satisfies this new equation, it may not satisfy the
original equation because, to satisfy the original equation, we would need to be able to divide both sidesof the equation by cosθ. However, if cosθ = 0, we cannot divide both sides of the equation by cosθ.
Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to checkfor extraneous solutions. Returning to the equation, it is important that we factor sinθ out of both terms
on the left-hand side instead of dividing both sides of the equation by sinθ. Factoring the left-hand side
of the equation, we can rewrite this equation as


sinθ(2cos2 θ − 1) = 0.


Chapter 1 | Functions and Graphs 67




1.20


1.21


Therefore, the solutions are given by the angles θ such that sinθ = 0 or cos2 θ = 1/2. The solutions
of the first equation are θ = 0, ± π, ± 2π,…. The solutions of the second equation are
θ = π/4, (π/4) ± (π/2), (π/4) ± π,…. After checking for extraneous solutions, the set of solutions to the
equation is


θ = nπ and θ = π
4
+ nπ


2
, n = 0, ± 1, ± 2,….


Find all solutions to the equation cos(2θ) = sinθ.


Example 1.26
Proving a Trigonometric Identity
Prove the trigonometric identity 1 + tan2 θ = sec2 θ.
Solution
We start with the identity


sin2 θ + cos2 θ = 1.


Dividing both sides of this equation by cos2 θ, we obtain
sin2 θ
cos2 θ


+ 1 = 1
cos2 θ


.


Since sinθ/cosθ = tanθ and 1/cosθ = secθ, we conclude that
tan2 θ + 1 = sec2 θ.


Prove the trigonometric identity 1 + cot2 θ = csc2 θ.


Graphs and Periods of the Trigonometric Functions
We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see thispattern in the graphs of the functions. Let P = (x, y) be a point on the unit circle and let θ be the corresponding angle
. Since the angle θ and θ + 2π correspond to the same point P, the values of the trigonometric functions at θ and
at θ + 2π are the same. Consequently, the trigonometric functions are periodic functions. The period of a function f is
defined to be the smallest positive value p such that f (x + p) = f (x) for all values x in the domain of f . The sine,
cosine, secant, and cosecant functions have a period of 2π. Since the tangent and cotangent functions repeat on an interval
of length π, their period is π (Figure 1.34).


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Figure 1.34 The six trigonometric functions are periodic.


Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider thefollowing function:
(1.10)f (x) = Asin⎛⎝B(x − α)⎞⎠+ C.


In Figure 1.35, the constant α causes a horizontal or phase shift. The factor B changes the period. This transformed
sine function will have a period 2π/|B|. The factor A results in a vertical stretch by a factor of |A|. We say |A| is the
“amplitude of f . ” The constant C causes a vertical shift.


Figure 1.35 A graph of a general sine function.


Notice in Figure 1.34 that the graph of y = cosx is the graph of y = sinx shifted to the left π/2 units. Therefore, we


Chapter 1 | Functions and Graphs 69




can write cosx = sin(x + π/2). Similarly, we can view the graph of y = sinx as the graph of y = cosx shifted right π/2
units, and state that sinx = cos(x − π/2).
A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function ofthe day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.7 hours and
December 21 is the shortest day of the year with 8.3 hours. It can be shown that the function


h(t) = 3.7sin⎛⎝

365


(x − 80.5)⎞⎠+ 12


is a model for the number of hours of daylight h as a function of day of the year t (Figure 1.36).


Figure 1.36 The hours of daylight as a function of day of the year can be modeledby a shifted sine curve.


Example 1.27
Sketching the Graph of a Transformed Sine Curve
Sketch a graph of f (x) = 3sin⎛⎝2⎛⎝x − π4⎞⎠⎞⎠+ 1.


Solution
This graph is a phase shift of y = sin(x) to the right by π/4 units, followed by a horizontal compression by a
factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of f is π.


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1.22 Describe the relationship between the graph of f (x) = 3sin(4x) − 5 and the graph of y = sin(x).


Chapter 1 | Functions and Graphs 71




1.3 EXERCISES
For the following exercises, convert each angle in degreesto radians. Write the answer as a multiple of π.
113. 240°
114. 15°
115. −60°
116. −225°
117. 330°
For the following exercises, convert each angle in radiansto degrees.
118. π


2
rad


119. 7π
6


rad


120. 11π
2


rad


121. −3π rad
122. 5π


12
rad


Evaluate the following functional values.
123. cos⎛⎝4π3 ⎞⎠
124. tan⎛⎝19π4 ⎞⎠
125. sin⎛⎝−3π4 ⎞⎠
126. sec⎛⎝π6⎞⎠
127. sin⎛⎝ π12⎞⎠
128. cos⎛⎝5π12⎞⎠
For the following exercises, consider triangle ABC, a righttriangle with a right angle at C. a. Find the missing side ofthe triangle. b. Find the six trigonometric function valuesfor the angle at A. Where necessary, round to one decimalplace.


129. a = 4, c = 7
130. a = 21, c = 29
131. a = 85.3, b = 125.5
132. b = 40, c = 41
133. a = 84, b = 13
134. b = 28, c = 35
For the following exercises, P is a point on the unit circle.
a. Find the (exact) missing coordinate value of each pointand b. find the values of the six trigonometric functions forthe angle θ with a terminal side that passes through point
P. Rationalize denominators.
135. P⎛⎝ 725, y⎞⎠, y > 0


136. P⎛⎝−1517 , y⎞⎠, y < 0


137. P⎛⎝x, 73 ⎞⎠, x < 0


138. P⎛⎝x, − 154 ⎞⎠, x > 0
For the following exercises, simplify each expression bywriting it in terms of sines and cosines, then simplify. Thefinal answer does not have to be in terms of sine and cosineonly.
139. tan2 x + sinxcscx
140. secxsinxcotx
141. tan2 x


sec2 x


142. secx − cosx
143. (1 + tanθ)2 − 2tanθ


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144. sinx(cscx − sinx)
145. cos t


sin t
+ sin t


1 + cos t


146. 1 + tan2α
1 + cot2α


For the following exercises, verify that each equation is anidentity.
147. tanθcotθ


cscθ
= sinθ


148. sec2 θ
tanθ


= secθcscθ


149. sin tcsc t + cos tsec t = 1
150. sinx


cosx + 1
+ cosx − 1


sinx
= 0


151. cotγ + tanγ = secγcscγ
152. sin2 β + tan2 β + cos2 β = sec2 β
153. 1


1 − sinα
+ 1


1 + sinα
= 2sec2α


154. tanθ − cotθ
sinθcosθ


= sec2 θ − csc2 θ


For the following exercises, solve the trigonometricequations on the interval 0 ≤ θ < 2π.
155. 2sinθ − 1 = 0
156. 1 + cosθ = 1


2


157. 2tan2 θ = 2
158. 4sin2 θ − 2 = 0
159. 3cotθ + 1 = 0
160. 3secθ − 2 3 = 0
161. 2cosθsinθ = sinθ
162. csc2 θ + 2cscθ + 1 = 0
For the following exercises, each graph is of the form
y = AsinBx or y = AcosBx, where B > 0. Write the
equation of the graph.


163.


164.


165.


Chapter 1 | Functions and Graphs 73




166.


For the following exercises, find a. the amplitude, b. theperiod, and c. the phase shift with direction for eachfunction.
167. y = sin⎛⎝x − π4⎞⎠
168. y = 3cos(2x + 3)
169. y = −1


2
sin⎛⎝


1
4
x⎞⎠


170. y = 2cos⎛⎝x − π3⎞⎠
171. y = −3sin(πx + 2)
172. y = 4cos⎛⎝2x − π2⎞⎠
173. [T] The diameter of a wheel rolling on the groundis 40 in. If the wheel rotates through an angle of 120°,
how many inches does it move? Approximate to the nearestwhole inch.
174. [T] Find the length of the arc intercepted by centralangle θ in a circle of radius r. Round to the nearest
hundredth. a. r = 12.8 cm, θ = 5π


6
rad b. r = 4.378 cm,


θ = 7π
6


rad c. r = 0.964 cm, θ = 50° d. r = 8.55 cm,
θ = 325°


175. [T] As a point P moves around a circle, the measureof the angle changes. The measure of how fast the angleis changing is called angular speed, ω, and is given by
ω = θ/t, where θ is in radians and t is time. Find the
angular speed for the given data. Round to the nearest
thousandth. a. θ = 7π


4
rad, t = 10 sec b.


θ = 3π
5
rad, t = 8 sec c. θ = 2π


9
rad, t = 1 min d.


θ = 23.76rad,t = 14 min
176. [T] A total of 250,000 m2 of land is needed to build anuclear power plant. Suppose it is decided that the area onwhich the power plant is to be built should be circular.a. Find the radius of the circular land area.b. If the land area is to form a 45° sector of a circle


instead of a whole circle, find the length of thecurved side.
177. [T] The area of an isosceles triangle with equal sides
of length x is 1


2
x2 sinθ, where θ is the angle formed by


the two sides. Find the area of an isosceles triangle withequal sides of length 8 in. and angle θ = 5π/12 rad.
178. [T] A particle travels in a circular path at a constantangular speed ω. The angular speed is modeled by the
function ω = 9|cos(πt − π/12)|. Determine the angular
speed at t = 9 sec.
179. [T] An alternating current for outlets in a home hasvoltage given by the function V(t) = 150cos368t, where
V is the voltage in volts at time t in seconds.a. Find the period of the function and interpret itsmeaning.b. Determine the number of periods that occur when 1sec has passed.
180. [T] The number of hours of daylight in a northeastcity is modeled by the function


N(t) = 12 + 3sin⎡⎣

365


(t − 79)⎤⎦,


where t is the number of days after January 1.a. Find the amplitude and period.b. Determine the number of hours of daylight on thelongest day of the year.c. Determine the number of hours of daylight on theshortest day of the year.d. Determine the number of hours of daylight 90 daysafter January 1.e. Sketch the graph of the function for one periodstarting on January 1.


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181. [T] Suppose that T = 50 + 10sin⎡⎣ π12(t − 8)⎤⎦ is a
mathematical model of the temperature (in degreesFahrenheit) at t hours after midnight on a certain day of theweek.a. Determine the amplitude and period.b. Find the temperature 7 hours after midnight.c. At what time does T = 60°?


d. Sketch the graph of T over 0 ≤ t ≤ 24.
182. [T] The function H(t) = 8sin⎛⎝π6 t⎞⎠ models the height
H (in feet) of the tide t hours after midnight. Assume that
t = 0 is midnight.


a. Find the amplitude and period.b. Graph the function over one period.c. What is the height of the tide at 4:30 a.m.?


Chapter 1 | Functions and Graphs 75




1.4 | Inverse Functions
Learning Objectives


1.4.1 Determine the conditions for when a function has an inverse.
1.4.2 Use the horizontal line test to recognize when a function is one-to-one.
1.4.3 Find the inverse of a given function.
1.4.4 Draw the graph of an inverse function.
1.4.5 Evaluate inverse trigonometric functions.


An inverse function reverses the operation done by a particular function. In other words, whatever a function does, theinverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for aninverse function to exist. We examine how to find an inverse function and study the relationship between the graph of afunction and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometricfunctions.
Existence of an Inverse Function
We begin with an example. Given a function f and an output y = f (x), we are often interested in finding what
value or values x were mapped to y by f . For example, consider the function f (x) = x3 + 4. Since any output
y = x3 + 4, we can solve this equation for x to find that the input is x = y − 43 . This equation defines x as a function
of y. Denoting this function as f −1, and writing x = f −1 (y) = y − 43 , we see that for any x in the domain of
f , f −1 ⎛⎝ f (x)⎞⎠ = f −1 ⎛⎝x


3 + 4⎞⎠ = x. Thus, this new function, f −1, “undid” what the original function f did. A function
with this property is called the inverse function of the original function.
Definition
Given a function f with domain D and range R, its inverse function (if it exists) is the function f −1 with domain
R and range D such that f −1 (y) = x if f (x) = y. In other words, for a function f and its inverse f −1,


(1.11)f −1 ⎛⎝ f (x)⎞⎠ = x for all x inD, and f ⎛⎝ f −1 (y)⎞⎠ = y for all y in R.


Note that f −1 is read as “f inverse.” Here, the −1 is not used as an exponent and f −1 (x) ≠ 1/ f (x). Figure 1.37 shows
the relationship between the domain and range of f and the domain and range of f −1.


Figure 1.37 Given a function f and its inverse
f −1, f −1 (y) = x if and only if f (x) = y. The range of f
becomes the domain of f −1 and the domain of f becomes the
range of f −1.


Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each


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input to exactly one output. For example, let’s try to find the inverse function for f (x) = x2. Solving the equation y = x2
for x, we arrive at the equation x = ± y. This equation does not describe x as a function of y because there are two
solutions to this equation for every y > 0. The problem with trying to find an inverse function for f (x) = x2 is that two
inputs are sent to the same output for each output y > 0. The function f (x) = x3 + 4 discussed earlier did not have this
problem. For that function, each input was sent to a different output. A function that sends each input to a different outputis called a one-to-one function.
Definition
We say a f is a one-to-one function if f (x1) ≠ f (x2) when x1 ≠ x2.


One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no twoinputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the xy -plane, according to the
horizontal line test, it cannot intersect the graph more than once. We note that the horizontal line test is different fromthe vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line testdetermines whether a function is one-to-one (Figure 1.38).
Rule: Horizontal Line Test
A function f is one-to-one if and only if every horizontal line intersects the graph of f no more than once.


Figure 1.38 (a) The function f (x) = x2 is not one-to-one
because it fails the horizontal line test. (b) The function
f (x) = x3 is one-to-one because it passes the horizontal line
test.


Example 1.28
Determining Whether a Function Is One-to-One
For each of the following functions, use the horizontal line test to determine whether it is one-to-one.


Chapter 1 | Functions and Graphs 77




a.


b.
Solution


a. Since the horizontal line y = n for any integer n ≥ 0 intersects the graph more than once, this function
is not one-to-one.


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1.23


b. Since every horizontal line intersects the graph once (at most), this function is one-to-one.


Is the function f graphed in the following image one-to-one?


Chapter 1 | Functions and Graphs 79




Finding a Function’s Inverse
We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements inthe domain of f to elements in the range of f . The inverse function maps each element from the range of f back to its
corresponding element from the domain of f . Therefore, to find the inverse function of a one-to-one function f , given
any y in the range of f , we need to determine which x in the domain of f satisfies f (x) = y. Since f is one-to-one,
there is exactly one such value x. We can find that value x by solving the equation f (x) = y for x. Doing so, we are
able to write x as a function of y where the domain of this function is the range of f and the range of this new function
is the domain of f . Consequently, this function is the inverse of f , and we write x = f −1(y). Since we typically use the
variable x to denote the independent variable and y to denote the dependent variable, we often interchange the roles of x
and y, and write y = f −1(x). Representing the inverse function in this way is also helpful later when we graph a function
f and its inverse f −1 on the same axes.


Problem-Solving Strategy: Finding an Inverse Function
1. Solve the equation y = f (x) for x.
2. Interchange the variables x and y and write y = f −1(x).


Example 1.29
Finding an Inverse Function
Find the inverse for the function f (x) = 3x − 4. State the domain and range of the inverse function. Verify that
f −1( f (x)) = x.


Solution
Follow the steps outlined in the strategy.
Step 1. If y = 3x − 4, then 3x = y + 4 and x = 1


3
y + 4


3
.


Step 2. Rewrite as y = 1
3
x + 4


3
and let y = f −1 (x).


Therefore, f −1 (x) = 1
3
x + 4


3
.


Since the domain of f is (−∞, ∞), the range of f −1 is (−∞, ∞). Since the range of f is (−∞, ∞), the
domain of f −1 is (−∞, ∞).
You can verify that f −1( f (x)) = x by writing


f −1( f (x)) = f −1(3x − 4) = 1
3
(3x − 4) + 4


3
= x − 4


3
+ 4


3
= x.


Note that for f −1(x) to be the inverse of f (x), both f −1( f (x)) = x and f ( f −1(x)) = x for all x in the domain
of the inside function.


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1.24 Find the inverse of the function f (x) = 3x/(x − 2). State the domain and range of the inverse function.


Graphing Inverse Functions
Let’s consider the relationship between the graph of a function f and the graph of its inverse. Consider the graph of f
shown in Figure 1.39 and a point (a, b) on the graph. Since b = f (a), then f −1 (b) = a. Therefore, when we graph
f −1, the point (b, a) is on the graph. As a result, the graph of f −1 is a reflection of the graph of f about the line
y = x.


Figure 1.39 (a) The graph of this function f shows point (a, b) on the graph of f . (b)
Since (a, b) is on the graph of f , the point (b, a) is on the graph of f −1. The graph of
f −1 is a reflection of the graph of f about the line y = x.


Example 1.30
Sketching Graphs of Inverse Functions
For the graph of f in the following image, sketch a graph of f −1 by sketching the line y = x and using
symmetry. Identify the domain and range of f −1.


Chapter 1 | Functions and Graphs 81




1.25


Solution
Reflect the graph about the line y = x. The domain of f −1 is [0, ∞). The range of f −1 is [−2, ∞). By using
the preceding strategy for finding inverse functions, we can verify that the inverse function is f −1 (x) = x2 − 2,
as shown in the graph.


Sketch the graph of f (x) = 2x + 3 and the graph of its inverse using the symmetry property of inverse
functions.


Restricting Domains
As we have seen, f (x) = x2 does not have an inverse function because it is not one-to-one. However, we can choose a
subset of the domain of f such that the function is one-to-one. This subset is called a restricted domain. By restricting the
domain of f , we can define a new function g such that the domain of g is the restricted domain of f and g(x) = f (x)
for all x in the domain of g. Then we can define an inverse function for g on that domain. For example, since f (x) = x2
is one-to-one on the interval [0, ∞), we can define a new function g such that the domain of g is [0, ∞) and g(x) = x2
for all x in its domain. Since g is a one-to-one function, it has an inverse function, given by the formula g−1(x) = x. On
the other hand, the function f (x) = x2 is also one-to-one on the domain (−∞, 0]. Therefore, we could also define a new
function h such that the domain of h is (−∞, 0] and h(x) = x2 for all x in the domain of h. Then h is a one-to-one
function and must also have an inverse. Its inverse is given by the formula h−1(x) = − x (Figure 1.40).


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Figure 1.40 (a) For g(x) = x2 restricted to [0, ∞), g−1 (x) = x. (b) For
h(x) = x2 restricted to (−∞, 0], h−1 (x) = − x.


Example 1.31
Restricting the Domain
Consider the function f (x) = (x + 1)2.


a. Sketch the graph of f and use the horizontal line test to show that f is not one-to-one.
b. Show that f is one-to-one on the restricted domain [−1, ∞). Determine the domain and range for the


inverse of f on this restricted domain and find a formula for f −1.
Solution


a. The graph of f is the graph of y = x2 shifted left 1 unit. Since there exists a horizontal line intersecting
the graph more than once, f is not one-to-one.


b. On the interval [−1, ∞), f is one-to-one.


Chapter 1 | Functions and Graphs 83




1.26


The domain and range of f −1 are given by the range and domain of f , respectively. Therefore, the
domain of f −1 is [0, ∞) and the range of f −1 is [−1, ∞). To find a formula for f −1, solve the
equation y = (x + 1)2 for x. If y = (x + 1)2, then x = −1 ± y. Since we are restricting the domain
to the interval where x ≥ −1, we need ± y ≥ 0. Therefore, x = −1 + y. Interchanging x and y,
we write y = −1 + x and conclude that f −1 (x) = −1 + x.


Consider f (x) = 1/x2 restricted to the domain (−∞, 0). Verify that f is one-to-one on this domain.
Determine the domain and range of the inverse of f and find a formula for f −1.


Inverse Trigonometric Functions
The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domainof a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function(Figure 1.34). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict
the domain to the interval ⎡⎣−π2, π2⎤⎦. By doing so, we define the inverse sine function on the domain [−1, 1] such that
for any x in the interval [−1, 1], the inverse sine function tells us which angle θ in the interval ⎡⎣−π2, π2⎤⎦ satisfies
sinθ = x. Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric
functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.
Definition
The inverse sine function, denoted sin−1 or arcsin, and the inverse cosine function, denoted cos−1 or arccos, are
defined on the domain D = {x| − 1 ≤ x ≤ 1} as follows:


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(1.12)sin−1 (x) = y if and only if sin(y) = x and − π
2
≤ y ≤ π


2
;


cos−1 (x) = y if and only if cos(y) = x and 0 ≤ y ≤ π.


The inverse tangent function, denoted tan−1 or arctan, and inverse cotangent function, denoted cot−1 or arccot, are
defined on the domain D = {x| − ∞ < x < ∞} as follows:


(1.13)tan−1 (x) = y if and only if tan(y) = x and − π
2
< y < π


2
;


cot−1 (x) = y if and only if cot(y) = x and 0 < y < π.


The inverse cosecant function, denoted csc−1 or arccsc, and inverse secant function, denoted sec−1 or arcsec, are
defined on the domain D = {x||x| ≥ 1} as follows:


(1.14)csc−1 (x) = y if and only if csc(y) = x and − π
2
≤ y ≤ π


2
, y ≠ 0;


sec−1 (x) = y if and only if sec(y) = x and 0 ≤ y ≤ π, y ≠ π/2.


To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domainsdefined earlier and reflect the graphs about the line y = x (Figure 1.41).


Figure 1.41 The graph of each of the inverse trigonometric functions is a reflection about the line y = x of
the corresponding restricted trigonometric function.
Go to the following site (http://www.openstaxcollege.org/l/20_inversefun) for more comparisons offunctions and their inverses.


Chapter 1 | Functions and Graphs 85




When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate cos−1 ⎛⎝12⎞⎠, we need to
find an angle θ such that cosθ = 1


2
. Clearly, many angles have this property. However, given the definition of cos−1, we


need the angle θ that not only solves this equation, but also lies in the interval [0, π]. We conclude that cos−1 ⎛⎝12⎞⎠ = π3.
We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions
sin⎛⎝sin


−1 ⎛


2
2



⎠ and sin−1(sin(π)). For the first one, we simplify as follows:


sin⎛⎝sin
−1 ⎛


2
2



⎠ = sin




π
4

⎠ =


2
2
.


For the second one, we have
sin−1 ⎛⎝sin(π)⎞⎠ = sin−1 (0) = 0.


The inverse function is supposed to “undo” the original function, so why isn’t sin−1 ⎛⎝sin(π)⎞⎠ = π ? Recalling our definition
of inverse functions, a function f and its inverse f −1 satisfy the conditions f ⎛⎝ f −1 (y)⎞⎠ = y for all y in the domain of
f −1 and f −1 ⎛⎝ f (x)⎞⎠ = x for all x in the domain of f , so what happened here? The issue is that the inverse sine function,
sin−1, is the inverse of the restricted sine function defined on the domain ⎡⎣−π2, π2⎤⎦. Therefore, for x in the interval

⎣−


π
2
, π
2

⎦, it is true that sin−1 (sinx) = x. However, for values of x outside this interval, the equation does not hold, even


though sin−1(sinx) is defined for all real numbers x.
What about sin(sin−1 y)? Does that have a similar issue? The answer is no. Since the domain of sin−1 is the interval
[−1, 1], we conclude that sin(sin−1 y) = y if −1 ≤ y ≤ 1 and the expression is not defined for other values of y. To
summarize,


sin(sin−1 y) = y if −1 ≤ y ≤ 1


and
sin−1 (sinx) = x if − π


2
≤ x ≤ π


2
.


Similarly, for the cosine function,
cos(cos−1 y) = y if −1 ≤ y ≤ 1


and
cos−1 (cosx) = x if 0 ≤ x ≤ π.


Similar properties hold for the other trigonometric functions and their inverses.
Example 1.32
Evaluating Expressions Involving Inverse Trigonometric Functions
Evaluate each of the following expressions.


a. sin−1 ⎛⎝− 32 ⎞⎠
b. tan⎛⎝tan−1 ⎛⎝− 13







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c. cos−1 ⎛⎝cos⎛⎝5π4 ⎞⎠⎞⎠
d. sin−1 ⎛⎝cos⎛⎝2π3 ⎞⎠⎞⎠


Solution
a. Evaluating sin−1 ⎛⎝− 3/2⎞⎠ is equivalent to finding the angle θ such that sinθ = − 3/2 and


−π/2 ≤ θ ≤ π/2. The angle θ = −π/3 satisfies these two conditions. Therefore,
sin−1 ⎛⎝− 3/2



⎠ = −π/3.


b. First we use the fact that tan−1 ⎛⎝−1/ 3⎞⎠ = −π/6. Then tan(π/6) = −1/ 3. Therefore,
tan⎛⎝tan


−1 ⎛
⎝−1/ 3





⎠ = −1/ 3.


c. To evaluate cos−1 ⎛⎝cos(5π/4)⎞⎠, first use the fact that cos(5π/4) = − 2/2. Then we need to find the
angle θ such that cos(θ) = − 2/2 and 0 ≤ θ ≤ π. Since 3π/4 satisfies both these conditions, we have
cos⎛⎝cos


−1 (5π/4)⎞⎠ = cos

⎝cos


−1 ⎛
⎝− 2/2





⎠ = 3π/4.


d. Since cos(2π/3) = −1/2, we need to evaluate sin−1 (−1/2). That is, we need to find the angle θ such
that sin(θ) = −1/2 and −π/2 ≤ θ ≤ π/2. Since −π/6 satisfies both these conditions, we can conclude
that sin−1 (cos(2π/3)) = sin−1 (−1/2) = −π/6.


Chapter 1 | Functions and Graphs 87




The Maximum Value of a Function
In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain,even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strengthof a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have afunction that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails.Safe design often depends on knowing maximum values.
This project describes a simple example of a function with a maximum value that depends on two equation coefficients.We will see that maximum values can depend on several factors other than the independent variable x.


1. Consider the graph in Figure 1.42 of the function y = sinx + cosx. Describe its overall shape. Is it periodic?
How do you know?


Figure 1.42 The graph of y = sinx + cosx.


Using a graphing calculator or other graphing device, estimate the x - and y -values of the maximum point for
the graph (the first such point where x > 0). It may be helpful to express the x -value as a multiple of π.


2. Now consider other graphs of the form y = Asinx + Bcosx for various values of A and B. Sketch the graph
when A = 2 and B = 1, and find the x - and y-values for the maximum point. (Remember to express the x-value
as a multiple of π, if possible.) Has it moved?


3. Repeat for A = 1, B = 2. Is there any relationship to what you found in part (2)?
4. Complete the following table, adding a few choices of your own for A and B:


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A B x y A B x y
0 1 3 1


1 0 1 3


1 1 12 5
1 2 5 12
2 1
2 2
3 4
4 3


5. Try to figure out the formula for the y-values.
6. The formula for the x -values is a little harder. The most helpful points from the table are


(1, 1), ⎛⎝1, 3

⎠,

⎝ 3, 1



⎠. (Hint: Consider inverse trigonometric functions.)


7. If you found formulas for parts (5) and (6), show that they work together. That is, substitute the x -value
formula you found into y = Asinx + Bcosx and simplify it to arrive at the y -value formula you found.


Chapter 1 | Functions and Graphs 89




1.4 EXERCISES
For the following exercises, use the horizontal line test todetermine whether each of the given graphs is one-to-one.
183.


184.


185.


186.


187.


188.


For the following exercises, a. find the inverse function,and b. find the domain and range of the inverse function.
189. f (x) = x2 − 4, x ≥ 0
190. f (x) = x − 43


191. f (x) = x3 + 1
192. f (x) = (x − 1)2, x ≤ 1


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193. f (x) = x − 1
194. f (x) = 1


x + 2


For the following exercises, use the graph of f to sketch
the graph of its inverse function.
195.


196.


197.


198.


For the following exercises, use composition to determinewhich pairs of functions are inverses.
199. f (x) = 8x, g(x) = x


8


200. f (x) = 8x + 3, g(x) = x − 3
8


201. f (x) = 5x − 7, g(x) = x + 5
7


202. f (x) = 2
3
x + 2, g(x) = 3


2
x + 3


203. f (x) = 1
x − 1


, x ≠ 1, g(x) = 1x + 1, x ≠ 0


204. f (x) = x3 + 1, g(x) = (x − 1)1/3
205.
f (x) = x2 + 2x + 1, x ≥ −1, g(x) = −1 + x, x ≥ 0


206.
f (x) = 4 − x2, 0 ≤ x ≤ 2, g(x) = 4 − x2, 0 ≤ x ≤ 2


For the following exercises, evaluate the functions. Givethe exact value.
207. tan−1 ⎛⎝ 33 ⎞⎠
208. cos−1 ⎛⎝− 22 ⎞⎠
209. cot−1 (1)
210. sin−1 (−1)
211. cos−1 ⎛⎝ 32 ⎞⎠


Chapter 1 | Functions and Graphs 91




212. cos⎛⎝tan−1 ⎛⎝ 3⎞⎠⎞⎠
213. sin⎛⎝cos−1 ⎛⎝ 22 ⎞⎠⎞⎠
214. sin−1 ⎛⎝sin⎛⎝π3⎞⎠⎞⎠
215. tan−1 ⎛⎝tan⎛⎝−π6⎞⎠⎞⎠
216. The function C = T(F) = (5/9)(F − 32) converts
degrees Fahrenheit to degrees Celsius.


a. Find the inverse function F = T−1(C)
b. What is the inverse function used for?


217. [T] The velocity V (in centimeters per second) ofblood in an artery at a distance x cm from the center ofthe artery can be modeled by the function
V = f (x) = 500(0.04 − x2) for 0 ≤ x ≤ 0.2.


a. Find x = f −1(V).
b. Interpret what the inverse function is used for.c. Find the distance from the center of an artery witha velocity of 15 cm/sec, 10 cm/sec, and 5 cm/sec.


218. A function that converts dress sizes in the UnitedStates to those in Europe is given by D(x) = 2x + 24.
a. Find the European dress sizes that correspond tosizes 6, 8, 10, and 12 in the United States.b. Find the function that converts European dresssizes to U.S. dress sizes.c. Use part b. to find the dress sizes in the UnitedStates that correspond to 46, 52, 62, and 70.


219. [T] The cost to remove a toxin from a lake ismodeled by the function C(p) = 75p/(85 − p), where
C is the cost (in thousands of dollars) and p is the amount
of toxin in a small lake (measured in parts per billion[ppb]). This model is valid only when the amount of toxinis less than 85 ppb.a. Find the cost to remove 25 ppb, 40 ppb, and 50 ppbof the toxin from the lake.b. Find the inverse function. c. Use part b. todetermine how much of the toxin is removed for$50,000.
220. [T] A race car is accelerating at a velocity given
by v(t) = 25


4
t + 54, where v is the velocity (in feet per


second) at time t.a. Find the velocity of the car at 10 sec.b. Find the inverse function.c. Use part b. to determine how long it takes for thecar to reach a speed of 150 ft/sec.


221. [T] An airplane’s Mach number M is the ratio ofits speed to the speed of sound. When a plane is flyingat a constant altitude, then its Mach angle is given by
µ = 2sin−1 ⎛⎝


1
M

⎠. Find the Mach angle (to the nearest


degree) for the following Mach numbers.


a. µ = 1.4
b. µ = 2.8
c. µ = 4.3


222. [T] Using µ = 2sin−1 ⎛⎝ 1M⎞⎠, find the Mach number
M for the following angles.a. µ = π


6


b. µ = 2π
7


c. µ = 3π
8


223. [T] The temperature (in degrees Celsius) of a city inthe northern United States can be modeled by the function
T(x) = 5 + 18sin⎡⎣


π
6
(x − 4.6)⎤⎦, where x is time in


months and x = 1.00 corresponds to January 1. Determine
the month and day when the temperature is 21°C.
224. [T] The depth (in feet) of water at a dock changeswith the rise and fall of tides. It is modeled by the function
D(t) = 5sin⎛⎝


π
6
t − 7π


6

⎠+ 8, where t is the number of


hours after midnight. Determine the first time aftermidnight when the depth is 11.75 ft.
225. [T] An object moving in simple harmonic motion
is modeled by the function s(t) = −6cos⎛⎝πt2 ⎞⎠, where s
is measured in inches and t is measured in seconds.
Determine the first time when the distance moved is 4.5 ft.


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226. [T] A local art gallery has a portrait 3 ft in heightthat is hung 2.5 ft above the eye level of an average person.The viewing angle θ can be modeled by the function
θ = tan−1 5.5x − tan


−1 2.5
x , where x is the distance (in


feet) from the portrait. Find the viewing angle when aperson is 4 ft from the portrait.
227. [T]Use a calculator to evaluate tan−1 (tan(2.1)) and
cos−1 (cos(2.1)). Explain the results of each.
228. [T] Use a calculator to evaluate sin(sin−1(−2)) and
tan(tan−1(−2)). Explain the results of each.


Chapter 1 | Functions and Graphs 93




1.5 | Exponential and Logarithmic Functions
Learning Objectives


1.5.1 Identify the form of an exponential function.
1.5.2 Explain the difference between the graphs of xb and bx.
1.5.3 Recognize the significance of the number e.
1.5.4 Identify the form of a logarithmic function.
1.5.5 Explain the relationship between exponential and logarithmic functions.
1.5.6 Describe how to calculate a logarithm to a different base.
1.5.7 Identify the hyperbolic functions, their graphs, and basic identities.


In this section we examine exponential and logarithmic functions. We use the properties of these functions to solveequations involving exponential or logarithmic terms, and we study the meaning and importance of the number e. We also
define hyperbolic and inverse hyperbolic functions, which involve combinations of exponential and logarithmic functions.(Note that we present alternative definitions of exponential and logarithmic functions in the chapter Applications ofIntegrations, and prove that the functions have the same properties with either definition.)
Exponential Functions
Exponential functions arise in many applications. One common example is population growth.
For example, if a population starts with P0 individuals and then grows at an annual rate of 2%, its population after 1 year
is


P(1) = P0 + 0.02P0 = P0(1 + 0.02) = P0(1.02).


Its population after 2 years is
P(2) = P(1) + 0.02P(1) = P(1)(1.02) = P0 (1.02)


2.


In general, its population after t years is
P(t) = P0 (1.02)


t,


which is an exponential function. More generally, any function of the form f (x) = bx, where b > 0, b ≠ 1, is an
exponential function with base b and exponent x. Exponential functions have constant bases and variable exponents. Note
that a function of the form f (x) = xb for some constant b is not an exponential function but a power function.
To see the difference between an exponential function and a power function, we compare the functions y = x2 and y = 2x.
In Table 1.10, we see that both 2x and x2 approach infinity as x → ∞. Eventually, however, 2x becomes larger than
x2 and grows more rapidly as x → ∞. In the opposite direction, as x → −∞, x2 → ∞, whereas 2x → 0. The line
y = 0 is a horizontal asymptote for y = 2x.


x −3 −2 −1 0 1 2 3 4 5 6


x2 9 4 1 0 1 4 9 16 25 36


2x 1/8 1/4 1/2 1 2 4 8 16 32 64


Table 1.10 Values of x2 and 2x


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In Figure 1.43, we graph both y = x2 and y = 2x to show how the graphs differ.


Figure 1.43 Both 2x and x2 approach infinity as x → ∞,
but 2x grows more rapidly than x2. As
x → −∞, x2 → ∞, whereas 2x → 0.


Evaluating Exponential Functions
Recall the properties of exponents: If x is a positive integer, then we define bx = b · b ⋯ b (with x factors of b). If x
is a negative integer, then x = −y for some positive integer y, and we define bx = b−y = 1/by. Also, b0 is defined
to be 1. If x is a rational number, then x = p/q, where p and q are integers and bx = b p/q = b pq . For example,
93/2 = 93 = 27. However, how is bx defined if x is an irrational number? For example, what do we mean by 2 2?
This is too complex a question for us to answer fully right now; however, we can make an approximation. In Table 1.11,
we list some rational numbers approaching 2, and the values of 2x for each rational number x are presented as well.
We claim that if we choose rational numbers x getting closer and closer to 2, the values of 2x get closer and closer to
some number L. We define that number L to be 2 2.


x 1.4 1.41 1.414 1.4142 1.41421 1.414213


2x 2.639 2.65737 2.66475 2.665119 2.665138 2.665143


Table 1.11 Values of 2x for a List of Rational Numbers Approximating 2


Example 1.33
Bacterial Growth
Suppose a particular population of bacteria is known to double in size every 4 hours. If a culture starts with
1000 bacteria, the number of bacteria after 4 hours is n(4) = 1000 · 2. The number of bacteria after 8 hours is
n(8) = n(4) · 2 = 1000 · 22. In general, the number of bacteria after 4m hours is n(4m) = 1000 · 2m. Letting


Chapter 1 | Functions and Graphs 95




1.27


t = 4m, we see that the number of bacteria after t hours is n(t) = 1000 · 2t/4. Find the number of bacteria
after 6 hours, 10 hours, and 24 hours.
Solution
The number of bacteria after 6 hours is given by n(6) = 1000 · 26/4 ≈ 2828 bacteria. The number of bacteria
after 10 hours is given by n(10) = 1000 · 210/4 ≈ 5657 bacteria. The number of bacteria after 24 hours is
given by n(24) = 1000 · 26 = 64,000 bacteria.


Given the exponential function f (x) = 100 · 3x/2, evaluate f (4) and f (10).


Go to World Population Balance (http://www.openstaxcollege.org/l/20_exponengrow) for anotherexample of exponential population growth.


Graphing Exponential Functions
For any base b > 0, b ≠ 1, the exponential function f (x) = bx is defined for all real numbers x and bx > 0. Therefore,
the domain of f (x) = bx is (−∞, ∞) and the range is (0, ∞). To graph bx, we note that for b > 1, bx is increasing
on (−∞, ∞) and bx → ∞ as x → ∞, whereas bx → 0 as x → −∞. On the other hand, if 0 < b < 1, f (x) = bx is
decreasing on (−∞, ∞) and bx → 0 as x → ∞ whereas bx → ∞ as x → −∞ (Figure 1.44).


Figure 1.44 If b > 1, then bx is increasing on (−∞, ∞).
If 0 < b < 1, then bx is decreasing on (−∞, ∞).


Visit this site (http://www.openstaxcollege.org/l/20_inverse) for more exploration of the graphs ofexponential functions.


Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.
Rule: Laws of Exponents
For any constants a > 0, b > 0, and for all x and y,


1. bx · by = bx + y
2. bx


by
= b


x − y


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1.28


3. (bx)y = bxy
4. (ab)x = ax bx
5. ax


bx
= ⎛⎝


a
b



x


Example 1.34
Using the Laws of Exponents
Use the laws of exponents to simplify each of the following expressions.


a.

⎝2x


2/3⎞


3



⎝4x


−1/3⎞


2


b.

⎝x


3 y−1⎞⎠
2



⎝xy


2⎞


−2


Solution
a. We can simplify as follows:



⎝2x


2/3⎞


3



⎝4x


−1/3⎞


2
=


23 ⎛⎝x
2/3⎞


3


42 ⎛⎝x
−1/3⎞


2
= 8x


2


16x−2/3
= x


2 x2/3


2
= x


8/3


2
.


b. We can simplify as follows:

⎝x


3 y−1⎞⎠
2



⎝xy


2⎞


−2
=

⎝x


3⎞


2 ⎛
⎝y


−1⎞


2


x−2 ⎛⎝y
2⎞


−2
=


x6 y−2


x−2 y−4
= x6 x2 y−2 y4 = x8 y2.


Use the laws of exponents to simplify ⎛⎝6x−3 y2⎞⎠/⎛⎝12x−4 y5⎞⎠.


The Number e
A special type of exponential function appears frequently in real-world applications. To describe it, consider the followingexample of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests P
dollars in a savings account with an annual interest rate r, compounded annually. The amount of money after 1 year is


A(1) = P + rP = P(1 + r).


The amount of money after 2 years is
A(2) = A(1) + rA(1) = P(1 + r) + rP(1 + r) = P(1 + r)2.


More generally, the amount after t years is
A(t) = P(1 + r)t.


Chapter 1 | Functions and Graphs 97




If the money is compounded 2 times per year, the amount of money after half a year is
A⎛⎝


1
2

⎠ = P +




r
2

⎠P = P



⎝1 +


r
2



⎠.


The amount of money after 1 year is
A(1) = A⎛⎝


1
2

⎠+


r
2

⎠A


1
2

⎠ = P



⎝1 +


r
2

⎠+


r
2

⎝P

⎝1 +


r
2



⎠ = P



⎝1 +


r
2



2
.


After t years, the amount of money in the account is
A(t) = P⎛⎝1 +


r
2



2t
.


More generally, if the money is compounded n times per year, the amount of money in the account after t years is given
by the function


A(t) = P⎛⎝1 + rn



nt
.


What happens as n → ∞? To answer this question, we let m = n/r and write

⎝1 + rn





nt
= ⎛⎝1 +


1
m



mrt
,


and examine the behavior of (1 + 1/m)m as m → ∞, using a table of values (Table 1.12).
m 10 100 1000 10,000 100,000 1,000,000



⎝1 + 1m




m


2.5937 2.7048 2.71692 2.71815 2.718268 2.718280


Table 1.12 Values of ⎛⎝1 + 1m⎞⎠m as m → ∞
Looking at this table, it appears that (1 + 1/m)m is approaching a number between 2.7 and 2.8 as m → ∞. In fact,
(1 + 1/m)m does approach some number as m → ∞. We call this number e . To six decimal places of accuracy,


e ≈ 2.718282.


The letter e was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although
Euler did not discover the number, he showed many important connections between e and logarithmic functions. We still
use the notation e today to honor Euler’s work because it appears in many areas of mathematics and because we can use it
in many practical applications.
Returning to our savings account example, we can conclude that if a person puts P dollars in an account at an annual
interest rate r, compounded continuously, then A(t) = Pert. This function may be familiar. Since functions involving
base e arise often in applications, we call the function f (x) = ex the natural exponential function. Not only is this
function interesting because of the definition of the number e, but also, as discussed next, its graph has an important
property.
Since e > 1, we know ex is increasing on (−∞, ∞). In Figure 1.45, we show a graph of f (x) = ex along with a
tangent line to the graph of at x = 0. We give a precise definition of tangent line in the next chapter; but, informally, we
say a tangent line to a graph of f at x = a is a line that passes through the point ⎛⎝a, f (a)⎞⎠ and has the same “slope” as
f at that point . The function f (x) = ex is the only exponential function bx with tangent line at x = 0 that has a slope
of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponentialfunction to use in many instances.


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1.29


Figure 1.45 The graph of f (x) = ex has a tangent line with
slope 1 at x = 0.


Example 1.35
Compounding Interest
Suppose $500 is invested in an account at an annual interest rate of r = 5.5%, compounded continuously.


a. Let t denote the number of years after the initial investment and A(t) denote the amount of money in
the account at time t. Find a formula for A(t).


b. Find the amount of money in the account after 10 years and after 20 years.
Solution


a. If P dollars are invested in an account at an annual interest rate r, compounded continuously, then
A(t) = Pert. Here P = $500 and r = 0.055. Therefore, A(t) = 500e0.055t.


b. After 10 years, the amount of money in the account is
A(10) = 500e0.055 · 10 = 500e0.55 ≈ $866.63.


After 20 years, the amount of money in the account is
A(20) = 500e0.055 · 20 = 500e1.1 ≈ $1, 502.08.


If $750 is invested in an account at an annual interest rate of 4%, compounded continuously, find a
formula for the amount of money in the account after t years. Find the amount of money after 30 years.


Logarithmic Functions
Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. Thesecome in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistryor decibels in sound levels.
The exponential function f (x) = bx is one-to-one, with domain (−∞, ∞) and range (0, ∞). Therefore, it has an inverse
function, called the logarithmic function with base b. For any b > 0, b ≠ 1, the logarithmic function with base b,
denoted logb, has domain (0, ∞) and range (−∞, ∞), and satisfies


logb (x) = y if and only if b
y = x.


For example,


Chapter 1 | Functions and Graphs 99




log2(8) = 3 since 2
3 = 8,


log10



1
100

⎠ = −2 since 10


−2 = 1
102


= 1
100


,


logb(1) = 0 since b
0 = 1 for any base b > 0.


Furthermore, since y = logb(x) and y = bx are inverse functions,
logb (b


x) = x and b
logb (x) = x.


The most commonly used logarithmic function is the function loge. Since this function uses natural e as its base, it is
called the natural logarithm. Here we use the notation ln(x) or lnx to mean loge (x). For example,


ln(e) = loge (e) = 1, ln

⎝e


3⎞
⎠ = loge



⎝e


3⎞
⎠ = 3, ln(1) = loge (1) = 0.


Since the functions f (x) = ex and g(x) = ln(x) are inverses of each other,
ln(ex) = x and elnx = x,


and their graphs are symmetric about the line y = x (Figure 1.46).


Figure 1.46 The functions y = ex and y = ln(x) are
inverses of each other, so their graphs are symmetric about theline y = x.


At this site (http://www.openstaxcollege.org/l/20_logscale) you can see an example of a base-10logarithmic scale.


In general, for any base b > 0, b ≠ 1, the function g(x) = logb(x) is symmetric about the line y = x with the function
f (x) = bx. Using this fact and the graphs of the exponential functions, we graph functions logb for several values of
b > 1 (Figure 1.47).


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Figure 1.47 Graphs of y = logb(x) are depicted for
b = 2, e, 10.


Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties oflogarithms.
Rule: Properties of Logarithms
If a, b, c > 0, b ≠ 1, and r is any real number, then


1. logb (ac) = logb (a) + logb (c) (Product property)


2. logb


a
c

⎠ = logb (a) − logb (c) (Quotient property)


3. logb (a
r) = rlogb (a) (Power property)


Example 1.36
Solving Equations Involving Exponential Functions
Solve each of the following equations for x.


a. 5x = 2
b. ex + 6e−x = 5


Solution
a. Applying the natural logarithm function to both sides of the equation, we have


ln5x = ln2.


Using the power property of logarithms,
x ln5 = ln2.


Therefore, x = ln2/ln5.
b. Multiplying both sides of the equation by ex, we arrive at the equation


e2x + 6 = 5ex.


Chapter 1 | Functions and Graphs 101




1.30


Rewriting this equation as
e2x − 5ex + 6 = 0,


we can then rewrite it as a quadratic equation in ex :
(ex)2 − 5(ex) + 6 = 0.


Now we can solve the quadratic equation. Factoring this equation, we obtain
(ex − 3)(ex − 2) = 0.


Therefore, the solutions satisfy ex = 3 and ex = 2. Taking the natural logarithm of both sides gives us
the solutions x = ln3, ln2.


Solve e2x /(3 + e2x) = 1/2.


Example 1.37
Solving Equations Involving Logarithmic Functions
Solve each of the following equations for x.


a. ln⎛⎝1x⎞⎠ = 4
b. log10 x + log10 x = 2
c. ln(2x) − 3ln⎛⎝x2⎞⎠ = 0


Solution
a. By the definition of the natural logarithm function,


ln⎛⎝
1
x

⎠ = 4 if and only if e


4 = 1x .


Therefore, the solution is x = 1/e4.
b. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equationas


log10 x + log10 x = log10 x x = log10 x
3/2 = 3


2
log10 x.


Therefore, the equation can be rewritten as
3
2
log10 x = 2 or log10 x =


4
3
.


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1.31


The solution is x = 104/3 = 10 103 .
c. Using the power property of logarithmic functions, we can rewrite the equation as ln(2x) − ln⎛⎝x6⎞⎠ = 0.


Using the quotient property, this becomes
ln


2
x5

⎠ = 0.


Therefore, 2/x5 = 1, which implies x = 25 . We should then check for any extraneous solutions.


Solve ln⎛⎝x3⎞⎠− 4ln(x) = 1.


When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log10 or log,
called the common logarithm, or ln, which is the natural logarithm. However, exponential functions and logarithm functionscan be expressed in terms of any desired base b. If you need to use a calculator to evaluate an expression with a different
base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential orlogarithmic function in terms of the natural exponential and natural logarithmic functions.
Rule: Change-of-Base Formulas
Let a > 0, b > 0, and a ≠ 1, b ≠ 1.


1. ax = bxlogba for any real number x.
If b = e, this equation reduces to ax = exloge a = ex lna.


2. loga x = logb xlogba for any real number x > 0.
If b = e, this equation reduces to loga x = lnxlna.


Proof
For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know thatfor any base b > 0, b ≠ 1, logb(ax) = xlogba. Therefore,


b
logb(a


x)
= b


xlogba.


In addition, we know that bx and logb(x) are inverse functions. Therefore,
b
logb(a


x)
= ax.


Combining these last two equalities, we conclude that ax = bxlogba.
To prove the second property, we show that


(logba) · (loga x) = logb x.


Let u = logba, v = loga x, and w = logb x. We will show that u · v = w. By the definition of logarithmic functions, we


Chapter 1 | Functions and Graphs 103




1.32


know that bu = a, av = x, and bw = x. From the previous equations, we see that
buv = (bu)v = av = x = bw.


Therefore, buv = bw. Since exponential functions are one-to-one, we can conclude that u · v = w.

Example 1.38
Changing Bases
Use a calculating utility to evaluate log37 with the change-of-base formula presented earlier.
Solution
Use the second equation with a = 3 and e = 3:
log37 =


ln7
ln3


≈ 1.77124.


Use the change-of-base formula and a calculating utility to evaluate log46.


Example 1.39
Chapter Opener: The Richter Scale for Earthquakes


Figure 1.48 (credit: modification of work by RobbHannawacker, NPS)


In 1935, Charles Richter developed a scale (now known as the Richter scale) to measure the magnitude of anearthquake. The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquakewith magnitude R1 on the Richter scale and a second earthquake with magnitude R2 on the Richter scale.
Suppose R1 > R2, which means the earthquake of magnitude R1 is stronger, but how much stronger is it than
the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measurethe amplitude of the earthquake waves. If A1 is the amplitude measured for the first earthquake and A2 is the
amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfythe following equation:


R1 − R2 = log10


A1
A2



⎠.


Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richterscale. Then,


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1.33


8 − 7 = log10


A1
A2



⎠.


Therefore,
log10


A1
A2



⎠ = 1,


which implies A1 /A2 = 10 or A1 = 10A2. Since A1 is 10 times the size of A2, we say that the first
earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 onthe Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation


log10


A1
A2



⎠ = 8 − 6 = 2.


Therefore, A1 = 100A2. That is, the first earthquake is 100 times more intense than the second earthquake.
How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in2011 with the magnitude 7.3 earthquake in Haiti in 2010?
Solution
To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:
9 − 7.3 = log10




A1
A2



⎠.


Therefore, A1 /A2 = 101.7, and we conclude that the earthquake in Japan was approximately 50 times more
intense than the earthquake in Haiti.


Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake.


Hyperbolic Functions
The hyperbolic functions are defined in terms of certain combinations of ex and e−x. These functions arise naturally
in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes.Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary(Figure 1.49). If we introduce a coordinate system so that the low point of the chain lies along the y -axis, we can describe
the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.


Chapter 1 | Functions and Graphs 105




Figure 1.49 The shape of a strand of silk in a spider’s webcan be described in terms of a hyperbolic function. The sameshape applies to a chain or cable hanging from two supports withonly its own weight. (credit: “Mtpaley”, Wikimedia Commons)


Definition
Hyperbolic cosine


coshx = e
x + e−x


2


Hyperbolic sine
sinhx = e


x − e−x
2


Hyperbolic tangent
tanhx = sinhx


coshx
= e


x − e−x


ex + e−x


Hyperbolic cosecant
csch x = 1


sinhx
= 2


ex − e−x


Hyperbolic secant
sech x = 1


coshx
= 2


ex + e−x


Hyperbolic cotangent
cothx = coshx


sinhx
= e


x + e−x


ex − e−x


The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh, sech, csch, and coth arepronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.
Using the definition of cosh(x) and principles of physics, it can be shown that the height of a hanging chain, such as the
one in Figure 1.49, can be described by the function h(x) = acosh(x/a) + c for certain constants a and c.
But why are these functions called hyperbolic functions? To answer this question, consider the quantity cosh2 t − sinh2 t.
Using the definition of cosh and sinh, we see that


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cosh2 t − sinh2 t = e
2t + 2 + e−2t


4
− e


2t − 2 + e−2t


4
= 1.


This identity is the analog of the trigonometric identity cos2 t + sin2 t = 1. Here, given a value t, the point
(x, y) = (cosh t, sinh t) lies on the unit hyperbola x2 − y2 = 1 (Figure 1.50).


Figure 1.50 The unit hyperbola cosh2 t − sinh2 t = 1.
Graphs of Hyperbolic Functions
To graph coshx and sinhx, we make use of the fact that both functions approach (1/2)ex as x → ∞, since e−x → 0
as x → ∞. As x → −∞, coshx approaches 1/2e−x, whereas sinhx approaches −1/2e−x. Therefore, using the graphs
of 1/2ex, 1/2e−x, and −1/2e−x as guides, we graph coshx and sinhx. To graph tanhx, we use the fact that
tanh(0) = 1, −1 < tanh(x) < 1 for all x, tanhx → 1 as x → ∞, and tanhx → −1 as x → −∞. The graphs of the
other three hyperbolic functions can be sketched using the graphs of coshx, sinhx, and tanhx (Figure 1.51).


Chapter 1 | Functions and Graphs 107




Figure 1.51 The hyperbolic functions involve combinations of ex and
e−x.


Identities Involving Hyperbolic Functions
The identity cosh2 t − sinh2 t, shown in Figure 1.50, is one of several identities involving the hyperbolic functions,
some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperboliccosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.
Rule: Identities Involving Hyperbolic Functions


1. cosh(−x) = coshx
2. sinh(−x) = −sinhx
3. coshx + sinhx = ex
4. coshx − sinhx = e−x


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1.34


5. cosh2 x − sinh2 x = 1
6. 1 − tanh2 x = sech2 x
7. coth2 x − 1 = csch2 x
8. sinh(x ± y) = sinhxcoshy ± coshxsinhy
9. cosh(x ± y) = coshxcoshy ± sinhxsinhy


Example 1.40
Evaluating Hyperbolic Functions


a. Simplify sinh(5lnx).
b. If sinhx = 3/4, find the values of the remaining five hyperbolic functions.


Solution
a. Using the definition of the sinh function, we write


sinh(5lnx) = e
5lnx − e−5lnx


2
= e


ln⎛⎝x
5⎞

− e


ln⎛⎝x
−5⎞


2
= x


5 − x−5
2


.


b. Using the identity cosh2 x − sinh2 x = 1, we see that
cosh2 x = 1 + ⎛⎝


3
4



2
= 25


16
.


Since coshx ≥ 1 for all x, we must have coshx = 5/4. Then, using the definitions for the other
hyperbolic functions, we conclude that tanhx = 3/5, csch x = 4/3, sech x = 4/5, and cothx = 5/3.


Simplify cosh(2lnx).


Inverse Hyperbolic Functions
From the graphs of the hyperbolic functions, we see that all of them are one-to-one except coshx and sech x. If we
restrict the domains of these two functions to the interval [0, ∞), then all the hyperbolic functions are one-to-one, and we
can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, theinverse hyperbolic functions involve logarithmic functions.
Definition
Inverse Hyperbolic Functions


sinh−1 x = arcsinhx = ln⎛⎝x + x
2 + 1⎞⎠ cosh


−1 x = arccoshx = ln⎛⎝x + x
2 − 1⎞⎠


tanh−1 x = arctanhx = 1
2
ln⎛⎝


1 + x
1 − x

⎠ coth


−1 x = arccotx = 1
2
ln⎛⎝


x + 1
x − 1



sech−1 x = arcsech x = ln



⎜1 + 1 − x


2


x



⎟ csch−1 x = arccsch x = ln





⎜1x +


1 + x2
|x|







Chapter 1 | Functions and Graphs 109




1.35


Let’s look at how to derive the first equation. The others follow similarly. Suppose y = sinh−1 x. Then, x = sinhy and,
by the definition of the hyperbolic sine function, x = ey − e−y


2
. Therefore,


ey − 2x − e
−y


= 0.


Multiplying this equation by ey, we obtain
e
2y


− 2xey − 1 = 0.


This can be solved like a quadratic equation, with the solution
ey = 2x ± 4x


2 + 4
2


= x ± x2 + 1.


Since ey > 0, the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the
equation, we conclude that


y = ln⎛⎝x + x
2 + 1⎞⎠.


Example 1.41
Evaluating Inverse Hyperbolic Functions
Evaluate each of the following expressions.


sinh−1 (2)


tanh−1(1/4)


Solution
sinh−1 (2) = ln⎛⎝2 + 2


2 + 1⎞⎠ = ln

⎝2 + 5



⎠ ≈ 1.4436


tanh−1(1/4) = 1
2
ln⎛⎝


1 + 1/4
1 − 1/4



⎠ =


1
2
ln⎛⎝


5/4
3/4

⎠ =


1
2
ln⎛⎝


5
3

⎠ ≈ 0.2554


Evaluate tanh−1(1/2).


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1.5 EXERCISES
For the following exercises, evaluate the given exponentialfunctions as indicated, accurate to two significant digitsafter the decimal.
229. f (x) = 5x a. x = 3 b. x = 1


2
c. x = 2


230. f (x) = (0.3)x a. x = −1 b. x = 4 c. x = −1.5
231. f (x) = 10x a. x = −2 b. x = 4 c. x = 5


3


232. f (x) = ex a. x = 2 b. x = −3.2 c. x = π
For the following exercises, match the exponential equationto the correct graph.


a. y = 4−x
b. y = 3x − 1
c. y = 2x + 1
d. y = ⎛⎝12⎞⎠


x
+ 2


e. y = −3−x
f. y = 1 − 5x


233.


234.


235.


Chapter 1 | Functions and Graphs 111




236.


237.


238.


For the following exercises, sketch the graph of theexponential function. Determine the domain, range, andhorizontal asymptote.
239. f (x) = ex + 2


240. f (x) = −2x


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241. f (x) = 3x + 1


242. f (x) = 4x − 1


243. f (x) = 1 − 2−x


244. f (x) = 5x + 1 + 2


Chapter 1 | Functions and Graphs 113




245. f (x) = e−x − 1


For the following exercises, write the equation inequivalent exponential form.
246. log381 = 4
247. log82 = 13
248. log51 = 0
249. log525 = 2
250. log0.1 = −1


251. ln⎛⎝ 1e3

⎠ = −3


252. log93 = 0.5
253. ln1 = 0
For the following exercises, write the equation inequivalent logarithmic form.
254. 23 = 8
255. 4−2 = 1


16


256. 102 = 100
257. 90 = 1
258. ⎛⎝13⎞⎠


3
= 1


27


259. 643 = 4
260. ex = y
261. 9y = 150
262. b3 = 45
263. 4−3/2 = 0.125
For the following exercises, sketch the graph of thelogarithmic function. Determine the domain, range, andvertical asymptote.
264. f (x) = 3 + lnx


265. f (x) = ln(x − 1)


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266. f (x) = ln(−x)


267. f (x) = 1 − lnx


268. f (x) = logx − 1


269. f (x) = ln(x + 1)


For the following exercises, use properties of logarithms towrite the expressions as a sum, difference, and/or productof logarithms.
270. logx4 y
271. log3 9a3b
272. lna b3


273. log5 125xy3


274. log4 xy364
275. ln⎛



⎜ 6


e3







For the following exercises, solve the exponential equationexactly.
276. 5x = 125
277. e3x − 15 = 0
278. 8x = 4
279. 4x + 1 − 32 = 0
280. 3x/14 = 1


10


281. 10x = 7.21
282. 4 · 23x − 20 = 0


Chapter 1 | Functions and Graphs 115




283. 73x − 2 = 11
For the following exercises, solve the logarithmic equationexactly, if possible.
284. log3 x = 0
285. log5 x = −2
286. log4 (x + 5) = 0
287. log(2x − 7) = 0
288. ln x + 3 = 2
289. log6 (x + 9) + log6 x = 2
290. log4 (x + 2) − log4 (x − 1) = 0
291. lnx + ln(x − 2) = ln4
For the following exercises, use the change-of-baseformula and either base 10 or base e to evaluate the givenexpressions. Answer in exact form and in approximateform, rounding to four decimal places.
292. log547
293. log782
294. log6103
295. log0.5211
296. log2π
297. log0.20.452
298. Rewrite the following expressions in terms ofexponentials and simplify. a. 2cosh(lnx) b.
cosh4x + sinh4x c. cosh2x − sinh2x d.
ln(coshx + sinhx) + ln(coshx − sinhx)


299. [T] The number of bacteria N in a culture after t days
can be modeled by the function N(t) = 1300 · (2)t/4. Find
the number of bacteria present after 15 days.
300. [T] The demand D (in millions of barrels) for oilin an oil-rich country is given by the function
D(p) = 150 · (2.7)


−0.25p
, where p is the price (in dollars)


of a barrel of oil. Find the amount of oil demanded (to thenearest million barrels) when the price is between $15 and$20.


301. [T] The amount A of a $100,000 investment payingcontinuously and compounded for t years is given by
A(t) = 100,000 · e0.055t. Find the amount A accumulated
in 5 years.
302. [T] An investment is compounded monthly,quarterly, or yearly and is given by the function
A = P⎛⎝1 +


j
n



nt


, where A is the value of the investment
at time t, P is the initial principle that was invested, j
is the annual interest rate, and n is the number of time
the interest is compounded per year. Given a yearly interestrate of 3.5% and an initial principle of $100,000, find theamount A accumulated in 5 years for interest that is
compounded a. daily, b., monthly, c. quarterly, and d.yearly.
303. [T] The concentration of hydrogen ions in a
substance is denoted by ⎡⎣H+ ⎤⎦, measured in moles per
liter. The pH of a substance is defined by the logarithmic
function pH = −log⎡⎣H+ ⎤⎦. This function is used to
measure the acidity of a substance. The pH of water is 7. Asubstance with a pH less than 7 is an acid, whereas one thathas a pH of more than 7 is a base.a. Find the pH of the following substances. Roundanswers to one digit.b. Determine whether the substance is an acid or abase.


i. Eggs: ⎡⎣H+ ⎤⎦ = 1.6 × 10−8 mol/L
ii. Beer: ⎡⎣H+ ⎤⎦ = 3.16 × 10−3 mol/L
iii. Tomato Juice: ⎡⎣H+ ⎤⎦ = 7.94 × 10−5 mol/L


304. [T] Iodine-131 is a radioactive substance that decays
according to the function Q(t) = Q0 · e−0.08664t, where
Q0 is the initial quantity of a sample of the substance and t
is in days. Determine how long it takes (to the nearest day)for 95% of a quantity to decay.
305. [T] According to the World Bank, at the end of2013 ( t = 0 ) the U.S. population was 316 million and
was increasing according to the following model:
P(t) = 316e0.0074t, where P is measured in millions of
people and t is measured in years after 2013.a. Based on this model, what will be the population ofthe United States in 2020?b. Determine when the U.S. population will be twicewhat it is in 2013.


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306. [T] The amount A accumulated after 1000 dollars isinvested for t years at an interest rate of 4% is modeled by
the function A(t) = 1000(1.04)t.


a. Find the amount accumulated after 5 years and 10years.b. Determine how long it takes for the originalinvestment to triple.
307. [T] A bacterial colony grown in a lab is known todouble in number in 12 hours. Suppose, initially, there are1000 bacteria present.


a. Use the exponential function Q = Q0 ekt to
determine the value k, which is the growth rate of
the bacteria. Round to four decimal places.b. Determine approximately how long it takes for200,000 bacteria to grow.


308. [T] The rabbit population on a game reserve doublesevery 6 months. Suppose there were 120 rabbits initially.
a. Use the exponential function P = P0at to


determine the growth rate constant a. Round to
four decimal places.b. Use the function in part a. to determineapproximately how long it takes for the rabbitpopulation to reach 3500.


309. [T] The 1906 earthquake in San Francisco had amagnitude of 8.3 on the Richter scale. At the same time,in Japan, an earthquake with magnitude 4.9 caused onlyminor damage. Approximately how much more energy wasreleased by the San Francisco earthquake than by theJapanese earthquake?


Chapter 1 | Functions and Graphs 117




absolute value function


algebraic function
base
composite function
cubic function
decreasing on the interval I


degree
dependent variable
domain
even function
exponent
function
graph of a function
horizontal line test
hyperbolic functions


increasing on the interval I
independent variable
inverse function
inverse hyperbolic functions


inverse trigonometric functions
linear function
logarithmic function


mathematical model
natural exponential function


CHAPTER 1 REVIEW
KEY TERMS


f (x) =




−x, x < 0
x, x ≥ 0


a function involving any combination of only the basic operations of addition, subtraction,multiplication, division, powers, and roots applied to an input variable x
the number b in the exponential function f (x) = bx and the logarithmic function f (x) = logb x


given two functions f and g, a new function, denoted g ∘ f , such that ⎛⎝g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠
a polynomial of degree 3; that is, a function of the form f (x) = ax3 + bx2 + cx + d, where a ≠ 0


a function decreasing on the interval I if, for all x1, x2 ∈ I, f (x1) ≥ f (x2) if
x1 < x2


for a polynomial function, the value of the largest exponent of any term
the output variable for a function


the set of inputs for a function
a function is even if f (−x) = f (x) for all x in the domain of f


the value x in the expression bx
a set of inputs, a set of outputs, and a rule for mapping each input to exactly one output


the set of points (x, y) such that x is in the domain of f and y = f (x)
a function f is one-to-one if and only if every horizontal line intersects the graph of f , at most,


once
the functions denoted sinh, cosh, tanh, csch, sech, and coth, which involve certain


combinations of ex and e−x
a function increasing on the interval I if for all x1, x2 ∈ I, f (x1) ≤ f (x2) if x1 < x2


the input variable for a function
for a function f , the inverse function f −1 satisfies f −1 (y) = x if f (x) = y


the inverses of the hyperbolic functions where cosh and sech are restricted to the
domain [0, ∞); each of these functions can be expressed in terms of a composition of the natural logarithm function
and an algebraic function


the inverses of the trigonometric functions are defined on restricted domains wherethey are one-to-one functions
a function that can be written in the form f (x) = mx + b


a function of the form f (x) = logb (x) for some base b > 0, b ≠ 1 such that y = logb(x) if
and only if by = x


A method of simulating real-life situations with mathematical equations
the function f (x) = ex


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natural logarithm
number e


odd function
one-to-one function
periodic function
piecewise-defined function
point-slope equation
polynomial function
power function
quadratic function
radians
range
rational function
restricted domain
root function
slope
slope-intercept form
symmetry about the origin


symmetry about the y-axis


table of values
transcendental function
transformation of a function
trigonometric functions
trigonometric identity
vertical line test
zeros of a function


the function lnx = loge x
as m gets larger, the quantity (1 + (1/m)m gets closer to some real number; we define that real number to be


e; the value of e is approximately 2.718282
a function is odd if f (−x) = − f (x) for all x in the domain of f


a function f is one-to-one if f (x1) ≠ f (x2) if x1 ≠ x2
a function is periodic if it has a repeating pattern as the values of x move from left to right


a function that is defined differently on different parts of its domain
equation of a linear function indicating its slope and a point on the graph of the function
a function of the form f (x) = an xn + an − 1 xn − 1 +… + a1 x + a0


a function of the form f (x) = xn for any positive integer n ≥ 1
a polynomial of degree 2; that is, a function of the form f (x) = ax2 + bx + c where a ≠ 0


for a circular arc of length s on a circle of radius 1, the radian measure of the associated angle θ is s
the set of outputs for a function


a function of the form f (x) = p(x)/q(x), where p(x) and q(x) are polynomials
a subset of the domain of a function f


a function of the form f (x) = x1/n for any integer n ≥ 2
the change in y for each unit change in x


equation of a linear function indicating its slope and y-intercept
the graph of a function f is symmetric about the origin if (−x, −y) is on the graph of f


whenever (x, y) is on the graph
the graph of a function f is symmetric about the y -axis if (−x, y) is on the graph of f


whenever (x, y) is on the graph
a table containing a list of inputs and their corresponding outputs


a function that cannot be expressed by a combination of basic arithmetic operations
a shift, scaling, or reflection of a function


functions of an angle defined as ratios of the lengths of the sides of a right triangle
an equation involving trigonometric functions that is true for all angles θ for which the


functions in the equation are defined
given the graph of a function, every vertical line intersects the graph, at most, once
when a real number x is a zero of a function f , f (x) = 0


KEY EQUATIONS
• Composition of two functions



⎝g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠


• Absolute value function
f (x) =






−x, x < 0
x, x ≥ 0


Chapter 1 | Functions and Graphs 119




• Point-slope equation of a line
y − y1 = m(x − x1)


• Slope-intercept form of a line
y = mx + b


• Standard form of a line
ax + by = c


• Polynomial function
f (x) = an x


n + an − 1 x
n − 1 +⋯+ a1 x + a0


• Generalized sine function
f (x) = Asin⎛⎝B(x − α)⎞⎠+ C


• Inverse functions
f −1 ⎛⎝ f (x)⎞⎠ = x for all x inD, and f ⎛⎝ f


−1 (y)⎞⎠ = y for all y in R.


KEY CONCEPTS
1.1 Review of Functions


• A function is a mapping from a set of inputs to a set of outputs with exactly one output for each input.
• If no domain is stated for a function y = f (x), the domain is considered to be the set of all real numbers x for
which the function is defined.


• When sketching the graph of a function f , each vertical line may intersect the graph, at most, once.
• A function may have any number of zeros, but it has, at most, one y-intercept.
• To define the composition g ∘ f , the range of f must be contained in the domain of g.
• Even functions are symmetric about the y -axis whereas odd functions are symmetric about the origin.


1.2 Basic Classes of Functions
• The power function f (x) = xn is an even function if n is even and n ≠ 0, and it is an odd function if n is odd.
• The root function f (x) = x1/n has the domain [0, ∞) if n is even and the domain (−∞, ∞) if n is odd. If n
is odd, then f (x) = x1/n is an odd function.


• The domain of the rational function f (x) = p(x)/q(x), where p(x) and q(x) are polynomial functions, is the set
of x such that q(x) ≠ 0.


• Functions that involve the basic operations of addition, subtraction, multiplication, division, and powers arealgebraic functions. All other functions are transcendental. Trigonometric, exponential, and logarithmic functionsare examples of transcendental functions.
• A polynomial function f with degree n ≥ 1 satisfies f (x) → ±∞ as x → ±∞. The sign of the output as


x → ∞ depends on the sign of the leading coefficient only and on whether n is even or odd.
• Vertical and horizontal shifts, vertical and horizontal scalings, and reflections about the x - and y -axes are
examples of transformations of functions.


1.3 Trigonometric Functions
• Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radianmeasure 1. An angle with a degree measure of 180° has a radian measure of π rad.


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• For acute angles θ, the values of the trigonometric functions are defined as ratios of two sides of a right triangle
in which one of the acute angles is θ.


• For a general angle θ, let (x, y) be a point on a circle of radius r corresponding to this angle θ. The
trigonometric functions can be written as ratios involving x, y, and r.


• The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period 2π. The
tangent and cotangent functions have period π.


1.4 Inverse Functions
• For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determinewhether the function is one-to-one by using the horizontal line test.
• If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one andthen define the inverse of the function on the smaller domain.
• For a function f and its inverse f −1, f ⎛⎝ f −1 (x)⎞⎠ = x for all x in the domain of f −1 and f −1 ⎛⎝ f (x)⎞⎠ = x for all


x in the domain of f .
• Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometricfunctions.
• The graph of a function f and its inverse f −1 are symmetric about the line y = x.


1.5 Exponential and Logarithmic Functions
• The exponential function y = bx is increasing if b > 1 and decreasing if 0 < b < 1. Its domain is (−∞, ∞)
and its range is (0, ∞).


• The logarithmic function y = logb(x) is the inverse of y = bx. Its domain is (0, ∞) and its range is (−∞, ∞).
• The natural exponential function is y = ex and the natural logarithmic function is y = lnx = loge x.
• Given an exponential function or logarithmic function in base a, we can make a change of base to convert this
function to any base b > 0, b ≠ 1. We typically convert to base e.


• The hyperbolic functions involve combinations of the exponential functions ex and e−x. As a result, the inverse
hyperbolic functions involve the natural logarithm.


CHAPTER 1 REVIEW EXERCISES
True or False? Justify your answer with a proof or acounterexample.
310. A function is always one-to-one.
311. f ∘g = g ∘ f , assuming f and g are functions.
312. A relation that passes the horizontal and vertical linetests is a one-to-one function.
313. A relation passing the horizontal line test is afunction.
For the following problems, state the domain and range ofthe given functions:


f = x2 + 2x − 3, g = ln(x − 5), h = 1
x + 4


314. h
315. g
316. h ∘ f
317. g ∘ f
Find the degree, y-intercept, and zeros for the followingpolynomial functions.
318. f (x) = 2x2 + 9x − 5


Chapter 1 | Functions and Graphs 121




319. f (x) = x3 + 2x2 − 2x
Simplify the following trigonometric expressions.
320. tan2 x


sec2 x
+ cos2 x


321. cos(2x) = sin2 x
Solve the following trigonometric equations on the interval
θ = [−2π, 2π] exactly.
322. 6cos2 x − 3 = 0
323. sec2 x − 2secx + 1 = 0
Solve the following logarithmic equations.
324. 5x = 16
325. log2(x + 4) = 3
Are the following functions one-to-one over their domainof existence? Does the function have an inverse? If so, find
the inverse f −1(x) of the function. Justify your answer.
326. f (x) = x2 + 2x + 1


327. f (x) = 1x
For the following problems, determine the largest domainon which the function is one-to-one and find the inverse onthat domain.
328. f (x) = 9 − x


329. f (x) = x2 + 3x + 4
330. A car is racing along a circular track with diameter of1 mi. A trainer standing in the center of the circle marks hisprogress every 5 sec. After 5 sec, the trainer has to turn 55°to keep up with the car. How fast is the car traveling?
For the following problems, consider a restaurant ownerwho wants to sell T-shirts advertising his brand. He recallsthat there is a fixed cost and variable cost, although he doesnot remember the values. He does know that the T-shirtprinting company charges $440 for 20 shirts and $1000 for100 shirts.


331. a. Find the equation C = f (x) that describes the
total cost as a function of number of shirts and b. determinehow many shirts he must sell to break even if he sells theshirts for $10 each.
332. a. Find the inverse function x = f −1(C) and
describe the meaning of this function. b. Determine howmany shirts the owner can buy if he has $8000 to spend.
For the following problems, consider the population ofOcean City, New Jersey, which is cyclical by season.
333. The population can be modeled by
P(t) = 82.5 − 67.5cos⎡⎣(π/6)t⎤⎦, where t is time in
months (t = 0 represents January 1) and P is population
(in thousands). During a year, in what intervals is thepopulation less than 20,000? During what intervals is thepopulation more than 140,000?
334. In reality, the overall population is most likelyincreasing or decreasing throughout each year. Let’sreformulate the model as
P(t) = 82.5 − 67.5cos⎡⎣(π/6)t⎤⎦+ t, where t is time in
months ( t = 0 represents January 1) and P is population
(in thousands). When is the first time the populationreaches 200,000?
For the following problems, consider radioactive dating. Ahuman skeleton is found in an archeological dig. Carbondating is implemented to determine how old the skeleton is
by using the equation y = ert, where y is the percentage
of radiocarbon still present in the material, t is the number
of years passed, and r = −0.0001210 is the decay rate of
radiocarbon.
335. If the skeleton is expected to be 2000 years old, whatpercentage of radiocarbon should be present?
336. Find the inverse of the carbon-dating equation. Whatdoes it mean? If there is 25% radiocarbon, how old is theskeleton?


