calculus
eighth edition
James Stewart
M c Master University
and
University of Toronto
Australia • Brazil • Mexico • Singapore • United Kingdom • United States
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Calculus, Eighth Edition
James Stewart
© 2016, 2012 Cengage Learning
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Contents
Preface
xi
To the Student
xxiii
Calculators, Computers, and other graphing devices
Diagnostic tests
xxiv
xxvi
A Preview of Calculus 1
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Four Ways to Represent a Function 10
Mathematical Models: A Catalog of Essential Functions 23
New Functions from Old Functions 36
The Tangent and Velocity Problems 45
The Limit of a Function 50
Calculating Limits Using the Limit Laws 62
The Precise Definition of a Limit 72
Continuity 82
Review 94
Principles of Problem Solving 98
2
2.1
2.2
2.3
95
2.4
2.5
2.6
Derivatives and Rates of Change 106
Writing Project • Early Methods for Finding Tangents 117
The Derivative as a Function 117
Differentiation Formulas 130
Applied Project • Building a Better Roller Coaster 144
Derivatives of Trigonometric Functions 144
The Chain Rule 152
Applied Project • Where Should a Pilot Start Descent? 161
Implicit Differentiation 161
Laboratory Project • Families of Implicit Curves 168
iii
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iv
Contents
2.7
2.8
2.9
Rates of Change in the Natural and Social Sciences 169
Related Rates 181
Linear Approximations and Differentials 188
Laboratory Project • Taylor Polynomials 194
Review 195
Problems Plus 200
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
Maximum and Minimum Values 204
Applied Project • The Calculus of Rainbows 213
The Mean Value Theorem 215
How Derivatives Affect the Shape of a Graph 221
Limits at Infinity; Horizontal Asymptotes 231
Summary of Curve Sketching 244
Graphing with Calculus and Calculators 251
Optimization Problems 258
Applied Project • The Shape of a Can 270
Applied Project • Planes and Birds: Minimizing Energy 271
Newton’s Method 272
Antiderivatives 278
Review 285
Problems Plus 289
4
4.1
4.2
Areas and Distances 294
The Definite Integral 306
Discovery Project • Area Functions 319
4.3 The Fundamental Theorem of Calculus 320
4.4 Indefinite Integrals and the Net Change Theorem 330
Writing Project • Newton, Leibniz, and the Invention of Calculus 339
4.5 The Substitution Rule 340
Review 348
Problems Plus 352
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Contents
5
5.1
5.2
5.3
5.4
5.5
Areas Between Curves 356
Applied Project • The Gini Index 364
Volumes 366
Volumes by Cylindrical Shells 377
Work 383
Average Value of a Function 389
Applied Project • Calculus and Baseball 392
Review 393
Problems Plus 395
6
6.1
Inverse Functions 400
Instructors may cover either Sections 6.2–6.4 or Sections 6.2*–6.4*. See the Preface.
6.2
Exponential Functions and
Their Derivatives 408
6.2* The Natural Logarithmic
Function 438
6.3
Logarithmic
Functions 421
6.3* The Natural Exponential
Function 447
6.4
Derivatives of Logarithmic
Functions 428
6.4* General Logarithmic and
Exponential Functions 455
6.5
Exponential Growth and Decay 466
Applied Project • Controlling Red Blood Cell Loss During Surgery 473
6.6
Inverse Trigonometric Functions 474
Applied Project • Where to Sit at the Movies 483
6.7
6.8
Hyperbolic Functions 484
Indeterminate Forms and l’Hospital’s Rule 491
Writing Project • The Origins of l’Hospital’s Rule 503
Review 503
Problems Plus 508
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v
vi
Contents
7
7.1
7.2
7.3
7.4
7.5
7.6
Integration by Parts 512
Trigonometric Integrals 519
Trigonometric Substitution 526
Integration of Rational Functions by Partial Fractions 533
Strategy for Integration 543
Integration Using Tables and Computer Algebra Systems 548
Discovery Project • Patterns in Integrals 553
7.7 Approximate Integration 554
7.8 Improper Integrals 567
Review 577
Problems Plus 580
8
8.1
8.2
8.3
8.4
8.5
Arc Length 584
Discovery Project • Arc Length Contest 590
Area of a Surface of Revolution 591
Discovery Project • Rotating on a Slant 597
Applications to Physics and Engineering 598
Discovery Project • Complementary Coffee Cups 608
Applications to Economics and Biology 609
Probability 613
Review 621
Problems Plus 623
9
9.1
9.2
9.3
Modeling with Differential Equations 626
Direction Fields and Euler’s Method 631
Separable Equations 639
Applied Project • How Fast Does a Tank Drain? 648
Applied Project • Which Is Faster, Going Up or Coming Down? 649
9.4 Models for Population Growth 650
9.5 Linear Equations 660
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Contents
9.6
Predator-Prey Systems 667
Review 674
Problems Plus 677
10
10.1 Curves Defined by Parametric Equations 680
Laboratory Project • Running Circles Around Circles 688
10.2 Calculus with Parametric Curves 689
Laboratory Project • Bézier Curves 697
10.3 Polar Coordinates 698
Laboratory Project • Families of Polar Curves 708
10.4 Areas and Lengths in Polar Coordinates 709
10.5 Conic Sections 714
10.6 Conic Sections in Polar Coordinates 722
Review 729
Problems Plus 732
11
11.1 Sequences 734
Laboratory Project • Logistic Sequences 747
11.2 Series 747
11.3 The Integral Test and Estimates of Sums 759
11.4 The Comparison Tests 767
11.5 Alternating Series 772
11.6 Absolute Convergence and the Ratio and Root Tests 777
11.7 Strategy for Testing Series 784
11.8 Power Series 786
11.9 Representations of Functions as Power Series 792
11.10 Taylor and Maclaurin Series 799
Laboratory Project • An Elusive Limit 813
Writing Project • How Newton Discovered the Binomial Series 813
11.11 Applications of Taylor Polynomials 814
Applied Project • Radiation from the Stars 823
Review 824
Problems Plus 827
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vii
viii
Contents
12
12.1 Three-Dimensional Coordinate Systems 832
12.2 Vectors 838
12.3 The Dot Product 847
12.4 The Cross Product 854
Discovery Project • The Geometry of a Tetrahedron 863
12.5 Equations of Lines and Planes 863
Laboratory Project • Putting 3D in Perspective 873
12.6 Cylinders and Quadric Surfaces 874
Review 881
7et1206un03
04/21/10
MasterID: 01462
Problems Plus 884
13
13.1 Vector Functions and Space Curves 888
13.2 Derivatives and Integrals of Vector Functions 895
13.3 Arc Length and Curvature 901
13.4 Motion in Space: Velocity and Acceleration 910
Applied Project • Kepler’s Laws 920
Review 921
Problems Plus 924
14
14.1 Functions of Several Variables 928
14.2 Limits and Continuity 943
14.3 Partial Derivatives 951
14.4 Tangent Planes and Linear Approximations 967
Applied Project • The Speedo LZR Racer 976
14.5 The Chain Rule 977
14.6 Directional Derivatives and the Gradient Vector 986
14.7 Maximum and Minimum Values 999
Applied Project • Designing a Dumpster 1010
Discovery Project • Quadratic Approximations and Critical Points 1010
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Contents
14.8 Lagrange Multipliers 1011
Applied Project • Rocket Science 1019
Applied Project • Hydro-Turbine Optimization 1020
Review 1021
Problems Plus 1025
15
15.1 Double Integrals over Rectangles 1028
15.2 Double Integrals over General Regions 1041
15.3 Double Integrals in Polar Coordinates 1050
15.4 Applications of Double Integrals 1056
15.5 Surface Area 1066
15.6 Triple Integrals 1069
Discovery Project • Volumes of Hyperspheres 1080
15.7 Triple Integrals in Cylindrical Coordinates 1080
Discovery Project • The Intersection of Three Cylinders 1084
15.8 Triple Integrals in Spherical Coordinates 1085
Applied Project • Roller Derby 1092
15.9 Change of Variables in Multiple Integrals 1092
Review 1101
Problems Plus 1105
16
16.1 Vector Fields 1108
16.2 Line Integrals 1115
16.3 The Fundamental Theorem for Line Integrals 1127
16.4 Green’s Theorem 1136
16.5 Curl and Divergence 1143
16.6 Parametric Surfaces and Their Areas 1151
16.7 Surface Integrals 1162
16.8 Stokes’ Theorem 1174
Writing Project • Three Men and Two Theorems 1180
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ix
x
Contents
16.9 The Divergence Theorem 1181
16.10 Summary 1187
Review 1188
Problems Plus 1191
17
17.1
17.2
17.3
17.4
Second-Order Linear Equations 1194
Nonhomogeneous Linear Equations 1200
Applications of Second-Order Differential Equations 1208
Series Solutions 1216
Review 1221
A
B
C
D
E
F
G
H
Numbers, Inequalities, and Absolute Values A2
Coordinate Geometry and Lines A10
Graphs of Second-Degree Equations A16
Trigonometry A24
Sigma Notation A34
Proofs of Theorems A39
Complex Numbers A48
Answers to Odd-Numbered Exercises A57
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Preface
A great discovery solves a great problem but there is a grain of discovery in the
solution of any problem. Your problem may be modest; but if it challenges your
curiosity and brings into play your inventive faculties, and if you solve it by your
own means, you may experience the tension and enjoy the triumph of discovery.
g e o r g e p o lya
The art of teaching, Mark Van Doren said, is the art of assisting discovery. I have tried
to write a book that assists students in discovering calculus—both for its practical power
and its surprising beauty. In this edition, as in the first seven editions, I aim to convey
to the student a sense of the utility of calculus and develop technical competence, but I
also strive to give some appreciation for the intrinsic beauty of the subject. Newton
undoubtedly experienced a sense of triumph when he made his great discoveries. I want
students to share some of that excitement.
The emphasis is on understanding concepts. I think that nearly everybody agrees that
this should be the primary goal of calculus instruction. In fact, the impetus for the current calculus reform movement came from the Tulane Conference in 1986, which formulated as their first recommendation:
Focus on conceptual understanding.
I have tried to implement this goal through the Rule of Three: “Topics should be presented geometrically, numerically, and algebraically.” Visualization, numerical and
graphical experimentation, and other approaches have changed how we teach conceptual reasoning in fundamental ways. More recently, the Rule of Three has been expanded
to become the Rule of Four by emphasizing the verbal, or descriptive, point of view as
well.
In writing the eighth edition my premise has been that it is possible to achieve conceptual understanding and still retain the best traditions of traditional calculus. The book
contains elements of reform, but within the context of a traditional curriculum.
I have written several other calculus textbooks that might be preferable for some instructors. Most of them also come in single variable and multivariable versions.
●
●
Calculus: Early Transcendentals, Eighth Edition, is similar to the present textbook
except that the exponential, logarithmic, and inverse trigonometric functions are
covered in the first semester.
Essential Calculus, Second Edition, is a much briefer book (840 pages), though it
contains almost all of the topics in Calculus, Eighth Edition. The relative brevity is
achieved through briefer exposition of some topics and putting some features on the
website.
xi
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xii
Preface
●
●
●
●
●
●
Essential Calculus: Early Transcendentals, Second Edition, resembles Essential
Calculus, but the exponential, logarithmic, and inverse trigonometric functions are
covered in Chapter 3.
Calculus: Concepts and Contexts, Fourth Edition, emphasizes conceptual understanding even more strongly than this book. The coverage of topics is not encyclopedic and the material on transcendental functions and on parametric equations is
woven throughout the book instead of being treated in separate chapters.
Calculus: Early Vectors introduces vectors and vector functions in the first semester
and integrates them throughout the book. It is suitable for students taking engineering and physics courses concurrently with calculus.
Brief Applied Calculus is intended for students in business, the social sciences, and
the life sciences.
Biocalculus: Calculus for the Life Sciences is intended to show students in the life
sciences how calculus relates to biology.
Biocalculus: Calculus, Probability, and Statistics for the Life Sciences contains all
the content of Biocalculus: Calculus for the Life Sciences as well as three additional chapters covering probability and statistics.
The changes have resulted from talking with my colleagues and students at the University of Toronto and from reading journals, as well as suggestions from users and reviewers. Here are some of the many improvements that I’ve incorporated into this edition:
●
●
●
●
●
The data in examples and exercises have been updated to be more timely.
New examples have been added (see Examples 5.1.5, 11.2.5, and 14.3.3, for
instance). And the solutions to some of the existing examples have been amplified.
Three new projects have been added: The project Planes and Birds: Minimizing
Energy (page 271) asks how birds can minimize power and energy by flapping their
wings versus gliding. The project Controlling Red Blood Cell Loss During Surgery
(page 473) describes the ANH procedure, in which blood is extracted from the
patient before an operation and is replaced by saline solution. This dilutes the
patient’s blood so that fewer red blood cells are lost during bleeding and the
extracted blood is returned to the patient after surgery. In the project The Speedo
LZR Racer (page 976) it is explained that this suit reduces drag in the water and, as
a result, many swimming records were broken. Students are asked why a small
decrease in drag can have a big effect on performance.
I have streamlined Chapter 15 (Multiple Integrals) by combining the first two sections so that iterated integrals are treated earlier.
More than 20% of the exercises in each chapter are new. Here are some of my
favorites: 2.1.61, 2.2.34–36, 3.3.30, 3.3.54, 3.7.39, 3.7.67, 4.1.19–20, 4.2.67–68,
4.4.63, 5.1.51, 6.2.79, 6.7.54, 6.8.90, 8.1.39, 12.5.81, 12.6.29–30, 14.6.65–66.
In addition, there are some good new Problems Plus. (See Problems 10–12 on
page 201, Problem 10 on page 290, Problems 14–15 on pages 353–54, and Problem 8 on page 1026.)
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Preface
xiii
Conceptual Exercises
The most important way to foster conceptual understanding is through the problems
that we assign. To that end I have devised various types of problems. Some exercise sets
begin with requests to explain the meanings of the basic concepts of the section. (See, for
instance, the first few exercises in Sections 1.5, 1.8, 11.2, 14.2, and 14.3.) Similarly, all
the review sections begin with a Concept Check and a True-False Quiz. Other exercises
test conceptual understanding through graphs or tables (see Exercises 2.1.17, 2.2.33–36,
2.2.45–50, 9.1.11–13, 10.1.24–27, 11.10.2, 13.2.1–2, 13.3.33–39, 14.1.1–2, 14.1.32–38,
14.1.41–44, 14.3.3–10, 14.6.1–2, 14.7.3–4, 15.1.6–8, 16.1.11–18, 16.2.17–18, and
16.3.1–2).
Another type of exercise uses verbal description to test conceptual understanding (see
Exercises 1.8.10, 2.2.64, 3.3.57–58, and 7.8.67). I particularly value problems that combine and compare graphical, numerical, and algebraic approaches (see Exercises 2.7.25,
3.4.33–34, and 9.4.4).
Graded Exercise Sets
Each exercise set is carefully graded, progressing from basic conceptual exercises and
skill-development problems to more challenging problems involving applications and
proofs.
Real-World Data
My assistants and I spent a great deal of time looking in libraries, contacting companies
and government agencies, and searching the Internet for interesting real-world data to
introduce, motivate, and illustrate the concepts of calculus. As a result, many of the
examples and exercises deal with functions defined by such numerical data or graphs.
See, for instance, Figure 1 in Section 1.1 (seismograms from the Northridge earthquake),
Exercise 2.2.33 (unemployment rates), Exercise 4.1.16 (velocity of the space shuttle
Endeavour), and Figure 4 in Section 4.4 (San Francisco power consumption). Functions
of two variables are illustrated by a table of values of the wind-chill index as a function
of air temperature and wind speed (Example 14.1.2). Partial derivatives are introduced
in Section 14.3 by examining a column in a table of values of the heat index (perceived
air temperature) as a function of the actual temperature and the relative humidity. This
example is pursued further in connection with linear approximations (Example 14.4.3).
Directional derivatives are introduced in Section 14.6 by using a temperature contour
map to estimate the rate of change of temperature at Reno in the direction of Las Vegas.
Double integrals are used to estimate the average snowfall in Colorado on December
20–21, 2006 (Example 15.1.9). Vector fields are introduced in Section 16.1 by depictions
of actual velocity vector fields showing San Francisco Bay wind patterns.
Projects
One way of involving students and making them active learners is to have them work
(perhaps in groups) on extended projects that give a feeling of substantial accomplishment when completed. I have included four kinds of projects: Applied Projects involve
applications that are designed to appeal to the imagination of students. The project after
Section 9.3 asks whether a ball thrown upward takes longer to reach its maximum height
or to fall back to its original height. (The answer might surprise you.) The project after
Section 14.8 uses Lagrange multipliers to determine the masses of the three stages of
a rocket so as to minimize the total mass while enabling the rocket to reach a desired
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xiv
Preface
velocity. Laboratory Projects involve technology; the one following Section 10.2 shows
how to use Bézier curves to design shapes that represent letters for a laser printer. Writing Projects ask students to compare present-day methods with those of the founders of
calculus—Fermat’s method for finding tangents, for instance. Suggested references are
supplied. Discovery Projects anticipate results to be discussed later or encourage discovery through pattern recognition (see the one following Section 7.6). Others explore
aspects of geometry: tetrahedra (after Section 12.4), hyperspheres (after Section 15.6),
and intersections of three cylinders (after Section 15.7). Additional projects can be found
in the Instructor’s Guide (see, for instance, Group Exercise 4.1: Position from Samples).
Problem Solving
Students usually have difficulties with problems for which there is no single well-defined
procedure for obtaining the answer. I think nobody has improved very much on George
Polya’s four-stage problem-solving strategy and, accordingly, I have included a version
of his problem-solving principles following Chapter 1. They are applied, both explicitly
and implicitly, throughout the book. After the other chapters I have placed sections called
Problems Plus, which feature examples of how to tackle challenging calculus problems.
In selecting the varied problems for these sections I kept in mind the following advice
from David Hilbert: “A mathematical problem should be difficult in order to entice us,
yet not inaccessible lest it mock our efforts.” When I put these challenging problems on
assignments and tests I grade them in a different way. Here I reward a student significantly for ideas toward a solution and for recognizing which problem-solving principles
are relevant.
Dual Treatment of Exponential and Logarithmic Functions
There are two possible ways of treating the exponential and logarithmic functions and
each method has its passionate advocates. Because one often finds advocates of both
approaches teaching the same course, I include full treatments of both methods. In Sections 6.2, 6.3, and 6.4 the exponential function is defined first, followed by the logarithmic function as its inverse. (Students have seen these functions introduced this way since
high school.) In the alternative approach, presented in Sections 6.2*, 6.3*, and 6.4*, the
logarithm is defined as an integral and the exponential function is its inverse. This latter
method is, of course, less intuitive but more elegant. You can use whichever treatment
you prefer.
If the first approach is taken, then much of Chapter 6 can be covered before Chapters 4 and 5, if desired. To accommodate this choice of presentation there are specially
identified problems involving integrals of exponential and logarithmic functions at the
end of the appropriate sections of Chapters 4 and 5. This order of presentation allows a
faster-paced course to teach the transcendental functions and the definite integral in the
first semester of the course.
For instructors who would like to go even further in this direction I have prepared an
alternate edition of this book, called Calculus: Early Transcendentals, Eighth Edition,
in which the exponential and logarithmic functions are introduced in the first chapter.
Their limits and derivatives are found in the second and third chapters at the same time
as polynomials and the other elementary functions.
Tools for Enriching Calculus
TEC is a companion to the text and is intended to enrich and complement its contents.
(It is now accessible in the eBook via CourseMate and Enhanced WebAssign. Selected
Visuals and Modules are available at www.stewartcalculus.com.) Developed by Harvey
Keynes, Dan Clegg, Hubert Hohn, and myself, TEC uses a discovery and exploratory
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Preface
xv
approach. In sections of the book where technology is particularly appropriate, marginal
icons direct students to TEC Modules that provide a laboratory environment in which
they can explore the topic in different ways and at different levels. Visuals are animations of figures in text; Modules are more elaborate activities and include exercises.
Instructors can choose to become involved at several different levels, ranging from simply encouraging students to use the Visuals and Modules for independent exploration,
to assigning specific exercises from those included with each Module, or to creating
additional exercises, labs, and projects that make use of the Visuals and Modules.
TEC also includes Homework Hints for representative exercises (usually odd-numbered) in every section of the text, indicated by printing the exercise number in red.
These hints are usually presented in the form of questions and try to imitate an effective
teaching assistant by functioning as a silent tutor. They are constructed so as not to reveal
any more of the actual solution than is minimally necessary to make further progress.
Enhanced WebAssign
Technology is having an impact on the way homework is assigned to students, particularly in large classes. The use of online homework is growing and its appeal depends
on ease of use, grading precision, and reliability. With the Eighth Edition we have been
working with the calculus community and WebAssign to develop an online homework
system. Up to 70% of the exercises in each section are assignable as online homework,
including free response, multiple choice, and multi-part formats.
The system also includes Active Examples, in which students are guided in step-bystep tutorials through text examples, with links to the textbook and to video solutions.
Website
Visit CengageBrain.com or stewartcalculus.com for these additional materials:
●
Homework Hints
●
Algebra Review
●
Lies My Calculator and Computer Told Me
●
History of Mathematics, with links to the better historical websites
●
●
Additional Topics (complete with exercise sets): Fourier Series, Formulas for the
Remainder Term in Taylor Series, Rotation of Axes
Archived Problems (drill exercises that appeared in previous editions, together with
their solutions)
●
Challenge Problems (some from the Problems Plus sections from prior editions)
●
Links, for particular topics, to outside Web resources
●
Selected Visuals and Modules from Tools for Enriching Calculus (TEC)
Diagnostic Tests
The book begins with four diagnostic tests, in Basic Algebra, Analytic Geometry, Functions, and Trigonometry.
A Preview of Calculus
This is an overview of the subject and includes a list of questions to motivate the study
of calculus.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xvi
Preface
1 Functions and Limits
From the beginning, multiple representations of functions are stressed: verbal, numerical, visual, and algebraic. A discussion of mathematical models leads to a review of the
standard functions from these four points of view. The material on limits is motivated
by a prior discussion of the tangent and velocity problems. Limits are treated from
descriptive, graphical, numerical, and algebraic points of view. Section 1.7, on the precise
epsilon-delta defintion of a limit, is an optional section.
2 Derivatives
The material on derivatives is covered in two sections in order to give students more
time to get used to the idea of a derivative as a function. The examples and exercises
explore the meanings of derivatives in various contexts. Higher derivatives are introduced in Section 2.2.
3 Applications of Differentiation
The basic facts concerning extreme values and shapes of curves are deduced from the
Mean Value Theorem. Graphing with technology emphasizes the interaction between
calculus and calculators and the analysis of families of curves. Some substantial optimization problems are provided, including an explanation of why you need to raise your
head 42° to see the top of a rainbow.
4 Integrals
The area problem and the distance problem serve to motivate the definite integral, with
sigma notation introduced as needed. (Full coverage of sigma notation is provided in
Appendix E.) Emphasis is placed on explaining the meanings of integrals in various contexts and on estimating their values from graphs and tables.
5 Applications of Integration
Here I present the applications of integration—area, volume, work, average value—that
can reasonably be done without specialized techniques of integration. General methods
are emphasized. The goal is for students to be able to divide a quantity into small pieces,
estimate with Riemann sums, and recognize the limit as an integral.
6 Inverse Functions:
As discussed more fully on page xiv, only one of the two treatments of these functions
need be covered. Exponential growth and decay are covered in this chapter.
7 Techniques of Integration
All the standard methods are covered but, of course, the real challenge is to be able to
recognize which technique is best used in a given situation. Accordingly, in Section 7.5,
I present a strategy for integration. The use of computer algebra systems is discussed in
Section 7.6.
8 Further Applications
of Integration
Here are the applications of integration—arc length and surface area—for which it is
useful to have available all the techniques of integration, as well as applications to biology, economics, and physics (hydrostatic force and centers of mass). I have also
included a section on probability. There are more applications here than can realistically
be covered in a given course. Instructors should select applications suitable for their
students and for which they themselves have enthusiasm.
9 Differential Equations
Modeling is the theme that unifies this introductory treatment of differential equations.
Direction fields and Euler’s method are studied before separable and linear equations are
solved explicitly, so that qualitative, numerical, and analytic approaches are given equal
Exponential, Logarithmic, and
Inverse Trigonometric Functions
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Preface
xvii
consideration. These methods are applied to the exponential, logistic, and other models
for population growth. The first four or five sections of this chapter serve as a good
introduction to first-order differential equations. An optional final section uses predatorprey models to illustrate systems of differential equations.
10 Parametric Equations
and Polar Coordinates
This chapter introduces parametric and polar curves and applies the methods of calculus
to them. Parametric curves are well suited to laboratory projects; the two presented here
involve families of curves and Bézier curves. A brief treatment of conic sections in polar
coordinates prepares the way for Kepler’s Laws in Chapter 13.
11 Infinite Sequences and Series
The convergence tests have intuitive justifications (see page 759) as well as formal
proofs. Numerical estimates of sums of series are based on which test was used to prove
convergence. The emphasis is on Taylor series and polynomials and their applications
to physics. Error estimates include those from graphing devices.
12 Vectors and the
Geometry of Space
The material on three-dimensional analytic geometry and vectors is divided into two
chapters. Chapter 12 deals with vectors, the dot and cross products, lines, planes, and
surfaces.
13 Vector Functions
This chapter covers vector-valued functions, their derivatives and integrals, the length
and curvature of space curves, and velocity and acceleration along space curves, culminating in Kepler’s laws.
14 Partial Derivatives
Functions of two or more variables are studied from verbal, numerical, visual, and algebraic points of view. In particular, I introduce partial derivatives by looking at a specific
column in a table of values of the heat index (perceived air temperature) as a function
of the actual temperature and the relative humidity.
15 Multiple Integrals
Contour maps and the Midpoint Rule are used to estimate the average snowfall and
average temperature in given regions. Double and triple integrals are used to compute
probabilities, surface areas, and (in projects) volumes of hyperspheres and volumes of
intersections of three cylinders. Cylindrical and spherical coordinates are introduced in
the context of evaluating triple integrals.
16 Vector Calculus
Vector fields are introduced through pictures of velocity fields showing San Francisco
Bay wind patterns. The similarities among the Fundamental Theorem for line integrals,
Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem are emphasized.
17 Second-Order
Differential Equations
Since first-order differential equations are covered in Chapter 9, this final chapter deals
with second-order linear differential equations, their application to vibrating springs and
electric circuits, and series solutions.
Calculus, Eighth Edition, is supported by a complete set of ancillaries developed
under my direction. Each piece has been designed to enhance student understanding
and to facilitate creative instruction. The tables on pages xxi–xxii describe each of these
ancillaries.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xviii
Preface
The preparation of this and previous editions has involved much time spent reading the
reasoned (but sometimes contradictory) advice from a large number of astute reviewers.
I greatly appreciate the time they spent to understand my motivation for the approach
taken. I have learned something from each of them.
Eighth Edition Reviewers
Jay Abramson, Arizona State University
Adam Bowers, University of California San Diego
Neena Chopra, The Pennsylvania State University
Edward Dobson, Mississippi State University
Isaac Goldbring, University of Illinois at Chicago
Lea Jenkins, Clemson University
Rebecca Wahl, Butler University
Technology Reviewers
Maria Andersen, Muskegon Community College
Eric Aurand, Eastfield College
Joy Becker, University of Wisconsin–Stout
Przemyslaw Bogacki, Old Dominion University
Amy Elizabeth Bowman, University of Alabama
in Huntsville
Monica Brown, University of Missouri–St. Louis
Roxanne Byrne, University of Colorado at Denver and
Health Sciences Center
Teri Christiansen, University of Missouri–Columbia
Bobby Dale Daniel, Lamar University
Jennifer Daniel, Lamar University
Andras Domokos, California State University, Sacramento
Timothy Flaherty, Carnegie Mellon University
Lee Gibson, University of Louisville
Jane Golden, Hillsborough Community College
Semion Gutman, University of Oklahoma
Diane Hoffoss, University of San Diego
Lorraine Hughes, Mississippi State University
Jay Jahangiri, Kent State University
John Jernigan, Community College of Philadelphia
Brian Karasek, South Mountain Community College
Jason Kozinski, University of Florida
Carole Krueger, The University of Texas at Arlington
Ken Kubota, University of Kentucky
John Mitchell, Clark College
Donald Paul, Tulsa Community College
Chad Pierson, University of Minnesota, Duluth
Lanita Presson, University of Alabama in Huntsville
Karin Reinhold, State University of New York at Albany
Thomas Riedel, University of Louisville
Christopher Schroeder, Morehead State University
Angela Sharp, University of Minnesota, Duluth
Patricia Shaw, Mississippi State University
Carl Spitznagel, John Carroll University
Mohammad Tabanjeh, Virginia State University
Capt. Koichi Takagi, United States Naval Academy
Lorna TenEyck, Chemeketa Community College
Roger Werbylo, Pima Community College
David Williams, Clayton State University
Zhuan Ye, Northern Illinois University
Previous Edition Reviewers
B. D. Aggarwala, University of Calgary
John Alberghini, Manchester Community College
Michael Albert, Carnegie-Mellon University
Daniel Anderson, University of Iowa
Amy Austin, Texas A&M University
Donna J. Bailey, Northeast Missouri State University
Wayne Barber, Chemeketa Community College
Marilyn Belkin, Villanova University
Neil Berger, University of Illinois, Chicago
David Berman, University of New Orleans
Anthony J. Bevelacqua, University of North Dakota
Richard Biggs, University of Western Ontario
Robert Blumenthal, Oglethorpe University
Martina Bode, Northwestern University
Barbara Bohannon, Hofstra University
Jay Bourland, Colorado State University
Philip L. Bowers, Florida State University
Amy Elizabeth Bowman, University of Alabama in Huntsville
Stephen W. Brady, Wichita State University
Michael Breen, Tennessee Technological University
Robert N. Bryan, University of Western Ontario
David Buchthal, University of Akron
Jenna Carpenter, Louisiana Tech University
Jorge Cassio, Miami-Dade Community College
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Preface
Jack Ceder, University of California, Santa Barbara
Scott Chapman, Trinity University
Zhen-Qing Chen, University of Washington—Seattle
James Choike, Oklahoma State University
Barbara Cortzen, DePaul University
Carl Cowen, Purdue University
Philip S. Crooke, Vanderbilt University
Charles N. Curtis, Missouri Southern State College
Daniel Cyphert, Armstrong State College
Robert Dahlin
M. Hilary Davies, University of Alaska Anchorage
Gregory J. Davis, University of Wisconsin–Green Bay
Elias Deeba, University of Houston–Downtown
Daniel DiMaria, Suffolk Community College
Seymour Ditor, University of Western Ontario
Greg Dresden, Washington and Lee University
Daniel Drucker, Wayne State University
Kenn Dunn, Dalhousie University
Dennis Dunninger, Michigan State University
Bruce Edwards, University of Florida
David Ellis, San Francisco State University
John Ellison, Grove City College
Martin Erickson, Truman State University
Garret Etgen, University of Houston
Theodore G. Faticoni, Fordham University
Laurene V. Fausett, Georgia Southern University
Norman Feldman, Sonoma State University
Le Baron O. Ferguson, University of California—Riverside
Newman Fisher, San Francisco State University
José D. Flores, The University of South Dakota
William Francis, Michigan Technological University
James T. Franklin, Valencia Community College, East
Stanley Friedlander, Bronx Community College
Patrick Gallagher, Columbia University–New York
Paul Garrett, University of Minnesota–Minneapolis
Frederick Gass, Miami University of Ohio
Bruce Gilligan, University of Regina
Matthias K. Gobbert, University of Maryland, Baltimore County
Gerald Goff, Oklahoma State University
Stuart Goldenberg, California Polytechnic State University
John A. Graham, Buckingham Browne & Nichols School
Richard Grassl, University of New Mexico
Michael Gregory, University of North Dakota
Charles Groetsch, University of Cincinnati
Paul Triantafilos Hadavas, Armstrong Atlantic State University
Salim M. Haïdar, Grand Valley State University
D. W. Hall, Michigan State University
Robert L. Hall, University of Wisconsin–Milwaukee
Howard B. Hamilton, California State University, Sacramento
Darel Hardy, Colorado State University
Shari Harris, John Wood Community College
Gary W. Harrison, College of Charleston
Melvin Hausner, New York University/Courant Institute
Curtis Herink, Mercer University
Russell Herman, University of North Carolina at Wilmington
Allen Hesse, Rochester Community College
xix
Randall R. Holmes, Auburn University
James F. Hurley, University of Connecticut
Amer Iqbal, University of Washington—Seattle
Matthew A. Isom, Arizona State University
Gerald Janusz, University of Illinois at Urbana-Champaign
John H. Jenkins, Embry-Riddle Aeronautical University,
Prescott Campus
Clement Jeske, University of Wisconsin, Platteville
Carl Jockusch, University of Illinois at Urbana-Champaign
Jan E. H. Johansson, University of Vermont
Jerry Johnson, Oklahoma State University
Zsuzsanna M. Kadas, St. Michael’s College
Nets Katz, Indiana University Bloomington
Matt Kaufman
Matthias Kawski, Arizona State University
Frederick W. Keene, Pasadena City College
Robert L. Kelley, University of Miami
Akhtar Khan, Rochester Institute of Technology
Marianne Korten, Kansas State University
Virgil Kowalik, Texas A&I University
Kevin Kreider, University of Akron
Leonard Krop, DePaul University
Mark Krusemeyer, Carleton College
John C. Lawlor, University of Vermont
Christopher C. Leary, State University of New York at Geneseo
David Leeming, University of Victoria
Sam Lesseig, Northeast Missouri State University
Phil Locke, University of Maine
Joyce Longman, Villanova University
Joan McCarter, Arizona State University
Phil McCartney, Northern Kentucky University
Igor Malyshev, San Jose State University
Larry Mansfield, Queens College
Mary Martin, Colgate University
Nathaniel F. G. Martin, University of Virginia
Gerald Y. Matsumoto, American River College
James McKinney, California State Polytechnic University, Pomona
Tom Metzger, University of Pittsburgh
Richard Millspaugh, University of North Dakota
Lon H. Mitchell, Virginia Commonwealth University
Michael Montaño, Riverside Community College
Teri Jo Murphy, University of Oklahoma
Martin Nakashima, California State Polytechnic University,
Pomona
Ho Kuen Ng, San Jose State University
Richard Nowakowski, Dalhousie University
Hussain S. Nur, California State University, Fresno
Norma Ortiz-Robinson, Virginia Commonwealth University
Wayne N. Palmer, Utica College
Vincent Panico, University of the Pacific
F. J. Papp, University of Michigan–Dearborn
Mike Penna, Indiana University–Purdue University Indianapolis
Mark Pinsky, Northwestern University
Lothar Redlin, The Pennsylvania State University
Joel W. Robbin, University of Wisconsin–Madison
Lila Roberts, Georgia College and State University
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xx
Preface
E. Arthur Robinson, Jr., The George Washington University
Richard Rockwell, Pacific Union College
Rob Root, Lafayette College
Richard Ruedemann, Arizona State University
David Ryeburn, Simon Fraser University
Richard St. Andre, Central Michigan University
Ricardo Salinas, San Antonio College
Robert Schmidt, South Dakota State University
Eric Schreiner, Western Michigan University
Mihr J. Shah, Kent State University–Trumbull
Qin Sheng, Baylor University
Theodore Shifrin, University of Georgia
Wayne Skrapek, University of Saskatchewan
Larry Small, Los Angeles Pierce College
Teresa Morgan Smith, Blinn College
William Smith, University of North Carolina
Donald W. Solomon, University of Wisconsin–Milwaukee
Edward Spitznagel, Washington University
Joseph Stampfli, Indiana University
Kristin Stoley, Blinn College
M. B. Tavakoli, Chaffey College
Magdalena Toda, Texas Tech University
Ruth Trygstad, Salt Lake Community College
Paul Xavier Uhlig, St. Mary’s University, San Antonio
Stan Ver Nooy, University of Oregon
Andrei Verona, California State University–Los Angeles
Klaus Volpert, Villanova University
Russell C. Walker, Carnegie Mellon University
William L. Walton, McCallie School
Peiyong Wang, Wayne State University
Jack Weiner, University of Guelph
Alan Weinstein, University of California, Berkeley
Theodore W. Wilcox, Rochester Institute of Technology
Steven Willard, University of Alberta
Robert Wilson, University of Wisconsin–Madison
Jerome Wolbert, University of Michigan–Ann Arbor
Dennis H. Wortman, University of Massachusetts, Boston
Mary Wright, Southern Illinois University–Carbondale
Paul M. Wright, Austin Community College
Xian Wu, University of South Carolina
In addition, I would like to thank R. B. Burckel, Bruce Colletti, David Behrman, John
Dersch, Gove Effinger, Bill Emerson, Dan Kalman, Quyan Khan, Alfonso Gracia-Saz,
Allan MacIsaac, Tami Martin, Monica Nitsche, Lamia Raffo, Norton Starr, and Jim Trefzger for their suggestions; Al Shenk and Dennis Zill for permission to use exercises from
their calculus texts; COMAP for permission to use project material; George Bergman,
David Bleecker, Dan Clegg, Victor Kaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz, Paul Sally, Lowell Smylie, and Larry Wallen for ideas for exercises; Dan Drucker
for the roller derby project; Thomas Banchoff, Tom Farmer, Fred Gass, John Ramsay,
Larry Riddle, Philip Straffin, and Klaus Volpert for ideas for projects; Dan Anderson,
Dan Clegg, Jeff Cole, Dan Drucker, and Barbara Frank for solving the new exercises
and suggesting ways to improve them; Marv Riedesel and Mary Johnson for accuracy in
proofreading; Andy Bulman-Fleming, Lothar Redlin, Gina Sanders, and Saleem Watson
for additional proofreading; and Jeff Cole and Dan Clegg for their careful preparation
and proofreading of the answer manuscript.
In addition, I thank those who have contributed to past editions: Ed Barbeau, Jordan Bell, George Bergman, Fred Brauer, Andy Bulman-Fleming, Bob Burton, David
Cusick, Tom DiCiccio, Garret Etgen, Chris Fisher, Leon Gerber, Stuart Goldenberg,
Arnold Good, Gene Hecht, Harvey Keynes, E. L. Koh, Zdislav Kovarik, Kevin Kreider,
Emile LeBlanc, David Leep, Gerald Leibowitz, Larry Peterson, Mary Pugh, Lothar Redlin, Carl Riehm, John Ringland, Peter Rosenthal, Dusty Sabo, Doug Shaw, Dan Silver,
Simon Smith, Norton Starr, Saleem Watson, Alan Weinstein, and Gail Wolkowicz.
I also thank Kathi Townes, Stephanie Kuhns, Kristina Elliott, and Kira Abdallah of
TECHarts for their production services and the following Cengage Learning staff:
Cheryll Linthicum, content project manager; Stacy Green, senior content developer;
Samantha Lugtu, associate content developer; Stephanie Kreuz, product assistant; Lynh
Pham, media developer; Ryan Ahern, marketing manager; and Vernon Boes, art director.
They have all done an outstanding job.
I have been very fortunate to have worked with some of the best mathematics editors
in the business over the past three decades: Ron Munro, Harry Campbell, Craig Barth,
Jeremy Hayhurst, Gary Ostedt, Bob Pirtle, Richard Stratton, Liz Covello, and now Neha
Taleja. All of them have contributed greatly to the success of this book.
james stewart
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Instructor’s Guide
by Douglas Shaw
ISBN 978-1-305-27178-4
Each section of the text is discussed from several viewpoints.
The Instructor’s Guide contains suggested time to allot, points
to stress, text discussion topics, core materials for lecture,
workshop/discussion suggestions, group work exercises in
a form suitable for handout, and suggested homework
assignments.
Complete Solutions Manual
Single Variable
By Daniel Anderson, Jeffery A. Cole, and Daniel Drucker
ISBN 978-1-305-27610-9
Multivariable
By Dan Clegg and Barbara Frank
ISBN 978-1-305-27611-6
Includes worked-out solutions to all exercises in the text.
Printed Test Bank
By William Steven Harmon
ISBN 978-1-305-27180-7
Contains text-specific multiple-choice and free response test
items.
Cengage Learning Testing Powered by Cognero
(login.cengage.com)
This flexible online system allows you to author, edit, and
manage test bank content from multiple Cengage Learning
solutions; create multiple test versions in an instant; and
deliver tests from your LMS, your classroom, or wherever you
want.