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2 | LIMITS


Figure 2.1 The vision of human exploration by the National Aeronautics and Space Administration (NASA) to distant parts ofthe universe illustrates the idea of space travel at high speeds. But, is there a limit to how fast a spacecraft can go? (credit:NASA)


Chapter Outline
2.1 A Preview of Calculus
2.2 The Limit of a Function
2.3 The Limit Laws
2.4 Continuity
2.5 The Precise Definition of a Limit


Introduction
Science fiction writers often imagine spaceships that can travel to far-off planets in distant galaxies. However, back in 1905,Albert Einstein showed that a limit exists to how fast any object can travel. The problem is that the faster an object moves,the more mass it attains (in the form of energy), according to the equation


m =
m0


1 − v
2


c2


,


Chapter 2 | Limits 123




where m0 is the object’s mass at rest, v is its speed, and c is the speed of light. What is this speed limit? (We explore thisproblem further in Example 2.12.)
The idea of a limit is central to all of calculus. We begin this chapter by examining why limits are so important. Then, wego on to describe how to find the limit of a function at a given point. Not all functions have limits at all points, and wediscuss what this means and how we can tell if a function does or does not have a limit at a particular value. This chapter hasbeen created in an informal, intuitive fashion, but this is not always enough if we need to prove a mathematical statementinvolving limits. The last section of this chapter presents the more precise definition of a limit and shows how to provewhether a function has a limit.
2.1 | A Preview of Calculus


Learning Objectives
2.1.1 Describe the tangent problem and how it led to the idea of a derivative.
2.1.2 Explain how the idea of a limit is involved in solving the tangent problem.
2.1.3 Recognize a tangent to a curve at a point as the limit of secant lines.
2.1.4 Identify instantaneous velocity as the limit of average velocity over a small time interval.
2.1.5 Describe the area problem and how it was solved by the integral.
2.1.6 Explain how the idea of a limit is involved in solving the area problem.
2.1.7 Recognize how the ideas of limit, derivative, and integral led to the studies of infinite seriesand multivariable calculus.


As we embark on our study of calculus, we shall see how its development arose from common solutions to practicalproblems in areas such as engineering physics—like the space travel problem posed in the chapter opener. Two keyproblems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangentto a curve at a point; and (2) the area problem, or how to determine the area under a curve.
The Tangent Problem and Differential Calculus
Rate of change is one of the most critical concepts in calculus. We begin our investigation of rates of change by looking at
the graphs of the three lines f (x) = −2x − 3, g(x) = 1


2
x + 1, and h(x) = 2, shown in Figure 2.2.


Figure 2.2 The rate of change of a linear function is constant in each of these three graphs, with the constant determined by theslope.


As we move from left to right along the graph of f (x) = −2x − 3, we see that the graph decreases at a constant rate. For
every 1 unit we move to the right along the x-axis, the y-coordinate decreases by 2 units. This rate of change is determinedby the slope (−2) of the line. Similarly, the slope of 1/2 in the function g(x) tells us that for every change in x of 1 unit
there is a corresponding change in y of 1/2 unit. The function h(x) = 2 has a slope of zero, indicating that the values of the
function remain constant. We see that the slope of each linear function indicates the rate of change of the function.


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Compare the graphs of these three functions with the graph of k(x) = x2 (Figure 2.3). The graph of k(x) = x2 starts from
the left by decreasing rapidly, then begins to decrease more slowly and level off, and then finally begins to increase—slowlyat first, followed by an increasing rate of increase as it moves toward the right. Unlike a linear function, no single numberrepresents the rate of change for this function. We quite naturally ask: How do we measure the rate of change of a nonlinearfunction?


Figure 2.3 The function k(x) = x2 does not have a constant
rate of change.


We can approximate the rate of change of a function f (x) at a point ⎛⎝a, f (a)⎞⎠ on its graph by taking another point ⎛⎝x, f (x)⎞⎠
on the graph of f (x), drawing a line through the two points, and calculating the slope of the resulting line. Such a line is
called a secant line. Figure 2.4 shows a secant line to a function f (x) at a point ⎛⎝a, f (a)⎞⎠.


Figure 2.4 The slope of a secant line through a point

⎝a, f (a)⎞⎠ estimates the rate of change of the function at the
point ⎛⎝a, f (a)⎞⎠.


We formally define a secant line as follows:


Chapter 2 | Limits 125




Definition
The secant to the function f (x) through the points ⎛⎝a, f (a)⎞⎠ and ⎛⎝x, f (x)⎞⎠ is the line passing through these points. Its
slope is given by


(2.1)msec = f (x) − f (a)x − a .
The accuracy of approximating the rate of change of the function with a secant line depends on how close x is to a. As wesee in Figure 2.5, if x is closer to a, the slope of the secant line is a better measure of the rate of change of f (x) at a.


Figure 2.5 As x gets closer to a, the slope of the secant linebecomes a better approximation to the rate of change of thefunction f (x) at a.


The secant lines themselves approach a line that is called the tangent to the function f (x) at a (Figure 2.6). The slope of
the tangent line to the graph at ameasures the rate of change of the function at a. This value also represents the derivative ofthe function f (x) at a, or the rate of change of the function at a. This derivative is denoted by f ′ (a). Differential calculus
is the field of calculus concerned with the study of derivatives and their applications.


For an interactive demonstration of the slope of a secant line that you can manipulate yourself, visit thisapplet (Note: this site requires a Java browser plugin): Math Insight (http://www.openstaxcollege.org/l/20_mathinsight) .


Figure 2.6 Solving the Tangent Problem: As x approaches a,the secant lines approach the tangent line.


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Example 2.1 illustrates how to find slopes of secant lines. These slopes estimate the slope of the tangent line or,equivalently, the rate of change of the function at the point at which the slopes are calculated.
Example 2.1
Finding Slopes of Secant Lines
Estimate the slope of the tangent line (rate of change) to f (x) = x2 at x = 1 by finding slopes of secant lines
through (1, 1) and each of the following points on the graph of f (x) = x2.


a. (2, 4)
b. ⎛⎝32, 94⎞⎠


Solution
Use the formula for the slope of a secant line from the definition.


a. msec = 4 − 12 − 1 = 3


b. msec =
9
4
− 1


3
2
− 1


= 5
2
= 2.5


The point in part b. is closer to the point (1, 1), so the slope of 2.5 is closer to the slope of the tangent line. A
good estimate for the slope of the tangent would be in the range of 2 to 2.5 (Figure 2.7).


Figure 2.7 The secant lines to f (x) = x2 at (1, 1) through
(a) (2, 4) and (b) ⎛⎝32, 94⎞⎠ provide successively closer
approximations to the tangent line to f (x) = x2 at (1, 1).


Chapter 2 | Limits 127




2.1 Estimate the slope of the tangent line (rate of change) to f (x) = x2 at x = 1 by finding slopes of secant
lines through (1, 1) and the point ⎛⎝54, 2516⎞⎠ on the graph of f (x) = x2.


We continue our investigation by exploring a related question. Keeping in mind that velocity may be thought of as the rateof change of position, suppose that we have a function, s(t), that gives the position of an object along a coordinate axis
at any given time t. Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a giventime t = a? We start by approximating the instantaneous velocity with an average velocity. First, recall that the speed of
an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled. We define theaverage velocity of an object over a time period to be the change in its position divided by the length of the time period.
Definition
Let s(t) be the position of an object moving along a coordinate axis at time t. The average velocity of the object over
a time interval [a, t] where a < t (or [t, a] if t < a) is


(2.2)vave = s(t) − s(a)t − a .


As t is chosen closer to a, the average velocity becomes closer to the instantaneous velocity. Note that finding the averagevelocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function.Furthermore, to find the slope of a tangent line at a point a, we let the x-values approach a in the slope of the secant line.Similarly, to find the instantaneous velocity at time a, we let the t-values approach a in the average velocity. This processof letting x or t approach a in an expression is called taking a limit. Thus, we may define the instantaneous velocity asfollows.
Definition
For a position function s(t), the instantaneous velocity at a time t = a is the value that the average velocities
approach on intervals of the form [a, t] and [t, a] as the values of t become closer to a, provided such a value exists.


Example 2.2 illustrates this concept of limits and average velocity.
Example 2.2
Finding Average Velocity
A rock is dropped from a height of 64 ft. It is determined that its height (in feet) above ground t seconds later (for
0 ≤ t ≤ 2) is given by s(t) = −16t2 + 64. Find the average velocity of the rock over each of the given time
intervals. Use this information to guess the instantaneous velocity of the rock at time t = 0.5.


a. ⎡⎣0.49, 0.5⎤⎦
b. ⎡⎣0.5, 0.51⎤⎦


Solution
Substitute the data into the formula for the definition of average velocity.


a. vave = s(0.49) − s(0.5)0.49 − 0.5 = −15.84


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2.2


b. vave = s(0.51) − s(0.5)0.51 − 0.5 = −16.016
The instantaneous velocity is somewhere between −15.84 and −16.16 ft/sec. A good guess might be −16 ft/sec.


An object moves along a coordinate axis so that its position at time t is given by s(t) = t3. Estimate its
instantaneous velocity at time t = 2 by computing its average velocity over the time interval [2, 2.001].


The Area Problem and Integral Calculus
We now turn our attention to a classic question from calculus. Many quantities in physics—for example, quantities ofwork—may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area betweenthe graph of a function and the x-axis over an interval (Figure 2.8)?


Figure 2.8 The Area Problem: How do we find the area of theshaded region?


As in the answer to our previous questions on velocity, we first try to approximate the solution. We approximate the area bydividing up the interval ⎡⎣a, b⎤⎦ into smaller intervals in the shape of rectangles. The approximation of the area comes from
adding up the areas of these rectangles (Figure 2.9).


Figure 2.9 The area of the region under the curve isapproximated by summing the areas of thin rectangles.


As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the areabetween the graph of f (x) and the x-axis over the interval ⎡⎣a, b⎤⎦. Once again, we find ourselves taking a limit. Limits
of this type serve as a basis for the definition of the definite integral. Integral calculus is the study of integrals and theirapplications.


Chapter 2 | Limits 129




Example 2.3
Estimation Using Rectangles
Estimate the area between the x-axis and the graph of f (x) = x2 + 1 over the interval [0, 3] by using the three
rectangles shown in Figure 2.10.


Figure 2.10 The area of the region under the curve of
f (x) = x2 + 1 can be estimated using rectangles.


Solution
The areas of the three rectangles are 1 unit2, 2 unit2, and 5 unit2. Using these rectangles, our area estimate is 8unit2.


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2.3 Estimate the area between the x-axis and the graph of f (x) = x2 + 1 over the interval [0, 3] by using
the three rectangles shown here:


Other Aspects of Calculus
So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in twodimensions; however, there is no good reason to restrict our investigation to two dimensions. Suppose, for example, thatinstead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of arock fired from a catapult at a given time, or of an airplane moving in three dimensions. We might want to graph real-valuefunctions of two variables or determine volumes of solids of the type shown in Figure 2.11. These are only a few of thetypes of questions that can be asked and answered using multivariable calculus. Informally, multivariable calculus can becharacterized as the study of the calculus of functions of two or more variables. However, before exploring these and otherideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.


Figure 2.11 We can use multivariable calculus to find thevolume between a surface defined by a function of two variablesand a plane.


Chapter 2 | Limits 131




2.1 EXERCISES
For the following exercises, points P(1, 2) and Q(x, y)
are on the graph of the function f (x) = x2 + 1.
1. [T] Complete the following table with the appropriatevalues: y-coordinate ofQ, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Roundyour answer to eight significant digits.


x y Q(x, y) msec
1.1 a. e. i.
1.01 b. f. j.
1.001 c. g. k.
1.0001 d. h. l.


2. Use the values in the right column of the table in thepreceding exercise to guess the value of the slope of the linetangent to f at x = 1.
3. Use the value in the preceding exercise to find theequation of the tangent line at point P. Graph f (x) and the
tangent line.
For the following exercises, points P(1, 1) and Q(x, y)
are on the graph of the function f (x) = x3.
4. [T] Complete the following table with the appropriatevalues: y-coordinate ofQ, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Roundyour answer to eight significant digits.


x y Q(x, y) msec
1.1 a. e. i.
1.01 b. f. j.
1.001 c. g. k.
1.0001 d. h. l.


5. Use the values in the right column of the table in thepreceding exercise to guess the value of the slope of thetangent line to f at x = 1.
6. Use the value in the preceding exercise to find theequation of the tangent line at point P. Graph f (x) and the
tangent line.
For the following exercises, points P(4, 2) and Q(x, y)
are on the graph of the function f (x) = x.
7. [T] Complete the following table with the appropriatevalues: y-coordinate ofQ, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Roundyour answer to eight significant digits.


x y Q(x, y) msec
4.1 a. e. i.
4.01 b. f. j.
4.001 c. g. k.
4.0001 d. h. l.


8. Use the values in the right column of the table in thepreceding exercise to guess the value of the slope of thetangent line to f at x = 4.
9. Use the value in the preceding exercise to find theequation of the tangent line at point P.
For the following exercises, points P(1.5, 0) and Q⎛⎝ϕ, y⎞⎠
are on the graph of the function f ⎛⎝ϕ⎞⎠ = cos ⎛⎝πϕ⎞⎠.


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10. [T] Complete the following table with the appropriatevalues: y-coordinate ofQ, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Roundyour answer to eight significant digits.


x y Q⎛⎝ϕ, y⎞⎠ msec
1.4 a. e. i.
1.49 b. f. j.
1.499 c. g. k.
1.4999 d. h. l.


11. Use the values in the right column of the table in thepreceding exercise to guess the value of the slope of thetangent line to f at x = 4.
12. Use the value in the preceding exercise to find theequation of the tangent line at point P.
For the following exercises, points P(−1, −1) and
Q(x, y) are on the graph of the function f (x) = 1x .
13. [T] Complete the following table with the appropriatevalues: y-coordinate ofQ, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Roundyour answer to eight significant digits.


x y Q(x, y) msec
−1.05 a. e. i.
−1.01 b. f. j.
−1.005 c. g. k.
−1.001 d. h. l.


14. Use the values in the right column of the table in thepreceding exercise to guess the value of the slope of the linetangent to f at x = −1.
15. Use the value in the preceding exercise to find theequation of the tangent line at point P.
For the following exercises, the position function of a balldropped from the top of a 200-meter tall building is given


by s(t) = 200 − 4.9t2, where position s is measured in
meters and time t is measured in seconds. Round youranswer to eight significant digits.
16. [T] Compute the average velocity of the ball over thegiven time intervals.a. ⎡⎣4.99, 5⎤⎦


b. ⎡⎣5, 5.01⎤⎦
c. ⎡⎣4.999, 5⎤⎦
d. ⎡⎣5, 5.001⎤⎦


17. Use the preceding exercise to guess the instantaneousvelocity of the ball at t = 5 sec.
For the following exercises, consider a stone tossed into theair from ground level with an initial velocity of 15 m/sec.
Its height in meters at time t seconds is h(t) = 15t − 4.9t2.
18. [T] Compute the average velocity of the stone over thegiven time intervals.a. ⎡⎣1, 1.05⎤⎦


b. [1, 1.01]
c. ⎡⎣1, 1.005⎤⎦
d. [1, 1.001]


19. Use the preceding exercise to guess the instantaneousvelocity of the stone at t = 1 sec.
For the following exercises, consider a rocket shot into theair that then returns to Earth. The height of the rocket in
meters is given by h(t) = 600 + 78.4t − 4.9t2, where t is
measured in seconds.
20. [T] Compute the average velocity of the rocket overthe given time intervals.a. [9, 9.01]


b. [8.99, 9]
c. [9, 9.001]
d. [8.999, 9]


21. Use the preceding exercise to guess the instantaneousvelocity of the rocket at t = 9 sec.
For the following exercises, consider an athlete runninga 40-m dash. The position of the athlete is given by
d(t) = t


3


6
+ 4t, where d is the position in meters and t is


the time elapsed, measured in seconds.


Chapter 2 | Limits 133




22. [T] Compute the average velocity of the runner overthe given time intervals.a. ⎡⎣1.95, 2.05⎤⎦
b. ⎡⎣1.995, 2.005⎤⎦
c. ⎡⎣1.9995, 2.0005⎤⎦
d. [2, 2.00001]


23. Use the preceding exercise to guess the instantaneousvelocity of the runner at t = 2 sec.
For the following exercises, consider the function
f (x) = |x|.


24. Sketch the graph of f over the interval [−1, 2] and
shade the region above the x-axis.
25. Use the preceding exercise to find the exact value ofthe area between the x-axis and the graph of f over theinterval [−1, 2] using rectangles. For the rectangles, use
the square units, and approximate both above and below thelines. Use geometry to find the exact answer.
For the following exercises, consider the function
f (x) = 1 − x2. (Hint: This is the upper half of a circle of
radius 1 positioned at (0, 0).)
26. Sketch the graph of f over the interval [−1, 1].
27. Use the preceding exercise to find the exact areabetween the x-axis and the graph of f over the interval
[−1, 1] using rectangles. For the rectangles, use squares
0.4 by 0.4 units, and approximate both above and below thelines. Use geometry to find the exact answer.
For the following exercises, consider the function
f (x) = −x2 + 1.


28. Sketch the graph of f over the interval [−1, 1].
29. Approximate the area of the region between the x-axisand the graph of f over the interval [−1, 1].


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2.2 | The Limit of a Function
Learning Objectives


2.2.1 Using correct notation, describe the limit of a function.
2.2.2 Use a table of values to estimate the limit of a function or to identify when the limit does notexist.
2.2.3 Use a graph to estimate the limit of a function or to identify when the limit does not exist.
2.2.4 Define one-sided limits and provide examples.
2.2.5 Explain the relationship between one-sided and two-sided limits.
2.2.6 Using correct notation, describe an infinite limit.
2.2.7 Define a vertical asymptote.


The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years.In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, theformal definition of a limit—as we know and understand it today—did not appear until the late 19th century. We thereforebegin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. At the end of thischapter, armed with a conceptual understanding of limits, we examine the formal definition of a limit.
We begin our exploration of limits by taking a look at the graphs of the functions


f (x) = x
2 − 4
x − 2


, g(x) = |x − 2|
x − 2


, and h(x) = 1
(x − 2)2


,


which are shown in Figure 2.12. In particular, let’s focus our attention on the behavior of each graph at and around x = 2.


Figure 2.12 These graphs show the behavior of three different functions around x = 2.


Each of the three functions is undefined at x = 2, but if we make this statement and no other, we give a very incomplete
picture of how each function behaves in the vicinity of x = 2. To express the behavior of each graph in the vicinity of 2
more completely, we need to introduce the concept of a limit.
Intuitive Definition of a Limit
Let’s first take a closer look at how the function f (x) = (x2 − 4)/(x − 2) behaves around x = 2 in Figure 2.12. As the
values of x approach 2 from either side of 2, the values of y = f (x) approach 4. Mathematically, we say that the limit of
f (x) as x approaches 2 is 4. Symbolically, we express this limit as


Chapter 2 | Limits 135




lim
x → 2


f (x) = 4.


From this very brief informal look at one limit, let’s start to develop an intuitive definition of the limit. We can think of thelimit of a function at a number a as being the one real number L that the functional values approach as the x-values approacha, provided such a real number L exists. Stated more carefully, we have the following definition:
Definition
Let f (x) be a function defined at all values in an open interval containing a, with the possible exception of a itself,
and let L be a real number. If all values of the function f (x) approach the real number L as the values of x( ≠ a)
approach the number a, then we say that the limit of f (x) as x approaches a is L. (More succinct, as x gets closer to a,
f (x) gets closer and stays close to L.) Symbolically, we express this idea as


(2.3)lim
x → a


f (x) = L.


We can estimate limits by constructing tables of functional values and by looking at their graphs. This process is describedin the following Problem-Solving Strategy.
Problem-Solving Strategy: Evaluating a Limit Using a Table of Functional Values


1. To evaluate lim
x → a


f (x), we begin by completing a table of functional values. We should choose two sets of
x-values—one set of values approaching a and less than a, and another set of values approaching a and greaterthan a. Table 2.1 demonstrates what your tables might look like.


x f(x) x f(x)
a − 0.1 f (a − 0.1) a + 0.1 f (a + 0.1)


a − 0.01 f (a − 0.01) a + 0.01 f (a + 0.01)


a − 0.001 f (a − 0.001) a + 0.001 f (a + 0.001)


a − 0.0001 f (a − 0.0001) a + 0.0001 f (a + 0.0001)


Use additional values as necessary. Use additional values as necessary.
Table 2.1 Table of Functional Values for lim


x → a
f (x)


2. Next, let’s look at the values in each of the f (x) columns and determine whether the values seem to
be approaching a single value as we move down each column. In our columns, we look at the sequence
f (a − 0.1), f (a − 0.01), f (a − 0.001)., f (a − 0.0001), and so on, and
f (a + 0.1), f (a + 0.01), f (a + 0.001), f (a + 0.0001), and so on. (Note: Although we have chosen the
x-values a ± 0.1, a ± 0.01, a ± 0.001, a ± 0.0001, and so forth, and these values will probably work nearly
every time, on very rare occasions we may need to modify our choices.)


3. If both columns approach a common y-value L, we state lim
x → a


f (x) = L. We can use the following strategy to
confirm the result obtained from the table or as an alternative method for estimating a limit.


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4. Using a graphing calculator or computer software that allows us graph functions, we can plot the function
f (x), making sure the functional values of f (x) for x-values near a are in our window. We can use the trace
feature to move along the graph of the function and watch the y-value readout as the x-values approach a. Ifthe y-values approach L as our x-values approach a from both directions, then lim


x → a
f (x) = L. We may need


to zoom in on our graph and repeat this process several times.


We apply this Problem-Solving Strategy to compute a limit in Example 2.4.
Example 2.4
Evaluating a Limit Using a Table of Functional Values 1
Evaluate lim


x → 0
sinx
x using a table of functional values.


Solution
We have calculated the values of f (x) = (sinx)/x for the values of x listed in Table 2.2.


x sinxx x sinxx
−0.1 0.998334166468 0.1 0.998334166468
−0.01 0.999983333417 0.01 0.999983333417
−0.001 0.999999833333 0.001 0.999999833333
−0.0001 0.999999998333 0.0001 0.999999998333


Table 2.2
Table of Functional Values for lim


x → 0
sinx
x


Note: The values in this table were obtained using a calculator and using all the places given in the calculatoroutput.
As we read down each (sinx)x column, we see that the values in each column appear to be approaching
one. Thus, it is fairly reasonable to conclude that lim


x → 0
sinx
x = 1. A calculator-or computer-generated graph of


f (x) = (sinx)x would be similar to that shown in Figure 2.13, and it confirms our estimate.


Chapter 2 | Limits 137




Figure 2.13 The graph of f (x) = (sinx)/x confirms the
estimate from Table 2.2.


Example 2.5
Evaluating a Limit Using a Table of Functional Values 2
Evaluate lim


x → 4
x − 2
x − 4


using a table of functional values.


Solution
As before, we use a table—in this case, Table 2.3—to list the values of the function for the given values of x.


x x−2x−4 x x−2x−4
3.9 0.251582341869 4.1 0.248456731317
3.99 0.25015644562 4.01 0.24984394501
3.999 0.250015627 4.001 0.249984377
3.9999 0.250001563 4.0001 0.249998438
3.99999 0.25000016 4.00001 0.24999984


Table 2.3
Table of Functional Values for lim


x → 4
x − 2
x − 4


After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We conclude that


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2.4


lim
x → 4


x − 2
x − 4


= 0.25. We confirm this estimate using the graph of f (x) = x − 2
x − 4


shown in Figure 2.14.


Figure 2.14 The graph of f (x) = x − 2
x − 4


confirms the
estimate from Table 2.3.


Estimate lim
x → 1


1
x − 1
x − 1


using a table of functional values. Use a graph to confirm your estimate.


At this point, we see from Example 2.4 and Example 2.5 that it may be just as easy, if not easier, to estimate a limit ofa function by inspecting its graph as it is to estimate the limit by using a table of functional values. In Example 2.6, weevaluate a limit exclusively by looking at a graph rather than by using a table of functional values.


Chapter 2 | Limits 139




Example 2.6
Evaluating a Limit Using a Graph
For g(x) shown in Figure 2.15, evaluate lim


x → −1
g(x).


Figure 2.15 The graph of g(x) includes one value not on a
smooth curve.


Solution
Despite the fact that g(−1) = 4, as the x-values approach −1 from either side, the g(x) values approach 3.
Therefore, lim


x → −1
g(x) = 3. Note that we can determine this limit without even knowing the algebraic expression


of the function.


Based on Example 2.6, we make the following observation: It is possible for the limit of a function to exist at a point, andfor the function to be defined at this point, but the limit of the function and the value of the function at the point may bedifferent.


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2.5 Use the graph of h(x) in Figure 2.16 to evaluate lim
x → 2


h(x), if possible.


Figure 2.16


Looking at a table of functional values or looking at the graph of a function provides us with useful insight into the valueof the limit of a function at a given point. However, these techniques rely too much on guesswork. We eventually need todevelop alternative methods of evaluating limits. These new methods are more algebraic in nature and we explore them inthe next section; however, at this point we introduce two special limits that are foundational to the techniques to come.
Theorem 2.1: Two Important Limits
Let a be a real number and c be a constant.


i. (2.4)lim
x → a


x = a


ii. (2.5)lim
x → a


c = c


We can make the following observations about these two limits.
i. For the first limit, observe that as x approaches a, so does f (x), because f (x) = x. Consequently, lim


x → a
x = a.


ii. For the second limit, consider Table 2.4.
x f(x) = c x f(x) = c
a − 0.1 c a + 0.1 c
a − 0.01 c a + 0.01 c
a − 0.001 c a + 0.001 c
a − 0.0001 c a + 0.0001 c


Table 2.4 Table of Functional Values for lim
x → a


c = c


Chapter 2 | Limits 141




Observe that for all values of x (regardless of whether they are approaching a), the values f (x) remain constant at c. We
have no choice but to conclude lim


x → a
c = c.


The Existence of a Limit
As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functionalvalues must approach a single real-number value at that point. If the functional values do not approach a single value, thenthe limit does not exist.
Example 2.7
Evaluating a Limit That Fails to Exist
Evaluate lim


x → 0
sin(1/x) using a table of values.


Solution
Table 2.5 lists values for the function sin(1/x) for the given values of x.


x sin⎛⎝1x⎞⎠ x sin⎛⎝1x⎞⎠
−0.1 0.544021110889 0.1 −0.544021110889
−0.01 0.50636564111 0.01 −0.50636564111
−0.001 −0.8268795405312 0.001 0.826879540532
−0.0001 0.305614388888 0.0001 −0.305614388888
−0.00001 −0.035748797987 0.00001 0.035748797987
−0.000001 0.349993504187 0.000001 −0.349993504187


Table 2.5
Table of Functional Values for lim


x → 0
sin⎛⎝


1
x



After examining the table of functional values, we can see that the y-values do not seem to approach any onesingle value. It appears the limit does not exist. Before drawing this conclusion, let’s take a more systematicapproach. Take the following sequence of x-values approaching 0:
2
π ,


2


, 2


, 2


, 2


, 2
11π


,….


The corresponding y-values are
1, −1, 1, −1, 1, −1,….


At this point we can indeed conclude that lim
x → 0


sin(1/x) does not exist. (Mathematicians frequently abbreviate


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2.6


“does not exist” as DNE. Thus, we would write lim
x → 0


sin(1/x) DNE.) The graph of f (x) = sin(1/x) is shown
in Figure 2.17 and it gives a clearer picture of the behavior of sin(1/x) as x approaches 0. You can see that
sin(1/x) oscillates ever more wildly between −1 and 1 as x approaches 0.


Figure 2.17 The graph of f (x) = sin(1/x) oscillates rapidly
between −1 and 1 as x approaches 0.


Use a table of functional values to evaluate lim
x → 2


|x2 − 4|
x − 2


, if possible.


One-Sided Limits
Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough informationabout the behavior of the function at that particular point. To see this, we now revisit the function g(x) = |x − 2|/(x − 2)
introduced at the beginning of the section (see Figure 2.12(b)). As we pick values of x close to 2, g(x) does not approach
a single value, so the limit as x approaches 2 does not exist—that is, lim


x → 2
g(x) DNE. However, this statement alone does


not give us a complete picture of the behavior of the function around the x-value 2. To provide a more accurate description,we introduce the idea of a one-sided limit. For all values to the left of 2 (or the negative side of 2), g(x) = −1. Thus, as x
approaches 2 from the left, g(x) approaches −1. Mathematically, we say that the limit as x approaches 2 from the left is −1.
Symbolically, we express this idea as


lim
x → 2−


g(x) = −1.


Similarly, as x approaches 2 from the right (or from the positive side), g(x) approaches 1. Symbolically, we express this
idea as


lim
x → 2+


g(x) = 1.


We can now present an informal definition of one-sided limits.
Definition
We define two types of one-sided limits.


Chapter 2 | Limits 143




Limit from the left: Let f (x) be a function defined at all values in an open interval of the form z, and let L be a real
number. If the values of the function f (x) approach the real number L as the values of x (where x < a) approach the
number a, then we say that L is the limit of f (x) as x approaches a from the left. Symbolically, we express this idea as


(2.6)lim
x → a−


f (x) = L.


Limit from the right: Let f (x) be a function defined at all values in an open interval of the form (a, c), and let L be a
real number. If the values of the function f (x) approach the real number L as the values of x (where x > a) approach
the number a, then we say that L is the limit of f (x) as x approaches a from the right. Symbolically, we express this
idea as


(2.7)lim
x → a+


f (x) = L.


Example 2.8
Evaluating One-Sided Limits
For the function f (x) = ⎧




x + 1 if x < 2


x2 − 4 if x ≥ 2
, evaluate each of the following limits.


a. lim
x → 2−


f (x)


b. lim
x → 2+


f (x)


Solution
We can use tables of functional values again Table 2.6. Observe that for values of x less than 2, we use
f (x) = x + 1 and for values of x greater than 2, we use f (x) = x2 − 4.


x f(x) = x+1 x f(x) = x2−4
1.9 2.9 2.1 0.41
1.99 2.99 2.01 0.0401
1.999 2.999 2.001 0.004001
1.9999 2.9999 2.0001 0.00040001
1.99999 2.99999 2.00001 0.0000400001


Table 2.6
Table of Functional Values for f (x) = ⎧




x + 1 if x < 2


x2 − 4 if x ≥ 2


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2.7


Based on this table, we can conclude that a. lim
x → 2−


f (x) = 3 and b. lim
x → 2+


f (x) = 0. Therefore, the (two-sided)
limit of f (x) does not exist at x = 2. Figure 2.18 shows a graph of f (x) and reinforces our conclusion about
these limits.


Figure 2.18 The graph of f (x) = ⎧


x + 1 if x < 2


x2 − 4 if x ≥ 2
has a


break at x = 2.


Use a table of functional values to estimate the following limits, if possible.
a. lim


x → 2−
|x2 − 4|
x − 2


b. lim
x → 2+


|x2 − 4|
x − 2


Let us now consider the relationship between the limit of a function at a point and the limits from the right and left at thatpoint. It seems clear that if the limit from the right and the limit from the left have a common value, then that common valueis the limit of the function at that point. Similarly, if the limit from the left and the limit from the right take on differentvalues, the limit of the function does not exist. These conclusions are summarized in Relating One-Sided and Two-Sided Limits.
Theorem 2.2: Relating One-Sided and Two-Sided Limits
Let f (x) be a function defined at all values in an open interval containing a, with the possible exception of a itself,
and let L be a real number. Then,


lim
x → a


f (x) = L. if and only if lim
x → a−


f (x) = L and lim
x → a+


f (x) = L.


Chapter 2 | Limits 145




Infinite Limits
Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us tocharacterize the behavior of a function around a given value. As we shall see, we can also describe the behavior of functionsthat do not have finite limits.
We now turn our attention to h(x) = 1/(x − 2)2, the third and final function introduced at the beginning of this section
(see Figure 2.12(c)). From its graph we see that as the values of x approach 2, the values of h(x) = 1/(x − 2)2 become
larger and larger and, in fact, become infinite. Mathematically, we say that the limit of h(x) as x approaches 2 is positive
infinity. Symbolically, we express this idea as


lim
x → 2


h(x) = +∞.


More generally, we define infinite limits as follows:
Definition
We define three types of infinite limits.
Infinite limits from the left: Let f (x) be a function defined at all values in an open interval of the form (b, a).


i. If the values of f (x) increase without bound as the values of x (where x < a) approach the number a, then
we say that the limit as x approaches a from the left is positive infinity and we write


(2.8)lim
x → a−


f (x) = +∞.


ii. If the values of f (x) decrease without bound as the values of x (where x < a) approach the number a, then
we say that the limit as x approaches a from the left is negative infinity and we write


(2.9)lim
x → a−


f (x) = −∞.


Infinite limits from the right: Let f (x) be a function defined at all values in an open interval of the form (a, c).
i. If the values of f (x) increase without bound as the values of x (where x > a) approach the number a, then


we say that the limit as x approaches a from the left is positive infinity and we write
(2.10)lim


x → a+
f (x) = +∞.


ii. If the values of f (x) decrease without bound as the values of x (where x > a) approach the number a, then
we say that the limit as x approaches a from the left is negative infinity and we write


(2.11)lim
x → a+


f (x) = −∞.


Two-sided infinite limit: Let f (x) be defined for all x ≠ a in an open interval containing a.
i. If the values of f (x) increase without bound as the values of x (where x ≠ a) approach the number a, then


we say that the limit as x approaches a is positive infinity and we write
(2.12)lim


x → a
f (x) = +∞.


ii. If the values of f (x) decrease without bound as the values of x (where x ≠ a) approach the number a, then
we say that the limit as x approaches a is negative infinity and we write


(2.13)lim
x → a


f (x) = −∞.


It is important to understand that when we write statements such as lim
x → a


f (x) = +∞ or lim
x → a


f (x) = −∞ we are
describing the behavior of the function, as we have just defined it. We are not asserting that a limit exists. For thelimit of a function f (x) to exist at a, it must approach a real number L as x approaches a. That said, if, for example,
lim
x → a


f (x) = +∞, we always write lim
x → a


f (x) = +∞ rather than lim
x → a


f (x) DNE.


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Example 2.9
Recognizing an Infinite Limit
Evaluate each of the following limits, if possible. Use a table of functional values and graph f (x) = 1/x to
confirm your conclusion.


a. lim
x → 0−


1
x


b. lim
x → 0+


1
x


c. lim
x → 0


1
x


Solution
Begin by constructing a table of functional values.


x 1x x 1x
−0.1 −10 0.1 10
−0.01 −100 0.01 100
−0.001 −1000 0.001 1000
−0.0001 −10,000 0.0001 10,000
−0.00001 −100,000 0.00001 100,000
−0.000001 −1,000,000 0.000001 1,000,000


Table 2.7
Table of Functional Values for f (x) = 1x


a. The values of 1/x decrease without bound as x approaches 0 from the left. We conclude that
lim


x → 0−
1
x = −∞.


b. The values of 1/x increase without bound as x approaches 0 from the right. We conclude that
lim


x → 0+
1
x = +∞.


c. Since lim
x → 0−


1
x = −∞ and lim


x → 0+
1
x = +∞ have different values, we conclude that


lim
x → 0


1
x DNE.


The graph of f (x) = 1/x in Figure 2.19 confirms these conclusions.


Chapter 2 | Limits 147




2.8


Figure 2.19 The graph of f (x) = 1/x confirms that the limit
as x approaches 0 does not exist.


Evaluate each of the following limits, if possible. Use a table of functional values and graph f (x) = 1/x2
to confirm your conclusion.


a. lim
x → 0−


1
x2


b. lim
x → 0+


1
x2


c. lim
x → 0


1
x2


It is useful to point out that functions of the form f (x) = 1/(x − a)n, where n is a positive integer, have infinite limits as x
approaches a from either the left or right (Figure 2.20). These limits are summarized in Infinite Limits from PositiveIntegers.


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Figure 2.20 The function f (x) = 1/(x − a)n has infinite limits at a.


Theorem 2.3: Infinite Limits from Positive Integers
If n is a positive even integer, then


lim
x → a


1
(x − a)n


= +∞.


If n is a positive odd integer, then
lim


x → a+
1


(x − a)n
= +∞


and
lim


x → a−
1


(x − a)n
= −∞.


We should also point out that in the graphs of f (x) = 1/(x − a)n, points on the graph having x-coordinates very near to a
are very close to the vertical line x = a. That is, as x approaches a, the points on the graph of f (x) are closer to the line
x = a. The line x = a is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows:


Definition
Let f (x) be a function. If any of the following conditions hold, then the line x = a is a vertical asymptote of f (x).


lim
x → a−


f (x) = +∞ or −∞


lim
x → a+


f (x) = +∞ or −∞


or
lim
x → a


f (x) = +∞ or −∞


Example 2.10


Chapter 2 | Limits 149




2.9


Finding a Vertical Asymptote
Evaluate each of the following limits using Infinite Limits from Positive Integers. Identify any vertical
asymptotes of the function f (x) = 1/(x + 3)4.


a. lim
x → −3−


1
(x + 3)4


b. lim
x → −3+


1
(x + 3)4


c. lim
x → −3


1
(x + 3)4


Solution
We can use Infinite Limits from Positive Integers directly.


a. lim
x → −3−


1
(x + 3)4


= +∞


b. lim
x → −3+


1
(x + 3)4


= +∞


c. lim
x → −3


1
(x + 3)4


= +∞


The function f (x) = 1/(x + 3)4 has a vertical asymptote of x = −3.


Evaluate each of the following limits. Identify any vertical asymptotes of the function f (x) = 1
(x − 2)3


.


a. lim
x → 2−


1
(x − 2)3


b. lim
x → 2+


1
(x − 2)3


c. lim
x → 2


1
(x − 2)3


In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at severaldifferent points.
Example 2.11
Behavior of a Function at Different Points
Use the graph of f (x) in Figure 2.21 to determine each of the following values:


a. lim
x → −4−


f (x); lim
x → −4+


f (x); lim
x → −4


f (x); f (−4)


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2.10


b. lim
x → −2−


f (x); lim
x → −2+


f (x); lim
x → −2


f (x); f (−2)


c. lim
x → 1−


f (x); lim
x → 1+


f (x); lim
x → 1


f (x); f (1)


d. lim
x → 3−


f (x); lim
x → 3+


f (x); lim
x → 3


f (x); f (3)


Figure 2.21 The graph shows f (x).


Solution
Using Infinite Limits from Positive Integers and the graph for reference, we arrive at the following values:


a. lim
x → −4−


f (x) = 0; lim
x → −4+


f (x) = 0; lim
x → −4


f (x) = 0; f (−4) = 0


b. lim
x → −2−


f (x) = 3.; lim
x → −2+


f (x) = 3; lim
x → −2


f (x) = 3; f (−2) is undefined
c. lim


x → 1−
f (x) = 6; lim


x → 1+
f (x) = 3; lim


x → 1
f (x) DNE; f (1) = 6


d. lim
x → 3−


f (x) = −∞; lim
x → 3+


f (x) = −∞; lim
x → 3


f (x) = −∞; f (3) is undefined


Evaluate lim
x → 1


f (x) for f (x) shown here:


Chapter 2 | Limits 151




Example 2.12
Chapter Opener: Einstein’s Equation


Figure 2.22 (credit: NASA)


In the chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any objectcan travel. Given Einstein’s equation for the mass of a moving object, what is the value of this bound?
Solution
Our starting point is Einstein’s equation for the mass of a moving object,


m =
m0


1 − v
2


c2


,


where m0 is the object’s mass at rest, v is its speed, and c is the speed of light. To see how the mass changes at
high speeds, we can graph the ratio of masses m/m0 as a function of the ratio of speeds, v/c (Figure 2.23).


Figure 2.23 This graph shows the ratio of masses as afunction of the ratio of speeds in Einstein’s equation for themass of a moving object.


We can see that as the ratio of speeds approaches 1—that is, as the speed of the object approaches the speedof light—the ratio of masses increases without bound. In other words, the function has a vertical asymptote at
v/c = 1. We can try a few values of this ratio to test this idea.


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v
c 1 −


v2


c2
m
m0


0.99 0.1411 7.089
0.999 0.0447 22.37
0.9999 0.0141 70.71


Table 2.8Ratio of Masses and Speeds for aMoving Object
Thus, according to Table 2.8, if an object with mass 100 kg is traveling at 0.9999c, its mass becomes 7071 kg.Since no object can have an infinite mass, we conclude that no object can travel at or more than the speed of light.


Chapter 2 | Limits 153




2.2 EXERCISES
For the following exercises, consider the function
f (x) = x


2 − 1
|x − 1|


.


30. [T] Complete the following table for the function.Round your solutions to four decimal places.
x f(x) x f(x)
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.


31. What do your results in the preceding exercise indicateabout the two-sided limit lim
x → 1


f (x)? Explain your
response.
For the following exercises, consider the function
f (x) = (1 + x)1/x.


32. [T] Make a table showing the values of f for
x = −0.01, −0.001, −0.0001, −0.00001 and for
x = 0.01, 0.001, 0.0001, 0.00001. Round your solutions
to five decimal places.


x f(x) x f(x)
−0.01 a. 0.01 e.
−0.001 b. 0.001 f.
−0.0001 c. 0.0001 g.
−0.00001 d. 0.00001 h.


33. What does the table of values in the preceding exercise
indicate about the function f (x) = (1 + x)1/x?
34. To which mathematical constant does the limit in thepreceding exercise appear to be getting closer?
In the following exercises, use the given values to set up a


table to evaluate the limits. Round your solutions to eightdecimal places.
35. [T] lim


x → 0
sin2x
x ; ±0.1, ±0.01, ±0.001, ±.0001


x sin2xx x sin2xx
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.


36. [T] lim
x → 0


sin3x
x ±0.1, ±0.01, ±0.001, ±0.0001


X sin3xx x sin3xx
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.


37. Use the preceding two exercises to conjecture (guess)
the value of the following limit: lim


x → 0
sinax
x for a, a


positive real value.
[T] In the following exercises, set up a table of values tofind the indicated limit. Round to eight digits.


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38. lim
x → 2


x2 − 4
x2 + x − 6


x x2−4
x2+ x−6


x x2−4
x2+ x−6


1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.


39. lim
x → 1


(1 − 2x)


x 1 −2x x 1 −2x
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.


40. lim
x → 0


5
1 − e1/x


x 51 − e1/x x 51 − e1/x
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.


41. lim
z → 0


z − 1
z2 (z + 3)


z z−1z2 (z+3) z z−1z2 (z+3)
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.


42. lim
t → 0+


cos t
t


t cos tt
0.1 a.
0.01 b.
0.001 c.
0.0001 d.


43. lim
x → 2


1 − 2x
x2 − 4


x 1 − 2x
x2−4


x 1 − 2x
x2−4


1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.


[T] In the following exercises, set up a table of valuesand round to eight significant digits. Based on the table ofvalues, make a guess about what the limit is. Then, use a


Chapter 2 | Limits 155




calculator to graph the function and determine the limit.Was the conjecture correct? If not, why does the method oftables fail?
44. lim


θ → 0
sin⎛⎝


π
θ



θ sin⎛⎝πθ⎞⎠ θ sin⎛⎝πθ⎞⎠
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.


45. lim
α → 0+


1
α cos




π
α



a 1α cos ⎛⎝πα⎞⎠
0.1 a.
0.01 b.
0.001 c.
0.0001 d.


In the following exercises, consider the graph of thefunction y = f (x) shown here. Which of the statements
about y = f (x) are true and which are false? Explain why
a statement is false.


46. lim
x → 10


f (x) = 0


47. lim
x → −2+


f (x) = 3


48. lim
x → −8


f (x) = f (−8)


49. lim
x → 6


f (x) = 5


In the following exercises, use the following graph of thefunction y = f (x) to find the values, if possible. Estimate
when necessary.


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50. lim
x → 1−


f (x)


51. lim
x → 1+


f (x)


52. lim
x → 1


f (x)


53. lim
x → 2


f (x)


54. f (1)
In the following exercises, use the graph of the function
y = f (x) shown here to find the values, if possible.
Estimate when necessary.


55. lim
x → 0−


f (x)


56. lim
x → 0+


f (x)


57. lim
x → 0


f (x)


58. lim
x → 2


f (x)


In the following exercises, use the graph of the function
y = f (x) shown here to find the values, if possible.
Estimate when necessary.


59. lim
x → −2−


f (x)


60. lim
x → −2+


f (x)


61. lim
x → −2


f (x)


62. lim
x → 2−


f (x)


63. lim
x → 2+


f (x)


64. lim
x → 2


f (x)


In the following exercises, use the graph of the function
y = g(x) shown here to find the values, if possible.
Estimate when necessary.


65. lim
x → 0−


g(x)


Chapter 2 | Limits 157




66. lim
x → 0+


g(x)


67. lim
x → 0


g(x)


In the following exercises, use the graph of the function
y = h(x) shown here to find the values, if possible.
Estimate when necessary.


68. lim
x → 0−


h(x)


69. lim
x → 0+


h(x)


70. lim
x → 0


h(x)


In the following exercises, use the graph of the function
y = f (x) shown here to find the values, if possible.
Estimate when necessary.


71. lim
x → 0−


f (x)


72. lim
x → 0+


f (x)


73. lim
x → 0


f (x)


74. lim
x → 1


f (x)


75. lim
x → 2


f (x)


In the following exercises, sketch the graph of a functionwith the given properties.
76.
lim
x → 2


f (x) = 1, lim
x → 4−


f (x) = 3, lim
x → 4+


f (x) = 6, x = 4


is not defined.
77. lim


x → −∞
f (x) = 0, lim


x → −1−
f (x) = −∞,


lim
x → −1+


f (x) = ∞, lim
x → 0


f (x) = f (0), f (0) = 1, lim
x → ∞


f (x) = −∞


78. lim
x → −∞


f (x) = 2, lim
x → 3−


f (x) = −∞,


lim
x → 3+


f (x) = ∞, lim
x → ∞


f (x) = 2, f (0) = −1
3


79. lim
x → −∞


f (x) = 2, lim
x → −2


f (x) = −∞,


lim
x → ∞


f (x) = 2, f (0) = 0


80.
lim


x → −∞
f (x) = 0, lim


x → −1−
f (x) = ∞, lim


x → −1+
f (x) = −∞,


f (0) = −1, lim
x → 1−


f (x) = −∞, lim
x → 1+


f (x) = ∞, lim
x → ∞


f (x) = 0


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81. Shock waves arise in many physical applications,ranging from supernovas to detonation waves. A graph ofthe density of a shock wave with respect to distance, x, isshown here. We are mainly interested in the location of thefront of the shock, labeled xSF in the diagram.


a. Evaluate lim
x → xSF


+
ρ(x).


b. Evaluate lim
x → xSF


− ρ(x).


c. Evaluate lim
x → xSF


ρ(x). Explain the physical
meanings behind your answers.


82. A track coach uses a camera with a fast shutter toestimate the position of a runner with respect to time. Atable of the values of position of the athlete versus time isgiven here, where x is the position in meters of the runnerand t is time in seconds. What is lim
t → 2


x(t)? What does it
mean physically?


t (sec) x (m)
1.75 4.5
1.95 6.1
1.99 6.42
2.01 6.58
2.05 6.9
2.25 8.5


Chapter 2 | Limits 159




2.3 | The Limit Laws
Learning Objectives


2.3.1 Recognize the basic limit laws.
2.3.2 Use the limit laws to evaluate the limit of a function.
2.3.3 Evaluate the limit of a function by factoring.
2.3.4 Use the limit laws to evaluate the limit of a polynomial or rational function.
2.3.5 Evaluate the limit of a function by factoring or by using conjugates.
2.3.6 Evaluate the limit of a function by using the squeeze theorem.


In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, weestablish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, youhave the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised bythe Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These tworesults, together with the limit laws, serve as a foundation for calculating many limits.
Evaluating Limits with the Limit Laws
The first two limit laws were stated in Two Important Limits and we repeat them here. These basic results, together withthe other limit laws, allow us to evaluate limits of many algebraic functions.
Theorem 2.4: Basic Limit Results
For any real number a and any constant c,


i. (2.14)lim
x → a


x = a


ii. (2.15)lim
x → a


c = c


Example 2.13
Evaluating a Basic Limit
Evaluate each of the following limits using Basic Limit Results.


a. lim
x → 2


x


b. lim
x → 2


5


Solution
a. The limit of x as x approaches a is a: lim


x → 2
x = 2.


b. The limit of a constant is that constant: lim
x → 2


5 = 5.


We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.


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Theorem 2.5: Limit Laws
Let f (x) and g(x) be defined for all x ≠ a over some open interval containing a. Assume that L and M are real
numbers such that lim


x → a
f (x) = L and lim


x → a
g(x) = M. Let c be a constant. Then, each of the following statements


holds:
Sum law for limits: lim


x → a

⎝ f (x) + g(x)⎞⎠ = limx → a f (x) + limx → ag(x) = L + M


Difference law for limits: lim
x → a



⎝ f (x) − g(x)⎞⎠ = limx → a f (x) − limx → ag(x) = L − M


Constant multiple law for limits: lim
x → a


c f (x) = c · lim
x → a


f (x) = cL


Product law for limits: lim
x → a



⎝ f (x) · g(x)⎞⎠ = limx → a f (x) · limx → ag(x) = L ·M


Quotient law for limits: lim
x → a


f (x)
g(x)


=
lim
x → a


f (x)


lim
x → a


g(x)
= L


M
for M ≠ 0


Power law for limits: lim
x → a



⎝ f (x)⎞⎠n = ⎛⎝ limx → a f (x)





n
= Ln for every positive integer n.


Root law for limits: lim
x → a


f (x)
n


= lim
x → a


f (x)n = L
n for all L if n is odd and for L ≥ 0 if n is even.


We now practice applying these limit laws to evaluate a limit.
Example 2.14
Evaluating a Limit Using Limit Laws
Use the limit laws to evaluate lim


x → −3
(4x + 2).


Solution
Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mindthe requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.
lim


x → −3
(4x + 2) = lim


x → −3
4x + lim


x → −3
2 Apply the sum law.


= 4 · lim
x → −3


x + lim
x → −3


2 Apply the constant multiple law.


= 4 · (−3) + 2 = −10. Apply the basic limit results and simplify.


Example 2.15
Using Limit Laws Repeatedly
Use the limit laws to evaluate lim


x → 2
2x2 − 3x + 1


x3 + 4
.


Solution


Chapter 2 | Limits 161




2.11


To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewritethe limit in terms of other limits, each new limit must exist for the limit law to be applied.


lim
x → 2


2x2 − 3x + 1
x3 + 4


=
lim
x → 2



⎝2x


2 − 3x + 1⎞⎠


lim
x → 2



⎝x


3 + 4⎞⎠
Apply the quotient law, making sure that. (2)3 + 4 ≠ 0


=
2 · lim


x → 2
x2 − 3 · lim


x → 2
x + lim


x → 2
1


lim
x → 2


x3 + lim
x → 2


4
Apply the sum law and constant multiple law.


=
2 · ⎛⎝ limx → 2


x⎞⎠


2
− 3 · lim


x → 2
x + lim


x → 2
1



⎝ limx → 2


x⎞⎠


3
+ lim


x → 2
4


Apply the power law.


= 2(4) − 3(2) + 1
(2)3 + 4


= 1
4
. Apply the basic limit laws and simplify.


Use the limit laws to evaluate lim
x → 6


(2x − 1) x + 4. In each step, indicate the limit law applied.


Limits of Polynomial and Rational Functions
By now you have probably noticed that, in each of the previous examples, it has been the case that lim


x → a
f (x) = f (a). This


is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a forwhich the rational function is defined.
Theorem 2.6: Limits of Polynomial and Rational Functions
Let p(x) and q(x) be polynomial functions. Let a be a real number. Then,


lim
x → a


p(x) = p(a)


lim
x → a


p(x)
q(x)


=
p(a)
q(a)


when q(a) ≠ 0.


To see that this theorem holds, consider the polynomial p(x) = cn xn + cn − 1 xn − 1 + ⋯ + c1 x + c0. By applying the
sum, constant multiple, and power laws, we end up with


lim
x → a


p(x) = lim
x → a



⎝cn x


n + cn − 1 x
n − 1 + ⋯ + c1 x + c0





= cn

⎝ limx → ax





n
+ cn − 1



⎝ limx → ax





n − 1
+ ⋯ + c1



⎝ limx → ax



⎠+ limx → ac0


= cna
n + cn − 1a


n − 1 + ⋯ + c1a + c0
= p(a).


It now follows from the quotient law that if p(x) and q(x) are polynomials for which q(a) ≠ 0, then
lim
x → a


p(x)
q(x)


=
p(a)
q(a)


.


Example 2.16 applies this result.


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2.12


Example 2.16
Evaluating a Limit of a Rational Function
Evaluate the lim


x → 3
2x2 − 3x + 1


5x + 4
.


Solution
Since 3 is in the domain of the rational function f (x) = 2x2 − 3x + 1


5x + 4
, we can calculate the limit by substituting


3 for x into the function. Thus,
lim
x → 3


2x2 − 3x + 1
5x + 4


= 10
19


.


Evaluate lim
x → −2



⎝3x


3 − 2x + 7⎞⎠.


Additional Limit Evaluation Techniques
As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions bydirect substitution. However, as we saw in the introductory section on limits, it is certainly possible for lim


x → a
f (x) to exist


when f (a) is undefined. The following observation allows us to evaluate many limits of this type:
If for all x ≠ a, f (x) = g(x) over some open interval containing a, then lim


x → a
f (x) = lim


x → a
g(x).


To understand this idea better, consider the limit lim
x → 1


x2 − 1
x − 1


.


The function
f (x) = x


2 − 1
x − 1


= (x − 1)(x + 1)
x − 1


and the function g(x) = x + 1 are identical for all values of x ≠ 1. The graphs of these two functions are shown in Figure
2.24.


Chapter 2 | Limits 163




Figure 2.24 The graphs of f (x) and g(x) are identical for all x ≠ 1. Their limits at 1 are equal.


We see that
lim
x → 1


x2 − 1
x − 1


= lim
x → 1


(x − 1)(x + 1)
x − 1


= lim
x → 1


(x + 1)


= 2.


The limit has the form lim
x → a


f (x)
g(x)


, where lim
x → a


f (x) = 0 and lim
x → a


g(x) = 0. (In this case, we say that f (x)/g(x) has the
indeterminate form 0/0.) The following Problem-Solving Strategy provides a general outline for evaluating limits of this
type.
Problem-Solving Strategy: Calculating a Limit When f(x)/g(x) has the Indeterminate Form 0/0


1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediatelyusing the limit laws.
2. We then need to find a function that is equal to h(x) = f (x)/g(x) for all x ≠ a over some interval containing


a. To do this, we may need to try one or more of the following steps:
a. If f (x) and g(x) are polynomials, we should factor each function and cancel out any common factors.
b. If the numerator or denominator contains a difference involving a square root, we should trymultiplying the numerator and denominator by the conjugate of the expression involving the squareroot.
c. If f (x)/g(x) is a complex fraction, we begin by simplifying it.


3. Last, we apply the limit laws.


The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-canceltechnique; Example 2.18 shows multiplying by a conjugate. In Example 2.19, we look at simplifying a complex fraction.


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2.13


Example 2.17
Evaluating a Limit by Factoring and Canceling
Evaluate lim


x → 3
x2 − 3x


2x2 − 5x − 3
.


Solution
Step 1. The function f (x) = x2 − 3x


2x2 − 5x − 3
is undefined for x = 3. In fact, if we substitute 3 into the function


we get 0/0, which is undefined. Factoring and canceling is a good strategy:
lim
x → 3


x2 − 3x
2x2 − 5x − 3


= lim
x → 3


x(x − 3)
(x − 3)(2x + 1)


Step 2. For all x ≠ 3, x2 − 3x
2x2 − 5x − 3


= x
2x + 1


. Therefore,


lim
x → 3


x(x − 3)
(x − 3)(2x + 1)


= lim
x → 3


x
2x + 1


.


Step 3. Evaluate using the limit laws:
lim
x → 3


x
2x + 1


= 3
7
.


Evaluate lim
x → −3


x2 + 4x + 3
x2 − 9


.


Example 2.18
Evaluating a Limit by Multiplying by a Conjugate
Evaluate lim


x → −1
x + 2 − 1
x + 1


.


Solution
Step 1. x + 2 − 1


x + 1
has the form 0/0 at −1. Let’s begin by multiplying by x + 2 + 1, the conjugate of


x + 2 − 1, on the numerator and denominator:
lim


x → −1
x + 2 − 1
x + 1


= lim
x → −1


x + 2 − 1
x + 1


· x + 2 + 1
x + 2 + 1


.


Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping thatthe (x + 1) in the denominator cancels out in the end:
= lim


x → −1
x + 1


(x + 1)⎛⎝ x + 2 + 1



.


Chapter 2 | Limits 165




2.14


Step 3. Then we cancel:
= lim


x → −1
1


x + 2 + 1
.


Step 4. Last, we apply the limit laws:
lim


x → −1
1


x + 2 + 1
= 1


2
.


Evaluate lim
x → 5


x − 1 − 2
x − 5


.


Example 2.19
Evaluating a Limit by Simplifying a Complex Fraction
Evaluate lim


x → 1


1
x + 1


− 1
2


x − 1
.


Solution
Step 1. 1x + 1 − 12


x − 1
has the form 0/0 at 1. We simplify the algebraic fraction by multiplying by


2(x + 1)/2(x + 1) :


lim
x → 1


1
x + 1


− 1
2


x − 1
= lim


x → 1


1
x + 1


− 1
2


x − 1
· 2(x + 1)
2(x + 1)


.


Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be ableto cancel the factor (x − 1):
= lim


x → 1


2 − (x + 1)
2(x − 1)(x + 1)


.


Step 3. Then, we simplify the numerator:
= lim


x → 1
−x + 1


2(x − 1)(x + 1)
.


Step 4. Now we factor out −1 from the numerator:
= lim


x → 1


−(x − 1)
2(x − 1)(x + 1)


.


Step 5. Then, we cancel the common factors of (x − 1):
= lim


x → 1
−1


2(x + 1)
.


Step 6. Last, we evaluate using the limit laws:
lim
x → 1


−1
2(x + 1)


= − 1
4
.


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2.15


2.16


Evaluate lim
x → −3


1
x + 2


+ 1


x + 3
.


Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a littlecreativity, we can still use these same techniques.
Example 2.20
Evaluating a Limit When the Limit Laws Do Not Apply
Evaluate lim


x → 0




1
x +


5
x(x − 5)



⎠.


Solution
Both 1/x and 5/x(x − 5) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we
cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performingaddition and then applying one of our previous strategies. Observe that


1
x +


5
x(x − 5)


= x − 5 + 5
x(x − 5)


= x
x(x − 5)


.


Thus,
lim
x → 0




1
x +


5
x(x − 5)



⎠ = limx → 0


x
x(x − 5)


= lim
x → 0


1
x − 5


= − 1
5
.


Evaluate lim
x → 3





1
x − 3


− 4
x2 − 2x − 3



⎠.


Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. Forexample, to apply the limit laws to a limit of the form lim
x → a−


h(x), we require the function h(x) to be defined over an
open interval of the form (b, a); for a limit of the form lim


x → a+
h(x), we require the function h(x) to be defined over an


open interval of the form (a, c). Example 2.21 illustrates this point.
Example 2.21
Evaluating a One-Sided Limit Using the Limit Laws
Evaluate each of the following limits, if possible.


a. lim
x → 3−


x − 3


Chapter 2 | Limits 167




b. lim
x → 3+


x − 3


Solution
Figure 2.25 illustrates the function f (x) = x − 3 and aids in our understanding of these limits.


Figure 2.25 The graph shows the function f (x) = x − 3.
a. The function f (x) = x − 3 is defined over the interval [3, +∞). Since this function is not defined to


the left of 3, we cannot apply the limit laws to compute lim
x → 3−


x − 3. In fact, since f (x) = x − 3 is
undefined to the left of 3, lim


x → 3−
x − 3 does not exist.


b. Since f (x) = x − 3 is defined to the right of 3, the limit laws do apply to lim
x → 3+


x − 3. By applying
these limit laws we obtain lim


x → 3+
x − 3 = 0.


In Example 2.22 we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusionabout a two-sided limit of the same function.
Example 2.22
Evaluating a Two-Sided Limit Using the Limit Laws
For f (x) = ⎧




4x − 3 if x < 2


(x − 3)2 if x ≥ 2
, evaluate each of the following limits:


a. lim
x → 2−


f (x)


b. lim
x → 2+


f (x)


c. lim
x → 2


f (x)


Solution
Figure 2.26 illustrates the function f (x) and aids in our understanding of these limits.


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2.17


Figure 2.26 This graph shows a function f (x).
a. Since f (x) = 4x − 3 for all x in (−∞, 2), replace f (x) in the limit with 4x − 3 and apply the limit


laws:
lim


x → 2−
f (x) = lim


x → 2−
(4x − 3) = 5.


b. Since f (x) = (x − 3)2 for all x in (2, +∞), replace f (x) in the limit with (x − 3)2 and apply the
limit laws:


lim
x → 2+


f (x) = lim
x → 2−


(x − 3)2 = 1.


c. Since lim
x → 2−


f (x) = 5 and lim
x → 2+


f (x) = 1, we conclude that lim
x → 2


f (x) does not exist.


Graph f (x) = ⎧


−x − 2 if x < −1


2 if x = −1


x3 if x > −1


and evaluate lim
x → −1−


f (x).


We now turn our attention to evaluating a limit of the form lim
x → a


f (x)
g(x)


, where lim
x → a


f (x) = K, where K ≠ 0 and
lim
x → a


g(x) = 0. That is, f (x)/g(x) has the form K/0, K ≠ 0 at a.
Example 2.23
Evaluating a Limit of the Form K/0, K ≠ 0 Using the Limit Laws
Evaluate lim


x → 2−
x − 3
x2 − 2x


.


Solution
Step 1. After substituting in x = 2, we see that this limit has the form −1/0. That is, as x approaches 2 from the


Chapter 2 | Limits 169




2.18


left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of x − 3
x(x − 2)


becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:
lim


x → 2−
x − 3
x2 − 2x


= lim
x → 2−


x − 3
x(x − 2)


.


Step 2. Since x − 2 is the only part of the denominator that is zero when 2 is substituted, we then separate
1/(x − 2) from the rest of the function:


= lim
x → 2−


x − 3
x ·


1
x − 2


.


Step 3. lim
x → 2−


x − 3
x = −


1
2
and lim


x → 2−
1


x − 2
= −∞. Therefore, the product of (x − 3)/x and 1/(x − 2) has


a limit of +∞:
lim


x → 2−
x − 3
x2 − 2x


= +∞.


Evaluate lim
x → 1


x + 2
(x − 1)2


.


The Squeeze Theorem
The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limitsof very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishingbasic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a thatis unknown, between two functions having a common known limit at a. Figure 2.27 illustrates this idea.


Figure 2.27 The Squeeze Theorem applies when
f (x) ≤ g(x) ≤ h(x) and lim


x → a
f (x) = lim


x → a
h(x).


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2.19


Theorem 2.7: The Squeeze Theorem
Let f (x), g(x), and h(x) be defined for all x ≠ a over an open interval containing a. If


f (x) ≤ g(x) ≤ h(x)


for all x ≠ a in an open interval containing a and
lim
x → a


f (x) = L = lim
x → a


h(x)


where L is a real number, then lim
x → a


g(x) = L.


Example 2.24
Applying the Squeeze Theorem
Apply the squeeze theorem to evaluate lim


x → 0
xcosx.


Solution
Because −1 ≤ cosx ≤ 1 for all x, we have −x ≤ xcosx ≤ x for x ≥ 0 and −x ≥ xcosx ≥ x for x ≤ 0 (if
x is negative the direction of the inequalities changes when we multiply). Since lim


x → 0
(−x) = 0 = lim


x → 0
x, from


the squeeze theorem, we obtain lim
x → 0


xcosx = 0. The graphs of f (x) = − x, g(x) = xcosx, and h(x) = x are
shown in Figure 2.28.


Figure 2.28 The graphs of f (x), g(x), and h(x) are shown
around the point x = 0.


Use the squeeze theorem to evaluate lim
x → 0


x2 sin1x .


We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy,these limits prove invaluable for the development of the material in both the next section and the next chapter. The first ofthese limits is lim
θ → 0


sinθ. Consider the unit circle shown in Figure 2.29. In the figure, we see that sinθ is the y-coordinate
on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of thearc it subtends on the unit circle. Therefore, we see that for 0 < θ < π


2
, 0 < sinθ < θ.


Chapter 2 | Limits 171




Figure 2.29 The sine function is shown as a line on the unitcircle.


Because lim
θ → 0+


0 = 0 and lim
x → 0+


θ = 0, by using the squeeze theorem we conclude that
lim


θ → 0+
sinθ = 0.


To see that lim
θ → 0−


sinθ = 0 as well, observe that for −π
2
< θ < 0, 0 < −θ < π


2
and hence, 0 < sin(−θ) < −θ.


Consequently, 0 < − sinθ < −θ. It follows that 0 > sinθ > θ. An application of the squeeze theorem produces the
desired limit. Thus, since lim


θ → 0+
sinθ = 0 and lim


θ → 0−
sinθ = 0,


(2.16)lim
θ → 0


sinθ = 0.


Next, using the identity cosθ = 1 − sin2 θ for −π
2
< θ < π


2
, we see that


(2.17)lim
θ → 0


cosθ = lim
θ → 0


1 − sin2 θ = 1.


We now take a look at a limit that plays an important role in later chapters—namely, lim
θ → 0


sinθ
θ


. To evaluate this limit,
we use the unit circle in Figure 2.30. Notice that this figure adds one additional triangle to Figure 2.30. We see that thelength of the side opposite angle θ in this new triangle is tanθ. Thus, we see that for 0 < θ < π


2
, sinθ < θ < tanθ.


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Figure 2.30 The sine and tangent functions are shown as lineson the unit circle.


By dividing by sinθ in all parts of the inequality, we obtain
1 < θ


sinθ
< 1


cosθ
.


Equivalently, we have
1 > sinθ


θ
> cosθ.


Since lim
θ → 0+


1 = 1 = lim
θ → 0+


cosθ, we conclude that lim
θ → 0+


sinθ
θ


= 1. By applying a manipulation similar to that used
in demonstrating that lim


θ → 0−
sinθ = 0, we can show that lim


θ → 0−
sinθ
θ


= 1. Thus,
(2.18)lim


θ → 0
sinθ
θ


= 1.


In Example 2.25 we use this limit to establish lim
θ → 0


1 − cosθ
θ


= 0. This limit also proves useful in later chapters.


Example 2.25
Evaluating an Important Trigonometric Limit
Evaluate lim


θ → 0
1 − cosθ


θ
.


Solution
In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine inthe numerator to a sine:


Chapter 2 | Limits 173




(2.19)


2.20


lim
θ → 0


1 − cosθ
θ


= lim
θ → 0


1 − cosθ
θ


· 1 + cosθ
1 + cosθ


= lim
θ → 0


1 − cos2 θ
θ(1 + cosθ)


= lim
θ → 0


sin2 θ
θ(1 + cosθ)


= lim
θ → 0


sinθ
θ


· sinθ
1 + cosθ


= 1 · 0
2
= 0.


Therefore,
lim
θ → 0


1 − cosθ
θ


= 0.


Evaluate lim
θ → 0


1 − cosθ
sinθ


.


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Deriving the Formula for the Area of a Circle
Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of themethods of calculus. The Greek mathematician Archimedes (ca. 287−212; BCE) was particularly inventive, usingpolygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased.He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could havepredicted about the limit.
We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regularpolygon as being made up of n triangles. By taking the limit as the vertex angle of these triangles goes to zero, you canobtain the area of the circle. To see this, carry out the following steps:


1. Express the height h and the base b of the isosceles triangle in Figure 2.31 in terms of θ and r.


Figure 2.31
2. Using the expressions that you obtained in step 1, express the area of the isosceles triangle in terms of θ and r.(Substitute (1 /2)sinθ for sin(θ /2)cos(θ /2) in your expression.)
3. If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between θ and n. Solve thisfor n. Keep in mind there are 2π radians in a circle. (Use radians, not degrees.)
4. Find an expression for the area of the n-sided polygon in terms of r and θ.
5. To find a formula for the area of the circle, find the limit of the expression in step 4 as θ goes to zero. (Hint:


lim
θ → 0


(sinθ)
θ


= 1).


The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration.


Chapter 2 | Limits 175




2.3 EXERCISES
In the following exercises, use the limit laws to evaluateeach limit. Justify each step by indicating the appropriatelimit law(s).
83. lim


x → 0

⎝4x


2 − 2x + 3⎞⎠


84. lim
x → 1


x3 + 3x2 + 5
4 − 7x


85. lim
x → −2


x2 − 6x + 3


86. lim
x → −1


(9x + 1)2


In the following exercises, use direct substitution toevaluate each limit.
87. lim


x → 7
x2


88. lim
x → −2



⎝4x


2 − 1⎞⎠


89. lim
x → 0


1
1 + sinx


90. lim
x → 2


e2x − x
2


91. lim
x → 1


2 − 7x
x + 6


92. lim
x → 3


lne3x


In the following exercises, use direct substitution to showthat each limit leads to the indeterminate form 0/0. Then,
evaluate the limit.
93. lim


x → 4
x2 − 16
x − 4


94. lim
x → 2


x − 2
x2 − 2x


95. lim
x → 6


3x − 18
2x − 12


96. lim
h → 0


(1 + h)2 − 1
h


97. lim
t → 9


t − 9
t − 3


98. lim
h → 0


1
a + h


− 1a
h


, where a is a real-valued constant


99. lim
θ → π


sinθ
tanθ


100. lim
x → 1


x3 − 1
x2 − 1


101. lim
x → 1/2


2x2 + 3x − 2
2x − 1


102. lim
x → −3


x + 4 − 1
x + 3


In the following exercises, use direct substitution to obtainan undefined expression. Then, use the method ofExample 2.23 to simplify the function to help determinethe limit.
103. lim


x → −2−
2x2 + 7x − 4
x2 + x − 2


104. lim
x → −2+


2x2 + 7x − 4
x2 + x − 2


105. lim
x → 1−


2x2 + 7x − 4
x2 + x − 2


106. lim
x → 1+


2x2 + 7x − 4
x2 + x − 2


In the following exercises, assume that
lim
x → 6


f (x) = 4, lim
x → 6


g(x) = 9, and lim
x → 6


h(x) = 6. Use
these three facts and the limit laws to evaluate each limit.
107. lim


x → 6
2 f (x)g(x)


108. lim
x → 6


g(x) − 1
f (x)


109. lim
x → 6



⎝ f (x) +


1
3
g(x)⎞⎠


110. lim
x → 6



⎝h(x)⎞⎠3


2


111. lim
x → 6


g(x) − f (x)


112. lim
x → 6


x · h(x)


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113. lim
x → 6



⎣(x + 1) · f (x)⎤⎦


114. lim
x → 6



⎝ f (x) · g(x) − h(x)⎞⎠


[T] In the following exercises, use a calculator to drawthe graph of each piecewise-defined function and study thegraph to evaluate the given limits.
115. f (x) = ⎧



⎨x


2, x ≤ 3
x + 4, x > 3


a. lim
x → 3−


f (x)


b. lim
x → 3+


f (x)


116. g(x) = ⎧

⎨x


3 − 1, x ≤ 0
1, x > 0


a. lim
x → 0−


g(x)


b. lim
x → 0+


g(x)


117. h(x) = ⎧

⎨x


2 − 2x + 1, x < 2
3 − x, x ≥ 2


a. lim
x → 2−


h(x)


b. lim
x → 2+


h(x)


In the following exercises, use the following graphs and thelimit laws to evaluate each limit.


118. lim
x → −3+



⎝ f (x) + g(x)⎞⎠


119. lim
x → −3−



⎝ f (x) − 3g(x)⎞⎠


120. lim
x → 0


f (x)g(x)
3


121. lim
x → −5


2 + g(x)
f (x)


122. lim
x → 1



⎝ f (x)⎞⎠2


123. lim
x → 1


f (x) − g(x)


124. lim
x → −7



⎝x · g(x)⎞⎠


Chapter 2 | Limits 177




125. lim
x → −9



⎣x · f (x) + 2 · g(x)⎤⎦


For the following problems, evaluate the limit using thesqueeze theorem. Use a calculator to graph the functions
f (x), g(x), and h(x) when possible.
126. [T] True or False? If
2x − 1 ≤ g(x) ≤ x2 − 2x + 3, then lim


x → 2
g(x) = 0.


127. [T] lim
θ → 0


θ2 cos⎛⎝
1
θ



128. lim
x → 0


f (x), where f (x) = ⎧


0, x rational


x2, x irrrational


129. [T] In physics, the magnitude of an electric fieldgenerated by a point charge at a distance r in vacuum
is governed by Coulomb’s law: E(r) = q


4πε0 r
2
, where


E represents the magnitude of the electric field, q is thecharge of the particle, r is the distance between the particle
and where the strength of the field is measured, and 1


4πε0


is Coulomb’s constant: 8.988 × 109 N ·m2 /C2.
a. Use a graphing calculator to graph E(r) given that


the charge of the particle is q = 10−10.
b. Evaluate lim


r → 0+
E(r). What is the physical


meaning of this quantity? Is it physically relevant?Why are you evaluating from the right?
130. [T] The density of an object is given by its massdivided by its volume: ρ = m/V .


a. Use a calculator to plot the volume as a function ofdensity ⎛⎝V = m/ρ⎞⎠, assuming you are examining
something of mass 8 kg (m = 8).


b. Evaluate lim
ρ → 0+


V(ρ) and explain the physical
meaning.


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2.4 | Continuity
Learning Objectives


2.4.1 Explain the three conditions for continuity at a point.
2.4.2 Describe three kinds of discontinuities.
2.4.3 Define continuity on an interval.
2.4.4 State the theorem for limits of composite functions.
2.4.5 Provide an example of the intermediate value theorem.


Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Suchfunctions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this propertyover intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at apoint where a break occurs.
We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively,a function is continuous at a particular point if there is no break in its graph at that point.
Continuity at a Point
Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider variousfunctions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditionsthat prevent such failures.
Our first function of interest is shown in Figure 2.32. We see that the graph of f (x) has a hole at a. In fact, f (a) is
undefined. At the very least, for f (x) to be continuous at a, we need the following condition:


i. f (a) is defined


Figure 2.32 The function f (x) is not continuous at a
because f (a) is undefined.


However, as we see in Figure 2.33, this condition alone is insufficient to guarantee continuity at the point a. Although
f (a) is defined, the function has a gap at a. In this example, the gap exists because lim


x → a
f (x) does not exist. We must add


another condition for continuity at a—namely,
ii. lim


x → a
f (x) exists.


Chapter 2 | Limits 179




Figure 2.33 The function f (x) is not continuous at a
because lim


x → a
f (x) does not exist.


However, as we see in Figure 2.34, these two conditions by themselves do not guarantee continuity at a point. The functionin this figure satisfies both of our first two conditions, but is still not continuous at a. We must add a third condition to ourlist:
iii. lim


x → a
f (x) = f (a).


Figure 2.34 The function f (x) is not continuous at a
because lim


x → a
f (x) ≠ f (a).


Now we put our list of conditions together and form a definition of continuity at a point.
Definition
A function f (x) is continuous at a point a if and only if the following three conditions are satisfied:


i. f (a) is defined
ii. lim


x → a
f (x) exists


iii. lim
x → a


f (x) = f (a)


A function is discontinuous at a point a if it fails to be continuous at a.


The following procedure can be used to analyze the continuity of a function at a point using this definition.


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Problem-Solving Strategy: Determining Continuity at a Point
1. Check to see if f (a) is defined. If f (a) is undefined, we need go no further. The function is not continuous


at a. If f (a) is defined, continue to step 2.
2. Compute lim


x → a
f (x). In some cases, we may need to do this by first computing lim


x → a−
f (x) and lim


x → a+
f (x).


If lim
x → a


f (x) does not exist (that is, it is not a real number), then the function is not continuous at a and the
problem is solved. If lim


x → a
f (x) exists, then continue to step 3.


3. Compare f (a) and lim
x → a


f (x). If lim
x → a


f (x) ≠ f (a), then the function is not continuous at a. If
lim
x → a


f (x) = f (a), then the function is continuous at a.


The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a givenpoint. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.
Example 2.26
Determining Continuity at a Point, Condition 1
Using the definition, determine whether the function f (x) = (x2 − 4)/(x − 2) is continuous at x = 2. Justify
the conclusion.
Solution
Let’s begin by trying to calculate f (2). We can see that f (2) = 0/0, which is undefined. Therefore,
f (x) = x


2 − 4
x − 2


is discontinuous at 2 because f (2) is undefined. The graph of f (x) is shown in Figure 2.35.


Figure 2.35 The function f (x) is discontinuous at 2 because
f (2) is undefined.


Chapter 2 | Limits 181




Example 2.27
Determining Continuity at a Point, Condition 2
Using the definition, determine whether the function f (x) = ⎧



⎨−x


2 + 4 if x ≤ 3
4x − 8 if x > 3


is continuous at x = 3. Justify
the conclusion.
Solution
Let’s begin by trying to calculate f (3).


f (3) = − (32) + 4 = −5.


Thus, f (3) is defined. Next, we calculate lim
x → 3


f (x). To do this, we must compute lim
x → 3−


f (x) and
lim


x → 3+
f (x):


lim
x → 3−


f (x) = − (32) + 4 = −5


and
lim


x → 3+
f (x) = 4(3) − 8 = 4.


Therefore, lim
x → 3


f (x) does not exist. Thus, f (x) is not continuous at 3. The graph of f (x) is shown in Figure
2.36.


Figure 2.36 The function f (x) is not continuous at 3
because lim


x → 3
f (x) does not exist.


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2.21


Example 2.28
Determining Continuity at a Point, Condition 3
Using the definition, determine whether the function f (x) = ⎧




sinx
x if x ≠ 0


1 if x = 0
is continuous at x = 0.


Solution
First, observe that


f (0) = 1.


Next,
lim
x → 0


f (x) = lim
x → 0


sinx
x = 1.


Last, compare f (0) and lim
x → 1


f (x). We see that
f (0) = 1 = lim


x → 0
f (x).


Since all three of the conditions in the definition of continuity are satisfied, f (x) is continuous at x = 0.


Using the definition, determine whether the function f (x) = ⎧


2x + 1 if x < 1


2 if x = 1
−x + 4 if x > 1


is continuous at x = 1.
If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.


By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we canstate the following theorem.
Theorem 2.8: Continuity of Polynomials and Rational Functions
Polynomials and rational functions are continuous at every point in their domains.


Proof
Previously, we showed that if p(x) and q(x) are polynomials, lim


x → a
p(x) = p(a) for every polynomial p(x) and


lim
x → a


p(x)
q(x)


=
p(a)
q(a)


as long as q(a) ≠ 0. Therefore, polynomials and rational functions are continuous on their domains.

We now apply Continuity of Polynomials and Rational Functions to determine the points at which a given rationalfunction is continuous.
Example 2.29
Continuity of a Rational Function


Chapter 2 | Limits 183




2.22


For what values of x is f (x) = x + 1
x − 5


continuous?


Solution
The rational function f (x) = x + 1


x − 5
is continuous for every value of x except x = 5.


For what values of x is f (x) = 3x4 − 4x2 continuous?


Types of Discontinuities
As we have seen in Example 2.26 and Example 2.27, discontinuities take on several different appearances. Weclassify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jumpdiscontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jumpdiscontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinitediscontinuity is a discontinuity located at a vertical asymptote. Figure 2.37 illustrates the differences in these types ofdiscontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep inmind that not all discontinuities fit neatly into these categories.


Figure 2.37 Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.


These three discontinuities are formally defined as follows:
Definition
If f (x) is discontinuous at a, then


1. f has a removable discontinuity at a if lim
x → a


f (x) exists. (Note: When we state that lim
x → a


f (x) exists, we
mean that lim


x → a
f (x) = L, where L is a real number.)


2. f has a jump discontinuity at a if lim
x → a−


f (x) and lim
x → a+


f (x) both exist, but lim
x → a−


f (x) ≠ lim
x → a+


f (x).


(Note: When we state that lim
x → a−


f (x) and lim
x → a+


f (x) both exist, we mean that both are real-valued and that
neither take on the values ±∞.)


3. f has an infinite discontinuity at a if lim
x → a−


f (x) = ±∞ or lim
x → a+


f (x) = ±∞.


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Example 2.30
Classifying a Discontinuity
In Example 2.26, we showed that f (x) = x2 − 4


x − 2
is discontinuous at x = 2. Classify this discontinuity as


removable, jump, or infinite.
Solution
To classify the discontinuity at 2 we must evaluate lim


x → 2
f (x):


lim
x → 2


f (x) = lim
x → 2


x2 − 4
x − 2


= lim
x → 2


(x − 2)(x + 2)
x − 2


= lim
x → 2


(x + 2)


= 4.


Since f is discontinuous at 2 and lim
x → 2


f (x) exists, f has a removable discontinuity at x = 2.


Example 2.31
Classifying a Discontinuity
In Example 2.27, we showed that f (x) = ⎧



⎨−x


2 + 4 if x ≤ 3
4x − 8 if x > 3


is discontinuous at x = 3. Classify this
discontinuity as removable, jump, or infinite.
Solution
Earlier, we showed that f is discontinuous at 3 because lim


x → 3
f (x) does not exist. However, since


lim
x → 3−


f (x) = −5 and lim
x → 3−


f (x) = 4 both exist, we conclude that the function has a jump discontinuity at 3.


Example 2.32
Classifying a Discontinuity
Determine whether f (x) = x + 2


x + 1
is continuous at −1. If the function is discontinuous at −1, classify the


discontinuity as removable, jump, or infinite.
Solution
The function value f (−1) is undefined. Therefore, the function is not continuous at −1. To determine the type of


Chapter 2 | Limits 185




2.23


discontinuity, we must determine the limit at −1. We see that lim
x → −1−


x + 2
x + 1


= −∞ and lim
x → −1+


x + 2
x + 1


= +∞.


Therefore, the function has an infinite discontinuity at −1.


For f (x) = ⎧

⎨x


2 if x ≠ 1
3 if x = 1


, decide whether f is continuous at 1. If f is not continuous at 1, classify the
discontinuity as removable, jump, or infinite.


Continuity over an Interval
Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. Aswe develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function iscontinuous over an interval if we can use a pencil to trace the function between any two points in the interval without liftingthe pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of whatit means for a function to be continuous from the right at a point and continuous from the left at a point.
Continuity from the Right and from the Left
A function f (x) is said to be continuous from the right at a if lim


x → a+
f (x) = f (a).


A function f (x) is said to be continuous from the left at a if lim
x → a−


f (x) = f (a).


A function is continuous over an open interval if it is continuous at every point in the interval. A function f (x) is continuous
over a closed interval of the form ⎡⎣a, b⎤⎦ if it is continuous at every point in (a, b) and is continuous from the right at a
and is continuous from the left at b. Analogously, a function f (x) is continuous over an interval of the form (a, b⎤⎦ if it is
continuous over (a, b) and is continuous from the left at b. Continuity over other types of intervals are defined in a similar
fashion.
Requiring that lim


x → a+
f (x) = f (a) and lim


x → b−
f (x) = f (b) ensures that we can trace the graph of the function from the


point ⎛⎝a, f (a)⎞⎠ to the point ⎛⎝b, f (b)⎞⎠ without lifting the pencil. If, for example, lim
x → a+


f (x) ≠ f (a), we would need to lift
our pencil to jump from f (a) to the graph of the rest of the function over (a, b⎤⎦.
Example 2.33
Continuity on an Interval
State the interval(s) over which the function f (x) = x − 1


x2 + 2x
is continuous.


Solution
Since f (x) = x − 1


x2 + 2x
is a rational function, it is continuous at every point in its domain. The domain of


f (x) is the set (−∞, −2) ∪ (−2, 0) ∪ (0, +∞). Thus, f (x) is continuous over each of the intervals


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2.24


(−∞, −2), (−2, 0), and (0, +∞).


Example 2.34
Continuity over an Interval
State the interval(s) over which the function f (x) = 4 − x2 is continuous.
Solution
From the limit laws, we know that lim


x → a
4 − x2 = 4 − a2 for all values of a in (−2, 2). We also know that


lim
x → −2+


4 − x2 = 0 exists and lim
x → 2−


4 − x2 = 0 exists. Therefore, f (x) is continuous over the interval
[−2, 2].


State the interval(s) over which the function f (x) = x + 3 is continuous.


The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theoremultimately allows us to demonstrate that trigonometric functions are continuous over their domains.
Theorem 2.9: Composite Function Theorem
If f (x) is continuous at L and lim


x → a
g(x) = L, then


lim
x → a


f ⎛⎝g(x)⎞⎠ = f ⎛⎝ limx → ag(x)

⎠ = f (L).


Before we move on to Example 2.35, recall that earlier, in the section on limit laws, we showed lim
x → 0


cosx = 1 = cos(0).


Consequently, we know that f (x) = cosx is continuous at 0. In Example 2.35 we see how to combine this result with the
composite function theorem.
Example 2.35
Limit of a Composite Cosine Function
Evaluate lim


x → π/2
cos⎛⎝x −


π
2

⎠.


Solution


Chapter 2 | Limits 187




2.25


The given function is a composite of cosx and x − π
2
. Since lim


x → π/2



⎝x −


π
2

⎠ = 0 and cosx is continuous at 0,


we may apply the composite function theorem. Thus,
lim


x → π/2
cos⎛⎝x −


π
2

⎠ = cos



⎝ limx → π/2



⎝x −


π
2



⎠ = cos(0) = 1.


Evaluate lim
x → π


sin(x − π).


The proof of the next theorem uses the composite function theorem as well as the continuity of f (x) = sinx and
g(x) = cosx at the point 0 to show that trigonometric functions are continuous over their entire domains.


Theorem 2.10: Continuity of Trigonometric Functions
Trigonometric functions are continuous over their entire domains.


Proof
We begin by demonstrating that cosx is continuous at every real number. To do this, we must show that lim


x → a
cosx = cosa


for all values of a.
lim
x → a


cosx = lim
x → a


cos((x − a) + a) rewrite x = x − a + a


= lim
x → a



⎝cos(x − a)cosa − sin(x − a)sina⎞⎠ apply the identity for the cosine of the sum of two angles


= cos⎛⎝ limx → a(x − a)

⎠cosa − sin



⎝ limx → a(x − a)



⎠sina limx → a(x − a) = 0, and sinx and cosx are continuous at 0


= cos(0)cosa − sin(0)sina evaluate cos(0) and sin(0) and simplify


= 1 · cosa − 0 · sina = cosa.


The proof that sinx is continuous at every real number is analogous. Because the remaining trigonometric functions may
be expressed in terms of sinx and cosx, their continuity follows from the quotient limit law.

As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions.As we continue our study of calculus, we revisit this theorem many times.
The Intermediate Value Theorem
Functions that are continuous over intervals of the form ⎡⎣a, b⎤⎦, where a and b are real numbers, exhibit many useful
properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. Thefirst of these theorems is the Intermediate Value Theorem.
Theorem 2.11: The Intermediate Value Theorem
Let f be continuous over a closed, bounded interval ⎡⎣a, b⎤⎦. If z is any real number between f (a) and f (b), then there
is a number c in ⎡⎣a, b⎤⎦ satisfying f (c) = z in Figure 2.38.


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Figure 2.38 There is a number c ∈ ⎡⎣a, b⎤⎦ that satisfies
f (c) = z.


Example 2.36
Application of the Intermediate Value Theorem
Show that f (x) = x − cosx has at least one zero.
Solution
Since f (x) = x − cosx is continuous over (−∞, +∞), it is continuous over any closed interval of the form

⎣a, b⎤⎦. If you can find an interval ⎡⎣a, b⎤⎦ such that f (a) and f (b) have opposite signs, you can use the
Intermediate Value Theorem to conclude there must be a real number c in (a, b) that satisfies f (c) = 0. Note
that


f (0) = 0 − cos(0) = −1 < 0


and
f ⎛⎝
π
2

⎠ =


π
2
− cosπ


2
= π


2
> 0.


Using the Intermediate Value Theorem, we can see that there must be a real number c in [0, π/2] that satisfies
f (c) = 0. Therefore, f (x) = x − cosx has at least one zero.


Example 2.37
When Can You Apply the Intermediate Value Theorem?
If f (x) is continuous over [0, 2], f (0) > 0 and f (2) > 0, can we use the Intermediate Value Theorem to
conclude that f (x) has no zeros in the interval ⎡⎣0, 2]? Explain.


Chapter 2 | Limits 189




2.26


Solution
No. The Intermediate Value Theorem only allows us to conclude that we can find a value between f (0) and
f (2); it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the
function f (x) = (x − 1)2. It satisfies f (0) = 1 > 0, f (2) = 1 > 0, and f (1) = 0.


Example 2.38
When Can You Apply the Intermediate Value Theorem?
For f (x) = 1/x, f (−1) = −1 < 0 and f (1) = 1 > 0. Can we conclude that f (x) has a zero in the interval
[−1, 1]?


Solution
No. The function is not continuous over [−1, 1]. The Intermediate Value Theorem does not apply here.


Show that f (x) = x3 − x2 − 3x + 1 has a zero over the interval [0, 1].


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2.4 EXERCISES
For the following exercises, determine the point(s), if any,at which each function is discontinuous. Classify anydiscontinuity as jump, removable, infinite, or other.
131. f (x) = 1x
132. f (x) = 2


x2 + 1


133. f (x) = x
x2 − x


134. g(t) = t−1 + 1
135. f (x) = 5


ex − 2


136. f (x) = |x − 2|
x − 2


137. H(x) = tan2x
138. f (t) = t + 3


t2 + 5t + 6


For the following exercises, decide if the functioncontinuous at the given point. If it is discontinuous, whattype of discontinuity is it?
139. 2x2 − 5x + 3


x − 1
at x = 1


140. h(θ) = sinθ − cosθ
tanθ


at θ = π


141. g(u) =









6u2 + u − 2
2u − 1


if u ≠ 1
2


7
2


if u = 1
2


, at u = 1
2


142. f (y) = sin(πy)
tan(πy)


, at y = 1


143. f (x) = ⎧

⎨x


2 − ex if x < 0
x − 1 if x ≥ 0


, at x = 0


144. f (x) = ⎧


xsin(x) if x ≤ π
x tan(x) if x > π


, at x = π
In the following exercises, find the value(s) of k that makeseach function continuous over the given interval.
145. f (x) = ⎧




3x + 2, x < k
2x − 3, k ≤ x ≤ 8


146. f (θ) = ⎧



sinθ, 0 ≤ θ < π
2


cos(θ + k), π
2
≤ θ ≤ π


147. f (x) = ⎧


x2 + 3x + 2


x + 2
, x ≠ − 2


k, x = −2


148. f (x) = ⎧

⎨ e


kx, 0 ≤ x < 4
x + 3, 4 ≤ x ≤ 8


149. f (x) = ⎧

⎨ kx, 0 ≤ x ≤ 3
x + 1, 3 < x ≤ 10


In the following exercises, use the Intermediate ValueTheorem (IVT).
150. Let h(x) = ⎧



⎨3x


2 − 4, x ≤ 2
5 + 4x, x > 2


Over the interval
[0, 4], there is no value of x such that h(x) = 10,
although h(0) < 10 and h(4) > 10. Explain why this
does not contradict the IVT.
151. A particle moving along a line has at each time ta position function s(t), which is continuous. Assume
s(2) = 5 and s(5) = 2. Another particle moves such that
its position is given by h(t) = s(t) − t. Explain why there
must be a value c for 2 < c < 5 such that h(c) = 0.
152. [T] Use the statement “The cosine of t is equal to tcubed.”a. Write a mathematical equation of the statement.b. Prove that the equation in part a. has at least onereal solution.c. Use a calculator to find an interval of length 0.01that contains a solution.
153. Apply the IVT to determine whether 2x = x3 has
a solution in one of the intervals ⎡⎣1.25, 1.375⎤⎦ or

⎣1.375, 1.5⎤⎦. Briefly explain your response for each
interval.


Chapter 2 | Limits 191




154. Consider the graph of the function y = f (x) shown
in the following graph.


a. Find all values for which the function isdiscontinuous.b. For each value in part a., state why the formaldefinition of continuity does not apply.c. Classify each discontinuity as either jump,removable, or infinite.
155. Let f (x) = ⎧




3x, x > 1


x3, x < 1
.


a. Sketch the graph of f.b. Is it possible to find a value k such that f (1) = k,
which makes f (x) continuous for all real
numbers? Briefly explain.


156. Let f (x) = x4 − 1
x2 − 1


for x ≠ − 1, 1.
a. Sketch the graph of f.b. Is it possible to find values k1 and k2 such that


f (−1) = k and f (1) = k2, and that makes f (x)
continuous for all real numbers? Briefly explain.


157. Sketch the graph of the function y = f (x) with
properties i. through vii.i. The domain of f is (−∞, +∞).


ii. f has an infinite discontinuity at x = −6.
iii. f (−6) = 3
iv. lim


x → −3−
f (x) = lim


x → −3+
f (x) = 2


v. f (−3) = 3
vi. f is left continuous but not right continuous at


x = 3.
vii. lim


x → −∞
f (x) = −∞ and lim


x → +∞
f (x) = +∞


158. Sketch the graph of the function y = f (x) with
properties i. through iv.i. The domain of f is ⎡⎣0, 5⎤⎦.


ii. lim
x → 1+


f (x) and lim
x → 1−


f (x) exist and are equal.
iii. f (x) is left continuous but not continuous at


x = 2, and right continuous but not continuous at
x = 3.


iv. f (x) has a removable discontinuity at x = 1, a
jump discontinuity at x = 2, and the following
limits hold: lim


x → 3−
f (x) = −∞ and


lim
x → 3+


f (x) = 2.


In the following exercises, suppose y = f (x) is defined for
all x. For each description, sketch a graph with the indicatedproperty.
159. Discontinuous at x = 1 with lim


x → −1
f (x) = −1 and


lim
x → 2


f (x) = 4


160. Discontinuous at x = 2 but continuous elsewhere
with lim


x → 0
f (x) = 1


2


Determine whether each of the given statements is true.Justify your response with an explanation orcounterexample.
161. f (t) = 2


et − e−t
is continuous everywhere.


162. If the left- and right-hand limits of f (x) as x → a
exist and are equal, then f cannot be discontinuous at
x = a.


163. If a function is not continuous at a point, then it is notdefined at that point.
164. According to the IVT, cosx − sinx − x = 2 has a
solution over the interval [−1, 1].
165. If f (x) is continuous such that f (a) and f (b) have
opposite signs, then f (x) = 0 has exactly one solution in

⎣a, b⎤⎦.


166. The function f (x) = x2 − 4x + 3
x2 − 1


is continuous
over the interval [0, 3].


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167. If f (x) is continuous everywhere and
f (a), f (b) > 0, then there is no root of f (x) in the
interval ⎡⎣a, b⎤⎦.
[T] The following problems consider the scalar form ofCoulomb’s law, which describes the electrostatic forcebetween two point charges, such as electrons. It is given by
the equation F(r) = ke |q1q2|


r2
, where ke is Coulomb’s


constant, qi are the magnitudes of the charges of the two
particles, and r is the distance between the two particles.
168. To simplify the calculation of a model with manyinteracting particles, after some threshold value r = R,
we approximate F as zero.a. Explain the physical reasoning behind thisassumption.b. What is the force equation?c. Evaluate the force F using both Coulomb’s lawand our approximation, assuming two protons witha charge magnitude of


1.6022 × 10−19 coulombs (C), and the
Coulomb constant ke = 8.988 × 109Nm2/C2 are
1 m apart. Also, assume R < 1m. How much
inaccuracy does our approximation generate? Is ourapproximation reasonable?d. Is there any finite value of R for which this systemremains continuous at R?


169. Instead of making the force 0 at R, instead we letthe force be 10−20 for r ≥ R. Assume two protons, which
have a magnitude of charge 1.6022 × 10−19 C, and the
Coulomb constant ke = 8.988 × 109Nm2/C2. Is there a
value R that can make this system continuous? If so, findit.
Recall the discussion on spacecraft from the chapteropener. The following problems consider a rocket launchfrom Earth’s surface. The force of gravity on the rocket is
given by F(d) = − mk/d2, where m is the mass of the
rocket, d is the distance of the rocket from the center ofEarth, and k is a constant.
170. [T] Determine the value and units of k given thatthe mass of the rocket on Earth is 3 million kg. (Hint: Thedistance from the center of Earth to its surface is 6378 km.)
171. [T] After a certain distance D has passed, thegravitational effect of Earth becomes quite negligible, sowe can approximate the force function by
F(d) =






−mk
d2


if d < D


10,000 if d ≥ D
. Find the necessary condition


D such that the force function remains continuous.


172. As the rocket travels away from Earth’s surface, thereis a distance D where the rocket sheds some of its mass,since it no longer needs the excess fuel storage. We can
write this function as F(d) =












m1 k


d2
if d < D



m2 k


d2
if d ≥ D


. Is there


a D value such that this function is continuous, assuming
m1 ≠ m2?


Prove the following functions are continuous everywhere
173. f (θ) = sinθ
174. g(x) = |x|
175. Where is f (x) = ⎧




0 if x is irrational
1 if x is rational


continuous?


Chapter 2 | Limits 193




2.5 | The Precise Definition of a Limit
Learning Objectives


2.5.1 Describe the epsilon-delta definition of a limit.
2.5.2 Apply the epsilon-delta definition to find the limit of a function.
2.5.3 Describe the epsilon-delta definitions of one-sided limits and infinite limits.
2.5.4 Use the epsilon-delta definition to prove the limit laws.


By now you have progressed from the very informal definition of a limit in the introduction of this chapter to the intuitiveunderstanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function meansand how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precisemathematical language. The formal definition of a limit is quite possibly one of the most challenging definitions you willencounter early in your study of calculus; however, it is well worth any effort you make to reconcile it with your intuitivenotion of a limit. Understanding this definition is the key that opens the door to a better understanding of calculus.
Quantifying Closeness
Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance betweentwo points a and b on a number line is given by |a − b|.


• The statement | f (x) − L| < ε may be interpreted as: The distance between f (x) and L is less than ε.
• The statement 0 < |x − a| < δ may be interpreted as: x ≠ a and the distance between x and a is less than δ.


It is also important to look at the following equivalences for absolute value:
• The statement | f (x) − L| < ε is equivalent to the statement L − ε < f (x) < L + ε.
• The statement 0 < |x − a| < δ is equivalent to the statement a − δ < x < a + δ and x ≠ a.


With these clarifications, we can state the formal epsilon-delta definition of the limit.
Definition
Let f (x) be defined for all x ≠ a over an open interval containing a. Let L be a real number. Then


lim
x → a


f (x) = L


if, for every ε > 0, there exists a δ > 0, such that if 0 < |x − a| < δ, then | f (x) − L| < ε.


This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if webreak it down phrase by phrase. The statement itself involves something called a universal quantifier (for every ε > 0), an
existential quantifier (there exists a δ > 0), and, last, a conditional statement (if 0 < |x − a| < δ, then | f (x) − L| < ε).
Let’s take a look at Table 2.9, which breaks down the definition and translates each part.


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Definition Translation
1. For every ε > 0, 1. For every positive distance ε from L,
2. there exists a δ > 0, 2. There is a positive distance δ from a,


3. such that 3. such that
4. if 0 < |x − a| < δ, then
| f (x) − L| < ε.


4. if x is closer than δ to a and x ≠ a, then f (x) is closer than
ε to L.


Table 2.9 Translation of the Epsilon-Delta Definition of the Limit
We can get a better handle on this definition by looking at the definition geometrically. Figure 2.39 shows possible valuesof δ for various choices of ε > 0 for a given function f (x), a number a, and a limit L at a. Notice that as we choose
smaller values of ε (the distance between the function and the limit), we can always find a δ small enough so that if we
have chosen an x value within δ of a, then the value of f (x) is within ε of the limit L.


Figure 2.39 These graphs show possible values of δ , given successively smaller choices of ε.
Visit the following applet to experiment with finding values of δ for selected values of ε:


• The epsilon-delta definition of limit (http://www.openstaxcollege.org/l/20_epsilondelt)
Example 2.39 shows how you can use this definition to prove a statement about the limit of a specific function at aspecified value.
Example 2.39
Proving a Statement about the Limit of a Specific Function
Prove that lim


x → 1
(2x + 1) = 3.


Solution


Chapter 2 | Limits 195




Let ε > 0.
The first part of the definition begins “For every ε > 0.” This means we must prove that whatever follows is true
no matter what positive value of ε is chosen. By stating “Let ε > 0,” we signal our intent to do so.
Choose δ = ε


2
.


The definition continues with “there exists a δ > 0. ” The phrase “there exists” in a mathematical statement is
always a signal for a scavenger hunt. In other words, we must go and find δ. So, where exactly did δ = ε/2
come from? There are two basic approaches to tracking down δ. One method is purely algebraic and the other is
geometric.
We begin by tackling the problem from an algebraic point of view. Since ultimately we want |(2x + 1) − 3| < ε,
we begin by manipulating this expression: |(2x + 1) − 3| < ε is equivalent to |2x − 2| < ε, which in turn
is equivalent to |2||x − 1| < ε. Last, this is equivalent to |x − 1| < ε/2. Thus, it would seem that δ = ε/2 is
appropriate.
We may also find δ through geometric methods. Figure 2.40 demonstrates how this is done.


Figure 2.40 This graph shows how we find δ geometrically.


Assume 0 < |x − 1| < δ. When δ has been chosen, our goal is to show that if 0 < |x − 1| < δ, then
|(2x + 1) − 3| < ε. To prove any statement of the form “If this, then that,” we begin by assuming “this” and
trying to get “that.”
Thus,
|(2x + 1) − 3| = |2x − 2| property of absolute value


= |2(x − 1)|
= |2||x − 1| |2| = 2
= 2|x − 1|
< 2 · δ here’s where we use the assumption that 0 < |x − 1| < δ


= 2 · ε
2
= ε here’s where we use our choice of δ = ε/2


Analysis


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In this part of the proof, we started with |(2x + 1) − 3| and used our assumption 0 < |x − 1| < δ in a key part
of the chain of inequalities to get |(2x + 1) − 3| to be less than ε. We could just as easily have manipulated the
assumed inequality 0 < |x − 1| < δ to arrive at |(2x + 1) − 3| < ε as follows:
0 < |x − 1| < δ ⇒ |x − 1| < δ


⇒ − δ < x − 1 < δ
⇒ − ε


2
< x − 1 < ε


2
⇒ − ε < 2x − 2 < ε
⇒ − ε < 2x − 2 < ε
⇒ |2x − 2| < ε
⇒ |(2x + 1) − 3| < ε.


Therefore, lim
x → 1


(2x + 1) = 3. (Having completed the proof, we state what we have accomplished.)
After removing all the remarks, here is a final version of the proof:
Let ε > 0.
Choose δ = ε/2.
Assume 0 < |x − 1| < δ.
Thus,
|(2x + 1) − 3| = |2x − 2|


= |2(x − 1)|
= |2||x − 1|
= 2|x − 1|
< 2 · δ
= 2 · ε


2
= ε.


Therefore, lim
x → 1


(2x + 1) = 3.


The following Problem-Solving Strategy summarizes the type of proof we worked out in Example 2.39.
Problem-Solving Strategy: Proving That lim


x→ a
f(x) = L for a Specific Function f(x)


1. Let’s begin the proof with the following statement: Let ε > 0.
2. Next, we need to obtain a value for δ. After we have obtained this value, we make the following statement,


filling in the blank with our choice of δ : Choose δ = _______.
3. The next statement in the proof should be (at this point, we fill in our given value for a):Assume 0 < |x − a| < δ.
4. Next, based on this assumption, we need to show that | f (x) − L| < ε, where f (x) and L are our function


f (x) and our limit L. At some point, we need to use 0 < |x − a| < δ.
5. We conclude our proof with the statement: Therefore, lim


x → a
f (x) = L.


Chapter 2 | Limits 197




2.27


Example 2.40
Proving a Statement about a Limit
Complete the proof that lim


x → −1
(4x + 1) = −3 by filling in the blanks.


Let _____.
Choose δ = _______.
Assume 0 < |x − _______| < δ.
Thus, |________ − ________| = _____________________________________ε.
Solution
We begin by filling in the blanks where the choices are specified by the definition. Thus, we have
Let ε > 0.
Choose δ = _______.
Assume 0 < |x − (−1)| < δ. (or equivalently, 0 < |x + 1| < δ.)
Thus, |(4x + 1) − (−3)| = |4x + 4| = |4||x + 1| < 4δ_______ε.
Focusing on the final line of the proof, we see that we should choose δ = ε


4
.


We now complete the final write-up of the proof:
Let ε > 0.
Choose δ = ε


4
.


Assume 0 < |x − (−1)| < δ (or equivalently, 0 < |x + 1| < δ.)
Thus, |(4x + 1) − (−3)| = |4x + 4| = |4||x + 1| < 4δ = 4(ε/4) = ε.


Complete the proof that lim
x → 2


(3x − 2) = 4 by filling in the blanks.
Let _______.
Choose δ = _______.
Assume 0 < |x − ____| < ____.
Thus,
|_______ − ____| = ______________________________ε.


Therefore, lim
x → 2


(3x − 2) = 4.


In Example 2.39 and Example 2.40, the proofs were fairly straightforward, since the functions with which we wereworking were linear. In Example 2.41, we see how to modify the proof to accommodate a nonlinear function.
Example 2.41


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Proving a Statement about the Limit of a Specific Function (Geometric Approach)
Prove that lim


x → 2
x2 = 4.


Solution
1. Let ε > 0. The first part of the definition begins “For every ε > 0,” so we must prove that whatever


follows is true no matter what positive value of ε is chosen. By stating “Let ε > 0,” we signal our intent
to do so.


2. Without loss of generality, assume ε ≤ 4. Two questions present themselves: Why do we want ε ≤ 4
and why is it okay to make this assumption? In answer to the first question: Later on, in the process of
solving for δ, we will discover that δ involves the quantity 4 − ε. Consequently, we need ε ≤ 4. In
answer to the second question: If we can find δ > 0 that “works” for ε ≤ 4, then it will “work” for any
ε > 4 as well. Keep in mind that, although it is always okay to put an upper bound on ε, it is never okay
to put a lower bound (other than zero) on ε.


3. Choose δ = min⎧⎩⎨2 − 4 − ε, 4 + ε − 2⎫⎭⎬. Figure 2.41 shows how we made this choice of δ.


Figure 2.41 This graph shows how we find δ geometrically for a given εfor the proof in Example 2.41.
4. We must show: If 0 < |x − 2| < δ, then |x2 − 4| < ε, so we must begin by assuming


0 < |x − 2| < δ.


We don’t really need 0 < |x − 2| (in other words, x ≠ 2) for this proof. Since
0 < |x − 2| < δ ⇒ |x − 2| < δ, it is okay to drop 0 < |x − 2|.


|x − 2| < δ.


Hence,
−δ < x − 2 < δ.


Recall that δ = min⎧⎩⎨2 − 4 − ε, 4 + ε − 2⎫⎭⎬. Thus, δ ≥ 2 − 4 − ε and consequently
−⎛⎝2 − 4 − ε



⎠ ≤ − δ. We also use δ ≤ 4 + ε − 2 here. We might ask at this point: Why did we


substitute 2 − 4 − ε for δ on the left-hand side of the inequality and 4 + ε − 2 on the right-hand
side of the inequality? If we look at Figure 2.41, we see that 2 − 4 − ε corresponds to the distance on


Chapter 2 | Limits 199




2.28


the left of 2 on the x-axis and 4 + ε − 2 corresponds to the distance on the right. Thus,
−⎛⎝2 − 4 − ε



⎠ ≤ − δ < x − 2 < δ ≤ 4 + ε − 2.


We simplify the expression on the left:
−2 + 4 − ε < x − 2 < 4 + ε − 2.


Then, we add 2 to all parts of the inequality:
4 − ε < x < 4 + ε.


We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:
4 − ε < x2 < 4 + ε.


We subtract 4 from all parts of the inequality:
−ε < x2 − 4 < ε.


Last,
|x2 − 4| < ε.


5. Therefore,
lim
x → 2


x2 = 4.


Find δ corresponding to ε > 0 for a proof that lim
x → 9


x = 3.


The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions.Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may alsoapproach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provideus with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is theprimary tool used in proofs of statements about limits. For Example 2.42, we take on a purely algebraic approach.
Example 2.42
Proving a Statement about the Limit of a Specific Function (Algebraic Approach)
Prove that lim


x → −1

⎝x


2 − 2x + 3⎞⎠ = 6.


Solution
Let’s use our outline from the Problem-Solving Strategy:


1. Let ε > 0.
2. Choose δ = min{1, ε/5}. This choice of δ may appear odd at first glance, but it was obtained by


200 Chapter 2 | Limits


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2.29


taking a look at our ultimate desired inequality: |⎛⎝x2 − 2x + 3⎞⎠− 6| < ε. This inequality is equivalent
to |x + 1| · |x − 3| < ε. At this point, the temptation simply to choose δ = εx − 3 is very strong.
Unfortunately, our choice of δ must depend on ε only and no other variable. If we can replace |x − 3| by
a numerical value, our problem can be resolved. This is the place where assuming δ ≤ 1 comes into play.
The choice of δ ≤ 1 here is arbitrary. We could have just as easily used any other positive number. In
some proofs, greater care in this choice may be necessary. Now, since δ ≤ 1 and |x + 1| < δ ≤ 1, we
are able to show that |x − 3| < 5. Consequently, |x + 1| · |x − 3| < |x + 1| · 5. At this point we realize
that we also need δ ≤ ε/5. Thus, we choose δ = min{1, ε/5}.


3. Assume 0 < |x + 1| < δ. Thus,
|x + 1| < 1 and |x + 1| < ε5.


Since |x + 1| < 1, we may conclude that −1 < x + 1 < 1. Thus, by subtracting 4 from all parts of the
inequality, we obtain −5 < x − 3 < −1. Consequently, |x − 3| < 5. This gives us


|⎛⎝x2 − 2x + 3⎞⎠− 6| = |x + 1| · |x − 3| < ε5 · 5 = ε.


Therefore,
lim


x → −1

⎝x


2 − 2x + 3⎞⎠ = 6.


Complete the proof that lim
x → 1


x2 = 1.


Let ε > 0; choose δ = min{1, ε/3}; assume 0 < |x − 1| < δ.
Since |x − 1| < 1, we may conclude that −1 < x − 1 < 1. Thus, 1 < x + 1 < 3. Hence, |x + 1| < 3.


You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest toapply. The algebraic approach is also more useful in proving statements about limits.
Proving Limit Laws
We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws.The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.
Definition
The triangle inequality states that if a and b are any real numbers, then |a + b| ≤ |a| + |b|.


Proof
We prove the following limit law: If lim


x → a
f (x) = L and lim


x → a
g(x) = M, then lim


x → a

⎝ f (x) + g(x)⎞⎠ = L + M.


Let ε > 0.
Choose δ1 > 0 so that if 0 < |x − a| < δ1, then | f (x) − L| < ε/2.
Choose δ2 > 0 so that if 0 < |x − a| < δ2, then |g(x) − M| < ε/2.


Chapter 2 | Limits 201




Choose δ = min⎧⎩⎨δ1, δ2⎫⎭⎬.
Assume 0 < |x − a| < δ.
Thus,


0 < |x − a| < δ1 and 0 < |x − a| < δ2.


Hence,
|⎛⎝ f (x) + g(x)⎞⎠− (L + M)| = |⎛⎝ f (x) − L⎞⎠+ ⎛⎝g(x) − M⎞⎠|


≤ | f (x) − L| + |g(x) − M|
< ε


2
+ ε


2
= ε.



We now explore what it means for a limit not to exist. The limit lim


x → a
f (x) does not exist if there is no real number L for


which lim
x → a


f (x) = L. Thus, for all real numbers L, lim
x → a


f (x) ≠ L. To understand what this means, we look at each part
of the definition of lim


x → a
f (x) = L together with its opposite. A translation of the definition is given in Table 2.10.


Definition Opposite
1. For every ε > 0, 1. There exists ε > 0 so that


2. there exists a δ > 0, so that 2. for every δ > 0,


3. if 0 < |x − a| < δ, then
| f (x) − L| < ε.


3. There is an x satisfying 0 < |x − a| < δ so that
| f (x) − L| ≥ ε.


Table 2.10 Translation of the Definition of lim
x → a


f (x) = L and its Opposite
Finally, we may state what it means for a limit not to exist. The limit lim


x → a
f (x) does not exist if for every real number L,


there exists a real number ε > 0 so that for all δ > 0, there is an x satisfying 0 < |x − a| < δ, so that | f (x) − L| ≥ ε.
Let’s apply this in Example 2.43 to show that a limit does not exist.
Example 2.43
Showing That a Limit Does Not Exist
Show that lim


x → 0
|x|
x does not exist. The graph of f (x) = |x|/x is shown here:


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Solution
Suppose that L is a candidate for a limit. Choose ε = 1/2.
Let δ > 0. Either L ≥ 0 or L < 0. If L ≥ 0, then let x = − δ/2. Thus,


|x − 0| = |− δ2 − 0| = δ2 < δ
and


| |− δ2 |− δ2 − L| = |−1 − L| = L + 1 ≥ 1 > 12 = ε.
On the other hand, if L < 0, then let x = δ/2. Thus,


|x − 0| = |δ2 − 0| = δ2 < δ
and


| |δ2 |δ2 − L| = |1 − L| = |L| + 1 ≥ 1 > 12 = ε.
Thus, for any value of L, lim


x → 0
|x|
x ≠ L.


One-Sided and Infinite Limits
Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit,we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-deltadefinitions for limits from the right and left at a point. These definitions only require slight modifications from the definitionof the limit. In the definition of the limit from the right, the inequality 0 < x − a < δ replaces 0 < |x − a| < δ, which
ensures that we only consider values of x that are greater than (to the right of) a. Similarly, in the definition of the limit fromthe left, the inequality −δ < x − a < 0 replaces 0 < |x − a| < δ, which ensures that we only consider values of x that
are less than (to the left of) a.
Definition
Limit from the Right: Let f (x) be defined over an open interval of the form (a, b) where a < b. Then,


Chapter 2 | Limits 203




2.30


lim
x → a+


f (x) = L


if for every ε > 0, there exists a δ > 0 such that if 0 < x − a < δ, then | f (x) − L| < ε.
Limit from the Left: Let f (x) be defined over an open interval of the form (b, c) where b < c. Then,


lim
x → a−


f (x) = L


if for every ε > 0, there exists a δ > 0 such that if −δ < x − a < 0, then | f (x) − L| < ε.


Example 2.44
Proving a Statement about a Limit From the Right
Prove that lim


x → 4+
x − 4 = 0.


Solution
Let ε > 0.
Choose δ = ε2. Since we ultimately want | x − 4 − 0| < ε, we manipulate this inequality to get x − 4 < ε
or, equivalently, 0 < x − 4 < ε2, making δ = ε2 a clear choice. We may also determine δ geometrically, as
shown in Figure 2.42.


Figure 2.42 This graph shows how we find δ for the proof inExample 2.44.


Assume 0 < x − 4 < δ. Thus, 0 < x − 4 < ε2. Hence, 0 < x − 4 < ε. Finally, | x − 4 − 0| < ε.
Therefore, lim


x → 4+
x − 4 = 0.


Find δ corresponding to ε for a proof that lim
x → 1−


1 − x = 0.


We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by


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pursuing a formal definition of infinite limits. To have lim
x → a


f (x) = +∞, we want the values of the function f (x)
to get larger and larger as x approaches a. Instead of the requirement that | f (x) − L| < ε for arbitrarily small ε when
0 < |x − a| < δ for small enough δ, we want f (x) > M for arbitrarily large positiveM when 0 < |x − a| < δ for small
enough δ. Figure 2.43 illustrates this idea by showing the value of δ for successively larger values of M.


Figure 2.43 These graphs plot values of δ for M to show that lim
x → a


f (x) = +∞.


Definition
Let f (x) be defined for all x ≠ a in an open interval containing a. Then, we have an infinite limit


lim
x → a


f (x) = +∞


if for every M > 0, there exists δ > 0 such that if 0 < |x − a| < δ, then f (x) > M.
Let f (x) be defined for all x ≠ a in an open interval containing a. Then, we have a negative infinite limit


lim
x → a


f (x) = −∞


if for every M > 0, there exists δ > 0 such that if 0 < |x − a| < δ, then f (x) < −M.


Chapter 2 | Limits 205




2.5 EXERCISES
In the following exercises, write the appropriate ε − δ
definition for each of the given statements.
176. lim


x → a
f (x) = N


177. lim
t → b


g(t) = M


178. lim
x → c


h(x) = L


179. lim
x → a


φ(x) = A


The following graph of the function f satisfies
lim
x → 2


f (x) = 2. In the following exercises, determine a
value of δ > 0 that satisfies each statement.


180. If 0 < |x − 2| < δ, then | f (x) − 2| < 1.
181. If 0 < |x − 2| < δ, then | f (x) − 2| < 0.5.
The following graph of the function f satisfies
lim
x → 3


f (x) = −1. In the following exercises, determine a
value of δ > 0 that satisfies each statement.


182. If 0 < |x − 3| < δ, then | f (x) + 1| < 1.
183. If 0 < |x − 3| < δ, then | f (x) + 1| < 2.
The following graph of the function f satisfies
lim
x → 3


f (x) = 2. In the following exercises, for each value
of ε, find a value of δ > 0 such that the precise definition
of limit holds true.


184. ε = 1.5
185. ε = 3
[T] In the following exercises, use a graphing calculator tofind a number δ such that the statements hold true.
186. |sin(2x) − 12| < 0.1, whenever |x − π12| < δ


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187. | x − 4 − 2| < 0.1, whenever |x − 8| < δ
In the following exercises, use the precise definition oflimit to prove the given limits.
188. lim


x → 2
(5x + 8) = 18


189. lim
x → 3


x2 − 9
x − 3


= 6


190. lim
x → 2


2x2 − 3x − 2
x − 2


= 5


191. lim
x → 0


x4 = 0


192. lim
x → 2


(x2 + 2x) = 8


In the following exercises, use the precise definition oflimit to prove the given one-sided limits.
193. lim


x → 5−
5 − x = 0


194.
lim


x → 0+
f (x) = −2, where f (x) =






8x − 3, if x < 0
4x − 2, if x ≥ 0


.


195. lim
x → 1−


f (x) = 3, where f (x) =




5x − 2, if x < 1
7x − 1, if x ≥ 1


.


In the following exercises, use the precise definition oflimit to prove the given infinite limits.
196. lim


x → 0
1
x2


= ∞


197. lim
x → −1


3
(x + 1)2


= ∞


198. lim
x → 2


− 1
(x − 2)2


= −∞


199. An engineer is using a machine to cut a flat squareof Aerogel of area 144 cm2. If there is a maximum errortolerance in the area of 8 cm2, how accurately must theengineer cut on the side, assuming all sides have the samelength? How do these numbers relate to δ, ε, a, and L?
200. Use the precise definition of limit to prove that the
following limit does not exist: lim


x → 1
|x − 1|
x − 1


.


201. Using precise definitions of limits, prove that
lim
x → 0


f (x) does not exist, given that f (x) is the ceiling
function. (Hint: Try any δ < 1.)
202. Using precise definitions of limits, prove that
lim
x → 0


f (x) does not exist: f (x) = ⎧


1 if x is rational
0 if x is irrational


.


(Hint: Think about how you can always choose a rationalnumber 0 < r < d, but | f (r) − 0| = 1.)
203. Using precise definitions of limits, determine
lim
x → 0


f (x) for f (x) = ⎧


x if x is rational
0 if x is irrational


. (Hint: Break
into two cases, x rational and x irrational.)
204. Using the function from the previous exercise, usethe precise definition of limits to show that lim


x → a
f (x) does


not exist for a ≠ 0.
For the following exercises, suppose that lim


x → a
f (x) = L


and lim
x → a


g(x) = M both exist. Use the precise definition
of limits to prove the following limit laws:
205. lim


x → a

⎝ f (x) − g(x)⎞⎠ = L − M


206. lim
x → a



⎣c f (x)⎤⎦ = cL for any real constant c (Hint:


Consider two cases: c = 0 and c ≠ 0.)
207. lim


x → a

⎣ f (x)g(x)⎤⎦ = LM. (Hint: | f (x)g(x) − LM| =


| f (x)g(x) − f (x)M + f (x)M − LM| ≤ | f (x)||g(x) − M| + |M|| f (x) − L|.)


Chapter 2 | Limits 207




average velocity


constant multiple law for limits
continuity at a point


continuity from the left
continuity from the right
continuity over an interval


difference law for limits
differential calculus
discontinuity at a point
epsilon-delta definition of the limit


infinite discontinuity
infinite limit
instantaneous velocity


integral calculus
Intermediate Value Theorem


intuitive definition of the limit


jump discontinuity


limit


CHAPTER 2 REVIEW
KEY TERMS


the change in an object’s position divided by the length of a time period; the average velocity of anobject over a time interval [t, a] (if t < a or [a, t] if t > a) , with a position given by s(t), that is
vave =


s(t) − s(a)
t − a


the limit law lim
x → a


c f (x) = c · lim
x → a


f (x) = cL


A function f (x) is continuous at a point a if and only if the following three conditions are
satisfied: (1) f (a) is defined, (2) lim


x → a
f (x) exists, and (3) lim


x → a
f (x) = f (a)


A function is continuous from the left at b if lim
x → b−


f (x) = f (b)


A function is continuous from the right at a if lim
x → a+


f (x) = f (a)


a function that can be traced with a pencil without lifting the pencil; a function iscontinuous over an open interval if it is continuous at every point in the interval; a function f (x) is continuous over a
closed interval of the form ⎡⎣a, b⎤⎦ if it is continuous at every point in (a, b), and it is continuous from the right at a
and from the left at b


the limit law lim
x → a



⎝ f (x) − g(x)⎞⎠ = limx → a f (x) − limx → ag(x) = L − M


the field of calculus concerned with the study of derivatives and their applications
A function is discontinuous at a point or has a discontinuity at a point if it is not continuous atthe point


lim
x → a


f (x) = L if for every ε > 0, there exists a δ > 0 such that if
0 < |x − a| < δ, then | f (x) − L| < ε


An infinite discontinuity occurs at a point a if lim
x → a−


f (x) = ±∞ or lim
x → a+


f (x) = ±∞


A function has an infinite limit at a point a if it either increases or decreases without bound as it approachesa
The instantaneous velocity of an object with a position function that is given by s(t) is the


value that the average velocities on intervals of the form [t, a] and [a, t] approach as the values of tmove closer to
a, provided such a value exists


the study of integrals and their applications
Let f be continuous over a closed bounded interval ⎡⎣a, b⎤⎦; if z is any real number


between f (a) and f (b), then there is a number c in ⎡⎣a, b⎤⎦ satisfying f (c) = z
If all values of the function f (x) approach the real number L as the values of x( ≠ a)


approach a, f (x) approaches L
A jump discontinuity occurs at a point a if lim


x → a−
f (x) and lim


x → a+
f (x) both exist, but


lim
x → a−


f (x) ≠ lim
x → a+


f (x)


the process of letting x or t approach a in an expression; the limit of a function f (x) as x approaches a is the value


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limit laws


multivariable calculus
one-sided limit
power law for limits
product law for limits
quotient law for limits


removable discontinuity


root law for limits
secant


squeeze theorem


sum law for limits
tangent


triangle inequality
vertical asymptote


that f (x) approaches as x approaches a
the individual properties of limits; for each of the individual laws, let f (x) and g(x) be defined for all x ≠ a


over some open interval containing a; assume that L and M are real numbers so that lim
x → a


f (x) = L and
lim
x → a


g(x) = M; let c be a constant
the study of the calculus of functions of two or more variables


A one-sided limit of a function is a limit taken from either the left or the right
the limit law lim


x → a

⎝ f (x)⎞⎠n = ⎛⎝ limx → a f (x)





n
= Ln for every positive integer n


the limit law lim
x → a



⎝ f (x) · g(x)⎞⎠ = limx → a f (x) · limx → ag(x) = L ·M


the limit law lim
x → a


f (x)
g(x)


=
lim
x → a


f (x)


lim
x → a


g(x)
= L


M
for M ≠ 0


A removable discontinuity occurs at a point a if f (x) is discontinuous at a, but lim
x → a


f (x)


exists
the limit law lim


x → a
f (x)


n
= lim


x → a
f (x)n = L


n for all L if n is odd and for L ≥ 0 if n is even
A secant line to a function f (x) at a is a line through the point ⎛⎝a, f (a)⎞⎠ and another point on the function; the


slope of the secant line is given by msec = f (x) − f (a)x − a
states that if f (x) ≤ g(x) ≤ h(x) for all x ≠ a over an open interval containing a and


lim
x → a


f (x) = L = lim
x → a


h(x) where L is a real number, then lim
x → a


g(x) = L


The limit law lim
x → a



⎝ f (x) + g(x)⎞⎠ = limx → a f (x) + limx → ag(x) = L + M


A tangent line to the graph of a function at a point ⎛⎝a, f (a)⎞⎠ is the line that secant lines through ⎛⎝a, f (a)⎞⎠
approach as they are taken through points on the function with x-values that approach a; the slope of the tangent lineto a graph at a measures the rate of change of the function at a


If a and b are any real numbers, then |a + b| ≤ |a| + |b|
A function has a vertical asymptote at x = a if the limit as x approaches a from the right or left is


infinite
KEY EQUATIONS


• Slope of a Secant Line
msec =


f (x) − f (a)
x − a


• Average Velocity over Interval [a, t]
vave =


s(t) − s(a)
t − a


• Intuitive Definition of the Limit
lim
x → a


f (x) = L


• Two Important Limits
lim
x → a


x = a lim
x → a


c = c


• One-Sided Limits


Chapter 2 | Limits 209




lim
x → a−


f (x) = L lim
x → a+


f (x) = L


• Infinite Limits from the Left
lim


x → a−
f (x) = +∞ lim


x → a−
f (x) = −∞


• Infinite Limits from the Right
lim


x → a+
f (x) = +∞ lim


x → a+
f (x) = −∞


• Two-Sided Infinite Limits
lim
x → a


f (x) = +∞ : lim
x → a−


f (x) = +∞ and lim
x → a+


f (x) = +∞


lim
x → a


f (x) = −∞ : lim
x → a−


f (x) = −∞ and lim
x → a+


f (x) = −∞


• Basic Limit Results
lim
x → a


x = a lim
x → a


c = c


• Important Limits
lim
θ → 0


sinθ = 0


lim
θ → 0


cosθ = 1


lim
θ → 0


sinθ
θ


= 1


lim
θ → 0


1 − cosθ
θ


= 0


KEY CONCEPTS
2.1 A Preview of Calculus


• Differential calculus arose from trying to solve the problem of determining the slope of a line tangent to a curve at apoint. The slope of the tangent line indicates the rate of change of the function, also called the derivative. Calculatinga derivative requires finding a limit.
• Integral calculus arose from trying to solve the problem of finding the area of a region between the graph of afunction and the x-axis. We can approximate the area by dividing it into thin rectangles and summing the areas ofthese rectangles. This summation leads to the value of a function called the integral. The integral is also calculatedby finding a limit and, in fact, is related to the derivative of a function.
• Multivariable calculus enables us to solve problems in three-dimensional space, including determining motion inspace and finding volumes of solids.


2.2 The Limit of a Function
• A table of values or graph may be used to estimate a limit.
• If the limit of a function at a point does not exist, it is still possible that the limits from the left and right at that pointmay exist.
• If the limits of a function from the left and right exist and are equal, then the limit of the function is that commonvalue.
• We may use limits to describe infinite behavior of a function at a point.


2.3 The Limit Laws
• The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes eachtime.
• For polynomials and rational functions, lim


x → a
f (x) = f (a).


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• You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifyinga complex fraction.
• The squeeze theorem allows you to find the limit of a function if the function is always greater than one functionand less than another function with limits that are known.


2.4 Continuity
• For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and thevalue of the function at that point must equal the value of the limit at that point.
• Discontinuities may be classified as removable, jump, or infinite.
• A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous overa closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
• The composite function theorem states: If f (x) is continuous at L and lim


x → a
g(x) = L, then


lim
x → a


f ⎛⎝g(x)⎞⎠ = f ⎛⎝ limx → ag(x)

⎠ = f (L).


• The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the functiontakes on every value between the values at its endpoints.
2.5 The Precise Definition of a Limit


• The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the epsilon-deltadefinition of the limit.
• The epsilon-delta definition may be used to prove statements about limits.
• The epsilon-delta definition of a limit may be modified to define one-sided limits.


CHAPTER 2 REVIEW EXERCISES
True or False. In the following exercises, justify youranswer with a proof or a counterexample.
208. A function has to be continuous at x = a if the
lim
x → a


f (x) exists.


209. You can use the quotient rule to evaluate lim
x → 0


sinx
x .


210. If there is a vertical asymptote at x = a for the
function f (x), then f is undefined at the point x = a.
211. If lim


x → a
f (x) does not exist, then f is undefined at the


point x = a.


212. Using the graph, find each limit or explain why thelimit does not exist.a. lim
x → −1


f (x)


b. lim
x → 1


f (x)


c. lim
x → 0+


f (x)


d. lim
x → 2


f (x)


In the following exercises, evaluate the limit algebraicallyor explain why the limit does not exist.
213. lim


x → 2
2x2 − 3x − 2


x − 2


Chapter 2 | Limits 211




214. lim
x → 0


3x2 − 2x + 4


215. lim
x → 3


x3 − 2x2 − 1
3x − 2


216. lim
x → π/2


cotx
cosx


217. lim
x → −5


x2 + 25
x + 5


218. lim
x → 2


3x2 − 2x − 8
x2 − 4


219. lim
x → 1


x2 − 1
x3 − 1


220. lim
x → 1


x2 − 1
x − 1


221. lim
x → 4


4 − x
x − 2


222. lim
x → 4


1
x − 2


In the following exercises, use the squeeze theorem toprove the limit.
223. lim


x → 0
x2 cos(2πx) = 0


224. lim
x → 0


x3 sin ⎛⎝πx

⎠ = 0


225. Determine the domain such that the function
f (x) = x − 2 + xex is continuous over its domain.
In the following exercises, determine the value of c suchthat the function remains continuous. Draw your resultingfunction to ensure it is continuous.
226. f (x) = ⎧



⎨x


2 + 1, x > c
2x, x ≤ c


227. f (x) = ⎧



x + 1, x > −1


x2 + c, x ≤ − 1


In the following exercises, use the precise definition oflimit to prove the limit.


228. lim
x → 1


(8x + 16) = 24


229. lim
x → 0


x3 = 0


230. A ball is thrown into the air and the vertical position
is given by x(t) = −4.9t2 + 25t + 5. Use the Intermediate
Value Theorem to show that the ball must land on theground sometime between 5 sec and 6 sec after the throw.
231. A particle moving along a line has a displacement
according to the function x(t) = t2 − 2t + 4, where x is
measured in meters and t is measured in seconds. Find theaverage velocity over the time period t = [0, 2].
232. From the previous exercises, estimate theinstantaneous velocity at t = 2 by checking the average
velocity within t = 0.01 sec.


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3 | DERIVATIVES


Figure 3.1 The Hennessey Venom GT can go from 0 to 200 mph in 14.51 seconds. (credit: modification of work by Codex41,Flickr)


Chapter Outline
3.1 Defining the Derivative
3.2 The Derivative as a Function
3.3 Differentiation Rules
3.4 Derivatives as Rates of Change
3.5 Derivatives of Trigonometric Functions
3.6 The Chain Rule
3.7 Derivatives of Inverse Functions
3.8 Implicit Differentiation
3.9 Derivatives of Exponential and Logarithmic Functions


Introduction
The Hennessey Venom GT is one of the fastest cars in the world. In 2014, it reached a record-setting speed of 270.49 mph.It can go from 0 to 200 mph in 14.51 seconds. The techniques in this chapter can be used to calculate the acceleration theVenom achieves in this feat (see Example 3.8.)
Calculating velocity and changes in velocity are important uses of calculus, but it is far more widespread than that. Calculusis important in all branches of mathematics, science, and engineering, and it is critical to analysis in business and health as


Chapter 3 | Derivatives 213




well. In this chapter, we explore one of the main tools of calculus, the derivative, and show convenient ways to calculatederivatives. We apply these rules to a variety of functions in this chapter so that we can then explore applications of thesetechniques.
3.1 | Defining the Derivative


Learning Objectives
3.1.1 Recognize the meaning of the tangent to a curve at a point.
3.1.2 Calculate the slope of a tangent line.
3.1.3 Identify the derivative as the limit of a difference quotient.
3.1.4 Calculate the derivative of a given function at a point.
3.1.5 Describe the velocity as a rate of change.
3.1.6 Explain the difference between average velocity and instantaneous velocity.
3.1.7 Estimate the derivative from a table of values.


Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have establishedthe foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals.Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton
(1643–1727) and the German Gottfried Leibniz (1646–1716), whose images appear in Figure 3.2. When we credit
Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the firstto understand the relationship between the derivative and the integral. Both mathematicians benefited from the work ofpredecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears tohave been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although itseems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notationthat we commonly use today.


Figure 3.2 Newton and Leibniz are credited with developing calculus independently.


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Tangent Lines
We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope ofa secant line to a function at a point (a, f (a)) to estimate the rate of change, or the rate at which one variable changes in
relation to another variable. We can obtain the slope of the secant by choosing a value of x near a and drawing a line
through the points (a, f (a)) and ⎛⎝x, f (x)⎞⎠, as shown in Figure 3.3. The slope of this line is given by an equation in the
form of a difference quotient:


msec =
f (x) − f (a)


x − a .


We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing x with
a + h, where h is a value close to a. We can then calculate the slope of the line through the points (a, f (a)) and
(a + h, f (a + h)). In this case, we find the secant line has a slope given by the following difference quotient with
increment h:


msec =
f (a + h) − f (a)


a + h − a
=


f (a + h) − f (a)
h


.


Definition
Let f be a function defined on an interval I containing a. If x ≠ a is in I, then


(3.1)Q = f (x) − f (a)x − a
is a difference quotient.
Also, if h ≠ 0 is chosen so that a + h is in I, then


(3.2)Q = f (a + h) − f (a)
h


is a difference quotient with increment h.


View several Java applets (http://www.openstaxcollege.org/l/20_calcapplets) on the development of thederivative.


These two expressions for calculating the slope of a secant line are illustrated in Figure 3.3. We will see that each of thesetwo methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. Theprimary consideration in our choice usually depends on ease of calculation.


Chapter 3 | Derivatives 215




Figure 3.3 We can calculate the slope of a secant line in either of two ways.


In Figure 3.4(a) we see that, as the values of x approach a, the slopes of the secant lines provide better estimates of the
rate of change of the function at a. Furthermore, the secant lines themselves approach the tangent line to the function at
a, which represents the limit of the secant lines. Similarly, Figure 3.4(b) shows that as the values of h get closer to 0,
the secant lines also approach the tangent line. The slope of the tangent line at a is the rate of change of the function at a,
as shown in Figure 3.4(c).


Figure 3.4 The secant lines approach the tangent line (shown in green) as the second point approaches the first.
You can use this site (http://www.openstaxcollege.org/l/20_diffmicros) to explore graphs to see if theyhave a tangent line at a point.


In Figure 3.5 we show the graph of f (x) = x and its tangent line at (1, 1) in a series of tighter intervals about x = 1.
As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values onthe tangent line a good approximation to the values of the function for choices of x close to 1. In fact, the graph of f (x)
itself appears to be locally linear in the immediate vicinity of x = 1.


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Figure 3.5 For values of x close to 1, the graph of f (x) = x and its tangent line appear to coincide.


Formally we may define the tangent line to the graph of a function as follows.
Definition
Let f (x) be a function defined in an open interval containing a. The tangent line to f (x) at a is the line passing
through the point ⎛⎝a, f (a)⎞⎠ having slope


(3.3)mtan = limx → a f (x) − f (a)x − a
provided this limit exists.
Equivalently, we may define the tangent line to f (x) at a to be the line passing through the point ⎛⎝a, f (a)⎞⎠ having
slope


(3.4)mtan = limh → 0 f (a + h) − f (a)h
provided this limit exists.


Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define theslope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will dependon the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to findequations of tangent lines.


Chapter 3 | Derivatives 217




Example 3.1
Finding a Tangent Line
Find the equation of the line tangent to the graph of f (x) = x2 at x = 3.
Solution
First find the slope of the tangent line. In this example, use Equation 3.3.


mtan = limx → 3
f (x) − f (3)


x − 3
Apply the definition


= lim
x → 3


x2 − 9
x − 3


Substitute f (x) = x2 and f (3) = 9.


= lim
x → 3


(x − 3)(x + 3)
x − 3


= lim
x → 3


(x + 3) = 6 Factor the numerator to evaluate the limit.


Next, find a point on the tangent line. Since the line is tangent to the graph of f (x) at x = 3, it passes through
the point ⎛⎝3, f (3)⎞⎠. We have f (3) = 9, so the tangent line passes through the point (3, 9).
Using the point-slope equation of the line with the slope m = 6 and the point (3, 9), we obtain the line
y − 9 = 6(x − 3). Simplifying, we have y = 6x − 9. The graph of f (x) = x2 and its tangent line at 3 are
shown in Figure 3.6.


Figure 3.6 The tangent line to f (x) at x = 3.


Example 3.2
The Slope of a Tangent Line Revisited
Use Equation 3.4 to find the slope of the line tangent to the graph of f (x) = x2 at x = 3.
Solution
The steps are very similar to Example 3.1. See Equation 3.4 for the definition.


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mtan = limh → 0
f (3 + h) − f (3)


h
Apply the definition


= lim
h → 0


(3 + h)2 − 9
h


Substitute f (3 + h) = (3 + h)2 and f (3) = 9.


= lim
h → 0


9 + 6h + h2 − 9
h


Expand and simplify to evaluate the limit.


= lim
h → 0


h(6 + h)
h


= lim
h → 0


(6 + h) = 6


We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that theformulas can be interchanged.


Example 3.3
Finding the Equation of a Tangent Line
Find the equation of the line tangent to the graph of f (x) = 1/x at x = 2.
Solution
We can use Equation 3.3, but as we have seen, the results are the same if we use Equation 3.4.


mtan = limx → 2
f (x) − f (2)


x − 2
Apply the definition


= lim
x → 2


1
x −


1
2


x − 2
Substitute f (x) = 1x and f (2) =


1
2
.


= lim
x → 2


1
x −


1
2


x − 2
· 2x
2x


Multiply numerator and denominator by 2x to


simplify fractions.


= lim
x → 2


(2 − x)
(x − 2)(2x)


Simplify.


= lim
x → 2


−1
2x


Simplify using 2 − x
x − 2


= −1, for x ≠ 2.


= − 1
4


Evaluate the limit.


We now know that the slope of the tangent line is −1
4
. To find the equation of the tangent line, we also need a


point on the line. We know that f (2) = 1
2
. Since the tangent line passes through the point (2, 1


2
) we can use


the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation
y = − 1


4
x + 1. The graphs of f (x) = 1x and y = − 14x + 1 are shown in Figure 3.7.


Chapter 3 | Derivatives 219




3.1


Figure 3.7 The line is tangent to f (x) at x = 2.


Find the slope of the line tangent to the graph of f (x) = x at x = 4.


The Derivative of a Function at a Point
The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applicationsacross many disciplines. These applications include velocity and acceleration in physics, marginal profit functions inbusiness, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative.The process of finding a derivative is called differentiation.
Definition
Let f (x) be a function defined in an open interval containing a. The derivative of the function f (x) at a, denoted
by f ′ (a), is defined by


(3.5)f ′ (a) = lim
x → a


f (x) − f (a)
x − a


provided this limit exists.
Alternatively, we may also define the derivative of f (x) at a as


(3.6)f ′ (a) = lim
h → 0


f (a + h) − f (a)
h


.


Example 3.4
Estimating a Derivative
For f (x) = x2, use a table to estimate f ′(3) using Equation 3.5.
Solution


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3.2


Create a table using values of x just below 3 and just above 3.
x x2−9


x−3


2.9 5.9


2.99 5.99


2.999 5.999


3.001 6.001


3.01 6.01


3.1 6.1


After examining the table, we see that a good estimate is f ′ (3) = 6.


For f (x) = x2, use a table to estimate f ′(3) using Equation 3.6.


Example 3.5
Finding a Derivative
For f (x) = 3x2 − 4x + 1, find f ′(2) by using Equation 3.5.
Solution
Substitute the given function and value directly into the equation.


f ′ (x) = lim
x → 2


f (x) − f (2)
x − 2


Apply the definition


= lim
x → 2



⎝3x


2 − 4x + 1⎞⎠− 5
x − 2


Substitute f (x) = 3x2 − 4x + 1 and f (2) = 5.


= lim
x → 2


(x − 2)(3x + 2)
x − 2


Simplify and factor the numerator.


= lim
x → 2


(3x + 2) Cancel the common factor.


= 8 Evaluate the limit.


Chapter 3 | Derivatives 221




3.3


Example 3.6
Revisiting the Derivative
For f (x) = 3x2 − 4x + 1, find f ′(2) by using Equation 3.6.
Solution
Using this equation, we can substitute two values of the function into the equation, and we should get the samevalue as in Example 3.5.


f ′ (2) = lim
h → 0


f (2 + h) − f (2)
h


Apply the definition


= lim
h → 0


(3(2 + h)2 − 4(2 + h) + 1) − 5
h


Substitute f (2) = 5 and


f (2 + h) = 3(2 + h)2 − 4(2 + h) + 1.


= lim
h → 0


3h2 + 8h
h


Simplify the numerator.


= lim
h → 0


h(3h + 8)
h


Factor the numerator.


= lim
h → 0


(3h + 8) Cancel the common factor.


= 8 Evaluate the limit.


The results are the same whether we use Equation 3.5 or Equation 3.6.


For f (x) = x2 + 3x + 2, find f ′ (1).


Velocities and Rates of Change
Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s(t) is the position of an object
moving along a coordinate axis, the average velocity of the object over a time interval [a, t] if t > a or [t, a] if t < a is
given by the difference quotient


(3.7)vave = s(t) − s(a)t − a .
As the values of t approach a, the values of vave approach the value we call the instantaneous velocity at a. That is,
instantaneous velocity at a, denoted v(a), is given by


(3.8)v(a) = s′ (a) = lim
t → a


s(t) − s(a)
t − a .


To better understand the relationship between average velocity and instantaneous velocity, see Figure 3.8. In this figure,the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t = a whose position at time
t is given by the function s(t). The slope of the secant line (shown in green) is the average velocity of the object over the
time interval [a, t].


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Figure 3.8 The slope of the secant line is the average velocityover the interval [a, t]. The slope of the tangent line is the
instantaneous velocity.


We can use Equation 3.5 to calculate the instantaneous velocity, or we can estimate the velocity of a moving object byusing a table of values. We can then confirm the estimate by using Equation 3.7.
Example 3.7
Estimating Velocity
A lead weight on a spring is oscillating up and down. Its position at time t with respect to a fixed horizontal
line is given by s(t) = sin t (Figure 3.9). Use a table of values to estimate v(0). Check the estimate by using
Equation 3.5.


Figure 3.9 A lead weight suspended from a spring in verticaloscillatory motion.
Solution
We can estimate the instantaneous velocity at t = 0 by computing a table of average velocities using values of t
approaching 0, as shown in Table 3.2.


Chapter 3 | Derivatives 223




3.4


t sint −sin0
t −0


= sintt


−0.1 0.998334166


−0.01 0.9999833333


−0.001 0.999999833


0.001 0.999999833


0.01 0.9999833333


0.1 0.998334166


Table 3.2Average velocities using values of tapproaching 0
From the table we see that the average velocity over the time interval [−0.1, 0] is 0.998334166, the average
velocity over the time interval [−0.01, 0] is 0.9999833333, and so forth. Using this table of values, it appears
that a good estimate is v(0) = 1.
By using Equation 3.5, we can see that


v(0) = s′ (0) = lim
t → 0


sin t − sin0
t − 0


= lim
t → 0


sin t
t = 1.


Thus, in fact, v(0) = 1.


A rock is dropped from a height of 64 feet. Its height above ground at time t seconds later is given by
s(t) = −16t2 + 64, 0 ≤ t ≤ 2. Find its instantaneous velocity 1 second after it is dropped, using Equation
3.5.


As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are relatedconcepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, orthe rate of change of a function at any point along the function.
Definition
The instantaneous rate of change of a function f (x) at a value a is its derivative f ′(a).


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Example 3.8
Chapter Opener: Estimating Rate of Change of Velocity


Figure 3.10 (credit: modification of work by Codex41,Flickr)


Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it
went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51
seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change
of velocity (that is, its acceleration) as it approaches 229.9 mph. Does the rate at which the car is accelerating
appear to be increasing, decreasing, or constant?
Solution
First observe that 60 mph = 88 ft/s, 100 mph ≈ 146.67 ft/s, 200 mph ≈ 293.33 ft/s, and 229.9 mph
≈ 337.19 ft/s. We can summarize the information in a table.


t v(t)


0 0


3.05 88


5.88 147.67


14.51 293.33


19.96 337.19


Table 3.3
v(t) at different values
of t


Now compute the average acceleration of the car in feet per second on intervals of the form ⎡⎣t, 19.96⎤⎦ as t
approaches 19.96, as shown in the following table.


Chapter 3 | Derivatives 225




t v(t)− v(19.96)
t −19.96


= v(t)−337.19
t −19.96


0.0 16.89


3.05 14.74


5.88 13.46


14.51 8.05


Table 3.4Average acceleration
The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s).


Example 3.9
Rate of Change of Temperature
A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F at 9 p.m.,
reaches a low of 60° during the night, and rises back to 70° by 7 a.m. the next morning. Suppose that the
temperature in the house is given by T(t) = 0.4t2 − 4t + 70 for 0 ≤ t ≤ 10, where t is the number of hours
past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.
Solution
Since midnight is 3 hours past 9 p.m., we want to compute T′(3). Refer to Equation 3.5.


T′ (3) = lim
t → 3


T(t) − T(3)
t − 3


Apply the definition


= lim
t → 3


0.4t2 − 4t + 70 − 61.6
t − 3


Substitute T(t) = 0.4t2 − 4t + 70 and
T(3) = 61.6.


= lim
t → 3


0.4t2 − 4t + 8.4
t − 3


Simplify.


= lim
t → 3


0.4(t − 3)(t − 7)
t − 3


= lim
t → 3


0.4(t − 3)(t − 7)
t − 3


= lim
t → 3


0.4(t − 7) Cancel.


= −1.6 Evaluate the limit.


The instantaneous rate of change of the temperature at midnight is −1.6°F per hour.


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3.5


Example 3.10
Rate of Change of Profit
A toy company can sell x electronic gaming systems at a price of p = −0.01x + 400 dollars per gaming
system. The cost of manufacturing x systems is given by C(x) = 100x + 10,000 dollars. Find the rate of change
of profit when 10,000 games are produced. Should the toy company increase or decrease production?
Solution
The profit P(x) earned by producing x gaming systems is R(x) − C(x), where R(x) is the revenue obtained
from the sale of x games. Since the company can sell x games at p = −0.01x + 400 per game,


R(x) = xp = x(−0.01x + 400) = −0.01x2 + 400x.


Consequently,
P(x) = −0.01x2 + 300x − 10,000.


Therefore, evaluating the rate of change of profit gives
P′ (10000) = lim


x → 10000


P(x) − P(10000)
x − 10000


= lim
x → 10000


−0.01x2 + 300x − 10000 − 1990000
x − 10000


= lim
x → 10000


−0.01x2 + 300x − 2000000
x − 10000


= 100.


Since the rate of change of profit P′ (10,000) > 0 and P(10,000) > 0, the company should increase
production.


A coffee shop determines that the daily profit on scones obtained by charging s dollars per scone is
P(s) = −20s2 + 150s − 10. The coffee shop currently charges $3.25 per scone. Find P′(3.25), the rate of
change of profit when the price is $3.25 and decide whether or not the coffee shop should consider raising or
lowering its prices on scones.


Chapter 3 | Derivatives 227




3.1 EXERCISES
For the following exercises, use Equation 3.3 to find theslope of the secant line between the values x1 and x2 for
each function y = f (x).
1. f (x) = 4x + 7; x1 = 2, x2 = 5
2. f (x) = 8x − 3; x1 = −1, x2 = 3
3. f (x) = x2 + 2x + 1; x1 = 3, x2 = 3.5
4. f (x) = −x2 + x + 2; x1 = 0.5, x2 = 1.5
5. f (x) = 4


3x − 1
; x1 = 1, x2 = 3


6. f (x) = x − 7
2x + 1


; x1 = −2, x2 = 0


7. f (x) = x; x1 = 1, x2 = 16
8. f (x) = x − 9; x1 = 10, x2 = 13
9. f (x) = x1/3 + 1; x1 = 0, x2 = 8
10. f (x) = 6x2/3 + 2x1/3; x1 = 1, x2 = 27
For the following functions,


a. use Equation 3.4 to find the slope of the tangentline mtan = f ′ (a), and
b. find the equation of the tangent line to f at x = a.


11. f (x) = 3 − 4x, a = 2
12. f (x) = x


5
+ 6, a = −1


13. f (x) = x2 + x, a = 1
14. f (x) = 1 − x − x2, a = 0
15. f (x) = 7x , a = 3
16. f (x) = x + 8, a = 1
17. f (x) = 2 − 3x2, a = −2
18. f (x) = −3


x − 1
, a = 4


19. f (x) = 2
x + 3


, a = −4


20. f (x) = 3
x2


, a = 3


For the following functions y = f (x), find f ′ (a) using
Equation 3.3.
21. f (x) = 5x + 4, a = −1
22. f (x) = −7x + 1, a = 3
23. f (x) = x2 + 9x, a = 2
24. f (x) = 3x2 − x + 2, a = 1
25. f (x) = x, a = 4
26. f (x) = x − 2, a = 6
27. f (x) = 1x , a = 2
28. f (x) = 1


x − 3
, a = −1


29. f (x) = 1
x3


, a = 1


30. f (x) = 1x, a = 4
For the following exercises, given the function y = f (x),


a. find the slope of the secant line PQ for each point
Q⎛⎝x, f (x)⎞⎠ with x value given in the table.


b. Use the answers from a. to estimate the value of theslope of the tangent line at P.
c. Use the answer from b. to find the equation of thetangent line to f at point P.