TEC TOOLS FOR ENRICHING™ CALCULUS
By James Stewart, Harvey Keynes, Dan Clegg, and developer
Hubert Hohn
Tools for Enriching Calculus (TEC) functions as both a
powerful tool for instructors and as a tutorial environment
in which students can explore and review selected topics. The
Flash simulation modules in TEC include instructions, written
and audio explanations of the concepts, and exercises. TEC
is accessible in the eBook via CourseMate and Enhanced
WebAssign. Selected Visuals and Modules are available at
www.stewartcalculus.com.
Enhanced WebAssign®
www.webassign.net
Printed Access Code: ISBN 978-1-285-85826-5
Instant Access Code ISBN: 978-1-285-85825-8
Exclusively from Cengage Learning, Enhanced WebAssign
offers an extensive online program for Stewart’s Calculus
to encourage the practice that is so critical for concept
mastery. The meticulously crafted pedagogy and exercises
in our proven texts become even more effective in Enhanced
WebAssign, supplemented by multimedia tutorial support and
immediate feedback as students complete their assignments.
Key features include:
n T
housands of homework problems that match your textbook’s end-of-section exercises
Opportunities for students to review prerequisite skills and
content both at the start of the course and at the beginning
of each section
n
Read It eBook pages, Watch It videos, Master It tutorials,
and Chat About It links
n
A customizable Cengage YouBook with highlighting, notetaking, and search features, as well as links to multimedia
resources
n
Personal Study Plans (based on diagnostic quizzing) that
identify chapter topics that students will need to master
n
A WebAssign Answer Evaluator that recognizes and accepts
equivalent mathematical responses in the same way an
instructor grades
n
Stewart Website
www.stewartcalculus.com
Contents: Homework Hints n Algebra Review n Additional
Topics n Drill exercises n Challenge Problems n Web
Links n History of Mathematics n Tools for Enriching
Calculus (TEC)
■ Electronic items ■ Printed items
A Show My Work feature that gives instructors the option
of seeing students’ detailed solutions
n
Visualizing Calculus Animations, Lecture Videos, and more
n
(Table continues on page xxii)
xxi
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Cengage Customizable YouBook
YouBook is an eBook that is both interactive and customizable. Containing all the content from Stewart’s Calculus,
YouBook features a text edit tool that allows instructors to
modify the textbook narrative as needed. With YouBook,
instructors can quickly reorder entire sections and chapters
or hide any content they don’t teach to create an eBook that
perfectly matches their syllabus. Instructors can further
customize the text by adding instructor-created or YouTube
video links. Additional media assets include animated figures,
video clips, highlighting and note-taking features, and more.
YouBook is available within Enhanced WebAssign.
CourseMate
CourseMate is a perfect self-study tool for students, and
requires no set up from instructors. CourseMate brings course
concepts to life with interactive learning, study, and exam
preparation tools that support the printed textbook. CourseMate for Stewart’s Calculus includes an interactive eBook,
Tools for Enriching Calculus, videos, quizzes, flashcards,
and more. For instructors, CourseMate includes Engagement
Tracker, a first-of-its-kind tool that monitors student
engagement.
CengageBrain.com
To access additional course materials, please visit
www.cengagebrain.com. At the CengageBrain.com home
page, search for the ISBN of your title (from the back cover of
your book) using the search box at the top of the page. This
will take you to the product page where these resources can
be found.
Student Solutions Manual
Single Variable
By Daniel Anderson, Jeffery A. Cole, and Daniel Drucker
ISBN 978-1-305-27181-4
Multivariable
By Dan Clegg and Barbara Frank
ISBN 978-1-305-27182-1
Provides completely worked-out solutions to all oddnumbered exercises in the text, giving students a chance to
■ Electronic items
check their answer and ensure they took the correct steps
to arrive at the answer. The Student Solutions Manual
can be ordered or accessed online as an eBook at
www.cengagebrain.com by searching the ISBN.
Study Guide
Single Variable
By Richard St. Andre
ISBN 978-1-305-27913-1
Multivariable
By Richard St. Andre
ISBN 978-1-305-27184-5
For each section of the text, the Study Guide provides students
with a brief introduction, a short list of concepts to master,
and summary and focus questions with explained answers.
The Study Guide also contains self-tests with exam-style
questions. The Study Guide can be ordered or accessed online
as an eBook at www.cengagebrain.com by searching the
ISBN.
A Companion to Calculus
By Dennis Ebersole, Doris Schattschneider, Alicia Sevilla,
and Kay Somers
ISBN 978-0-495-01124-8
Written to improve algebra and problem-solving skills of
students taking a calculus course, every chapter in this
companion is keyed to a calculus topic, providing conceptual background and specific algebra techniques needed to
understand and solve calculus problems related to that topic.
It is designed for calculus courses that integrate the review of
precalculus concepts or for individual use. Order a copy of
the text or access the eBook online at www.cengagebrain.com
by searching the ISBN.
Linear Algebra for Calculus
by Konrad J. Heuvers, William P. Francis, John H. Kuisti,
Deborah F. Lockhart, Daniel S. Moak, and Gene M. Ortner
ISBN 978-0-534-25248-9
This comprehensive book, designed to supplement the calculus course, provides an introduction to and review of the basic
ideas of linear algebra. Order a copy of the text or access
the eBook online at www.cengagebrain.com by searching the
ISBN.
■ Printed items
xxii
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
To the Student
Reading a calculus textbook is different from reading a
newspaper or a novel, or even a physics book. Don’t be discouraged if you have to read a passage more than once
in order to understand it. You should have pencil and paper
and calculator at hand to sketch a diagram or make a
calculation.
Some students start by trying their homework problems
and read the text only if they get stuck on an exercise. I suggest that a far better plan is to read and understand a section
of the text before attempting the exercises. In particular, you
should look at the definitions to see the exact meanings of
the terms. And before you read each example, I suggest that
you cover up the solution and try solving the problem yourself. You’ll get a lot more from looking at the solution if
you do so.
Part of the aim of this course is to train you to think logically. Learn to write the solutions of the exercises in a connected, step-by-step fashion with explanatory sentences—
not just a string of disconnected equations or formulas.
The answers to the odd-numbered exercises appear at the
back of the book, in Appendix H. Some exercises ask for a
verbal explanation or interpretation or description. In such
cases there is no single correct way of expressing the
answer, so don’t worry that you haven’t found the definitive
answer. In addition, there are often several different forms
in which to express a numerical or algebraic answer, so if
your answer differs from mine, don’t immediately assume
you’re wrong. For example, if the answer given in the back
of the book is s2 2 1 and you obtain 1y (1 1 s2 ), then
you’re right and rationalizing the denominator will show
that the answers are equivalent.
The icon ; indicates an exercise that definitely requires
the use of either a graphing calculator or a computer with
graphing software. But that doesn’t mean that graphing
devices can’t be used to check your work on the other exercises as well. The symbol CAS is reserved for problems in
which the full resources of a computer algebra system (like
Maple, Mathematica, or the TI-89) are required.
You will also encounter the symbol |, which warns you
against committing an error. I have placed this symbol in
the margin in situations where I have observed that a large
proportion of my students tend to make the same mistake.
Tools for Enriching Calculus, which is a companion to
this text, is referred to by means of the symbol TEC and can
be accessed in the eBook via Enhanced WebAssign and
CourseMate (selected Visuals and Modules are available at
www.stewartcalculus.com). It directs you to modules in
which you can explore aspects of calculus for which the
computer is particularly useful.
You will notice that some exercise numbers are printed
in red: 5. This indicates that Homework Hints are available
for the exercise. These hints can be found on stewartcalculus.com as well as Enhanced WebAssign and CourseMate.
The homework hints ask you questions that allow you to
make progress toward a solution without actually giving
you the answer. You need to pursue each hint in an active
manner with pencil and paper to work out the details. If a
particular hint doesn’t enable you to solve the problem, you
can click to reveal the next hint.
I recommend that you keep this book for reference purposes after you finish the course. Because you will likely
forget some of the specific details of calculus, the book will
serve as a useful reminder when you need to use calculus in
subsequent courses. And, because this book contains more
material than can be covered in any one course, it can also
serve as a valuable resource for a working scientist or
engineer.
Calculus is an exciting subject, justly considered to be
one of the greatest achievements of the human intellect. I
hope you will discover that it is not only useful but also
intrinsically beautiful.
james stewart
xxiii
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Calculators, Computers, and
Other Graphing Devices
xxiv
© Dan Clegg
You can also use computer software such
as Graphing Calculator by Pacific Tech
(www.pacifict.com) to perform many of these
functions, as well as apps for phones and
tablets, like Quick Graph (Colombiamug) or
Math-Studio (Pomegranate Apps). Similar
functionality is available using a web interface
at WolframAlpha.com.
© Dan Clegg
© Dan Clegg
Advances in technology continue to bring a wider variety of tools for
doing mathematics. Handheld calculators are becoming more powerful, as are software programs and Internet resources. In addition,
many mathematical applications have been released for smartphones
and tablets such as the iPad.
Some exercises in this text are marked with a graphing icon ; ,
which indicates that the use of some technology is required. Often this
means that we intend for a graphing device to be used in drawing the
graph of a function or equation. You might also need technology to
find the zeros of a graph or the points of intersection of two graphs.
In some cases we will use a calculating device to solve an equation or
evaluate a definite integral numerically. Many scientific and graphing
calculators have these features built in, such as the Texas Instruments
TI-84 or TI-Nspire CX. Similar calculators are made by Hewlett Packard, Casio, and Sharp.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
The CAS icon is reserved for problems in which the full resources of
a computer algebra system (CAS) are required. A CAS is capable of
doing mathematics (like solving equations, computing derivatives or
integrals) symbolically rather than just numerically.
Examples of well-established computer algebra systems are the computer software packages Maple and Mathematica. The WolframAlpha
website uses the Mathematica engine to provide CAS functionality
via the Web.
Many handheld graphing calculators have CAS capabilities, such
as the TI-89 and TI-Nspire CX CAS from Texas Instruments. Some
tablet and smartphone apps also provide these capabilities, such as the
previously mentioned MathStudio.
© Dan Clegg
© Dan Clegg
© Dan Clegg
In general, when we use the term “calculator” in this book, we mean
the use of any of the resources we have mentioned.
xxv
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Diagnostic Tests
Success in calculus depends to a large extent on knowledge of the mathematics that
precedes calculus: algebra, analytic geometry, functions, and trigonometry. The following tests are intended to diagnose weaknesses that you might have in these areas.
After taking each test you can check your answers against the given answers and, if
necessary, refresh your skills by referring to the review materials that are provided.
A
1.Evaluate each expression without using a calculator.
(a) s23d4
(b) 234
(c) 324
(d)
5 23
5 21
(e)
SD
2
3
22
(f) 16 23y4
2.
Simplify each expression. Write your answer without negative exponents.
(a) s200 2 s32
(b) s3a 3b 3 ds4ab 2 d 2
(c)
S
3x 3y2 y 3
x 2 y21y2
D
22
3.Expand and simplify.
(a) 3sx 1 6d 1 4s2x 2 5d
(b) sx 1 3ds4x 2 5d
(c) ssa 1 sb dssa 2 sb d
(d) s2x 1 3d2
(e) sx 1 2d3
4.Factor each expression.
(a) 4x 2 2 25
(c) x 3 2 3x 2 2 4x 1 12
(e) 3x 3y2 2 9x 1y2 1 6x 21y2
(b) 2x 2 1 5x 2 12
(d) x 4 1 27x
(f) x 3 y 2 4xy
5. Simplify the rational expression.
(a)
x 2 1 3x 1 2
x2 2 x 2 2
(c)
x2
x11
2
x 24
x12
2
2x 2 2 x 2 1
x13
?
x2 2 9
2x 1 1
y
x
2
x
y
(d)
1
1
2
y
x
(b)
xxvi
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Diagnostic Tests
6.Rationalize the expression and simplify.
s10
s5 2 2
(a)
(b)
7.Rewrite by completing the square.
(a) x 2 1 x 1 1
s4 1 h 2 2
h
(b) 2x 2 2 12x 1 11
8.Solve the equation. (Find only the real solutions.)
(c) x 2 2 x 2 12 − 0
2x
2x 2 1
−
x11
x
(d) 2x 2 1 4x 1 1 − 0
(e) x 4 2 3x 2 1 2 − 0
(f) 3 x 2 4 − 10
(a) x 1 5 − 14 2 12 x
(b)
|
(g) 2xs4 2 xd21y2 2 3 s4 2 x − 0
|
9.
Solve each inequality. Write your answer using interval notation.
(a) 24 , 5 2 3x < 17
(b) x 2 , 2x 1 8
(c) xsx 2 1dsx 1 2d . 0
(d) x 2 4 , 3
|
2x 2 3
(e)
<1
x11
10. State whether each equation is true or false.
(a) s p 1 qd2 − p 2 1 q 2
(c) sa 2 1 b 2 − a 1 b
1
1
1
− 2
x2y
x
y
(e)
|
(b) sab − sa sb
1 1 TC
(d)
−11T
C
1yx
1
(f)
−
ayx 2 byx
a2b
answers to diagnostic test a: algebra
1.(a) 81
(d) 25
2.(a) 6s2
(b) 281
(c)
9
4
(f)
(e)
(b) 48a 5b7
(c)
1
81
1
8
x
9y7
3.(a) 11x 2 2
(b) 4x 2 1 7x 2 15
(c) a 2 b
(d) 4x 2 1 12x 1 9
(e) x 3 1 6x 2 1 12x 1 8
4.(a) s2x 2 5ds2x 1 5d
(c) sx 2 3dsx 2 2dsx 1 2d
(e) 3x21y2sx 2 1dsx 2 2d
x12
x22
1
(c)
x22
5.(a)
(b) s2x 2 3dsx 1 4d
(d) xsx 1 3dsx 2 2 3x 1 9d
(f) xysx 2 2dsx 1 2d
(b)
x21
x23
(d) 2sx 1 yd
1
6.(a) 5s2 1 2s10
(b)
7.(a) s x 1 12 d 1 34
(b)­
2sx 2 3d2 2 7
2
8.(a) 6
(d) 21 6 12 s2
(g)
s4 1 h 1 2
(b) 1
(c) 23, 4
(e) 61, 6s2
(f) 23 , 22
3
12
5
9.(a) f24, 3d
(c) s22, 0d ø s1, `d
(e) s21, 4g
10.(a) False
(d) False
(b) s22, 4d
(d) s1, 7d
(b) True
(e) False
(c) False
(f) True
If you had difficulty with these problems, you may wish to consult the
Review of Algebra on the website www.stewartcalculus.com.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xxvii
xxviii
Diagnostic Tests
B
1.Find an equation for the line that passes through the point s2, 25d and
(a) has slope 23
(b) is parallel to the x-axis
(c) is parallel to the y-axis
(d) is parallel to the line 2x 2 4y − 3
2.Find an equation for the circle that has center s21, 4d and passes through the point s3, 22d.
3.Find the center and radius of the circle with equation x 2 1 y 2 2 6x 1 10y 1 9 − 0.
4.Let As27, 4d and Bs5, 212d be points in the plane.
(a)Find the slope of the line that contains A and B.
(b)Find an equation of the line that passes through A and B. What are the intercepts?
(c) Find the midpoint of the segment AB.
(d) Find the length of the segment AB.
(e) Find an equation of the perpendicular bisector of AB.
(f) Find an equation of the circle for which AB is a diameter.
5.Sketch the region in the xy-plane defined by the equation or inequalities.
(a) 21 < y < 3
(c) y , 1 2
(b)
1
2x
| x | , 4 and | y | , 2
(d) y > x 2 2 1
(e) x 2 1 y 2 , 4
(f) 9x 2 1 16y 2 − 144
answers to diagnostic test b: analytic geometry
1.(a) y − 23x 1 1
(c) x − 2
(b) y − 25
(d) y −
1
2x
26
5.
(a)
(b)
3
2. sx 1 1d2 1 s y 2 4d2 − 52
x
_1
234
4x 1 3y 1 16 − 0; x-intercept 24, y-intercept 2 16
3
s21, 24d
20
3x 2 4y − 13
sx 1 1d2 1 s y 1 4d2 − 100
(d)
y
_4
1
1
4x
0
(e)
y
2
_1
y
0
y=1- 2 x
2
x
_2
y
0
(c)
2
0
3.Center s3, 25d, radius 5
4.(a)
(b)
(c)
(d)
(e)
(f)
y
1
x
(f )
≈+¥=4
0
y=≈-1
2
x
y
3
0
4 x
If you had difficulty with these problems,
you may wish to consult
6et-dtba05a-f
5.20.06
the review of analytic geometry in Appendixes
B and C.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Diagnostic Tests
xxix
C
y
1. The graph of a function f is given at the left.
(a) State the value of f s21d.
(b) Estimate the value of f s2d.
(c) For what values of x is f sxd − 2?
1
(d) Estimate the values of x such that f sxd − 0.
0
x
1
(e) State the domain and range of f.
2. If f sxd − x 3, evaluate the difference quotient
3. Find the domain of the function.
Figure For Problem 1
(a) f sxd −
2x 1 1
x 1x22
(b) tsxd −
2
f s2 1 hd 2 f s2d
and simplify your answer.
h
3
x
s
x 11
(c) hsxd − s4 2 x 1 sx 2 2 1
2
4. How are graphs of the functions obtained from the graph of f ?
(a) y − 2f sxd
(b) y − 2 f sxd 2 1
(c) y − f sx 2 3d 1 2
5. Without using a calculator, make a rough sketch of the graph.
(a) y − x 3
(b) y − sx 1 1d3
(c) y − sx 2 2d3 1 3
2
(d) y − 4 2 x
(e) y − sx
(f) y − 2 sx
(g) y − 22 x
(h) y − 1 1 x21
H
1 2 x 2 if x < 0
6. Let f sxd −
2x 1 1 if x . 0
(a) Evaluate f s22d and f s1d.
(b) Sketch the graph of f.
7.If f sxd − x 2 1 2x 2 1 and tsxd − 2x 2 3, find each of the following functions.
(a) f 8 t
(b) t 8 f
(c) t 8 t 8 t
answers to diagnostic test C: functions
1.(a) 22
(c) 23, 1
(e) f23, 3g, f22, 3g
(b) 2.8
(d) 22.5, 0.3
5. (a)
0
4.(a) Reflect about the x-axis
(b)Stretch vertically by a factor of 2, then shift 1 unit
downward
(c) Shift 3 units to the right and 2 units upward
(d)
(g)
1
x
_1
(e)
2
x
(2, 3)
x
0
1
x
1
x
x
0
(f)
y
0
(h)
y
y
0
1
y
1
0
_1
y
1
y
4
0
(c)
y
1
2. 12 1 6h 1 h 2
3.(a) s2`, 22d ø s22, 1d ø s1, `d
(b) s2`, `d
(c) s2`, 21g ø f1, 4g
(b)
y
1
x
0
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
xxx
Diagnostic Tests
6. (a) 23, 3(b)
7. (a) s f 8 tdsxd − 4x 2 2 8x 1 2
y
(b) s t 8 f dsxd − 2x 2 1 4x 2 5
1
_1
x
0
(c) s t 8 t 8 tdsxd − 8x 2 21
If you had difficulty
with these problems, you should look at sections 1.1–1.3 of this book.
4c3DTCax06b
10/30/08
D
1.Convert from degrees to radians.
(a) 3008
(b) 2188
2. Convert from radians to degrees.
(a) 5y6
(b) 2
3.Find the length of an arc of a circle with radius 12 cm if the arc subtends a central angle
of 308.
4. Find the exact values.
(a) tansy3d
(b) sins7y6d
(c) secs5y3d
5.Express the lengths a and b in the figure in terms of .
24
a
6.If sin x − 13 and sec y − 54, where x and y lie between 0 and y2, evaluate sinsx 1 yd.
¨
7. Prove the identities.
b
(a) tan sin 1 cos − sec (b)
Figure For Problem 5
2 tan x
− sin 2x
1 1 tan 2x
8.Find all values of x such that sin 2x − sin x and 0 < x < 2.
9.Sketch the graph of the function y − 1 1 sin 2x without using a calculator.
answers to diagnostic test D: trigonometry
s4 1 6 s2 d
1.(a) 5y3
(b) 2y10
6.
2.(a) 1508
(b) 3608y < 114.68
8. 0, y3, , 5y3, 2
3. 2 cm
1
15
y
2
9.
4.(a) s3
(b) 221
5.(a) 24 sin (b) 24 cos (c) 2
_π
0
π
x
4c3DTDax09
If you had difficulty with these problems, you should look
at Appendix D of this book.