228 Chapter 3 | Derivatives


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31. [T] f (x) = x2 + 3x + 4, P(1, 8) (Round to 6
decimal places.)


x Slope
mPQ


x Slope
mPQ


1.1 (i) 0.9 (vii)
1.01 (ii) 0.99 (viii)
1.001 (iii) 0.999 (ix)
1.0001 (iv) 0.9999 (x)
1.00001 (v) 0.99999 (xi)
1.000001 (vi) 0.999999 (xii)


32. [T] f (x) = x + 1
x2 − 1


, P(0, −1)


x Slope
mPQ


x Slope
mPQ


0.1 (i) −0.1 (vii)


0.01 (ii) −0.01 (viii)


0.001 (iii) −0.001 (ix)


0.0001 (iv) −0.0001 (x)


0.00001 (v) −0.00001 (xi)


0.000001 (vi) −0.000001 (xii)


33. [T] f (x) = 10e0.5x, P(0, 10) (Round to 4 decimal
places.)


x Slope mPQ
−0.1 (i)


−0.01 (ii)


−0.001 (iii)


−0.0001 (iv)


−0.00001 (v)


−0.000001 (vi)


34. [T] f (x) = tan(x), P(π, 0)
x Slope mPQ
3.1 (i)
3.14 (ii)
3.141 (iii)
3.1415 (iv)
3.14159 (v)
3.141592 (vi)


[T] For the following position functions y = s(t), an
object is moving along a straight line, where t is in seconds
and s is in meters. Find


a. the simplified expression for the average velocityfrom t = 2 to t = 2 + h;
b. the average velocity between t = 2 and


t = 2 + h, where (i) h = 0.1, (ii) h = 0.01,
(iii) h = 0.001, and (iv) h = 0.0001; and


c. use the answer from a. to estimate the instantaneous


Chapter 3 | Derivatives 229




velocity at t = 2 second.
35. s(t) = 1


3
t + 5


36. s(t) = t2 − 2t
37. s(t) = 2t3 + 3
38. s(t) = 16


t2
− 4t


39. Use the following graph to evaluate a. f ′ (1) and b.
f ′ (6).


40. Use the following graph to evaluate a. f ′ (−3) and b.
f ′ (1.5).


For the following exercises, use the limit definition ofderivative to show that the derivative does not exist at
x = a for each of the given functions.
41. f (x) = x1/3, x = 0
42. f (x) = x2/3, x = 0
43. f (x) = ⎧




1, x < 1
x, x ≥ 1


, x = 1


44. f (x) = |x|x , x = 0


45. [T] The position in feet of a race car along a straighttrack after t seconds is modeled by the function
s(t) = 8t2 − 1


16
t3.


a. Find the average velocity of the vehicle over thefollowing time intervals to four decimal places:i. [4, 4.1]ii. [4, 4.01]iii. [4, 4.001]iv. [4, 4.0001]b. Use a. to draw a conclusion about the instantaneousvelocity of the vehicle at t = 4 seconds.
46. [T] The distance in feet that a ball rolls down an
incline is modeled by the function s(t) = 14t2, where t is
seconds after the ball begins rolling.a. Find the average velocity of the ball over thefollowing time intervals:i. [5, 5.1]ii. [5, 5.01]iii. [5, 5.001]iv. [5, 5.0001]b. Use the answers from a. to draw a conclusion aboutthe instantaneous velocity of the ball at t = 5


seconds.
47. Two vehicles start out traveling side by side alonga straight road. Their position functions, shown in thefollowing graph, are given by s = f (t) and s = g(t),
where s is measured in feet and t is measured in seconds.


a. Which vehicle has traveled farther at t = 2
seconds?b. What is the approximate velocity of each vehicle at
t = 3 seconds?


c. Which vehicle is traveling faster at t = 4 seconds?
d. What is true about the positions of the vehicles at


t = 4 seconds?


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48. [T] The total cost C(x), in hundreds of dollars,
to produce x jars of mayonnaise is given by
C(x) = 0.000003x3 + 4x + 300.


a. Calculate the average cost per jar over thefollowing intervals:i. [100, 100.1]ii. [100, 100.01]iii. [100, 100.001]iv. [100, 100.0001]b. Use the answers from a. to estimate the averagecost to produce 100 jars of mayonnaise.
49. [T] For the function f (x) = x3 − 2x2 − 11x + 12,
do the following.a. Use a graphing calculator to graph f in anappropriate viewing window.b. Use the ZOOM feature on the calculator toapproximate the two values of x = a for which


mtan = f ′ (a) = 0.


50. [T] For the function f (x) = x
1 + x2


, do the
following.a. Use a graphing calculator to graph f in an


appropriate viewing window.b. Use the ZOOM feature on the calculator toapproximate the values of x = a for which
mtan = f ′ (a) = 0.


51. Suppose that N(x) computes the number of gallons
of gas used by a vehicle traveling x miles. Suppose the
vehicle gets 30 mpg.


a. Find a mathematical expression for N(x).
b. What is N(100)? Explain the physical meaning.
c. What is N′(100)? Explain the physical meaning.


52. [T] For the function f (x) = x4 − 5x2 + 4, do the
following.a. Use a graphing calculator to graph f in an


appropriate viewing window.b. Use the nDeriv function, which numerically finds
the derivative, on a graphing calculator to estimate
f ′ (−2), f ′(−0.5), f ′(1.7), and f ′(2.718).


53. [T] For the function f (x) = x2
x2 + 1


, do the
following.a. Use a graphing calculator to graph f in an


appropriate viewing window.b. Use the nDeriv function on a graphing calculator
to find f ′ (−4), f ′(−2), f ′(2), and f ′(4).


Chapter 3 | Derivatives 231




3.2 | The Derivative as a Function
Learning Objectives


3.2.1 Define the derivative function of a given function.
3.2.2 Graph a derivative function from the graph of a given function.
3.2.3 State the connection between derivatives and continuity.
3.2.4 Describe three conditions for when a function does not have a derivative.
3.2.5 Explain the meaning of a higher-order derivative.


As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to thefunction at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seemsreasonable to conclude that knowing the derivative of the function at every point would produce valuable information aboutthe behavior of the function. However, the process of finding the derivative at even a handful of values using the techniquesof the preceding section would quickly become quite tedious. In this section we define the derivative function and learn aprocess for finding it.
Derivative Functions
The derivative function gives the derivative of a function at each point in the domain of the original function for which thederivative is defined. We can formally define a derivative function as follows.
Definition
Let f be a function. The derivative function, denoted by f ′, is the function whose domain consists of those values
of x such that the following limit exists:


(3.9)f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


.


A function f (x) is said to be differentiable at a if f ′(a) exists. More generally, a function is said to be differentiable
on S if it is differentiable at every point in an open set S, and a differentiable function is one in which f ′(x) exists on
its domain.
In the next few examples we use Equation 3.9 to find the derivative of a function.
Example 3.11
Finding the Derivative of a Square-Root Function
Find the derivative of f (x) = x.
Solution
Start directly with the definition of the derivative function. Use Equation 3.1.


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3.6


f ′ (x) = lim
h → 0


x + h − x
h


Substitute f (x + h) = x + h and f (x) = x


into f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


.


= lim
h → 0


x + h − x
h


· x + h + x
x + h + x


Multiply numerator and denominator by


x + h + xwithout distributing in the


denominator.


= lim
h → 0


h
h( x + h + x)


Multiply the numerators and simplify.


= lim
h → 0


1
( x + h + x)


Cancel the h.


= 1
2 x


Evaluate the limit.


Example 3.12
Finding the Derivative of a Quadratic Function
Find the derivative of the function f (x) = x2 − 2x.
Solution
Follow the same procedure here, but without having to multiply by the conjugate.


f ′ (x) = lim
h → 0


((x + h)2 − 2(x + h)) − (x2 − 2x)
h


Substitute f (x + h) = (x + h)2 − 2(x + h) and


f (x) = x2 − 2x into


f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


.


= lim
h → 0


x2 + 2xh + h2 − 2x − 2h − x2 + 2x
h


Expand (x + h)2 − 2(x + h).


= lim
h → 0


2xh − 2h + h2


h
Simplify.


= lim
h → 0


h(2x − 2 + h)
h


Factor out h from the numerator.


= lim
h → 0


(2x − 2 + h) Cancel the common factor of h.


= 2x − 2 Evaluate the limit.


Find the derivative of f (x) = x2.


We use a variety of different notations to express the derivative of a function. In Example 3.12 we showed that if
f (x) = x2 − 2x, then f ′ (x) = 2x − 2. If we had expressed this function in the form y = x2 − 2x, we could have
expressed the derivative as y′ = 2x − 2 or dy


dx
= 2x − 2. We could have conveyed the same information by writing


d
dx

⎝x


2 − 2x⎞⎠ = 2x − 2. Thus, for the function y = f (x), each of the following notations represents the derivative of
f (x):


Chapter 3 | Derivatives 233




f ′ (x),
dy
dx


, y′, d
dx

⎝ f (x)⎞⎠.


In place of f ′ (a) we may also use dy
dx |x = a Use of the dydx notation (called Leibniz notation) is quite common in


engineering and physics. To understand this notation better, recall that the derivative of a function at a point is the limit ofthe slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed
in the form Δy


Δx
where Δy is the difference in the y values corresponding to the difference in the x values, which are


expressed as Δx (Figure 3.11). Thus the derivative, which can be thought of as the instantaneous rate of change of y
with respect to x, is expressed as


dy
dx


= lim
Δx → 0


Δy
Δx


.


Figure 3.11 The derivative is expressed as dy
dx


= lim
Δx → 0


Δy
Δx


.


Graphing a Derivative
We have already discussed how to graph a function, so given the equation of a function or the equation of a derivativefunction, we could graph it. Given both, we would expect to see a correspondence between the graphs of these twofunctions, since f ′(x) gives the rate of change of a function f (x) (or slope of the tangent line to f (x)).
In Example 3.11 we found that for f (x) = x, f ′(x) = 1/2 x. If we graph these functions on the same axes, as in Figure
3.12, we can use the graphs to understand the relationship between these two functions. First, we notice that f (x) is
increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently,we expect f ′ (x) > 0 for all values of x in its domain. Furthermore, as x increases, the slopes of the tangent lines to f (x)
are decreasing and we expect to see a corresponding decrease in f ′(x). We also observe that f (0) is undefined and that
lim


x → 0+
f ′ (x) = +∞, corresponding to a vertical tangent to f (x) at 0.


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Figure 3.12 The derivative f ′(x) is positive everywhere
because the function f (x) is increasing.


In Example 3.12 we found that for f (x) = x2 − 2x, f ′ (x) = 2x − 2. The graphs of these functions are shown in Figure
3.13. Observe that f (x) is decreasing for x < 1. For these same values of x, f ′ (x) < 0. For values of x > 1, f (x) is
increasing and f ′ (x) > 0. Also, f (x) has a horizontal tangent at x = 1 and f ′ (1) = 0.


Figure 3.13 The derivative f ′ (x) < 0 where the function
f (x) is decreasing and f ′ (x) > 0 where f (x) is increasing.
The derivative is zero where the function has a horizontaltangent.


Example 3.13
Sketching a Derivative Using a Function
Use the following graph of f (x) to sketch a graph of f ′ (x).


Chapter 3 | Derivatives 235




3.7


Solution
The solution is shown in the following graph. Observe that f (x) is increasing and f ′ (x) > 0 on ( – 2, 3). Also,
f (x) is decreasing and f ′ (x) < 0 on (−∞, −2) and on (3, +∞). Also note that f (x) has horizontal tangents
at – 2 and 3, and f ′ (−2) = 0 and f ′ (3) = 0.


Sketch the graph of f (x) = x2 − 4. On what interval is the graph of f ′ (x) above the x -axis?


Derivatives and Continuity
Now that we can graph a derivative, let’s examine the behavior of the graphs. First, we consider the relationship betweendifferentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there;


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however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuousat a point and fail to be differentiable at the point for one of several reasons.
Theorem 3.1: Differentiability Implies Continuity
Let f (x) be a function and a be in its domain. If f (x) is differentiable at a, then f is continuous at a.


Proof
If f (x) is differentiable at a, then f ′(a) exists and


f ′ (a) = lim
x → a


f (x) − f (a)
x − a .


We want to show that f (x) is continuous at a by showing that lim
x → a


f (x) = f (a). Thus,
lim
x → a


f (x) = lim
x → a



⎝ f (x) − f (a) + f (a)⎞⎠


= lim
x → a


f (x) − f (a)


x − a · (x − a) + f (a)

⎠ Multiply and divide f (x) − f (a) by x − a.


=

⎝ limx → a


f (x) − f (a)
x − a



⎠ ·

⎝ limx → a(x − a)



⎠+ limx → a f (a)


= f ′(a) · 0 + f (a)


= f (a).


Therefore, since f (a) is defined and lim
x → a


f (x) = f (a), we conclude that f is continuous at a.

We have just proven that differentiability implies continuity, but now we consider whether continuity impliesdifferentiability. To determine an answer to this question, we examine the function f (x) = |x|. This function is continuous
everywhere; however, f ′(0) is undefined. This observation leads us to believe that continuity does not imply
differentiability. Let’s explore further. For f (x) = |x|,


f ′ (0) = lim
x → 0


f (x) − f (0)
x − 0


= lim
x → 0


|x| − |0|
x − 0


= lim
x → 0


|x|
x .


This limit does not exist because
lim


x → 0−
|x|
x = −1 and lim


x → 0+
|x|
x = 1.


See Figure 3.14.


Chapter 3 | Derivatives 237




Figure 3.14 The function f (x) = |x| is continuous at 0 but
is not differentiable at 0.


Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function
f (x) = x3 :


f ′ (0) = lim
x → 0


x3 − 0
x − 0


= lim
x → 0


1


x2
3


= +∞.


Thus f ′ (0) does not exist. A quick look at the graph of f (x) = x3 clarifies the situation. The function has a vertical
tangent line at 0 (Figure 3.15).


Figure 3.15 The function f (x) = x3 has a vertical tangent at
x = 0. It is continuous at 0 but is not differentiable at 0.


The function f (x) = ⎧

⎨xsin




1
x

⎠ if x ≠ 0


0 if x = 0
also has a derivative that exhibits interesting behavior at 0. We see that


f ′ (0) = lim
x → 0


xsin(1/x) − 0
x − 0


= lim
x → 0


sin⎛⎝
1
x

⎠.


This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approachzero (Figure 3.16).


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Figure 3.16 The function f (x) = ⎧

⎨xsin




1
x

⎠ if x ≠ 0


0 if x = 0
is not


differentiable at 0.


In summary:
1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function mustbe continuous. However, if a function is continuous, it may still fail to be differentiable.
2. We saw that f (x) = |x| failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the


left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From
this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point.


3. As we saw in the example of f (x) = x3 , a function fails to be differentiable at a point where there is a vertical
tangent line.


4. As we saw with f (x) = ⎧

⎨xsin




1
x

⎠ if x ≠ 0


0 if x = 0
a function may fail to be differentiable at a point in more complicated


ways as well.
Example 3.14
A Piecewise Function that is Continuous and Differentiable
A toy company wants to design a track for a toy car that starts out along a parabolic curve and then convertsto a straight line (Figure 3.17). The function that describes the track is to have the form
f (x) =





1
10


x2 + bx + c if x < −10


−1
4
x + 5


2
if x ≥ −10


where x and f (x) are in inches. For the car to move smoothly along the
track, the function f (x) must be both continuous and differentiable at −10. Find values of b and c that make
f (x) both continuous and differentiable.


Chapter 3 | Derivatives 239




3.8


Figure 3.17 For the car to move smoothly along the track, thefunction must be both continuous and differentiable.
Solution
For the function to be continuous at x = −10, lim


x → 10−
f (x) = f (−10). Thus, since


lim
x → −10−


f (x) = 1
10


(−10)2 − 10b + c = 10 − 10b + c


and f (−10) = 5, we must have 10 − 10b + c = 5. Equivalently, we have c = 10b − 5.
For the function to be differentiable at −10,


f ′ (10) = lim
x → −10


f (x) − f (−10)
x + 10


must exist. Since f (x) is defined using different rules on the right and the left, we must evaluate this limit from
the right and the left and then set them equal to each other:


lim
x → −10−


f (x) − f (−10)
x + 10


= lim
x → −10−


1
10
x2 + bx + c − 5


x + 10


= lim
x → −10−


1
10
x2 + bx + (10b − 5) − 5


x + 10
Substitute c = 10b − 5.


= lim
x → −10−


x2 − 100 + 10bx + 100b
10(x + 10)


= lim
x → −10−


(x + 10)(x − 10 + 10b)
10(x + 10)


Factor by grouping.


= b − 2.


We also have
lim


x → −10+
f (x) − f (−10)


x + 10
= lim


x → −10+


− 1
4
x + 5


2
− 5


x + 10


= lim
x → −10+


−(x + 10)
4(x + 10)


= − 1
4
.


This gives us b − 2 = − 1
4
. Thus b = 7


4
and c = 10⎛⎝74⎞⎠− 5 = 252 .


Find values of a and b that make f (x) = ⎧


ax + b if x < 3


x2 if x ≥ 3
both continuous and differentiable at 3.


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Higher-Order Derivatives
The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivativeof a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change ofvelocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative.Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively,these are referred to as higher-order derivatives. The notation for the higher-order derivatives of y = f (x) can be
expressed in any of the following forms:


f ″(x), f‴(x), f (4) (x),…, f (n) (x)


y″(x), y‴(x), y(4) (x),…, y(n) (x)


d2 y


dx2
,
d3 y


dy3
,
d4 y


dy4
,…,


dn y
dyn


.


It is interesting to note that the notation for d2 y
dx2


may be viewed as an attempt to express d
dx


dy
dx

⎠ more compactly.


Analogously, d
dx


d
dx


dy
dx



⎠ =


d
dx





⎜ d


2 y


dx2





⎟ =


d3 y


dx3
.


Example 3.15
Finding a Second Derivative
For f (x) = 2x2 − 3x + 1, find f ″(x).
Solution
First find f ′(x).


f ′ (x) = lim
h → 0



⎝2(x + h)


2 − 3(x + h) + 1⎞⎠− (2x
2 − 3x + 1)


h


Substitute f (x) = 2x2 − 3x + 1
and


f (x + h) = 2(x + h)2 − 3(x + h) + 1


into f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


.


= lim
h → 0


4xh + h2 − 3h
h


Simplify the numerator.


= lim
h → 0


(4x + h − 3)
Factor out the h in the numerator
and cancel with the h in the
denominator.


= 4x − 3 Take the limit.


Next, find f ″(x) by taking the derivative of f ′ (x) = 4x − 3.


f ″(x) = lim
h → 0


f ′ (x + h) − f ′(x)
h


Use f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


with f ′(x) in


place of f (x).


= lim
h → 0



⎝4(x + h) − 3⎞⎠− (4x − 3)


h
Substitute f ′ (x + h) = 4(x + h) − 3 and
f ′ (x) = 4x − 3.


= lim
h → 0


4 Simplify.


= 4 Take the limit.


Chapter 3 | Derivatives 241




3.9


3.10


Find f ″(x) for f (x) = x2.


Example 3.16
Finding Acceleration
The position of a particle along a coordinate axis at time t (in seconds) is given by s(t) = 3t2 − 4t + 1 (in
meters). Find the function that describes its acceleration at time t.
Solution
Since v(t) = s′(t) and a(t) = v′ (t) = s″(t), we begin by finding the derivative of s(t) :


s′ (t) = lim
h → 0


s(t + h) − s(t)
h


= lim
h → 0


3(t + h)2 − 4(t + h) + 1 − ⎛⎝3t
2 − 4t + 1⎞⎠


h
= 6t − 4.


Next,
s″(t) = lim


h → 0


s′ (t + h) − s′(t)
h


= lim
h → 0


6(t + h) − 4 − (6t − 4)
h


= 6.


Thus, a = 6m/s2.


For s(t) = t3, find a(t).


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3.2 EXERCISES
For the following exercises, use the definition of aderivative to find f ′ (x).
54. f (x) = 6
55. f (x) = 2 − 3x
56. f (x) = 2x


7
+ 1


57. f (x) = 4x2
58. f (x) = 5x − x2
59. f (x) = 2x
60. f (x) = x − 6
61. f (x) = 9x
62. f (x) = x + 1x
63. f (x) = 1x
For the following exercises, use the graph of y = f (x) to
sketch the graph of its derivative f ′ (x).
64.


65.


66.


Chapter 3 | Derivatives 243




67.


For the following exercises, the given limit represents thederivative of a function y = f (x) at x = a. Find f (x)
and a.
68. lim


h → 0


(1 + h)2/3 − 1
h


69. lim
h → 0



⎣3(2 + h)


2 + 2⎤⎦− 14
h


70. lim
h → 0


cos(π + h) + 1
h


71. lim
h → 0


(2 + h)4 − 16
h


72. lim
h → 0


[2(3 + h)2 − (3 + h)] − 15
h


73. lim
h → 0


eh − 1
h


For the following functions,
a. sketch the graph and
b. use the definition of a derivative to show that thefunction is not differentiable at x = 1.


74. f (x) = ⎧


2 x, 0 ≤ x ≤ 1
3x − 1, x > 1


75. f (x) = ⎧


3, x < 1
3x, x ≥ 1


76. f (x) = ⎧

⎨−x


2 + 2, x ≤ 1
x, x > 1


77. f (x) = ⎧


2x, x ≤ 1
2
x , x > 1


For the following graphs,
a. determine for which values of x = a the


lim
x → a


f (x) exists but f is not continuous at
x = a, and


b. determine for which values of x = a the function
is continuous but not differentiable at x = a.


78.


79.


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80. Use the graph to evaluate a. f ′ (−0.5), b. f ′ (0), c.
f ′ (1), d. f ′ (2), and e. f ′ (3), if it exists.


For the following functions, use
f ″(x) = lim


h → 0


f ′ (x + h) − f ′(x)
h


to find f ″(x).
81. f (x) = 2 − 3x
82. f (x) = 4x2


83. f (x) = x + 1x
For the following exercises, use a calculator to graph f (x).
Determine the function f ′ (x), then use a calculator to
graph f ′ (x).
84. [T] f (x) = − 5x
85. [T] f (x) = 3x2 + 2x + 4.
86. [T] f (x) = x + 3x
87. [T] f (x) = 1


2x


88. [T] f (x) = 1 + x + 1x
89. [T] f (x) = x3 + 1
For the following exercises, describe what the twoexpressions represent in terms of each of the givensituations. Be sure to include units.


a. f (x + h) − f (x)
h


b. f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


90. P(x) denotes the population of a city at time x in
years.
91. C(x) denotes the total amount of money (in
thousands of dollars) spent on concessions by x customers
at an amusement park.
92. R(x) denotes the total cost (in thousands of dollars)
of manufacturing x clock radios.
93. g(x) denotes the grade (in percentage points) received
on a test, given x hours of studying.
94. B(x) denotes the cost (in dollars) of a sociology
textbook at university bookstores in the United States in x
years since 1990.
95. p(x) denotes atmospheric pressure at an altitude of x
feet.
96. Sketch the graph of a function y = f (x) with all of
the following properties:a. f ′ (x) > 0 for −2 ≤ x < 1


b. f ′ (2) = 0
c. f ′ (x) > 0 for x > 2
d. f (2) = 2 and f (0) = 1
e. lim


x → −∞
f (x) = 0 and lim


x → ∞
f (x) = ∞


f. f ′ (1) does not exist.
97. Suppose temperature T in degrees Fahrenheit at a
height x in feet above the ground is given by y = T(x).


a. Give a physical interpretation, with units, of T′(x).
b. If we know that T′ (1000) = −0.1, explain the


physical meaning.
98. Suppose the total profit of a company is y = P(x)
thousand dollars when x units of an item are sold.


a. What does P(b) − P(a)
b − a


for 0 < a < b measure,
and what are the units?b. What does P′(x) measure, and what are the units?


c. Suppose that P′ (30) = 5, what is the
approximate change in profit if the number of itemssold increases from 30 to 31?


Chapter 3 | Derivatives 245




99. The graph in the following figure models the numberof people N(t) who have come down with the flu t weeks
after its initial outbreak in a town with a population of
50,000 citizens.


a. Describe what N′(t) represents and how it behaves
as t increases.


b. What does the derivative tell us about how thistown is affected by the flu outbreak?


For the following exercises, use the following table, whichshows the height h of the Saturn V rocket for the Apollo
11 mission t seconds after launch.


Time (seconds) Height (meters)
0 0


1 2


2 4


3 13


4 25


5 32


100. What is the physical meaning of h′ (t)? What are
the units?


101. [T] Construct a table of values for h′ (t) and graph
both h(t) and h′ (t) on the same graph. (Hint: for interior
points, estimate both the left limit and right limit andaverage them.)
102. [T] The best linear fit to the data is given by
H(t) = 7.229t − 4.905, where H is the height of the
rocket (in meters) and t is the time elapsed since takeoff.
From this equation, determine H′ (t). Graph H(t) with
the given data and, on a separate coordinate plane, graph
H′ (t).


103. [T] The best quadratic fit to the data is given by
G(t) = 1.429t2 + 0.0857t − 0.1429, where G is the
height of the rocket (in meters) and t is the time elapsed
since takeoff. From this equation, determine G′ (t). Graph
G(t) with the given data and, on a separate coordinate
plane, graph G′ (t).
104. [T] The best cubic fit to the data is given by
F(t) = 0.2037t3 + 2.956t2 − 2.705t + 0.4683, where
F is the height of the rocket (in m) and t is the time
elapsed since take off. From this equation, determine
F′ (t). Graph F(t) with the given data and, on a separate
coordinate plane, graph F′ (t). Does the linear, quadratic,
or cubic function fit the data best?
105. Using the best linear, quadratic, and cubic fits tothe data, determine what H″(t), G″(t) and F″(t) are. What
are the physical meanings of H″(t), G″(t) and F″(t), and
what are their units?


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3.3 | Differentiation Rules
Learning Objectives


3.3.1 State the constant, constant multiple, and power rules.
3.3.2 Apply the sum and difference rules to combine derivatives.
3.3.3 Use the product rule for finding the derivative of a product of functions.
3.3.4 Use the quotient rule for finding the derivative of a quotient of functions.
3.3.5 Extend the power rule to functions with negative exponents.
3.3.6 Combine the differentiation rules to find the derivative of a polynomial or rational function.


Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather
challenging process. For example, previously we found that d


dx
( x) = 1


2 x
by using a process that involved multiplying an


expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate d
dx

⎝ x
3 ⎞
⎠ using the definition,


while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass thisprocess. We begin with the basics.
The Basic Rules
The functions f (x) = c and g(x) = xn where n is a positive integer are the building blocks from which all polynomials
and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resortingto the limit definition of the derivative, we must first develop formulas for differentiating these basic functions.
The Constant Rule
We first apply the limit definition of the derivative to find the derivative of the constant function, f (x) = c. For this
function, both f (x) = c and f (x + h) = c, so we obtain the following result:


f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


= lim
h → 0


c − c
h


= lim
h → 0


0
h


= lim
h → 0


0 = 0.


The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant functionis zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is 0. We
restate this rule in the following theorem.
Theorem 3.2: The Constant Rule
Let c be a constant.
If f (x) = c, then f ′(c) = 0.
Alternatively, we may express this rule as


d
dx


(c) = 0.


Example 3.17


Chapter 3 | Derivatives 247




3.11


Applying the Constant Rule
Find the derivative of f (x) = 8.
Solution
This is just a one-step application of the rule:


f ′(8) = 0.


Find the derivative of g(x) = −3.


The Power Rule
We have shown that


d
dx

⎝x


2⎞
⎠ = 2x and


d
dx

⎝x


1/2⎞
⎠ =


1
2
x−1/2.


At this point, you might see a pattern beginning to develop for derivatives of the form d
dx


(xn). We continue our
examination of derivative formulas by differentiating power functions of the form f (x) = xn where n is a positive integer.
We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating
and proving the general rule for derivatives of functions of this form, we take a look at a specific case, d


dx
(x3). As we go


through this derivation, pay special attention to the portion of the expression in boldface, as the technique used in this caseis essentially the same as the technique used to prove the general case.
Example 3.18
Differentiating x3
Find d


dx

⎝x


3⎞
⎠.


Solution


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3.12


d
dx

⎝x


3⎞
⎠ = limh → 0


(x + h)3 − x3


h


= lim
h → 0


x3 + 3x2h + 3xh2 + h3 − x3


h


Notice that the fir t term in the expansion of


(x + h)3 is x3 and the second term is 3x2h. All
other terms contain powers of h that are two or
greater.


= lim
h → 0


3x2h + 3xh2 + h3


h
In this step the x3 terms have been cancelled,


leaving only terms containing h.


= lim
h → 0


h(3x2 + 3xh + h2)
h


Factor out the common factor of h.


= lim
h → 0


(3x2 + 3xh + h2)
After cancelling the common factor of h, the


only term not containing h is 3x2.


= 3x2 Let h go to 0.


Find d
dx

⎝x


4⎞
⎠.


As we shall see, the procedure for finding the derivative of the general form f (x) = xn is very similar. Although it is often
unwise to draw general conclusions from specific examples, we note that when we differentiate f (x) = x3, the power on
x becomes the coefficient of x2 in the derivative and the power on x in the derivative decreases by 1. The following
theorem states that the power rule holds for all positive integer powers of x. We will eventually extend this result to
negative integer powers. Later, we will see that this rule may also be extended first to rational powers of x and then to
arbitrary powers of x. Be aware, however, that this rule does not apply to functions in which a constant is raised to a
variable power, such as f (x) = 3x.


Theorem 3.3: The Power Rule
Let n be a positive integer. If f (x) = xn, then


f ′ (x) = nxn − 1.


Alternatively, we may express this rule as
d
dx


xn = nxn − 1.


Proof
For f (x) = xn where n is a positive integer, we have


f ′ (x) = lim
h → 0


(x + h)n − xn


h
.


Since (x + h)n = xn + nxn − 1h + ⎛⎝
n
2

⎠ x


n − 2h2 + ⎛⎝
n
3

⎠ x


n − 3h3 +… + nxhn − 1 + hn,


we see that


Chapter 3 | Derivatives 249




3.13


(x + h)n − xn = nxn − 1h + ⎛⎝
n
2

⎠ x


n − 2h2 + ⎛⎝
n
3

⎠ x


n − 3h3 +… + nxhn − 1 + hn.


Next, divide both sides by h:
(x + h)n − xn


h
=


nxn − 1h + ⎛⎝
n
2

⎠ x


n − 2h2 + ⎛⎝
n
3

⎠ x


n − 3h3 +… + nxhn − 1 + hn


h
.


Thus,
(x + h)n − xn


h
= nxn − 1 + ⎛⎝


n
2

⎠ x


n − 2h + ⎛⎝
n
3

⎠ x


n − 3h2 +… + nxhn − 2 + hn − 1.


Finally,
f ′ (x) = lim


h → 0



⎝nx


n − 1 + ⎛⎝
n
2

⎠ x


n − 2h + ⎛⎝
n
3

⎠ x


n − 3h2 +… + nxhn − 1 + hn⎞⎠


= nxn − 1.



Example 3.19
Applying the Power Rule
Find the derivative of the function f (x) = x10 by applying the power rule.
Solution
Using the power rule with n = 10, we obtain


f ′(x) = 10x10 − 1 = 10x9.


Find the derivative of f (x) = x7.


The Sum, Difference, and Constant Multiple Rules
We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions.Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract,or multiply by a constant. These rules are summarized in the following theorem.
Theorem 3.4: Sum, Difference, and Constant Multiple Rules
Let f (x) and g(x) be differentiable functions and k be a constant. Then each of the following equations holds.
Sum Rule. The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f
and the derivative of g.


d
dx

⎝ f (x) + g(x)⎞⎠ = ddx



⎝ f (x)⎞⎠+ ddx



⎝g(x)⎞⎠;


that is,
for j(x) = f (x) + g(x), j′ (x) = f ′ (x) + g′(x).


Difference Rule. The derivative of the difference of a function f and a function g is the same as the difference of the


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derivative of f and the derivative of g:
d
dx

⎝ f (x) − g(x)⎞⎠ = ddx



⎝ f (x)⎞⎠− ddx



⎝g(x)⎞⎠;


that is,
for j(x) = f (x) − g(x), j′ (x) = f ′ (x) − g′(x).


Constant Multiple Rule. The derivative of a constant cmultiplied by a function f is the same as the constant multipliedby the derivative:
d
dx

⎝k f (x)⎞⎠ = k ddx



⎝ f (x)⎞⎠;


that is,
for j(x) = k f (x), j′ (x) = k f ′(x).


Proof
We provide only the proof of the sum rule here. The rest follow in a similar manner.
For differentiable functions f (x) and g(x), we set j(x) = f (x) + g(x). Using the limit definition of the derivative we
have


j′ (x) = lim
h → 0


j(x + h) − j(x)
h


.


By substituting j(x + h) = f (x + h) + g(x + h) and j(x) = f (x) + g(x), we obtain
j′(x) = lim


h → 0



⎝ f (x + h) + g(x + h)⎞⎠− ⎛⎝ f (x) + g(x)⎞⎠


h
.


Rearranging and regrouping the terms, we have
j′(x) = lim


h → 0




f (x + h) − f (x)


h
+
g(x + h) − g(x)


h

⎠.


We now apply the sum law for limits and the definition of the derivative to obtain
j′(x) = lim


h → 0




f (x + h) − f (x)


h

⎠+ limh → 0




g(x + h) − g(x)


h

⎠ = f ′ (x) + g′ (x).



Example 3.20
Applying the Constant Multiple Rule
Find the derivative of g(x) = 3x2 and compare it to the derivative of f (x) = x2.
Solution
We use the power rule directly:


g′ (x) = d
dx

⎝3x


2⎞
⎠ = 3


d
dx

⎝x


2⎞
⎠ = 3(2x) = 6x.


Since f (x) = x2 has derivative f ′ (x) = 2x, we see that the derivative of g(x) is 3 times the derivative of


Chapter 3 | Derivatives 251




3.14


f (x). This relationship is illustrated in Figure 3.18.


Figure 3.18 The derivative of g(x) is 3 times the derivative of f (x).


Example 3.21
Applying Basic Derivative Rules
Find the derivative of f (x) = 2x5 + 7.
Solution
We begin by applying the rule for differentiating the sum of two functions, followed by the rules fordifferentiating constant multiples of functions and the rule for differentiating powers. To better understand thesequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:


f ′ (x) = d
dx

⎝2x


5 + 7⎞⎠


= d
dx

⎝2x


5⎞
⎠+


d
dx


(7) Apply the sum rule.


= 2 d
dx

⎝x


5⎞
⎠+


d
dx


(7) Apply the constant multiple rule.


= 2⎛⎝5x
4⎞
⎠+ 0 Apply the power rule and the constant rule.


= 10x4. Simplify.


Find the derivative of f (x) = 2x3 − 6x2 + 3.


Example 3.22
Finding the Equation of a Tangent Line


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3.15


Find the equation of the line tangent to the graph of f (x) = x2 − 4x + 6 at x = 1.
Solution
To find the equation of the tangent line, we need a point and a slope. To find the point, compute


f (1) = 12 − 4(1) + 6 = 3.


This gives us the point (1, 3). Since the slope of the tangent line at 1 is f ′ (1), we must first find f ′ (x). Using
the definition of a derivative, we have


f ′ (x) = 2x − 4


so the slope of the tangent line is f ′ (1) = −2. Using the point-slope formula, we see that the equation of the
tangent line is


y − 3 = −2(x − 1).


Putting the equation of the line in slope-intercept form, we obtain
y = −2x + 5.


Find the equation of the line tangent to the graph of f (x) = 3x2 − 11 at x = 2. Use the point-slope
form.


The Product Rule
Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first oneexamines the derivative of the product of two functions. Although it might be tempting to assume that the derivative ofthe product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this
pattern. To see why we cannot use this pattern, consider the function f (x) = x2, whose derivative is f ′ (x) = 2x and not
d
dx


(x) · d
dx


(x) = 1 · 1 = 1.


Theorem 3.5: Product Rule
Let f (x) and g(x) be differentiable functions. Then


d
dx

⎝ f (x)g(x)⎞⎠ = ddx



⎝ f (x)⎞⎠ · g(x) + ddx



⎝g(x)⎞⎠ · f (x).


That is,
if j(x) = f (x)g(x), then j′ (x) = f ′ (x)g(x) + g′ (x) f (x).


This means that the derivative of a product of two functions is the derivative of the first function times the secondfunction plus the derivative of the second function times the first function.


Proof
We begin by assuming that f (x) and g(x) are differentiable functions. At a key point in this proof we need to use the
fact that, since g(x) is differentiable, it is also continuous. In particular, we use the fact that since g(x) is continuous,
lim
h → 0


g(x + h) = g(x).


Chapter 3 | Derivatives 253




By applying the limit definition of the derivative to j(x) = f (x)g(x), we obtain
j′ (x) = lim


h → 0


f (x + h)g(x + h) − f (x)g(x)
h


.


By adding and subtracting f (x)g(x + h) in the numerator, we have
j′ (x) = lim


h → 0


f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)
h


.


After breaking apart this quotient and applying the sum law for limits, the derivative becomes
j′ (x) = lim


h → 0




f (x + h)g(x + h) − f (x)g(x + h)


h

⎠+ limh → 0




f (x)g(x + h) − f (x)g(x)


h

⎠.


Rearranging, we obtain
j′ (x) = lim


h → 0




f (x + h) − f (x)


h
· g(x + h)



⎠+ limh → 0




g(x + h) − g(x)


h
· f (x)

⎠.


By using the continuity of g(x), the definition of the derivatives of f (x) and g(x), and applying the limit laws, we arrive
at the product rule,


j′ (x) = f ′ (x)g(x) + g′ (x) f (x).



Example 3.23
Applying the Product Rule to Constant Functions
For j(x) = f (x)g(x), use the product rule to find j′(2) if f (2) = 3, f ′ (2) = −4, g(2) = 1, and g′ (2) = 6.
Solution
Since j(x) = f (x)g(x), j′ (x) = f ′ (x)g(x) + g′ (x) f (x), and hence


j′ (2) = f ′ (2)g(2) + g′ (2) f (2) = (−4)(1) + (6)(3) = 14.


Example 3.24
Applying the Product Rule to Binomials
For j(x) = (x2 + 2)(3x3 − 5x), find j′(x) by applying the product rule. Check the result by first finding the
product and then differentiating.
Solution
If we set f (x) = x2 + 2 and g(x) = 3x3 − 5x, then f ′ (x) = 2x and g′ (x) = 9x2 − 5. Thus,


j′ (x) = f ′ (x)g(x) + g′ (x) f (x) = (2x)⎛⎝3x
3 − 5x⎞⎠+ (9x


2 − 5)(x2 + 2).


Simplifying, we have
j′ (x) = 15x4 + 3x2 − 10.


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3.16


To check, we see that j(x) = 3x5 + x3 − 10x and, consequently, j′ (x) = 15x4 + 3x2 − 10.


Use the product rule to obtain the derivative of j(x) = 2x5 ⎛⎝4x2 + x⎞⎠.


The Quotient Rule
Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in thefollowing theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of thefunction in the numerator times the function in the denominator minus the derivative of the function in the denominatortimes the function in the numerator, all divided by the square of the function in the denominator. In order to better graspwhy we cannot simply take the quotient of the derivatives, keep in mind that


d
dx

⎝x


2⎞
⎠ = 2x, not


d
dx

⎝x


3⎞


d
dx
(x)


= 3x
2


1
= 3x2.


Theorem 3.6: The Quotient Rule
Let f (x) and g(x) be differentiable functions. Then


d
dx


f (x)
g(x)

⎠ =


d
dx
( f (x)) · g(x) − d


dx
(g(x)) · f (x)


(g(x))2
.


That is,
if j(x) =


f (x)
g(x)


, then j′ (x) =
f ′ (x)g(x) − g′ (x) f (x)


(g(x))2
.


The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply thisnew rule for finding derivatives in the next example.
Example 3.25
Applying the Quotient Rule
Use the quotient rule to find the derivative of k(x) = 5x2


4x + 3
.


Solution
Let f (x) = 5x2 and g(x) = 4x + 3. Thus, f ′ (x) = 10x and g′ (x) = 4. Substituting into the quotient rule, we
have


k′ (x) =
f ′ (x)g(x) − g′ (x) f (x)


(g(x))2
= 10x(4x + 3) − 4(5x


2)
(4x + 3)2


.


Chapter 3 | Derivatives 255




3.17


Simplifying, we obtain
k′ (x) = 20x


2 + 30x
(4x + 3)2


.


Find the derivative of h(x) = 3x + 1
4x − 3


.


It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form xk where k
is a negative integer.
Theorem 3.7: Extended Power Rule
If k is a negative integer, then


d
dx

⎝x


k⎞
⎠ = kx


k − 1.


Proof
If k is a negative integer, we may set n = −k, so that n is a positive integer with k = −n. Since for each positive integer
n, x−n = 1


xn
, we may now apply the quotient rule by setting f (x) = 1 and g(x) = xn. In this case, f ′ (x) = 0 and


g′ (x) = nxn − 1. Thus,
d
d
(x−n) =


0(xn) − 1⎛⎝nx
n − 1⎞


(xn)2
.


Simplifying, we see that
d
d
(x−n) = −nx


n − 1


x2n
= −nx(n − 1) − 2n = −nx−n − 1.


Finally, observe that since k = −n, by substituting we have
d
dx

⎝x


k⎞
⎠ = kx


k − 1.



Example 3.26
Using the Extended Power Rule
Find d


dx

⎝x


−4⎞
⎠.


Solution
By applying the extended power rule with k = −4, we obtain


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3.18


d
dx

⎝x


−4⎞
⎠ = −4x


−4 − 1 = −4x−5.


Example 3.27
Using the Extended Power Rule and the Constant Multiple Rule
Use the extended power rule and the constant multiple rule to find f (x) = 6


x2
.


Solution
It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do
so. However, it is far easier to differentiate this function by first rewriting it as f (x) = 6x−2.


f ′ (x) = d
dx


6
x2

⎠ =


d
dx

⎝6x


−2⎞
⎠ Rewrite


6
x2


as 6x−2.


= 6 d
dx


(x−2) Apply the constant multiple rule.


= 6(−2x−3) Use the extended power rule to diffe entiate x−2.


= −12x−3 Simplify.


Find the derivative of g(x) = 1
x7


using the extended power rule.


Combining Differentiation Rules
As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just onedifferentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we mayfind the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations ofdifferentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order inwhich we would evaluate the function.
Example 3.28
Combining Differentiation Rules
For k(x) = 3h(x) + x2g(x), find k′(x).
Solution
Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.


Chapter 3 | Derivatives 257




k′ (x) = d
dx

⎝3h(x) + x


2g(x)⎞⎠ =
d
dx

⎝3h(x)⎞⎠+ d


dx

⎝x


2g(x)⎞⎠ Apply the sum rule.


= 3 d
dx

⎝h(x)⎞⎠+




d
dx

⎝x


2⎞
⎠g(x) +


d
dx

⎝g(x)⎞⎠x2





Apply the constant multiple rule to


diffe entiate 3h(x) and the product


rule to diffe entiate x2g(x).


= 3h′ (x) + 2xg(x) + g′(x)x2


Example 3.29
Extending the Product Rule
For k(x) = f (x)g(x)h(x), express k′ (x) in terms of f (x), g(x), h(x), and their derivatives.
Solution
We can think of the function k(x) as the product of the function f (x)g(x) and the function h(x). That is,
k(x) = ⎛⎝ f (x)g(x)⎞⎠ · h(x). Thus,


k′(x) = d
dx

⎝ f (x)g(x)⎞⎠ · h(x) + ddx



⎝h(x)⎞⎠ · ⎛⎝ f (x)g(x)⎞⎠


Apply the product rule to the product


of f (x)g(x) and h(x).


= ⎛⎝ f ′ (x)g(x) + g′ (x) f (x)h⎞⎠(x) + h′ (x) f (x)g(x) Apply the product rule to f (x)g(x).


= f ′ (x)g(x)h(x) + f (x)g′ (x)h(x) + f (x)g(x)h′ (x). Simplify.


Example 3.30
Combining the Quotient Rule and the Product Rule
For h(x) = 2x3 k(x)


3x + 2
, find h′(x).


Solution
This procedure is typical for finding the derivative of a rational function.
h′ (x) =


d
dx

⎝2x


3 k(x)⎞⎠ · (3x + 2) −
d
dx
(3x + 2) · ⎛⎝2x


3 k(x)⎞⎠


(3x + 2)2
Apply the quotient rule.


=

⎝6x


2 k(x) + k′ (x) · 2x3⎞⎠(3x + 2) − 3

⎝2x


3 k(x)⎞⎠


(3x + 2)2


Apply the product rule to fin


d
dx

⎝2x


3 k(x)⎞⎠. Use
d
dx


(3x + 2) = 3.


= −6x
3 k(x) + 18x3 k(x) + 12x2 k(x) + 6x4 k′ (x) + 4x3 k′ (x)


(3x + 2)2
Simplify.


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3.19 Find d
dx

⎝3 f (x) − 2g(x)⎞⎠.


Example 3.31
Determining Where a Function Has a Horizontal Tangent
Determine the values of x for which f (x) = x3 − 7x2 + 8x + 1 has a horizontal tangent line.
Solution
To find the values of x for which f (x) has a horizontal tangent line, we must solve f ′ (x) = 0. Since


f ′ (x) = 3x2 − 14x + 8 = (3x − 2)(x − 4),


we must solve (3x − 2)(x − 4) = 0. Thus we see that the function has horizontal tangent lines at x = 2
3
and


x = 4 as shown in the following graph.


Figure 3.19 This function has horizontal tangent lines at x =2/3 and x = 4.


Example 3.32
Finding a Velocity
The position of an object on a coordinate axis at time t is given by s(t) = t


t2 + 1
. What is the initial velocity of


the object?
Solution
Since the initial velocity is v(0) = s′ (0), begin by finding s′(t) by applying the quotient rule:


s′ (t) =
1⎛⎝t


2 + 1⎞⎠− 2t(t)



⎝t
2 + 1⎞⎠


2
= 1 − t


2



⎝t
2 + 1⎞⎠


2
.


Chapter 3 | Derivatives 259




3.20


After evaluating, we see that v(0) = 1.


Find the values of x for which the line tangent to the graph of f (x) = 4x2 − 3x + 2 has a tangent line
parallel to the line y = 2x + 3.


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Formula One Grandstands
Formula One car races can be very exciting to watch and attract a lot of spectators. Formula One track designers haveto ensure sufficient grandstand space is available around the track to accommodate these viewers. However, car racingcan be dangerous, and safety considerations are paramount. The grandstands must be placed where spectators will notbe in danger should a driver lose control of a car (Figure 3.20).


Figure 3.20 The grandstand next to a straightaway of the Circuit de Barcelona-Catalunya race track, located wherethe spectators are not in danger.


Safety is especially a concern on turns. If a driver does not slow down enough before entering the turn, the car mayslide off the racetrack. Normally, this just results in a wider turn, which slows the driver down. But if the driver losescontrol completely, the car may fly off the track entirely, on a path tangent to the curve of the racetrack.
Suppose you are designing a new Formula One track. One section of the track can be modeled by the function
f (x) = x3 + 3x + x (Figure 3.21). The current plan calls for grandstands to be built along the first straightaway
and around a portion of the first curve. The plans call for the front corner of the grandstand to be located at the point
(−1.9, 2.8). We want to determine whether this location puts the spectators in danger if a driver loses control of the
car.


Chapter 3 | Derivatives 261




Figure 3.21 (a) One section of the racetrack can be modeled by the function f (x) = x3 + 3x + x. (b) The
front corner of the grandstand is located at (−1.9, 2.8).
1. Physicists have determined that drivers are most likely to lose control of their cars as they are coming into aturn, at the point where the slope of the tangent line is 1. Find the (x, y) coordinates of this point near the turn.
2. Find the equation of the tangent line to the curve at this point.
3. To determine whether the spectators are in danger in this scenario, find the x-coordinate of the point where thetangent line crosses the line y = 2.8. Is this point safely to the right of the grandstand? Or are the spectators


in danger?
4. What if a driver loses control earlier than the physicists project? Suppose a driver loses control at the point


(−2.5, 0.625). What is the slope of the tangent line at this point?
5. If a driver loses control as described in part 4, are the spectators safe?
6. Should you proceed with the current design for the grandstand, or should the grandstands be moved?


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3.3 EXERCISES
For the following exercises, find f ′(x) for each function.
106. f (x) = x7 + 10
107. f (x) = 5x3 − x + 1
108. f (x) = 4x2 − 7x
109. f (x) = 8x4 + 9x2 − 1
110. f (x) = x4 + 2x
111. f (x) = 3x⎛⎝18x4 + 13x + 1⎞⎠
112. f (x) = (x + 2)⎛⎝2x2 − 3⎞⎠


113. f (x) = x2⎛⎝ 2x2 + 5x3



114. f (x) = x3 + 2x2 − 4
3


115. f (x) = 4x3 − 2x + 1
x2


116. f (x) = x2 + 4
x2 − 4


117. f (x) = x + 9
x2 − 7x + 1


For the following exercises, find the equation of the tangentline T(x) to the graph of the given function at the indicated
point. Use a graphing calculator to graph the function andthe tangent line.
118. [T] y = 3x2 + 4x + 1 at (0, 1)
119. [T] y = 2 x + 1 at (4, 5)
120. [T] y = 2x


x − 1
at (−1, 1)


121. [T] y = 2x − 3x2 at (1, −1)
For the following exercises, assume that f (x) and g(x)
are both differentiable functions for all x. Find the
derivative of each of the functions h(x).


122. h(x) = 4 f (x) + g(x)
7


123. h(x) = x3 f (x)
124. h(x) = f (x)g(x)


2


125. h(x) = 3 f (x)
g(x) + 2


For the following exercises, assume that f (x) and g(x)
are both differentiable functions with values as given inthe following table. Use the following table to calculate thefollowing derivatives.


x 1 2 3 4


f(x) 3 5 −2 0


g(x) 2 3 −4 6


f′(x) −1 7 8 −3


g′(x) 4 1 2 9


126. Find h′(1) if h(x) = x f (x) + 4g(x).
127. Find h′ (2) if h(x) = f (x)


g(x)
.


128. Find h′ (3) if h(x) = 2x + f (x)g(x).
129. Find h′ (4) if h(x) = 1x + g(x)f (x).
For the following exercises, use the following figure to findthe indicated derivatives, if they exist.


Chapter 3 | Derivatives 263




130. Let h(x) = f (x) + g(x). Find
a. h′ (1),
b. h′ (3), and
c. h′ (4).


131. Let h(x) = f (x)g(x). Find
a. h′ (1),
b. h′ (3), and
c. h′ (4).


132. Let h(x) = f (x)
g(x)


. Find
a. h′ (1),
b. h′ (3), and
c. h′ (4).


For the following exercises,
a. evaluate f ′ (a), and
b. graph the function f (x) and the tangent line at


x = a.


133. [T] f (x) = 2x3 + 3x − x2, a = 2
134. [T] f (x) = 1x − x2, a = 1
135. [T] f (x) = x2 − x12 + 3x + 2, a = 0
136. [T] f (x) = 1x − x2/3, a = −1
137. Find the equation of the tangent line to the graph of
f (x) = 2x3 + 4x2 − 5x − 3 at x = −1.
138. Find the equation of the tangent line to the graph of
f (x) = x2 + 4x − 10 at x = 8.
139. Find the equation of the tangent line to the graph of
f (x) = (3x − x2)(3 − x − x2) at x = 1.


140. Find the point on the graph of f (x) = x3 such that
the tangent line at that point has an x intercept of 6.
141. Find the equation of the line passing through the
point P(3, 3) and tangent to the graph of f (x) = 6


x − 1
.


142. Determine all points on the graph of
f (x) = x3 + x2 − x − 1 for which the slope of the tangent
line isa. horizontalb. −1.
143. Find a quadratic polynomial such that
f (1) = 5, f ′ (1) = 3 and f ″(1) = −6.
144. A car driving along a freeway with traffic has
traveled s(t) = t3 − 6t2 + 9t meters in t seconds.


a. Determine the time in seconds when the velocity ofthe car is 0.b. Determine the acceleration of the car when thevelocity is 0.
145. [T] A herring swimming along a straight line has
traveled s(t) = t2


t2 + 2
feet in t seconds. Determine the


velocity of the herring when it has traveled 3 seconds.
146. The population in millions of arctic flounder in theAtlantic Ocean is modeled by the function
P(t) = 8t + 3


0.2t2 + 1
, where t is measured in years.


a. Determine the initial flounder population.b. Determine P′ (10) and briefly interpret the result.
147. [T] The concentration of antibiotic in thebloodstream t hours after being injected is given by the
function C(t) = 2t2 + t


t3 + 50
, where C is measured in


milligrams per liter of blood.a. Find the rate of change of C(t).
b. Determine the rate of change for t = 8, 12, 24,


and 36.
c. Briefly describe what seems to be occurring as thenumber of hours increases.


148. A book publisher has a cost function given by
C(x) = x


3 + 2x + 3
x2


, where x is the number of copies of
a book in thousands and C is the cost, per book, measuredin dollars. Evaluate C′ (2) and explain its meaning.


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149. [T] According to Newton’s law of universalgravitation, the force F between two bodies of constant
mass m1 and m2 is given by the formula F = Gm1m2


d2
,


where G is the gravitational constant and d is the distance
between the bodies.a. Suppose that G, m1, andm2 are constants. Find


the rate of change of force F with respect to
distance d.


b. Find the rate of change of force F with
gravitational constant G = 6.67 × 10−11
Nm2 /kg2, on two bodies 10 meters apart, each
with a mass of 1000 kilograms.


Chapter 3 | Derivatives 265




3.4 | Derivatives as Rates of Change
Learning Objectives


3.4.1 Determine a new value of a quantity from the old value and the amount of change.
3.4.2 Calculate the average rate of change and explain how it differs from the instantaneous rateof change.
3.4.3 Apply rates of change to displacement, velocity, and acceleration of an object moving alonga straight line.
3.4.4 Predict the future population from the present value and the population growth rate.
3.4.5 Use derivatives to calculate marginal cost and revenue in a business situation.


In this section we look at some applications of the derivative by focusing on the interpretation of the derivative as the rate ofchange of a function. These applications include acceleration and velocity in physics, population growth rates in biology,and marginal functions in economics.
Amount of Change Formula
One application for derivatives is to estimate an unknown value of a function at a point by using a known value of afunction at some given point together with its rate of change at the given point. If f (x) is a function defined on an interval

⎣a, a + h⎤⎦, then the amount of change of f (x) over the interval is the change in the y values of the function over that
interval and is given by


f (a + h) − f (a).


The average rate of change of the function f over that same interval is the ratio of the amount of change over that interval
to the corresponding change in the x values. It is given by


f (a + h) − f (a)
h


.


As we already know, the instantaneous rate of change of f (x) at a is its derivative
f ′ (a) = lim


h → 0


f (a + h) − f (a)
h


.


For small enough values of h, f ′ (a) ≈ f (a + h) − f (a)
h


. We can then solve for f (a + h) to get the amount of change
formula:


(3.10)f (a + h) ≈ f (a) + f ′(a)h.
We can use this formula if we know only f (a) and f ′(a) and wish to estimate the value of f (a + h). For example, we
may use the current population of a city and the rate at which it is growing to estimate its population in the near future. Aswe can see in Figure 3.22, we are approximating f (a + h) by the y coordinate at a + h on the line tangent to f (x) at
x = a. Observe that the accuracy of this estimate depends on the value of h as well as the value of f ′ (a).


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3.21


Figure 3.22 The new value of a changed quantity equals theoriginal value plus the rate of change times the interval ofchange: f (a + h) ≈ f (a) + f ′ (a)h.
Here is an interesting demonstration (http://www.openstaxcollege.org/l/20_chainrule) of rate of change.


Example 3.33
Estimating the Value of a Function
If f (3) = 2 and f ′ (3) = 5, estimate f (3.2).
Solution
Begin by finding h. We have h = 3.2 − 3 = 0.2. Thus,


f (3.2) = f (3 + 0.2) ≈ f (3) + (0.2) f ′ (3) = 2 + 0.2(5) = 3.


Given f (10) = −5 and f ′ (10) = 6, estimate f (10.1).


Motion along a Line
Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position.If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important tointroduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.
Definition
Let s(t) be a function giving the position of an object at time t.
The velocity of the object at time t is given by v(t) = s′ (t).
The speed of the object at time t is given by |v(t)|.
The acceleration of the object at t is given by a(t) = v′ (t) = s″(t).


Chapter 3 | Derivatives 267




Example 3.34
Comparing Instantaneous Velocity and Average Velocity
A ball is dropped from a height of 64 feet. Its height above ground (in feet) t seconds later is given by
s(t) = −16t2 + 64.


a. What is the instantaneous velocity of the ball when it hits the ground?
b. What is the average velocity during its fall?


Solution
The first thing to do is determine how long it takes the ball to reach the ground. To do this, set s(t) = 0. Solving
−16t2 + 64 = 0, we get t = 2, so it take 2 seconds for the ball to reach the ground.


a. The instantaneous velocity of the ball as it strikes the ground is v(2). Since v(t) = s′ (t) = −32t, we
obtain v(t) = −64 ft/s.


b. The average velocity of the ball during its fall is
vave =


s(2) − s(0)
2 − 0


= 0 − 64
2


= −32 ft/s.


Example 3.35
Interpreting the Relationship between v(t) and a(t)
A particle moves along a coordinate axis in the positive direction to the right. Its position at time t is given by
s(t) = t3 − 4t + 2. Find v(1) and a(1) and use these values to answer the following questions.


a. Is the particle moving from left to right or from right to left at time t = 1?
b. Is the particle speeding up or slowing down at time t = 1?


Solution
Begin by finding v(t) and a(t).


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and a(t) = v′ (t) = s″(t) = 6t.
Evaluating these functions at t = 1, we obtain v(1) = −1 and a(1) = 6.


a. Because v(1) < 0, the particle is moving from right to left.
b. Because v(1) < 0 and a(1) > 0, velocity and acceleration are acting in opposite directions. In other


words, the particle is being accelerated in the direction opposite the direction in which it is traveling,causing |v(t)| to decrease. The particle is slowing down.


Example 3.36
Position and Velocity
The position of a particle moving along a coordinate axis is given by s(t) = t3 − 9t2 + 24t + 4, t ≥ 0.


a. Find v(t).
b. At what time(s) is the particle at rest?
c. On what time intervals is the particle moving from left to right? From right to left?
d. Use the information obtained to sketch the path of the particle along a coordinate axis.


Solution
a. The velocity is the derivative of the position function:


v(t) = s′ (t) = 3t2 − 18t + 24.


b. The particle is at rest when v(t) = 0, so set 3t2 − 18t + 24 = 0. Factoring the left-hand side of the
equation produces 3(t − 2)(t − 4) = 0. Solving, we find that the particle is at rest at t = 2 and t = 4.


c. The particle is moving from left to right when v(t) > 0 and from right to left when v(t) < 0. Figure
3.23 gives the analysis of the sign of v(t) for t ≥ 0, but it does not represent the axis along which the
particle is moving.


Figure 3.23 The sign of v(t) determines the direction of theparticle.


Since 3t2 − 18t + 24 > 0 on [0, 2) ∪ (2, +∞), the particle is moving from left to right on these
intervals.
Since 3t2 − 18t + 24 < 0 on (2, 4), the particle is moving from right to left on this interval.


d. Before we can sketch the graph of the particle, we need to know its position at the time it startsmoving ⎛⎝t = 0) and at the times that it changes direction (t = 2, 4). We have s(0) = 4, s(2) = 24, and
s(4) = 20. This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and


Chapter 3 | Derivatives 269




3.22


3.23


20 on the coordinate axis. The path of the particle is shown on a coordinate axis in Figure 3.24.


Figure 3.24 The path of the particle can be determined byanalyzing v(t).


A particle moves along a coordinate axis. Its position at time t is given by s(t) = t2 − 5t + 1. Is the
particle moving from right to left or from left to right at time t = 3?


Population Change
In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types ofpopulations, including those as diverse as bacteria colonies and cities. We can use a current population, together with agrowth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a populationand consequently can be represented by the derivative of the size of the population.
Definition
If P(t) is the number of entities present in a population, then the population growth rate of P(t) is defined to be
P′ (t).


Example 3.37
Estimating a Population
The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximatepopulation 2 years from now?
Solution
Let P(t) be the population (in thousands) t years from now. Thus, we know that P(0) = 10 and based on the
information, we anticipate P(5) = 30. Now estimate P′ (0), the current growth rate, using


P′ (0) ≈ P(5) − P(0)
5 − 0


= 30 − 10
5


= 4.


By applying Equation 3.10 to P(t), we can estimate the population 2 years from now by writing
P(2) ≈ P(0) + (2)P′ (0) ≈ 10 + 2(4) = 18;


thus, in 2 years the population will be 18,000.


The current population of a mosquito colony is known to be 3,000; that is, P(0) = 3,000. If
P′ (0) = 100, estimate the size of the population in 3 days, where t is measured in days.


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Changes in Cost and Revenue
In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost,revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies theuse of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of therevenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and therevenue function.
Definition
If C(x) is the cost of producing x items, then the marginal cost MC(x) is MC(x) = C′ (x).
If R(x) is the revenue obtained from selling x items, then the marginal revenue MR(x) is MR(x) = R′ (x).
If P(x) = R(x) − C(x) is the profit obtained from selling x items, then the marginal profit MP(x) is defined to be
MP(x) = P′ (x) = MR(x) − MC(x) = R′ (x) − C′ (x).


We can roughly approximate
MC(x) = C′ (x) = lim


h → 0


C(x + h) − C(x)
h


by choosing an appropriate value for h. Since x represents objects, a reasonable and small value for h is 1. Thus, by
substituting h = 1, we get the approximation MC(x) = C′ (x) ≈ C(x + 1) − C(x). Consequently, C′ (x) for a given
value of x can be thought of as the change in cost associated with producing one additional item. In a similar way,
MR(x) = R′ (x) approximates the revenue obtained by selling one additional item, and MP(x) = P′ (x) approximates the
profit obtained by producing and selling one additional item.
Example 3.38
Applying Marginal Revenue
Assume that the number of barbeque dinners that can be sold, x, can be related to the price charged, p, by the
equation p(x) = 9 − 0.03x, 0 ≤ x ≤ 300.
In this case, the revenue in dollars obtained by selling x barbeque dinners is given by


R(x) = xp(x) = x(9 − 0.03x) = −0.03x2 + 9x for 0 ≤ x ≤ 300.


Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner.Compare this to the actual revenue obtained from the sale of this dinner.
Solution
First, find the marginal revenue function: MR(x) = R′ (x) = −0.06x + 9.
Next, use R′ (100) to approximate R(101) − R(100), the revenue obtained from the sale of the 101st dinner.
Since R′ (100) = 3, the revenue obtained from the sale of the 101st dinner is approximately $3.
The actual revenue obtained from the sale of the 101st dinner is


R(101) − R(100) = 602.97 − 600 = 2.97, or $2.97.


The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.


Chapter 3 | Derivatives 271




3.24 Suppose that the profit obtained from the sale of x fish-fry dinners is given by
P(x) = −0.03x2 + 8x − 50. Use the marginal profit function to estimate the profit from the sale of the 101st
fish-fry dinner.


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3.4 EXERCISES
For the following exercises, the given functions representthe position of a particle traveling along a horizontal line.


a. Find the velocity and acceleration functions.
b. Determine the time intervals when the object isslowing down or speeding up.


150. s(t) = 2t3 − 3t2 − 12t + 8
151. s(t) = 2t3 − 15t2 + 36t − 10
152. s(t) = t


1 + t2


153. A rocket is fired vertically upward from the ground.The distance s in feet that the rocket travels from the
ground after t seconds is given by s(t) = −16t2 + 560t.


a. Find the velocity of the rocket 3 seconds after beingfired.b. Find the acceleration of the rocket 3 seconds afterbeing fired.
154. A ball is thrown downward with a speed of 8 ft/s from the top of a 64-foot-tall building. After t seconds,its height above the ground is given by
s(t) = −16t2 − 8t + 64.


a. Determine how long it takes for the ball to hit theground.b. Determine the velocity of the ball when it hits theground.
155. The position function s(t) = t2 − 3t − 4 represents
the position of the back of a car backing out of a drivewayand then driving in a straight line, where s is in feet and
t is in seconds. In this case, s(t) = 0 represents the time
at which the back of the car is at the garage door, so
s(0) = −4 is the starting position of the car, 4 feet inside
the garage.a. Determine the velocity of the car when s(t) = 0.


b. Determine the velocity of the car when s(t) = 14.
156. The position of a hummingbird flying along a straight
line in t seconds is given by s(t) = 3t3 − 7t meters.


a. Determine the velocity of the bird at t = 1 sec.
b. Determine the acceleration of the bird at t = 1 sec.
c. Determine the acceleration of the bird when thevelocity equals 0.


157. A potato is launched vertically upward with an initialvelocity of 100 ft/s from a potato gun at the top of an85-foot-tall building. The distance in feet that the potatotravels from the ground after t seconds is given by
s(t) = −16t2 + 100t + 85.


a. Find the velocity of the potato after 0.5 s and
5.75 s.


b. Find the speed of the potato at 0.5 s and 5.75 s.c. Determine when the potato reaches its maximumheight.d. Find the acceleration of the potato at 0.5 s and 1.5s.e. Determine how long the potato is in the air.f. Determine the velocity of the potato upon hittingthe ground.
158. The position function s(t) = t3 − 8t gives the
position in miles of a freight train where east is the positivedirection and t is measured in hours.


a. Determine the direction the train is traveling when
s(t) = 0.


b. Determine the direction the train is traveling when
a(t) = 0.


c. Determine the time intervals when the train isslowing down or speeding up.
159. The following graph shows the position y = s(t) of
an object moving along a straight line.


a. Use the graph of the position function to determinethe time intervals when the velocity is positive,negative, or zero.b. Sketch the graph of the velocity function.c. Use the graph of the velocity function to determinethe time intervals when the acceleration is positive,negative, or zero.d. Determine the time intervals when the object isspeeding up or slowing down.


Chapter 3 | Derivatives 273




160. The cost function, in dollars, of a company thatmanufactures food processors is given by
C(x) = 200 + 7x +


x2
7
, where x is the number of food


processors manufactured.a. Find the marginal cost function.b. Find the marginal cost of manufacturing 12 foodprocessors.c. Find the actual cost of manufacturing the thirteenthfood processor.
161. The price p (in dollars) and the demand x for a
certain digital clock radio is given by the price–demandfunction p = 10 − 0.001x.


a. Find the revenue function R(x).
b. Find the marginal revenue function.c. Find the marginal revenue at x = 2000 and 5000.


162. [T] A profit is earned when revenue exceeds cost.Suppose the profit function for a skateboard manufacturer
is given by P(x) = 30x − 0.3x2 − 250, where x is the
number of skateboards sold.a. Find the exact profit from the sale of the thirtiethskateboard.b. Find the marginal profit function and use it toestimate the profit from the sale of the thirtiethskateboard.
163. [T] In general, the profit function is the differencebetween the revenue and cost functions:
P(x) = R(x) − C(x). Suppose the price-demand and cost
functions for the production of cordless drills is givenrespectively by p = 143 − 0.03x and
C(x) = 75,000 + 65x, where x is the number of
cordless drills that are sold at a price of p dollars per drill
and C(x) is the cost of producing x cordless drills.


a. Find the marginal cost function.b. Find the revenue and marginal revenue functions.c. Find R′(1000) and R′(4000). Interpret the
results.d. Find the profit and marginal profit functions.e. Find P′(1000) and P′(4000). Interpret the
results.


164. A small town in Ohio commissioned an actuarialfirm to conduct a study that modeled the rate of changeof the town’s population. The study found that the town’spopulation (measured in thousands of people) can be
modeled by the function P(t) = − 1


3
t3 + 64t + 3000,


where t is measured in years.
a. Find the rate of change function P′ (t) of the


population function.b. Find P′ (1), P′ (2), P′ (3), and P′ (4). Interpret
what the results mean for the town.c. Find P″(1), P″(2), P″(3), and P″(4). Interpret
what the results mean for the town’s population.


165. [T] A culture of bacteria grows in number according
to the function N(t) = 3000⎛⎝1 + 4tt2 + 100



⎠, where t is


measured in hours.a. Find the rate of change of the number of bacteria.b. Find N′ (0), N′ (10), N′ (20), and N′ (30).
c. Interpret the results in (b).d. Find N″(0), N″(10), N″(20), and N″(30).


Interpret what the answers imply about the bacteriapopulation growth.
166. The centripetal force of an object of mass m is given
by F(r) = mv2r , where v is the speed of rotation and r
is the distance from the center of rotation.a. Find the rate of change of centripetal force withrespect to the distance from the center of rotation.b. Find the rate of change of centripetal force of anobject with mass 1000 kilograms, velocity of 13.89m/s, and a distance from the center of rotation of200 meters.
The following questions concern the population (inmillions) of London by decade in the 19th century, which islisted in the following table.


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Years since 1800 Population (millions)
1 0.8795
11 1.040
21 1.264
31 1.516
41 1.661
51 2.000
61 2.634
71 3.272
81 3.911
91 4.422


Table 3.5 Population of London Source:http://en.wikipedia.org/wiki/Demographics of London.
167. [T]a. Using a calculator or a computer program, find thebest-fit linear function to measure the population.b. Find the derivative of the equation in a. and explainits physical meaning.c. Find the second derivative of the equation andexplain its physical meaning.
168. [T]a. Using a calculator or a computer program, find thebest-fit quadratic curve through the data.b. Find the derivative of the equation and explain itsphysical meaning.c. Find the second derivative of the equation andexplain its physical meaning.
For the following exercises, consider an astronaut on alarge planet in another galaxy. To learn more about thecomposition of this planet, the astronaut drops an electronicsensor into a deep trench. The sensor transmits its verticalposition every second in relation to the astronaut’s position.The summary of the falling sensor data is displayed in thefollowing table.


Time after dropping (s) Position (m)
0 0
1 −1
2 −2
3 −5
4 −7
5 −14


169. [T]a. Using a calculator or computer program, find thebest-fit quadratic curve to the data.b. Find the derivative of the position function andexplain its physical meaning.c. Find the second derivative of the position functionand explain its physical meaning.
170. [T]a. Using a calculator or computer program, find thebest-fit cubic curve to the data.b. Find the derivative of the position function andexplain its physical meaning.c. Find the second derivative of the position functionand explain its physical meaning.d. Using the result from c. explain why a cubicfunction is not a good choice for this problem.
The following problems deal with the Holling type I, II,and III equations. These equations describe the ecologicalevent of growth of a predator population given the amountof prey available for consumption.
171. [T] The Holling type I equation is described by
f (x) = ax, where x is the amount of prey available and
a > 0 is the rate at which the predator meets the prey for
consumption.a. Graph the Holling type I equation, given a = 0.5.


b. Determine the first derivative of the Holling type Iequation and explain physically what the derivativeimplies.c. Determine the second derivative of the Holling typeI equation and explain physically what thederivative implies.d. Using the interpretations from b. and c. explainwhy the Holling type I equation may not berealistic.


Chapter 3 | Derivatives 275




172. [T] The Holling type II equation is described by
f (x) = axn + x, where x is the amount of prey available
and a > 0 is the maximum consumption rate of the
predator.a. Graph the Holling type II equation given a = 0.5


and n = 5. What are the differences between the
Holling type I and II equations?b. Take the first derivative of the Holling type IIequation and interpret the physical meaning of thederivative.


c. Show that f (n) = 1
2
a and interpret the meaning of


the parameter n.
d. Find and interpret the meaning of the secondderivative. What makes the Holling type II functionmore realistic than the Holling type I function?


173. [T] The Holling type III equation is described by
f (x) = ax


2


n2 + x2
, where x is the amount of prey available


and a > 0 is the maximum consumption rate of the
predator.a. Graph the Holling type III equation given a = 0.5


and n = 5. What are the differences between the
Holling type II and III equations?b. Take the first derivative of the Holling type IIIequation and interpret the physical meaning of thederivative.c. Find and interpret the meaning of the secondderivative (it may help to graph the secondderivative).d. What additional ecological phenomena does theHolling type III function describe compared withthe Holling type II function?


174. [T] The populations of the snowshoe hare (inthousands) and the lynx (in hundreds) collected over 7years from 1937 to 1943 are shown in the following table.The snowshoe hare is the primary prey of the lynx.
Population of snowshoehare (thousands) Population oflynx (hundreds)
20 10
55 15
65 55
95 60


Table 3.6 Snowshoe Hare and LynxPopulations Source: http://www.biotopics.co.uk/newgcse/predatorprey.html.
a. Graph the data points and determine whichHolling-type function fits the data best.b. Using the meanings of the parameters a and n,


determine values for those parameters byexamining a graph of the data. Recall that n
measures what prey value results in the half-maximum of the predator value.c. Plot the resulting Holling-type I, II, and IIIfunctions on top of the data. Was the result frompart a. correct?


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3.5 | Derivatives of Trigonometric Functions
Learning Objectives


3.5.1 Find the derivatives of the sine and cosine function.
3.5.2 Find the derivatives of the standard trigonometric functions.
3.5.3 Calculate the higher-order derivatives of the sine and cosine.


One of the most important types of motion in physics is simple harmonic motion, which is associated with such systemsas an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosinefunctions. In this section we expand our knowledge of derivative formulas to include derivatives of these and othertrigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulasfor the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine andcosine functions will enable us to find the velocity and acceleration of simple harmonic motion.
Derivatives of the Sine and Cosine Functions
We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at itsderivative. Recall that for a function f (x),


f ′ (x) = lim
h → 0


f (x + h) − f (x)
h


.


Consequently, for values of h very close to 0, f ′ (x) ≈ f (x + h) − f (x)
h


. We see that by using h = 0.01,
d
dx


(sinx) ≈ sin(x + 0.01) − sinx
0.01


By setting D(x) = sin(x + 0.01) − sinx
0.01


and using a graphing utility, we can get a graph of an approximation to the
derivative of sinx (Figure 3.25).


Figure 3.25 The graph of the function D(x) looks a lot like a
cosine curve.


Upon inspection, the graph of D(x) appears to be very close to the graph of the cosine function. Indeed, we will show that
d
dx


(sinx) = cosx.


If we were to follow the same steps to approximate the derivative of the cosine function, we would find that


Chapter 3 | Derivatives 277




d
dx


(cosx) = −sinx.


Theorem 3.8: The Derivatives of sin x and cos x
The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.


(3.11)d
dx


(sinx) = cosx


(3.12)d
dx


(cosx) = −sinx


Proof
Because the proofs for d


dx
(sinx) = cosx and d


dx
(cosx) = −sinx use similar techniques, we provide only the proof for


d
dx


(sinx) = cosx. Before beginning, recall two important trigonometric limits we learned in Introduction to Limits:
lim
h → 0


sinh
h


= 1 and lim
h → 0


coshh − 1
h


= 0.


The graphs of y = (sinh)
h


and y = (cosh − 1)
h


are shown in Figure 3.26.


Figure 3.26 These graphs show two important limits needed to establish the derivative formulas for thesine and cosine functions.


We also recall the following trigonometric identity for the sine of the sum of two angles:
sin(x + h) = sinxcosh + cosxsinh.


Now that we have gathered all the necessary equations and identities, we proceed with the proof.
d
dx


sinx = lim
h → 0


sin(x + h) − sinx
h


Apply the definition of he derivative.


= lim
h → 0


sinxcosh + cosxsinh − sinx
h


Use trig identity for the sine of the sum of two angles.


= lim
h → 0




sinxcosh − sinx


h
+ cosxsinh


h

⎠ Regroup.


= lim
h → 0



⎝sinx


cosxh − 1


h

⎠+ cosx




sinh
h



⎠ Factor out sinx and cosx.


= sinx(0) + cosx(1) Apply trig limit formulas.


= cosx Simplify.


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3.25



Figure 3.27 shows the relationship between the graph of f (x) = sinx and its derivative f ′ (x) = cosx. Notice that at the
points where f (x) = sinx has a horizontal tangent, its derivative f ′ (x) = cosx takes on the value zero. We also see that
where f (x) = sinx is increasing, f ′ (x) = cosx > 0 and where f (x) = sinx is decreasing, f ′ (x) = cosx < 0.


Figure 3.27 Where f (x) has a maximum or a minimum,
f ′(x) = 0 that is, f ′(x) = 0 where f (x) has a horizontal
tangent. These points are noted with dots on the graphs.


Example 3.39
Differentiating a Function Containing sin x
Find the derivative of f (x) = 5x3 sinx.
Solution
Using the product rule, we have


f ′(x) = d
dx

⎝5x


3⎞
⎠ · sinx +


d
dx


(sinx) · 5x3


= 15x2 · sinx + cosx · 5x3.