10/30/08
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A Preview of Calculus
By the time you finish this course, you will be able to calculate the length of the curve used to design the Gateway Arch
in St. Louis, determine where a pilot should start descent
for a smooth landing, compute the force on a baseball bat
when it strikes the ball, and measure the amount of light
sensed by the human eye as the pupil changes size.
calculus is fundamentally different from the mathematics that you have studied previously: calculus is less static and more dynamic. It is concerned with change and motion; it deals
with quantities that approach other quantities. For that reason it may be useful to have an overview
of the subject before beginning its intensive study. Here we give a glimpse of some of the main
ideas of calculus by showing how the concept of a limit arises when we attempt to solve a variety
of problems.
1
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
2
a preview of calculus
The Area Problem
A¡
The origins of calculus go back at least 2500 years to the ancient Greeks, who found
areas using the “method of exhaustion.” They knew how to find the area A of any polygon by dividing it into triangles as in Figure 1 and adding the areas of these triangles.
It is a much more difficult problem to find the area of a curved figure. The Greek
method of exhaustion was to inscribe polygons in the figure and circumscribe polygons
about the figure and then let the number of sides of the polygons increase. Figure 2 illustrates this process for the special case of a circle with inscribed regular polygons.
A∞
A™
A¢
A£
A=A¡+A™+A£+A¢+A∞
FIGURE 1
A£
A¢
A∞
Aß
A¶
A¡™
FIGURE 2
Let An be the area of the inscribed polygon with n sides. As n increases, it appears that
An becomes closer and closer to the area of the circle. We say that the area of the circle
is the limit of the areas of the inscribed polygons, and we write
TEC In the Preview Visual, you
can see how areas of inscribed and
circumscribed polygons approximate
the area of a circle.
y
A − lim An
nl`
The Greeks themselves did not use limits explicitly. However, by indirect reasoning,
Eudoxus (fifth century bc) used exhaustion to prove the familiar formula for the area of
a circle: A − r 2.
We will use a similar idea in Chapter 4 to find areas of regions of the type shown in
Figure 3. We will approximate the desired area A by areas of rectangles (as in Figure 4),
let the width of the rectangles decrease, and then calculate A as the limit of these sums
of areas of rectangles.
y
y
(1, 1)
y
(1, 1)
(1, 1)
(1, 1)
y=≈
A
0
1
x
0
1
4
1
2
3
4
1
x
0
1
x
0
1
n
1
x
FIGURE 3
The area problem is the central problem in the branch of calculus called integral calculus. The techniques that we will develop in Chapter 4 for finding areas will also enable
us to compute the volume of a solid, the length of a curve, the force of water against a
dam, the mass and center of gravity of a rod, and the work done in pumping water out
of a tank.
The Tangent Problem
Consider the problem of trying to find an equation of the tangent line t to a curve with
equation y − f sxd at a given point P. (We will give a precise definition of a tangent line in
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
a preview of calculus
y
Chapter 1. For now you can think of it as a line that touches the curve at P as in Figure 5.)
Since we know that the point P lies on the tangent line, we can find the equation of t if we
know its slope m. The problem is that we need two points to compute the slope and we
know only one point, P, on t. To get around the problem we first find an approximation
to m by taking a nearby point Q on the curve and computing the slope mPQ of the secant
line PQ. From Figure 6 we see that
t
y=ƒ
P
0
x
FIGURE 5
1
mPQ −
t
m − lim mPQ
Q lP
Q { x, ƒ}
ƒ-f(a)
P { a, f(a)}
and we say that m is the limit of mPQ as Q approaches P along the curve. Because x
approaches a as Q approaches P, we could also use Equation 1 to write
x-a
a
0
x
x
FIGURE 6
The secant line at PQ
y
f sxd 2 f sad
x2a
Now imagine that Q moves along the curve toward P as in Figure 7. You can see that
the secant line rotates and approaches the tangent line as its limiting position. This means
that the slope mPQ of the secant line becomes closer and closer to the slope m of the tangent line. We write
The tangent line at P
y
3
t
Q
P
0
FIGURE 7
Secant lines approaching the
tangent line
x
2
f sxd 2 f sad
x2a
m − lim
xla
Specific examples of this procedure will be given in Chapter 1.
The tangent problem has given rise to the branch of calculus called differential calculus, which was not invented until more than 2000 years after integral calculus. The main
ideas behind differential calculus are due to the French mathematician Pierre Fermat (1601–1665) and were developed by the English mathematicians John Wallis
(1616–1703), Isaac Barrow (1630–1677), and Isaac Newton (1642–1727) and the German mathematician Gottfried Leibniz (1646–1716).
The two branches of calculus and their chief problems, the area problem and the tangent problem, appear to be very different, but it turns out that there is a very close connection between them. The tangent problem and the area problem are inverse problems
in a sense that will be described in Chapter 4.
Velocity
When we look at the speedometer of a car and read that the car is traveling at 48 miyh,
what does that information indicate to us? We know that if the velocity remains constant,
then after an hour we will have traveled 48 mi. But if the velocity of the car varies, what
does it mean to say that the velocity at a given instant is 48 miyh?
In order to analyze this question, let’s examine the motion of a car that travels along a
straight road and assume that we can measure the distance traveled by the car (in feet) at
l-second intervals as in the following chart:
t − Time elapsed ssd
0
1
2
3
4
5
d − Distance sftd
0
2
9
24
42
71
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
4
a preview of calculus
As a first step toward finding the velocity after 2 seconds have elapsed, we find the average velocity during the time interval 2 < t < 4:
change in position
time elapsed
average velocity −
42 2 9
422
−
− 16.5 ftys
Similarly, the average velocity in the time interval 2 < t < 3 is
average velocity −
24 2 9
− 15 ftys
322
We have the feeling that the velocity at the instant t − 2 can’t be much different from the
average velocity during a short time interval starting at t − 2. So let’s imagine that the distance traveled has been measured at 0.l-second time intervals as in the following chart:
t
2.0
2.1
2.2
2.3
2.4
2.5
d
9.00
10.02
11.16
12.45
13.96
15.80
Then we can compute, for instance, the average velocity over the time interval f2, 2.5g:
average velocity −
15.80 2 9.00
− 13.6 ftys
2.5 2 2
The results of such calculations are shown in the following chart:
Time interval
f2, 3g
f2, 2.5g
f2, 2.4g
f2, 2.3g
f2, 2.2g
f2, 2.1g
Average velocity sftysd
15.0
13.6
12.4
11.5
10.8
10.2
The average velocities over successively smaller intervals appear to be getting closer to
a number near 10, and so we expect that the velocity at exactly t − 2 is about 10 ftys. In
Chapter 2 we will define the instantaneous velocity of a moving object as the limiting value
of the average velocities over smaller and smaller time intervals.
In Figure 8 we show a graphical representation of the motion of the car by plotting the
distance traveled as a function of time. If we write d − f std, then f std is the number of
feet traveled after t seconds. The average velocity in the time interval f2, tg is
d
Q { t, f(t)}
average velocity −
which is the same as the slope of the secant line PQ in Figure 8. The velocity v when
t − 2 is the limiting value of this average velocity as t approaches 2; that is,
20
10
0
change in position
f std 2 f s2d
−
time elapsed
t22
P { 2, f(2)}
1
2
FIGURE 8
3
4
5
t
v − lim
tl2
f std 2 f s2d
t22
and we recognize from Equation 2 that this is the same as the slope of the tangent line
to the curve at P.
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5
a preview of calculus
Thus, when we solve the tangent problem in differential calculus, we are also solving
problems concerning velocities. The same techniques also enable us to solve problems
involving rates of change in all of the natural and social sciences.
The Limit of a Sequence
In the fifth century bc the Greek philosopher Zeno of Elea posed four problems, now
known as Zeno’s paradoxes, that were intended to challenge some of the ideas concerning
space and time that were held in his day. Zeno’s second paradox concerns a race between
the Greek hero Achilles and a tortoise that has been given a head start. Zeno argued, as follows, that Achilles could never pass the tortoise: Suppose that Achil­les starts at position
a 1 and the tortoise starts at position t1. (See Figure 9.) When Achilles reaches the point
a 2 − t1, the tortoise is farther ahead at position t2. When Achilles reaches a 3 − t2, the tortoise is at t3. This process continues indefinitely and so it appears that the tortoise will
always be ahead! But this defies common sense.
Achilles
FIGURE 9
a¡
tortoise
a™
a£
a¢
a∞
...
t¡
t™
t£
t¢
...
One way of explaining this paradox is with the idea of a sequence. The successive positions of Achilles sa 1, a 2 , a 3 , . . .d or the successive positions of the tortoise st1, t2 , t3 , . . .d
form what is known as a sequence.
In general, a sequence ha nj is a set of numbers written in a definite order. For instance,
the sequence
h1, 12 , 13 , 14 , 15 , . . . j
can be described by giving the following formula for the nth term:
an −
a¢ a £
a™
0
We can visualize this sequence by plotting its terms on a number line as in Figure 10(a) or by drawing its graph as in Figure 10(b). Observe from either picture that the
terms of the sequence a n − 1yn are becoming closer and closer to 0 as n increases. In
fact, we can find terms as small as we please by making n large enough. We say that the
limit of the sequence is 0, and we indicate this by writing
a¡
1
(a)
1
lim
nl`
1 2 3 4 5 6 7 8
(b)
FIGURE 10
1
n
n
1
−0
n
In general, the notation
lim a n − L
nl`
is used if the terms a n approach the number L as n becomes large. This means that the numbers a n can be made as close as we like to the number L by taking n sufficiently large.
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6
a preview of calculus
The concept of the limit of a sequence occurs whenever we use the decimal representation of a real number. For instance, if
a 1 − 3.1
a 2 − 3.14
a 3 − 3.141
a 4 − 3.1415
a 5 − 3.14159
a 6 − 3.141592
a 7 − 3.1415926
f
then
lim a n − nl`
The terms in this sequence are rational approximations to .
Let’s return to Zeno’s paradox. The successive positions of Achilles and the tortoise
form sequences ha nj and htn j, where a n , tn for all n. It can be shown that both sequences
have the same limit:
lim a n − p − lim tn
nl`
nl`
It is precisely at this point p that Achilles overtakes the tortoise.
The Sum of a Series
Another of Zeno’s paradoxes, as passed on to us by Aristotle, is the following: “A man
standing in a room cannot walk to the wall. In order to do so, he would first have to
go half the distance, then half the remaining distance, and then again half of what still
remains. This process can always be continued and can never be ended.” (See Figure 11.)
1
2
FIGURE 11
1
4
1
8
1
16
Of course, we know that the man can actually reach the wall, so this suggests that perhaps the total distance can be expressed as the sum of infinitely many smaller distances
as follows:
3
1−
1
1
1
1
1
1 1 1
1 ∙∙∙ 1 n 1 ∙∙∙
2
4
8
16
2
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a preview of calculus
7
Zeno was arguing that it doesn’t make sense to add infinitely many numbers together.
But there are other situations in which we implicitly use infinite sums. For instance, in
decimal notation, the symbol 0.3 − 0.3333 . . . means
3
3
3
3
1
1
1
1 ∙∙∙
10
100
1000
10,000
and so, in some sense, it must be true that
3
3
3
3
1
1
1
1
1 ∙∙∙ −
10
100
1000
10,000
3
More generally, if dn denotes the nth digit in the decimal representation of a number, then
0.d1 d2 d3 d4 . . . −
d1
d2
d3
dn
1 2 1 3 1 ∙∙∙ 1 n 1 ∙∙∙
10
10
10
10
Therefore some infinite sums, or infinite series as they are called, have a meaning. But
we must define carefully what the sum of an infinite series is.
Returning to the series in Equation 3, we denote by sn the sum of the first n terms of
the series. Thus
s1 − 12 − 0.5
s2 − 12 1 14 − 0.75
s3 − 12 1 14 1 18 − 0.875
1
s4 − 12 1 14 1 18 1 16
− 0.9375
1
1
s5 − 12 1 14 1 18 1 16
1 32
− 0.96875
1
1
1
s6 − 12 1 14 1 18 1 16
1 32
1 64
− 0.984375
1
1
1
1
s7 − 12 1 14 1 18 1 16
1 32
1 64
1 128
− 0.9921875
f
1
s10 − 12 1 14 1 ∙ ∙ ∙ 1 1024
< 0.99902344
f
s16 −
1
1
1
1 1 ∙ ∙ ∙ 1 16 < 0.99998474
2
4
2
Observe that as we add more and more terms, the partial sums become closer and closer
to 1. In fact, it can be shown that by taking n large enough (that is, by adding sufficiently
many terms of the series), we can make the partial sum sn as close as we please to the number 1. It therefore seems reasonable to say that the sum of the infinite series is 1 and to
write
1
1
1
1
1 1 1 ∙∙∙ 1 n 1 ∙∙∙ − 1
2
4
8
2
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8
a preview of calculus
In other words, the reason the sum of the series is 1 is that
lim sn − 1
nl`
In Chapter 11 we will discuss these ideas further. We will then use Newton’s idea of
combining infinite series with differential and integral calculus.
Summary
We have seen that the concept of a limit arises in trying to find the area of a region, the
slope of a tangent to a curve, the velocity of a car, or the sum of an infinite series. In
each case the common theme is the calculation of a quantity as the limit of other, easily
calculated quantities. It is this basic idea of a limit that sets calculus apart from other
areas of mathematics. In fact, we could define calculus as the part of mathematics that
deals with limits.
After Sir Isaac Newton invented his version of calculus, he used it to explain the
motion of the planets around the sun. Today calculus is used in calculating the orbits of
satellites and spacecraft, in predicting population sizes, in estimating how fast oil prices
rise or fall, in forecasting weather, in measuring the cardiac output of the heart, in calculating life insurance premiums, and in a great variety of other areas. We will explore
some of these uses of calculus in this book.
In order to convey a sense of the power of the subject, we end this preview with a list
of some of the questions that you will be able to answer using calculus:
1.How can we explain the fact, illustrated in Figure 12, that the angle of elevation
from an observer up to the highest point in a rainbow is 42°? (See page 213.)
rays from sun
138°
rays from sun
42°
2.How can we explain the shapes of cans on supermarket shelves? (See page 270.)
3. Where is the best place to sit in a movie theater? (See page 483.)
4.How can we design a roller coaster for a smooth ride? (See page 144.)
5. How far away from an airport should a pilot start descent? (See page 161.)
observer
FIGURE 12
6.How can we fit curves together to design shapes to represent letters on a laser
printer? (See page 697.)
7.How can we estimate the number of workers that were needed to build the Great
Pyramid of Khufu in ancient Egypt? (See page 388.)
8.Where should an infielder position himself to catch a baseball thrown by an
outfielder and relay it to home plate? (See page 392.)
9.Does a ball thrown upward take longer to reach its maximum height or to fall
back to its original height? (See page 649.)
10.How can we explain the fact that planets and satellites move in elliptical orbits?
(See page 916.)
11.How can we distribute water flow among turbines at a hydroelectric station so
as to maximize the total energy production? (See page 1020.)
12.If a marble, a squash ball, a steel bar, and a lead pipe roll down a slope, which
of them reaches the bottom first? (See page 1092.)
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1
Often a graph is the best
way to represent a function
because it conveys so much
information at a glance.
Shown is a graph of the
vertical ground acceleration
created by the 2011
earthquake near Tohoku,
Japan. The earthquake
had a magnitude of 9.0 on
the Richter scale and was
so powerful that it moved
northern Japan 8 feet closer
to North America.
Functions and Limits
Pictura Collectus/Alamy
(cm/s@)
2000
1000
0
time
_1000
_2000
0
50
100
150
200
Seismological Society of America
The fundamental objects that we deal with in calculus are functions. We stress that a
function can be represented in different ways: by an equation, in a table, by a graph, or in words.
We look at the main types of functions that occur in calculus and describe the process of using
these functions as mathematical models of real-world phenomena.
In A Preview of Calculus (page 1) we saw how the idea of a limit underlies the various
branches of calculus. It is therefore appropriate to begin our study of calculus by investigating
limits of functions and their properties.
9
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10
Chapter 1 Functions and Limits
Year
Population
(millions)
1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000
2010
1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080
6870
Functions arise whenever one quantity depends on another. Consider the following four
situations.
A.The area A of a circle depends on the radius r of the circle. The rule that connects r
and A is given by the equation A − r 2. With each positive number r there is associated one value of A, and we say that A is a function of r.
B.The human population of the world P depends on the time t. The table gives estimates of the world population Pstd at time t, for certain years. For instance,
Ps1950d < 2,560,000,000
But for each value of the time t there is a corresponding value of P, and we say that
P is a function of t.
C.The cost C of mailing an envelope depends on its weight w. Although there is no
simple formula that connects w and C, the post office has a rule for determining C
when w is known.
D.The vertical acceleration a of the ground as measured by a seismograph during an
earthquake is a function of the elapsed time t. Figure 1 shows a graph generated by
seismic activity during the Northridge earthquake that shook Los Angeles in 1994.
For a given value of t, the graph provides a corresponding value of a.
a
{cm/s@}
100
50
5
FIGURE 1
Vertical ground acceleration
during the Northridge earthquake
10
15
20
25
30
t (seconds)
_50
Calif. Dept. of Mines and Geology
Each of these examples describes a rule whereby, given a number (r, t, w, or t), another
number (A, P, C, or a) is assigned. In each case we say that the second number is a
function of the first number.
A function f is a rule that assigns to each element x in a set D exactly one
element, called f sxd, in a set E.
We usually consider functions for which the sets D and E are sets of real numbers.
The set D is called the domain of the function. The number f sxd is the value of f at x
and is read “ f of x.” The range of f is the set of all possible values of f sxd as x varies
throughout the domain. A symbol that represents an arbitrary number in the domain of a
function f is called an independent variable. A symbol that represents a number in the
range of f is called a dependent variable. In Example A, for instance, r is the independent variable and A is the dependent variable.
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11
Section 1.1 Four Ways to Represent a Function
x
(input)
f
ƒ
(output)
FIGURE 2
Machine diagram for a function f
x
ƒ
a
f(a)
f
D
It’s helpful to think of a function as a machine (see Figure 2). If x is in the domain of
the function f, then when x enters the machine, it’s accepted as an input and the machine
produces an output f sxd according to the rule of the function. Thus we can think of the
domain as the set of all possible inputs and the range as the set of all possible outputs.
The preprogrammed functions in a calculator are good examples of a function as a
machine. For example, the square root key on your calculator computes such a function.
You press the key labeled s (or s x ) and enter the input x. If x , 0, then x is not in the
domain of this function; that is, x is not an acceptable input, and the calculator will indicate an error. If x > 0, then an approximation to s x will appear in the display. Thus the
s x key on your calculator is not quite the same as the exact mathematical function f
defined by f sxd − s x .
Another way to picture a function is by an arrow diagram as in Figure 3. Each arrow
connects an element of D to an element of E. The arrow indicates that f sxd is associated
with x, f sad is associated with a, and so on.
The most common method for visualizing a function is its graph. If f is a function
with domain D, then its graph is the set of ordered pairs
|
hsx, f sxdd x [ Dj
E
(Notice that these are input-output pairs.) In other words, the graph of f consists of all
points sx, yd in the coordinate plane such that y − f sxd and x is in the domain of f.
The graph of a function f gives us a useful picture of the behavior or “life history”
of a function. Since the y-coordinate of any point sx, yd on the graph is y − f sxd, we can
read the value of f sxd from the graph as being the height of the graph above the point x
(see Figure 4). The graph of f also allows us to picture the domain of f on the x-axis and
its range on the y-axis as in Figure 5.
FIGURE 3
Arrow diagram for f
y
y
{ x, ƒ}
range
ƒ
f (2)
f (1)
0
1
2
x
x
FIGURE 4
y
0
domain
x
FIGURE 5
Example 1 The graph of a function f is shown in Figure 6.
(a) Find the values of f s1d and f s5d.
(b) What are the domain and range of f ?
1
0
y ƒ(x)
Solution
1
x
FIGURE 6
The notation for intervals is given in
Appendix A.
(a) We see from Figure 6 that the point s1, 3d lies on the graph of f, so the value of f
at 1 is f s1d − 3. (In other words, the point on the graph that lies above x − 1 is 3 units
above the x-axis.)
When x − 5, the graph lies about 0.7 units below the x-axis, so we estimate that
f s5d < 20.7.
(b) We see that f sxd is defined when 0 < x < 7, so the domain of f is the closed interval f0, 7g. Notice that f takes on all values from 22 to 4, so the range of f is
|
hy 22 < y < 4j − f22, 4g
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■
12
Chapter 1 Functions and Limits
y
Example 2 Sketch the graph and find the domain and range of each function.
(a) fsxd − 2x 2 1
(b) tsxd − x 2
Solution
y=2x-1
0
-1
x
1
2
FIGURE 7
y
(2, 4)
y=≈
(_1, 1)
(a) The equation of the graph is y − 2x 2 1, and we recognize this as being the equation of a line with slope 2 and y-intercept 21. (Recall the slope-intercept form of the
equation of a line: y − mx 1 b. See Appendix B.) This enables us to sketch a portion
of the graph of f in Figure 7. The expression 2x 2 1 is defined for all real numbers, so
the domain of f is the set of all real numbers, which we denote by R. The graph shows
that the range is also R.
(b) Since ts2d − 2 2 − 4 and ts21d − s21d2 − 1, we could plot the points s2, 4d and
s21, 1d, together with a few other points on the graph, and join them to produce the
graph (Figure 8). The equation of the graph is y − x 2, which represents a parabola (see
Appendix C). The domain of t is R. The range of t consists of all values of tsxd, that is,
all numbers of the form x 2. But x 2 > 0 for all numbers x and any positive number y is a
square. So the range of t is hy y > 0j − f0, `d. This can also be seen from Figure 8. ■
|
1
0
1
x
Example 3 If f sxd − 2x 2 2 5x 1 1 and h ± 0, evaluate
f sa 1 hd 2 f sad
.
h
Solution We first evaluate f sa 1 hd by replacing x by a 1 h in the expression for f sxd:
FIGURE 8
f sa 1 hd − 2sa 1 hd2 2 5sa 1 hd 1 1
− 2sa 2 1 2ah 1 h 2 d 2 5sa 1 hd 1 1
− 2a 2 1 4ah 1 2h 2 2 5a 2 5h 1 1
The expression
Then we substitute into the given expression and simplify:
f sa 1 hd 2 f sad
s2a 2 1 4ah 1 2h 2 2 5a 2 5h 1 1d 2 s2a 2 2 5a 1 1d
−
h
h
f sa 1 hd 2 f sad
h
in Example 3 is called a difference
quotient and occurs frequently in
calculus. As we will see in Chapter
2, it represents the average rate of
change of f sxd between x − a and
x − a 1 h.
−
2a 2 1 4ah 1 2h 2 2 5a 2 5h 1 1 2 2a 2 1 5a 2 1
h
−
4ah 1 2h 2 2 5h
− 4a 1 2h 2 5
h
■
Representations of Functions
There are four possible ways to represent a function:
●
●
●
●
verbally
numerically
visually
algebraically (by a description in words)
(by a table of values)
(by a graph)
(by an explicit formula)
If a single function can be represented in all four ways, it’s often useful to go from one
representation to another to gain additional insight into the function. (In Example 2, for
instance, we started with algebraic formulas and then obtained the graphs.) But certain
functions are described more naturally by one method than by another. With this in mind,
let’s reexamine the four situations that we considered at the beginning of this section.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
13
Section 1.1 Four Ways to Represent a Function
A.The most useful representation of the area of a circle as a function of its radius is
probably the algebraic formula Asrd − r 2, though it is possible to compile a table
of values or to sketch a graph (half a parabola). Because a circle has to have a positive radius, the domain is hr r . 0j − s0, `d, and the range is also s0, `d.
B.We are given a description of the function in words: Pstd is the human population of
the world at time t. Let’s measure t so that t − 0 corresponds to the year 1900. The
table of values of world population provides a convenient representation of this function. If we plot these values, we get the graph (called a scatter plot) in Figure 9. It
too is a useful representation; the graph allows us to absorb all the data at once. What
about a formula? Of course, it’s impossible to devise an explicit formula that gives
the exact human population Pstd at any time t. But it is possible to find an expression
for a function that approximates Pstd. In fact, using methods explained in Section
1.2, we obtain the approximation
|
t
(years
since 1900)
Population
(millions)
0
10
20
30
40
50
60
70
80
90
100
110
1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080
6870
Pstd < f std − s1.43653 3 10 9 d s1.01395d t
Figure 10 shows that it is a reasonably good “fit.” The function f is called a mathematical model for population growth. In other words, it is a function with an explicit
formula that approximates the behavior of our given function. We will see, however,
that the ideas of calculus can be applied to a table of values; an explicit formula is
not necessary.
P
P
5x10'
0
5x10'
20
40
60
80
Years since 1900
FIGURE 9
100
120
t
0
20
40
60
80
Years since 1900
100
120
t
FIGURE 10
A function defined by a table of
values is called a tabular function.
w (ounces)
Cswd (dollars)
0,w<1
1,w<2
2,w<3
3,w<4
4,w<5
∙
∙
∙
0.98
1.19
1.40
1.61
1.82
∙
∙
∙
The function P is typical of the functions that arise whenever we attempt to apply
calculus to the real world. We start with a verbal description of a function. Then we
may be able to construct a table of values of the function, perhaps from instrument
readings in a scientific experiment. Even though we don’t have complete knowledge
of the values of the function, we will see throughout the book that it is still possible
to perform the operations of calculus on such a function.
C.Again the function is described in words: Let Cswd be the cost of mailing a large envelope with weight w. The rule that the US Postal Service used as of 2015 is as follows:
The cost is 98 cents for up to 1 oz, plus 21 cents for each additional ounce (or less)
up to 13 oz. The table of values shown in the margin is the most convenient representation for this function, though it is possible to sketch a graph (see Example 10).
D.The graph shown in Figure 1 is the most natural representation of the vertical acceleration function astd. It’s true that a table of values could be compiled, and it is
even possible to devise an approximate formula. But everything a geologist needs to
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14
Chapter 1 Functions and Limits
know— amplitudes and patterns — can be seen easily from the graph. (The same is
true for the patterns seen in electrocardiograms of heart patients and polygraphs for
lie-detection.)
In the next example we sketch the graph of a function that is defined verbally.
T
Example 4 When you turn on a hot-water faucet, the temperature T of the water
depends on how long the water has been running. Draw a rough graph of T as a function of the time t that has elapsed since the faucet was turned on.
0
SOLUTION The initial temperature of the running water is close to room temperature
because the water has been sitting in the pipes. When the water from the hot-water tank
starts flowing from the faucet, T increases quickly. In the next phase, T is constant at
the tempera­ture of the heated water in the tank. When the tank is drained, T decreases
to the temperature of the water supply. This enables us to make the rough sketch of T
as a function of t in Figure 11.
■
t
FIGURE 11
In the following example we start with a verbal description of a function in a physical
situation and obtain an explicit algebraic formula. The ability to do this is a useful skill
in solving calculus problems that ask for the maximum or minimum values of quantities.
Example 5 A rectangular storage container with an open top has a volume of 10 m3.
The length of its base is twice its width. Material for the base costs $10 per square
meter; material for the sides costs $6 per square meter. Express the cost of materials as
a function of the width of the base.
h
w
SOLUTION We draw a diagram as in Figure 12 and introduce notation by letting w and
2w be the width and length of the base, respectively, and h be the height.
The area of the base is s2wdw − 2w 2, so the cost, in dollars, of the material for the
base is 10s2w 2 d. Two of the sides have area wh and the other two have area 2wh, so the
cost of the material for the sides is 6f2swhd 1 2s2whdg. The total cost is therefore
C − 10s2w 2 d 1 6f2swhd 1 2s2whdg − 20 w 2 1 36 wh
2w
FIGURE 12
To express C as a function of w alone, we need to eliminate h and we do so by using
the fact that the volume is 10 m3. Thus
w s2wdh − 10
which gives
PS In setting up applied functions as
in Example 5, it may be useful to review
the principles of problem solving as
discussed on page 98, particularly
Step 1: Understand the Problem.
10
5
2 −
2w
w2
h−
Substituting this into the expression for C, we have
S D
C − 20w 2 1 36w
5
w
2
− 20w 2 1
180
w
Therefore the equation
Cswd − 20w 2 1
180
w
w . 0
expresses C as a function of w.
Example 6 Find the domain of each function.
(a) f sxd − sx 1 2 (b) tsxd −
■
1
x2 2 x
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Section 1.1 Four Ways to Represent a Function
Domain Convention
If a function is given by a formula
and the domain is not stated explicitly, the convention is that the domain
is the set of all numbers for which
the formula makes sense and defines
a real number.
15
SOLUTION
(a) Because the square root of a negative number is not defined (as a real number),
the domain of f consists of all values of x such that x 1 2 > 0. This is equivalent to
x > 22, so the domain is the interval f22, `d.
(b) Since
1
1
tsxd − 2
−
x 2x
xsx 2 1d
and division by 0 is not allowed, we see that tsxd is not defined when x − 0 or x − 1.
Thus the domain of t is
|
hx x ± 0, x ± 1j
which could also be written in interval notation as
s2`, 0d ø s0, 1d ø s1, `d
y
The graph of a function is a curve in the xy-plane. But the question arises: which
curves in the xy-plane are graphs of functions? This is answered by the following test.
x=a
(a, b)
0
The Vertical Line Test A curve in the xy-plane is the graph of a function of x if
and only if no vertical line intersects the curve more than once.
x
a
(a) This curve represents a function.
y
(a, c)
x=a
(a, b)
0
■
a
x
(b) This curve doesn’t represent
a function.
FIGURE 13
The reason for the truth of the Vertical Line Test can be seen in Figure 13. If each
vertical line x − a intersects a curve only once, at sa, bd, then exactly one function value
is defined by f sad − b. But if a line x − a intersects the curve twice, at sa, bd and sa, cd,
then the curve can’t represent a function because a function can’t assign two different
values to a.
For example, the parabola x − y 2 2 2 shown in Figure 14(a) is not the graph of a
function of x because, as you can see, there are vertical lines that intersect the parabola
twice. The parabola, however, does contain the graphs of two functions of x. Notice
that the equation x − y 2 2 2 implies y 2 − x 1 2, so y − 6sx 1 2 . Thus the upper
and lower halves of the parabola are the graphs of the functions f sxd − s x 1 2 [from
Example 6(a)] and tsxd − 2s x 1 2 . [See Figures 14(b) and (c).]
We observe that if we reverse the roles of x and y, then the equation x − hsyd − y 2 2 2
does define x as a function of y (with y as the independent variable and x as the dependent variable) and the parabola now appears as the graph of the function h.
y
(_2, 0)
FIGURE 14
0
y
x
_2 0
(a) x=¥-2
(b) y=œ„„„„
x+2
y
x
_2
0
x
(c) y=_ œ„„„„
x+2
Piecewise Defined Functions
The functions in the following four examples are defined by different formulas in dif­
ferent parts of their domains. Such functions are called piecewise defined functions.
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16
Chapter 1 Functions and Limits
Example 7 A function f is defined by
f sxd −
H
1 2 x if x < 21
x2
if x . 21
Evaluate f s22d, f s21d, and f s0d and sketch the graph.
Solution Remember that a function is a rule. For this particular function the rule is
the following: First look at the value of the input x. If it happens that x < 21, then the
value of f sxd is 1 2 x. On the other hand, if x . 21, then the value of f sxd is x 2.
Since 22 < 21, we have f s22d − 1 2 s22d − 3.
Since 21 < 21, we have f s21d − 1 2 s21d − 2.
y
Since 0 . 21, we have f s0d − 0 2 − 0.
1
_1
0
1
x
FIGURE 15
How do we draw the graph of f ? We observe that if x < 21, then f sxd − 1 2 x,
so the part of the graph of f that lies to the left of the vertical line x − 21 must coincide with the line y − 1 2 x, which has slope 21 and y-intercept 1. If x . 21,
then f sxd − x 2, so the part of the graph of f that lies to the right of the line x − 21
must coincide with the graph of y − x 2, which is a parabola. This enables us to sketch
the graph in Figure 15. The solid dot indicates that the point s21, 2d is included on the
graph; the open dot indicates that the point s21, 1d is excluded from the graph.
■
The next example of a piecewise defined function is the absolute value function.
Recall that the absolute value of a number a, denoted by a , is the distance from a to 0
on the real number line. Distances are always positive or 0, so we have
| |
For a more extensive review of
absolute values, see Appendix A.
| a | > 0 for every number a
For example,
| 3 | − 3 | 23 | − 3 | 0 | − 0 | s2 2 1 | − s2 2 1 | 3 2 | − 2 3
In general, we have
| a | − a if
| a | − 2a if
a>0
a,0
(Remember that if a is negative, then 2a is positive.)
Example 8 Sketch the graph of the absolute value function f sxd − | x |.
y
SOLUTION From the preceding discussion we know that
y=| x |
|x| −
0
FIGURE 16
x
H
x
if x > 0
2x if x , 0
Using the same method as in Example 7, we see that the graph of f coincides with the
line y − x to the right of the y-axis and coincides with the line y − 2x to the left of the
y-axis (see Figure 16).
■
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Section 1.1 Four Ways to Represent a Function
17
y
Example 9 Find a formula for the function f graphed in Figure 17.
1
SOLUTION The line through s0, 0d and s1, 1d has slope m − 1 and y-intercept b − 0,
so its equation is y − x. Thus, for the part of the graph of f that joins s0, 0d to s1, 1d,
we have
0
x
1
f sxd − x if 0 < x < 1
FIGURE 17
The line through s1, 1d and s2, 0d has slope m − 21, so its point-slope form is
Point-slope form of the equation of
a line:
y 2 y1 − msx 2 x 1 d
y 2 0 − s21dsx 2 2d or y − 2 2 x
So we have
See Appendix B.
f sxd − 2 2 x if 1 , x < 2
We also see that the graph of f coincides with the x-axis for x . 2. Putting this information together, we have the following three-piece formula for f :
H
x
if 0 < x < 1
f sxd − 2 2 x if 1 , x < 2
0
if x . 2
■
Example 10 In Example C at the beginning of this section we considered the cost
Cswd of mailing a large envelope with weight w. In effect, this is a piecewise defined
function because, from the table of values on page 13, we have
C
1.50
1.00
Cswd −
0.50
0
1
2
3
4
5
figure 18
w
0.98
1.19
1.40
1.61
∙
∙
∙
if
if
if
if
0,w<1
1,w<2
2,w<3
3,w<4
The
graph is shown in Figure 18. You can see why functions similar to this one are
called step functions—they jump from one value to the next. Such functions will be
■
studied in Chapter 2.
Symmetry
If a function f satisfies f s2xd − f sxd for every number x in its domain, then f is called
an even function. For instance, the function f sxd − x 2 is even because
y
f(_x)
f s2xd − s2xd2 − x 2 − f sxd
ƒ
_x
0
FIGURE 19
An even function
x
x
The geometric significance of an even function is that its graph is symmetric with respect
to the y-axis (see Figure 19). This means that if we have plotted the graph of f for x > 0,
we obtain the entire graph simply by reflecting this portion about the y-axis.
If f satisfies f s2xd − 2f sxd for every number x in its domain, then f is called an odd
function. For example, the function f sxd − x 3 is odd because
f s2xd − s2xd3 − 2x 3 − 2f sxd
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
18
Chapter 1 Functions and Limits
The graph of an odd function is symmetric about the origin (see Figure 20). If we already
have the graph of f for x > 0, we can obtain the entire graph by rotating this portion
through 1808 about the origin.
y
_x
0
ƒ
x
x
Example 11 Determine whether each of the following functions is even, odd, or
neither even nor odd.
(a) f sxd − x 5 1 x (b) tsxd − 1 2 x 4 (c) hsxd − 2x 2 x 2
SOLUTION
f s2xd − s2xd5 1 s2xd − s21d5x 5 1 s2xd
(a)
FIGURE 20
An odd function
− 2x 5 2 x − 2sx 5 1 xd
− 2f sxd
Therefore f is an odd function.
ts2xd − 1 2 s2xd4 − 1 2 x 4 − tsxd
(b)
So t is even.
hs2xd − 2s2xd 2 s2xd2 − 22x 2 x 2
(c)
Since hs2xd ± hsxd and hs2xd ± 2hsxd, we conclude that h is neither even nor odd. ■
The graphs of the functions in Example 11 are shown in Figure 21. Notice that the
graph of h is symmetric neither about the y-axis nor about the origin.
1
1
f
1
_1
y
y
y
g
1
x
h
1
x
1
x
_1
figure 21
( b)
(a)
(c)
Increasing and Decreasing Functions
The graph shown in Figure 22 rises from A to B, falls from B to C, and rises again from C
to D. The function f is said to be increasing on the interval fa, bg, decreasing on fb, cg,
and increasing again on fc, dg. Notice that if x 1 and x 2 are any two numbers between
a and b with x 1 , x 2, then f sx 1 d , f sx 2 d. We use this as the defining property of an
increasing function.
y
B
D
y=ƒ
A
figure 22
0 a x¡
f(x¡)
C
f(x™)
x™
b
c
d
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
19
Section 1.1 Four Ways to Represent a Function
A function f is called increasing on an interval I if
f sx 1 d , f sx 2 d whenever x 1 , x 2 in I
y
It is called decreasing on I if
y=≈
f sx 1 d . f sx 2 d whenever x 1 , x 2 in I
In the definition of an increasing function it is important to realize that the inequality
f sx 1 d , f sx 2 d must be satisfied for every pair of numbers x 1 and x 2 in I with x 1 , x 2.
You can see from Figure 23 that the function f sxd − x 2 is decreasing on the interval
s2`, 0g and increasing on the interval f0, `d.
x
0
figure 23
1.1 Exercises
1.If f sxd − x 1 s2 2 x and tsud − u 1 s2 2 u , is it true
that f − t?
2.If
f sxd −
x2 2 x
and tsxd − x
x21
is it true that f − t?
3.The graph of a function f is given.
(a) State the value of f s1d.
(b) Estimate the value of f s21d.
(c) For what values of x is f sxd − 1?
(d) Estimate the value of x such that f sxd − 0.
(e) State the domain and range of f.
(f) On what interval is f increasing?
y
x
1
5.Figure 1 was recorded by an instrument operated by the
California Department of Mines and Geology at the University
Hospital of the University of Southern California in Los
Angeles. Use it to estimate the range of the vertical ground
acceleration function at USC during the Northridge earthquake.
6.In this section we discussed examples of ordinary, everyday
functions: Population is a function of time, postage cost is a
function of weight, water temperature is a function of time. Give
three other examples of functions from everyday life that are
described verbally. What can you say about the domain and
range of each of your functions? If possible, sketch a rough
graph of each function.
7.
8.
y
0
y
y
1
1
4.The graphs of f and t are given.
f
Estimate the solution of the equation f sxd − 21.
On what interval is f decreasing?
State the domain and range of f.
State the domain and range of t.
7–10 D
etermine whether the curve is the graph of a function of x.
If it is, state the domain and range of the function.
1
0
(c)
(d)
(e)
(f)
1
x
0
1
x
1
x
g
2
0
y
9.10.
2
x
1
0
y
1
1
x
0
(a) State the values of f s24d and ts3d.
(b) For what values of x is f sxd − tsxd?
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
20
Chapter 1 Functions and Limits
11.Shown is a graph of the global average temperature T during
the 20th century. Estimate the following.
(a) The global average temperature in 1950
(b) The year when the average temperature was 14.2°C
(c)The year when the temperature was smallest; the year it
was largest
(d) The range of T
B
C
20
15.The graph shows the power consumption for a day in September in San Francisco. (P is measured in megawatts; t is mea­
sured in hours starting at midnight.)
(a) What was the power consumption at 6 am? At 6 pm?
(b)When was the power consumption the lowest? When was
it the highest? Do these times seem reasonable?
14
13
1900
2000 t
1950
Source: Adapted from Globe and Mail [Toronto], 5 Dec. 2009. Print.
12.Trees grow faster and form wider rings in warm years and
grow more slowly and form narrower rings in cooler years. The
figure shows ring widths of a Siberian pine from 1500 to 2000.
(a) What is the range of the ring width function?
(b)What does the graph tend to say about the temperature
of the earth? Does the graph reflect the volcanic eruptions of the mid-19th century?
R
Ring width (mm)
A
100
0
T (•C)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
y
P
800
600
400
200
0
3
6
9
12
15
18
21
t
Pacific Gas & Electric
16.Sketch a rough graph of the number of hours of daylight as a
function of the time of year.
17.Sketch a rough graph of the outdoor temperature as a function
of time during a typical spring day.