After simplifying, we obtain
f ′ (x) = 15x2 sinx + 5x3 cosx.


Find the derivative of f (x) = sinxcosx.


Example 3.40
Finding the Derivative of a Function Containing cos x


Chapter 3 | Derivatives 279




3.26


3.27


Find the derivative of g(x) = cosx
4x2


.


Solution
By applying the quotient rule, we have


g′ (x) = (−sinx)4x
2 − 8x(cosx)

⎝4x


2⎞


2
.


Simplifying, we obtain
g′ (x) = −4x


2 sinx − 8xcosx
16x4


= −xsinx − 2cosx
4x3


.


Find the derivative of f (x) = xcosx.


Example 3.41
An Application to Velocity
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t) = 2sin t − t
for 0 ≤ t ≤ 2π. At what times is the particle at rest?
Solution
To determine when the particle is at rest, set s′ (t) = v(t) = 0. Begin by finding s′ (t). We obtain


s′ (t) = 2cos t − 1,


so we must solve
2cos t − 1 = 0 for 0 ≤ t ≤ 2π.


The solutions to this equation are t = π
3
and t = 5π


3
. Thus the particle is at rest at times t = π


3
and t = 5π


3
.


A particle moves along a coordinate axis. Its position at time t is given by s(t) = 3t + 2cos t for
0 ≤ t ≤ 2π. At what times is the particle at rest?


Derivatives of Other Trigonometric Functions
Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can usethe quotient rule to find formulas for their derivatives.


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3.28


Example 3.42
The Derivative of the Tangent Function
Find the derivative of f (x) = tanx.
Solution
Start by expressing tanx as the quotient of sinx and cosx :


f (x) = tanx = sinxcosx.


Now apply the quotient rule to obtain
f ′ (x) = cosxcosx − (−sinx)sinx


(cosx)2
.


Simplifying, we obtain
f ′ (x) = cos


2 x + sin2 x
cos2 x


.


Recognizing that cos2 x + sin2 x = 1, by the Pythagorean theorem, we now have
f ′ (x) = 1


cos2 x
.


Finally, use the identity secx = 1cosx to obtain
f ′ (x) = sec2 x.


Find the derivative of f (x) = cotx.


The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide theseformulas in the following theorem.
Theorem 3.9: Derivatives of tanx, cotx, secx, and cscx
The derivatives of the remaining trigonometric functions are as follows:


(3.13)d
dx


(tanx) = sec2 x


(3.14)d
dx


(cotx) = −csc2 x


(3.15)d
dx


(secx) = secx tanx


(3.16)d
dx


(cscx) = −cscxcotx.


Example 3.43


Chapter 3 | Derivatives 281




3.29


Finding the Equation of a Tangent Line
Find the equation of a line tangent to the graph of f (x) = cotx at x = π


4
.


Solution
To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute


f ⎛⎝
π
4

⎠ = cot


π
4
= 1.


Thus the tangent line passes through the point ⎛⎝π4, 1⎞⎠. Next, find the slope by finding the derivative of
f (x) = cotx and evaluating it at π


4
:


f ′ (x) = −csc2 x and f ′ ⎛⎝
π
4

⎠ = −csc


2 ⎛

π
4

⎠ = −2.


Using the point-slope equation of the line, we obtain
y − 1 = −2⎛⎝x −


π
4



or equivalently,
y = −2x + 1 + π


2
.


Example 3.44
Finding the Derivative of Trigonometric Functions
Find the derivative of f (x) = cscx + x tanx.
Solution
To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find


f ′ (x) = d
dx


(cscx) + d
dx


(x tanx).


In the first term, d
dx


(cscx) = −cscxcotx, and by applying the product rule to the second term we obtain
d
dx


(x tanx) = (1)(tanx) + (sec2 x)(x).


Therefore, we have
f ′ (x) = −cscxcotx + tanx + xsec2 x.


Find the derivative of f (x) = 2tanx − 3cotx.


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3.30


3.31


Find the slope of the line tangent to the graph of f (x) = tanx at x = π
6
.


Higher-Order Derivatives
The higher-order derivatives of sinx and cosx follow a repeating pattern. By following the pattern, we can find any
higher-order derivative of sinx and cosx.
Example 3.45
Finding Higher-Order Derivatives of y = sinx
Find the first four derivatives of y = sinx.
Solution
Each step in the chain is straightforward:


y = sinx
dy
dx


= cosx


d2 y


dx2
= −sinx


d3 y


dx3
= −cosx


d4 y


dx4
= sinx.


Analysis
Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step inthe pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x, so


d4


dx4
(sinx) = d


8


dx8
(sinx) = d


12


dx12
(sinx) = … = d


4n


dx4n
(sinx) = sinx


d5


dx5
(sinx) = d


9


dx9
(sinx) = d


13


dx13
(sinx) = … = d


4n + 1


dx4n + 1
(sinx) = cosx.


For y = cosx, find d4 y
dx4


.


Example 3.46
Using the Pattern for Higher-Order Derivatives of y = sinx
Find d74


dx74
(sinx).


Chapter 3 | Derivatives 283




3.32


3.33


Solution
We can see right away that for the 74th derivative of sinx, 74 = 4(18) + 2, so


d74


dx74
(sinx) = d


72 + 2


dx72 + 2
(sinx) = d


2


dx2
(sinx) = −sinx.


For y = sinx, find d59
dx59


(sinx).


Example 3.47
An Application to Acceleration
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t) = 2 − sin t.
Find v(π/4) and a(π/4). Compare these values and decide whether the particle is speeding up or slowing down.
Solution
First find v(t) = s′ (t):


v(t) = s′ (t) = −cos t.


Thus,
v⎛⎝
π
4

⎠ = −


1
2
.


Next, find a(t) = v′(t). Thus, a(t) = v′ (t) = sin t and we have
a⎛⎝
π
4

⎠ =


1
2
.


Since v⎛⎝π4⎞⎠ = − 12 < 0 and a⎛⎝π4⎞⎠ = 12 > 0, we see that velocity and acceleration are acting in opposite
directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling.Consequently, the particle is slowing down.


A block attached to a spring is moving vertically. Its position at time t is given by s(t) = 2sin t. Find
v⎛⎝

6

⎠ and a⎛⎝5π6 ⎞⎠. Compare these values and decide whether the block is speeding up or slowing down.


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3.5 EXERCISES
For the following exercises, find dy


dx
for the given


functions.
175. y = x2 − secx + 1
176. y = 3cscx + 5x
177. y = x2 cotx
178. y = x − x3 sinx
179. y = secxx
180. y = sinx tanx
181. y = (x + cosx)(1 − sinx)
182. y = tanx


1 − secx


183. y = 1 − cotx
1 + cotx


184. y = cosx(1 + cscx)
For the following exercises, find the equation of the tangentline to each of the given functions at the indicated valuesof x. Then use a calculator to graph both the function and
the tangent line to ensure the equation for the tangent lineis correct.
185. [T] f (x) = −sinx, x = 0
186. [T] f (x) = cscx, x = π


2


187. [T] f (x) = 1 + cosx, x = 3π
2


188. [T] f (x) = secx, x = π
4


189. [T] f (x) = x2 − tanx x = 0
190. [T] f (x) = 5cotx x = π


4


For the following exercises, find d2 y
dx2


for the given
functions.
191. y = xsinx − cosx


192. y = sinxcosx
193. y = x − 1


2
sinx


194. y = 1x + tanx
195. y = 2cscx
196. y = sec2 x
197. Find all x values on the graph of
f (x) = −3sinxcosx where the tangent line is horizontal.
198. Find all x values on the graph of f (x) = x − 2cosx
for 0 < x < 2π where the tangent line has slope 2.
199. Let f (x) = cotx. Determine the points on the graph
of f for 0 < x < 2π where the tangent line(s) is (are)
parallel to the line y = −2x.
200. [T] A mass on a spring bounces up and down insimple harmonic motion, modeled by the function
s(t) = −6cos t where s is measured in inches and t is
measured in seconds. Find the rate at which the spring isoscillating at t = 5 s.
201. Let the position of a swinging pendulum in simpleharmonic motion be given by s(t) = acos t + bsin t. Find
the constants a and b such that when the velocity is 3 cm/
s, s = 0 and t = 0.
202. After a diver jumps off a diving board, the edge ofthe board oscillates with position given by s(t) = −5cos t
cm at t seconds after the jump.


a. Sketch one period of the position function for
t ≥ 0.


b. Find the velocity function.c. Sketch one period of the velocity function for
t ≥ 0.


d. Determine the times when the velocity is 0 over oneperiod.e. Find the acceleration function.f. Sketch one period of the acceleration function for
t ≥ 0.


Chapter 3 | Derivatives 285




203. The number of hamburgers sold at a fast-foodrestaurant in Pasadena, California, is given by
y = 10 + 5sinx where y is the number of hamburgers
sold and x represents the number of hours after the
restaurant opened at 11 a.m. until 11 p.m., when the storecloses. Find y′ and determine the intervals where the
number of burgers being sold is increasing.
204. [T] The amount of rainfall per month in Phoenix,Arizona, can be approximated by y(t) = 0.5 + 0.3cos t,
where t is months since January. Find y′ and use a
calculator to determine the intervals where the amount ofrain falling is decreasing.
For the following exercises, use the quotient rule to derivethe given equations.
205. d


dx
(cotx) = −csc2 x


206. d
dx


(secx) = secx tanx


207. d
dx


(cscx) = −cscxcotx


208. Use the definition of derivative and the identity
cos(x + h) = cosxcosh − sinxsinh to prove that
d(cosx)


dx
= −sinx.


For the following exercises, find the requested higher-orderderivative for the given functions.
209. d3 y


dx3
of y = 3cosx


210. d2 y
dx2


of y = 3sinx + x2 cosx


211. d4 y
dx4


of y = 5cosx


212. d2 y
dx2


of y = secx + cotx


213. d3 y
dx3


of y = x10 − secx


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3.6 | The Chain Rule
Learning Objectives


3.6.1 State the chain rule for the composition of two functions.
3.6.2 Apply the chain rule together with the power rule.
3.6.3 Apply the chain rule and the product/quotient rules correctly in combination when both arenecessary.
3.6.4 Recognize the chain rule for a composition of three or more functions.
3.6.5 Describe the proof of the chain rule.


We have seen the techniques for differentiating basic functions (xn, sinx, cosx, etc.) as well as sums, differences,
products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate
compositions of functions, such as h(x) = sin⎛⎝x3⎞⎠ or k(x) = 3x2 + 1. In this section, we study the rule for finding the
derivative of the composition of two or more functions.
Deriving the Chain Rule
When we have a function that is a composition of two or more functions, we could use all of the techniques we have alreadylearned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we areable to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a compositefunction is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
To put this rule into context, let’s take a look at an example: h(x) = sin ⎛⎝x3⎞⎠. We can think of the derivative of this function
with respect to x as the rate of change of sin⎛⎝x3⎞⎠ relative to the change in x. Consequently, we want to know how sin⎛⎝x3⎞⎠
changes as x changes. We can think of this event as a chain reaction: As x changes, x3 changes, which leads to a change
in sin ⎛⎝x3⎞⎠. This chain reaction gives us hints as to what is involved in computing the derivative of sin⎛⎝x3⎞⎠. First of all, a
change in x forcing a change in x3 suggests that somehow the derivative of x3 is involved. In addition, the change in x3
forcing a change in sin⎛⎝x3⎞⎠ suggests that the derivative of sin(u) with respect to u, where u = x3, is also part of the
final derivative.
We can take a more formal look at the derivative of h(x) = sin ⎛⎝x3⎞⎠ by setting up the limit that would give us the derivative
at a specific value a in the domain of h(x) = sin ⎛⎝x3⎞⎠.


h′ (a) = lim
x → a


sin ⎛⎝x
3⎞
⎠− sin



⎝a


3⎞


x − a .


This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the
expression x3 − a3 to obtain


h′ (a) = lim
x → a


sin ⎛⎝x
3⎞
⎠− sin



⎝a


3⎞


x3 − a3
· x


3 − a3
x − a .


From the definition of the derivative, we can see that the second factor is the derivative of x3 at x = a. That is,
lim
x → a


x3 − a3
x − a =


d
dx

⎝x


3⎞
⎠ = 3a


2.


However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting
u = x3 and observing that as x → a, u → a3 :


Chapter 3 | Derivatives 287




lim
x → a


sin ⎛⎝x
3⎞
⎠− sin



⎝a


3⎞


x3 − a3
= lim


u → a3


sinu − sin ⎛⎝a
3⎞


u − a3


= d
du


(sinu)
u = a3


= cos ⎛⎝a
3).


Thus, h′ (a) = cos⎛⎝a3⎞⎠ · 3a2.
In other words, if h(x) = sin ⎛⎝x3⎞⎠, then h′ (x) = cos ⎛⎝x3⎞⎠ · 3x2. Thus, if we think of h(x) = sin⎛⎝x3⎞⎠ as the composition

⎝ f ∘g⎞⎠(x) = f ⎛⎝g(x)⎞⎠ where f (x) = sin x and g(x) = x3, then the derivative of h(x) = sin⎛⎝x3⎞⎠ is the product of the
derivative of g(x) = x3 and the derivative of the function f (x) = sinx evaluated at the function g(x) = x3. At this point,
we anticipate that for h(x) = sin ⎛⎝g(x)⎞⎠, it is quite likely that h′(x) = cos(g(x))g′(x). As we determined above, this is the
case for h(x) = sin ⎛⎝x3⎞⎠.
Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form toother composite functions. An informal proof is provided at the end of the section.
Rule: The Chain Rule
Let f and g be functions. For all x in the domain of g for which g is differentiable at x and f is differentiable at
g(x), the derivative of the composite function


h(x) = ⎛⎝ f ∘g⎞⎠(x) = f ⎛⎝g(x)⎞⎠


is given by
(3.17)h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x).


Alternatively, if y is a function of u, and u is a function of x, then
dy
dx


=
dy
du


· du
dx


.


Watch an animation (http://www.openstaxcollege.org/l/20_chainrule2) of the chain rule.


Problem-Solving Strategy: Applying the Chain Rule
1. To differentiate h(x) = f ⎛⎝g(x)⎞⎠, begin by identifying f (x) and g(x).
2. Find f ′(x) and evaluate it at g(x) to obtain f ′ ⎛⎝g(x)⎞⎠.
3. Find g′(x).
4. Write h′ (x) = f ′ ⎛⎝g(x)⎞⎠ · g′ (x).


Note: When applying the chain rule to the composition of two or more functions, keep in mind that we work our wayfrom the outside function in. It is also useful to remember that the derivative of the composition of two functions canbe thought of as having two parts; the derivative of the composition of three functions has three parts; and so on. Also,remember that we never evaluate a derivative at a derivative.


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3.34


The Chain and Power Rules Combined
We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example,to find derivatives of functions of the form h(x) = (g(x))n, we need to use the chain rule combined with the power rule. To
do so, we can think of h(x) = ⎛⎝g(x)⎞⎠n as f ⎛⎝g(x)⎞⎠ where f (x) = xn. Then f ′ (x) = nxn − 1. Thus, f ′ ⎛⎝g(x)⎞⎠ = n⎛⎝g(x)⎞⎠n − 1.
This leads us to the derivative of a power function using the chain rule,


h′ (x) = n⎛⎝g(x)⎞⎠n − 1g′ (x)


Rule: Power Rule for Composition of Functions
For all values of x for which the derivative is defined, if


h(x) = ⎛⎝g(x)⎞⎠n.


Then
(3.18)h′ (x) = n⎛⎝g(x)⎞⎠n − 1g′ (x).


Example 3.48
Using the Chain and Power Rules
Find the derivative of h(x) = 1



⎝3x


2 + 1⎞⎠
2
.


Solution
First, rewrite h(x) = 1



⎝3x


2 + 1⎞⎠
2
= ⎛⎝3x


2 + 1⎞⎠
−2


.


Applying the power rule with g(x) = 3x2 + 1, we have
h′ (x) = −2⎛⎝3x


2 + 1⎞⎠
−3


(6x).


Rewriting back to the original form gives us
h′ (x) = −12x


(3x2 + 1)3
.


Find the derivative of h(x) = ⎛⎝2x3 + 2x − 1⎞⎠4.


Example 3.49
Using the Chain and Power Rules with a Trigonometric Function
Find the derivative of h(x) = sin3 x.


Chapter 3 | Derivatives 289




3.35


Solution
First recall that sin3 x = (sinx)3, so we can rewrite h(x) = sin3 x as h(x) = (sinx)3.
Applying the power rule with g(x) = sinx, we obtain


h′ (x) = 3(sinx)2 cosx = 3sin2 xcosx.


Example 3.50
Finding the Equation of a Tangent Line
Find the equation of a line tangent to the graph of h(x) = 1


(3x − 5)2
at x = 2.


Solution
Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the
y-coordinate, substitute 2 into h(x). Since h(2) = 1



⎝3(2) − 5⎞⎠2


= 1, the point is (2, 1).
For the slope, we need h′(2). To find h′(x), first we rewrite h(x) = (3x − 5)−2 and apply the power rule to
obtain


h′ (x) = −2(3x − 5)−3 (3) = −6(3x − 5)−3.


By substituting, we have h′ (2) = −6⎛⎝3(2) − 5⎞⎠−3 = −6. Therefore, the line has equation y − 1 = −6(x − 2).
Rewriting, the equation of the line is y = −6x + 13.


Find the equation of the line tangent to the graph of f (x) = ⎛⎝x2 − 2⎞⎠3 at x = −2.


Combining the Chain Rule with Other Rules
Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other ruleswe have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with theproduct rule.
Example 3.51
Using the Chain Rule on a General Cosine Function
Find the derivative of h(x) = cos ⎛⎝g(x)⎞⎠.
Solution


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3.36


Think of h(x) = cos(g(x)) as f ⎛⎝g(x)⎞⎠ where f (x) = cosx. Since f ′ (x) = −sinx. we have
f ′ ⎛⎝g(x)⎞⎠ = −sin ⎛⎝g(x)⎞⎠. Then we do the following calculation.


h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x) Apply the chain rule.


= −sin ⎛⎝g(x)⎞⎠g′ (x) Substitute f ′ ⎛⎝g(x)⎞⎠ = −sin ⎛⎝g(x)⎞⎠.


Thus, the derivative of h(x) = cos ⎛⎝g(x)⎞⎠ is given by h′ (x) = −sin ⎛⎝g(x)⎞⎠g′ (x).


In the following example we apply the rule that we have just derived.
Example 3.52
Using the Chain Rule on a Cosine Function
Find the derivative of h(x) = cos ⎛⎝5x2⎞⎠.
Solution
Let g(x) = 5x2. Then g′ (x) = 10x. Using the result from the previous example,


h′ (x) = −sin ⎛⎝5x
2⎞
⎠ · 10x


= −10xsin ⎛⎝5x
2⎞
⎠.


Example 3.53
Using the Chain Rule on Another Trigonometric Function
Find the derivative of h(x) = sec⎛⎝4x5 + 2x⎞⎠.
Solution
Apply the chain rule to h(x) = sec ⎛⎝g(x)⎞⎠ to obtain


h′ (x) = sec(g(x) tan ⎛⎝g(x)⎞⎠g′ (x).


In this problem, g(x) = 4x5 + 2x, so we have g′ (x) = 20x4 + 2. Therefore, we obtain
h′ (x) = sec⎛⎝4x


5 + 2x⎞⎠tan

⎝4x


5 + 2x⎞⎠

⎝20x


4 + 2⎞⎠


= (20x4 + 2)sec⎛⎝4x
5 + 2x⎞⎠tan



⎝4x


5 + 2x⎞⎠.


Find the derivative of h(x) = sin(7x + 2).


Chapter 3 | Derivatives 291




3.37


At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunctionwith the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in Example 3.51and Example 3.53. For convenience, formulas are also given in Leibniz’s notation, which some students find easier toremember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary tomemorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.
Theorem 3.10: Using the Chain Rule with Trigonometric Functions
For all values of x for which the derivative is defined,


d
dx

⎝sin(g(x)⎞⎠ = cos ⎛⎝g(x)⎞⎠g′(x) ddx


sinu = cosudu
dx


d
dx

⎝cos(g(x)⎞⎠ = −sin ⎛⎝g(x)⎞⎠g′(x) ddx


cosu = −sinudu
dx


d
dx

⎝tan(g(x)⎞⎠ = sec2 ⎛⎝g(x)⎞⎠g′(x) ddx


tanu = sec2udu
dx


d
dx

⎝cot(g(x)⎞⎠ = −csc2 ⎛⎝g(x)⎞⎠g′(x) ddx


cotu = −csc2udu
dx


d
dx

⎝sec(g(x)⎞⎠ = sec(g(x) tan ⎛⎝g(x)⎞⎠g′(x) ddx


secu = secu tanudu
dx


d
dx

⎝csc(g(x)⎞⎠ = −csc(g(x))cot ⎛⎝g(x)⎞⎠g′(x) ddx


cscu = −cscucotudu
dx


.


Example 3.54
Combining the Chain Rule with the Product Rule
Find the derivative of h(x) = (2x + 1)5 (3x − 2)7.
Solution
First apply the product rule, then apply the chain rule to each term of the product.


h′ (x) = d
dx

⎝(2x + 1)


5⎞
⎠ · (3x − 2)


7 + d
dx

⎝(3x − 2)


7⎞
⎠ · (2x + 1)


5 Apply the product rule.


= 5(2x + 1)4 · 2 · (3x − 2)7 + 7(3x − 2)6 · 3 · (2x + 1)5 Apply the chain rule.


= 10(2x + 1)4 (3x − 2)7 + 21(3x − 2)6 (2x + 1)5 Simplify.


= (2x + 1)4 (3x − 2)6 (10(3x − 7) + 21(2x + 1)) Factor out (2x + 1)4 (3x − 2)6.


= (2x + 1)4 (3x − 2)6(72x − 49) Simplify.


Find the derivative of h(x) = x
(2x + 3)3


.


Composites of Three or More Functions
We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating thecomposition of three or more functions, we need to apply the chain rule more than once. If we look at this situation ingeneral terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multipletimes.
In general terms, first we let


k(x) = h⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠.


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3.38


Then, applying the chain rule once we obtain
k′ (x) = d


dx

⎝h( f ⎛⎝g(x)⎞⎠⎞⎠ = h′⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠ · ddx


f ⎛⎝⎛⎝g(x)⎞⎠⎞⎠.


Applying the chain rule again, we obtain
k′ (x) = h′ ⎛⎝ f ⎛⎝g(x)⎞⎠ f ′ ⎛⎝g(x)⎞⎠g′ (x)⎞⎠.


Rule: Chain Rule for a Composition of Three Functions
For all values of x for which the function is differentiable, if


k(x) = h⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠,


then
k′ (x) = h′ ⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠ f ′ ⎛⎝g(x)⎞⎠g′ (x).


In other words, we are applying the chain rule twice.


Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the compositionof four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivativeat a time.
Example 3.55
Differentiating a Composite of Three Functions
Find the derivative of k(x) = cos4 ⎛⎝7x2 + 1⎞⎠.
Solution
First, rewrite k(x) as


k(x) = ⎛⎝cos

⎝7x


2 + 1⎞⎠



4
.


Then apply the chain rule several times.
k′ (x) = 4⎛⎝cos



⎝7x


2 + 1⎞⎠



3 ⎛

d
dx


(cos ⎛⎝7x
2 + 1⎞⎠



⎠ Apply the chain rule.


= 4⎛⎝cos

⎝7x


2 + 1⎞⎠



3 ⎛
⎝−sin



⎝7x


2 + 1⎞⎠




d
dx

⎝7x


2 + 1⎞⎠

⎠ Apply the chain rule.


= 4⎛⎝cos

⎝7x


2 + 1⎞⎠



3 ⎛
⎝−sin



⎝7x


2 + 1⎞⎠

⎠(14x) Apply the chain rule.


= −56xsin(7x2 + 1)cos3(7x2 + 1) Simplify.


Find the derivative of h(x) = sin6⎛⎝x3⎞⎠.


Example 3.56


Chapter 3 | Derivatives 293




3.39


Using the Chain Rule in a Velocity Problem
A particle moves along a coordinate axis. Its position at time t is given by s(t) = sin(2t) + cos(3t). What is the
velocity of the particle at time t = π


6
?


Solution
To find v(t), the velocity of the particle at time t, we must differentiate s(t). Thus,


v(t) = s′ (t) = 2cos(2t) − 3sin(3t).


Substituting t = π
6
into v(t), we obtain v⎛⎝π6⎞⎠ = −2.


A particle moves along a coordinate axis. Its position at time t is given by s(t) = sin(4t). Find its
acceleration at time t.


Proof
At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example,we assume that g(x) ≠ g(a) for x ≠ a in some open interval containing a. We begin by applying the limit definition of
the derivative to the function h(x) to obtain h′(a):


h′ (a) = lim
x → a


f ⎛⎝g(x)⎞⎠− f ⎛⎝g(a)⎞⎠
x − a .


Rewriting, we obtain
h′ (a) = lim


x → a
f ⎛⎝g(x)⎞⎠− f ⎛⎝g(a)⎞⎠
g(x) − g(a)


·
g(x) − g(a)


x − a .


Although it is clear that
lim
x → a


g(x) − g(a)
x − a = g′(a),


it is not obvious that
lim
x → a


f ⎛⎝g(x)⎞⎠− f ⎛⎝g(a)⎞⎠
g(x) − g(a)


= f ′ ⎛⎝g(a)⎞⎠.


To see that this is true, first recall that since g is differentiable at a, g is also continuous at a. Thus,
lim
x → a


g(x) = g(a).


Next, make the substitution y = g(x) and b = g(a) and use change of variables in the limit to obtain
lim
x → a


f ⎛⎝g(x)⎞⎠− f ⎛⎝g(a)⎞⎠
g(x) − g(a)


= lim
y → b


f (y) − f (b)
y − b


= f ′ (b) = f ′ ⎛⎝g(a)⎞⎠.


Finally,
h′ (a) = lim


x → a
f ⎛⎝g(x)⎞⎠− f ⎛⎝g(a)⎞⎠
g(x) − g(a)


·
g(x) − g(a)


x − a = f ′

⎝g(a)⎞⎠g′ (a).




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3.40


Example 3.57
Using the Chain Rule with Functional Values
Let h(x) = f ⎛⎝g(x)⎞⎠. If g(1) = 4, g′ (1) = 3, and f ′ (4) = 7, find h′ (1).
Solution
Use the chain rule, then substitute.


h′ (1) = f ′ ⎛⎝g(1)⎞⎠g′ (1) Apply the chain rule.


= f ′ (4) · 3 Substitute g(1) = 4 and g′ (1) = 3.


= 7 · 3 Substitute f ′(4) = 7.


= 21 Simplify.


Given h(x) = f ⎛⎝g(x)⎞⎠. If g(2) = −3, g′ (2) = 4, and f ′ (−3) = 7, find h′ (2).


The Chain Rule Using Leibniz’s Notation
As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for thechain rule is used heavily in physics applications.
For h(x) = f ⎛⎝g(x)⎞⎠, let u = g(x) and y = h(x) = g(u). Thus,


h′ (x) =
dy
dx


, f ′ ⎛⎝g(x)⎞⎠ = f ′ (u) =
dy
du


and g′ (x) = du
dx


.


Consequently,
dy
dx


= h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x) =
dy
du


· du
dx


.


Rule: Chain Rule Using Leibniz’s Notation
If y is a function of u, and u is a function of x, then


dy
dx


=
dy
du


· du
dx


.


Example 3.58
Taking a Derivative Using Leibniz’s Notation, Example 1
Find the derivative of y = ⎛⎝ x3x + 2⎞⎠


5
.


Solution
First, let u = x


3x + 2
. Thus, y = u5. Next, find du


dx
and dy


du
. Using the quotient rule,


Chapter 3 | Derivatives 295




3.41


du
dx


= 2
(3x + 2)2


and
dy
du


= 5u4.


Finally, we put it all together.
dy
dx


=
dy
du


· du
dx


Apply the chain rule.


= 5u4 · 2
(3x + 2)2


Substitute
dy
du


= 5u4 and du
dx


= 2
(3x + 2)2


.


= 5⎛⎝
x


3x + 2



4
· 2
(3x + 2)2


Substitute u = x
3x + 2


.


= 10x
4


(3x + 2)6
Simplify.


It is important to remember that, when using the Leibniz form of the chain rule, the final answer must beexpressed entirely in terms of the original variable given in the problem.


Example 3.59
Taking a Derivative Using Leibniz’s Notation, Example 2
Find the derivative of y = tan⎛⎝4x2 − 3x + 1⎞⎠.
Solution
First, let u = 4x2 − 3x + 1. Then y = tanu. Next, find du


dx
and dy


du
:


du
dx


= 8x − 3 and
dy
du


= sec2u.


Finally, we put it all together.
dy
dx


=
dy
du


· du
dx


Apply the chain rule.


= sec2u · (8x − 3) Use du
dx


= 8x − 3 and
dy
du


= sec2u.


= sec2(4x2 − 3x + 1) · (8x − 3) Substitute u = 4x2 − 3x + 1.


Use Leibniz’s notation to find the derivative of y = cos ⎛⎝x3⎞⎠. Make sure that the final answer is
expressed entirely in terms of the variable x.


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3.6 EXERCISES
For the following exercises, given y = f (u) and
u = g(x), find dy


dx
by using Leibniz’s notation for the


chain rule: dy
dx


=
dy
du


du
dx


.


214. y = 3u − 6, u = 2x2
215. y = 6u3, u = 7x − 4
216. y = sinu, u = 5x − 1
217. y = cosu, u = −x


8


218. y = tanu, u = 9x + 2
219. y = 4u + 3, u = x2 − 6x
For each of the following exercises,


a. decompose each function in the form y = f (u)
and u = g(x), and


b. find dy
dx


as a function of x.
220. y = (3x − 2)6


221. y = ⎛⎝3x2 + 1⎞⎠3


222. y = sin5 (x)


223. y = ⎛⎝x7 + 7x⎞⎠
7


224. y = tan(secx)
225. y = csc(πx + 1)
226. y = cot2 x
227. y = −6sin−3 x
For the following exercises, find dy


dx
for each function.


228. y = ⎛⎝3x2 + 3x − 1⎞⎠4


229. y = (5 − 2x)−2


230. y = cos3 (πx)


231. y = ⎛⎝2x3 − x2 + 6x + 1⎞⎠3


232. y = 1
sin2(x)


233. y = (tanx + sinx)−3
234. y = x2 cos4 x
235. y = sin(cos7x)
236. y = 6 + secπx2
237. y = cot3 (4x + 1)
238. Let y = ⎡⎣ f (x)⎤⎦3 and suppose that f ′ (1) = 4 and
dy
dx


= 10 for x = 1. Find f (1).


239. Let y = ⎛⎝ f (x) + 5x2⎞⎠4 and suppose that
f (−1) = −4 and dy


dx
= 3 when x = −1. Find f ′ (−1)


240. Let y = ⎛⎝ f (u) + 3x⎞⎠2 and u = x3 − 2x. If
f (4) = 6 and dy


dx
= 18 when x = 2, find f ′ (4).


241. [T] Find the equation of the tangent line to
y = −sin⎛⎝


x
2

⎠ at the origin. Use a calculator to graph the


function and the tangent line together.
242. [T] Find the equation of the tangent line to
y = ⎛⎝3x +


1
x



2 at the point (1, 16). Use a calculator to
graph the function and the tangent line together.
243. Find the x -coordinates at which the tangent line to
y = ⎛⎝x −


6
x



8 is horizontal.
244. [T] Find an equation of the line that is normal to
g(θ) = sin2 (πθ) at the point ⎛⎝14, 12⎞⎠. Use a calculator to
graph the function and the normal line together.
For the following exercises, use the information in thefollowing table to find h′(a) at the given value for a.


Chapter 3 | Derivatives 297




x f(x) f′(x) g(x) g′(x)


0 2 5 0 2
1 1 −2 3 0
2 4 4 1 −1
3 3 −3 2 3


245. h(x) = f ⎛⎝g(x)⎞⎠; a = 0
246. h(x) = g⎛⎝ f (x)⎞⎠; a = 0


247. h(x) = ⎛⎝x4 + g(x)⎞⎠−2; a = 1


248. h(x) = ⎛⎝ f (x)g(x)⎞⎠
2


; a = 3


249. h(x) = f ⎛⎝x + f (x)⎞⎠; a = 1
250. h(x) = ⎛⎝1 + g(x)⎞⎠3; a = 2
251. h(x) = g⎛⎝2 + f ⎛⎝x2⎞⎠⎞⎠; a = 1
252. h(x) = f ⎛⎝g(sinx)⎞⎠; a = 0
253. [T] The position function of a freight train is given by
s(t) = 100(t + 1)−2, with s in meters and t in seconds.
At time t = 6 s, find the train’s


a. velocity andb. acceleration.c. Using a. and b. is the train speeding up or slowingdown?
254. [T] A mass hanging from a vertical spring is insimple harmonic motion as given by the following positionfunction, where t is measured in seconds and s is in
inches: s(t) = −3cos⎛⎝πt + π4⎞⎠.


a. Determine the position of the spring at t = 1.5 s.
b. Find the velocity of the spring at t = 1.5 s.


255. [T] The total cost to produce x boxes of Thin Mint
Girl Scout cookies is C dollars, where
C = 0.0001x3 − 0.02x2 + 3x + 300. In t weeks
production is estimated to be x = 1600 + 100t boxes.


a. Find the marginal cost C′ (x).
b. Use Leibniz’s notation for the chain rule,


dC
dt


= dC
dx


· dx
dt


, to find the rate with respect to
time t that the cost is changing.


c. Use b. to determine how fast costs are increasingwhen t = 2 weeks. Include units with the answer.
256. [T] The formula for the area of a circle is A = πr2,
where r is the radius of the circle. Suppose a circle is
expanding, meaning that both the area A and the radius r
(in inches) are expanding.


a. Suppose r = 2 − 100
(t + 7)2


where t is time in
seconds. Use the chain rule dA


dt
= dA


dr
· dr
dt


to find
the rate at which the area is expanding.b. Use a. to find the rate at which the area isexpanding at t = 4 s.


257. [T] The formula for the volume of a sphere is
S = 4


3
πr3, where r (in feet) is the radius of the sphere.


Suppose a spherical snowball is melting in the sun.
a. Suppose r = 1


(t + 1)2
− 1


12
where t is time in


minutes. Use the chain rule dS
dt


= dS
dr


· dr
dt


to find
the rate at which the snowball is melting.b. Use a. to find the rate at which the volume ischanging at t = 1 min.


258. [T] The daily temperature in degrees Fahrenheit ofPhoenix in the summer can be modeled by the function
T(x) = 94 − 10cos⎡⎣


π
12


(x − 2)⎤⎦, where x is hours after
midnight. Find the rate at which the temperature ischanging at 4 p.m.
259. [T] The depth (in feet) of water at a dock changeswith the rise and fall of tides. The depth is modeled by
the function D(t) = 5sin⎛⎝π6 t − 7π6 ⎞⎠+ 8, where t is the
number of hours after midnight. Find the rate at which thedepth is changing at 6 a.m.


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3.7 | Derivatives of Inverse Functions
Learning Objectives


3.7.1 Calculate the derivative of an inverse function.
3.7.2 Recognize the derivatives of the standard inverse trigonometric functions.


In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functionswhose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use thelimit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometricfunctions. This formula may also be used to extend the power rule to rational exponents.
The Derivative of an Inverse Function
We begin by considering a function and its inverse. If f (x) is both invertible and differentiable, it seems reasonable that
the inverse of f (x) is also differentiable. Figure 3.28 shows the relationship between a function f (x) and its inverse
f −1 (x). Look at the point ⎛⎝a, f −1 (a)⎞⎠ on the graph of f −1(x) having a tangent line with a slope of ⎛⎝ f −1⎞⎠′ (a) = pq . This
point corresponds to a point ⎛⎝ f −1 (a), a⎞⎠ on the graph of f (x) having a tangent line with a slope of f ′ ⎛⎝ f −1 (a)⎞⎠ = qp.
Thus, if f −1(x) is differentiable at a, then it must be the case that



⎝ f


−1⎞
⎠′ (a) =


1
f ′ ⎛⎝ f


−1 (a)⎞⎠
.


Figure 3.28 The tangent lines of a function and its inverse arerelated; so, too, are the derivatives of these functions.


We may also derive the formula for the derivative of the inverse by first recalling that x = f ⎛⎝ f −1 (x)⎞⎠. Then by
differentiating both sides of this equation (using the chain rule on the right), we obtain


1 = f ′ ⎛⎝ f
−1 (x)⎞⎠



⎝ f


−1 )′(x)⎞⎠.


Solving for ( f −1 )′(x), we obtain
(3.19)⎛


⎝ f
−1⎞
⎠′ (x) =


1
f ′ ⎛⎝ f


−1 (x)⎞⎠
.


We summarize this result in the following theorem.


Chapter 3 | Derivatives 299




3.42


Theorem 3.11: Inverse Function Theorem
Let f (x) be a function that is both invertible and differentiable. Let y = f −1 (x) be the inverse of f (x). For all x
satisfying f ′ ⎛⎝ f −1 (x)⎞⎠ ≠ 0,


dy
dx


= d
dx

⎝ f


−1(x)⎞⎠ =

⎝ f


−1⎞
⎠′ (x) =


1
f ′ ⎛⎝ f


−1 (x)⎞⎠
.


Alternatively, if y = g(x) is the inverse of f (x), then
g(x) = 1


f ′ ⎛⎝g(x)⎞⎠
.


Example 3.60
Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of g(x) = x + 2x . Compare the resulting derivative to that
obtained by differentiating the function directly.
Solution
The inverse of g(x) = x + 2x is f (x) = 2x − 1. Since g′ (x) = 1f ′ ⎛⎝g(x)⎞⎠, begin by finding f ′ (x). Thus,


f ′ (x) = −2
(x − 1)2


and f ′ ⎛⎝g(x)⎞⎠ = −2

⎝g(x) − 1⎞⎠2


= −2


x + 2
x − 1





2
= − x


2


2
.


Finally,
g′ (x) = 1


f ′ ⎛⎝g(x)⎞⎠
= − 2


x2
.


We can verify that this is the correct derivative by applying the quotient rule to g(x) to obtain
g′ (x) = − 2


x2
.


Use the inverse function theorem to find the derivative of g(x) = 1
x + 2


. Compare the result obtained
by differentiating g(x) directly.


Example 3.61
Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of g(x) = x3 .


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3.43


Solution
The function g(x) = x3 is the inverse of the function f (x) = x3. Since g′ (x) = 1


f ′ ⎛⎝g(x)⎞⎠
, begin by finding


f ′ (x). Thus,
f ′ (x) = 3x3 and f ′ ⎛⎝g(x)⎞⎠ = 3⎛⎝ x


3 ⎞


2
= 3x2/3.


Finally,
g′ (x) = 1


3x2/3
= 1


3
x−2/3.


Find the derivative of g(x) = x5 by applying the inverse function theorem.


From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of
the form 1n, where n is a positive integer. This extension will ultimately allow us to differentiate xq, where q is any
rational number.
Theorem 3.12: Extending the Power Rule to Rational Exponents
The power rule may be extended to rational exponents. That is, if n is a positive integer, then


(3.20)d
dx

⎝x


1/n⎞
⎠ =


1
nx


(1/n) − 1.


Also, if n is a positive integer and m is an arbitrary integer, then
(3.21)d


dx

⎝x


m/n⎞
⎠ =


m
n x


(m/n) − 1.


Proof
The function g(x) = x1/n is the inverse of the function f (x) = xn. Since g′ (x) = 1


f ′ ⎛⎝g(x)⎞⎠
, begin by finding f ′ (x).


Thus,
f ′ (x) = nxn − 1 and f ′ ⎛⎝g(x)⎞⎠ = n(x1/n)n − 1 = nx


(n − 1)/n.


Finally,
g′ (x) = 1


nx
(n − 1)/n


= 1nx
(1 − n)/n = 1nx


(1/n) − 1.


To differentiate xm/n we must rewrite it as ⎛⎝x1/n⎞⎠m and apply the chain rule. Thus,
d
dx

⎝x


m/n⎞
⎠ =


d
dx



⎝x


1/n⎞


m⎞
⎠ = m



⎝x


1/n⎞


m − 1
· 1nx


(1/n) − 1 = mn x
(m/n) − 1.




Chapter 3 | Derivatives 301




3.44


Example 3.62
Applying the Power Rule to a Rational Power
Find the equation of the line tangent to the graph of y = x2/3 at x = 8.
Solution
First find dy


dx
and evaluate it at x = 8. Since


dy
dx


= 2
3
x−1/3 and


dy
dx |x = 8 = 13


the slope of the tangent line to the graph at x = 8 is 1
3
.


Substituting x = 8 into the original function, we obtain y = 4. Thus, the tangent line passes through the point
(8, 4). Substituting into the point-slope formula for a line, we obtain the tangent line


y = 1
3
x + 4


3
.


Find the derivative of s(t) = 2t + 1.


Derivatives of Inverse Trigonometric Functions
We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluablein the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in thattheir derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraicfunctions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time,we see that the derivative of a function need not be of the same type as the original function.
Example 3.63
Derivative of the Inverse Sine Function
Use the inverse function theorem to find the derivative of g(x) = sin−1 x.
Solution
Since for x in the interval ⎡⎣−π2, π2⎤⎦, f (x) = sinx is the inverse of g(x) = sin−1 x, begin by finding f ′(x).
Since


f ′ (x) = cosx and f ′ ⎛⎝g(x)⎞⎠ = cos ⎛⎝sin
−1 x⎞⎠ = 1 − x


2,


we see that
g′ (x) = d


dx

⎝sin


−1 x⎞⎠ =
1


f ′ ⎛⎝g(x)⎞⎠
= 1


1 − x2
.


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Analysis
To see that cos ⎛⎝sin−1 x⎞⎠ = 1 − x2, consider the following argument. Set sin−1 x = θ. In this case, sinθ = x
where −π


2
≤ θ ≤ π


2
. We begin by considering the case where 0 < θ < π


2
. Since θ is an acute angle, we may


construct a right triangle having acute angle θ, a hypotenuse of length 1 and the side opposite angle θ having
length x. From the Pythagorean theorem, the side adjacent to angle θ has length 1 − x2. This triangle is
shown in Figure 3.29. Using the triangle, we see that cos ⎛⎝sin−1 x⎞⎠ = cosθ = 1 − x2.


Figure 3.29 Using a right triangle having acute angle θ, a
hypotenuse of length 1, and the side opposite angle θ having
length x, we can see that cos ⎛⎝sin−1 x⎞⎠ = cosθ = 1 − x2.


In the case where −π
2
< θ < 0, we make the observation that 0 < −θ < π


2
and hence


cos ⎛⎝sin
−1 x⎞⎠ = cosθ = cos(−θ) = 1 − x


2.


Now if θ = π
2
or θ = − π


2
, x = 1 or x = −1, and since in either case cosθ = 0 and 1 − x2 = 0, we have


cos ⎛⎝sin
−1 x⎞⎠ = cosθ = 1 − x


2.


Consequently, in all cases, cos ⎛⎝sin−1 x⎞⎠ = 1 − x2.


Example 3.64
Applying the Chain Rule to the Inverse Sine Function
Apply the chain rule to the formula derived in Example 3.61 to find the derivative of h(x) = sin−1 ⎛⎝g(x)⎞⎠ and
use this result to find the derivative of h(x) = sin−1 ⎛⎝2x3⎞⎠.
Solution
Applying the chain rule to h(x) = sin−1 ⎛⎝g(x)⎞⎠, we have


h′ (x) = 1
1 − ⎛⎝g(x)⎞⎠2


g′ (x).


Now let g(x) = 2x3, so g′ (x) = 6x. Substituting into the previous result, we obtain


Chapter 3 | Derivatives 303




3.45


h′ (x) = 1


1 − 4x6
· 6x


= 6x


1 − 4x6
.


Use the inverse function theorem to find the derivative of g(x) = tan−1 x.


The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem.These formulas are provided in the following theorem.
Theorem 3.13: Derivatives of Inverse Trigonometric Functions


(3.22)d
dx


sin−1 x = 1
1 − (x)2


(3.23)d
dx


cos−1 x = −1
1 − (x)2


(3.24)d
dx


tan−1 x = 1
1 + (x)2


(3.25)d
dx


cot−1 x = −1
1 + (x)2


(3.26)d
dx


sec−1 x = 1


|x| (x)2 − 1


(3.27)d
dx


csc−1 x = −1


|x| (x)2 − 1


Example 3.65
Applying Differentiation Formulas to an Inverse Tangent Function
Find the derivative of f (x) = tan−1 ⎛⎝x2⎞⎠.
Solution
Let g(x) = x2, so g′ (x) = 2x. Substituting into Equation 3.24, we obtain


f ′ (x) = 1


1 + ⎛⎝x
2⎞


2
· (2x).


Simplifying, we have
f ′ (x) = 2x


1 + x4
.


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3.46


3.47


Example 3.66
Applying Differentiation Formulas to an Inverse Sine Function
Find the derivative of h(x) = x2 sin−1 x.
Solution
By applying the product rule, we have


h′ (x) = 2xsin−1 x + 1
1 − x2


· x2.


Find the derivative of h(x) = cos−1 (3x − 1).


Example 3.67
Applying the Inverse Tangent Function
The position of a particle at time t is given by s(t) = tan−1 ⎛⎝1t ⎞⎠ for t ≥ 12. Find the velocity of the particle at
time t = 1.
Solution
Begin by differentiating s(t) in order to find v(t). Thus,


v(t) = s′ (t) = 1


1 + ⎛⎝1t



2
· −1
t2


.


Simplifying, we have
v(t) = − 1


t2 + 1
.


Thus, v(1) = − 1
2
.


Find the equation of the line tangent to the graph of f (x) = sin−1 x at x = 0.


Chapter 3 | Derivatives 305




3.7 EXERCISES
For the following exercises, use the graph of y = f (x) to


a. sketch the graph of y = f −1 (x), and
b. use part a. to estimate ⎛⎝ f −1⎞⎠′ (1).


260.


261.


262.


263.


For the following exercises, use the functions y = f (x) to
find


a. d f
dx


at x = a and
b. x = f −1 (y).
c. Then use part b. to find d f −1


dy
at y = f (a).


264. f (x) = 6x − 1, x = −2
265. f (x) = 2x3 − 3, x = 1
266. f (x) = 9 − x2, 0 ≤ x ≤ 3, x = 2
267. f (x) = sinx, x = 0
For each of the following functions, find ⎛⎝ f −1⎞⎠′ (a).
268. f (x) = x2 + 3x + 2, x ≥ −1, a = 2
269. f (x) = x3 + 2x + 3, a = 0
270. f (x) = x + x, a = 2
271. f (x) = x − 2x , x < 0, a = 1
272. f (x) = x + sinx, a = 0
273. f (x) = tanx + 3x2, a = 0
For each of the given functions y = f (x),


a. find the slope of the tangent line to its inverse
function f −1 at the indicated point P, and


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b. find the equation of the tangent line to the graph of
f −1 at the indicated point.


274. f (x) = 4
1 + x2


, P(2, 1)


275. f (x) = x − 4, P(2, 8)


276. f (x) = ⎛⎝x3 + 1⎞⎠4, P(16, 1)
277. f (x) = −x3 − x + 2, P(−8, 2)
278. f (x) = x5 + 3x3 − 4x − 8, P(−8, 1)
For the following exercises, find dy


dx
for the given


function.
279. y = sin−1 ⎛⎝x2⎞⎠
280. y = cos−1 ( x)
281. y = sec−1 ⎛⎝1x⎞⎠
282. y = csc−1 x


283. y = ⎛⎝1 + tan−1 x⎞⎠3


284. y = cos−1 (2x) · sin−1 (2x)
285. y = 1


tan−1 (x)


286. y = sec−1 (−x)
287. y = cot−1 4 − x2
288. y = x · csc−1 x
For the following exercises, use the given values to find

⎝ f


−1⎞
⎠′ (a).


289. f (π) = 0, f ′(π) = −1, a = 0
290. f (6) = 2, f ′ (6) = 1


3
, a = 2


291. f ⎛⎝13⎞⎠ = −8, f ′⎛⎝13⎞⎠ = 2, a = −8


292. f ⎛⎝ 3⎞⎠ = 12, f ′⎛⎝ 3⎞⎠ = 23, a = 12
293. f (1) = −3, f ′(1) = 10, a = −3
294. f (1) = 0, f ′(1) = −2, a = 0
295. [T] The position of a moving hockey puck after t
seconds is s(t) = tan−1 t where s is in meters.


a. Find the velocity of the hockey puck at any time t.
b. Find the acceleration of the puck at any time t.
c. Evaluate a. and b. for t = 2, 4, and 6 seconds.
d. What conclusion can be drawn from the results inc.?


296. [T] A building that is 225 feet tall casts a shadowof various lengths x as the day goes by. An angle of
elevation θ is formed by lines from the top and bottom
of the building to the tip of the shadow, as seen in thefollowing figure. Find the rate of change of the angle of
elevation dθ


dx
when x = 272 feet.


297. [T] A pole stands 75 feet tall. An angle θ is formed
when wires of various lengths of x feet are attached from
the ground to the top of the pole, as shown in the following
figure. Find the rate of change of the angle dθ


dx
when a wire


of length 90 feet is attached.


Chapter 3 | Derivatives 307




298. [T] A television camera at ground level is 2000 feetaway from the launching pad of a space rocket that isset to take off vertically, as seen in the following figure.The angle of elevation of the camera can be found by
θ = tan−1 ⎛⎝


x
2000

⎠, where x is the height of the rocket.


Find the rate of change of the angle of elevation afterlaunch when the camera and the rocket are 5000 feet apart.


299. [T] A local movie theater with a 30-foot-high screenthat is 10 feet above a person’s eye level when seatedhas a viewing angle θ (in radians) given by
θ = cot−1 x


40
− cot−1 x


10
, where x is the distance in


feet away from the movie screen that the person is sitting,as shown in the following figure.


a. Find dθ
dx


.


b. Evaluate dθ
dx


for x = 5, 10, 15, and 20.
c. Interpret the results in b..
d. Evaluate dθ


dx
for x = 25, 30, 35, and 40


e. Interpret the results in d. At what distance x should
the person stand to maximize his or her viewingangle?


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3.8 | Implicit Differentiation
Learning Objectives


3.8.1 Find the derivative of a complicated function by using implicit differentiation.
3.8.2 Use implicit differentiation to determine the equation of a tangent line.


We have already studied how to find equations of tangent lines to functions and the rate of change of a function at a specificpoint. In all these cases we had the explicit equation for the function and differentiated these functions explicitly. Supposeinstead that we want to determine the equation of a tangent line to an arbitrary curve or the rate of change of an arbitrarycurve at a point. In this section, we solve these problems by finding the derivatives of functions that define y implicitly in
terms of x.
Implicit Differentiation
In most discussions of math, if the dependent variable y is a function of the independent variable x, we express y in terms
of x. If this is the case, we say that y is an explicit function of x. For example, when we write the equation y = x2 + 1,
we are defining y explicitly in terms of x. On the other hand, if the relationship between the function y and the variable x
is expressed by an equation where y is not expressed entirely in terms of x, we say that the equation defines y implicitly
in terms of x. For example, the equation y − x2 = 1 defines the function y = x2 + 1 implicitly.
Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical linetest). We are using the idea that portions of y are functions that satisfy the given equation, but that y is not actually a
function of x.
In general, an equation defines a function implicitly if the function satisfies that equation. An equation may define manydifferent functions implicitly. For example, the functions
y = 25 − x2 and y = ⎧



⎨ 25 − x


2 if − 25 ≤ x < 0


− 25 − x2 if 0 ≤ x ≤ 25
, which are illustrated in Figure 3.30, are just three of the many


functions defined implicitly by the equation x2 + y2 = 25.


Chapter 3 | Derivatives 309




Figure 3.30 The equation x2 + y2 = 25 defines many functions implicitly.


If we want to find the slope of the line tangent to the graph of x2 + y2 = 25 at the point (3, 4), we could evaluate
the derivative of the function y = 25 − x2 at x = 3. On the other hand, if we want the slope of the tangent line at the
point (3, −4), we could use the derivative of y = − 25 − x2. However, it is not always easy to solve for a function
defined implicitly by an equation. Fortunately, the technique of implicit differentiation allows us to find the derivative of
an implicitly defined function without ever solving for the function explicitly. The process of finding dy


dx
using implicit


differentiation is described in the following problem-solving strategy.


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Problem-Solving Strategy: Implicit Differentiation
To perform implicit differentiation on an equation that defines a function y implicitly in terms of a variable x, use
the following steps:


1. Take the derivative of both sides of the equation. Keep in mind that y is a function of x. Consequently, whereas
d
dx


(sinx) = cosx, d
dx


(siny) = cosy
dy
dx


because we must use the chain rule to differentiate siny with respect
to x.


2. Rewrite the equation so that all terms containing dy
dx


are on the left and all terms that do not contain dy
dx


are
on the right.


3. Factor out dy
dx


on the left.
4. Solve for dy


dx
by dividing both sides of the equation by an appropriate algebraic expression.


Example 3.68
Using Implicit Differentiation
Assuming that y is defined implicitly by the equation x2 + y2 = 25, find dy


dx
.


Solution
Follow the steps in the problem-solving strategy.


d
dx

⎝x


2 + y2⎞⎠ =
d
dx


(25) Step 1. Diffe entiate both sides of the equation.


d
dx

⎝x


2⎞
⎠+


d
dx

⎝y


2⎞
⎠ = 0


Step 1.1. Use the sum rule on the left.


On the right d
dx


(25) = 0.


2x + 2y
dy
dx


= 0
Step 1.2. Take the derivatives, so d


dx

⎝x


2⎞
⎠ = 2x


and d
dx

⎝y


2⎞
⎠ = 2y


dy
dx


.


2y
dy
dx


= −2x
Step 2. Keep the terms with


dy
dx


on the left.


Move the remaining terms to the right.


dy
dx


= − xy
Step 4. Divide both sides of the equation by


2y. (Step 3 does not apply in this case.)


Analysis
Note that the resulting expression for dy


dx
is in terms of both the independent variable x and the dependent


variable y. Although in some cases it may be possible to express dy
dx


in terms of x only, it is generally not
possible to do so.


Chapter 3 | Derivatives 311




Example 3.69
Using Implicit Differentiation and the Product Rule
Assuming that y is defined implicitly by the equation x3 siny + y = 4x + 3, find dy


dx
.


Solution
d
dx

⎝x


3 siny + y⎞⎠ =
d
dx


(4x + 3) Step 1: Diffe entiate both sides of the equation.


d
dx

⎝x


3 siny⎞⎠+
d
dx


(y) = 4
Step 1.1: Apply the sum rule on the left.


On the right, d
dx


(4x + 3) = 4.




d
dx

⎝x


3⎞
⎠ · siny +


d
dx

⎝siny⎞⎠ · x3



⎠+


dy
dx


= 4
Step 1.2: Use the product rule to fin


d
dx

⎝x


3 siny⎞⎠. Observe that
d
dx


(y) =
dy
dx


.


3x2 siny +

⎝cosy


dy
dx

⎠ · x


3 +
dy
dx


= 4
Step 1.3: We know d


dx

⎝x


3⎞
⎠ = 3x


2. Use the


chain rule to obtain d
dx

⎝siny⎞⎠ = cosy


dy
dx


.


x3 cosy
dy
dx


+
dy
dx


= 4 − 3x2 siny
Step 2: Keep all terms containing


dy
dx


on the


left. Move all other terms to the right.


dy
dx

⎝x


3 cosy + 1⎞⎠ = 4 − 3x
2 siny Step 3: Factor out


dy
dx


on the left.


dy
dx


=
4 − 3x2 siny


x3 cosy + 1


Step 4: Solve for
dy
dx


by dividing both sides of


the equation by x3 cosy + 1.


Example 3.70
Using Implicit Differentiation to Find a Second Derivative
Find d2 y


dx2
if x2 + y2 = 25.


Solution
In Example 3.68, we showed that dy


dx
= − xy. We can take the derivative of both sides of this equation to find


d2 y


dx2
.


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3.48


d2 y


dx2
= d


dy

⎝−


x
y

⎠ Diffe entiate both sides of


dy
dx


= − xy.


= −



⎝1 · y − x


dy
dx



y2
Use the quotient rule to fin d


dy

⎝−


x
y

⎠.


=
−y + xdy


dx


y2
Simplify.


=
−y + x⎛⎝−


x
y



y2
Substitute


dy
dx


= − xy.


=
−y2 − x2


y3
Simplify.


At this point we have found an expression for d2 y
dx2


. If we choose, we can simplify the expression further by
recalling that x2 + y2 = 25 and making this substitution in the numerator to obtain d2 y


dx2
= − 25


y3
.


Find dy
dx


for y defined implicitly by the equation 4x5 + tany = y2 + 5x.


Finding Tangent Lines Implicitly
Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations oftangent lines to curves described by equations.
Example 3.71
Finding a Tangent Line to a Circle
Find the equation of the line tangent to the curve x2 + y2 = 25 at the point (3, −4).
Solution
Although we could find this equation without using implicit differentiation, using that method makes it much
easier. In Example 3.68, we found dy


dx
= − xy.


The slope of the tangent line is found by substituting (3, −4) into this expression. Consequently, the slope of the
tangent line is dy


dx |(3, −4) = − 3−4 = 34.
Using the point (3, −4) and the slope 3


4
in the point-slope equation of the line, we obtain the equation


y = 3
4
x − 25


4
(Figure 3.31).


Chapter 3 | Derivatives 313




Figure 3.31 The line y = 3
4
x − 25


4
is tangent to


x2 + y2 = 25 at the point (3, −4).


Example 3.72
Finding the Equation of the Tangent Line to a Curve
Find the equation of the line tangent to the graph of y3 + x3 − 3xy = 0 at the point ⎛⎝32, 32⎞⎠ (Figure 3.32). This
curve is known as the folium (or leaf) of Descartes.


Figure 3.32 Finding the tangent line to the folium of
Descartes at ⎛⎝32, 32⎞⎠.


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Solution
Begin by finding dy


dx
.


d
dx

⎝y


3 + x3 − 3xy⎞⎠ =
d
dx


(0)


3y2
dy
dx


+ 3x2 −

⎝3y +


dy
dx


3x

⎠ = 0


dy
dx


=
3y − 3x2


3y2 − 3x
.


Next, substitute ⎛⎝32, 32⎞⎠ into dydx = 3y − 3x
2


3y2 − 3x
to find the slope of the tangent line:


dy
dx |⎛⎝32, 32⎞⎠ = −1.


Finally, substitute into the point-slope equation of the line to obtain
y = −x + 3.


Example 3.73
Applying Implicit Differentiation
In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation
4x2 + 25y2 = 100. The rocket can fire missiles along lines tangent to its path. The object of the game is to
destroy an incoming asteroid traveling along the positive x-axis toward (0, 0). If the rocket fires a missile when
it is located at ⎛⎝3, 83⎞⎠, where will it intersect the x-axis?


Solution
To solve this problem, we must determine where the line tangent to the graph of
4x2 + 25y2 = 100 at ⎛⎝3, 83⎞⎠ intersects the x-axis. Begin by finding dydx implicitly.
Differentiating, we have


8x + 50y
dy
dx


= 0.


Solving for dy
dx


, we have
dy
dx


= − 4x
25y


.


The slope of the tangent line is dy
dx |⎛⎝3, 83⎞⎠ = − 950. The equation of the tangent line is y = − 950x + 183200. To


Chapter 3 | Derivatives 315




3.49


determine where the line intersects the x-axis, solve 0 = − 9
50


x + 183
200


. The solution is x = 61
3
. The missile


intersects the x-axis at the point ⎛⎝613 , 0⎞⎠.


Find the equation of the line tangent to the hyperbola x2 − y2 = 16 at the point (5, 3).


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3.8 EXERCISES
For the following exercises, use implicit differentiation to
find dy


dx
.


300. x2 − y2 = 4
301. 6x2 + 3y2 = 12
302. x2 y = y − 7
303. 3x3 + 9xy2 = 5x3
304. xy − cos(xy) = 1
305. y x + 4 = xy + 8
306. −xy − 2 = x


7


307. ysin(xy) = y2 + 2
308. (xy)2 + 3x = y2
309. x3 y + xy3 = −8
For the following exercises, find the equation of the tangentline to the graph of the given equation at the indicatedpoint. Use a calculator or computer software to graph thefunction and the tangent line.
310. [T] x4 y − xy3 = −2, (−1, −1)
311. [T] x2 y2 + 5xy = 14, (2, 1)
312. [T] tan(xy) = y, ⎛⎝π4, 1⎞⎠
313. [T] xy2 + sin(πy) − 2x2 = 10, (2, −3)
314. [T] xy + 5x − 7 = − 34y, (1, 2)


315. [T] xy + sin(x) = 1, ⎛⎝π2, 0⎞⎠


316. [T] The graph of a folium of Descartes with equation
2x3 + 2y3 − 9xy = 0 is given in the following graph.


a. Find the equation of the tangent line at the point
(2, 1). Graph the tangent line along with the
folium.b. Find the equation of the normal line to the tangentline in a. at the point (2, 1).


317. For the equation x2 + 2xy − 3y2 = 0,
a. Find the equation of the normal to the tangent lineat the point (1, 1).
b. At what other point does the normal line in a.intersect the graph of the equation?


318. Find all points on the graph of y3 − 27y = x2 − 90
at which the tangent line is vertical.
319. For the equation x2 + xy + y2 = 7,


a. Find the x -intercept(s).
b. Find the slope of the tangent line(s) at thex-intercept(s).c. What does the value(s) in b. indicate about thetangent line(s)?


320. Find the equation of the tangent line to the graph of
the equation sin−1 x + sin−1 y = π


6
at the point ⎛⎝0, 12⎞⎠.


321. Find the equation of the tangent line to the graph of
the equation tan−1 (x + y) = x2 + π


4
at the point (0, 1).


322. Find y′ and y″ for x2 + 6xy − 2y2 = 3.


Chapter 3 | Derivatives 317




323. [T] The number of cell phones produced when x
dollars is spent on labor and y dollars is spent on capital
invested by a manufacturer can be modeled by the equation
60x3/4 y1/4 = 3240.


a. Find dy
dx


and evaluate at the point (81, 16).
b. Interpret the result of a.


324. [T] The number of cars produced when x dollars is
spent on labor and y dollars is spent on capital invested
by a manufacturer can be modeled by the equation
30x1/3 y2/3 = 360. (Both x and y are measured in
thousands of dollars.)


a. Find dy
dx


and evaluate at the point (27, 8).
b. Interpret the result of a.


325. The volume of a right circular cone of radius x
and height y is given by V = 1


3
πx2 y. Suppose that the


volume of the cone is 85πcm3. Find dy
dx


when x = 4 and
y = 16.


For the following exercises, consider a closed rectangularbox with a square base with side x and height y.
326. Find an equation for the surface area of therectangular box, S(x, y).
327. If the surface area of the rectangular box is 78 square
feet, find dy


dx
when x = 3 feet and y = 5 feet.


For the following exercises, use implicit differentiation todetermine y′. Does the answer agree with the formulas we
have previously determined?
328. x = siny
329. x = cosy
330. x = tany


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3.9 | Derivatives of Exponential and Logarithmic
Functions


Learning Objectives
3.9.1 Find the derivative of exponential functions.
3.9.2 Find the derivative of logarithmic functions.
3.9.3 Use logarithmic differentiation to determine the derivative of a function.


So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions.In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction toFunctions and Graphs, exponential functions play an important role in modeling population growth and the decayof radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewritingcomplicated expressions.
Derivative of the Exponential Function
Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmicfunctions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that theseassumptions hold are beyond the scope of this course.
First of all, we begin with the assumption that the function B(x) = bx, b > 0, is defined for every real number and is
continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginningwith the definition of bn, where n is a positive integer—as the product of b multiplied by itself n times. Later,
we defined b0 = 1, b−n = 1


bn
, for a positive integer n, and bs/t = ( bt )s for positive integers s and t. These


definitions leave open the question of the value of br where r is an arbitrary real number. By assuming the continuity of
B(x) = bx, b > 0, we may interpret br as lim


x → r
bx where the values of x as we take the limit are rational. For example,


we may view 4π as the number satisfying
43 < 4π < 44, 43.1 < 4π < 43.2, 43.14 < 4π < 43.15,


43.141 < 4π < 43.142, 43.1415 < 4π < 43.1416 ,….


As we see in the following table, 4π ≈ 77.88.


Chapter 3 | Derivatives 319




x 4x x 4x


43 64 43.141593 77.8802710486


43.1 73.5166947198 43.1416 77.8810268071


43.14 77.7084726013 43.142 77.9242251944


43.141 77.8162741237 43.15 78.7932424541


43.1415 77.8702309526 43.2 84.4485062895


43.14159 77.8799471543 44 256
Table 3.7 Approximating a Value of 4π


We also assume that for B(x) = bx, b > 0, the value B′ (0) of the derivative exists. In this section, we show that by
making this one additional assumption, it is possible to prove that the function B(x) is differentiable everywhere.
We make one final assumption: that there is a unique value of b > 0 for which B′ (0) = 1. We define e to be this
unique value, as we did in Introduction to Functions and Graphs. Figure 3.33 provides graphs of the functions
y = 2x, y = 3x, y = 2.7x, and y = 2.8x. A visual estimate of the slopes of the tangent lines to these functions at 0
provides evidence that the value of e lies somewhere between 2.7 and 2.8. The function E(x) = ex is called the natural
exponential function. Its inverse, L(x) = loge x = lnx is called the natural logarithmic function.


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Figure 3.33 The graph of E(x) = ex is between y = 2x and y = 3x.


For a better estimate of e, we may construct a table of estimates of B′ (0) for functions of the form B(x) = bx. Before
doing this, recall that


B′ (0) = lim
x → 0


bx − b0
x − 0


= lim
x → 0


bx − 1
x ≈


bx − 1
x


for values of x very close to zero. For our estimates, we choose x = 0.00001 and x = −0.00001 to obtain the estimate
b−0.00001 − 1
−0.00001


< B′ (0) < b
0.00001 − 1
0.00001


.


See the following table.


Chapter 3 | Derivatives 321




b b−0.00001−1
−0.00001


< B′ (0) < b0.00001−1
0.00001


b b−0.00001−1
−0.00001


< B′ (0) < b0.00001−1
0.00001


2 0.693145 < B′ (0) < 0.69315 2.7183 1.000002 < B′ (0) < 1.000012


2.7 0.993247 < B′ (0) < 0.993257 2.719 1.000259 < B′ (0) < 1.000269


2.71 0.996944 < B′ (0) < 0.996954 2.72 1.000627 < B′ (0) < 1.000637


2.718 0.999891 < B′ (0) < 0.999901 2.8 1.029614 < B′ (0) < 1.029625


2.7182 0.999965 < B′ (0) < 0.999975 3 1.098606 < B′ (0) < 1.098618


Table 3.8 Estimating a Value of e
The evidence from the table suggests that 2.7182 < e < 2.7183.
The graph of E(x) = ex together with the line y = x + 1 are shown in Figure 3.34. This line is tangent to the graph of
E(x) = ex at x = 0.


Figure 3.34 The tangent line to E(x) = ex at x = 0 has
slope 1.


Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of
B(x) = bx, b > 0. Recall that we have assumed that B′ (0) exists. By applying the limit definition to the derivative we
conclude that


(3.28)
B′ (0) = lim


h → 0
b0 + h − b0


h
= lim


h → 0
bh − 1


h
.


Turning to B′ (x), we obtain the following.


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B′ (x) = lim
h → 0


bx + h − bx
h


Apply the limit definition of he derivative.


= lim
h → 0


bx bh − bx
h


Note that bx + h = bx bh.


= lim
h → 0


bx(bh − 1)
h


Factor out bx.


= bx lim
h → 0


bh − 1
h


Apply a property of limits.


= bxB′ (0) Use B′ (0) = lim
h → 0


b0 + h − b0
h


= lim
h → 0


bh − 1
h


.


We see that on the basis of the assumption that B(x) = bx is differentiable at 0, B(x) is not only differentiable everywhere,
but its derivative is


(3.29)B′ (x) = bxB′ (0).
For E(x) = ex, E′ (0) = 1. Thus, we have E′ (x) = ex. (The value of B′ (0) for an arbitrary function of the form
B(x) = bx, b > 0, will be derived later.)


Theorem 3.14: Derivative of the Natural Exponential Function
Let E(x) = ex be the natural exponential function. Then


E′ (x) = ex.


In general,
d
dx

⎝e


g(x)⎞
⎠ = e


g(x)
g′ (x).


Example 3.74
Derivative of an Exponential Function
Find the derivative of f (x) = etan(2x).
Solution
Using the derivative formula and the chain rule,


f ′ (x) = etan(2x) d
dx

⎝tan(2x)⎞⎠


= etan(2x) sec2 (2x) · 2.


Example 3.75
Combining Differentiation Rules
Find the derivative of y = ex2x .


Chapter 3 | Derivatives 323




3.50


3.51


Solution
Use the derivative of the natural exponential function, the quotient rule, and the chain rule.


y′ =



⎝e


x2 · 2

⎠x · x − 1 · e


x2


x2
Apply the quotient rule.


=
ex


2 ⎛
⎝2x


2 − 1⎞⎠


x2
Simplify.


Find the derivative of h(x) = xe2x.


Example 3.76
Applying the Natural Exponential Function
A colony of mosquitoes has an initial population of 1000. After t days, the population is given by
A(t) = 1000e0.3t. Show that the ratio of the rate of change of the population, A′ (t), to the population, A(t) is
constant.
Solution
First find A′ (t). By using the chain rule, we have A′ (t) = 300e0.3t. Thus, the ratio of the rate of change of the
population to the population is given by


A′ (t) = 300e
0.3t


1000e0.3t
= 0.3.


The ratio of the rate of change of the population to the population is the constant 0.3.


If A(t) = 1000e0.3t describes the mosquito population after t days, as in the preceding example, what
is the rate of change of A(t) after 4 days?


Derivative of the Logarithmic Function
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivativeof its inverse, the natural logarithmic function.
Theorem 3.15: The Derivative of the Natural Logarithmic Function
If x > 0 and y = lnx, then


(3.30)dy
dx


= 1x .


More generally, let g(x) be a differentiable function. For all values of x for which g′ (x) > 0, the derivative of


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h(x) = ln ⎛⎝g(x)⎞⎠ is given by
(3.31)h′ (x) = 1


g(x)
g′ (x).


Proof
If x > 0 and y = lnx, then ey = x. Differentiating both sides of this equation results in the equation


ey
dy
dx


= 1.


Solving for dy
dx


yields
dy
dx


= 1
ey


.


Finally, we substitute x = ey to obtain
dy
dx


= 1x .


We may also derive this result by applying the inverse function theorem, as follows. Since y = g(x) = lnx is the inverse
of f (x) = ex, by applying the inverse function theorem we have


dy
dx


= 1
f ′ ⎛⎝g(x)⎞⎠


= 1
elnx


= 1x .


Using this result and applying the chain rule to h(x) = ln ⎛⎝g(x)⎞⎠ yields
h′ (x) = 1


g(x)
g′ (x).



The graph of y = lnx and its derivative dy


dx
= 1x are shown in Figure 3.35.


Figure 3.35 The function y = lnx is increasing on
(0, +∞). Its derivative y′ = 1x is greater than zero on
(0, +∞).


Example 3.77
Taking a Derivative of a Natural Logarithm


Chapter 3 | Derivatives 325




3.52


Find the derivative of f (x) = ln ⎛⎝x3 + 3x − 4⎞⎠.
Solution
Use Equation 3.31 directly.


f ′ (x) = 1
x3 + 3x − 4


· ⎛⎝3x
2 + 3⎞⎠ Use g(x) = x


3 + 3x − 4 in h′ (x) = 1
g(x)


g′ (x).


= 3x
2 + 3


x3 + 3x − 4
Rewrite.


Example 3.78
Using Properties of Logarithms in a Derivative
Find the derivative of f (x) = ln⎛⎝x2 sinx2x + 1 ⎞⎠.


Solution
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithmsprior to finding the derivative, we can make the problem much simpler.


f (x) = ln


x2 sinx
2x + 1



⎠ = 2lnx + ln(sinx) − ln(2x + 1) Apply properties of logarithms.


f ′ (x) = 2x + cotx −
2


2x + 1
Apply sum rule and h′ (x) = 1


g(x)
g′ (x).


Differentiate: f (x) = ln(3x + 2)5.


Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y = logb x
and y = bx for b > 0, b ≠ 1.


Theorem 3.16: Derivatives of General Exponential and Logarithmic Functions
Let b > 0, b ≠ 1, and let g(x) be a differentiable function.


i. If, y = logb x, then
(3.32)dy


dx
= 1


x lnb
.


More generally, if h(x) = logb ⎛⎝g(x)⎞⎠, then for all values of x for which g(x) > 0,
(3.33)h′ (x) = g′ (x)


g(x) lnb
.


ii. If y = bx, then


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(3.34)dy
dx


= bx lnb.


More generally, if h(x) = bg(x), then
(3.35)h′ (x) = bg(x)g″(x) lnb.


Proof
If y = logb x, then by = x. It follows that ln(by) = ln x. Thus y ln b = ln x. Solving for y, we have y = lnxlnb.
Differentiating and keeping in mind that lnb is a constant, we see that


dy
dx


= 1
x lnb


.


The derivative in Equation 3.33 now follows from the chain rule.
If y = bx, then ln y = x lnb. Using implicit differentiation, again keeping in mind that lnb is constant, it follows that
1
y
dy
dx


= lnb. Solving for dy
dx


and substituting y = bx, we see that
dy
dx


= y lnb = bx lnb.


The more general derivative (Equation 3.35) follows from the chain rule.

Example 3.79
Applying Derivative Formulas
Find the derivative of h(x) = 3x


3x + 2
.


Solution
Use the quotient rule and Derivatives of General Exponential and Logarithmic Functions.


h′ (x) = 3
x ln3(3x + 2) − 3x ln3(3x)


(3x + 2)2
Apply the quotient rule.


= 2 · 3
x ln3


(3x + 2)2
Simplify.


Example 3.80
Finding the Slope of a Tangent Line
Find the slope of the line tangent to the graph of y = log2 (3x + 1) at x = 1.
Solution


Chapter 3 | Derivatives 327




3.53


To find the slope, we must evaluate dy
dx


at x = 1. Using Equation 3.33, we see that
dy
dx


= 3
ln2(3x + 1)


.


By evaluating the derivative at x = 1, we see that the tangent line has slope
dy
dx |x = 1 = 34ln2 = 3ln16.


Find the slope for the line tangent to y = 3x at x = 2.


Logarithmic Differentiation
At this point, we can take derivatives of functions of the form y = ⎛⎝g(x)⎞⎠n for certain values of n, as well as functions
of the form y = bg(x), where b > 0 and b ≠ 1. Unfortunately, we still do not know the derivatives of functions such as
y = xx or y = xπ. These functions require a technique called logarithmic differentiation, which allows us to differentiate
any function of the form h(x) = g(x) f (x). It can also be used to convert a very complex differentiation problem into a
simpler one, such as finding the derivative of y = x 2x + 1


ex sin3 x
. We outline this technique in the following problem-solving


strategy.
Problem-Solving Strategy: Using Logarithmic Differentiation


1. To differentiate y = h(x) using logarithmic differentiation, take the natural logarithm of both sides of the
equation to obtain ln y = ln ⎛⎝h(x)⎞⎠.


2. Use properties of logarithms to expand ln ⎛⎝h(x)⎞⎠ as much as possible.
3. Differentiate both sides of the equation. On the left we will have 1y dydx.
4. Multiply both sides of the equation by y to solve for dy


dx
.