18.Sketch a rough graph of the market value of a new car as a
function of time for a period of 20 years. Assume the car is
well maintained.
19.Sketch the graph of the amount of a particular brand of coffee
sold by a store as a function of the price of the coffee.
1500
1600
1700
1800
1900
2000 t
Year
Source: Adapted from G. Jacoby et al., “Mongolian Tree Rings and 20thCentury Warming,” Science 273 (1996): 771–73.
13.You put some ice cubes in a glass, fill the glass with cold water,
and then let the glass sit on a table. Describe how the temperature of the water changes as time passes. Then sketch a rough
graph of the temperature of the water as a function of the
elapsed time.
14.Three runners compete in a 100-meter race. The graph
depicts the distance run as a function of time for each runner.
Describe in words what the graph tells you about this race.
Who won the race? Did each runner finish the race?
20.You place a frozen pie in an oven and bake it for an hour.
Then you take it out and let it cool before eating it. Describe
how the temperature of the pie changes as time passes.
Then sketch a rough graph of the temperature of the pie as a
function of time.
21.A homeowner mows the lawn every Wednesday afternoon.
Sketch a rough graph of the height of the grass as a function
of time over the course of a four-week period.
22.An airplane takes off from an airport and lands an hour later
at another airport, 400 miles away. If t represents the time in
minutes since the plane has left the terminal building, let xstd
be the horizontal distance traveled and ystd be the altitude of
the plane.
(a) Sketch a possible graph of xstd.
(b) Sketch a possible graph of ystd.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
21
Section 1.1 Four Ways to Represent a Function
(c) Sketch a possible graph of the ground speed.
(d) Sketch a possible graph of the vertical velocity.
30.f sxd −
23.Temperature readings T (in °F) were recorded every two hours
from midnight to 2:00 pm in Atlanta on June 4, 2013. The time
t was measured in hours from midnight.
31–37 Find the domain of the function.
31. f sxd −
t
0
2
4
6
8
10
12
14
T
74
69
68
66
70
78
82
86
x13
f sxd 2 f s1d
, x11
x21
x14
2x 3 2 5
32. f sxd − 2
2
x 29
x 1x26
3
33. f std − s
2t 2 1 34. tstd − s3 2 t 2 s2 1 t
1
35. hsxd −
(a)Use the readings to sketch a rough graph of T as a function
of t.
(b)Use your graph to estimate the temperature at 9:00 am.
24.Researchers measured the blood alcohol concentration (BAC)
of eight adult male subjects after rapid consumption of 30 mL
of ethanol (corresponding to two standard alcoholic drinks).
The table shows the data they obtained by averaging the BAC
(in gydL) of the eight men.
(a)Use the readings to sketch the graph of the BAC as a
function of t.
(b)Use your graph to describe how the effect of alcohol
varies with time.
t (hours)
BAC
t (hours)
BAC
0
0.2
0.5
0.75
1.0
1.25
1.5
0
0.025
0.041
0.040
0.033
0.029
0.024
1.75
2.0
2.25
2.5
3.0
3.5
4.0
0.022
0.018
0.015
0.012
0.007
0.003
0.001
Source: Adapted from P. Wilkinson et al., “Pharmacokinetics of Ethanol after
Oral Administration in the Fasting State,” Journal of Pharmacokinetics and
Biopharmaceutics 5 (1977): 207–24.
25.If f sxd − 3x 2 2 x 1 2, find f s2d, f s22d, f sad, f s2ad,
f sa 1 1d, 2 f sad, f s2ad, f sa 2 d, [ f sad] 2, and f sa 1 hd.
26.A spherical balloon with radius r inches has volume
Vsrd − 43 r 3. Find a function that represents the amount of
air required to inflate the balloon from a radius of r inches
to a radius of r 1 1 inches.
27–30 Evaluate the difference quotient for the given function.
Simplify your answer.
27. f sxd − 4 1 3x 2 x 2, 28.f sxd − x 3, 29. f sxd −
f s3 1 hd 2 f s3d
h
f sa 1 hd 2 f sad
h
1
f sxd 2 f sad
, x
x2a
sx 2 5x
4
2
36. f sud −
37. Fs pd − s2 2 s p
u11
1
11
u11
38.Find the domain and range and sketch the graph of the
function hsxd − s4 2 x 2 .
39–40 Find the domain and sketch the graph of the function.
39. f sxd − 1.6x 2 2.4
40. tstd −
t2 2 1
t11
41–44 Evaluate f s23d, f s0d, and f s2d for the piecewise defined
function. Then sketch the graph of the function.
41. f sxd −
42. f sxd −
43. f sxd −
44. f sxd −
H
H
H
H
x 1 2 if x , 0
1 2 x if x > 0
3 2 12 x if x , 2
2x 2 5 if x > 2
x 1 1 if x < 21
x2
if x . 21
21
if x < 1
7 2 2x if x . 1
45–50 Sketch the graph of the function.
| |
|
45. f sxd − x 1 x
|
47. tstd − 1 2 3t
49. f sxd −
H| |
x
1
46. f sxd − x 1 2
|
|
|| |
|
48. hstd − t 1 t 1 1
| |
| |
if x < 1
50. tsxd −
if x . 1
|| x | 2 1|
51–56 Find an expression for the function whose graph is the
given curve.
51.The line segment joining the points s1, 23d and s5, 7d
52.The line segment joining the points s25, 10d and s7, 210d
53.The bottom half of the parabola x 1 s y 2 1d2 − 0
54.The top half of the circle x 2 1 s y 2 2d 2 − 4
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
22
Chapter 1 Functions and Limits
55.
56.
y
1
1
0
64.A cell phone plan has a basic charge of $35 a month. The
plan includes 400 free minutes and charges 10 cents for each
additional minute of usage. Write the monthly cost C as a
function of the number x of minutes used and graph C as a
function of x for 0 < x < 600.
y
0
x
1
1
x
57–61 Find a formula for the described function and state its
domain.
57.A rectangle has perimeter 20 m. Express the area of the
rectangle as a function of the length of one of its sides.
2
58.A rectangle has area 16 m . Express the perimeter of the rect­
angle as a function of the length of one of its sides.
59.Express the area of an equilateral triangle as a function of the
length of a side.
60.A closed rectangular box with volume 8 ft3 has length twice the
width. Express the height of the box as a function of the width.
61.An open rectangular box with volume 2 m3 has a square base.
Express the surface area of the box as a function of the length
of a side of the base.
62.A Norman window has the shape of a rectangle surmounted
by a semicircle. If the perimeter of the window is 30 ft,
express the area A of the window as a function of the width
x of the window.
65.In a certain state the maximum speed permitted on freeways
is 65 miyh and the minimum speed is 40 miyh. The fine for
violating these limits is $15 for every mile per hour above the
maximum speed or below the minimum speed. Express the
amount of the fine F as a function of the driving speed x and
graph Fsxd for 0 < x < 100.
66.An electricity company charges its customers a base rate
of $10 a month, plus 6 cents per kilowatt-hour (kWh) for
the first 1200 kWh and 7 cents per kWh for all usage over
1200 kWh. Express the monthly cost E as a function of the
amount x of electricity used. Then graph the function E for
0 < x < 2000.
67.In a certain country, income tax is assessed as follows. There
is no tax on income up to $10,000. Any income over $10,000
is taxed at a rate of 10%, up to an income of $20,000. Any
income over $20,000 is taxed at 15%.
(a)Sketch the graph of the tax rate R as a function of the
income I.
(b)How much tax is assessed on an income of $14,000?
On $26,000?
(c)Sketch the graph of the total assessed tax T as a function
of the income I.
68.The functions in Example 10 and Exercise 67 are called step
functions because their graphs look like stairs. Give two other
examples of step functions that arise in everyday life.
69–70 Graphs of f and t are shown. Decide whether each function is even, odd, or neither. Explain your reasoning.
y
69.
70.
g
f
f
x
x
63.A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in.
by cutting out equal squares of side x at each corner and
then folding up the sides as in the figure. Express the vol­
ume V of the box as a function of x.
20
x
12
x
x
x
x
x
x
x
y
g
x
71. (a)If the point s5, 3d is on the graph of an even function,
what other point must also be on the graph?
(b)If the point s5, 3d is on the graph of an odd function, what
other point must also be on the graph?
72. A
function f has domain f25, 5g and a portion of its graph
is shown.
(a) Complete the graph of f if it is known that f is even.
(b) Complete the graph of f if it is known that f is odd.
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Section 1.2 Mathematical Models: A Catalog of Essential Functions
y
75. f sxd −
x
x11
23
| |
76. f sxd − x x
77. f sxd − 1 1 3x 2 2 x 4
78. f sxd − 1 1 3x 3 2 x 5
_5
0
5
x
79.If f and t are both even functions, is f 1 t even? If f and t
are both odd functions, is f 1 t odd? What if f is even and t is
odd? Justify your answers.
73–78 Determine whether f is even, odd, or neither. If you have
a graphing calculator, use it to check your answer visually.
80.If f and t are both even functions, is the product ft even? If f
and t are both odd functions, is ft odd? What if f is even and
t is odd? Justify your answers.
x2
74. f sxd − 4
x 11
x
73. f sxd − 2
x 11
A mathematical model is a mathematical description (often by means of a function or
an equation) of a real-world phenomenon such as the size of a population, the demand
for a product, the speed of a falling object, the concentration of a product in a chemical
reaction, the life expectancy of a person at birth, or the cost of emission reductions. The
purpose of the model is to understand the phenomenon and perhaps to make predictions
about future behavior.
Figure 1 illustrates the process of mathematical modeling. Given a real-world problem, our first task is to formulate a mathematical model by identifying and naming the
independent and dependent variables and making assumptions that simplify the phenomenon enough to make it mathematically tractable. We use our knowledge of the physical
situation and our mathematical skills to obtain equations that relate the variables. In
situations where there is no physical law to guide us, we may need to collect data (either
from a library or the Internet or by conducting our own experiments) and examine the
data in the form of a table in order to discern patterns. From this numeri­cal representation
of a function we may wish to obtain a graphical representation by plotting the data. The
graph might even suggest a suitable algebraic formula in some cases.
Real-world
problem
Formulate
Mathematical
model
Solve
Mathematical
conclusions
Interpret
Real-world
predictions
Test
FIGURE 1
The modeling process
The second stage is to apply the mathematics that we know (such as the calculus
that will be developed throughout this book) to the mathematical model that we have
formulated in order to derive mathematical conclusions. Then, in the third stage, we take
those mathematical conclusions and interpret them as information about the original
real-world phenomenon by way of offering explanations or making predictions. The final
step is to test our predictions by checking against new real data. If the predictions don’t
compare well with reality, we need to refine our model or to formulate a new model and
start the cycle again.
A mathematical model is never a completely accurate representation of a physical
situation—it is an idealization. A good model simplifies reality enough to permit math-
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24
Chapter 1 Functions and Limits
ematical calculations but is accurate enough to provide valuable conclusions. It is important to realize the limitations of the model. In the end, Mother Nature has the final say.
There are many different types of functions that can be used to model relationships
observed in the real world. In what follows, we discuss the behavior and graphs of these
functions and give examples of situations appropriately modeled by such functions.
Linear Models
The coordinate geometry of lines is
reviewed in Appendix B.
When we say that y is a linear function of x, we mean that the graph of the function is
a line, so we can use the slope-intercept form of the equation of a line to write a formula
for the function as
y − f sxd − mx 1 b
where m is the slope of the line and b is the y-intercept.
A characteristic feature of linear functions is that they grow at a constant rate. For
instance, Figure 2 shows a graph of the linear function f sxd − 3x 2 2 and a table of
sample values. Notice that whenever x increases by 0.1, the value of f sxd increases by
0.3. So f sxd increases three times as fast as x. Thus the slope of the graph of y − 3x 2 2,
namely 3, can be interpreted as the rate of change of y with respect to x.
y
y=3x-2
0
1
x
_2
figure 2
x
f sxd − 3x 2 2
1.0
1.1
1.2
1.3
1.4
1.5
1.0
1.3
1.6
1.9
2.2
2.5
Example 1 (a) As dry air moves upward, it expands and cools. If the ground temperature is 20°C
and the temperature at a height of 1 km is 10°C, express the temperature T (in °C) as a
function of the height h (in kilometers), assuming that a linear model is appropriate.
(b) Draw the graph of the function in part (a). What does the slope represent?
(c) What is the temperature at a height of 2.5 km?
SOLUTION
(a) Because we are assuming that T is a linear function of h, we can write
T − mh 1 b
We are given that T − 20 when h − 0, so
20 − m 0 1 b − b
In other words, the y-intercept is b − 20.
We are also given that T − 10 when h − 1, so
10 − m 1 1 20
The slope of the line is therefore m − 10 2 20 − 210 and the required linear function is
T − 210h 1 20
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25
Section 1.2 Mathematical Models: A Catalog of Essential Functions
T
(b) The graph is sketched in Figure 3. The slope is m − 210°Cykm, and this represents the rate of change of temperature with respect to height.
(c) At a height of h − 2.5 km, the temperature is
20
10
0
T=_10h+20
1
T − 210s2.5d 1 20 − 25°C
h
3
■
If there is no physical law or principle to help us formulate a model, we construct an
empirical model, which is based entirely on collected data. We seek a curve that “fits”
the data in the sense that it captures the basic trend of the data points.
FIGURE 3
Example 2 Table 1 lists the average carbon dioxide level in the atmosphere, measured in parts per million at Mauna Loa Observatory from 1980 to 2012. Use the data
in Table 1 to find a model for the carbon dioxide level.
SOLUTION We use the data in Table 1 to make the scatter plot in Figure 4, where t represents time (in years) and C represents the CO2 level (in parts per million, ppm).
C (ppm)
400
Table 1
Year
CO 2 level
(in ppm)
1980
1982
1984
1986
1988
1990
1992
1994
1996
338.7
341.2
344.4
347.2
351.5
354.2
356.3
358.6
362.4
Year
CO 2 level
(in ppm)
1998
2000
2002
2004
2006
2008
2010
2012
366.5
369.4
373.2
377.5
381.9
385.6
389.9
393.8
390
380
370
360
350
340
1980
1985
1990
1995
2000
2005
2010
t
FIGURE 4 Scatter plot for the average CO2 level Notice that the data points appear to lie close to a straight line, so it’s natural to
choose a linear model in this case. But there are many possible lines that approximate
these data points, so which one should we use? One possibility is the line that passes
through the first and last data points. The slope of this line is
393.8 2 338.7
55.1
−
− 1.721875 < 1.722
2012 2 1980
32
We write its equation as
C 2 338.7 − 1.722st 2 1980d
or
1
C − 1.722t 2 3070.86
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26
Chapter 1 Functions and Limits
Equation 1 gives one possible linear model for the carbon dioxide level; it is graphed
in Figure 5.
C (ppm)
400
390
380
370
360
350
FIGURE 5
340
Linear model through first
and last data points
A computer or graphing calculator
finds the regression line by the method
of least squares, which is to minimize
the sum of the squares of the vertical
distances between the data points and
the line. The details are explained in
Section 14.7.
1980
1985
1990
1995
2000
2005
2010
t
Notice that our model gives values higher than most of the actual CO2 levels. A
better linear model is obtained by a procedure from statistics called linear regression.
If we use a graphing calculator, we enter the data from Table 1 into the data editor and
choose the linear regression command. (With Maple we use the fit[leastsquare] command in the stats package; with Mathematica we use the Fit command.) The machine
gives the slope and y-intercept of the regression line as
m − 1.71262 b − 23054.14
So our least squares model for the CO2 level is
2
C − 1.71262t 2 3054.14
In Figure 6 we graph the regression line as well as the data points. Comparing with
Figure 5, we see that it gives a better fit than our previous linear model.
C (ppm)
400
390
380
370
360
350
Figure 6
The regression line
340
1980
1985
1990
1995
2000
2005
2010
t
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.2 Mathematical Models: A Catalog of Essential Functions
27
Example 3 Use the linear model given by Equa­tion 2 to estimate the average CO2
level for 1987 and to predict the level for the year 2020. According to this model, when
will the CO2 level exceed 420 parts per million?
Solution Using Equation 2 with t − 1987, we estimate that the average CO2 level in
1987 was
Cs1987d − s1.71262ds1987d 2 3054.14 < 348.84
This is an example of interpolation because we have estimated a value between
observed values. (In fact, the Mauna Loa Observatory reported that the average CO2
level in 1987 was 348.93 ppm, so our estimate is quite accurate.)
With t − 2020, we get
Cs2020d − s1.71262ds2020d 2 3054.14 < 405.35
So we predict that the average CO2 level in the year 2020 will be 405.4 ppm. This is an
example of extrapolation because we have predicted a value outside the time frame of
observations. Consequently, we are far less certain about the accuracy of our prediction.
Using Equation 2, we see that the CO2 level exceeds 420 ppm when
1.71262t 2 3054.14 . 420
Solving this inequality, we get
t.
3474.14
< 2028.55
1.71262
We therefore predict that the CO2 level will exceed 420 ppm by the year 2029. This
pre­diction is risky because it involves a time quite remote from our observations. In
fact, we see from Figure 6 that the trend has been for CO2 levels to increase rather more
rapidly in recent years, so the level might exceed 420 ppm well before 2029.
■
y
2
Polynomials
0
1
x
(a) y=≈+x+1
Psxd − a n x n 1 a n21 x n21 1 ∙ ∙ ∙ 1 a 2 x 2 1 a 1 x 1 a 0
where n is a nonnegative integer and the numbers a 0 , a 1, a 2 , . . . , a n are constants called
the coefficients of the polynomial. The domain of any polynomial is R − s2`, `d.
If the leading coefficient a n ± 0, then the degree of the polynomial is n. For example,
the function
Psxd − 2x 6 2 x 4 1 25 x 3 1 s2
y
2
1
A function P is called a polynomial if
x
(b) y=_2≈+3x+1
FIGURE 7
The graphs of quadratic functions
are parabolas.
is a polynomial of degree 6.
A polynomial of degree 1 is of the form Psxd − mx 1 b and so it is a linear function.
A polynomial of degree 2 is of the form Psxd − ax 2 1 bx 1 c and is called a quadratic
function. Its graph is always a parabola obtained by shifting the parabola y − ax 2, as we
will see in the next section. The parabola opens upward if a . 0 and downward if a , 0.
(See Figure 7.)
A polynomial of degree 3 is of the form
Psxd − ax 3 1 bx 2 1 cx 1 d a ± 0
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28
Chapter 1 Functions and Limits
and is called a cubic function. Figure 8 shows the graph of a cubic function in part (a)
and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see later why
the graphs have these shapes.
y
y
1
2
0
FIGURE 8
y
x
1
20
1
x
(a) y=˛-x+1
x
1
(b) y=x$-3≈+x
(c) y=3x%-25˛+60x
Polynomials are commonly used to model various quantities that occur in the natural
and social sciences. For instance, in Section 2.7 we will explain why economists often use
a polynomial Psxd to represent the cost of producing x units of a commodity. In the following example we use a quadratic function to model the fall of a ball.
Table 2
Time
(seconds)
Height
(meters)
0
1
2
3
4
5
6
7
8
9
450
445
431
408
375
332
279
216
143
61
Example 4 A ball is dropped from the upper observation deck of the CN Tower, 450 m
above the ground, and its height h above the ground is recorded at 1-second intervals in
Table 2. Find a model to fit the data and use the model to predict the time at which the
ball hits the ground.
Solution We draw a scatter plot of the data in Figure 9 and observe that a linear
model is inappropriate. But it looks as if the data points might lie on a parabola, so we
try a quadratic model instead. Using a graphing calculator or computer algebra system
(which uses the least squares method), we obtain the following quadratic model:
3
h − 449.36 1 0.96t 2 4.90t 2
h (meters)
h
400
400
200
200
0
2
4
6
8
t
(seconds)
0
2
4
6
8
FIGURE 9 FIGURE 10 Scatter plot for a falling ball
Quadratic model for a falling ball
t
In Figure 10 we plot the graph of Equation 3 together with the data points and see
that the quadratic model gives a very good fit.
The ball hits the ground when h − 0, so we solve the quadratic equation
24.90t 2 1 0.96t 1 449.36 − 0
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29
Section 1.2 Mathematical Models: A Catalog of Essential Functions
The quadratic formula gives
t−
20.96 6 ss0.96d2 2 4s24.90ds449.36d
2s24.90d
The positive root is t < 9.67, so we predict that the ball will hit the ground after about
9.7 seconds.
■
Power Functions
A function of the form f sxd − x a, where a is a constant, is called a power function. We
consider several cases.
(i) a − n, where n is a positive integer
The graphs of f sxd − x n for n − 1, 2, 3, 4, and 5 are shown in Figure 11. (These are polynomials with only one term.) We already know the shape of the graphs of y − x (a line
through the origin with slope 1) and y − x 2 [a parabola, see Example 1.1.2(b)].
y
y=≈
y
y=x
y
1
1
0
1
x
0
y=x #
y
x
0
1
x
0
y=x%
y
1
1
1
y=x$
1
1
x
0
1
FIGURE 11 Graphs of f sxd − x n for n − 1, 2, 3, 4, 5
The general shape of the graph of f sxd − x n depends on whether n is even or odd.
If n is even, then f sxd − x n is an even function and its graph is similar to the parabola
y − x 2. If n is odd, then f sxd − x n is an odd function and its graph is similar to that
of y − x 3. Notice from Figure 12, however, that as n increases, the graph of y − x n
becomes flatter near 0 and steeper when x > 1. (If x is small, then x 2 is smaller, x 3
is even smaller, x 4 is smaller still, and so on.)
| |
y
A family of functions is a collection
of functions whose equations are
related. Figure 12 shows two families
of power functions, one with even
powers and one with odd powers.
y=x ^
(_1, 1)
FIGURE 12 y
y=x $
y=≈
(1, 1)
0
y=x #
(1, 1)
y=x %
0
x
x
(_1, _1)
(ii) a − 1yn, where n is a positive integer
n
x is a root function. For n − 2 it is the square root
The function f sxd − x 1yn − s
function f sxd − sx , whose domain is f0, `d and whose graph is the upper half of the
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
30
Chapter 1 Functions and Limits
n
parabola x − y 2. [See Figure 13(a).] For other even values of n, the graph of y − s
x is
3
similar to that of y − sx . For n − 3 we have the cube root function f sxd − sx whose
domain is R (recall that every real number has a cube root) and whose graph is shown
n
3
in Figure 13(b). The graph of y − s
x for n odd sn . 3d is similar to that of y − s
x.
y
y
(1, 1)
(1, 1)
0
FIGURE 13 y=∆
1
0
x
1
0
x
(a) ƒ=œ„
Graphs of root functions
y
x
x
x
(b) ƒ=Œ„
(iii) a − 21
The graph of the reciprocal function f sxd − x 21 − 1yx is shown in Figure 14. Its
graph has the equation y − 1yx, or xy − 1, and is a hyperbola with the coordinate axes
as its asymptotes. This function arises in physics and chemistry in connection with
Boyle’s Law, which says that, when the temperature is constant, the volume V of a gas
is inversely proportional to the pressure P:
V−
C
P
where C is a constant. Thus the graph of V as a function of P (see Figure 15) has the
same general shape as the right half of Figure 14.
Power functions are also used to model species-area relationships (Exercises 30–31),
illumination as a function of distance from a light source (Exercise 29), and the period
of revolution of a planet as a function of its distance from the sun (Exercise 32).
Figure 14
The reciprocal function
V
Rational Functions
A rational function f is a ratio of two polynomials:
0
f sxd −
P
Figure 15
Volume as a function of pressure
at constant temperature
where P and Q are polynomials. The domain consists of all values of x such that Qsxd ± 0.
A simple example of a rational function is the function f sxd − 1yx, whose domain is
hx x ± 0j; this is the reciprocal function graphed in Figure 14. The function
|
y
f sxd −
FIGURE 16 ƒ=
2x 4 2 x 2 1 1
x2 2 4
|
is a rational function with domain hx x ± 62j. Its graph is shown in Figure 16.
20
0
Psxd
Qsxd
2
2x$-≈+1
≈-4
x
Algebraic Functions
A function f is called an algebraic function if it can be constructed using algebraic
operations (such as addition, subtraction, multiplication, division, and taking roots) starting with polynomials. Any rational function is automatically an algebraic function. Here
are two more examples:
f sxd − sx 2 1 1 tsxd −
x 4 2 16x 2
3
1 sx 2 2ds
x11
x 1 sx
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31
Section 1.2 Mathematical Models: A Catalog of Essential Functions
When we sketch algebraic functions in Chapter 3, we will see that their graphs can
assume a variety of shapes. Figure 17 illustrates some of the possibilities.
y
y
y
1
1
2
1
_3
x
0
(a) ƒ=xœ„„„„
x+3
FIGURE 17
x
5
0
(b) ©=$œ„„„„„„
≈-25
x
1
(c) h(x)=x@?#(x-2)@
An example of an algebraic function occurs in the theory of relativity. The mass of a
particle with velocity v is
m − f svd −
m0
s1 2 v 2yc 2
where m 0 is the rest mass of the particle and c − 3.0 3 10 5 kmys is the speed of light in a
vacuum.
Trigonometric Functions
Trigonometry and the trigonometric functions are reviewed on Reference Page 2 and also
in Appendix D. In calculus the convention is that radian measure is always used (except
when otherwise indicated). For example, when we use the function f sxd − sin x, it is
understood that sin x means the sine of the angle whose radian measure is x. Thus the
graphs of the sine and cosine functions are as shown in Figure 18.
The Reference Pages are located at
the back of the book.
y
_
_π
π
2
y
3π
2
1
_1
0
π
2
π
_π
2π
5π
2
3π
x
π
2
1
_1
(a) ƒ=sin x
FIGURE 18
_
π
0
π
2
3π
3π
2
2π
5π
2
x
(b) ©=cos x
Notice that for both the sine and cosine functions the domain is s2`, `d and the range
is the closed interval f21, 1g. Thus, for all values of x, we have
21 < sin x < 1 21 < cos x < 1
or, in terms of absolute values,
| sin x | < 1 | cos x | < 1
Also, the zeros of the sine function occur at the integer multiples of ; that is,
sin x − 0 when x − n n an integer
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32
Chapter 1 Functions and Limits
An important property of the sine and cosine functions is that they are periodic functions and have period 2. This means that, for all values of x,
sinsx 1 2d − sin x cossx 1 2d − cos x
The periodic nature of these functions makes them suitable for modeling repetitive phenomena such as tides, vibrating springs, and sound waves. For instance, in Example 1.3.4
we will see that a reasonable model for the number of hours of daylight in Philadelphia
t days after January 1 is given by the function
F
Lstd − 12 1 2.8 sin
2
st 2 80d
365
G
1
?
1 2 2 cos x
Solution This function is defined for all values of x except for those that make the
denominator 0. But
Example 5 What is the domain of the function f sxd −
1 2 2 cos x − 0
&?
cos x −
1
2
&?
x−
5
1 2n or x −
1 2n
3
3
where n is any integer (because the cosine function has period 2). So the domain of f
is the set of all real numbers except for the ones noted above.
■
y
The tangent function is related to the sine and cosine functions by the equation
tan x −
1
_
3π
2
0
_π _ π
2
π
2
π
3π
2
x
sin x
cos x
and its graph is shown in Figure 19. It is undefined whenever cos x − 0, that is, when
x − 6y2, 63y2, . . . . Its range is s2`, `d. Notice that the tangent function has per­iod :
tansx 1 d − tan x for all x
The remaining three trigonometric functions (cosecant, secant, and cotangent) are
the reciprocals of the sine, cosine, and tangent functions. Their graphs are shown in
Appendix D.
figure 19
y − tanxx
y=tan
y
1
0
Exponential Functions
y
1
(a) y=2®
figure 20
x
1
0
1
(b) y=(0.5)®
x
The exponential functions are the functions of the form f sxd − b x, where the base b is
a positive constant. The graphs of y − 2 x and y − s0.5d x are shown in Figure 20. In both
cases the domain is s2`, `d and the range is s0, `d.
Exponential functions will be studied in detail in Chapter 6, and we will see that they
are useful for modeling many natural phenomena, such as population growth (if b . 1)
and radioactive decay (if b , 1d.
Logarithmic Functions
The logarithmic functions f sxd − log b x, where the base b is a positive constant, are the
inverse functions of the exponential functions. They will be studied in Chapter 6. Figure
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33
Section 1.2 Mathematical Models: A Catalog of Essential Functions
y
21 shows the graphs of four logarithmic functions with various bases. In each case the
domain is s0, `d, the range is s2`, `d, and the function increases slowly when x . 1.
y=log™ x
y=log£ x
1
0
1
y=log∞ x
x
y=log¡¸ x
Example 6 Classify the following functions as one of the types of functions that we
have discussed.
(a) f sxd − 5 x
(b) tsxd − x 5
11x
(c) hsxd −
(d) ustd − 1 2 t 1 5t 4
1 2 sx
SOLUTION (a) f sxd − 5 x is an exponential function. (The x is the exponent.)
(b) tsxd − x 5 is a power function. (The x is the base.) We could also consider it to be a
polynomial of degree 5.
11x
(c) hsxd −
is an algebraic function.
1 2 sx
figure 21
(d) ustd − 1 2 t 1 5t 4 is a polynomial of degree 4.
■
1. 2 Exercises
1–2 Classify each function as a power function, root function,
polynomial (state its degree), rational function, algebraic function,
trigonometric function, exponential function, or logarithmic function.
(b) tsxd − sx
1.(a) f sxd − log 2 x
(c) hsxd −
2x 3
1 2 x2
F
(d) ustd − 1 2 1.1t 1 2.54t 2
g
(f ) wsd − sin cos 2
(e) vstd − 5 t
2.(a) y − x
f
(b) y − x (c) y − x 2 s2 2 x 3 d
(e) y −
y
4
3
(d) y − s
x
(c) y − x 3
(b) y − 3 x
4. (a) y − 3x
(d) y − tan t 2 cos t
x
sx 2 1
3
11s
x
3
s
11s
(f ) y −
3–4 Match each equation with its graph. Explain your choices.
(Don’t use a computer or graphing calculator.)
3. (a) y − x 2 (b) y − x 5 (c) y − x 8
y
0
f
g
G
5–6 Find the domain of the function.
5. f sxd −
cos x
1
6. tsxd −
1 2 sin x
1 2 tan x
h
x
7. (a)Find an equation for the family of linear functions with
slope 2 and sketch several members of the family.
(b)Find an equation for the family of linear functions such
that f s2d − 1 and sketch several members of the family.
(c)Which function belongs to both families?
8.
What do all members of the family of linear functions
f sxd − 1 1 msx 1 3d have in common? Sketch several
members of the family.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
34
Chapter 1 Functions and Limits
9.
What do all members of the family of linear functions
f sxd − c 2 x have in common? Sketch several members of
the family.
10.Find expressions for the quadratic functions whose graphs
are shown.
y
(_2, 2)
f
(4, 2)
0
3
x
g
y
(0, 1)
0
x
(1, _2.5)
11.Find an expression for a cubic function f if f s1d − 6 and
f s21d − f s0d − f s2d − 0.
12. R
ecent studies indicate that the average surface temperature of the earth has been rising steadily. Some scientists
have modeled the temperature by the linear function
T − 0.02t 1 8.50, where T is temperature in °C and t
represents years since 1900.
(a)What do the slope and T-intercept represent?
(b)Use the equation to predict the average global surface
temperature in 2100.
13.If the recommended adult dosage for a drug is D (in mg),
then to determine the appropriate dosage c for a child of
age a, pharmacists use the equation c − 0.0417Dsa 1 1d.
Suppose the dosage for an adult is 200 mg.
(a)Find the slope of the graph of c. What does it represent?
(b)What is the dosage for a newborn?
14.The manager of a weekend flea market knows from past
experience that if he charges x dollars for a rental space at
the market, then the number y of spaces he can rent is given
by the equation y − 200 2 4x.
(a)Sketch a graph of this linear function. (Remember that
the rental charge per space and the number of spaces
rented can’t be negative quantities.)
(b)What do the slope, the y-intercept, and the x-intercept of
the graph represent?
15.The relationship between the Fahrenheit sFd and Celsius
sCd temperature scales is given by the linear function
F − 95 C 1 32.
(a) Sketch a graph of this function.
(b)What is the slope of the graph and what does it represent? What is the F-intercept and what does it represent?
16. J ason leaves Detroit at 2:00 pm and drives at a constant speed
west along I-94. He passes Ann Arbor, 40 mi from Detroit, at
2:50 pm.
(a)Express the distance traveled in terms of the time
elapsed.
(b)Draw the graph of the equation in part (a).
(c)What is the slope of this line? What does it represent?
17. Biologists
have noticed that the chirping rate of crickets of
a certain species is related to temperature, and the relationship appears to be very nearly linear. A cricket produces
113 chirps per minute at 70°F and 173 chirps per minute
at 80°F.
(a)Find a linear equation that models the temperature T as
a function of the number of chirps per minute N.
(b)What is the slope of the graph? What does it represent?
(c)If the crickets are chirping at 150 chirps per minute,
estimate the temperature.
18. T
he manager of a furniture factory finds that it costs $2200
to manufacture 100 chairs in one day and $4800 to produce
300 chairs in one day.
(a)Express the cost as a function of the number of chairs
produced, assuming that it is linear. Then sketch the
graph.
(b)What is the slope of the graph and what does it represent?
(c)What is the y-intercept of the graph and what does it
represent?
19. A
t the surface of the ocean, the water pressure is the same
as the air pressure above the water, 15 lbyin2. Below the surface, the water pressure increases by 4.34 lbyin2 for every
10 ft of descent.
(a)Express the water pressure as a function of the depth
below the ocean surface.
(b)At what depth is the pressure 100 lbyin2?
20. T
he monthly cost of driving a car depends on the number
of miles driven. Lynn found that in May it cost her $380 to
drive 480 mi and in June it cost her $460 to drive 800 mi.
(a)Express the monthly cost C as a function of the distance
driven d, assuming that a linear relationship gives a
suitable model.
(b)Use part (a) to predict the cost of driving 1500 miles per
month.
(c)Draw the graph of the linear function. What does the
slope represent?
(d)What does the C-intercept represent?
(e)Why does a linear function give a suitable model in this
situation?
21–22 For each scatter plot, decide what type of function you
might choose as a model for the data. Explain your choices.
21.
(a)
y
0
(b)
x
y
0
x
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
35
Section 1.2 Mathematical Models: A Catalog of Essential Functions
22.
(a) y
(b) y
0
x
0
x
he table shows (lifetime) peptic ulcer rates (per 100 popula; 23. T
tion) for various family incomes as reported by the National
Health Interview Survey.
Income
Ulcer rate
(per 100 population)
$4,000
$6,000
$8,000
$12,000
$16,000
$20,000
$30,000
$45,000
$60,000
14.1
13.0
13.4
12.5
12.0
12.4
10.5
9.4
8.2
; 25.Anthropologists use a linear model that relates human femur
(thighbone) length to height. The model allows an anthropologist to determine the height of an individual when only a
partial skeleton (including the femur) is found. Here we find
the model by analyzing the data on femur length and height
for the eight males given in the following table.
(a)Make a scatter plot of the data.
(b) Find and graph the regression line that models the data.
(c)An anthropologist finds a human femur of length
53 cm. How tall was the person?
Femur length
(cm)
Height
(cm)
Femur length
(cm)
Height
(cm)
50.1
48.3
45.2
44.7
178.5
173.6
164.8
163.7
44.5
42.7
39.5
38.0
168.3
165.0
155.4
155.8
; 26.When laboratory rats are exposed to asbestos fibers, some
of them develop lung tumors. The table lists the results of
several experiments by different scientists.
(a)Find the regression line for the data.
(b)Make a scatter plot and graph the regression line.
Does the regression line appear to be a suitable model
for the data?
(c)What does the y-intercept of the regression line represent?
(a)Make a scatter plot of these data and decide whether a
linear model is appropriate.
(b)Find and graph a linear model using the first and last
data points.
(c) Find and graph the least squares regression line.
(d)Use the linear model in part (c) to estimate the ulcer
rate for an income of $25,000.
(e)According to the model, how likely is someone with an
income of $80,000 to suffer from peptic ulcers?
(f )Do you think it would be reasonable to apply the model
to someone with an income of $200,000?
iologists have observed that the chirping rate of crickets of
; 24. B
a certain species appears to be related to temperature. The
table shows the chirping rates for various temperatures.
(a) Make a scatter plot of the data.
(b) Find and graph the regression line.
(c)Use the linear model in part (b) to estimate the chirping
rate at 100°F.
Temperature
(°F)
Chirping rate
(chirpsymin)
Temperature
(°F)
Chirping rate
(chirpsymin)
50
55
60
65
70
20
46
79
91
113
75
80
85
90
140
173
198
211
Asbestos
Percent of mice
exposure
that develop
(fibersymL)
lung tumors
50
400
500
900
1100
2
6
5
10
26
Asbestos
Percent of mice
exposure
that develop
(fibersymL)
lung tumors
1600
1800
2000
3000
42
37
38
50
; 27.The table shows world average daily oil consumption from
1985 to 2010 measured in thousands of barrels per day.
(a)Make a scatter plot and decide whether a linear model
is appropriate.
(b)Find and graph the regression line.
(c)Use the linear model to estimate the oil consumption in
2002 and 2012.
Years
since 1985
Thousands of barrels
of oil per day
0
5
10
15
20
25
60,083
66,533
70,099
76,784
84,077
87,302
Source: US Energy Information Administration
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
36
Chapter 1 Functions and Limits
he table shows average US retail residential prices of
; 28. T
electricity from 2000 to 2012, measured in cents per
kilowatt hour.
(a)Make a scatter plot. Is a linear model appropriate?
(b)Find and graph the regression line.
(c)Use your linear model from part (b) to estimate the
average retail price of electricity in 2005 and 2013.
he table shows the number N of species of reptiles and
; 31. T
amphibians inhabiting Caribbean islands and the area A of
the island in square miles.
(a)Use a power function to model N as a function of A.
(b)The Caribbean island of Dominica has area 291 mi 2.
How many species of reptiles and amphibians would
you expect to find on Dominica?
Years since 2000
CentsykWh
Island
A
N
0
2
4
6
8
10
12
8.24
8.44
8.95
10.40
11.26
11.54
11.58
Saba
Monserrat
Puerto Rico
Jamaica
Hispaniola
Cuba
4
40
3,459
4,411
29,418
44,218
5
9
40
39
84
76
Source: US Energy Information Administration
29.Many physical quantities are connected by inverse square
laws, that is, by power functions of the form f sxd − kx 22.
In particular, the illumination of an object by a light source
is inversely proportional to the square of the distance from
the source. Suppose that after dark you are in a room with
just one lamp and you are trying to read a book. The light is
too dim and so you move halfway to the lamp. How much
brighter is the light?
he table shows the mean (average) distances d of the
; 32. T
planets from the sun (taking the unit of measurement to be
the distance from planet Earth to the sun) and their periods
T (time of revolution in years).
(a) Fit a power model to the data.
(b)Kepler’s Third Law of Planetary Motion states that
“ The square of the period of revolution of a planet
is propor­tional to the cube of its mean distance from
the sun.”
Does your model corroborate Kepler’s Third Law?
30. I t makes sense that the larger the area of a region, the larger
the number of species that inhabit the region. Many ecologists have modeled the species-area relation with a power
function and, in particular, the number of species S of bats
living in caves in central Mexico has been related to the
surface area A of the caves by the equation S − 0.7A0.3.
(a)The cave called Misión Imposible near Puebla,
Mexico, has a surface area of A − 60 m2. How many
species of bats would you expect to find in that cave?
(b)If you discover that four species of bats live in a cave,
estimate the area of the cave.
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
d
T
0.387
0.723
1.000
1.523
5.203
9.541
19.190
30.086
0.241
0.615
1.000
1.881
11.861
29.457
84.008
164.784
In this section we start with the basic functions we discussed in Section 1.2 and obtain
new functions by shifting, stretching, and reflecting their graphs. We also show how to
combine pairs of functions by the standard arithmetic operations and by composition.
Transformations of Functions
By applying certain transformations to the graph of a given function we can obtain
the graphs of related functions. This will give us the ability to sketch the graphs of
many functions quickly by hand. It will also enable us to write equations for given graphs.
Let’s first consider translations. If c is a positive number, then the graph of y − f sxd 1 c
is just the graph of y − f sxd shifted upward a distance of c units (because each y-coordinate is increased by the same number c). Likewise, if tsxd − f sx 2 cd, where c . 0, then
the value of t at x is the same as the value of f at x 2 c (c units to the left of x). ThereCopyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
37
Section 1.3 New Functions from Old Functions
fore the graph of y − f sx 2 cd is just the graph of y − f sxd shifted c units to the right
(see Figure 1).