5. Replace y by h(x).


Example 3.81
Using Logarithmic Differentiation
Find the derivative of y = ⎛⎝2x4 + 1⎞⎠tanx.


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Solution
Use logarithmic differentiation to find this derivative.
lny = ln⎛⎝2x


4 + 1⎞⎠
tanx


Step 1. Take the natural logarithm of both sides.


lny = tanx ln ⎛⎝2x
4 + 1⎞⎠ Step 2. Expand using properties of logarithms.


1
y
dy
dx


= sec2 x ln ⎛⎝2x
4 + 1⎞⎠+


8x3


2x4 + 1
· tanx


Step 3. Diffe entiate both sides. Use the


product rule on the right.


dy
dx


= y ·

⎝sec


2 x ln ⎛⎝2x
4 + 1⎞⎠+


8x3


2x4 + 1
· tanx

⎠ Step 4. Multiply by y on both sides.


dy
dx


= ⎛⎝2x
4 + 1⎞⎠


tanx⎛
⎝sec


2 x ln ⎛⎝2x
4 + 1⎞⎠+


8x3


2x4 + 1
· tanx

⎠ Step 5. Substitute y =



⎝2x


4 + 1⎞⎠
tanx


.


Example 3.82
Using Logarithmic Differentiation
Find the derivative of y = x 2x + 1


ex sin3 x
.


Solution
This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.


lny = ln x 2x + 1
ex sin3 x


Step 1. Take the natural logarithm of both sides.


lny = lnx + 1
2
ln(2x + 1) − x lne − 3lnsinx Step 2. Expand using properties of logarithms.


1
y
dy
dx


= 1x +
1


2x + 1
− 1 − 3cosx


sinx
Step 3. Diffe entiate both sides.


dy
dx


= y⎛⎝
1
x +


1
2x + 1


− 1 − 3cotx⎞⎠ Step 4. Multiply by y on both sides.


dy
dx


= x 2x + 1
ex sin3 x




1
x +


1
2x + 1


− 1 − 3cotx⎞⎠ Step 5. Substitute y =
x 2x + 1
ex sin3 x


.


Example 3.83
Extending the Power Rule
Find the derivative of y = xr where r is an arbitrary real number.
Solution
The process is the same as in Example 3.82, though with fewer complications.


Chapter 3 | Derivatives 329




3.54


3.55


lny = lnxr Step 1. Take the natural logarithm of both sides.


lny = r lnx Step 2. Expand using properties of logarithms.


1
y
dy
dx


= r1x Step 3. Diffe entiate both sides.


dy
dx


= yrx Step 4. Multiply by y on both sides.


dy
dx


= xr rx Step 5. Substitute y = x
r.


dy
dx


= rxr − 1 Simplify.


Use logarithmic differentiation to find the derivative of y = xx.


Find the derivative of y = (tanx)π.


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3.9 EXERCISES
For the following exercises, find f ′ (x) for each function.
331. f (x) = x2 ex


332. f (x) = e−xx
333. f (x) = ex3 lnx


334. f (x) = e2x + 2x
335. f (x) = ex − e−x


ex + e−x


336. f (x) = 10x
ln10


337. f (x) = 24x + 4x2
338. f (x) = 3sin3x
339. f (x) = xπ · π x
340. f (x) = ln ⎛⎝4x3 + x⎞⎠
341. f (x) = ln 5x − 7
342. f (x) = x2 ln9x
343. f (x) = log(secx)


344. f (x) = log7 ⎛⎝6x4 + 3⎞⎠5


345. f (x) = 2x · log37x2 − 4
For the following exercises, use logarithmic differentiation
to find dy


dx
.


346. y = x x
347. y = (sin2x)4x


348. y = (lnx)lnx


349. y = x log2 x


350. y = ⎛⎝x2 − 1⎞⎠lnx


351. y = xcotx


352. y = x + 11
x2 − 4


3


353. y = x−1/2 ⎛⎝x2 + 3⎞⎠2/3 (3x − 4)4
354. [T] Find an equation of the tangent line to the graph
of f (x) = 4xe



⎝x
2 − 1⎞⎠ at the point where x = −1. Graph


both the function and the tangent line.
355. [T] Find the equation of the line that is normal to thegraph of f (x) = x · 5x at the point where x = 1. Graph
both the function and the normal line.
356. [T] Find the equation of the tangent line to the graph
of x3 − x lny + y3 = 2x + 5 at the point where x = 2.
(Hint: Use implicit differentiation to find dy


dx
.) Graph both


the curve and the tangent line.
357. Consider the function y = x1/x for x > 0.


a. Determine the points on the graph where thetangent line is horizontal.b. Determine the points on the graph where y′ > 0
and those where y′ < 0.


Chapter 3 | Derivatives 331




358. The formula I(t) = sin t
et


is the formula for a
decaying alternating current.a. Complete the following table with the appropriatevalues.


t sint
et


0 (i)
π
2


(ii)


π (iii)



2


(iv)


2π (v)


2π (vi)


3π (vii)



2


(viii)


4π (ix)


b. Using only the values in the table, determine wherethe tangent line to the graph of I(t) is horizontal.
359. [T] The population of Toledo, Ohio, in 2000 wasapproximately 500,000. Assume the population isincreasing at a rate of 5% per year.a. Write the exponential function that relates the totalpopulation as a function of t.


b. Use a. to determine the rate at which the populationis increasing in t years.
c. Use b. to determine the rate at which the populationis increasing in 10 years.


360. [T] An isotope of the element erbium has a half-lifeof approximately 12 hours. Initially there are 9 grams of theisotope present.a. Write the exponential function that relates theamount of substance remaining as a function of t,
measured in hours.b. Use a. to determine the rate at which the substanceis decaying in t hours.


c. Use b. to determine the rate of decay at t = 4
hours.


361. [T] The number of cases of influenza in New YorkCity from the beginning of 1960 to the beginning of 1961is modeled by the function
N(t) = 5.3e0.093t


2 − 0.87t, (0 ≤ t ≤ 4), where N(t)
gives the number of cases (in thousands) and t is measuredin years, with t = 0 corresponding to the beginning of
1960.a. Show work that evaluates N(0) and N(4). Briefly


describe what these values indicate about thedisease in New York City.b. Show work that evaluates N′ (0) and N′ (3).
Briefly describe what these values indicate aboutthe disease in the United States.


362. [T] The relative rate of change of a differentiable
function y = f (x) is given by 100 · f ′ (x)


f (x)
%. One model


for population growth is a Gompertz growth function,
given by P(x) = ae−b · e−cx where a, b, and c are
constants.a. Find the relative rate of change formula for thegeneric Gompertz function.b. Use a. to find the relative rate of change of apopulation in x = 20 months when


a = 204, b = 0.0198, and c = 0.15.
c. Briefly interpret what the result of b. means.


For the following exercises, use the population of NewYork City from 1790 to 1860, given in the following table.


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Years since 1790 Population
0 33,131
10 60,515
20 96,373
30 123,706
40 202,300
50 312,710
60 515,547
70 813,669


Table 3.9 New York City Population OverTime Source: http://en.wikipedia.org/wiki/Largest_cities_in_the_United_States_by_population_by_decade.
363. [T] Using a computer program or a calculator, fit a
growth curve to the data of the form p = abt.
364. [T] Using the exponential best fit for the data, writea table containing the derivatives evaluated at each year.
365. [T] Using the exponential best fit for the data, writea table containing the second derivatives evaluated at eachyear.
366. [T] Using the tables of first and second derivativesand the best fit, answer the following questions:a. Will the model be accurate in predicting the futurepopulation of New York City? Why or why not?b. Estimate the population in 2010. Was the predictioncorrect from a.?


Chapter 3 | Derivatives 333




acceleration
amount of change
average rate of change
chain rule
constant multiple rule


constant rule
derivative
derivative function
difference quotient


difference rule


differentiable at a
differentiable function
differentiable on S
differentiation
higher-order derivative
implicit differentiation


instantaneous rate of change


logarithmic differentiation
marginal cost
marginal profit
marginal revenue


CHAPTER 3 REVIEW
KEY TERMS


is the rate of change of the velocity, that is, the derivative of velocity
the amount of a function f (x) over an interval ⎡⎣x, x + h⎤⎦ is f (x + h) − f (x)


is a function f (x) over an interval ⎡⎣x, x + h⎤⎦ is f (x + h) − f (a)
b − a


the chain rule defines the derivative of a composite function as the derivative of the outer function evaluatedat the inner function times the derivative of the inner function
the derivative of a constant cmultiplied by a function f is the same as the constant multiplied by


the derivative: d
dx

⎝c f (x)⎞⎠ = c f ′ (x)


the derivative of a constant function is zero: d
dx


(c) = 0, where c is a constant
the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, isthe derivative


gives the derivative of a function at each point in the domain of the original function for which thederivative is defined
of a function f (x) at a is given by


f (a + h) − f (a)
h


or
f (x) − f (a)


x − a


the derivative of the difference of a function f and a function g is the same as the difference of the
derivative of f and the derivative of g: d


dx

⎝ f (x) − g(x)⎞⎠ = f ′ (x) − g′ (x)


a function for which f ′(a) exists is differentiable at a
a function for which f ′(x) exists is a differentiable function


a function for which f ′(x) exists for each x in the open set S is differentiable on S
the process of taking a derivative


a derivative of a derivative, from the second derivative to the nth derivative, is called a higher-order derivative
is a technique for computing dy


dx
for a function defined by an equation, accomplished by


differentiating both sides of the equation (remembering to treat the variable y as a function) and solving for dy
dx


the rate of change of a function at any point along the function a, also called f ′(a),
or the derivative of the function at a


is a technique that allows us to differentiate a function by first taking the natural logarithmof both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly
is the derivative of the cost function, or the approximate cost of producing one more item
is the derivative of the profit function, or the approximate profit obtained by producing and selling onemore item
is the derivative of the revenue function, or the approximate revenue obtained by selling one moreitem


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population growth rate
power rule


product rule


quotient rule


speed
sum rule


is the derivative of the population with respect to time
the derivative of a power function is a function in which the power on x becomes the coefficient of the term


and the power on x in the derivative decreases by 1: If n is an integer, then d
dx


xn = nxn − 1


the derivative of a product of two functions is the derivative of the first function times the second function
plus the derivative of the second function times the first function: d


dx

⎝ f (x)g(x)⎞⎠ = f ′ (x)g(x) + g′ (x) f (x)


the derivative of the quotient of two functions is the derivative of the first function times the secondfunction minus the derivative of the second function times the first function, all divided by the square of the second
function: d


dx


f (x)
g(x)

⎠ =


f ′ (x)g(x) − g′ (x) f (x)

⎝g(x)⎞⎠2


is the absolute value of velocity, that is, |v(t)| is the speed of an object at time t whose velocity is given by v(t)
the derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the


derivative of g: d
dx

⎝ f (x) + g(x)⎞⎠ = f ′ (x) + g′ (x)


KEY EQUATIONS
• Difference quotient


Q =
f (x) − f (a)


x − a


• Difference quotient with increment h
Q =


f (a + h) − f (a)
a + h − a


=
f (a + h) − f (a)


h


• Slope of tangent line
mtan = limx → a


f (x) − f (a)
x − a


mtan = limh → 0
f (a + h) − f (a)


h


• Derivative of f (x) at a
f ′ (a) = lim


x → a
f (x) − f (a)


x − a


f ′(a) = lim
h → 0


f (a + h) − f (a)
h


• Average velocity
vave =


s(t) − s(a)
t − a


• Instantaneous velocity
v(a) = s′ (a) = lim


t → a
s(t) − s(a)


t − a


• The derivative function
f ′ (x) = lim


h → 0


f (x + h) − f (x)
h


• Derivative of sine function
d
dx


(sinx) = cosx


• Derivative of cosine function
d
dx


(cosx) = −sinx


• Derivative of tangent function


Chapter 3 | Derivatives 335




d
dx


(tanx) = sec2 x


• Derivative of cotangent function
d
dx


(cotx) = −csc2 x


• Derivative of secant function
d
dx


(secx) = secx tanx


• Derivative of cosecant function
d
dx


(cscx) = −cscxcotx


• The chain rule
h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x)


• The power rule for functions
h′ (x) = n⎛⎝g(x)⎞⎠n − 1g′ (x)


• Inverse function theorem

⎝ f


−1⎞
⎠′ (x) =


1
f ′ ⎛⎝ f


−1 (x)⎞⎠
whenever f ′ ⎛⎝ f −1 (x)⎞⎠ ≠ 0 and f (x) is differentiable.


• Power rule with rational exponents
d
dx

⎝x


m/n⎞
⎠ =


m
n x


(m/n) − 1.


• Derivative of inverse sine function
d
dx


sin−1 x = 1
1 − (x)2


• Derivative of inverse cosine function
d
dx


cos−1 x = −1
1 − (x)2


• Derivative of inverse tangent function
d
dx


tan−1 x = 1
1 + (x)2


• Derivative of inverse cotangent function
d
dx


cot−1 x = −1
1 + (x)2


• Derivative of inverse secant function
d
dx


sec−1 x = 1


|x| (x)2 − 1


• Derivative of inverse cosecant function
d
dx


csc−1 x = −1


|x| (x)2 − 1


• Derivative of the natural exponential function
d
dx

⎝e


g(x)⎞
⎠ = e


g(x)
g′ (x)


• Derivative of the natural logarithmic function
d
dx

⎝lng(x)⎞⎠ = 1g(x)


g′ (x)


• Derivative of the general exponential function


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d
dx

⎝b


g(x)⎞
⎠ = b


g(x)
g′ (x) lnb


• Derivative of the general logarithmic function
d
dx

⎝logbg(x)



⎠ =


g′ (x)
g(x) lnb


KEY CONCEPTS
3.1 Defining the Derivative


• The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it byfinding the limit of the difference quotient or the difference quotient with increment h.
• The derivative of a function f (x) at a value a is found using either of the definitions for the slope of the tangent
line.


• Velocity is the rate of change of position. As such, the velocity v(t) at time t is the derivative of the position s(t)
at time t. Average velocity is given by


vave =
s(t) − s(a)


t − a .


Instantaneous velocity is given by
v(a) = s′ (a) = lim


t → a
s(t) − s(a)


t − a .


• We may estimate a derivative by using a table of values.
3.2 The Derivative as a Function


• The derivative of a function f (x) is the function whose value at x is f ′(x).
• The graph of a derivative of a function f (x) is related to the graph of f (x). Where f (x) has a tangent line with
positive slope, f ′ (x) > 0. Where f (x) has a tangent line with negative slope, f ′ (x) < 0. Where f (x) has a
horizontal tangent line, f ′ (x) = 0.


• If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a pointif it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner orcusp.
• Higher-order derivatives are derivatives of derivatives, from the second derivative to the nth derivative.


3.3 Differentiation Rules
• The derivative of a constant function is zero.
• The derivative of a power function is a function in which the power on x becomes the coefficient of the term and
the power on x in the derivative decreases by 1.


• The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative.
• The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and thederivative of g.
• The derivative of the difference of a function f and a function g is the same as the difference of the derivative of fand the derivative of g.
• The derivative of a product of two functions is the derivative of the first function times the second function plus thederivative of the second function times the first function.
• The derivative of the quotient of two functions is the derivative of the first function times the second function minus


Chapter 3 | Derivatives 337




the derivative of the second function times the first function, all divided by the square of the second function.
• We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resortingto the definition of the derivative. These formulas can be used singly or in combination with each other.


3.4 Derivatives as Rates of Change
• Using f (a + h) ≈ f (a) + f ′ (a)h, it is possible to estimate f (a + h) given f ′ (a) and f (a).
• The rate of change of position is velocity, and the rate of change of velocity is acceleration. Speed is the absolutevalue, or magnitude, of velocity.
• The population growth rate and the present population can be used to predict the size of a future population.
• Marginal cost, marginal revenue, and marginal profit functions can be used to predict, respectively, the cost ofproducing one more item, the revenue obtained by selling one more item, and the profit obtained by producing andselling one more item.


3.5 Derivatives of Trigonometric Functions
• We can find the derivatives of sin x and cos x by using the definition of derivative and the limit formulas foundearlier. The results are


d
dx


sinx = cosx d
dx


cosx = −sinx.


• With these two formulas, we can determine the derivatives of all six basic trigonometric functions.
3.6 The Chain Rule


• The chain rule allows us to differentiate compositions of two or more functions. It states that for h(x) = f ⎛⎝g(x)⎞⎠,
h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x).


In Leibniz’s notation this rule takes the form
dy
dx


=
dy
du


· du
dx


.


• We can use the chain rule with other rules that we have learned, and we can derive formulas for some of them.
• The chain rule combines with the power rule to form a new rule:


If h(x) = ⎛⎝g(x)⎞⎠n, then h′ (x) = n⎛⎝g(x)⎞⎠n − 1g′ (x).


• When applied to the composition of three functions, the chain rule can be expressed as follows: If
h(x) = f ⎛⎝g⎛⎝k(x)⎞⎠⎞⎠, then h′ (x) = f ′(g⎛⎝k(x)⎞⎠g′ ⎛⎝k(x)⎞⎠k′ (x).


3.7 Derivatives of Inverse Functions
• The inverse function theorem allows us to compute derivatives of inverse functions without using the limitdefinition of the derivative.
• We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.


3.8 Implicit Differentiation
• We use implicit differentiation to find derivatives of implicitly defined functions (functions defined by equations).
• By using implicit differentiation, we can find the equation of a tangent line to the graph of a curve.


3.9 Derivatives of Exponential and Logarithmic Functions
• On the basis of the assumption that the exponential function y = bx, b > 0 is continuous everywhere and


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differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
• We can use a formula to find the derivative of y = lnx, and the relationship logb x = lnxlnb allows us to extend
our differentiation formulas to include logarithms with arbitrary bases.


• Logarithmic differentiation allows us to differentiate functions of the form y = g(x) f (x) or very complex functions
by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.


CHAPTER 3 REVIEW EXERCISES
True or False? Justify the answer with a proof or acounterexample.
367. Every function has a derivative.
368. A continuous function has a continuous derivative.
369. A continuous function has a derivative.
370. If a function is differentiable, it is continuous.
Use the limit definition of the derivative to exactly evaluatethe derivative.
371. f (x) = x + 4


372. f (x) = 3x
Find the derivatives of the following functions.
373. f (x) = 3x3 − 4


x2


374. f (x) = ⎛⎝4 − x2⎞⎠3


375. f (x) = esinx


376. f (x) = ln(x + 2)


377. f (x) = x2 cosx + x tan(x)


378. f (x) = 3x2 + 2


379. f (x) = x
4
sin−1 (x)


380. x2 y = ⎛⎝y + 2⎞⎠+ xysin(x)
Find the following derivatives of various orders.
381. First derivative of y = x ln(x)cosx


382. Third derivative of y = (3x + 2)2


383. Second derivative of y = 4x + x2 sin(x)
Find the equation of the tangent line to the followingequations at the specified point.
384. y = cos−1 (x) + x at x = 0


385. y = x + ex − 1x at x = 1
Draw the derivative for the following graphs.
386.


387.


The following questions concern the water level in OceanCity, New Jersey, in January, which can be approximated
by w(t) = 1.9 + 2.9cos⎛⎝π6 t⎞⎠, where t is measured in
hours after midnight, and the height is measured in feet.
388. Find and graph the derivative. What is the physicalmeaning?


Chapter 3 | Derivatives 339




389. Find w′ (3). What is the physical meaning of this
value?
The following questions consider the wind speeds ofHurricane Katrina, which affected New Orleans, Louisiana,in August 2005. The data are displayed in a table.
Hours after Midnight,August 26 Wind Speed(mph)
1 45
5 75
11 100
29 115
49 145
58 175
73 155
81 125
85 95
107 35


Table 3.10Wind Speeds of HurricaneKatrina Source:http://news.nationalgeographic.com/news/2005/09/0914_050914_katrina_timeline.html.
390. Using the table, estimate the derivative of the windspeed at hour 39. What is the physical meaning?
391. Estimate the derivative of the wind speed at hour 83.What is the physical meaning?


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4 | APPLICATIONS OFDERIVATIVES


Figure 4.1 As a rocket is being launched, at what rate should the angle of a video camera change to continue viewing therocket? (credit: modification of work by Steve Jurvetson, Wikimedia Commons)


Chapter Outline
4.1 Related Rates
4.2 Linear Approximations and Differentials
4.3 Maxima and Minima
4.4 The Mean Value Theorem
4.5 Derivatives and the Shape of a Graph
4.6 Limits at Infinity and Asymptotes
4.7 Applied Optimization Problems
4.8 L’Hôpital’s Rule
4.9 Newton’s Method
4.10 Antiderivatives


Introduction
A rocket is being launched from the ground and cameras are recording the event. A video camera is located on the grounda certain distance from the launch pad. At what rate should the angle of inclination (the angle the camera makes with the


Chapter 4 | Applications of Derivatives 341




ground) change to allow the camera to record the flight of the rocket as it heads upward? (See Example 4.3.)
A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is justone application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum andminimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizingrevenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, toapproximate roots of functions, and to provide accurate graphs of functions.
4.1 | Related Rates


Learning Objectives
4.1.1 Express changing quantities in terms of derivatives.
4.1.2 Find relationships among the derivatives in a given problem.
4.1.3 Use the chain rule to find the rate of change of one quantity that depends on the rate ofchange of other quantities.


We have seen that for quantities that are changing over time, the rates at which these quantities change are given byderivatives. If two related quantities are changing over time, the rates at which the quantities change are related. Forexample, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing.In this section, we consider several problems in which two or more related quantities are changing and we study how todetermine the relationship between the rates of change of these quantities.
Setting up Related-Rates Problems
In many real-world applications, related quantities are changing with respect to time. For example, if we consider theballoon example again, we can say that the rate of change in the volume, V , is related to the rate of change in the radius,
r. In this case, we say that dV


dt
and dr


dt
are related rates because V is related to r. Here we study several examples of


related quantities that are changing with respect to time and we look at how to calculate one rate of change given anotherrate of change.
Example 4.1
Inflating a Balloon
A spherical balloon is being filled with air at the constant rate of 2 cm3 /sec (Figure 4.2). How fast is the radius
increasing when the radius is 3 cm?


Figure 4.2 As the balloon is being filled with air, both the radius and the volume are increasing with respect to time.
Solution


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4.1


The volume of a sphere of radius r centimeters is
V = 4


3
πr3 cm3.


Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, t
seconds after beginning to fill the balloon with air, the volume of air in the balloon is


V(t) = 4
3
π[r(t)]3 cm3.


Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate ofchange in the volume is related to the rate of change in the radius by the equation
V′(t) = 4π⎡⎣r(t)⎤⎦2 r′ (t).


The balloon is being filled with air at the constant rate of 2 cm3/sec, so V′(t) = 2 cm3 /sec. Therefore,
2cm3 /sec = ⎛⎝4π



⎣r(t)⎤⎦2 cm2⎞⎠ ·



⎝r′(t)cm/s⎞⎠,


which implies
r′(t) = 1


2π⎡⎣r(t)⎤⎦2
cm/sec.


When the radius r = 3 cm,
r′(t) = 1


18π
cm/sec.


What is the instantaneous rate of change of the radius when r = 6 cm?


Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems.
Problem-Solving Strategy: Solving a Related-Rates Problem


1. Assign symbols to all variables involved in the problem. Draw a figure if applicable.
2. State, in terms of the variables, the information that is given and the rate to be determined.
3. Find an equation relating the variables introduced in step 1.
4. Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independentvariable. This new equation will relate the derivatives.
5. Substitute all known values into the equation from step 4, then solve for the unknown rate of change.


Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if thevalue for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then thatquantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine thispotential error in the following example.
Examples of the Process
Let’s now implement the strategy just described to solve several related-rates problems. The first example involves a planeflying overhead. The relationship we are studying is between the speed of the plane and the rate at which the distancebetween the plane and a person on the ground is changing.


Chapter 4 | Applications of Derivatives 343




Example 4.2
An Airplane Flying at a Constant Elevation
An airplane is flying overhead at a constant elevation of 4000 ft. A man is viewing the plane from a position
3000 ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is
flying at the rate of 600 ft/sec, at what rate is the distance between the man and the plane increasing when the
plane passes over the radio tower?
Solution
Step 1. Draw a picture, introducing variables to represent the different quantities involved.


Figure 4.3 An airplane is flying at a constant height of 4000 ft. The distance between theperson and the airplane and the person and the place on the ground directly below the airplaneare changing. We denote those quantities with the variables s and x, respectively.


As shown, x denotes the distance between the man and the position on the ground directly below the airplane.
The variable s denotes the distance between the man and the plane. Note that both x and s are functions of
time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of
4000 ft. Since an object’s height above the ground is measured as the shortest distance between the object and
the ground, the line segment of length 4000 ft is perpendicular to the line segment of length x feet, creating a
right triangle.
Step 2. Since x denotes the horizontal distance between the man and the point on the ground below the plane,
dx/dt represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, dx


dt
= 600


ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the planeis directly above the radio tower, we need to find ds/dt when x = 3000 ft.
Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating x and s:



⎣x(t)⎤⎦2 + 40002 = ⎡⎣s(t)⎤⎦2.


Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant iszero, we arrive at the equation


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4.2


xdx
dt


= sds
dt


.


Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly
over the radio tower. That is, find ds


dt
when x = 3000 ft. Since the speed of the plane is 600 ft/sec, we know


that dx
dt


= 600 ft/sec. We are not given an explicit value for s; however, since we are trying to find ds
dt


when
x = 3000 ft, we can use the Pythagorean theorem to determine the distance s when x = 3000 and the height
is 4000 ft. Solving the equation


30002 + 40002 = s2


for s, we have s = 5000 ft at the time of interest. Using these values, we conclude that ds/dt is a solution of
the equation


(3000)(600) = (5000) · ds
dt


.


Therefore,
ds
dt


= 3000 · 600
5000


= 360 ft/sec.


Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. Forexample, in step 3, we related the variable quantities x(t) and s(t) by the equation

⎣x(t)⎤⎦2 + 40002 = ⎡⎣s(t)⎤⎦2.


Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we areallowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If wemistakenly substituted x(t) = 3000 into the equation before differentiating, our equation would have been
30002 + 40002 = ⎡⎣s(t)⎤⎦2.


After differentiating, our equation would become
0 = s(t)ds


dt
.


As a result, we would incorrectly conclude that ds
dt


= 0.


What is the speed of the plane if the distance between the person and the plane is increasing at the rate of
300 ft/sec?


We now return to the problem involving the rocket launch from the beginning of the chapter.
Example 4.3
Chapter Opener: A Rocket Launch


Chapter 4 | Applications of Derivatives 345




Figure 4.4 (credit: modification of work by Steve Jurvetson,Wikimedia Commons)


A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the
rocket is 1000 ft above the launch pad, its velocity is 600 ft/sec. Find the necessary rate of change of the
camera’s angle as a function of time so that it stays focused on the rocket.
Solution
Step 1. Draw a picture introducing the variables.


Figure 4.5 A camera is positioned 5000 ft from the launch pad of the rocket. The height of therocket and the angle of the camera are changing with respect to time. We denote those quantitieswith the variables h and θ, respectively.


Let h denote the height of the rocket above the launch pad and θ be the angle between the camera lens and the


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ground.
Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is
1000 ft off the ground. That is, we need to find dθ


dt
when h = 1000 ft. At that time, we know the velocity of the


rocket is dh
dt


= 600 ft/sec.


Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: h and
θ. How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think
about trigonometric functions. Recall that tanθ is the ratio of the length of the opposite side of the triangle to the
length of the adjacent side. Thus, we have


tanθ = h
5000


.


This gives us the equation
h = 5000tanθ.


Step 4. Differentiating this equation with respect to time t, we obtain
dh
dt


= 5000sec2 θ dθ
dt


.


Step 5. We want to find dθ
dt


when h = 1000 ft. At this time, we know that dh
dt


= 600 ft/sec. We need to
determine sec2 θ. Recall that secθ is the ratio of the length of the hypotenuse to the length of the adjacent
side. We know the length of the adjacent side is 5000 ft. To determine the length of the hypotenuse, we use the
Pythagorean theorem, where the length of one leg is 5000 ft, the length of the other leg is h = 1000 ft, and
the length of the hypotenuse is c feet as shown in the following figure.


We see that
10002 + 50002 = c2


and we conclude that the hypotenuse is
c = 1000 26 ft.


Therefore, when h = 1000, we have
sec2 θ = ⎛⎝


1000 26
5000





2
= 26


25
.


Recall from step 4 that the equation relating dθ
dt


to our known values is
dh
dt


= 5000sec2 θ dθ
dt


.


When h = 1000 ft, we know that dh
dt


= 600 ft/sec and sec2 θ = 26
25


. Substituting these values into the


Chapter 4 | Applications of Derivatives 347




4.3


previous equation, we arrive at the equation
600 = 5000⎛⎝


26
25



dt


.


Therefore, dθ
dt


= 3
26


rad/sec.


What rate of change is necessary for the elevation angle of the camera if the camera is placed on theground at a distance of 4000 ft from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket
is 2000 ft off the ground?


In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of waterin the cone is decreasing with the rate at which the volume of water is decreasing.
Example 4.4
Water Draining from a Funnel
Water is draining from the bottom of a cone-shaped funnel at the rate of 0.03 ft3 /sec. The height of the funnel
is 2 ft and the radius at the top of the funnel is 1 ft. At what rate is the height of the water in the funnel changing
when the height of the water is 1


2
ft?


Solution
Step 1: Draw a picture introducing the variables.


Figure 4.6 Water is draining from a funnel of height 2 ft andradius 1 ft. The height of the water and the radius of water arechanging over time. We denote these quantities with thevariables h and r, respectively.


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4.4


Let h denote the height of the water in the funnel, r denote the radius of the water at its surface, and V denote
the volume of the water.
Step 2: We need to determine dh


dt
when h = 1


2
ft. We know that dV


dt
= −0.03 ft/sec.


Step 3: The volume of water in the cone is
V = 1


3
πr2h.


From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the
same. Therefore, r


h
= 1


2
or r = h


2
. Using this fact, the equation for volume can be simplified to


V = 1
3
π⎛⎝
h
2



2
h = π


12
h3.


Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time t, we obtain
dV
dt


= π
4
h2 dh


dt
.


Step 5: We want to find dh
dt


when h = 1
2
ft. Since water is leaving at the rate of 0.03 ft3 /sec, we know that


dV
dt


= −0.03 ft3 /sec. Therefore,


−0.03 = π
4


1
2



2
dh
dt


,


which implies
−0.03 = π


16
dh
dt


.


It follows that
dh
dt


= − 0.48π = −0.153 ft/sec.


At what rate is the height of the water changing when the height of the water is 1
4
ft?


Chapter 4 | Applications of Derivatives 349




4.1 EXERCISES
For the following exercises, find the quantities for the givenequation.
1. Find dy


dt
at x = 1 and y = x2 + 3 if dx


dt
= 4.


2. Find dx
dt


at x = −2 and y = 2x2 + 1 if dy
dt


= −1.


3. Find dz
dt


at (x, y) = (1, 3) and z2 = x2 + y2 if
dx
dt


= 4 and dy
dt


= 3.


For the following exercises, sketch the situation ifnecessary and used related rates to solve for the quantities.
4. [T] If two electrical resistors are connected in parallel,the total resistance (measured in ohms, denoted by theGreek capital letter omega, Ω) is given by the equation
1
R


= 1
R1


+ 1
R2


. If R1 is increasing at a rate of 0.5 Ω/min
and R2 decreases at a rate of 1.1Ω/min, at what rate
does the total resistance change when R1 = 20Ω and
R2 = 50Ω/min?


5. A 10-ft ladder is leaning against a wall. If the top of theladder slides down the wall at a rate of 2 ft/sec, how fastis the bottom moving along the ground when the bottom ofthe ladder is 5 ft from the wall?


6. A 25-ft ladder is leaning against a wall. If we push theladder toward the wall at a rate of 1 ft/sec, and the bottomof the ladder is initially 20 ft away from the wall, how
fast does the ladder move up the wall 5 sec after we start
pushing?


7. Two airplanes are flying in the air at the same height:airplane A is flying east at 250 mi/h and airplane B is flyingnorth at 300 mi/h. If they are both heading to the same
airport, located 30 miles east of airplane A and 40 milesnorth of airplane B, at what rate is the distance between theairplanes changing?


8. You and a friend are riding your bikes to a restaurantthat you think is east; your friend thinks the restaurant isnorth. You both leave from the same point, with you ridingat 16 mph east and your friend riding 12 mph north. After
you traveled 4 mi, at what rate is the distance between you
changing?
9. Two buses are driving along parallel freeways that are
5 mi apart, one heading east and the other heading west.
Assuming that each bus drives a constant 55 mph, find the
rate at which the distance between the buses is changingwhen they are 13 mi apart, heading toward each other.
10. A 6-ft-tall person walks away from a 10-ft lamppost ata constant rate of 3 ft/sec. What is the rate that the tip of
the shadow moves away from the pole when the person is
10 ft away from the pole?


11. Using the previous problem, what is the rate at whichthe tip of the shadow moves away from the person when theperson is 10 ft from the pole?


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12. A 5-ft-tall person walks toward a wall at a rate of 2ft/sec. A spotlight is located on the ground 40 ft from thewall. How fast does the height of the person’s shadow onthe wall change when the person is 10 ft from the wall?
13. Using the previous problem, what is the rate at whichthe shadow changes when the person is 10 ft from the wall,if the person is walking away from the wall at a rate of 2 ft/sec?
14. A helicopter starting on the ground is rising directlyinto the air at a rate of 25 ft/sec. You are running on theground starting directly under the helicopter at a rate of 10ft/sec. Find the rate of change of the distance between thehelicopter and yourself after 5 sec.
15. Using the previous problem, what is the rate at whichthe distance between you and the helicopter is changingwhen the helicopter has risen to a height of 60 ft in the air,assuming that, initially, it was 30 ft above you?
For the following exercises, draw and label diagrams tohelp solve the related-rates problems.
16. The side of a cube increases at a rate of 1


2
m/sec. Find


the rate at which the volume of the cube increases when theside of the cube is 4 m.
17. The volume of a cube decreases at a rate of 10 m/sec.
Find the rate at which the side of the cube changes whenthe side of the cube is 2 m.
18. The radius of a circle increases at a rate of 2 m/sec.
Find the rate at which the area of the circle increases whenthe radius is 5 m.
19. The radius of a sphere decreases at a rate of 3 m/sec.
Find the rate at which the surface area decreases when theradius is 10 m.
20. The radius of a sphere increases at a rate of 1 m/sec.
Find the rate at which the volume increases when the radiusis 20 m.
21. The radius of a sphere is increasing at a rate of 9 cm/sec. Find the radius of the sphere when the volume and theradius of the sphere are increasing at the same numericalrate.
22. The base of a triangle is shrinking at a rate of 1 cm/minand the height of the triangle is increasing at a rate of 5 cm/min. Find the rate at which the area of the triangle changeswhen the height is 22 cm and the base is 10 cm.
23. A triangle has two constant sides of length 3 ft and 5ft. The angle between these two sides is increasing at a rateof 0.1 rad/sec. Find the rate at which the area of the triangleis changing when the angle between the two sides is π/6.


24. A triangle has a height that is increasing at a rate of 2cm/sec and its area is increasing at a rate of 4 cm2/sec. Findthe rate at which the base of the triangle is changing whenthe height of the triangle is 4 cm and the area is 20 cm2.
For the following exercises, consider a right cone that isleaking water. The dimensions of the conical tank are aheight of 16 ft and a radius of 5 ft.
25. How fast does the depth of the water change when thewater is 10 ft high if the cone leaks water at a rate of 10ft3/min?
26. Find the rate at which the surface area of the waterchanges when the water is 10 ft high if the cone leaks waterat a rate of 10 ft3/min.
27. If the water level is decreasing at a rate of 3 in./minwhen the depth of the water is 8 ft, determine the rate atwhich water is leaking out of the cone.
28. A vertical cylinder is leaking water at a rate of 1ft3/sec. If the cylinder has a height of 10 ft and a radius of 1ft, at what rate is the height of the water changing when theheight is 6 ft?
29. A cylinder is leaking water but you are unable todetermine at what rate. The cylinder has a height of 2 mand a radius of 2 m. Find the rate at which the water isleaking out of the cylinder if the rate at which the height isdecreasing is 10 cm/min when the height is 1 m.
30. A trough has ends shaped like isosceles triangles,with width 3 m and height 4 m, and the trough is 10m long. Water is being pumped into the trough at a rate
of 5 m3 /min. At what rate does the height of the water
change when the water is 1 m deep?


Chapter 4 | Applications of Derivatives 351




31. A tank is shaped like an upside-down square pyramid,with base of 4 m by 4 m and a height of 12 m (see thefollowing figure). How fast does the height increase whenthe water is 2 m deep if water is being pumped in at a rate
of 2


3
m/sec?


For the following problems, consider a pool shaped like thebottom half of a sphere, that is being filled at a rate of 25ft3/min. The radius of the pool is 10 ft.
32. Find the rate at which the depth of the water ischanging when the water has a depth of 5 ft.
33. Find the rate at which the depth of the water ischanging when the water has a depth of 1 ft.
34. If the height is increasing at a rate of 1 in./sec whenthe depth of the water is 2 ft, find the rate at which water isbeing pumped in.
35. Gravel is being unloaded from a truck and falls into apile shaped like a cone at a rate of 10 ft3/min. The radius ofthe cone base is three times the height of the cone. Find therate at which the height of the gravel changes when the pilehas a height of 5 ft.
36. Using a similar setup from the preceding problem, findthe rate at which the gravel is being unloaded if the pile is5 ft high and the height is increasing at a rate of 4 in./min.
For the following exercises, draw the situations and solvethe related-rate problems.
37. You are stationary on the ground and are watchinga bird fly horizontally at a rate of 10 m/sec. The bird is
located 40 m above your head. How fast does the angle ofelevation change when the horizontal distance between youand the bird is 9 m?
38. You stand 40 ft from a bottle rocket on the ground andwatch as it takes off vertically into the air at a rate of 20 ft/sec. Find the rate at which the angle of elevation changeswhen the rocket is 30 ft in the air.


39. A lighthouse, L, is on an island 4 mi away from theclosest point, P, on the beach (see the following image). Ifthe lighthouse light rotates clockwise at a constant rate of10 revolutions/min, how fast does the beam of light moveacross the beach 2 mi away from the closest point on thebeach?


40. Using the same setup as the previous problem,determine at what rate the beam of light moves across thebeach 1 mi away from the closest point on the beach.
41. You are walking to a bus stop at a right-angle corner.You move north at a rate of 2 m/sec and are 20 m southof the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus!What is the rate at which the angle between you and the busis changing when you are 20 m south of the intersection andthe bus is 10 m west of the intersection?
For the following exercises, refer to the figure of baseballdiamond, which has sides of 90 ft.


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42. [T] A batter hits a ball toward third base at 75 ft/secand runs toward first base at a rate of 24 ft/sec. At what ratedoes the distance between the ball and the batter changewhen 2 sec have passed?
43. [T] A batter hits a ball toward second base at 80 ft/secand runs toward first base at a rate of 30 ft/sec. At what ratedoes the distance between the ball and the batter changewhen the runner has covered one-third of the distance tofirst base? (Hint: Recall the law of cosines.)
44. [T] A batter hits the ball and runs toward first base ata speed of 22 ft/sec. At what rate does the distance betweenthe runner and second base change when the runner has run30 ft?
45. [T] Runners start at first and second base. When thebaseball is hit, the runner at first base runs at a speed of18 ft/sec toward second base and the runner at second baseruns at a speed of 20 ft/sec toward third base. How fast isthe distance between runners changing 1 sec after the ballis hit?


Chapter 4 | Applications of Derivatives 353




4.2 | Linear Approximations and Differentials
Learning Objectives


4.2.1 Describe the linear approximation to a function at a point.
4.2.2 Write the linearization of a given function.
4.2.3 Draw a graph that illustrates the use of differentials to approximate the change in aquantity.
4.2.4 Calculate the relative error and percentage error in using a differential approximation.


We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, weexamine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functionsare the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition,the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions (http://cnx.org/content/m53760/latest/) .
Linear Approximation of a Function at a Point
Consider a function f that is differentiable at a point x = a. Recall that the tangent line to the graph of f at a is given
by the equation


y = f (a) + f ′(a)(x − a).


For example, consider the function f (x) = 1x at a = 2. Since f is differentiable at x = 2 and f ′(x) = − 1x2, we see
that f ′(2) = − 1


4
. Therefore, the tangent line to the graph of f at a = 2 is given by the equation


y = 1
2
− 1


4
(x − 2).


Figure 4.7(a) shows a graph of f (x) = 1x along with the tangent line to f at x = 2. Note that for x near 2, the graph of
the tangent line is close to the graph of f . As a result, we can use the equation of the tangent line to approximate f (x) for
x near 2. For example, if x = 2.1, the y value of the corresponding point on the tangent line is


y = 1
2
− 1


4
(2.1 − 2) = 0.475.


The actual value of f (2.1) is given by
f (2.1) = 1


2.1
≈ 0.47619.


Therefore, the tangent line gives us a fairly good approximation of f (2.1) (Figure 4.7(b)). However, note that for values
of x far from 2, the equation of the tangent line does not give us a good approximation. For example, if x = 10, the y
-value of the corresponding point on the tangent line is


y = 1
2
− 1


4
(10 − 2) = 1


2
− 2 = −1.5,


whereas the value of the function at x = 10 is f (10) = 0.1.


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Figure 4.7 (a) The tangent line to f (x) = 1/x at x = 2 provides a good approximation to f for x near 2.
(b) At x = 2.1, the value of y on the tangent line to f (x) = 1/x is 0.475. The actual value of f (2.1) is
1/2.1, which is approximately 0.47619.


In general, for a differentiable function f , the equation of the tangent line to f at x = a can be used to approximate
f (x) for x near a. Therefore, we can write


f (x) ≈ f (a) + f ′(a)(x − a) for x near a.


We call the linear function
(4.1)L(x) = f (a) + f ′(a)(x − a)


the linear approximation, or tangent line approximation, of f at x = a. This function L is also known as the
linearization of f at x = a.
To show how useful the linear approximation can be, we look at how to find the linear approximation for f (x) = x at
x = 9.


Example 4.5
Linear Approximation of x
Find the linear approximation of f (x) = x at x = 9 and use the approximation to estimate 9.1.
Solution
Since we are looking for the linear approximation at x = 9, using Equation 4.1 we know the linear
approximation is given by


L(x) = f (9) + f ′(9)(x − 9).


We need to find f (9) and f ′(9).


Chapter 4 | Applications of Derivatives 355




4.5


f (x) = x ⇒ f (9) = 9 = 3


f ′(x) = 1
2 x


⇒ f ′(9) = 1
2 9


= 1
6


Therefore, the linear approximation is given by Figure 4.8.
L(x) = 3 + 1


6
(x − 9)


Using the linear approximation, we can estimate 9.1 by writing
9.1 = f (9.1) ≈ L(9.1) = 3 + 1


6
(9.1 − 9) ≈ 3.0167.


Figure 4.8 The local linear approximation to f (x) = x at
x = 9 provides an approximation to f for x near 9.


Analysis
Using a calculator, the value of 9.1 to four decimal places is 3.0166. The value given by the linear
approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linearapproximation is a good way to estimate x, at least for x near 9. At the same time, it may seem odd to use
a linear approximation when we can just push a few buttons on a calculator to evaluate 9.1. However, how
does the calculator evaluate 9.1? The calculator uses an approximation! In fact, calculators and computers use
approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.


Find the local linear approximation to f (x) = x3 at x = 8. Use it to approximate 8.13 to five decimal
places.


Example 4.6
Linear Approximation of sinx
Find the linear approximation of f (x) = sinx at x = π


3
and use it to approximate sin(62°).


Solution
First we note that since π


3
rad is equivalent to 60°, using the linear approximation at x = π/3 seems


reasonable. The linear approximation is given by
L(x) = f ⎛⎝


π
3

⎠+ f ′




π
3



⎝x −


π
3

⎠.


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4.6


We see that
f (x) = sinx ⇒ f ⎛⎝


π
3

⎠ = sin




π
3

⎠ =


3
2


f ′(x) = cosx ⇒ f ′⎛⎝
π
3

⎠ = cos




π
3

⎠ =


1
2


Therefore, the linear approximation of f at x = π/3 is given by Figure 4.9.
L(x) = 3


2
+ 1


2

⎝x −


π
3



To estimate sin(62°) using L, we must first convert 62° to radians. We have 62° = 62π
180


radians, so the
estimate for sin(62°) is given by


sin(62°) = f ⎛⎝
62π
180

⎠ ≈ L


62π
180

⎠ =


3
2


+ 1
2


62π
180


− π
3

⎠ =


3
2


+ 1
2



180

⎠ =


3
2


+ π
180


≈ 0.88348.


Figure 4.9 The linear approximation to f (x) = sinx at x = π/3 provides an approximation
to sinx for x near π/3.


Find the linear approximation for f (x) = cosx at x = π
2
.


Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximationfor f (x) = (1 + x)n at x = 0, which can be used to estimate roots and powers for real numbers near 1. The same idea
can be extended to a function of the form f (x) = (m + x)n to estimate roots and powers near a different number m.
Example 4.7
Approximating Roots and Powers
Find the linear approximation of f (x) = (1 + x)n at x = 0. Use this approximation to estimate (1.01)3.
Solution
The linear approximation at x = 0 is given by


Chapter 4 | Applications of Derivatives 357




4.7


L(x) = f (0) + f ′(0)(x − 0).


Because
f (x) = (1 + x)n ⇒ f (0) = 1


f ′(x) = n(1 + x)n − 1 ⇒ f ′(0) = n,


the linear approximation is given by Figure 4.10(a).
L(x) = 1 + n(x − 0) = 1 + nx


We can approximate (1.01)3 by evaluating L(0.01) when n = 3. We conclude that
(1.01)3 = f (1.01) ≈ L(1.01) = 1 + 3(0.01) = 1.03.


Figure 4.10 (a) The linear approximation of f (x) at x = 0 is L(x). (b) The actual value of 1.013 is
1.030301. The linear approximation of f (x) at x = 0 estimates 1.013 to be 1.03.


Find the linear approximation of f (x) = (1 + x)4 at x = 0 without using the result from the preceding
example.


Differentials
We have seen that linear approximations can be used to estimate function values. They can also be used to estimate theamount a function value changes as a result of a small change in the input. To discuss this more formally, we define a relatedconcept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a smallchange in input values.
When we first looked at derivatives, we used the Leibniz notation dy/dx to represent the derivative of y with respect to
x. Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a
meaning to the expressions dy and dx. Suppose y = f (x) is a differentiable function. Let dx be an independent variable that
can be assigned any nonzero real number, and define the dependent variable dy by


(4.2)dy = f ′(x)dx.
It is important to notice that dy is a function of both x and dx. The expressions dy and dx are called differentials. We can


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4.8


divide both sides of Equation 4.2 by dx, which yields
(4.3)dy


dx
= f ′(x).


This is the familiar expression we have used to denote a derivative. Equation 4.2 is known as the differential form ofEquation 4.3.
Example 4.8
Computing differentials
For each of the following functions, find dy and evaluate when x = 3 and dx = 0.1.


a. y = x2 + 2x
b. y = cosx


Solution
The key step is calculating the derivative. When we have that, we can obtain dy directly.


a. Since f (x) = x2 + 2x, we know f ′(x) = 2x + 2, and therefore
dy = (2x + 2)dx.


When x = 3 and dx = 0.1,
dy = (2 · 3 + 2)(0.1) = 0.8.


b. Since f (x) = cosx, f ′(x) = −sin(x). This gives us
dy = −sinxdx.


When x = 3 and dx = 0.1,
dy = −sin(3)(0.1) = −0.1sin(3).


For y = ex2, find dy.


We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of afunction resulting from a small change in input values. Consider a function f that is differentiable at point a. Suppose
the input x changes by a small amount. We are interested in how much the output y changes. If x changes from a to
a + dx, then the change in x is dx (also denoted Δx), and the change in y is given by


Δy = f (a + dx) − f (a).


Instead of calculating the exact change in y, however, it is often easier to approximate the change in y by using a linear
approximation. For x near a, f (x) can be approximated by the linear approximation


L(x) = f (a) + f ′(a)(x − a).


Therefore, if dx is small,
f (a + dx) ≈ L(a + dx) = f (a) + f ′(a)(a + dx − a).


Chapter 4 | Applications of Derivatives 359




4.9


That is,
f (a + dx) − f (a) ≈ L(a + dx) − f (a) = f ′(a)dx.


In other words, the actual change in the function f if x increases from a to a + dx is approximately the difference
between L(a + dx) and f (a), where L(x) is the linear approximation of f at a. By definition of L(x), this difference
is equal to f ′(a)dx. In summary,


Δy = f (a + dx) − f (a) ≈ L(a + dx) − f (a) = f ′(a)dx = dy.


Therefore, we can use the differential dy = f ′(a)dx to approximate the change in y if x increases from x = a to
x = a + dx. We can see this in the following graph.


Figure 4.11 The differential dy = f ′(a)dx is used to approximate the actual
change in y if x increases from a to a + dx.


We now take a look at how to use differentials to approximate the change in the value of the function that results from asmall change in the value of the input. Note the calculation with differentials is much simpler than calculating actual valuesof functions and the result is very close to what we would obtain with the more exact calculation.
Example 4.9
Approximating Change with Differentials
Let y = x2 + 2x. Compute Δy and dy at x = 3 if dx = 0.1.
Solution
The actual change in y if x changes from x = 3 to x = 3.1 is given by


Δy = f (3.1) − f (3) = [(3.1)2 + 2(3.1)] − [32 + 2(3)] = 0.81.


The approximate change in y is given by dy = f ′(3)dx. Since f ′(x) = 2x + 2, we have
dy = f ′(3)dx = (2(3) + 2)(0.1) = 0.8.


For y = x2 + 2x, find Δy and dy at x = 3 if dx = 0.2.


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Calculating the Amount of Error
Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated basedon measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in themeasurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and studyhow differentials can be used to estimate the error.
Consider a function f with an input that is a measured quantity. Suppose the exact value of the measured quantity is a,
but the measured value is a + dx. We say the measurement error is dx (or Δx). As a result, an error occurs in the calculated
quantity f (x). This type of error is known as a propagated error and is given by


Δy = f (a + dx) − f (a).


Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so wecannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can usedifferentials to approximate the propagated error Δy. Specifically, if f is a differentiable function at a, the propagated
error is


Δy ≈ dy = f ′(a)dx.


Unfortunately, we do not know the exact value a. However, we can use the measured value a + dx, and estimate
Δy ≈ dy ≈ f ′(a + dx)dx.


In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if weassume the measurement of the side length is made with a certain amount of accuracy.
Example 4.10
Volume of a Cube
Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.


a. Use differentials to estimate the error in the computed volume of the cube.
b. Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimatederror with the actual potential error.


Solution
a. The measurement of the side length is accurate to within ±0.1 cm. Therefore,


−0.1 ≤ dx ≤ 0.1.


The volume of a cube is given by V = x3, which leads to
dV = 3x2dx.


Using the measured side length of 5 cm, we can estimate that
−3(5)2(0.1) ≤ dV ≤ 3(5)2(0.1).


Therefore,
−7.5 ≤ dV ≤ 7.5.


b. If the side length is actually 4.9 cm, then the volume of the cube is
V(4.9) = (4.9)3 = 117.649 cm3.


Chapter 4 | Applications of Derivatives 361




4.10


If the side length is actually 5.1 cm, then the volume of the cube is
V(5.1) = (5.1)3 = 132.651 cm3.


Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is
measured to be 5 cm, the computed volume is V(5) = 53 = 125. Therefore, the error in the computed
volume is


117.649 − 125 ≤ ΔV ≤ 132.651 − 125.


That is,
−7.351 ≤ ΔV ≤ 7.651.


We see the estimated error dV is relatively close to the actual potential error in the computed volume.


Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with anaccuracy of 0.2 cm.


The measurement error dx (=Δx) and the propagated error Δy are absolute errors. We are typically interested in the size
of an error relative to the size of the quantity being measured or calculated. Given an absolute error Δq for a particular
quantity, we define the relative error as Δqq , where q is the actual value of the quantity. The percentage error is the
relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual
height is 62 in., the absolute error is 1 in. but the relative error is 1


62
= 0.016, or 1.6%. By comparison, if we measure the


width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is 1
4
in., whereas the relative


error is 0.25
8


= 1
32


, or 3.1%. Therefore, the percentage error in the measurement of the cardboard is larger, even though
0.25 in. is less than 1 in.
Example 4.11
Relative and Percentage Error
An astronaut using a camera measures the radius of Earth as 4000 mi with an error of ±80 mi. Let’s use
differentials to estimate the relative and percentage error of using this radius measurement to calculate the volumeof Earth, assuming the planet is a perfect sphere.
Solution
If the measurement of the radius is accurate to within ±80, we have


−80 ≤ dr ≤ 80.


Since the volume of a sphere is given by V = ⎛⎝43⎞⎠πr3, we have


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4.11


dV = 4πr2dr.


Using the measured radius of 4000 mi, we can estimate
−4π(4000)2(80) ≤ dV ≤ 4π(4000)2(80).


To estimate the relative error, consider dV
V


. Since we do not know the exact value of the volume V , use the
measured radius r = 4000 mi to estimate V . We obtain V ≈ ⎛⎝43⎞⎠π(4000)3. Therefore the relative error satisfies


−4π(4000)2(80)


4π(4000)3 /3
≤ dV


V
≤ 4π(4000)


2(80)


4π(4000)3 /3
,


which simplifies to
−0.06 ≤ dV


V
≤ 0.06.


The relative error is 0.06 and the percentage error is 6%.


Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of ±100
mi.


Chapter 4 | Applications of Derivatives 363




4.2 EXERCISES
46. What is the linear approximation for any generic linearfunction y = mx + b?
47. Determine the necessary conditions such that thelinear approximation function is constant. Use a graph toprove your result.
48. Explain why the linear approximation becomes lessaccurate as you increase the distance between x and a.
Use a graph to prove your argument.
49. When is the linear approximation exact?
For the following exercises, find the linear approximation
L(x) to y = f (x) near x = a for the function.
50. [T] f (x) = x + x4, a = 0
51. [T] f (x) = 1x , a = 2
52. [T] f (x) = tanx, a = π


4


53. [T] f (x) = sinx, a = π
2


54. [T] f (x) = xsinx, a = 2π
55. [T] f (x) = sin2 x, a = 0
For the following exercises, compute the values givenwithin 0.01 by deciding on the appropriate f (x) and a,
and evaluating L(x) = f (a) + f ′(a)(x − a). Check your
answer using a calculator.
56. [T] (2.001)6
57. [T] sin(0.02)
58. [T] cos(0.03)
59. [T] (15.99)1/4


60. [T] 1
0.98


61. [T] sin(3.14)
For the following exercises, determine the appropriate
f (x) and a, and evaluate L(x) = f (a) + f ′ (a)(x − a).
Calculate the numerical error in the linear approximationsthat follow.


62. (1.01)3
63. cos(0.01)
64. ⎛⎝sin(0.01)⎞⎠2
65. (1.01)−3


66. ⎛⎝1 + 110⎞⎠
10


67. 8.99
For the following exercises, find the differential of thefunction.
68. y = 3x4 + x2 − 2x + 1
69. y = xcosx
70. y = 1 + x
71. y = x2 + 2


x − 1


For the following exercises, find the differential andevaluate for the given x and dx.
72. y = 3x2 − x + 6, x = 2, dx = 0.1
73. y = 1


x + 1
, x = 1, dx = 0.25


74. y = tanx, x = 0, dx = π
10


75. y = 3x2 + 2
x + 1


, x = 0, dx = 0.1


76. y = sin(2x)x , x = π, dx = 0.25
77. y = x3 + 2x + 1x , x = 1, dx = 0.05
For the following exercises, find the change in volume dV
or in surface area dA.
78. dV if the sides of a cube change from 10 to 10.1.
79. dA if the sides of a cube change from x to x + dx.
80. dA if the radius of a sphere changes from r by dr.


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81. dV if the radius of a sphere changes from r by dr.
82. dV if a circular cylinder with r = 2 changes height
from 3 cm to 3.05 cm.
83. dV if a circular cylinder of height 3 changes from
r = 2 to r = 1.9 cm.
For the following exercises, use differentials to estimate themaximum and relative error when computing the surfacearea or volume.
84. A spherical golf ball is measured to have a radius of
5 mm, with a possible measurement error of 0.1 mm.
What is the possible change in volume?
85. A pool has a rectangular base of 10 ft by 20 ft and adepth of 6 ft. What is the change in volume if you only fillit up to 5.5 ft?
86. An ice cream cone has height 4 in. and radius 1 in. Ifthe cone is 0.1 in. thick, what is the difference between thevolume of the cone, including the shell, and the volume ofthe ice cream you can fit inside the shell?
For the following exercises, confirm the approximations byusing the linear approximation at x = 0.
87. 1 − x ≈ 1 − 1


2
x


88. 1
1 − x2


≈ 1


89. c2 + x2 ≈ c


Chapter 4 | Applications of Derivatives 365




4.3 | Maxima and Minima
Learning Objectives


4.3.1 Define absolute extrema.
4.3.2 Define local extrema.
4.3.3 Explain how to find the critical points of a function over a closed interval.
4.3.4 Describe how to use critical points to locate absolute extrema over a closed interval.


Given a particular function, we are often interested in determining the largest and smallest values of the function. Thisinformation is important in creating accurate graphs. Finding the maximum and minimum values of a function alsohas practical significance because we can use this method to solve optimization problems, such as maximizing profit,minimizing the amount of material used in manufacturing an aluminum can, or finding the maximum height a rocket canreach. In this section, we look at how to use derivatives to find the largest and smallest values for a function.
Absolute Extrema
Consider the function f (x) = x2 + 1 over the interval (−∞, ∞). As x → ±∞, f (x) → ∞. Therefore, the function
does not have a largest value. However, since x2 + 1 ≥ 1 for all real numbers x and x2 + 1 = 1 when x = 0, the
function has a smallest value, 1, when x = 0. We say that 1 is the absolute minimum of f (x) = x2 + 1 and it occurs at
x = 0. We say that f (x) = x2 + 1 does not have an absolute maximum (see the following figure).


Figure 4.12 The given function has an absolute minimum of 1at x = 0. The function does not have an absolute maximum.


Definition
Let f be a function defined over an interval I and let c ∈ I. We say f has an absolute maximum on I at c if
f (c) ≥ f (x) for all x ∈ I. We say f has an absolute minimum on I at c if f (c) ≤ f (x) for all x ∈ I. If f has
an absolute maximum on I at c or an absolute minimum on I at c, we say f has an absolute extremum on I at
c.


Before proceeding, let’s note two important issues regarding this definition. First, the term absolute here does not refer toabsolute value. An absolute extremum may be positive, negative, or zero. Second, if a function f has an absolute extremum
over an interval I at c, the absolute extremum is f (c). The real number c is a point in the domain at which the absolute
extremum occurs. For example, consider the function f (x) = 1/(x2 + 1) over the interval (−∞, ∞). Since


f (0) = 1 ≥ 1
x2 + 1


= f (x)


for all real numbers x, we say f has an absolute maximum over (−∞, ∞) at x = 0. The absolute maximum is


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f (0) = 1. It occurs at x = 0, as shown in Figure 4.13(b).
A function may have both an absolute maximum and an absolute minimum, just one extremum, or neither. Figure 4.13shows several functions and some of the different possibilities regarding absolute extrema. However, the following theorem,called the Extreme Value Theorem, guarantees that a continuous function f over a closed, bounded interval [a, b] has
both an absolute maximum and an absolute minimum.


Figure 4.13 Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of
(−∞, ∞). Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a
bounded interval.


Theorem 4.1: Extreme Value Theorem
If f is a continuous function over the closed, bounded interval [a, b], then there is a point in [a, b] at which f has
an absolute maximum over [a, b] and there is a point in [a, b] at which f has an absolute minimum over [a, b].


The proof of the extreme value theorem is beyond the scope of this text. Typically, it is proved in a course on real analysis.There are a couple of key points to note about the statement of this theorem. For the extreme value theorem to apply, the


Chapter 4 | Applications of Derivatives 367




function must be continuous over a closed, bounded interval. If the interval I is open or the function has even one point
of discontinuity, the function may not have an absolute maximum or absolute minimum over I. For example, consider the
functions shown in Figure 4.13(d), (e), and (f). All three of these functions are defined over bounded intervals. However,the function in graph (e) is the only one that has both an absolute maximum and an absolute minimum over its domain.The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions iscontinuous over a closed, bounded interval. Although the function in graph (d) is defined over the closed interval [0, 4],
the function is discontinuous at x = 2. The function has an absolute maximum over [0, 4] but does not have an absolute
minimum. The function in graph (f) is continuous over the half-open interval [0, 2), but is not defined at x = 2, and
therefore is not continuous over a closed, bounded interval. The function has an absolute minimum over [0, 2), but does
not have an absolute maximum over [0, 2). These two graphs illustrate why a function over a bounded interval may fail to
have an absolute maximum and/or absolute minimum.
Before looking at how to find absolute extrema, let’s examine the related concept of local extrema. This idea is useful indetermining where absolute extrema occur.
Local Extrema and Critical Points
Consider the function f shown in Figure 4.14. The graph can be described as two mountains with a valley in the middle.
The absolute maximum value of the function occurs at the higher peak, at x = 2. However, x = 0 is also a point of
interest. Although f (0) is not the largest value of f , the value f (0) is larger than f (x) for all x near 0. We say f has a
local maximum at x = 0. Similarly, the function f does not have an absolute minimum, but it does have a local minimum
at x = 1 because f (1) is less than f (x) for x near 1.


Figure 4.14 This function f has two local maxima and one
local minimum. The local maximum at x = 2 is also the
absolute maximum.


Definition
A function f has a local maximum at c if there exists an open interval I containing c such that I is contained
in the domain of f and f (c) ≥ f (x) for all x ∈ I. A function f has a local minimum at c if there exists an open
interval I containing c such that I is contained in the domain of f and f (c) ≤ f (x) for all x ∈ I. A function f
has a local extremum at c if f has a local maximum at c or f has a local minimum at c.


Note that if f has an absolute extremum at c and f is defined over an interval containing c, then f (c) is also
considered a local extremum. If an absolute extremum for a function f occurs at an endpoint, we do not consider that to be


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a local extremum, but instead refer to that as an endpoint extremum.
Given the graph of a function f , it is sometimes easy to see where a local maximum or local minimum occurs. However,
it is not always easy to see, since the interesting features on the graph of a function may not be visible because they occur ata very small scale. Also, we may not have a graph of the function. In these cases, how can we use a formula for a functionto determine where these extrema occur?
To answer this question, let’s look at Figure 4.14 again. The local extrema occur at x = 0, x = 1, and x = 2. Notice
that at x = 0 and x = 1, the derivative f ′(x) = 0. At x = 2, the derivative f ′(x) does not exist, since the function
f has a corner there. In fact, if f has a local extremum at a point x = c, the derivative f ′(c) must satisfy one of the
following conditions: either f ′(c) = 0 or f ′(c) is undefined. Such a value c is known as a critical point and it is important
in finding extreme values for functions.
Definition
Let c be an interior point in the domain of f . We say that c is a critical point of f if f ′(c) = 0 or f ′(c) is
undefined.


As mentioned earlier, if f has a local extremum at a point x = c, then c must be a critical point of f . This fact is known
as Fermat’s theorem.
Theorem 4.2: Fermat’s Theorem
If f has a local extremum at c and f is differentiable at c, then f ′(c) = 0.


Proof
Suppose f has a local extremum at c and f is differentiable at c. We need to show that f ′(c) = 0. To do this, we
will show that f ′(c) ≥ 0 and f ′(c) ≤ 0, and therefore f ′(c) = 0. Since f has a local extremum at c, f has a local
maximum or local minimum at c. Suppose f has a local maximum at c. The case in which f has a local minimum
at c can be handled similarly. There then exists an open interval I such that f (c) ≥ f (x) for all x ∈ I. Since f is
differentiable at c, from the definition of the derivative, we know that


f ′(c) = lim
x → c


f (x) − f (c)
x − c .


Since this limit exists, both one-sided limits also exist and equal f ′(c). Therefore,
(4.4)f ′(c) = lim


x → c+
f (x) − f (c)


x − c ,


and
(4.5)f ′(c) = lim


x → c−
f (x) − f (c)


x − c .


Since f (c) is a local maximum, we see that f (x) − f (c) ≤ 0 for x near c. Therefore, for x near c, but x > c,
we have f (x) − f (c)x − c ≤ 0. From Equation 4.4 we conclude that f ′(c) ≤ 0. Similarly, it can be shown that f ′(c) ≥ 0.
Therefore, f ′(c) = 0.

From Fermat’s theorem, we conclude that if f has a local extremum at c, then either f ′(c) = 0 or f ′(c) is undefined.
In other words, local extrema can only occur at critical points.


Chapter 4 | Applications of Derivatives 369




Note this theorem does not claim that a function f must have a local extremum at a critical point. Rather, it states that
critical points are candidates for local extrema. For example, consider the function f (x) = x3. We have f ′(x) = 3x2 = 0
when x = 0. Therefore, x = 0 is a critical point. However, f (x) = x3 is increasing over (−∞, ∞), and thus f does
not have a local extremum at x = 0. In Figure 4.15, we see several different possibilities for critical points. In some of
these cases, the functions have local extrema at critical points, whereas in other cases the functions do not. Note that thesegraphs do not show all possibilities for the behavior of a function at a critical point.


Figure 4.15 (a–e) A function f has a critical point at c if f ′(c) = 0 or f ′(c) is undefined. A function may or may not
have a local extremum at a critical point.


Later in this chapter we look at analytical methods for determining whether a function actually has a local extremum at acritical point. For now, let’s turn our attention to finding critical points. We will use graphical observations to determinewhether a critical point is associated with a local extremum.
Example 4.12
Locating Critical Points
For each of the following functions, find all critical points. Use a graphing utility to determine whether thefunction has a local extremum at each of the critical points.


a. f (x) = 1
3
x3 − 5


2
x2 + 4x


b. f (x) = ⎛⎝x2 − 1⎞⎠3


c. f (x) = 4x
1 + x2


Solution


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a. The derivative f ′(x) = x2 − 5x + 4 is defined for all real numbers x. Therefore, we only need to find
the values for x where f ′(x) = 0. Since f ′(x) = x2 − 5x + 4 = (x − 4)(x − 1), the critical points are
x = 1 and x = 4. From the graph of f in Figure 4.16, we see that f has a local maximum at x = 1
and a local minimum at x = 4.


Figure 4.16 This function has a local maximum and a localminimum.
b. Using the chain rule, we see the derivative is


f ′(x) = 3⎛⎝x
2 − 1⎞⎠


2
(2x) = 6x⎛⎝x


2 − 1⎞⎠
2
.


Therefore, f has critical points when x = 0 and when x2 − 1 = 0. We conclude that the critical points
are x = 0, ±1. From the graph of f in Figure 4.17, we see that f has a local (and absolute) minimum
at x = 0, but does not have a local extremum at x = 1 or x = −1.


Figure 4.17 This function has three critical points: x = 0,
x = 1, and x = −1. The function has a local (and absolute)
minimum at x = 0, but does not have extrema at the other two
critical points.


c. By the chain rule, we see that the derivative is
f ′(x) =



⎝1 + x


24⎞⎠− 4x(2x)



⎝1 + x


2⎞


2
= 4 − 4x


2



⎝1 + x


2⎞


2
.


The derivative is defined everywhere. Therefore, we only need to find values for x where f ′(x) = 0.
Solving f ′(x) = 0, we see that 4 − 4x2 = 0, which implies x = ±1. Therefore, the critical points
are x = ±1. From the graph of f in Figure 4.18, we see that f has an absolute maximum at x = 1


Chapter 4 | Applications of Derivatives 371




4.12


and an absolute minimum at x = −1. Hence, f has a local maximum at x = 1 and a local minimum at
x = −1. (Note that if f has an absolute extremum over an interval I at a point c that is not an endpoint
of I, then f has a local extremum at c.)


Figure 4.18 This function has an absolute maximum and anabsolute minimum.


Find all critical points for f (x) = x3 − 1
2
x2 − 2x + 1.


Locating Absolute Extrema
The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum andan absolute minimum. As shown in Figure 4.13, one or both of these absolute extrema could occur at an endpoint. If anabsolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absoluteextremum is a local extremum. Therefore, by Fermat’s Theorem, the point c at which the local extremum occurs must
be a critical point. We summarize this result in the following theorem.
Theorem 4.3: Location of Absolute Extrema
Let f be a continuous function over a closed, bounded interval I. The absolute maximum of f over I and the
absolute minimum of f over I must occur at endpoints of I or at critical points of f in I.


With this idea in mind, let’s examine a procedure for locating absolute extrema.
Problem-Solving Strategy: Locating Absolute Extrema over a Closed Interval
Consider a continuous function f defined over the closed interval [a, b].


1. Evaluate f at the endpoints x = a and x = b.
2. Find all critical points of f that lie over the interval (a, b) and evaluate f at those critical points.
3. Compare all values found in (1) and (2). From Location of Absolute Extrema, the absolute extrema mustoccur at endpoints or critical points. Therefore, the largest of these values is the absolute maximum of f . The


smallest of these values is the absolute minimum of f .


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Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuousfunctions.
Example 4.13
Locating Absolute Extrema
For each of the following functions, find the absolute maximum and absolute minimum over the specified intervaland state where those values occur.


a. f (x) = −x2 + 3x − 2 over [1, 3].
b. f (x) = x2 − 3x2/3 over [0, 2].


Solution
a. Step 1. Evaluate f at the endpoints x = 1 and x = 3.


f (1) = 0 and f (3) = −2


Step 2. Since f ′(x) = −2x + 3, f ′ is defined for all real numbers x. Therefore, there are no critical
points where the derivative is undefined. It remains to check where f ′(x) = 0. Since
f ′(x) = −2x + 3 = 0 at x = 3


2
and 3


2
is in the interval [1, 3], f ⎛⎝32⎞⎠ is a candidate for an absolute


extremum of f over [1, 3]. We evaluate f ⎛⎝32⎞⎠ and find
f ⎛⎝
3
2

⎠ =


1
4
.


Step 3. We set up the following table to compare the values found in steps 1 and 2.
x f(x) Conclusion
0 0


3
2


1
4


Absolute maximum


3 −2 Absolute minimum


From the table, we find that the absolute maximum of f over the interval [1, 3] is 1
4
, and it occurs at


x = 3
2
. The absolute minimum of f over the interval [1, 3] is −2, and it occurs at x = 3 as shown in


the following graph.


Chapter 4 | Applications of Derivatives 373




Figure 4.19 This function has both an absolute maximum and an absolute minimum.
b. Step 1. Evaluate f at the endpoints x = 0 and x = 2.


f (0) = 0 and f (2) = 4 − 3 4
3


≈ − 0.762


Step 2. The derivative of f is given by
f ′(x) = 2x − 2


x1/3
= 2x


4/3 − 2
x1/3


for x ≠ 0. The derivative is zero when 2x4/3 − 2 = 0, which implies x = ±1. The derivative is
undefined at x = 0. Therefore, the critical points of f are x = 0, 1, −1. The point x = 0 is an
endpoint, so we already evaluated f (0) in step 1. The point x = −1 is not in the interval of interest, so
we need only evaluate f (1). We find that


f (1) = −2.


Step 3. We compare the values found in steps 1 and 2, in the following table.
x f(x) Conclusion
0 0 Absolute maximum


1 −2 Absolute minimum


2 −0.762


We conclude that the absolute maximum of f over the interval [0, 2] is zero, and it occurs at x = 0. The
absolute minimum is −2, and it occurs at x = 1 as shown in the following graph.


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4.13


Figure 4.20 This function has an absolute maximum at anendpoint of the interval.


Find the absolute maximum and absolute minimum of f (x) = x2 − 4x + 3 over the interval [1, 4].


At this point, we know how to locate absolute extrema for continuous functions over closed intervals. We have also definedlocal extrema and determined that if a function f has a local extremum at a point c, then c must be a critical point of f .
However, c being a critical point is not a sufficient condition for f to have a local extremum at c. Later in this chapter,
we show how to determine whether a function actually has a local extremum at a critical point. First, however, we need tointroduce the Mean Value Theorem, which will help as we analyze the behavior of the graph of a function.


Chapter 4 | Applications of Derivatives 375




4.3 EXERCISES
90. In precalculus, you learned a formula for the positionof the maximum or minimum of a quadratic equation
y = ax2 + bx + c, which was m = − b


(2a)
. Prove this


formula using calculus.
91. If you are finding an absolute minimum over aninterval [a, b], why do you need to check the endpoints?
Draw a graph that supports your hypothesis.
92. If you are examining a function over an interval
(a, b), for a and b finite, is it possible not to have an
absolute maximum or absolute minimum?
93. When you are checking for critical points, explain whyyou also need to determine points where f (x) is undefined.
Draw a graph to support your explanation.
94. Can you have a finite absolute maximum for
y = ax2 + bx + c over (−∞, ∞)? Explain why or why
not using graphical arguments.
95. Can you have a finite absolute maximum for
y = ax3 + bx2 + cx + d over (−∞, ∞) assuming a is
non-zero? Explain why or why not using graphicalarguments.
96. Let m be the number of local minima and M be the
number of local maxima. Can you create a function where
M > m + 2? Draw a graph to support your explanation.
97. Is it possible to have more than one absolutemaximum? Use a graphical argument to prove yourhypothesis.
98. Is it possible to have no absolute minimum ormaximum for a function? If so, construct such a function.If not, explain why this is not possible.
99. [T] Graph the function y = eax. For which values
of a, on any infinite domain, will you have an absolute
minimum and absolute maximum?
For the following exercises, determine where the local andabsolute maxima and minima occur on the graph given.Assume domains are closed intervals unless otherwisespecified.


100.


101.


102.


103.


For the following problems, draw graphs of f (x), which
is continuous, over the interval [−4, 4] with the following
properties:


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104. Absolute maximum at x = 2 and absolute minima at
x = ±3


105. Absolute minimum at x = 1 and absolute maximum
at x = 2
106. Absolute maximum at x = 4, absolute minimum at
x = −1, local maximum at x = −2, and a critical point
that is not a maximum or minimum at x = 2
107. Absolute maxima at x = 2 and x = −3, local
minimum at x = 1, and absolute minimum at x = 4
For the following exercises, find the critical points in thedomains of the following functions.
108. y = 4x3 − 3x
109. y = 4 x − x2


110. y = 1
x − 1


111. y = ln(x − 2)
112. y = tan(x)
113. y = 4 − x2


114. y = x3/2 − 3x5/2


115. y = x2 − 1
x2 + 2x − 3


116. y = sin2(x)
117. y = x + 1x
For the following exercises, find the local and/or absolutemaxima for the functions over the specified domain.
118. f (x) = x2 + 3 over [−1, 4]
119. y = x2 + 2x over [1, 4]


120. y = ⎛⎝x − x2⎞⎠2 over [−1, 1]
121. y = 1⎛


⎝x − x
2⎞


over [0, 1]


122. y = 9 − x over [1, 9]


123. y = x + sin(x) over [0, 2π]
124. y = x


1 + x
over [0, 100]


125. y = |x + 1| + |x − 1| over [−3, 2]
126. y = x − x3 over [0, 4]
127. y = sinx + cosx over [0, 2π]
128. y = 4sinθ − 3cosθ over [0, 2π]
For the following exercises, find the local and absoluteminima and maxima for the functions over (−∞, ∞).
129. y = x2 + 4x + 5
130. y = x3 − 12x
131. y = 3x4 + 8x3 − 18x2
132. y = x3 (1 − x)6


133. y = x2 + x + 6
x − 1


134. y = x2 − 1
x − 1


For the following functions, use a calculator to graph thefunction and to estimate the absolute and local maxima andminima. Then, solve for them explicitly.
135. [T] y = 3x 1 − x2
136. [T] y = x + sin(x)
137. [T] y = 12x5 + 45x4 + 20x3 − 90x2 − 120x + 3
138. [T] y = x3 + 6x2 − x − 30


x − 2


139. [T] y = 4 − x2
4 + x2


140. A company that produces cell phones has a cost
function of C = x2 − 1200x + 36,400, where C is cost
in dollars and x is number of cell phones produced (in
thousands). How many units of cell phone (in thousands)minimizes this cost function?


Chapter 4 | Applications of Derivatives 377




141. A ball is thrown into the air and its position is given
by h(t) = −4.9t2 + 60t + 5m. Find the height at which
the ball stops ascending. How long after it is thrown doesthis happen?
For the following exercises, consider the production ofgold during the California gold rush (1848–1888). The
production of gold can be modeled by G(t) = (25t)⎛


⎝t
2 + 16⎞⎠


,


where t is the number of years since the rush began
(0 ≤ t ≤ 40) and G is ounces of gold produced (in
millions). A summary of the data is shown in the followingfigure.


142. Find when the maximum (local and global) goldproduction occurred, and the amount of gold producedduring that maximum.
143. Find when the minimum (local and global) goldproduction occurred. What was the amount of goldproduced during this minimum?
Find the critical points, maxima, and minima for thefollowing piecewise functions.
144. y = ⎧



⎨x


2 − 4x 0 ≤ x ≤ 1


x2 − 4 1 < x ≤ 2


145. y = ⎧

⎨ x


2 + 1 x ≤ 1


x2 − 4x + 5 x > 1


For the following exercises, find the critical points of thefollowing generic functions. Are they maxima, minima, orneither? State the necessary conditions.
146. y = ax2 + bx + c, given that a > 0
147. y = (x − 1)a, given that a > 1


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4.4 | The Mean Value Theorem
Learning Objectives


4.4.1 Explain the meaning of Rolle’s theorem.
4.4.2 Describe the significance of the Mean Value Theorem.
4.4.3 State three important consequences of the Mean Value Theorem.


The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at theend of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.
Rolle’s Theorem
Informally, Rolle’s theorem states that if the outputs of a differentiable function f are equal at the endpoints of an interval,
then there must be an interior point c where f ′(c) = 0. Figure 4.21 illustrates this theorem.


Figure 4.21 If a differentiable function f satisfies f (a) = f (b), then its derivative must be zero at some point(s)
between a and b.


Theorem 4.4: Rolle’s Theorem
Let f be a continuous function over the closed interval [a, b] and differentiable over the open interval (a, b) such
that f (a) = f (b). There then exists at least one c ∈ (a, b) such that f ′(c) = 0.


Proof
Let k = f (a) = f (b). We consider three cases:


1. f (x) = k for all x ∈ (a, b).
2. There exists x ∈ (a, b) such that f (x) > k.
3. There exists x ∈ (a, b) such that f (x) < k.


Case 1: If f (x) = 0 for all x ∈ (a, b), then f ′(x) = 0 for all x ∈ (a, b).
Case 2: Since f is a continuous function over the closed, bounded interval [a, b], by the extreme value theorem, it has
an absolute maximum. Also, since there is a point x ∈ (a, b) such that f (x) > k, the absolute maximum is greater than
k. Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an
interior point c ∈ (a, b). Because f has a maximum at an interior point c, and f is differentiable at c, by Fermat’s
theorem, f ′(c) = 0.


Chapter 4 | Applications of Derivatives 379




Case 3: The case when there exists a point x ∈ (a, b) such that f (x) < k is analogous to case 2, with maximum replaced
by minimum.

An important point about Rolle’s theorem is that the differentiability of the function f is critical. If f is not differentiable,
even at a single point, the result may not hold. For example, the function f (x) = |x| − 1 is continuous over [−1, 1] and
f (−1) = 0 = f (1), but f ′(c) ≠ 0 for any c ∈ (−1, 1) as shown in the following figure.


Figure 4.22 Since f (x) = |x| − 1 is not differentiable at
x = 0, the conditions of Rolle’s theorem are not satisfied. In
fact, the conclusion does not hold here; there is no c ∈ (−1, 1)
such that f ′(c) = 0.


Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c where
f ′(c) = 0.


Example 4.14
Using Rolle’s Theorem
For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and findall values c in the given interval where f ′(c) = 0.


a. f (x) = x2 + 2x over [−2, 0]
b. f (x) = x3 − 4x over [−2, 2]


Solution
a. Since f is a polynomial, it is continuous and differentiable everywhere. In addition, f (−2) = 0 = f (0).


Therefore, f satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value
c ∈ (−2, 0) such that f ′(c) = 0. Since f ′(x) = 2x + 2 = 2(x + 1), we see that
f ′(c) = 2(c + 1) = 0 implies c = −1 as shown in the following graph.


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4.14


Figure 4.23 This function is continuous and differentiableover [−2, 0], f ′(c) = 0 when c = −1.
b. As in part a. f is a polynomial and therefore is continuous and differentiable everywhere. Also,


f (−2) = 0 = f (2). That said, f satisfies the criteria of Rolle’s theorem. Differentiating, we find that
f ′(x) = 3x2 − 4. Therefore, f ′(c) = 0 when x = ± 2


3
. Both points are in the interval [−2, 2], and,


therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph.


Figure 4.24 For this polynomial over [−2, 2], f ′(c) = 0
at x = ±2/ 3.


Verify that the function f (x) = 2x2 − 8x + 6 defined over the interval [1, 3] satisfies the conditions of
Rolle’s theorem. Find all points c guaranteed by Rolle’s theorem.


Chapter 4 | Applications of Derivatives 381




The Mean Value Theorem and Its Meaning
Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions
f that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are
not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’stheorem (Figure 4.25). The Mean Value Theorem states that if f is continuous over the closed interval [a, b] and
differentiable over the open interval (a, b), then there exists a point c ∈ (a, b) such that the tangent line to the graph of
f at c is parallel to the secant line connecting ⎛⎝a, f (a)⎞⎠ and ⎛⎝b, f (b)⎞⎠.


Figure 4.25 The Mean Value Theorem says that for a functionthat meets its conditions, at some point the tangent line has thesame slope as the secant line between the ends. For thisfunction, there are two values c1 and c2 such that the tangent
line to f at c1 and c2 has the same slope as the secant line.


Theorem 4.5: Mean Value Theorem
Let f be continuous over the closed interval [a, b] and differentiable over the open interval (a, b). Then, there
exists at least one point c ∈ (a, b) such that


f ′(c) =
f (b) − f (a)


b − a
.


Proof
The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem.Consider the line connecting ⎛⎝a, f (a)⎞⎠ and ⎛⎝b, f (b)⎞⎠. Since the slope of that line is


f (b) − f (a)
b − a


and the line passes through the point ⎛⎝a, f (a)⎞⎠, the equation of that line can be written as
y =


f (b) − f (a)
b − a


(x − a) + f (a).


Let g(x) denote the vertical difference between the point ⎛⎝x, f (x)⎞⎠ and the point (x, y) on that line. Therefore,
g(x) = f (x) −




f (b) − f (a)


b − a
(x − a) + f (a)



⎦.


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Figure 4.26 The value g(x) is the vertical difference
between the point (x, f (x)) and the point (x, y) on the secant
line connecting (a, f (a)) and (b, f (b)).


Since the graph of f intersects the secant line when x = a and x = b, we see that g(a) = 0 = g(b). Since f is a
differentiable function over (a, b), g is also a differentiable function over (a, b). Furthermore, since f is continuous
over [a, b], g is also continuous over [a, b]. Therefore, g satisfies the criteria of Rolle’s theorem. Consequently, there
exists a point c ∈ (a, b) such that g′(c) = 0. Since


g′(x) = f ′(x) −
f (b) − f (a)


b − a
,


we see that
g′(c) = f ′(c) −


f (b) − f (a)
b − a


.


Since g′(c) = 0, we conclude that
f ′(c) =


f (b) − f (a)
b − a


.



In the next example, we show how the Mean Value Theorem can be applied to the function f (x) = x over the interval
[0, 9]. The method is the same for other functions, although sometimes with more interesting consequences.
Example 4.15
Verifying that the Mean Value Theorem Applies
For f (x) = x over the interval [0, 9], show that f satisfies the hypothesis of the Mean Value Theorem, and
therefore there exists at least one value c ∈ (0, 9) such that f ′(c) is equal to the slope of the line connecting

⎝0, f (0)⎞⎠ and ⎛⎝9, f (9)⎞⎠. Find these values c guaranteed by the Mean Value Theorem.
Solution
We know that f (x) = x is continuous over [0, 9] and differentiable over (0, 9). Therefore, f satisfies the
hypotheses of the Mean Value Theorem, and there must exist at least one value c ∈ (0, 9) such that f ′ (c) is
equal to the slope of the line connecting ⎛⎝0, f (0)⎞⎠ and ⎛⎝9, f (9)⎞⎠ (Figure 4.27). To determine which value(s)


Chapter 4 | Applications of Derivatives 383




of c are guaranteed, first calculate the derivative of f . The derivative f ′ (x) = 1
(2 x)


. The slope of the line
connecting (0, f (0)) and (9, f (9)) is given by


f (9) − f (0)
9 − 0


= 9 − 0
9 − 0


= 3
9
= 1


3
.


We want to find c such that f ′(c) = 1
3
. That is, we want to find c such that


1
2 c


= 1
3
.


Solving this equation for c, we obtain c = 9
4
. At this point, the slope of the tangent line equals the slope of the


line joining the endpoints.


Figure 4.27 The slope of the tangent line at c = 9/4 is the same as the slope of the line segment
connecting (0, 0) and (9, 3).


One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for1 h down a straight road with an average velocity of 45 mph. Let s(t) and v(t) denote the position and velocity of the
car, respectively, for 0 ≤ t ≤ 1 h. Assuming that the position function s(t) is differentiable, we can apply the Mean Value
Theorem to conclude that, at some time c ∈ (0, 1), the speed of the car was exactly


v(c) = s′ (c) = s(1) − s(0)
1 − 0


= 45 mph.


Example 4.16
Mean Value Theorem and Velocity
If a rock is dropped from a height of 100 ft, its position t seconds after it is dropped until it hits the ground is


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4.15


given by the function s(t) = −16t2 + 100.
a. Determine how long it takes before the rock hits the ground.
b. Find the average velocity vavg of the rock for when the rock is released and the rock hits the ground.
c. Find the time t guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is


vavg.


Solution
a. When the rock hits the ground, its position is s(t) = 0. Solving the equation −16t2 + 100 = 0 for t,


we find that t = ±5
2
sec. Since we are only considering t ≥ 0, the ball will hit the ground 5


2
sec after


it is dropped.
b. The average velocity is given by


vavg =
s(5/2) − s(0)


5/2 − 0
= 1 − 100


5/2
= −40 ft/sec.


c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to finda time t such that v(t) = s′(t) = vavg = −40 ft/sec. Since s(t) is continuous over the interval [0, 5/2]
and differentiable over the interval (0, 5/2), by the Mean Value Theorem, there is guaranteed to be a
point c ∈ (0, 5/2) such that


s′ (c) = s(5/2) − s(0)
5/2 − 0


= −40.


Taking the derivative of the position function s(t), we find that s′ (t) = −32t. Therefore, the equation
reduces to s′ (c) = −32c = −40. Solving this equation for c, we have c = 5


4
. Therefore, 5


4
sec after


the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall:
−40 ft/sec.


Figure 4.28 At time t = 5/4 sec, the velocity of the rock is
equal to its average velocity from the time it is dropped until ithits the ground.


Suppose a ball is dropped from a height of 200 ft. Its position at time t is s(t) = −16t2 + 200. Find the
time t when the instantaneous velocity of the ball equals its average velocity.


Chapter 4 | Applications of Derivatives 385




Corollaries of the Mean Value Theorem
Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we usein upcoming sections.
At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to concludethat the converse is also true. In particular, if f ′ (x) = 0 for all x in some interval I, then f (x) is constant over that
interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discussthem shortly.
Theorem 4.6: Corollary 1: Functions with a Derivative of Zero
Let f be differentiable over an interval I. If f ′ (x) = 0 for all x ∈ I, then f (x) = constant for all x ∈ I.


Proof
Since f is differentiable over I, f must be continuous over I. Suppose f (x) is not constant for all x in I. Then there
exist a, b ∈ I, where a ≠ b and f (a) ≠ f (b). Choose the notation so that a < b. Therefore,


f (b) − f (a)
b − a


≠ 0.


Since f is a differentiable function, by the Mean Value Theorem, there exists c ∈ (a, b) such that
f ′ (c) =


f (b) − f (a)
b − a


.


Therefore, there exists c ∈ I such that f ′ (c) ≠ 0, which contradicts the assumption that f ′ (x) = 0 for all x ∈ I.

From Corollary 1: Functions with a Derivative of Zero, it follows that if two functions have the same derivative,they differ by, at most, a constant.
Theorem 4.7: Corollary 2: Constant Difference Theorem
If f and g are differentiable over an interval I and f ′ (x) = g′ (x) for all x ∈ I, then f (x) = g(x) + C for some
constant C.


Proof
Let h(x) = f (x) − g(x). Then, h′ (x) = f ′ (x) − g′ (x) = 0 for all x ∈ I. By Corollary 1, there is a constant C such that
h(x) = C for all x ∈ I. Therefore, f (x) = g(x) + C for all x ∈ I.

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recallthat a function f is increasing over I if f (x1) < f (x2) whenever x1 < x2, whereas f is decreasing over I if
f (x)1 > f (x2) whenever x1 < x2. Using the Mean Value Theorem, we can show that if the derivative of a function is
positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure 4.29). Wemake use of this fact in the next section, where we show how to use the derivative of a function to locate local maximumand minimum values of the function, and how to determine the shape of the graph.
This fact is important because it means that for a given function f , if there exists a function F such that F′ (x) = f (x);
then, the only other functions that have a derivative equal to f are F(x) + C for some constant C. We discuss this result
in more detail later in the chapter.


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Figure 4.29 If a function has a positive derivative over some interval I, then the function
increases over that interval I; if the derivative is negative over some interval I, then the
function decreases over that interval I.


Theorem 4.8: Corollary 3: Increasing and Decreasing Functions
Let f be continuous over the closed interval [a, b] and differentiable over the open interval (a, b).


i. If f ′ (x) > 0 for all x ∈ (a, b), then f is an increasing function over [a, b].
ii. If f ′(x) < 0 for all x ∈ (a, b), then f is a decreasing function over [a, b].


Proof
We will prove i.; the proof of ii. is similar. Suppose f is not an increasing function on I. Then there exist a and b in I
such that a < b, but f (a) ≥ f (b). Since f is a differentiable function over I, by the Mean Value Theorem there exists
c ∈ (a, b) such that


f ′ (c) =
f (b) − f (a)


b − a
.


Since f (a) ≥ f (b), we know that f (b) − f (a) ≤ 0. Also, a < b tells us that b − a > 0. We conclude that
f ′ (c) =


f (b) − f (a)
b − a


≤ 0.


However, f ′ (x) > 0 for all x ∈ I. This is a contradiction, and therefore f must be an increasing function over I.


Chapter 4 | Applications of Derivatives 387




4.4 EXERCISES
148. Why do you need continuity to apply the Mean ValueTheorem? Construct a counterexample.
149. Why do you need differentiability to apply the MeanValue Theorem? Find a counterexample.
150. When are Rolle’s theorem and the Mean ValueTheorem equivalent?
151. If you have a function with a discontinuity, is it stillpossible to have f ′ (c)(b − a) = f (b) − f (a)? Draw such
an example or prove why not.
For the following exercises, determine over what intervals(if any) the Mean Value Theorem applies. Justify youranswer.
152. y = sin(πx)
153. y = 1


x3


154. y = 4 − x2


155. y = x2 − 4
156. y = ln(3x − 5)
For the following exercises, graph the functions on acalculator and draw the secant line that connects theendpoints. Estimate the number of points c such that
f ′(c)(b − a) = f (b) − f (a).


157. [T] y = 3x3 + 2x + 1 over [−1, 1]
158. [T] y = tan⎛⎝π4x⎞⎠ over ⎡⎣−32, 32⎤⎦
159. [T] y = x2 cos(πx) over [−2, 2]
160. [T]
y = x6 − 3


4
x5 − 9


8
x4 + 15


16
x3 + 3


32
x2 + 3


16
x + 1


32
over


[−1, 1]


For the following exercises, use the Mean Value Theoremand find all points 0 < c < 2 such that
f (2) − f (0) = f ′ (c)(2 − 0).


161. f (x) = x3
162. f (x) = sin(πx)


163. f (x) = cos(2πx)
164. f (x) = 1 + x + x2
165. f (x) = (x − 1)10


166. f (x) = (x − 1)9
For the following exercises, show there is no c such that
f (1) − f (−1) = f ′ (c)(2). Explain why the Mean Value
Theorem does not apply over the interval [−1, 1].
167. f (x) = |x − 12|
168. f (x) = 1


x2


169. f (x) = |x|
170. f (x) = ⌊x⌋ (Hint: This is called the floor function
and it is defined so that f (x) is the largest integer less than
or equal to x.)
For the following exercises, determine whether the MeanValue Theorem applies for the functions over the giveninterval [a, b]. Justify your answer.
171. y = ex over [0, 1]
172. y = ln(2x + 3) over ⎡⎣−32, 0⎤⎦
173. f (x) = tan(2πx) over [0, 2]
174. y = 9 − x2 over [−3, 3]
175. y = 1


|x + 1|
over [0, 3]


176. y = x3 + 2x + 1 over [0, 6]
177. y = x2 + 3x + 2x over [−1, 1]
178. y = x


sin(πx) + 1
over [0, 1]


179. y = ln(x + 1) over [0, e − 1]
180. y = xsin(πx) over [0, 2]


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181. y = 5 + |x| over [−1, 1]
For the following exercises, consider the roots of theequation.
182. Show that the equation y = x3 + 3x2 + 16 has
exactly one real root. What is it?
183. Find the conditions for exactly one root (double root)
for the equation y = x2 + bx + c
184. Find the conditions for y = ex − b to have one root.
Is it possible to have more than one root?
For the following exercises, use a calculator to graph thefunction over the interval [a, b] and graph the secant line
from a to b. Use the calculator to estimate all values of c
as guaranteed by the Mean Value Theorem. Then, find theexact value of c, if possible, or write the final equation
and use a calculator to estimate to four digits.
185. [T] y = tan(πx) over ⎡⎣−14, 14⎤⎦
186. [T] y = 1


x + 1
over [0, 3]


187. [T] y = |x2 + 2x − 4| over [−4, 0]
188. [T] y = x + 1x over ⎡⎣12, 4⎤⎦
189. [T] y = x + 1 + 1


x2
over [3, 8]


190. At 10:17 a.m., you pass a police car at 55 mph thatis stopped on the freeway. You pass a second police car at55 mph at 10:53 a.m., which is located 39 mi from the firstpolice car. If the speed limit is 60 mph, can the police citeyou for speeding?
191. Two cars drive from one spotlight to the next, leavingat the same time and arriving at the same time. Is thereever a time when they are going the same speed? Prove ordisprove.
192. Show that y = sec2 x and y = tan2 x have the same
derivative. What can you say about y = sec2 x − tan2 x?
193. Show that y = csc2 x and y = cot2 x have the same
derivative. What can you say about y = csc2 x − cot2 x?


Chapter 4 | Applications of Derivatives 389




4.5 | Derivatives and the Shape of a Graph
Learning Objectives


4.5.1 Explain how the sign of the first derivative affects the shape of a function’s graph.
4.5.2 State the first derivative test for critical points.
4.5.3 Use concavity and inflection points to explain how the sign of the second derivative affectsthe shape of a function’s graph.
4.5.4 Explain the concavity test for a function over an open interval.
4.5.5 Explain the relationship between a function and its first and second derivatives.
4.5.6 State the second derivative test for local extrema.


Earlier in this chapter we stated that if a function f has a local extremum at a point c, then c must be a critical point
of f . However, a function is not guaranteed to have a local extremum at a critical point. For example, f (x) = x3 has a
critical point at x = 0 since f ′(x) = 3x2 is zero at x = 0, but f does not have a local extremum at x = 0. Using the
results from the previous section, we are now able to determine whether a critical point of a function actually correspondsto a local extreme value. In this section, we also see how the second derivative provides information about the shape of agraph by describing whether the graph of a function curves upward or curves downward.
The First Derivative Test
Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval I then the
function is increasing over I. On the other hand, if the derivative of the function is negative over an interval I, then the
function is decreasing over I as shown in the following figure.


Figure 4.30 Both functions are increasing over the interval
(a, b). At each point x, the derivative f ′(x) > 0. Both
functions are decreasing over the interval (a, b). At each point
x, the derivative f ′(x) < 0.


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A continuous function f has a local maximum at point c if and only if f switches from increasing to decreasing at
point c. Similarly, f has a local minimum at c if and only if f switches from decreasing to increasing at c. If f is a
continuous function over an interval I containing c and differentiable over I, except possibly at c, the only way f
can switch from increasing to decreasing (or vice versa) at point c is if f ′ changes sign as x increases through c. If
f is differentiable at c, the only way that f ′. can change sign as x increases through c is if f ′ (c) = 0. Therefore,
for a function f that is continuous over an interval I containing c and differentiable over I, except possibly at c, the
only way f can switch from increasing to decreasing (or vice versa) is if f ′(c) = 0 or f ′ (c) is undefined. Consequently,
to locate local extrema for a function f , we look for points c in the domain of f such that f ′(c) = 0 or f ′ (c) is
undefined. Recall that such points are called critical points of f .
Note that f need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In
Figure 4.31, we show that if a continuous function f has a local extremum, it must occur at a critical point, but a function
may not have a local extremum at a critical point. We show that if f has a local extremum at a critical point, then the sign
of f ′ switches as x increases through that point.


Figure 4.31 The function f has four critical points: a, b, c, and d. The function f has local maxima at a
and d, and a local minimum at b. The function f does not have a local extremum at c. The sign of f ′
changes at all local extrema.


Using Figure 4.31, we summarize the main results regarding local extrema.
• If a continuous function f has a local extremum, it must occur at a critical point c.
• The function has a local extremum at the critical point c if and only if the derivative f ′ switches sign as x
increases through c.


• Therefore, to test whether a function has a local extremum at a critical point c, we must determine the sign of
f ′ (x) to the left and right of c.


This result is known as the first derivative test.


Chapter 4 | Applications of Derivatives 391




Theorem 4.9: First Derivative Test
Suppose that f is a continuous function over an interval I containing a critical point c. If f is differentiable over
I, except possibly at point c, then f (c) satisfies one of the following descriptions:


i. If f ′ changes sign from positive when x < c to negative when x > c, then f (c) is a local maximum of f .
ii. If f ′ changes sign from negative when x < c to positive when x > c, then f (c) is a local minimum of f .
iii. If f ′ has the same sign for x < c and x > c, then f (c) is neither a local maximum nor a local minimum of


f .


We can summarize the first derivative test as a strategy for locating local extrema.
Problem-Solving Strategy: Using the First Derivative Test
Consider a function f that is continuous over an interval I.


1. Find all critical points of f and divide the interval I into smaller intervals using the critical points as
endpoints.


2. Analyze the sign of f ′ in each of the subintervals. If f ′ is continuous over a given subinterval (which is
typically the case), then the sign of f ′ in that subinterval does not change and, therefore, can be determined
by choosing an arbitrary test point x in that subinterval and by evaluating the sign of f ′ at that test point. Use
the sign analysis to determine whether f is increasing or decreasing over that interval.


3. Use First Derivative Test and the results of step 2 to determine whether f has a local maximum, a local
minimum, or neither at each of the critical points.


Now let’s look at how to use this strategy to locate all local extrema for particular functions.
Example 4.17
Using the First Derivative Test to Find Local Extrema
Use the first derivative test to find the location of all local extrema for f (x) = x3 − 3x2 − 9x − 1. Use a
graphing utility to confirm your results.
Solution
Step 1. The derivative is f ′ (x) = 3x2 − 6x − 9. To find the critical points, we need to find where f ′ (x) = 0.
Factoring the polynomial, we conclude that the critical points must satisfy


3(x2 − 2x − 3) = 3(x − 3)(x + 1) = 0.


Therefore, the critical points are x = 3, −1. Now divide the interval (−∞, ∞) into the smaller intervals
(−∞, −1), (−1, 3) and (3, ∞).


Step 2. Since f ′ is a continuous function, to determine the sign of f ′ (x) over each subinterval, it suffices to
choose a point over each of the intervals (−∞, −1), (−1, 3) and (3, ∞) and determine the sign of f ′ at each


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of these points. For example, let’s choose x = −2, x = 0, and x = 4 as test points.
Interval Test Point Sign of f′ (x) = 3(x−3)(x+1) at Test Point Conclusion
(−∞, −1) x = −2 (+)(−)(−) = + f is increasing.


(−1, 3) x = 0 (+)(−)(+) = − f is decreasing.


(3, ∞) x = 4 (+)(+)(+) = + f is increasing.


Step 3. Since f ′ switches sign from positive to negative as x increases through 1, f has a local maximum at
x = −1. Since f ′ switches sign from negative to positive as x increases through 3, f has a local minimum at
x = 3. These analytical results agree with the following graph.


Figure 4.32 The function f has a maximum at x = −1 and
a minimum at x = 3


Chapter 4 | Applications of Derivatives 393




4.16 Use the first derivative test to locate all local extrema for f (x) = −x3 + 3
2
x2 + 18x.


Example 4.18
Using the First Derivative Test
Use the first derivative test to find the location of all local extrema for f (x) = 5x1/3 − x5/3. Use a graphing
utility to confirm your results.
Solution
Step 1. The derivative is


f ′ (x) = 5
3
x−2/3 − 5


3
x2/3 = 5


3x2/3
− 5x


2/3


3
= 5 − 5x


4/3


3x2/3
=


5⎛⎝1 − x
4/3⎞


3x2/3
.


The derivative f ′ (x) = 0 when 1 − x4/3 = 0. Therefore, f ′ (x) = 0 at x = ±1. The derivative f ′ (x) is
undefined at x = 0. Therefore, we have three critical points: x = 0, x = 1, and x = −1. Consequently,
divide the interval (−∞, ∞) into the smaller intervals (−∞, −1), (−1, 0), (0, 1), and (1, ∞).
Step 2: Since f ′ is continuous over each subinterval, it suffices to choose a test point x in each of the
intervals from step 1 and determine the sign of f ′ at each of these points. The points
x = −2, x = − 1


2
, x = 1


2
, and x = 2 are test points for these intervals.


Interval Test Point Sign of f′ (x) = 5⎛⎝1 − x4/3⎞⎠
3x2/3


at Test Point Conclusion


(−∞, −1) x = −2 (+)(−)
+ = −


f is decreasing.


(−1, 0) x = − 1
2


(+)(+)
+ = +


f is increasing.


(0, 1) x = 1
2


(+)(+)
+ = +


f is increasing.


(1, ∞) x = 2 (+)(−)
+ = −


f is decreasing.


Step 3: Since f is decreasing over the interval (−∞, −1) and increasing over the interval (−1, 0), f has a
local minimum at x = −1. Since f is increasing over the interval (−1, 0) and the interval (0, 1), f does not
have a local extremum at x = 0. Since f is increasing over the interval (0, 1) and decreasing over the interval
(1, ∞), f has a local maximum at x = 1. The analytical results agree with the following graph.


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4.17


Figure 4.33 The function f has a local minimum at x = −1
and a local maximum at x = 1.


Use the first derivative test to find all local extrema for f (x) = x − 13 .


Concavity and Points of Inflection
We now know how to determine where a function is increasing or decreasing. However, there is another issue to considerregarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion iscalled the concavity of the function.
Figure 4.34(a) shows a function f with a graph that curves upward. As x increases, the slope of the tangent line
increases. Thus, since the derivative increases as x increases, f ′ is an increasing function. We say this function f is
concave up. Figure 4.34(b) shows a function f that curves downward. As x increases, the slope of the tangent line
decreases. Since the derivative decreases as x increases, f ′ is a decreasing function. We say this function f is concave
down.
Definition
Let f be a function that is differentiable over an open interval I. If f ′ is increasing over I, we say f is concave
up over I. If f ′ is decreasing over I, we say f is concave down over I.


Chapter 4 | Applications of Derivatives 395




Figure 4.34 (a), (c) Since f ′ is increasing over the interval (a, b), we say f
is concave up over (a, b). (b), (d) Since f ′ is decreasing over the interval
(a, b), we say f is concave down over (a, b).


In general, without having the graph of a function f , how can we determine its concavity? By definition, a function f is
concave up if f ′ is increasing. From Corollary 3, we know that if f ′ is a differentiable function, then f ′ is increasing
if its derivative f ″(x) > 0. Therefore, a function f that is twice differentiable is concave up when f ″(x) > 0. Similarly,
a function f is concave down if f ′ is decreasing. We know that a differentiable function f ′ is decreasing if its derivative
f ″(x) < 0. Therefore, a twice-differentiable function f is concave down when f ″(x) < 0. Applying this logic is known
as the concavity test.
Theorem 4.10: Test for Concavity
Let f be a function that is twice differentiable over an interval I.


i. If f ″(x) > 0 for all x ∈ I, then f is concave up over I.
ii. If f ″(x) < 0 for all x ∈ I, then f is concave down over I.


We conclude that we can determine the concavity of a function f by looking at the second derivative of f . In addition, we
observe that a function f can switch concavity (Figure 4.35). However, a continuous function can switch concavity only
at a point x if f ″(x) = 0 or f ″(x) is undefined. Consequently, to determine the intervals where a function f is concave
up and concave down, we look for those values of x where f ″(x) = 0 or f ″(x) is undefined. When we have determined


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these points, we divide the domain of f into smaller intervals and determine the sign of f ″ over each of these smaller
intervals. If f ″ changes sign as we pass through a point x, then f changes concavity. It is important to remember that a
function f may not change concavity at a point x even if f ″(x) = 0 or f ″(x) is undefined. If, however, f does change
concavity at a point a and f is continuous at a, we say the point ⎛⎝a, f (a)⎞⎠ is an inflection point of f .


Definition
If f is continuous at a and f changes concavity at a, the point ⎛⎝a, f (a)⎞⎠ is an inflection point of f .


Figure 4.35 Since f ″(x) > 0 for x < a, the function f is concave up over the interval
(−∞, a). Since f ″(x) < 0 for x > a, the function f is concave down over the interval
(a, ∞). The point ⎛⎝a, f (a)⎞⎠ is an inflection point of f .


Example 4.19
Testing for Concavity
For the function f (x) = x3 − 6x2 + 9x + 30, determine all intervals where f is concave up and all intervals
where f is concave down. List all inflection points for f . Use a graphing utility to confirm your results.
Solution
To determine concavity, we need to find the second derivative f ″(x). The first derivative is
f ′(x) = 3x2 − 12x + 9, so the second derivative is f ″(x) = 6x − 12. If the function changes concavity, it
occurs either when f ″(x) = 0 or f ″(x) is undefined. Since f ″ is defined for all real numbers x, we need only
find where f ″(x) = 0. Solving the equation 6x − 12 = 0, we see that x = 2 is the only place where f could
change concavity. We now test points over the intervals (−∞, 2) and (2, ∞) to determine the concavity of f .
The points x = 0 and x = 3 are test points for these intervals.


Chapter 4 | Applications of Derivatives 397




4.18


Interval Test Point Sign of f″(x) = 6x−12 at Test Point Conclusion
(−∞, 2) x = 0 − f is concave down


(2, ∞) x = 3 + f is concave up.


We conclude that f is concave down over the interval (−∞, 2) and concave up over the interval (2, ∞). Since
f changes concavity at x = 2, the point ⎛⎝2, f (2)⎞⎠ = (2, 32) is an inflection point. Figure 4.36 confirms the
analytical results.


Figure 4.36 The given function has a point of inflection at
(2, 32) where the graph changes concavity.


For f (x) = −x3 + 3
2
x2 + 18x, find all intervals where f is concave up and all intervals where f is


concave down.


We now summarize, in Table 4.6, the information that the first and second derivatives of a function f provide about the
graph of f , and illustrate this information in Figure 4.37.


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Sign of f′ Sign of f″ Is f increasing or decreasing? Concavity
Positive Positive Increasing Concave up
Positive Negative Increasing Concave down
Negative Positive Decreasing Concave up
Negative Negative Decreasing Concave down


Table 4.6What Derivatives Tell Us about Graphs


Figure 4.37 Consider a twice-differentiable function f over an open interval I. If f ′(x) > 0 for all x ∈ I, the
function is increasing over I. If f ′(x) < 0 for all x ∈ I, the function is decreasing over I. If f ″(x) > 0 for all
x ∈ I, the function is concave up. If f ″(x) < 0 for all x ∈ I, the function is concave down on I.


The Second Derivative Test
The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used tolocate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.
We know that if a continuous function has a local extrema, it must occur at a critical point. However, a function need nothave a local extrema at a critical point. Here we examine how the second derivative test can be used to determine whethera function has a local extremum at a critical point. Let f be a twice-differentiable function such that f ′ (a) = 0 and f ″
is continuous over an open interval I containing a. Suppose f ″(a) < 0. Since f ″ is continuous over I, f ″(x) < 0 for
all x ∈ I (Figure 4.38). Then, by Corollary 3, f ′ is a decreasing function over I. Since f ′ (a) = 0, we conclude that
for all x ∈ I, f ′ (x) > 0 if x < a and f ′ (x) < 0 if x > a. Therefore, by the first derivative test, f has a local maximum
at x = a. On the other hand, suppose there exists a point b such that f ′ (b) = 0 but f ″(b) > 0. Since f ″ is continuous
over an open interval I containing b, then f ″(x) > 0 for all x ∈ I (Figure 4.38). Then, by Corollary 3, f ′ is an
increasing function over I. Since f ′ (b) = 0, we conclude that for all x ∈ I, f ′ (x) < 0 if x < b and f ′ (x) > 0 if
x > b. Therefore, by the first derivative test, f has a local minimum at x = b.


Chapter 4 | Applications of Derivatives 399




Figure 4.38 Consider a twice-differentiable function f such
that f ″ is continuous. Since f ′(a) = 0 and f ″(a) < 0,
there is an interval I containing a such that for all x in I, f
is increasing if x < a and f is decreasing if x > a. As a
result, f has a local maximum at x = a. Since f ′(b) = 0
and f ″(b) > 0, there is an interval I containing b such that
for all x in I, f is decreasing if x < b and f is increasing
if x > b. As a result, f has a local minimum at x = b.


Theorem 4.11: Second Derivative Test
Suppose f ′ (c) = 0, f ″ is continuous over an interval containing c.


i. If f ″(c) > 0, then f has a local minimum at c.
ii. If f ″(c) < 0, then f has a local maximum at c.
iii. If f ″(c) = 0, then the test is inconclusive.


Note that for case iii. when f ″(c) = 0, then f may have a local maximum, local minimum, or neither at c. For
example, the functions f (x) = x3, f (x) = x4, and f (x) = −x4 all have critical points at x = 0. In each case, the
second derivative is zero at x = 0. However, the function f (x) = x4 has a local minimum at x = 0 whereas the function
f (x) = −x4 has a local maximum at x, and the function f (x) = x3 does not have a local extremum at x = 0.
Let’s now look at how to use the second derivative test to determine whether f has a local maximum or local minimum at
a critical point c where f ′ (c) = 0.
Example 4.20
Using the Second Derivative Test
Use the second derivative to find the location of all local extrema for f (x) = x5 − 5x3.
Solution
To apply the second derivative test, we first need to find critical points c where f ′ (c) = 0. The derivative is


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f ′ (x) = 5x4 − 15x2. Therefore, f ′ (x) = 5x4 − 15x2 = 5x2 ⎛⎝x2 − 3⎞⎠ = 0 when x = 0, ± 3.
To determine whether f has a local extrema at any of these points, we need to evaluate the sign of f ″ at these
points. The second derivative is


f ″(x) = 20x3 − 30x = 10x⎛⎝2x
2 − 3⎞⎠.


In the following table, we evaluate the second derivative at each of the critical points and use the secondderivative test to determine whether f has a local maximum or local minimum at any of these points.
x f″(x) Conclusion
− 3 −30 3 Local maximum


0 0 Second derivative test is inconclusive


3 30 3 Local minimum


By the second derivative test, we conclude that f has a local maximum at x = − 3 and f has a local minimum
at x = 3. The second derivative test is inconclusive at x = 0. To determine whether f has a local extrema at
x = 0, we apply the first derivative test. To evaluate the sign of f ′ (x) = 5x2 ⎛⎝x2 − 3⎞⎠ for x ∈ ⎛⎝− 3, 0⎞⎠ and
x ∈ ⎛⎝0, 3



⎠, let x = −1 and x = 1 be the two test points. Since f ′ (−1) < 0 and f ′ (1) < 0, we conclude


that f is decreasing on both intervals and, therefore, f does not have a local extrema at x = 0 as shown in the
following graph.


Chapter 4 | Applications of Derivatives 401




4.19


Figure 4.39 The function f has a local maximum at x = − 3 and a local minimum at x = 3


Consider the function f (x) = x3 − ⎛⎝32⎞⎠x2 − 18x. The points c = 3, −2 satisfy f ′ (c) = 0. Use the
second derivative test to determine whether f has a local maximum or local minimum at those points.


We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquiredan understanding of the basic shape of the graph. In the next section we discuss what happens to a function as x → ±∞.
At that point, we have enough tools to provide accurate graphs of a large variety of functions.


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4.5 EXERCISES
194. If c is a critical point of f (x), when is there no
local maximum or minimum at c? Explain.
195. For the function y = x3, is x = 0 both an
inflection point and a local maximum/minimum?
196. For the function y = x3, is x = 0 an inflection
point?
197. Is it possible for a point c to be both an inflection
point and a local extrema of a twice differentiablefunction?
198. Why do you need continuity for the first derivativetest? Come up with an example.
199. Explain whether a concave-down function has tocross y = 0 for some value of x.
200. Explain whether a polynomial of degree 2 can have
an inflection point.
For the following exercises, analyze the graphs of f ′,
then list all intervals where f is increasing or decreasing.
201.


202.


203.


204.


Chapter 4 | Applications of Derivatives 403




205.


For the following exercises, analyze the graphs of f ′,
then list all intervals where


a. f is increasing and decreasing and
b. the minima and maxima are located.


206.


207.


208.


209.


210.


For the following exercises, analyze the graphs of f ′,
then list all inflection points and intervals f that are
concave up and concave down.
211.


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212.


213.


214.


215.


For the following exercises, draw a graph that satisfiesthe given specifications for the domain x = [−3, 3]. The
function does not have to be continuous or differentiable.
216. f (x) > 0, f ′ (x) > 0 over
x > 1, −3 < x < 0, f ′ (x) = 0 over 0 < x < 1
217. f ′ (x) > 0 over x > 2, −3 < x < −1, f ′ (x) < 0
over −1 < x < 2, f ″(x) < 0 for all x
218. f ″(x) < 0 over
−1 < x < 1, f ″(x) > 0, −3 < x < −1, 1 < x < 3,


local maximum at x = 0, local minima at x = ±2
219. There is a local maximum at x = 2, local minimum
at x = 1, and the graph is neither concave up nor concave
down.
220. There are local maxima at x = ±1, the function is
concave up for all x, and the function remains positive for
all x.
For the following exercises, determine


a. intervals where f is increasing or decreasing and
b. local minima and maxima of f .


221. f (x) = sinx + sin3 x over −π < x < π
222. f (x) = x2 + cosx
For the following exercises, determine a. intervals where f
is concave up or concave down, and b. the inflection pointsof f .
223. f (x) = x3 − 4x2 + x + 2


Chapter 4 | Applications of Derivatives 405




For the following exercises, determine
a. intervals where f is increasing or decreasing,
b. local minima and maxima of f ,
c. intervals where f is concave up and concave


down, and
d. the inflection points of f .


224. f (x) = x2 − 6x
225. f (x) = x3 − 6x2
226. f (x) = x4 − 6x3


227. f (x) = x11 − 6x10


228. f (x) = x + x2 − x3
229. f (x) = x2 + x + 1
230. f (x) = x3 + x4
For the following exercises, determine


a. intervals where f is increasing or decreasing,
b. local minima and maxima of f ,
c. intervals where f is concave up and concave


down, and
d. the inflection points of f . Sketch the curve, then


use a calculator to compare your answer. If youcannot determine the exact answer analytically, usea calculator.
231. [T] f (x) = sin(πx) − cos(πx) over x = [−1, 1]
232. [T] f (x) = x + sin(2x) over x = ⎡⎣−π2, π2⎤⎦
233. [T] f (x) = sinx + tanx over ⎛⎝−π2, π2⎞⎠
234. [T] f (x) = (x − 2)2 (x − 4)2


235. [T] f (x) = 1
1 − x


, x ≠ 1


236. [T] f (x) = sinxx over x = [−2π, 2π]
[2π, 0) ∪ (0, 2π]


237. f (x) = sin(x)ex over x = [−π, π]


238. f (x) = lnx x, x > 0
239. f (x) = 1


4
x + 1x , x > 0


240. f (x) = exx , x ≠ 0
For the following exercises, interpret the sentences in termsof f , f ′, and f ″.
241. The population is growing more slowly. Here f is
the population.
242. A bike accelerates faster, but a car goes faster. Here
f = Bike’s position minus Car’s position.
243. The airplane lands smoothly. Here f is the plane’s
altitude.
244. Stock prices are at their peak. Here f is the stock
price.
245. The economy is picking up speed. Here f is a
measure of the economy, such as GDP.
For the following exercises, consider a third-degreepolynomial f (x), which has the properties
f ′ (1) = 0, f ′ (3) = 0. Determine whether the following
statements are true or false. Justify your answer.
246. f (x) = 0 for some 1 ≤ x ≤ 3
247. f ″(x) = 0 for some 1 ≤ x ≤ 3
248. There is no absolute maximum at x = 3
249. If f (x) has three roots, then it has 1 inflection
point.
250. If f (x) has one inflection point, then it has three real
roots.


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4.6 | Limits at Infinity and Asymptotes
Learning Objectives


4.6.1 Calculate the limit of a function as x increases or decreases without bound.
4.6.2 Recognize a horizontal asymptote on the graph of a function.
4.6.3 Estimate the end behavior of a function as x increases or decreases without bound.
4.6.4 Recognize an oblique asymptote on the graph of a function.
4.6.5 Analyze a function and its derivatives to draw its graph.


We have shown how to use the first and second derivatives of a function to describe the shape of a graph. To graph afunction f defined on an unbounded domain, we also need to know the behavior of f as x → ±∞. In this section, we
define limits at infinity and show how these limits affect the graph of a function. At the end of this section, we outline astrategy for graphing an arbitrary function f .
Limits at Infinity
We begin by examining what it means for a function to have a finite limit at infinity. Then we study the idea of a functionwith an infinite limit at infinity. Back in Introduction to Functions and Graphs, we looked at vertical asymptotes; inthis section we deal with horizontal and oblique asymptotes.
Limits at Infinity and Horizontal Asymptotes
Recall that lim


x → a
f (x) = L means f (x) becomes arbitrarily close to L as long as x is sufficiently close to a. We can


extend this idea to limits at infinity. For example, consider the function f (x) = 2 + 1x . As can be seen graphically in
Figure 4.40 and numerically in Table 4.8, as the values of x get larger, the values of f (x) approach 2. We say the limit
as x approaches ∞ of f (x) is 2 and write lim


x → ∞
f (x) = 2. Similarly, for x < 0, as the values |x| get larger, the values


of f (x) approaches 2. We say the limit as x approaches −∞ of f (x) is 2 and write lim
x → a


f (x) = 2.


Figure 4.40 The function approaches the asymptote y = 2 as x approaches ±∞.


Chapter 4 | Applications of Derivatives 407




x 10 100 1,000 10,000


2 + 1x 2.1 2.01 2.001 2.0001


x −10 −100 −1000 −10,000


2 + 1x 1.9 1.99 1.999 1.9999


Table 4.8 Values of a function f as x → ±∞
More generally, for any function f , we say the limit as x → ∞ of f (x) is L if f (x) becomes arbitrarily close to
L as long as x is sufficiently large. In that case, we write lim


x → a
f (x) = L. Similarly, we say the limit as x → −∞ of


f (x) is L if f (x) becomes arbitrarily close to L as long as x < 0 and |x| is sufficiently large. In that case, we write
lim


x → −∞
f (x) = L. We now look at the definition of a function having a limit at infinity.


Definition
(Informal) If the values of f (x) become arbitrarily close to L as x becomes sufficiently large, we say the function
f has a limit at infinity and write


lim
x → ∞


f (x) = L.


If the values of f (x) becomes arbitrarily close to L for x < 0 as |x| becomes sufficiently large, we say that the
function f has a limit at negative infinity and write


lim
x → ∞


f (x) = L.


If the values f (x) are getting arbitrarily close to some finite value L as x → ∞ or x → −∞, the graph of f approaches
the line y = L. In that case, the line y = L is a horizontal asymptote of f (Figure 4.41). For example, for the function
f (x) = 1x , since limx → ∞ f (x) = 0, the line y = 0 is a horizontal asymptote of f (x) = 1x .


Definition
If lim


x → ∞
f (x) = L or lim


x → −∞
f (x) = L, we say the line y = L is a horizontal asymptote of f .


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Figure 4.41 (a) As x → ∞, the values of f are getting arbitrarily close to L. The line y = L
is a horizontal asymptote of f . (b) As x → −∞, the values of f are getting arbitrarily close to
M. The line y = M is a horizontal asymptote of f .


A function cannot cross a vertical asymptote because the graph must approach infinity (or −∞) from at least one direction
as x approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may
cross a horizontal asymptote an unlimited number of times. For example, the function f (x) = (cosx)x + 1 shown in Figure
4.42 intersects the horizontal asymptote y = 1 an infinite number of times as it oscillates around the asymptote with ever-
decreasing amplitude.


Figure 4.42 The graph of f (x) = (cosx)/x + 1 crosses its
horizontal asymptote y = 1 an infinite number of times.


The algebraic limit laws and squeeze theorem we introduced in Introduction to Limits also apply to limits at infinity. Weillustrate how to use these laws to compute several limits at infinity.
Example 4.21
Computing Limits at Infinity
For each of the following functions f , evaluate lim


x → ∞
f (x) and lim


x → −∞
f (x). Determine the horizontal


asymptote(s) for f .
a. f (x) = 5 − 2


x2


b. f (x) = sinxx
c. f (x) = tan−1 (x)


Chapter 4 | Applications of Derivatives 409




Solution
a. Using the algebraic limit laws, we have


lim
x → ∞

⎝5 −


2
x2

⎠ = limx → ∞5 − 2



⎝ limx → ∞


1
x

⎠.

⎝ limx → ∞


1
x

⎠ = 5 − 2 · 0 = 5.


Similarly, lim
x → ∞


f (x) = 5. Therefore, f (x) = 5 − 2
x2


has a horizontal asymptote of y = 5 and f
approaches this horizontal asymptote as x → ±∞ as shown in the following graph.


Figure 4.43 This function approaches a horizontal asymptoteas x → ±∞.
b. Since −1 ≤ sinx ≤ 1 for all x, we have


−1
x ≤


sinx
x ≤


1
x


for all x ≠ 0. Also, since
lim


x → ∞
−1
x = 0 = limx → ∞


1
x ,


we can apply the squeeze theorem to conclude that
lim


x → ∞
sinx
x = 0.


Similarly,
lim


x → −∞
sinx
x = 0.


Thus, f (x) = sinxx has a horizontal asymptote of y = 0 and f (x) approaches this horizontal asymptote
as x → ±∞ as shown in the following graph.


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Figure 4.44 This function crosses its horizontal asymptote multiple times.
c. To evaluate lim


x → ∞
tan−1 (x) and lim


x → −∞
tan−1 (x), we first consider the graph of y = tan(x) over the


interval (−π/2, π/2) as shown in the following graph.


Figure 4.45 The graph of tanx has vertical asymptotes at
x = ±π


2


Since
lim


x → (π/2)−
tanx = ∞,


it follows that
lim


x → ∞
tan−1(x) = π


2
.


Similarly, since
lim


x → (π/2)+
tanx = −∞,


it follows that
lim


x → −∞
tan−1 (x) = − π


2
.


As a result, y = π
2
and y = − π


2
are horizontal asymptotes of f (x) = tan−1 (x) as shown in the following


graph.


Chapter 4 | Applications of Derivatives 411




4.20


Figure 4.46 This function has two horizontal asymptotes.


Evaluate lim
x → −∞



⎝3 +


4
x

⎠ and limx → ∞⎛⎝3 + 4x⎞⎠. Determine the horizontal asymptotes of f (x) = 3 + 4x , if


any.


Infinite Limits at Infinity
Sometimes the values of a function f become arbitrarily large as x → ∞ (or as x → −∞). In this case, we write
lim


x → ∞
f (x) = ∞ (or lim


x → −∞
f (x) = ∞). On the other hand, if the values of f are negative but become arbitrarily large in


magnitude as x → ∞ (or as x → −∞), we write lim
x → ∞


f (x) = −∞ (or lim
x → −∞


f (x) = −∞).


For example, consider the function f (x) = x3. As seen in Table 4.9 and Figure 4.47, as x → ∞ the values f (x)
become arbitrarily large. Therefore, lim


x → ∞
x3 = ∞. On the other hand, as x → −∞, the values of f (x) = x3 are


negative but become arbitrarily large in magnitude. Consequently, lim
x → −∞


x3 = −∞.


x 10 20 50 100 1000


x3 1000 8000 125,000 1,000,000 1,000,000,000


x −10 −20 −50 −100 −1000


x3 −1000 −8000 −125,000 −1,000,000 −1,000,000,000


Table 4.9 Values of a power function as x → ±∞


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Figure 4.47 For this function, the functional values approach infinity as
x → ±∞.


Definition
(Informal) We say a function f has an infinite limit at infinity and write


lim
x → ∞


f (x) = ∞.


if f (x) becomes arbitrarily large for x sufficiently large. We say a function has a negative infinite limit at infinity and
write


lim
x → ∞


f (x) = −∞.


if f (x) < 0 and | f (x)| becomes arbitrarily large for x sufficiently large. Similarly, we can define infinite limits as
x → −∞.


Formal Definitions
Earlier, we used the terms arbitrarily close, arbitrarily large, and sufficiently large to define limits at infinity informally.Although these terms provide accurate descriptions of limits at infinity, they are not precise mathematically. Here are moreformal definitions of limits at infinity. We then look at how to use these definitions to prove results involving limits atinfinity.
Definition
(Formal) We say a function f has a limit at infinity, if there exists a real number L such that for all ε > 0, there
exists N > 0 such that


| f (x) − L| < ε


for all x > N. In that case, we write
lim


x → ∞
f (x) = L


(see Figure 4.48).
We say a function f has a limit at negative infinity if there exists a real number L such that for all ε > 0, there
exists N < 0 such that


| f (x) − L| < ε


for all x < N. In that case, we write
lim


x → −∞
f (x) = L.


Chapter 4 | Applications of Derivatives 413




4.21


Figure 4.48 For a function with a limit at infinity, for all
x > N, | f (x) − L| < ε.


Earlier in this section, we used graphical evidence in Figure 4.40 and numerical evidence in Table 4.8 to conclude that
lim


x → ∞


2 + 1
x

⎠ = 2. Here we use the formal definition of limit at infinity to prove this result rigorously.


Example 4.22 A Finite Limit at Infinity Example
Use the formal definition of limit at infinity to prove that lim


x → ∞


2 + 1
x

⎠ = 2.


Solution
Let ε > 0. Let N = 1ε . Therefore, for all x > N, we have


|2 + 1x − 2| = |1x | = 1x < 1N = ε.


Use the formal definition of limit at infinity to prove that lim
x → ∞


3 − 1
x2

⎠ = 3.


We now turn our attention to a more precise definition for an infinite limit at infinity.
Definition
(Formal) We say a function f has an infinite limit at infinity and write


lim
x → ∞


f (x) = ∞


if for all M > 0, there exists an N > 0 such that
f (x) > M


for all x > N (see Figure 4.49).
We say a function has a negative infinite limit at infinity and write


lim
x → ∞


f (x) = −∞


if for all M < 0, there exists an N > 0 such that
f (x) < M


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4.22


for all x > N.
Similarly we can define limits as x → −∞.


Figure 4.49 For a function with an infinite limit at infinity, forall x > N, f (x) > M.


Earlier, we used graphical evidence (Figure 4.47) and numerical evidence (Table 4.9) to conclude that lim
x → ∞


x3 = ∞.


Here we use the formal definition of infinite limit at infinity to prove that result.
Example 4.23 An Infinite Limit at Infinity
Use the formal definition of infinite limit at infinity to prove that lim


x → ∞
x3 = ∞.


Solution
Let M > 0. Let N = M3 . Then, for all x > N, we have


x3 > N 3 = ⎛⎝ M
3 ⎞


3
= M.


Therefore, lim
x → ∞


x3 = ∞.


Use the formal definition of infinite limit at infinity to prove that lim
x → ∞


3x2 = ∞.


End Behavior
The behavior of a function as x → ±∞ is called the function’s end behavior. At each of the function’s ends, the function
could exhibit one of the following types of behavior:


1. The function f (x) approaches a horizontal asymptote y = L.
2. The function f (x) → ∞ or f (x) → −∞.
3. The function does not approach a finite limit, nor does it approach ∞ or −∞. In this case, the function may have


some oscillatory behavior.


Chapter 4 | Applications of Derivatives 415




Let’s consider several classes of functions here and look at the different types of end behaviors for these functions.
End Behavior for Polynomial Functions
Consider the power function f (x) = xn where n is a positive integer. From Figure 4.50 and Figure 4.51, we see that


lim
x → ∞


xn = ∞; n = 1, 2, 3,…


and
lim


x → −∞
xn =






∞; n = 2, 4, 6,…
−∞; n = 1, 3, 5,…


.


Figure 4.50 For power functions with an even power of n,
lim


x → ∞
xn = ∞ = lim


x → −∞
xn.


Figure 4.51 For power functions with an odd power of n,
lim


x → ∞
xn = ∞ and lim


x → −∞
xn = −∞.


Using these facts, it is not difficult to evaluate lim
x → ∞


cxn and lim
x → −∞


cxn, where c is any constant and n is a positive
integer. If c > 0, the graph of y = cxn is a vertical stretch or compression of y = xn, and therefore


lim
x → ∞


cxn = lim
x → ∞


xn and lim
x → −∞


cxn = lim
x → −∞


xn if c > 0.


If c < 0, the graph of y = cxn is a vertical stretch or compression combined with a reflection about the x -axis, and
therefore


lim
x → ∞


cxn = − lim
x → ∞


xn and lim
x → −∞


cxn = − lim
x → −∞


xn if c < 0.


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4.23


If c = 0, y = cxn = 0, in which case lim
x → ∞


cxn = 0 = lim
x → −∞


cxn.


Example 4.24
Limits at Infinity for Power Functions
For each function f , evaluate lim


x → ∞
f (x) and lim


x → −∞
f (x).


a. f (x) = −5x3
b. f (x) = 2x4


Solution
a. Since the coefficient of x3 is −5, the graph of f (x) = −5x3 involves a vertical stretch and reflection


of the graph of y = x3 about the x -axis. Therefore, lim
x → ∞



⎝−5x


3⎞
⎠ = −∞ and limx → −∞⎛⎝−5x3⎞⎠ = ∞.


b. Since the coefficient of x4 is 2, the graph of f (x) = 2x4 is a vertical stretch of the graph of y = x4.
Therefore, lim


x → ∞
2x4 = ∞ and lim


x → −∞
2x4 = ∞.


Let f (x) = −3x4. Find lim
x → ∞


f (x).


We now look at how the limits at infinity for power functions can be used to determine lim
x → ±∞


f (x) for any polynomial
function f . Consider a polynomial function


f (x) = an x
n + an − 1 x


n − 1 +… + a1 x + a0


of degree n ≥ 1 so that an ≠ 0. Factoring, we see that
f (x) = an x


n⎛
⎝1 +


an − 1
an


1
x +… +


a1
an


1
xn − 1


+
a0
an

⎠.


As x → ±∞, all the terms inside the parentheses approach zero except the first term. We conclude that
lim


x → ±∞
f (x) = lim


x → ±∞
an x


n.


For example, the function f (x) = 5x3 − 3x2 + 4 behaves like g(x) = 5x3 as x → ±∞ as shown in Figure 4.52 and
Table 4.10.


Chapter 4 | Applications of Derivatives 417




Figure 4.52 The end behavior of a polynomial is determinedby the behavior of the term with the largest exponent.
x 10 100 1000


f(x) = 5x3−3x2+4 4704 4,970,004 4,997,000,004


g(x) = 5x3 5000 5,000,000 5,000,000,000


x −10 −100 −1000


f(x) = 5x3−3x2+4 −5296 −5,029,996 −5,002,999,996


g(x) = 5x3 −5000 −5,000,000 −5,000,000,000


Table 4.10 A polynomial’s end behavior is determined by the term with thelargest exponent.
End Behavior for Algebraic Functions
The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials. In
Example 4.25, we show that the limits at infinity of a rational function f (x) = p(x)


q(x)
depend on the relationship between


the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function,we divide the numerator and denominator by the highest power of x appearing in the denominator. This determines which
term in the overall expression dominates the behavior of the function at large values of x.
Example 4.25
Determining End Behavior for Rational Functions
For each of the following functions, determine the limits as x → ∞ and x → −∞. Then, use this information
to describe the end behavior of the function.


a. f (x) = 3x − 1
2x + 5


(Note: The degree of the numerator and the denominator are the same.)


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b. f (x) = 3x2 + 2x
4x3 − 5x + 7


(Note: The degree of numerator is less than the degree of the denominator.)


c. f (x) = 3x2 + 4x
x + 2


(Note: The degree of numerator is greater than the degree of the denominator.)


Solution
a. The highest power of x in the denominator is x. Therefore, dividing the numerator and denominator by


x and applying the algebraic limit laws, we see that
lim


x → ±∞
3x − 1
2x + 5


= lim
x → ±∞


3 − 1/x
2 + 5/x


=
lim


x → ±∞
(3 − 1/x)


lim
x → ±∞


(2 + 5/x)


=
lim


x → ±∞
3 − lim


x → ±∞
1/x


lim
x → ±∞


2 + lim
x → ±∞


5/x


= 3 − 0
2 + 0


= 3
2
.


Since lim
x → ±∞


f (x) = 3
2
, we know that y = 3


2
is a horizontal asymptote for this function as shown in


the following graph.


Figure 4.53 The graph of this rational function approaches ahorizontal asymptote as x → ±∞.
b. Since the largest power of x appearing in the denominator is x3, divide the numerator and denominator


by x3. After doing so and applying algebraic limit laws, we obtain
lim


x → ±∞
3x2 + 2x


4x3 − 5x + 7
= lim


x → ±∞
3/x + 2/x2


4 − 5/x2 + 7/x3
= 3.0 + 2.0


4 − 5.0 + 7.0
= 0.


Therefore f has a horizontal asymptote of y = 0 as shown in the following graph.


Chapter 4 | Applications of Derivatives 419




Figure 4.54 The graph of this rational function approachesthe horizontal asymptote y = 0 as x → ±∞.
c. Dividing the numerator and denominator by x, we have


lim
x → ±∞


3x2 + 4x
x + 2


= lim
x → ±∞


3x + 4
1 + 2/x


.


As x → ±∞, the denominator approaches 1. As x → ∞, the numerator approaches +∞. As
x → −∞, the numerator approaches −∞. Therefore lim


x → ∞
f (x) = ∞, whereas lim


x → −∞
f (x) = −∞


as shown in the following figure.


Figure 4.55 As x → ∞, the values f (x) → ∞. As x → −∞, the
values f (x) → −∞.


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4.24 Evaluate lim
x → ±∞


3x2 + 2x − 1
5x2 − 4x + 7


and use these limits to determine the end behavior of
f (x) = 3x


2 + 2x − 2
5x2 − 4x + 7


.


Before proceeding, consider the graph of f (x) = ⎛⎝3x2 + 4x⎞⎠
(x + 2)


shown in Figure 4.56. As x → ∞ and x → −∞, the
graph of f appears almost linear. Although f is certainly not a linear function, we now investigate why the graph of f
seems to be approaching a linear function. First, using long division of polynomials, we can write


f (x) = 3x
2 + 4x
x + 2


= 3x − 2 + 4
x + 2


.


Since 4
(x + 2)


→ 0 as x → ±∞, we conclude that
lim


x → ±∞

⎝ f (x) − (3x − 2)⎞⎠ = limx → ±∞


4
x + 2


= 0.


Therefore, the graph of f approaches the line y = 3x − 2 as x → ±∞. This line is known as an oblique asymptote for
f (Figure 4.56).


Figure 4.56 The graph of the rational function
f (x) = ⎛⎝3x


2 + 4x⎞⎠/(x + 2) approaches the oblique asymptote
y = 3x − 2 as x → ±∞.


We can summarize the results of Example 4.25 to make the following conclusion regarding end behavior for rationalfunctions. Consider a rational function
f (x) =


p(x)
q(x)


=
an x


n + an − 1 x
n − 1 +… + a1 x + a0


bm x
m + bm − 1 x


m − 1 +… + b1 x + b0
,


where an ≠ 0 and bm ≠ 0.
1. If the degree of the numerator is the same as the degree of the denominator (n = m), then f has a horizontal


asymptote of y = an /bm as x → ±∞.
2. If the degree of the numerator is less than the degree of the denominator (n < m), then f has a horizontal


asymptote of y = 0 as x → ±∞.
3. If the degree of the numerator is greater than the degree of the denominator (n > m), then f does not have a


Chapter 4 | Applications of Derivatives 421




horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of theleading terms. In addition, using long division, the function can be rewritten as
f (x) =


p(x)
q(x)


= g(x) + r(x)
q(x)


,


where the degree of r(x) is less than the degree of q(x). As a result, lim
x → ±∞


r(x)/q(x) = 0. Therefore, the values
of ⎡⎣ f (x) − g(x)⎤⎦ approach zero as x → ±∞. If the degree of p(x) is exactly one more than the degree of q(x)
(n = m + 1), the function g(x) is a linear function. In this case, we call g(x) an oblique asymptote.
Now let’s consider the end behavior for functions involving a radical.


Example 4.26
Determining End Behavior for a Function Involving a Radical
Find the limits as x → ∞ and x → −∞ for f (x) = 3x − 2


4x2 + 5
and describe the end behavior of f .


Solution
Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of
x. To determine the appropriate power of x, consider the expression 4x2 + 5 in the denominator. Since


4x2 + 5 ≈ 4x2 = 2|x|


for large values of x in effect x appears just to the first power in the denominator. Therefore, we divide the
numerator and denominator by |x|. Then, using the fact that |x| = x for x > 0, |x| = −x for x < 0, and
|x| = x2 for all x, we calculate the limits as follows:


lim
x → ∞


3x − 2


4x2 + 5
= lim


x → ∞
(1/|x|)(3x − 2)


(1/|x|) 4x2 + 5


= lim
x → ∞


(1/x)(3x − 2)

⎝1/x


2⎞


⎝4x


2 + 5⎞⎠


= lim
x → ∞


3 − 2/x


4 + 5/x2
= 3


4
= 3


2


lim
x → −∞


3x − 2


4x2 + 5
= lim


x → −∞
(1/|x|)(3x − 2)


(1/|x|) 4x2 + 5


= lim
x → −∞


(−1/x)(3x − 2)

⎝1/x


2⎞


⎝4x


2 + 5⎞⎠


= lim
x → −∞


−3 + 2/x


4 + 5/x2
= −3


4
= −3


2
.


Therefore, f (x) approaches the horizontal asymptote y = 3
2
as x → ∞ and the horizontal asymptote y = − 3


2
as x → −∞ as shown in the following graph.


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4.25


Figure 4.57 This function has two horizontal asymptotes and it crosses oneof the asymptotes.


Evaluate lim
x → ∞


3x2 + 4
x + 6


.


Determining End Behavior for Transcendental Functions
The six basic trigonometric functions are periodic and do not approach a finite limit as x → ±∞. For example, sinx
oscillates between 1 and −1 (Figure 4.58). The tangent function x has an infinite number of vertical asymptotes as
x → ±∞; therefore, it does not approach a finite limit nor does it approach ±∞ as x → ±∞ as shown in Figure 4.59.


Figure 4.58 The function f (x) = sinx oscillates between
1 and −1 as x → ±∞


Figure 4.59 The function f (x) = tanx does not approach a
limit and does not approach ±∞ as x → ±∞


Recall that for any base b > 0, b ≠ 1, the function y = bx is an exponential function with domain (−∞, ∞) and range
(0, ∞). If b > 1, y = bx is increasing over `(−∞, ∞). If 0 < b < 1, y = bx is decreasing over (−∞, ∞). For
the natural exponential function f (x) = ex, e ≈ 2.718 > 1. Therefore, f (x) = ex is increasing on `(−∞, ∞) and the


Chapter 4 | Applications of Derivatives 423




range is `(0, ∞). The exponential function f (x) = ex approaches ∞ as x → ∞ and approaches 0 as x → −∞ asshown in Table 4.11 and Figure 4.60.
x −5 −2 0 2 5


ex 0.00674 0.135 1 7.389 148.413


Table 4.11 End behavior of the natural exponential function


Figure 4.60 The exponential function approaches zero as
x → −∞ and approaches ∞ as x → ∞.


Recall that the natural logarithm function f (x) = ln(x) is the inverse of the natural exponential function y = ex.
Therefore, the domain of f (x) = ln(x) is (0, ∞) and the range is (−∞, ∞). The graph of f (x) = ln(x) is the reflection
of the graph of y = ex about the line y = x. Therefore, ln(x) → −∞ as x → 0+ and ln(x) → ∞ as x → ∞ as shown
in Figure 4.61 and Table 4.12.


x 0.01 0.1 1 10 100


ln(x) −4.605 −2.303 0 2.303 4.605


Table 4.12 End behavior of the natural logarithm function


Figure 4.61 The natural logarithm function approaches ∞ as
x → ∞.


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4.26


Example 4.27
Determining End Behavior for a Transcendental Function
Find the limits as x → ∞ and x → −∞ for f (x) = (2 + 3ex)


(7 − 5ex)
and describe the end behavior of f .


Solution
To find the limit as x → ∞, divide the numerator and denominator by ex :


lim
x → ∞


f (x) = lim
x → ∞


2 + 3ex


7 − 5ex


= lim
x → ∞


(2/ex) + 3
(7/ex) − 5


.


As shown in Figure 4.60, ex → ∞ as x → ∞. Therefore,
lim


x → ∞
2
ex


= 0 = lim
x → ∞


7
ex


.


We conclude that lim
x → ∞


f (x) = − 3
5
, and the graph of f approaches the horizontal asymptote y = − 3


5


as x → ∞. To find the limit as x → −∞, use the fact that ex → 0 as x → −∞ to conclude that
lim


x → ∞
f (x) = 2


7
, and therefore the graph of approaches the horizontal asymptote y = 2


7
as x → −∞.


Find the limits as x → ∞ and x → −∞ for f (x) = (3ex − 4)
(5ex + 2)


.


Guidelines for Drawing the Graph of a Function
We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Beforeshowing how to graph specific functions, let’s look at a general strategy to use when graphing any function.
Problem-Solving Strategy: Drawing the Graph of a Function
Given a function f , use the following steps to sketch a graph of f :


1. Determine the domain of the function.
2. Locate the x - and y -intercepts.
3. Evaluate lim


x → ∞
f (x) and lim


x → −∞
f (x) to determine the end behavior. If either of these limits is a finite number


L, then y = L is a horizontal asymptote. If either of these limits is ∞ or −∞, determine whether f has
an oblique asymptote. If f is a rational function such that f (x) = p(x)


q(x)
, where the degree of the numerator


is greater than the degree of the denominator, then f can be written as
f (x) =


p(x)
q(x)


= g(x) + r(x)
q(x)


,


where the degree of r(x) is less than the degree of q(x). The values of f (x) approach the values of g(x) as


Chapter 4 | Applications of Derivatives 425




x → ±∞. If g(x) is a linear function, it is known as an oblique asymptote.
4. Determine whether f has any vertical asymptotes.
5. Calculate f ′. Find all critical points and determine the intervals where f is increasing and where f is


decreasing. Determine whether f has any local extrema.
6. Calculate f ″. Determine the intervals where f is concave up and where f is concave down. Use this


information to determine whether f has any inflection points. The second derivative can also be used as an
alternate means to determine or verify that f has a local extremum at a critical point.


Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Example 4.28
Sketching a Graph of a Polynomial
Sketch a graph of f (x) = (x − 1)2 (x + 2).
Solution
Step 1. Since f is a polynomial, the domain is the set of all real numbers.
Step 2. When x = 0, f (x) = 2. Therefore, the y -intercept is (0, 2). To find the x -intercepts, we need to solve
the equation (x − 1)2 (x + 2) = 0, gives us the x -intercepts (1, 0) and (−2, 0)
Step 3. We need to evaluate the end behavior of f . As x → ∞, (x − 1)2 → ∞ and (x + 2) → ∞. Therefore,
lim


x → ∞
f (x) = ∞. As x → −∞, (x − 1)2 → ∞ and (x + 2) → −∞. Therefore, lim


x → ∞
f (x) = −∞. To get


even more information about the end behavior of f , we can multiply the factors of f . When doing so, we see
that


f (x) = (x − 1)2 (x + 2) = x3 − 3x + 2.


Since the leading term of f is x3, we conclude that f behaves like y = x3 as x → ±∞.
Step 4. Since f is a polynomial function, it does not have any vertical asymptotes.
Step 5. The first derivative of f is


f ′ (x) = 3x2 − 3.


Therefore, f has two critical points: x = 1, −1. Divide the interval (−∞, ∞) into the three smaller intervals:
(−∞, −1), (−1, 1), and (1, ∞). Then, choose test points x = −2, x = 0, and x = 2 from these
intervals and evaluate the sign of f ′ (x) at each of these test points, as shown in the following table.


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Interval TestPoint Sign of Derivativef′(x) = 3x2−3 = 3(x−1)(x+1)
Conclusion


(−∞, −1) x = −2 (+)(−)(−) = + f is
increasing.


(−1, 1) x = 0 (+)(−)(+) = − f is
decreasing.


(1, ∞) x = 2 (+)(+)(+) = + f is
increasing.


From the table, we see that f has a local maximum at x = −1 and a local minimum at x = 1. Evaluating
f (x) at those two points, we find that the local maximum value is f (−1) = 4 and the local minimum value is
f (1) = 0.


Step 6. The second derivative of f is
f ″(x) = 6x.


The second derivative is zero at x = 0. Therefore, to determine the concavity of f , divide the interval
(−∞, ∞) into the smaller intervals (−∞, 0) and (0, ∞), and choose test points x = −1 and x = 1 to
determine the concavity of f on each of these smaller intervals as shown in the following table.


Interval Test Point Sign of f″(x) = 6x Conclusion
(−∞, 0) x = −1 − f is concave down.


(0, ∞) x = 1 + f is concave up.


We note that the information in the preceding table confirms the fact, found in step 5, that f has a local
maximum at x = −1 and a local minimum at x = 1. In addition, the information found in step 5—namely, f
has a local maximum at x = −1 and a local minimum at x = 1, and f ′ (x) = 0 at those points—combined
with the fact that f ″ changes sign only at x = 0 confirms the results found in step 6 on the concavity of f .
Combining this information, we arrive at the graph of f (x) = (x − 1)2 (x + 2) shown in the following graph.


Chapter 4 | Applications of Derivatives 427




4.27 Sketch a graph of f (x) = (x − 1)3 (x + 2).


Example 4.29
Sketching a Rational Function
Sketch the graph of f (x) = x2⎛


⎝1 − x
2⎞


.


Solution
Step 1. The function f is defined as long as the denominator is not zero. Therefore, the domain is the set of all
real numbers x except x = ±1.
Step 2. Find the intercepts. If x = 0, then f (x) = 0, so 0 is an intercept. If y = 0, then x2⎛


⎝1 − x
2⎞


= 0,


which implies x = 0. Therefore, (0, 0) is the only intercept.
Step 3. Evaluate the limits at infinity. Since f is a rational function, divide the numerator and denominator by
the highest power in the denominator: x2. We obtain


lim
x → ±∞


x2


1 − x2
= lim


x → ±∞
1


1


x2
− 1


= −1.


Therefore, f has a horizontal asymptote of y = −1 as x → ∞ and x → −∞.
Step 4. To determine whether f has any vertical asymptotes, first check to see whether the denominator has any
zeroes. We find the denominator is zero when x = ±1. To determine whether the lines x = 1 or x = −1 are
vertical asymptotes of f , evaluate lim


x → 1
f (x) and lim


x → −1
f (x). By looking at each one-sided limit as x → 1,


we see that


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lim
x → 1+


x2


1 − x2
= −∞ and lim


x → 1−
x2


1 − x2
= ∞.


In addition, by looking at each one-sided limit as x → −1, we find that
lim


x → −1+
x2


1 − x2
= ∞ and lim


x → −1−
x2


1 − x2
= −∞.


Step 5. Calculate the first derivative:
f ′ (x) =



⎝1 − x


2⎞
⎠(2x) − x


2 (−2x)



⎝1 − x


2⎞


2
= 2x

⎝1 − x


2⎞


2
.


Critical points occur at points x where f ′ (x) = 0 or f ′ (x) is undefined. We see that f ′ (x) = 0 when x = 0.
The derivative f ′ is not undefined at any point in the domain of f . However, x = ±1 are not in the domain of
f . Therefore, to determine where f is increasing and where f is decreasing, divide the interval (−∞, ∞) into
four smaller intervals: (−∞, −1), (−1, 0), (0, 1), and (1, ∞), and choose a test point in each interval to
determine the sign of f ′ (x) in each of these intervals. The values x = −2, x = − 1


2
, x = 1


2
, and x = 2


are good choices for test points as shown in the following table.
Interval Test Point Sign of f′ (x) = 2x



⎝1 − x2




2


Conclusion


(−∞, −1) x = −2 −/ + = − f is decreasing.


(−1, 0) x = −1/2 −/ + = − f is decreasing.


(0, 1) x = 1/2 +/ + = + f is increasing.


(1, ∞) x = 2 +/ + = + f is increasing.


From this analysis, we conclude that f has a local minimum at x = 0 but no local maximum.
Step 6. Calculate the second derivative:


Chapter 4 | Applications of Derivatives 429




f ″(x) =

⎝1 − x


2⎞


2
(2) − 2x⎛⎝2



⎝1 − x


2⎞
⎠(−2x)






⎝1 − x


2⎞


4


=

⎝1 − x


2⎞


⎣2

⎝1 − x


2⎞
⎠+ 8x


2⎤



⎝1 − x


2⎞


4


=
2⎛⎝1 − x


2⎞
⎠+ 8x


2



⎝1 − x


2⎞


3


= 6x
2 + 2



⎝1 − x


2⎞


3
.


To determine the intervals where f is concave up and where f is concave down, we first need to find all points
x where f ″(x) = 0 or f ″(x) is undefined. Since the numerator 6x2 + 2 ≠ 0 for any x, f ″(x) is never zero.
Furthermore, f ″ is not undefined for any x in the domain of f . However, as discussed earlier, x = ±1 are
not in the domain of f . Therefore, to determine the concavity of f , we divide the interval (−∞, ∞) into the
three smaller intervals (−∞, −1), (−1, −1), and (1, ∞), and choose a test point in each of these intervals
to evaluate the sign of f ″(x). in each of these intervals. The values x = −2, x = 0, and x = 2 are possible
test points as shown in the following table.


Interval Test Point Sign of f″(x) = 6x2+2

⎝1 − x2




3


Conclusion


(−∞, −1) x = −2 +/ − = − f is concave down.


(−1, −1) x = 0 +/ + = + f is concave up.


(1, ∞) x = 2 +/ − = − f is concave down.


Combining all this information, we arrive at the graph of f shown below. Note that, although f changes
concavity at x = −1 and x = 1, there are no inflection points at either of these places because f is not
continuous at x = −1 or x = 1.


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4.28 Sketch a graph of f (x) = (3x + 5)
(8 + 4x)


.


Example 4.30
Sketching a Rational Function with an Oblique Asymptote
Sketch the graph of f (x) = x2


(x − 1)


Solution
Step 1. The domain of f is the set of all real numbers x except x = 1.
Step 2. Find the intercepts. We can see that when x = 0, f (x) = 0, so (0, 0) is the only intercept.
Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of thedenominator, f must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials
to write


f (x) = x
2


x − 1
= x + 1 + 1


x − 1
.


Since 1/(x − 1) → 0 as x → ±∞, f (x) approaches the line y = x + 1 as x → ±∞. The line y = x + 1 is
an oblique asymptote for f .
Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at
x = 1. Looking at both one-sided limits as x → 1, we find


lim
x → 1+


x2
x − 1


= ∞ and lim
x → 1−


x2
x − 1


= −∞.


Therefore, x = 1 is a vertical asymptote, and we have determined the behavior of f as x approaches 1 from
the right and the left.


Chapter 4 | Applications of Derivatives 431




Step 5. Calculate the first derivative:
f ′ (x) = (x − 1)(2x) − x


2 (1)
(x − 1)2


= x
2 − 2x


(x − 1)2
.


We have f ′ (x) = 0 when x2 − 2x = x(x − 2) = 0. Therefore, x = 0 and x = 2 are critical points. Since f
is undefined at x = 1, we need to divide the interval (−∞, ∞) into the smaller intervals (−∞, 0), (0, 1),
(1, 2), and (2, ∞), and choose a test point from each interval to evaluate the sign of f ′ (x) in each of these
smaller intervals. For example, let x = −1, x = 1


2
, x = 3


2
, and x = 3 be the test points as shown in the


following table.
Interval Test Point Sign of f′(x) = x2−2x(x−1)2 = x(x−2)(x−1)2


Conclusion


(−∞, 0) x = −1 (−)(−)/ + = + f is increasing.


(0, 1) x = 1/2 (+)(−)/ + = − f is decreasing.


(1, 2) x = 3/2 (+)(−)/ + = − f is decreasing.


(2, ∞) x = 3 (+)(+)/ + = + f is increasing.


From this table, we see that f has a local maximum at x = 0 and a local minimum at x = 2. The value of f
at the local maximum is f (0) = 0 and the value of f at the local minimum is f (2) = 4. Therefore, (0, 0) and
(2, 4) are important points on the graph.
Step 6. Calculate the second derivative:


f ″(x) =
(x − 1)2 (2x − 2) − ⎛⎝x


2 − 2x⎞⎠

⎝2(x − 1)⎞⎠


(x − 1)4


=
(x − 1)⎡⎣(x − 1)(2x − 2) − 2



⎝x


2 − 2x⎞⎠



(x − 1)4


=
(x − 1)(2x − 2) − 2⎛⎝x


2 − 2x⎞⎠


(x − 1)3


=
2x2 − 4x + 2 − ⎛⎝2x


2 − 4x⎞⎠


(x − 1)3


= 2
(x − 1)3


.


We see that f ″(x) is never zero or undefined for x in the domain of f . Since f is undefined at x = 1, to
check concavity we just divide the interval (−∞, ∞) into the two smaller intervals (−∞, 1) and (1, ∞), and
choose a test point from each interval to evaluate the sign of f ″(x) in each of these intervals. The values x = 0


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4.29


and x = 2 are possible test points as shown in the following table.
Interval Test Point Sign of f″(x) = 2(x−1)3 Conclusion


(−∞, 1) x = 0 +/ − = − f is concave down.


(1, ∞) x = 2 +/ + = + f is concave up.


From the information gathered, we arrive at the following graph for f .


Find the oblique asymptote for f (x) = ⎛⎝3x3 − 2x + 1⎞⎠⎛
⎝2x


2 − 4⎞⎠
.


Example 4.31
Sketching the Graph of a Function with a Cusp
Sketch a graph of f (x) = (x − 1)2/3.
Solution
Step 1. Since the cube-root function is defined for all real numbers x and (x − 1)2/3 = ⎛⎝ x − 13 ⎞⎠


2
, the domain


of f is all real numbers.
Step 2: To find the y -intercept, evaluate f (0). Since f (0) = 1, the y -intercept is (0, 1). To find the x
-intercept, solve (x − 1)2/3 = 0. The solution of this equation is x = 1, so the x -intercept is (1, 0).


Chapter 4 | Applications of Derivatives 433




Step 3: Since lim
x → ±∞


(x − 1)2/3 = ∞, the function continues to grow without bound as x → ∞ and x → −∞.
Step 4: The function has no vertical asymptotes.
Step 5: To determine where f is increasing or decreasing, calculate f ′. We find


f ′ (x) = 2
3
(x − 1)−1/3 = 2


3(x − 1)1/3
.


This function is not zero anywhere, but it is undefined when x = 1. Therefore, the only critical point is x = 1.
Divide the interval (−∞, ∞) into the smaller intervals (−∞, 1) and (1, ∞), and choose test points in each
of these intervals to determine the sign of f ′ (x) in each of these smaller intervals. Let x = 0 and x = 2 be the
test points as shown in the following table.


Interval Test Point Sign of f′ (x) = 2
3(x−1)1/3 Conclusion


(−∞, 1) x = 0 +/ − = − f is decreasing.


(1, ∞) x = 2 +/ + = + f is increasing.


We conclude that f has a local minimum at x = 1. Evaluating f at x = 1, we find that the value of f at the
local minimum is zero. Note that f ′ (1) is undefined, so to determine the behavior of the function at this critical
point, we need to examine lim


x → 1
f ′ (x). Looking at the one-sided limits, we have


lim
x → 1+


2
3(x − 1)1/3


= ∞ and lim
x → 1−


2
3(x − 1)1/3


= −∞.


Therefore, f has a cusp at x = 1.
Step 6: To determine concavity, we calculate the second derivative of f :


f ″(x) = − 2
9
(x − 1)−4/3 = −2


9(x − 1)4/3
.


We find that f ″(x) is defined for all x, but is undefined when x = 1. Therefore, divide the interval (−∞, ∞)
into the smaller intervals (−∞, 1) and (1, ∞), and choose test points to evaluate the sign of f ″(x) in each of
these intervals. As we did earlier, let x = 0 and x = 2 be test points as shown in the following table.


Interval Test Point Sign of f″(x) = −2
9(x−1)4/3 Conclusion


(−∞, 1) x = 0 −/ + = − f is concave down.


(1, ∞) x = 2 −/ + = − f is concave down.


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4.30


From this table, we conclude that f is concave down everywhere. Combining all of this information, we arrive
at the following graph for f .


Consider the function f (x) = 5 − x2/3. Determine the point on the graph where a cusp is located.
Determine the end behavior of f .


Chapter 4 | Applications of Derivatives 435




4.6 EXERCISES
For the following exercises, examine the graphs. Identifywhere the vertical asymptotes are located.
251.


252.


253.


254.


255.


For the following functions f (x), determine whether
there is an asymptote at x = a. Justify your answer
without graphing on a calculator.
256. f (x) = x + 1


x2 + 5x + 4
, a = −1


257. f (x) = x
x − 2


, a = 2


258. f (x) = (x + 2)3/2, a = −2
259. f (x) = (x − 1)−1/3, a = 1
260. f (x) = 1 + x−2/5, a = 1
For the following exercises, evaluate the limit.
261. lim


x → ∞
1


3x + 6


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262. lim
x → ∞


2x − 5
4x


263. lim
x → ∞


x2 − 2x + 5
x + 2


264. lim
x → −∞


3x3 − 2x
x2 + 2x + 8


265. lim
x → −∞


x4 − 4x3 + 1
2 − 2x2 − 7x4


266. lim
x → ∞


3x


x2 + 1


267. lim
x → −∞


4x2 − 1
x + 2


268. lim
x → ∞


4x
x2 − 1


269. lim
x → −∞


4x
x2 − 1


270. lim
x → ∞


2 x
x − x + 1


For the following exercises, find the horizontal and verticalasymptotes.
271. f (x) = x − 9x
272. f (x) = 1


1 − x2


273. f (x) = x3
4 − x2


274. f (x) = x2 + 3
x2 + 1


275. f (x) = sin(x)sin(2x)
276. f (x) = cosx + cos(3x) + cos(5x)
277. f (x) = xsin(x)


x2 − 1


278. f (x) = x
sin(x)


279. f (x) = 1
x3 + x2


280. f (x) = 1
x − 1


− 2x


281. f (x) = x3 + 1
x3 − 1


282. f (x) = sinx + cosx
sinx − cosx


283. f (x) = x − sinx
284. f (x) = 1x − x
For the following exercises, construct a function f (x) that
has the given asymptotes.
285. x = 1 and y = 2
286. x = 1 and y = 0
287. y = 4, x = −1
288. x = 0
For the following exercises, graph the function on agraphing calculator on the window x = ⎡⎣−5, 5⎤⎦ and
estimate the horizontal asymptote or limit. Then, calculatethe actual horizontal asymptote or limit.
289. [T] f (x) = 1


x + 10


290. [T] f (x) = x + 1
x2 + 7x + 6


291. [T] lim
x → −∞


x2 + 10x + 25


292. [T] lim
x → −∞


x + 2
x2 + 7x + 6


293. [T] lim
x → ∞


3x + 2
x + 5


For the following exercises, draw a graph of the functionswithout using a calculator. Be sure to notice all importantfeatures of the graph: local maxima and minima, inflectionpoints, and asymptotic behavior.
294. y = 3x2 + 2x + 4
295. y = x3 − 3x2 + 4
296. y = 2x + 1


x2 + 6x + 5


297. y = x3 + 4x2 + 3x
3x + 9


Chapter 4 | Applications of Derivatives 437




298. y = x2 + x − 2
x2 − 3x − 4


299. y = x2 − 5x + 4
300. y = 2x 16 − x2
301. y = cosxx , on x = [−2π, 2π]
302. y = ex − x3
303. y = x tanx, x = [−π, π]
304. y = x ln(x), x > 0
305. y = x2 sin(x), x = [−2π, 2π]
306. For f (x) = P(x)


Q(x)
to have an asymptote at y = 2


then the polynomials P(x) and Q(x) must have what
relation?
307. For f (x) = P(x)


Q(x)
to have an asymptote at x = 0,


then the polynomials P(x) and Q(x). must have what
relation?
308. If f ′ (x) has asymptotes at y = 3 and x = 1, then
f (x) has what asymptotes?
309. Both f (x) = 1


(x − 1)
and g(x) = 1


(x − 1)2
have


asymptotes at x = 1 and y = 0. What is the most obvious
difference between these two functions?
310. True or false: Every ratio of polynomials has verticalasymptotes.


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4.7 | Applied Optimization Problems
Learning Objectives


4.7.1 Set up and solve optimization problems in several applied fields.
One common application of calculus is calculating the minimum or maximum value of a function. For example, companiesoften want to minimize production costs or maximize revenue. In manufacturing, it is often desirable to minimize theamount of material used to package a product with a certain volume. In this section, we show how to set up these types ofminimization and maximization problems and solve them by using the tools developed in this chapter.
Solving Optimization Problems over a Closed, Bounded Interval
The basic idea of the optimization problems that follow is the same. We have a particular quantity that we are interestedin maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For example, inExample 4.32, we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the sidelengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on howmuch fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let’s look at howwe can maximize the area of a rectangle subject to some constraint on the perimeter.
Example 4.32
Maximizing the Area of a Garden
A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for theother three sides (Figure 4.62). Given 100 ft of wire fencing, determine the dimensions that would create a
garden of maximum area. What is the maximum area?


Figure 4.62 We want to determine the measurements x and
y that will create a garden with a maximum area using 100 ft
of fencing.


Solution
Let x denote the length of the side of the garden perpendicular to the rock wall and y denote the length of the
side parallel to the rock wall. Then the area of the garden is


A = x · y.


Chapter 4 | Applications of Derivatives 439




We want to find the maximum possible area subject to the constraint that the total fencing is 100 ft. From Figure
4.62, the total amount of fencing used will be 2x + y. Therefore, the constraint equation is


2x + y = 100.


Solving this equation for y, we have y = 100 − 2x. Thus, we can write the area as
A(x) = x · (100 − 2x) = 100x − 2x2.


Before trying to maximize the area function A(x) = 100x − 2x2, we need to determine the domain under
consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive.Therefore, we need x > 0 and y > 0. Since y = 100 − 2x, if y > 0, then x < 50. Therefore, we are trying
to determine the maximum value of A(x) for x over the open interval (0, 50). We do not know that a function
necessarily has a maximum value over an open interval. However, we do know that a continuous function hasan absolute maximum (and absolute minimum) over a closed interval. Therefore, let’s consider the function
A(x) = 100x − 2x2 over the closed interval ⎡⎣0, 50⎤⎦. If the maximum value occurs at an interior point, then
we have found the value x in the open interval (0, 50) that maximizes the area of the garden. Therefore, we
consider the following problem:
Maximize A(x) = 100x − 2x2 over the interval ⎡⎣0, 50⎤⎦.
As mentioned earlier, since A is a continuous function on a closed, bounded interval, by the extreme value
theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. Atthe endpoints, A(x) = 0. Since the area is positive for all x in the open interval (0, 50), the maximum must
occur at a critical point. Differentiating the function A(x), we obtain


A′ (x) = 100 − 4x.


Therefore, the only critical point is x = 25 (Figure 4.63). We conclude that the maximum area must occur when
x = 25. Then we have y = 100 − 2x = 100 − 2(25) = 50. To maximize the area of the garden, let x = 25 ft
and y = 50 ft. The area of this garden is 1250 ft2.


Figure 4.63 To maximize the area of the garden, we need to find the
maximum value of the function A(x) = 100x − 2x2.


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4.31 Determine the maximum area if we want to make the same rectangular garden as in Figure 4.63, butwe have 200 ft of fencing.


Now let’s look at a general strategy for solving optimization problems similar to Example 4.32.
Problem-Solving Strategy: Solving Optimization Problems


1. Introduce all variables. If applicable, draw a figure and label all variables.
2. Determine which quantity is to be maximized or minimized, and for what range of values of the other variables(if this can be determined at this time).
3. Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula mayinvolve more than one variable.
4. Write any equations relating the independent variables in the formula from step 3. Use these equations to


write the quantity to be maximized or minimized as a function of one variable.
5. Identify the domain of consideration for the function in step 4 based on the physical problem to be solved.
6. Locate the maximum or minimum value of the function from step 4. This step typically involves looking for


critical points and evaluating a function at endpoints.


Now let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to beused.
Example 4.33
Maximizing the Volume of a Box
An open-top box is to be made from a 24 in. by 36 in. piece of cardboard by removing a square from each
corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to geta box with the maximum volume?
Solution
Step 1: Let x be the side length of the square to be removed from each corner (Figure 4.64). Then, the
remaining four flaps can be folded up to form an open-top box. Let V be the volume of the resulting box.


Chapter 4 | Applications of Derivatives 441




Figure 4.64 A square with side length x inches is removed from each
corner of the piece of cardboard. The remaining flaps are folded to form anopen-top box.


Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize V .
Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is
V = L ·W ·H, where L, W, andH are the length, width, and height, respectively.
Step 4: From Figure 4.64, we see that the height of the box is x inches, the length is 36 − 2x inches, and the
width is 24 − 2x inches. Therefore, the volume of the box is


V(x) = (36 − 2x)(24 − 2x)x = 4x3 − 120x2 + 864x.


Step 5: To determine the domain of consideration, let’s examine Figure 4.64. Certainly, we need x > 0.
Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24
in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether thereis a maximum volume of the box for x over the open interval (0, 12). Since V is a continuous function over
the closed interval [0, 12], we know V will have an absolute maximum over the closed interval. Therefore,
we consider V over the closed interval [0, 12] and check whether the absolute maximum occurs at an interior
point.
Step 6: Since V(x) is a continuous function over the closed, bounded interval [0, 12], V must have an absolute
maximum (and an absolute minimum). Since V(x) = 0 at the endpoints and V(x) > 0 for 0 < x < 12, the
maximum must occur at a critical point. The derivative is


V′ (x) = 12x2 − 240x + 864.


To find the critical points, we need to solve the equation
12x2 − 240x + 864 = 0.


Dividing both sides of this equation by 12, the problem simplifies to solving the equation
x2 − 20x + 72 = 0.


Using the quadratic formula, we find that the critical points are
x = 20± (−20)


2 − 4(1)(72)
2


= 20± 112
2


= 20±4 7
2


= 10±2 7.


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4.32


Since 10 + 2 7 is not in the domain of consideration, the only critical point we need to consider is 10 − 2 7.
Therefore, the volume is maximized if we let x = 10 − 2 7 in. The maximum volume is
V(10 − 2 7) = 640 + 448 7 ≈ 1825 in.3 as shown in the following graph.


Figure 4.65 Maximizing the volume of the box leads to finding the maximum value of acubic polynomial.


Watch a video (http://www.openstaxcollege.org/l/20_boxvolume) about optimizing the volume of a box.


Suppose the dimensions of the cardboard in Example 4.33 are 20 in. by 30 in. Let x be the side length
of each square and write the volume of the open-top box as a function of x. Determine the domain of
consideration for x.


Example 4.34
Minimizing Travel Time
An island is 2 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the
shore that is 6 mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the
visitor runs at a rate of 8 mph and swims at a rate of 3 mph. How far should the visitor run before swimming
to minimize the time it takes to reach the island?
Solution
Step 1: Let x be the distance running and let y be the distance swimming (Figure 4.66). Let T be the time it
takes to get from the cabin to the island.


Chapter 4 | Applications of Derivatives 443




Figure 4.66 How can we choose x and y to minimize the travel time from
the cabin to the island?


Step 2: The problem is to minimize T .
Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spentswimming. Since Distance = Rate × Time (D = R × T), the time spent running is


Trunning =
Drunning
Rrunning


= x
8
,


and the time spent swimming is
Tswimming =


Dswimming
Rswimming


=
y
3
.


Therefore, the total time spent traveling is
T = x


8
+


y
3
.


Step 4: From Figure 4.66, the line segment of y miles forms the hypotenuse of a right triangle with legs
of length 2 mi and 6 − xmi. Therefore, by the Pythagorean theorem, 22 + (6 − x)2 = y2, and we obtain
y = (6 − x)2 + 4. Thus, the total time spent traveling is given by the function


T(x) = x
8
+ (6 − x)


2 + 4
3


.


Step 5: From Figure 4.66, we see that 0 ≤ x ≤ 6. Therefore, ⎡⎣0, 6⎤⎦ is the domain of consideration.
Step 6: Since T(x) is a continuous function over a closed, bounded interval, it has a maximum and a minimum.
Let’s begin by looking for any critical points of T over the interval ⎡⎣0, 6⎤⎦. The derivative is


T′ (x) = 1
8
− 1


2



⎣(6 − x)


2 + 4⎤⎦
−1/2


3
· 2(6 − x) = 1


8
− (6 − x)


3 (6 − x)2 + 4
.


If T′ (x) = 0, then


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(4.6)


4.33


1
8
= 6 − x


3 (6 − x)2 + 4
.


Therefore,
3 (6 − x)2 + 4 = 8(6 − x).


Squaring both sides of this equation, we see that if x satisfies this equation, then x must satisfy
9⎡⎣(6 − x)


2 + 4⎤⎦ = 64(6 − x)
2,


which implies
55(6 − x)2 = 36.


We conclude that if x is a critical point, then x satisfies
(x − 6)2 = 36


55
.


Therefore, the possibilities for critical points are
x = 6± 6


55
.


Since x = 6 + 6/ 55 is not in the domain, it is not a possibility for a critical point. On the other hand,
x = 6 − 6/ 55 is in the domain. Since we squared both sides of Equation 4.6 to arrive at the possible critical
points, it remains to verify that x = 6 − 6/ 55 satisfies Equation 4.6. Since x = 6 − 6/ 55 does satisfy that
equation, we conclude that x = 6 − 6/ 55 is a critical point, and it is the only one. To justify that the time is
minimized for this value of x, we just need to check the values of T(x) at the endpoints x = 0 and x = 6,
and compare them with the value of T(x) at the critical point x = 6 − 6/ 55. We find that T(0) ≈ 2.108 h
and T(6) ≈ 1.417 h, whereas T⎛⎝6 − 6/ 55⎞⎠ ≈ 1.368 h. Therefore, we conclude that T has a local minimum at
x ≈ 5.19 mi.


Suppose the island is 1 mi from shore, and the distance from the cabin to the point on the shore closest
to the island is 15 mi. Suppose a visitor swims at the rate of 2.5 mph and runs at a rate of 6 mph. Let x
denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor toget from the cabin to the island.


In business, companies are interested in maximizing revenue. In the following example, we consider a scenario in which acompany has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent acar. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in.
Example 4.35
Maximizing Revenue
Owners of a car rental company have determined that if they charge customers p dollars per day to rent a
car, where 50 ≤ p ≤ 200, the number of cars n they rent per day can be modeled by the linear function


Chapter 4 | Applications of Derivatives 445




4.34


n(p) = 1000 − 5p. If they charge $50 per day or less, they will rent all their cars. If they charge $200 per
day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and
$200 per day to rent a car, how much should they charge to maximize their revenue?
Solution
Step 1: Let p be the price charged per car per day and let n be the number of cars rented per day. Let R be the
revenue per day.
Step 2: The problem is to maximize R.
Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car perday—that is, R = n × p.
Step 4: Since the number of cars rented per day is modeled by the linear function n(p) = 1000 − 5p, the
revenue R can be represented by the function


R(p) = n × p = ⎛⎝1000 − 5p⎞⎠p = −5p2 + 1000p.


Step 5: Since the owners plan to charge between $50 per car per day and $200 per car per day, the problem is
to find the maximum revenue R(p) for p in the closed interval ⎡⎣50, 200⎤⎦.
Step 6: Since R is a continuous function over the closed, bounded interval ⎡⎣50, 200⎤⎦, it has an absolute
maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points.The derivative is R′ (p) = −10p + 1000. Therefore, the critical point is p = 100 When p = 100,
R(100) = $50,000. When p = 50, R(p) = $37,500. When p = 200, R(p) = $0. Therefore, the absolute
maximum occurs at p = $100. The car rental company should charge $100 per day per car to maximize
revenue as shown in the following figure.


Figure 4.67 To maximize revenue, a car rental company has tobalance the price of a rental against the number of cars peoplewill rent at that price.


A car rental company charges its customers p dollars per day, where 60 ≤ p ≤ 150. It has found that
the number of cars rented per day can be modeled by the linear function n(p) = 750 − 5p. How much should
the company charge each customer to maximize revenue?


Example 4.36


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Maximizing the Area of an Inscribed Rectangle
A rectangle is to be inscribed in the ellipse


x2
4


+ y2 = 1.


What should the dimensions of the rectangle be to maximize its area? What is the maximum area?
Solution
Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let L
be the length of the rectangle and W be its width. Let A be the area of the rectangle.


Figure 4.68 We want to maximize the area of a rectangle inscribed in anellipse.


Step 2: The problem is to maximize A.
Step 3: The area of the rectangle is A = LW.
Step 4: Let (x, y) be the corner of the rectangle that lies in the first quadrant, as shown in Figure 4.68. We can
write length L = 2x and width W = 2y. Since x2


4 + y2 = 1
and y > 0, we have y = 1 − x2


4
. Therefore, the


area is
A = LW = (2x)⎛⎝2y⎞⎠ = 4x 1 − x


2


4
= 2x 4 − x2.


Step 5: From Figure 4.68, we see that to inscribe a rectangle in the ellipse, the x -coordinate of the corner in
the first quadrant must satisfy 0 < x < 2. Therefore, the problem reduces to looking for the maximum value of
A(x) over the open interval (0, 2). Since A(x) will have an absolute maximum (and absolute minimum) over
the closed interval [0, 2], we consider A(x) = 2x 4 − x2 over the interval [0, 2]. If the absolute maximum
occurs at an interior point, then we have found an absolute maximum in the open interval.
Step 6: As mentioned earlier, A(x) is a continuous function over the closed, bounded interval [0, 2]. Therefore,
it has an absolute maximum (and absolute minimum). At the endpoints x = 0 and x = 2, A(x) = 0. For
0 < x < 2, A(x) > 0. Therefore, the maximum must occur at a critical point. Taking the derivative of A(x),
we obtain


Chapter 4 | Applications of Derivatives 447




(4.7)


4.35


A′(x) = 2 4 − x2 + 2x · 1
2 4 − x2


(−2x)


= 2 4 − x2 − 2x
2


4 − x2


= 8 − 4x
2


4 − x2
.


To find critical points, we need to find where A′(x) = 0. We can see that if x is a solution of
8 − 4x2


4 − x2
= 0,


then x must satisfy
8 − 4x2 = 0.


Therefore, x2 = 2. Thus, x = ± 2 are the possible solutions of Equation 4.7. Since we are considering x
over the interval [0, 2], x = 2 is a possibility for a critical point, but x = − 2 is not. Therefore, we check
whether 2 is a solution of Equation 4.7. Since x = 2 is a solution of Equation 4.7, we conclude that 2
is the only critical point of A(x) in the interval [0, 2]. Therefore, A(x) must have an absolute maximum at the
critical point x = 2. To determine the dimensions of the rectangle, we need to find the length L and the width
W. If x = 2 then


y = 1 − ( 2)
2


4
= 1 − 1


2
= 1


2
.


Therefore, the dimensions of the rectangle are L = 2x = 2 2 and W = 2y = 2
2
= 2. The area of this


rectangle is A = LW = (2 2)( 2) = 4.


Modify the area function A if the rectangle is to be inscribed in the unit circle x2 + y2 = 1. What is the
domain of consideration?


Solving Optimization Problems when the Interval Is Not Closed or IsUnbounded
In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme valuetheorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domainis neither closed nor bounded.
Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For
example, the function f (x) = x2 + 4 over (−∞, ∞) has an absolute minimum of 4 at x = 0. Therefore, we can still
consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. Inthe next example, we try to minimize a function over an unbounded domain. We will see that, although the domain ofconsideration is (0, ∞), the function has an absolute minimum.
In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult toshow that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surfacearea. Consequently, we consider the modified problem of determining which open-topped box with a specified volume hasthe smallest surface area.


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Example 4.37
Minimizing Surface Area
A rectangular box with a square base, an open top, and a volume of 216 in.3 is to be constructed. What should
the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?
Solution
Step 1: Draw a rectangular box and introduce the variable x to represent the length of each side of the square
base; let y represent the height of the box. Let S denote the surface area of the open-top box.


Figure 4.69 We want to minimize the surface area of asquare-based box with a given volume.


Step 2: We need to minimize the surface area. Therefore, we need to minimize S.
Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base.
The area of each of the four vertical sides is x · y. The area of the base is x2. Therefore, the surface area of the
box is


S = 4xy + x2.


Step 4: Since the volume of this box is x2 y and the volume is given as 216 in.3, the constraint equation is
x2 y = 216.


Solving the constraint equation for y, we have y = 216
x2


. Therefore, we can write the surface area as a function
of x only:


S(x) = 4x


216
x2

⎠+ x


2.


Therefore, S(x) = 864x + x2.
Step 5: Since we are requiring that x2 y = 216, we cannot have x = 0. Therefore, we need x > 0. On the
other hand, x is allowed to have any positive value. Note that as x becomes large, the height of the box
y becomes correspondingly small so that x2 y = 216. Similarly, as x becomes small, the height of the box
becomes correspondingly large. We conclude that the domain is the open, unbounded interval (0, ∞). Note that,
unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absoluteminimum over a closed, bounded interval. However, in the next step, we discover why this function must have anabsolute minimum over the interval (0, ∞).
Step 6: Note that as x → 0+ , S(x) → ∞. Also, as x → ∞, S(x) → ∞. Since S is a continuous function


Chapter 4 | Applications of Derivatives 449




4.36


that approaches infinity at the ends, it must have an absolute minimum at some x ∈ (0, ∞). This minimum must
occur at a critical point of S. The derivative is


S′ (x) = − 864
x2


+ 2x.


Therefore, S′ (x) = 0 when 2x = 864
x2


. Solving this equation for x, we obtain x3 = 432, so
x = 432


3
= 6 2


3
. Since this is the only critical point of S, the absolute minimum must occur at x = 6 23


(see Figure 4.70). When x = 6 23 , y = 216

⎝6 2


3 ⎞


2
= 3 2


3
in. Therefore, the dimensions of the box should be


x = 6 2
3


in. and y = 3 23 in. With these dimensions, the surface area is
S⎛⎝6 2


3 ⎞
⎠ =


864


6 2
3


+ ⎛⎝6 2
3 ⎞


2
= 108 4


3
in.2


Figure 4.70 We can use a graph to determine the dimensionsof a box of given the volume and the minimum surface area.


Consider the same open-top box, which is to have volume 216 in.3. Suppose the cost of the material for
the base is 20 ¢ /in.2 and the cost of the material for the sides is 30 ¢ /in.2 and we are trying to minimize the
cost of this box. Write the cost as a function of the side lengths of the base. (Let x be the side length of the base
and y be the height of the box.)


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4.7 EXERCISES
For the following exercises, answer by proof,counterexample, or explanation.
311. When you find the maximum for an optimizationproblem, why do you need to check the sign of thederivative around the critical points?
312. Why do you need to check the endpoints foroptimization problems?
313. True or False. For every continuous nonlinearfunction, you can find the value x that maximizes the
function.
314. True or False. For every continuous nonconstantfunction on a closed, finite domain, there exists at least one
x that minimizes or maximizes the function.
For the following exercises, set up and evaluate eachoptimization problem.
315. To carry a suitcase on an airplane, the length
+width + height of the box must be less than or equal
to 62 in. Assuming the height is fixed, show that the
maximum volume is V = h⎛⎝31 − ⎛⎝12⎞⎠h⎞⎠


2
. What height


allows you to have the largest volume?
316. You are constructing a cardboard box with thedimensions 2 m by 4 m. You then cut equal-size squares
from each corner so you may fold the edges. What are thedimensions of the box with the largest volume?


317. Find the positive integer that minimizes the sum ofthe number and its reciprocal.
318. Find two positive integers such that their sum is 10,
and minimize and maximize the sum of their squares.
For the following exercises, consider the construction of apen to enclose an area.
319. You have 400 ft of fencing to construct a
rectangular pen for cattle. What are the dimensions of thepen that maximize the area?


320. You have 800 ft of fencing to make a pen for hogs.
If you have a river on one side of your property, what is thedimension of the rectangular pen that maximizes the area?
321. You need to construct a fence around an area of
1600 ft. What are the dimensions of the rectangular pen to
minimize the amount of material needed?
322. Two poles are connected by a wire that is alsoconnected to the ground. The first pole is 20 ft tall and
the second pole is 10 ft tall. There is a distance of 30 ft
between the two poles. Where should the wire be anchoredto the ground to minimize the amount of wire needed?


323. [T] You are moving into a new apartment and noticethere is a corner where the hallway narrows from
8 ft to 6 ft. What is the length of the longest item that can
be carried horizontally around the corner?


324. A patient’s pulse measures
70 bpm, 80 bpm, then 120 bpm. To determine an
accurate measurement of pulse, the doctor wants to knowwhat value minimizes the expression
(x − 70)2 + (x − 80)2 + (x − 120)2? What value
minimizes it?


Chapter 4 | Applications of Derivatives 451




325. In the previous problem, assume the patient wasnervous during the third measurement, so we only weightthat value half as much as the others. What is the value that
minimizes (x − 70)2 + (x − 80)2 + 1


2
(x − 120)2?


326. You can run at a speed of 6 mph and swim at a speed
of 3 mph and are located on the shore, 4 miles east of
an island that is 1 mile north of the shoreline. How far
should you run west to minimize the time needed to reachthe island?


For the following problems, consider a lifeguard at acircular pool with diameter 40 m. He must reach someone
who is drowning on the exact opposite side of the pool, atposition C. The lifeguard swims with a speed v and runs
around the pool at speed w = 3v.


327. Find a function that measures the total amount oftime it takes to reach the drowning person as a function ofthe swim angle, θ.
328. Find at what angle θ the lifeguard should swim to
reach the drowning person in the least amount of time.
329. A truck uses gas as g(v) = av + bv , where v
represents the speed of the truck and g represents the
gallons of fuel per mile. At what speed is fuel consumptionminimized?
For the following exercises, consider a limousine that gets
m(v) = (120 − 2v)


5
mi/gal at speed v, the chauffeur


costs $15/h, and gas is $3.5/gal.
330. Find the cost per mile at speed v.
331. Find the cheapest driving speed.
For the following exercises, consider a pizzeria that sell


pizzas for a revenue of R(x) = ax and costs
C(x) = b + cx + dx2, where x represents the number of
pizzas.
332. Find the profit function for the number of pizzas.How many pizzas gives the largest profit per pizza?
333. Assume that R(x) = 10x and C(x) = 2x + x2.
How many pizzas sold maximizes the profit?
334. Assume that R(x) = 15x, and
C(x) = 60 + 3x + 1


2
x2. How many pizzas sold


maximizes the profit?
For the following exercises, consider a wire 4 ft long cut
into two pieces. One piece forms a circle with radius r and
the other forms a square of side x.
335. Choose x to maximize the sum of their areas.
336. Choose x to minimize the sum of their areas.
For the following exercises, consider two nonnegativenumbers x and y such that x + y = 10. Maximize and
minimize the quantities.
337. xy
338. x2 y2


339. y − 1x
340. x2 − y
For the following exercises, draw the given optimizationproblem and solve.
341. Find the volume of the largest right circular cylinderthat fits in a sphere of radius 1.
342. Find the volume of the largest right cone that fits in asphere of radius 1.
343. Find the area of the largest rectangle that fits into the
triangle with sides x = 0, y = 0 and x


4
+


y
6
= 1.


344. Find the largest volume of a cylinder that fits into acone that has base radius R and height h.
345. Find the dimensions of the closed cylinder volume
V = 16π that has the least amount of surface area.
346. Find the dimensions of a right cone with surface area
S = 4π that has the largest volume.


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For the following exercises, consider the points on thegiven graphs. Use a calculator to graph the functions.
347. [T] Where is the line y = 5 − 2x closest to the
origin?
348. [T] Where is the line y = 5 − 2x closest to point
(1, 1)?


349. [T] Where is the parabola y = x2 closest to point
(2, 0)?


350. [T] Where is the parabola y = x2 closest to point
(0, 3)?


For the following exercises, set up, but do not evaluate,each optimization problem.
351. A window is composed of a semicircle placed ontop of a rectangle. If you have 20 ft of window-framing
materials for the outer frame, what is the maximum size ofthe window you can create? Use r to represent the radius
of the semicircle.


352. You have a garden row of 20 watermelon plants
that produce an average of 30 watermelons apiece. For
any additional watermelon plants planted, the output perwatermelon plant drops by one watermelon. How manyextra watermelon plants should you plant?


353. You are constructing a box for your cat to sleep in.The plush material for the square bottom of the box costs
$5/ft2 and the material for the sides costs $2/ft2. You
need a box with volume 4 ft2. Find the dimensions of the
box that minimize cost. Use x to represent the length of the
side of the box.
354. You are building five identical pens adjacent to each
other with a total area of 1000 m2, as shown in the
following figure. What dimensions should you use tominimize the amount of fencing?


355. You are the manager of an apartment complex with
50 units. When you set rent at $800/month, all
apartments are rented. As you increase rent by
$25/month, one fewer apartment is rented. Maintenance
costs run $50/month for each occupied unit. What is the
rent that maximizes the total amount of profit?


Chapter 4 | Applications of Derivatives 453




4.8 | L’Hôpital’s Rule
Learning Objectives


4.8.1 Recognize when to apply L’Hôpital’s rule.
4.8.2 Identify indeterminate forms produced by quotients, products, subtractions, and powers,and apply L’Hôpital’s rule in each case.
4.8.3 Describe the relative growth rates of functions.


In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule, uses derivatives tocalculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead ofrelying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and todetermine its exact value.
Applying L’Hôpital’s Rule
L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider


lim
x → a


f (x)
g(x)


.


If lim
x → a


f (x) = L1 and limx → ag(x) = L2 ≠ 0, then
lim
x → a


f (x)
g(x)


=
L1
L2


.


However, what happens if lim
x → a


f (x) = 0 and lim
x → a


g(x) = 0? We call this one of the indeterminate forms, of type 0
0
.


This is considered an indeterminate form because we cannot determine the exact behavior of f (x)
g(x)


as x → a without
further analysis. We have seen examples of this earlier in the text. For example, consider


lim
x → 2


x2 − 4
x − 2


and lim
x → 0


sinx
x .


For the first of these examples, we can evaluate the limit by factoring the numerator and writing
lim
x → 2


x2 − 4
x − 2


= lim
x → 2


(x + 2)(x − 2)
x − 2


= lim
x → 2


(x + 2) = 2 + 2 = 4.


For lim
x → 0


sinx
x we were able to show, using a geometric argument, that


lim
x → 0


sinx
x = 1.


Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way toevaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we couldnot calculate previously.
The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions
f and g such that lim


x → a
f (x) = 0 = lim


x → a
g(x) and such that g′ (a) ≠ 0 For x near a, we can write


f (x) ≈ f (a) + f ′ (a)(x − a)


and
g(x) ≈ g(a) + g′ (a)(x − a).


Therefore,
f (x)
g(x)



f (a) + f ′ (a)(x − a)
g(a) + g′ (a)(x − a)


.


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Figure 4.71 If lim
x → a


f (x) = lim
x → a


g(x), then the ratio f (x)/g(x) is
approximately equal to the ratio of their linear approximations near a.


Since f is differentiable at a, then f is continuous at a, and therefore f (a) = lim
x → a


f (x) = 0. Similarly,
g(a) = lim


x → a
g(x) = 0. If we also assume that f ′ and g′ are continuous at x = a, then f ′ (a) = lim


x → a
f ′ (x) and


g′ (a) = lim
x → a


g′ (x). Using these ideas, we conclude that
lim
x → a


f (x)
g(x)


= lim
x → a


f ′ (x)(x − a)
g′ (x)(x − a)


= lim
x → a


f ′ (x)
g′ (x)


.


Note that the assumption that f ′ and g′ are continuous at a and g′ (a) ≠ 0 can be loosened. We state L’Hôpital’s rule
formally for the indeterminate form 0


0
. Also note that the notation 0


0
does not mean we are actually dividing zero by zero.


Rather, we are using the notation 0
0
to represent a quotient of limits, each of which is zero.


Theorem 4.12: L’Hôpital’s Rule (0/0 Case)
Suppose f and g are differentiable functions over an open interval containing a, except possibly at a. If
lim
x → a


f (x) = 0 and lim
x → a


g(x) = 0, then
lim
x → a


f (x)
g(x)


= lim
x → a


f ′ (x)
g′ (x)


,


assuming the limit on the right exists or is ∞ or −∞. This result also holds if we are considering one-sided limits,
or if a = ∞ and −∞.


Proof
We provide a proof of this theorem in the special case when f , g, f ′, and g′ are all continuous over an open
interval containing a. In that case, since lim


x → a
f (x) = 0 = lim


x → a
g(x) and f and g are continuous at a, it follows that


f (a) = 0 = g(a). Therefore,


Chapter 4 | Applications of Derivatives 455




lim
x → a


f (x)
g(x)


= lim
x → a


f (x) − f (a)
g(x) − g(a)


since f (a) = 0 = g(a)


= lim
x → a


f (x) − f (a)
x − a


g(x) − g(a)
x − a


algebra


=
lim
x → a


f (x) − f (a)
x − a


lim
x → a


g(x) − g(a)
x − a


limit of a quotient


=
f ′ (a)
g′ (a)


definition of he derivative


=
lim
x → a


f ′ (x)


lim
x → a


g′ (x)
continuity of f ′ and g′


= lim
x → a


f ′ (x)
g′ (x)


. limit of a quotient


Note that L’Hôpital’s rule states we can calculate the limit of a quotient fg by considering the limit of the quotient of the
derivatives f ′


g′
. It is important to realize that we are not calculating the derivative of the quotient fg .



Example 4.38
Applying L’Hôpital’s Rule (0/0 Case)
Evaluate each of the following limits by applying L’Hôpital’s rule.


a. lim
x → 0


1 − cosx
x


b. lim
x → 1


sin(πx)
lnx


c. lim
x → ∞


e1/x − 1
1/x


d. lim
x → 0


sinx − x
x2


Solution
a. Since the numerator 1 − cosx → 0 and the denominator x → 0, we can apply L’Hôpital’s rule to


evaluate this limit. We have
lim
x → 0


1 − cosx
x = limx → 0


d
dx
(1 − cosx)


d
dx
(x)


= lim
x → 0


sinx
1


=
lim
x → 0


(sinx)


lim
x → 0


(1)


= 0
1
= 0.


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4.37


b. As x → 1, the numerator sin(πx) → 0 and the denominator ln(x) → 0. Therefore, we can apply
L’Hôpital’s rule. We obtain


lim
x → 1


sin(πx)
lnx


= lim
x → 1


πcos(πx)
1/x


= lim
x → 1


(πx)cos(πx)


= (π · 1)(−1) = −π.


c. As x → ∞, the numerator e1/x − 1 → 0 and the denominator ⎛⎝1x⎞⎠→ 0. Therefore, we can apply
L’Hôpital’s rule. We obtain


lim
x → ∞


e1/x − 1
1
x


= lim
x → ∞


e1/x


−1


x2





−1


x2



= lim
x → ∞


e1/x = e0 = 1 · lim
x → ∞


e1/x − 1
lnx


.


d. As x → 0, both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s
rule. We obtain


lim
x → 0


sinx − x
x2


= lim
x → 0


cosx − 1
2x


.


Since the numerator and denominator of this new quotient both approach zero as x → 0, we apply
L’Hôpital’s rule again. In doing so, we see that


lim
x → 0


cosx − 1
2x


= lim
x → 0


−sinx
2


= 0.


Therefore, we conclude that
lim
x → 0


sinx − x
x2


= 0.


Evaluate lim
x → 0


x
tanx.


We can also use L’Hôpital’s rule to evaluate limits of quotients f (x)
g(x)


in which f (x) → ±∞ and g(x) → ±∞. Limits of
this form are classified as indeterminate forms of type ∞/∞. Again, note that we are not actually dividing ∞ by ∞.
Since ∞ is not a real number, that is impossible; rather, ∞/∞. is used to represent a quotient of limits, each of which is
∞ or −∞.


Theorem 4.13: L’Hôpital’s Rule (∞/∞ Case)
Suppose f and g are differentiable functions over an open interval containing a, except possibly at a. Suppose
lim
x → a


f (x) = ∞ (or −∞) and lim
x → a


g(x) = ∞ (or −∞). Then,
lim
x → a


f (x)
g(x)


= lim
x → a


f ′ (x)
g′ (x)


,


assuming the limit on the right exists or is ∞ or −∞. This result also holds if the limit is infinite, if a = ∞ or


Chapter 4 | Applications of Derivatives 457




−∞, or the limit is one-sided.


Example 4.39
Applying L’Hôpital’s Rule (∞/∞ Case)
Evaluate each of the following limits by applying L’Hôpital’s rule.


a. lim
x → ∞


3x + 5
2x + 1


b. lim
x → 0+


lnx
cotx


Solution
a. Since 3x + 5 and 2x + 1 are first-degree polynomials with positive leading coefficients,


lim
x → ∞


(3x + 5) = ∞ and lim
x → ∞


(2x + 1) = ∞. Therefore, we apply L’Hôpital’s rule and obtain
lim


x → ∞
3x + 5
2x + 1


= lim
x → ∞


3 + 5/x
2x + 1


= lim
x → ∞


3
2
= 3


2
.


Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter weshowed how to evaluate such a limit by dividing the numerator and denominator by the highest power of
x in the denominator. In doing so, we saw that


lim
x → ∞


3x + 5
2x + 1


= lim
x → ∞


3 + 5/x
2x + 1


= 3
2
.


L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
b. Here, lim


x → 0+
lnx = −∞ and lim


x → 0+
cotx = ∞. Therefore, we can apply L’Hôpital’s rule and obtain


lim
x → 0+


lnx
cotx = limx → 0+


1/x
−csc2 x


= lim
x → 0+


1
−xcsc2 x


.


Now as x → 0+ , csc2 x → ∞. Therefore, the first term in the denominator is approaching zero and
the second term is getting really large. In such a case, anything can happen with the product. Therefore,we cannot make any conclusion yet. To evaluate the limit, we use the definition of cscx to write


lim
x → 0+


1
−xcsc2 x


= lim
x → 0+


sin2 x
−x .


Now lim
x → 0+


sin2 x = 0 and lim
x → 0+


x = 0, so we apply L’Hôpital’s rule again. We find


lim
x → 0+


sin2 x
−x = lim


x → 0+
2sinxcosx


−1
= 0


−1
= 0.


We conclude that
lim


x → 0+
lnx
cotx = 0.


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4.38


4.39


Evaluate lim
x → ∞


lnx
5x


.


As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that
to apply L’Hôpital’s rule to a quotient f (x)


g(x)
, it is essential that the limit of f (x)


g(x)
be of the form 0


0
or ∞/∞. Consider


the following example.
Example 4.40
When L’Hôpital’s Rule Does Not Apply
Consider lim


x → 1
x2 + 5
3x + 4


. Show that the limit cannot be evaluated by applying L’Hôpital’s rule.


Solution
Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot applyL’Hôpital’s rule. If we try to do so, we get


d
dx

⎝x


2 + 5⎞⎠ = 2x


and
d
dx


(3x + 4) = 3.


At which point we would conclude erroneously that
lim
x → 1


x2 + 5
3x + 4


= lim
x → 1


2x
3


= 2
3
.


However, since lim
x → 1



⎝x


2 + 5⎞⎠ = 6 and limx → 1(3x + 4) = 7, we actually have
lim
x → 1


x2 + 5
3x + 4


= 6
7
.


We can conclude that
lim
x → 1


x2 + 5
3x + 4


≠ lim
x → 1


d
dx

⎝x


2 + 5⎞⎠
d
dx
(3x + 4)


.


Explain why we cannot apply L’Hôpital’s rule to evaluate lim
x → 0+


cosx
x . Evaluate lim


x → 0+
cosx
x by other


means.


Other Indeterminate Forms
L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms 0


0
and ∞/∞. However, we can


also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The
expressions 0 ·∞, ∞ −∞, 1∞, ∞0, and 00 are all considered indeterminate forms. These expressions are not
real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these areindeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite


Chapter 4 | Applications of Derivatives 459




the indeterminate forms in such a way that we arrive at the indeterminate form 0
0
or ∞/∞.


Indeterminate Form of Type 0 ·∞
Suppose we want to evaluate lim


x → a

⎝ f (x) · g(x)⎞⎠, where f (x) → 0 and g(x) → ∞ (or −∞) as x → a. Since one term


in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen tothe product. We use the notation 0 ·∞ to denote the form that arises in this situation. The expression 0 ·∞ is considered
indeterminate because we cannot determine without further analysis the exact behavior of the product f (x)g(x) as x → ∞.
For example, let n be a positive integer and consider


f (x) = 1
(xn + 1)


and g(x) = 3x2.


As x → ∞, f (x) → 0 and g(x) → ∞. However, the limit as x → ∞ of f (x)g(x) = 3x2
(xn + 1)


varies, depending on n.
If n = 2, then lim


x → ∞
f (x)g(x) = 3. If n = 1, then lim


x → ∞
f (x)g(x) = ∞. If n = 3, then lim


x → ∞
f (x)g(x) = 0. Here we


consider another limit involving the indeterminate form 0 ·∞ and show how to rewrite the function as a quotient to use
L’Hôpital’s rule.
Example 4.41
Indeterminate Form of Type 0 ·∞
Evaluate lim


x → 0+
x lnx.


Solution
First, rewrite the function x lnx as a quotient to apply L’Hôpital’s rule. If we write


x lnx = lnx
1/x


,


we see that lnx → −∞ as x → 0+ and 1x → ∞ as x → 0+ . Therefore, we can apply L’Hôpital’s rule and
obtain


lim
x → 0+


lnx
1/x


= lim
x → 0+


d
dx
(lnx)


d
dx
(1/x)


= lim
x → 0+


1/x2
−1/x


= lim
x → 0+


(−x) = 0.


We conclude that
lim


x → 0+
x lnx = 0.


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4.40


Figure 4.72 Finding the limit at x = 0 of the function
f (x) = x lnx.


Evaluate lim
x → 0


xcotx.


Indeterminate Form of Type ∞−∞
Another type of indeterminate form is ∞−∞. Consider the following example. Let n be a positive integer and let
f (x) = 3xn and g(x) = 3x2 + 5. As x → ∞, f (x) → ∞ and g(x) → ∞. We are interested in lim


x → ∞

⎝ f (x) − g(x)⎞⎠.


Depending on whether f (x) grows faster, g(x) grows faster, or they grow at the same rate, as we see next, anything can
happen in this limit. Since f (x) → ∞ and g(x) → ∞, we write ∞−∞ to denote the form of this limit. As with our
other indeterminate forms, ∞−∞ has no meaning on its own and we must do more analysis to determine the value of the
limit. For example, suppose the exponent n in the function f (x) = 3xn is n = 3, then


lim
x → ∞



⎝ f (x) − g(x)⎞⎠ = limx → ∞



⎝3x


3 − 3x2 − 5⎞⎠ = ∞.


On the other hand, if n = 2, then
lim


x → ∞

⎝ f (x) − g(x)⎞⎠ = limx → ∞



⎝3x


2 − 3x2 − 5⎞⎠ = −5.


However, if n = 1, then
lim


x → ∞

⎝ f (x) − g(x)⎞⎠ = limx → ∞



⎝3x − 3x


2 − 5⎞⎠ = −∞.


Therefore, the limit cannot be determined by considering only ∞−∞. Next we see how to rewrite an expression involving
the indeterminate form ∞−∞ as a fraction to apply L’Hôpital’s rule.
Example 4.42
Indeterminate Form of Type ∞−∞


Chapter 4 | Applications of Derivatives 461




4.41


Evaluate lim
x → 0+




1
x2


− 1tanx

⎠.


Solution
By combining the fractions, we can write the function as a quotient. Since the least common denominator is
x2 tanx, we have


1
x2


− 1tanx =
(tanx) − x2


x2 tanx
.


As x → 0+ , the numerator tanx − x2 → 0 and the denominator x2 tanx → 0. Therefore, we can apply
L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have


lim
x → 0+


(tanx) − x2


x2 tanx
= lim


x → 0+



⎝sec


2 x⎞⎠− 2x


x2 sec2 x + 2x tanx
.


As x → 0+ , ⎛⎝sec2 x⎞⎠− 2x → 1 and x2 sec2 x + 2x tanx → 0. Since the denominator is positive as x
approaches zero from the right, we conclude that


lim
x → 0+



⎝sec


2 x⎞⎠− 2x


x2 sec2 x + 2x tanx
= ∞.


Therefore,
lim


x → 0+




1
x2


− 1tanx

⎠ = ∞.


Evaluate lim
x → 0+




1
x −


1
sinx

⎠.


Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions 00, ∞0, and
1∞ are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these
expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arisewhen finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminateforms.
Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problemevaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want
to evaluate lim


x → a
f (x)


g(x) and we arrive at the indeterminate form ∞0. (The indeterminate forms 00 and 1∞ can be
handled similarly.) We proceed as follows. Let


y = f (x)
g(x)


.


Then,
lny = ln⎛⎝ f (x)


g(x)⎞
⎠ = g(x)ln



⎝ f (x)⎞⎠.


Therefore,
lim
x → a



⎣ln(y)⎤⎦ = limx → a



⎣g(x)ln⎛⎝ f (x)⎞⎠⎤⎦.


Since lim
x → a


f (x) = ∞, we know that lim
x → a


ln⎛⎝ f (x)⎞⎠ = ∞. Therefore, limx → ag(x)ln⎛⎝ f (x)⎞⎠ is of the indeterminate form


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4.42


0 ·∞, and we can use the techniques discussed earlier to rewrite the expression g(x)ln⎛⎝ f (x)⎞⎠ in a form so that we can
apply L’Hôpital’s rule. Suppose lim


x → a
g(x)ln⎛⎝ f (x)⎞⎠ = L, where L may be ∞ or −∞. Then


lim
x → a



⎣ln(y)⎤⎦ = L.


Since the natural logarithm function is continuous, we conclude that
ln⎛⎝ limx → ay



⎠ = L,


which gives us
lim
x → a


y = lim
x → a


f (x)
g(x)


= eL.


Example 4.43
Indeterminate Form of Type ∞0
Evaluate lim


x → ∞
x1/x.


Solution
Let y = x1/x. Then,


ln⎛⎝x
1/x⎞
⎠ =


1
x lnx =


lnx
x .


We need to evaluate lim
x → ∞


lnx
x . Applying L’Hôpital’s rule, we obtain


lim
x → ∞


lny = lim
x → ∞


lnx
x = limx → ∞


1/x
1


= 0.


Therefore, lim
x → ∞


lny = 0. Since the natural logarithm function is continuous, we conclude that
ln⎛⎝ limx → ∞y



⎠ = 0,


which leads to
lim


x → ∞
y = lim


x → ∞
lnx
x = e


0 = 1.


Hence,
lim


x → ∞
x1/x = 1.


Evaluate lim
x → ∞


x1/ln
(x).


Example 4.44
Indeterminate Form of Type 00


Chapter 4 | Applications of Derivatives 463




4.43


Evaluate lim
x → 0+


xsinx.


Solution
Let


y = xsinx.


Therefore,
lny = ln⎛⎝x


sinx⎞
⎠ = sinx lnx.


We now evaluate lim
x → 0+


sinx lnx. Since lim
x → 0+


sinx = 0 and lim
x → 0+


lnx = −∞, we have the indeterminate
form 0 ·∞. To apply L’Hôpital’s rule, we need to rewrite sinx lnx as a fraction. We could write


sinx lnx = sinx
1/lnx


or
sinx lnx = lnx


1/sinx
= lnxcscx.


Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain
lim


x → 0+
sinx lnx = lim


x → 0+
sinx
1/lnx


= lim
x → 0+


cosx
−1/⎛⎝x(lnx)


2⎞


= lim
x → 0+



⎝−x(lnx)


2 cosx⎞⎠.


Unfortunately, we not only have another expression involving the indeterminate form 0 ·∞, but the new limit
is even more complicated to evaluate than the one with which we started. Instead, we try the second option. Bywriting


sinx lnx = lnx
1/sinx


= lnxcscx,


and applying L’Hôpital’s rule, we obtain
lim


x → 0+
sinx lnx = lim


x → 0+
lnx
cscx = lim


x → 0+
1/x


−cscxcotx = limx → 0+
−1


xcscxcotx.


Using the fact that cscx = 1
sinx


and cotx = cosx
sinx


, we can rewrite the expression on the right-hand side as
lim


x → 0+
−sin2 x
xcosx = lim


x → 0+


sinx
x · (−tanx)



⎦ =

⎝ limx → 0+


sinx
x

⎠ ·

⎝ limx → 0+


(−tanx)

⎠ = 1 · 0 = 0.


We conclude that lim
x → 0+


lny = 0. Therefore, ln⎛⎝ limx → 0+ y

⎠ = 0 and we have


lim
x → 0+


y = lim
x → 0+


xsinx = e0 = 1.


Hence,
lim


x → 0+
xsinx = 1.


Evaluate lim
x → 0+


xx.


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Growth Rates of Functions
Suppose the functions f and g both approach infinity as x → ∞. Although the values of both functions become
arbitrarily large as the values of x become sufficiently large, sometimes one function is growing more quickly than the
other. For example, f (x) = x2 and g(x) = x3 both approach infinity as x → ∞. However, as shown in the following
table, the values of x3 are growing much faster than the values of x2.


x 10 100 1000 10,000


f(x) = x2 100 10,000 1,000,000 100,000,000


g(x) = x3 1000 1,000,000 1,000,000,000 1,000,000,000,000


Table 4.21 Comparing the Growth Rates of x2 and x3
In fact,


lim
x → ∞


x3


x2
= lim


x → ∞
x = ∞. or, equivalently, lim


x → ∞
x2


x3
= lim


x → ∞
1
x = 0.


As a result, we say x3 is growing more rapidly than x2 as x → ∞. On the other hand, for f (x) = x2 and
g(x) = 3x2 + 4x + 1, although the values of g(x) are always greater than the values of f (x) for x > 0, each value of
g(x) is roughly three times the corresponding value of f (x) as x → ∞, as shown in the following table. In fact,


lim
x → ∞


x2


3x2 + 4x + 1
= 1


3
.


x 10 100 1000 10,000


f(x) = x2 100 10,000 1,000,000 100,000,000


g(x) = 3x2+4x+1 341 30,401 3,004,001 300,040,001


Table 4.22 Comparing the Growth Rates of x2 and 3x2 + 4x + 1
In this case, we say that x2 and 3x2 + 4x + 1 are growing at the same rate as x → ∞.
More generally, suppose f and g are two functions that approach infinity as x → ∞. We say g grows more rapidly than
f as x → ∞ if


lim
x → ∞


g(x)
f (x)


= ∞; or, equivalently, lim
x → ∞


f (x)
g(x)


= 0.


On the other hand, if there exists a constant M ≠ 0 such that
lim


x → ∞
f (x)
g(x)


= M,


we say f and g grow at the same rate as x → ∞.


Chapter 4 | Applications of Derivatives 465




Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.
Example 4.45
Comparing the Growth Rates of ln(x), x2, and ex


For each of the following pairs of functions, use L’Hôpital’s rule to evaluate lim
x → ∞


f (x)
g(x)

⎠.


a. f (x) = x2 and g(x) = ex
b. f (x) = ln(x) and g(x) = x2


Solution
a. Since lim


x → ∞
x2 = ∞ and lim


x → ∞
ex, we can use L’Hôpital’s rule to evaluate lim


x → ∞


x2


ex

⎦. We obtain


lim
x → ∞


x2


ex
= lim


x → ∞
2x
ex


.


Since lim
x → ∞


2x = ∞ and lim
x → ∞


ex = ∞, we can apply L’Hôpital’s rule again. Since
lim


x → ∞
2x
ex


= lim
x → ∞


2
ex


= 0,


we conclude that
lim


x → ∞
x2


ex
= 0.


Therefore, ex grows more rapidly than x2 as x → ∞ (See Figure 4.73 and Table 4.23).


Figure 4.73 An exponential function grows at a faster ratethan a power function.


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x 5 10 15 20


x2 25 100 225 400


ex 148 22,026 3,269,017 485,165,195


Table 4.23Growth rates of a power function and an exponential function.
b. Since lim


x → ∞
lnx = ∞ and lim


x → ∞
x2 = ∞, we can use L’Hôpital’s rule to evaluate lim


x → ∞
lnx
x2


. We
obtain


lim
x → ∞


lnx
x2


= lim
x → ∞


1/x
2x


= lim
x → ∞


1
2x2


= 0.


Thus, x2 grows more rapidly than lnx as x → ∞ (see Figure 4.74 and Table 4.24).


Figure 4.74 A power function grows at a faster rate than alogarithmic function.


x 10 100 1000 10,000


ln(x) 2.303 4.605 6.908 9.210


x2 100 10,000 1,000,000 100,000,000


Table 4.24Growth rates of a power function and a logarithmic function


Chapter 4 | Applications of Derivatives 467




4.44 Compare the growth rates of x100 and 2x.


Using the same ideas as in Example 4.45a. it is not difficult to show that ex grows more rapidly than x p for any p > 0.
In Figure 4.75 and Table 4.25, we compare ex with x3 and x4 as x → ∞.


Figure 4.75 The exponential function ex grows faster than x p for any p > 0. (a) A comparison of ex with
x3. (b) A comparison of ex with x4.


x 5 10 15 20


x3 125 1000 3375 8000


x4 625 10,000 50,625 160,000


ex 148 22,026 3,269,017 485,165,195


Table 4.25 An exponential function grows at a faster rate thanany power function
Similarly, it is not difficult to show that x p grows more rapidly than lnx for any p > 0. In Figure 4.76 and Table 4.26,
we compare lnx with x3 and x.


Figure 4.76 The function y = ln(x) grows more slowly than
x p for any p > 0 as x → ∞.


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x 10 100 1000 10,000


ln(x) 2.303 4.605 6.908 9.210


x3 2.154 4.642 10 21.544


x 3.162 10 31.623 100


Table 4.26 A logarithmic function grows at a slower ratethan any root function


Chapter 4 | Applications of Derivatives 469




4.8 EXERCISES
For the following exercises, evaluate the limit.
356. Evaluate the limit lim


x → ∞
ex
x .


357. Evaluate the limit lim
x → ∞


ex


xk
.


358. Evaluate the limit lim
x → ∞


lnx
xk


.


359. Evaluate the limit lim
x → a


x − a
x2 − a2


.


360. Evaluate the limit lim
x → a


x − a
x3 − a3


.


361. Evaluate the limit lim
x → a


x − a
xn − an


.


For the following exercises, determine whether you canapply L’Hôpital’s rule directly. Explain why or why not.Then, indicate if there is some way you can alter the limitso you can apply L’Hôpital’s rule.
362. lim


x → 0+
x2 lnx


363. lim
x → ∞


x1/x


364. lim
x → 0


x2/x


365. lim
x → 0


x2
1/x


366. lim
x → ∞


ex
x


For the following exercises, evaluate the limits with eitherL’Hôpital’s rule or previously learned methods.
367. lim


x → 3
x2 − 9
x − 3


368. lim
x → 3


x2 − 9
x + 3


369. lim
x → 0


(1 + x)−2 − 1
x


370. lim
x → π/2


cosx
2
π − x


371. lim
x → π


x − π
sinx


372. lim
x → 1


x − 1
sinx


373. lim
x → 0


(1 + x)n − 1
x


374. lim
x → 0


(1 + x)n − 1 − nx
x2


375. lim
x → 0


sinx − tanx
x3


376. lim
x → 0


1 + x − 1 − x
x


377. lim
x → 0


ex − x − 1
x2


378. lim
x → 0


tanx
x


379. lim
x → 1


x − 1
lnx


380. lim
x → 0


(x + 1)1/x


381. lim
x → 1


x − x3


x − 1


382. lim
x → 0+


x2x


383. lim
x → ∞


xsin⎛⎝
1
x



384. lim
x → 0


sinx − x
x2


385. lim
x → 0+


x ln⎛⎝x
4⎞


386. lim
x → ∞


(x − ex)


387. lim
x → ∞


x2 e−x


388. lim
x → 0


3x − 2x
x


389. lim
x → 0


1 + 1/x
1 − 1/x


390. lim
x → π/4


(1 − tanx)cotx


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391. lim
x → ∞


xe1/x


392. lim
x → 0


x1/cosx


393. lim
x → 0


x1/x


394. lim
x → 0



⎝1 −


1
x



x


395. lim
x → ∞



⎝1 −


1
x



x


For the following exercises, use a calculator to graph thefunction and estimate the value of the limit, then useL’Hôpital’s rule to find the limit directly.
396. [T] lim


x → 0
ex − 1


x


397. [T] lim
x → 0


xsin⎛⎝
1
x



398. [T] lim
x → 1


x − 1
1 − cos(πx)


399. [T] lim
x → 1


e
(x − 1) − 1
x − 1


400. [T] lim
x → 1


(x − 1)2


lnx


401. [T] lim
x → π


1 + cosx
sinx


402. [T] lim
x → 0



⎝cscx −


1
x



403. [T] lim
x → 0+


tan(xx)


404. [T] lim
x → 0+


lnx
sinx


405. [T] lim
x → 0


ex − e−x
x


Chapter 4 | Applications of Derivatives 471




4.9 | Newton’s Method
Learning Objectives


4.9.1 Describe the steps of Newton’s method.
4.9.2 Explain what an iterative process means.
4.9.3 Recognize when Newton’s method does not work.
4.9.4 Apply iterative processes to various situations.


In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form f (x) = 0.
For most functions, however, it is difficult—if not impossible—to calculate their zeroes explicitly. In this section, we takea look at a technique that provides a very efficient way of approximating the zeroes of functions. This technique makes useof tangent line approximations and is behind the method used often by calculators and computers to find zeroes.
Describing Newton’s Method
Consider the task of finding the solutions of f (x) = 0. If f is the first-degree polynomial f (x) = ax + b, then the
solution of f (x) = 0 is given by the formula x = − ba. If f is the second-degree polynomial f (x) = ax2 + bx + c,
the solutions of f (x) = 0 can be found by using the quadratic formula. However, for polynomials of degree 3 or more,
finding roots of f becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are
quite complicated. Also, if f is a polynomial of degree 5 or greater, it is known that no such formulas exist. For example,
consider the function


f (x) = x5 + 8x4 + 4x3 − 2x − 7.


No formula exists that allows us to find the solutions of f (x) = 0. Similar difficulties exist for nonpolynomial functions.
For example, consider the task of finding solutions of tan(x) − x = 0. No simple formula exists for the solutions of this
equation. In cases such as these, we can use Newton’s method to approximate the roots.
Newton’s method makes use of the following idea to approximate the solutions of f (x) = 0. By sketching a graph of
f , we can estimate a root of f (x) = 0. Let’s call this estimate x0. We then draw the tangent line to f at x0. If
f ′ (x0) ≠ 0, this tangent line intersects the x -axis at some point ⎛⎝x1, 0⎞⎠. Now let x1 be the next approximation to the
actual root. Typically, x1 is closer than x0 to an actual root. Next we draw the tangent line to f at x1. If f ′ (x1) ≠ 0,
this tangent line also intersects the x -axis, producing another approximation, x2. We continue in this way, deriving a list
of approximations: x0, x1, x2 ,…. Typically, the numbers x0, x1, x2 ,… quickly approach an actual root x * , as shown
in the following figure.


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Figure 4.77 The approximations x0, x1, x2 ,… approach the actual root x * . The
approximations are derived by looking at tangent lines to the graph of f .


Now let’s look at how to calculate the approximations x0, x1, x2 ,…. If x0 is our first approximation, the approximation
x1 is defined by letting ⎛⎝x1, 0⎞⎠ be the x -intercept of the tangent line to f at x0. The equation of this tangent line is given
by


y = f (x0) + f ′ (x0)(x − x0).


Therefore, x1 must satisfy
f (x0) + f ′ (x0)(x1 − x0) = 0.


Solving this equation for x1, we conclude that
x1 = x0 −


f (x0)
f ′(x0)


.


Similarly, the point ⎛⎝x2, 0⎞⎠ is the x -intercept of the tangent line to f at x1. Therefore, x2 satisfies the equation
x2 = x1 −


f (x1)
f ′(x1)


.


In general, for n > 0, xn satisfies
(4.8)


xn = xn − 1 −
f (xn − 1)
f ′(xn − 1)


.


Next we see how to make use of this technique to approximate the root of the polynomial f (x) = x3 − 3x + 1.


Chapter 4 | Applications of Derivatives 473




Example 4.46
Finding a Root of a Polynomial
Use Newton’s method to approximate a root of f (x) = x3 − 3x + 1 in the interval [1, 2]. Let x0 = 2 and find
x1, x2, x3, x4, and x5.
Solution
From Figure 4.78, we see that f has one root over the interval (1, 2). Therefore x0 = 2 seems like
a reasonable first approximation. To find the next approximation, we use Equation 4.8. Since
f (x) = x3 − 3x + 1, the derivative is f ′ (x) = 3x2 − 3. Using Equation 4.8 with n = 1 (and a calculator
that displays 10 digits), we obtain


x1 = x0 −
f (x0)
f ′(x0)


= 2 −
f (2)
f ′(2)


= 2 − 3
9
≈ 1.666666667.


To find the next approximation, x2, we use Equation 4.8 with n = 2 and the value of x1 stored on the
calculator. We find that


x2 = x1 =
f (x1)
f ′(x1)


≈ 1.548611111.


Continuing in this way, we obtain the following results:
x1 ≈ 1.666666667


x2 ≈ 1.548611111


x3 ≈ 1.532390162


x4 ≈ 1.532088989


x5 ≈ 1.532088886


x6 ≈ 1.532088886.


We note that we obtained the same value for x5 and x6. Therefore, any subsequent application of Newton’s
method will most likely give the same value for xn.


Figure 4.78 The function f (x) = x3 − 3x + 1 has one root
over the interval [1, 2].


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4.45 Letting x0 = 0, let’s use Newton’s method to approximate the root of f (x) = x3 − 3x + 1 over the
interval [0, 1] by calculating x1 and x2.


Newton’s method can also be used to approximate square roots. Here we show how to approximate 2. This method can
be modified to approximate the square root of any positive number.
Example 4.47
Finding a Square Root
Use Newton’s method to approximate 2 (Figure 4.79). Let f (x) = x2 − 2, let x0 = 2, and calculate
x1, x2, x3, x4, x5. (We note that since f (x) = x2 − 2 has a zero at 2, the initial value x0 = 2 is a
reasonable choice to approximate 2.)
Solution
For f (x) = x2 − 2, f ′ (x) = 2x. From Equation 4.8, we know that


xn = xn − 1 −
f (xn − 1)
f ′(xn − 1)


= xn − 1 −
x2 n − 1 − 2


2xn − 1


= 1
2
xn − 1 +


1
xn − 1


= 1
2

⎝xn − 1 +


2
xn − 1



⎠.


Therefore,
x1 =


1
2

⎝x0 +


2
x0

⎠ =


1
2

⎝2 +


2
2

⎠ = 1.5


x2 =
1
2

⎝x1 +


2
x1

⎠ =


1
2

⎝1.5 +


2
1.5

⎠ ≈ 1.416666667.


Continuing in this way, we find that
x1 = 1.5


x2 ≈ 1.416666667


x3 ≈ 1.414215686


x4 ≈ 1.414213562


x5 ≈ 1.414213562.


Since we obtained the same value for x4 and x5, it is unlikely that the value xn will change on any subsequent
application of Newton’s method. We conclude that 2 ≈ 1.414213562.


Chapter 4 | Applications of Derivatives 475




4.46


Figure 4.79 We can use Newton’s method to find 2.


Use Newton’s method to approximate 3 by letting f (x) = x2 − 3 and x0 = 3. Find x1 and x2.


When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation
by using the same formula. In particular, by defining the function F(x) = x − ⎡⎣ f (x)f ′ (x)⎤⎦, we can rewrite Equation 4.8 as
xn = F(xn − 1). This type of process, where each xn is defined in terms of xn − 1 by repeating the same function, is an
example of an iterative process. Shortly, we examine other iterative processes. First, let’s look at the reasons why Newton’smethod could fail to find a root.
Failures of Newton’s Method
Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’smethod might fail include the following:


1. At one of the approximations xn, the derivative f ′ is zero at xn, but f (xn) ≠ 0. As a result, the tangent line of
f at xn does not intersect the x -axis. Therefore, we cannot continue the iterative process.


2. The approximations x0, x1, x2 ,… may approach a different root. If the function f has more than one root, it is
possible that our approximations do not approach the one for which we are looking, but approach a different root(see Figure 4.80). This event most often occurs when we do not choose the approximation x0 close enough to the
desired root.


3. The approximations may fail to approach a root entirely. In Example 4.48, we provide an example of a functionand an initial guess x0 such that the successive approximations never approach a root because the successive
approximations continue to alternate back and forth between two values.


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Figure 4.80 If the initial guess x0 is too far from the root sought, it may lead
to approximations that approach a different root.


Example 4.48
When Newton’s Method Fails
Consider the function f (x) = x3 − 2x + 2. Let x0 = 0. Show that the sequence x1, x2 ,… fails to approach a
root of f .
Solution
For f (x) = x3 − 2x + 2, the derivative is f ′ (x) = 3x2 − 2. Therefore,


x1 = x0 −
f (x0)
f ′ (x0)


= 0 −
f (0)
f ′ (0)


= − 2
−2


= 1.


In the next step,
x2 = x1 −


f (x1)
f ′(x1)


= 1 −
f (1)
f ′ (1)


= 1 − 1
1
= 0.


Consequently, the numbers x0, x1, x2 ,… continue to bounce back and forth between 0 and 1 and never get
closer to the root of f which is over the interval [−2, −1] (see Figure 4.81). Fortunately, if we choose an
initial approximation x0 closer to the actual root, we can avoid this situation.


Chapter 4 | Applications of Derivatives 477




4.47


Figure 4.81 The approximations continue to alternatebetween 0 and 1 and never approach the root of f .


For f (x) = x3 − 2x + 2, let x0 = −1.5 and find x1 and x2.


From Example 4.48, we see that Newton’s method does not always work. However, when it does work, the sequence ofapproximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach aroot found using Newton’s method are included in texts on numerical analysis.
Other Iterative Processes
As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type ofiterative process.
Consider a function F and an initial number x0. Define the subsequent numbers xn by the formula xn = F(xn − 1). This
process is an iterative process that creates a list of numbers x0, x1, x2 ,…, xn ,…. This list of numbers may approach a
finite number x * as n gets larger, or it may not. In Example 4.49, we see an example of a function F and an initial
guess x0 such that the resulting list of numbers approaches a finite value.
Example 4.49
Finding a Limit for an Iterative Process
Let F(x) = 1


2
x + 4 and let x0 = 0. For all n ≥ 1, let xn = F(xn − 1). Find the values x1, x2, x3, x4, x5.


Make a conjecture about what happens to this list of numbers x1, x2, x3…, xn ,… as n → ∞. If the list of
numbers x1, x2, x3 ,… approaches a finite number x * , then x * satisfies x * = F(x * ), and x * is called
a fixed point of F.
Solution
If x0 = 0, then


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x1 =
1
2
(0) + 4 = 4


x2 =
1
2
(4) + 4 = 6


x3 =
1
2
(6) + 4 = 7


x4 =
1
2
(7) + 4 = 7.5


x5 =
1
2
(7.5) + 4 = 7.75


x6 =
1
2
(7.75) + 4 = 7.875


x7 =
1
2
(7.875) + 4 = 7.9375


x8 =
1
2
(7.9375) + 4 = 7.96875


x9 =
1
2
(7.96875) + 4 = 7.984375.


From this list, we conjecture that the values xn approach 8.
Figure 4.82 provides a graphical argument that the values approach 8 as n → ∞. Starting at the point
(x0, x0), we draw a vertical line to the point ⎛⎝x0, F(x0)⎞⎠. The next number in our list is x1 = F(x0). We use
x1 to calculate x2. Therefore, we draw a horizontal line connecting (x0, x1) to the point (x1, x1) on the line
y = x, and then draw a vertical line connecting (x1, x1) to the point ⎛⎝x1, F(x1)⎞⎠. The output F(x1) becomes
x2. Continuing in this way, we could create an infinite number of line segments. These line segments are trapped
between the lines F(x) = x


2
+ 4 and y = x. The line segments get closer to the intersection point of these two


lines, which occurs when x = F(x). Solving the equation x = x
2
+ 4, we conclude they intersect at x = 8.


Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers x0, x1, x2 ,…
approaches x * = 8 as n → ∞.


Figure 4.82 This iterative process approaches the value
x * = 8.


Chapter 4 | Applications of Derivatives 479




4.48 Consider the function F(x) = 1
3
x + 6. Let x0 = 0 and let xn = F(xn − 1) for n ≥ 2. Find


x1, x2, x3, x4, x5. Make a conjecture about what happens to the list of numbers x1, x2, x3 ,…xn ,… as
n → ∞.


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Iterative Processes and Chaos
Iterative processes can yield some v