Vertical and Horizontal Shifts Suppose c . 0. To obtain the graph of
y − f sxd 1 c, shift the graph of y − f sxd a distance c units upward
y − f sxd 2 c, shift the graph of y − f sxd a distance c units downward
y − f sx 2 cd, shift the graph of y − f sxd a distance c units to the right
y − f sx 1 cd, shift the graph of y − f sxd a distance c units to the left
y
y
y=ƒ+c
y=f(x+c)
c
y =ƒ
c
0
y=cƒ
(c>1)
y=f(_x)
y=f(x-c)
y=ƒ
y= 1c ƒ
c
x
c
x
0
y=ƒ-c
y=_ƒ
Figure 1 Translating the graph of f
Figure 2 Stretching and reflecting the graph of f
Now let’s consider the stretching and reflecting transformations. If c . 1, then the
graph of y − cf sxd is the graph of y − f sxd stretched by a factor of c in the vertical
direction (because each y-coordinate is multiplied by the same number c). The graph of
y − 2f sxd is the graph of y − f sxd reflected about the x-axis because the point sx, yd is
replaced by the point sx, 2yd. (See Figure 2 and the following chart, where the results of
other stretching, shrinking, and reflecting transformations are also given.)
Vertical and Horizontal Stretching and Reflecting Suppose c . 1. To obtain the
graph of
y − cf sxd, stretch the graph of y − f sxd vertically by a factor of c
y − s1ycdf sxd, shrink the graph of y − f sxd vertically by a factor of c
y − f scxd, shrink the graph of y − f sxd horizontally by a factor of c
y − f sxycd, stretch the graph of y − f sxd horizontally by a factor of c
y − 2f sxd, reflect the graph of y − f sxd about the x-axis
y − f s2xd, reflect the graph of y − f sxd about the y-axis
Figure 3 illustrates these stretching transformations when applied to the cosine function
with c − 2. For instance, in order to get the graph of y − 2 cos x we multiply the y-coordiCopyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
38
Chapter 1 Functions and Limits
nate of each point on the graph of y − cos x by 2. This means that the graph of y − cos x
gets stretched vertically by a factor of 2.
y
y=2 cos x
y
2
y=cos x
2
1
1
y=
2
0
y=cos 1 x
1
cos x
x
1
y=cos 2x
2
0
x
y=cos x
Figure 3
Example 1 Given the graph of y − sx , use transformations to graph y − sx 2 2,
y − sx 2 2 , y − 2sx , y − 2sx , and y − s2x .
SOLUTION The graph of the square root function y − sx , obtained from Figure 1.2.13(a), is shown in Figure 4(a). In the other parts of the figure we sketch
y − sx 2 2 by shifting 2 units downward, y − sx 2 2 by shifting 2 units to the
right, y − 2sx by reflecting about the x-axis, y − 2sx by stretching vertically by a
factor of 2, and y − s2x by reflecting about the y-axis.
y
y
y
y
y
y
1
0
x
1
x
0
0
x
2
x
0
x
0
x
0
_2
(a) y=œ„x
(b) y=œ„-2
x
(d) y=_ œ„x
(c) y=œ„„„„
x-2
(f ) y=œ„„
_x
(e) y=2 œ„x
Figure 4
■
Example 2 Sketch the graph of the function f sxd − x 2 1 6x 1 10.
SOLUTION Completing the square, we write the equation of the graph as
y − x 2 1 6x 1 10 − sx 1 3d2 1 1
This means we obtain the desired graph by starting with the parabola y − x 2 and shifting 3 units to the left and then 1 unit upward (see Figure 5).
y
y
1
(_3, 1)
0
Figure 5
(a) y=≈
x
_3
_1
0
(b) y=(x+3)@+1
x
■
Example 3 Sketch the graphs of the following functions.
(a) y − sin 2x
(b) y − 1 2 sin x
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
39
Section 1.3 New Functions from Old Functions
SOLUTION
(a) We obtain the graph of y − sin 2x from that of y − sin x by compressing horizontally by a factor of 2. (See Figures 6 and 7.) Thus, whereas the period of y − sin x is
2, the period of y − sin 2x is 2y2 − .
y
y
y=sin x
1
0
π
2
π
y=sin 2x
1
x
0 π π
4
x
π
2
FIGURE 7
FIGURE 6
(b) To obtain the graph of y − 1 2 sin x, we again start with y − sin x. We reflect
about the x-axis to get the graph of y − 2sin x and then we shift 1 unit upward to get
y − 1 2 sin x. (See Figure 8.)
y
y=1-sin x
2
1
0
FIGURE 8
π
2
π
3π
2
x
2π
■
Example 4 Figure 9 shows graphs of the number of hours of daylight as functions of
the time of the year at several latitudes. Given that Philadelphia is located at approximately 408N latitude, find a function that models the length of daylight at Philadelphia.
20
18
16
14
12
20° N
30° N
40° N
50° N
Hours 10
8
FIGURE 9
Graph of the length of daylight from
March 21 through December 21
at various latitudes
Source: Adapted from L. Harrison,
Daylight, Twilight, Darkness and Time
(New York: Silver, Burdett, 1935), 40.
6
60° N
4
2
0
Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
SOLUTION Notice that each curve resembles a shifted and stretched sine function.
By looking at the blue curve we see that, at the latitude of Philadelphia, daylight
lasts about 14.8 hours on June 21 and 9.2 hours on December 21, so the amplitude
of the curve (the factor by which we have to stretch the sine curve vertically) is
1
2 s14.8 2 9.2d − 2.8.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
40
Chapter 1 Functions and Limits
By what factor do we need to stretch the sine curve horizontally if we measure the
time t in days? Because there are about 365 days in a year, the period of our model
should be 365. But the period of y − sin t is 2, so the horizontal stretching factor
is 2y365.
We also notice that the curve begins its cycle on March 21, the 80th day of the
year, so we have to shift the curve 80 units to the right. In addition, we shift it 12 units
upward. Therefore we model the length of daylight in Philadelphia on the tth day of the
year by the function
F
y
Lstd − 12 1 2.8 sin
_1
0
1
x
(a) y=≈-1
y
2
st 2 80d
365
G
■
Another transformation of some interest is taking the absolute value of a function. If
y − f sxd , then according to the definition of absolute value, y − f sxd when f sxd > 0
and y − 2f sxd when f sxd , 0. This tells us how to get the graph of y − f sxd from the
graph of y − f sxd: the part of the graph that lies above the x-axis remains the same; the
part that lies below the x-axis is reflected about the x-axis.
|
|
|
|
Example 5 Sketch the graph of the function y − | x 2 2 1 |.
SOLUTION We first graph the parabola y − x 2 2 1 in Figure 10(a) by shifting the
_1
0
1
(b) y=| ≈-1 |
figure 10
x
parabola y − x 2 downward 1 unit. We see that the graph lies below the x-axis when
21 , x , 1, so we reflect that part of the graph about the x-axis to obtain the graph of
y − x 2 2 1 in Figure 10(b).
■
|
|
Combinations of Functions
Two functions f and t can be combined to form new functions f 1 t, f 2 t, ft, and fyt
in a manner similar to the way we add, subtract, multiply, and divide real numbers. The
sum and difference functions are defined by
s f 1 tdsxd − f sxd 1 tsxd s f 2 tdsxd − f sxd 2 tsxd
If the domain of f is A and the domain of t is B, then the domain of f 1 t is the intersection A > B because both f sxd and tsxd have to be defined. For example, the domain
of f sxd − sx is A − f0, `d and the domain of tsxd − s2 2 x is B − s2`, 2g, so the
domain of s f 1 tdsxd − sx 1 s2 2 x is A > B − f0, 2g.
Similarly, the product and quotient functions are defined by
SD
s ftdsxd − f sxd tsxd f
f sxd
sxd −
t
tsxd
The domain of ft is A > B. Because we can’t divide by 0, the domain of fyt is therefore
hx [ A > B tsxd ± 0j. For instance, if f sxd − x 2 and tsxd − x 2 1, then the domain
of the rational function s fytdsxd − x 2ysx 2 1d is hx x ± 1j, or s2`, 1d ø s1, `d.
There is another way of combining two functions to obtain a new function. For
example, suppose that y − f sud − su and u − tsxd − x 2 1 1. Since y is a function
of u and u is, in turn, a function of x, it follows that y is ultimately a function of x.
We compute this by substitution:
|
|
y − f sud − f stsxdd − f sx 2 1 1d − sx 2 1 1
The procedure is called composition because the new function is composed of the two
given functions f and t.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.3 New Functions from Old Functions
In general, given any two functions f and t, we start with a number x in the domain
of t and calculate tsxd. If this number tsxd is in the domain of f, then we can calculate
the value of f stsxdd. Notice that the output of one function is used as the input to the next
function. The result is a new function hsxd − f s tsxdd obtained by substituting t into f. It
is called the composition (or composite) of f and t and is denoted by f 8 t (“ f circle t”).
x (input)
g
©
41
f•g
f
Definition Given two functions f and t, the composite function f 8 t (also called
the composition of f and t) is defined by
s f 8 tdsxd − f stsxdd
f { ©} (output)
FIGURE 11 The f 8 t machine is composed of
the t machine (first) and then the
f machine.
The domain of f 8 t is the set of all x in the domain of t such that tsxd is in the domain
of f. In other words, s f 8 tdsxd is defined whenever both tsxd and f s tsxdd are defined.
Figure 11 shows how to picture f 8 t in terms of machines.
Example 6 If f sxd − x 2 and tsxd − x 2 3, find the composite functions f 8 t and t 8 f.
SOLUTION We have
s f 8 tdsxd − f stsxdd − f sx 2 3d − sx 2 3d2
st 8 f dsxd − ts f sxdd − tsx 2 d − x 2 2 3
n
NOTE You can see from Example 6 that, in general, f 8 t ± t 8 f. Remember, the
notation f 8 t means that the function t is applied first and then f is applied second. In
Example 6, f 8 t is the function that first subtracts 3 and then squares; t 8 f is the function
that first squares and then subtracts 3.
Example 7 If f sxd − sx and tsxd − s2 2 x , find each of the following functions
and their domains.
(a) f 8 t (b) t 8 f (c) f 8 f (d) t 8 t
SOLUTION
(a)
4
s f 8 tdsxd − f stsxdd − f (s2 2 x ) − ss2 2 x − s
22x
|
|
The domain of f 8 t is hx 2 2 x > 0j − hx x < 2j − s2`, 2g.
(b)
If 0 < a < b, then a 2 < b 2.
s t 8 f dsxd − ts f sxdd − t (sx ) − s2 2 sx
For sx to be defined we must have x > 0. For s2 2 sx to be defined we must have
2 2 sx > 0, that is, sx < 2, or x < 4. Thus we have 0 < x < 4, so the domain of
t 8 f is the closed interval f0, 4g.
(c)
4
s f 8 f dsxd − f s f sxdd − f (sx ) − ssx − s
x
The domain of f 8 f is f0, `d.
(d)
st 8 tdsxd − tstsxdd − t (s2 2 x ) − s2 2 s2 2 x
This expression is defined when both 2 2 x > 0 and 2 2 s2 2 x > 0. The first
inequality means x < 2, and the second is equivalent to s2 2 x < 2, or 2 2 x < 4, or
x > 22. Thus 22 < x < 2, so the domain of t 8 t is the closed interval f22, 2g.
■
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42
Chapter 1 Functions and Limits
It is possible to take the composition of three or more functions. For instance, the composite function f 8 t 8 h is found by first applying h, then t, and then f as follows:
s f 8 t 8 hdsxd − f stshsxddd
Example 8 Find f 8 t 8 h if f sxd − xysx 1 1d, tsxd − x 10, and hsxd − x 1 3.
SOLUTIOn
s f 8 t 8 hdsxd − f stshsxddd − f stsx 1 3dd
− f ssx 1 3d10 d −
sx 1 3d10
sx 1 3d10 1 1
■
So far we have used composition to build complicated functions from simpler ones.
But in calculus it is often useful to be able to decompose a complicated function into
simpler ones, as in the following example.
Example 9
Given Fsxd − cos2sx 1 9d, find functions f , t, and h such that F − f 8 t 8 h.
SOLUTION Since Fsxd − fcossx 1 9dg 2, the formula for F says: First add 9, then take
the cosine of the result, and finally square. So we let
hsxd − x 1 9 tsxd − cos x f sxd − x 2
Then
s f 8 t 8 hdsxd − f stshsxddd − f stsx 1 9dd − f scossx 1 9dd
− fcossx 1 9dg 2 − Fsxd
1. 3
■
Exercises
1. S
uppose the graph of f is given. Write equations for the
graphs that are obtained from the graph of f as follows.
(a) Shift 3 units upward.
(b) Shift 3 units downward.
(c) Shift 3 units to the right.
(d) Shift 3 units to the left.
(e) Reflect about the x-axis.
(f ) Reflect about the y-axis.
(g) Stretch vertically by a factor of 3.
(h) Shrink vertically by a factor of 3.
2.Explain how each graph is obtained from the graph
of y − f sxd.
(a) y − f sxd 1 8
(b) y − f sx 1 8d
(c) y − 8 f sxd
(e) y − 2f sxd 2 1
(d) y − f s8xd
(f ) y − 8 f s 81 xd
3.The graph of y − f sxd is given. Match each equation with
its graph and give reasons for your choices.
(a) y − f sx 2 4d
(b) y − f sxd 1 3
(c) y − 13 f sxd
(d) y − 2f sx 1 4d
y
@
6
!
f
3
#
$
_6
_3
%
0
3
6
x
_3
4.The graph of f is given. Draw the graphs of the following
functions.
(a) y − f sxd 2 3
(b) y − f sx 1 1d
1
(c) y − 2 f sxd
(d) y − 2f sxd
y
2
0
1
x
(e) y − 2 f sx 1 6d
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.3 New Functions from Old Functions
5. T
he graph of f is given. Use it to graph the following
functions.
(a) y − f s2xd
(b) y − f ( 12 x)
(c) y − f s2xd
(d) y − 2f s2xd
|
x
1
y=œ„„„„„„
3x-≈
1.5
0
x
3
7.
y
3
y
_1 0
_4
5
_1
x
_2.5
x
8.(a)How is the graph of y − 2 sin x related to the graph of
y − sin x? Use your answer and Figure 6 to sketch the
graph of y − 2 sin x.
(b)How is the graph of y − 1 1 sx related to the graph of
y − sx ? Use your answer and Figure 4(a) to sketch the
graph of y − 1 1 sx .
9–24 Graph the function by hand, not by plotting points, but by
starting with the graph of one of the standard functions given in
Section 1.2, and then applying the appropriate transformations.
9. y − 2x 2
| |
|
22. y −
|
S D
1
tan x 2
4
4
|
24. y − cos x
|
25.The city of New Orleans is located at latitude 30°N. Use
Figure 9 to find a function that models the number of hours
of daylight at New Orleans as a function of the time of year.
To check the accuracy of your model, use the fact that on
March 31 the sun rises at 5:51 am and sets at 6:18 pm in
New Orleans.
26.A variable star is one whose brightness alternately increases
and decreases. For the most visible variable star, Delta
Cephei, the time between periods of maximum brightness is
5.4 days, the average brightness (or magnitude) of the star
is 4.0, and its brightness varies by 60.35 magnitude. Find
a function that models the brightness of Delta Cephei as a
function of time.
y
2
20. y − x 2 2
23. y − sx 2 1
6–7 The graph of y − s3x 2 x 2 is given. Use transformations
to create a function whose graph is as shown.
0
19. y − sin ( 21 x )
|
1
6.
18. y − 3 2 2 cos x
21. y − x 2 2
y
0
17. y − 2 2 sx
43
10. y − sx 2 3d2
1
x
11. y − x 3 1 1
12. y − 1 2
13. y − 2 cos 3x
14. y − 2 sx 1 1
15. y − x 2 2 4x 1 5
16. y − 1 1 sin x
27. S
ome of the highest tides in the world occur in the Bay of
Fundy on the Atlantic Coast of Canada. At Hopewell Cape
the water depth at low tide is about 2.0 m and at high tide
it is about 12.0 m. The natural period of oscillation is
about 12 hours and on June 30, 2009, high tide occurred
at 6:45 am. Find a function involving the cosine function
that models the water depth Dstd (in meters) as a function
of time t (in hours after midnight) on that day.
28.In a normal respiratory cycle the volume of air that moves
into and out of the lungs is about 500 mL. The reserve and
residue volumes of air that remain in the lungs occupy
about 2000 mL and a single respiratory cycle for an average
human takes about 4 seconds. Find a model for the total
volume of air Vstd in the lungs as a function of time.
| |
| |
| |
29. (a)How is the graph of y − f ( x ) related to the graph of f ?
(b) Sketch the graph of y − sin x .
(c) Sketch the graph of y − s x .
30.Use the given graph of f to sketch the graph of y − 1yf sxd.
Which features of f are the most important in sketching
y − 1yf sxd? Explain how they are used.
y
1
0
1
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
44
Chapter 1 Functions and Limits
31–32 Find (a) f 1 t, (b) f 2 t, (c) f t, and (d) fyt and state their
domains.
31. f sxd − x 3 1 2x 2,
tsxd − 3x 2 2 1
32. f sxd − s3 2 x ,
tsxd − sx 2 2 1
(d) s t 8 f ds6d
(e) s t 8 tds22d
y
g
tsxd − x 2 1 x
34. f sxd − x 3 2 2,
tsxd − 1 2 4x
35. f sxd − sx 1 1,
37. f sxd − x 1
1
,
x
tsxd −
x
38. f sxd −
,
11x
x11
x12
g
tsxd − sin 2x
1
0
|
tsxd − 2 x, hsxd − sx
41. f sxd − sx 2 3 ,
tsxd − x 2, hsxd − x 3 1 2
tsxd −
x
3
, hsxd − s
x
x21
43. Fsxd − s2 x 1 x 2 d 444. Fsxd − cos2 x
3
x
s
46. Gsxd −
3
11s
x
47. vstd − secst 2 d tanst 2 d
48. ustd −
Î
3
x
11x
tan t
1 1 tan t
49–51 Express the function in the form f 8 t 8 h.
8
49. Rsxd − ssx 2 1 50. Hsxd − s
21 x
| |
51. Sstd − sin2scos td
52. Use the table to evaluate each expression.
(a)f s ts1dd
(b) ts f s1dd
(d) ts ts1dd
(e) s t 8 f ds3d
x
55.A stone is dropped into a lake, creating a circular ripple that
travels outward at a speed of 60 cmys.
(a)Express the radius r of this circle as a function of the
time t (in seconds).
(b)If A is the area of this circle as a function of the radius,
find A 8 r and interpret it.
43–48 Express the function in the form f 8 t.
45. Fsxd −
1
f
40. f sxd − x 2 4 ,
42. f sxd − tan x,
x
2
y
39–42 Find f 8 t 8 h.
39. f sxd − 3x 2 2, tsxd − sin x, hsxd − x 2
|
0
54. U
se the given graphs of f and t to estimate the value of
f s tsxdd for x − 25, 24, 23, . . . , 5. Use these estimates to
sketch a rough graph of f 8 t.
tsxd − 4x 2 3
tsxd − x 2 1 1
36. f sxd − sin x,
f
2
33–38 Find the functions (a) f 8 t, (b) t 8 f , (c) f 8 f , and (d) t 8 t
and their domains.
33. f sxd − 3x 1 5,
(f) s f 8 f ds4d
(c) f s f s1dd
(f ) s f 8 tds6d
x
1
2
3
4
5
6
f sxd
3
1
4
2
2
5
tsxd
6
3
2
1
2
3
53.Use the given graphs of f and t to evaluate each expression, or
explain why it is undefined.
(a) f s ts2dd
(b) ts f s0dd
(c) s f 8 tds0d
56.A spherical balloon is being inflated and the radius of the
balloon is increasing at a rate of 2 cmys.
(a)Express the radius r of the balloon as a function of the
time t (in seconds).
(b)If V is the volume of the balloon as a function of the
radius, find V 8 r and interpret it.
57.A ship is moving at a speed of 30 kmyh parallel to a straight
shoreline. The ship is 6 km from shore and it passes a lighthouse at noon.
(a)Express the distance s between the lighthouse and the ship
as a function of d, the distance the ship has traveled since
noon; that is, find f so that s − f sdd.
(b)Express d as a function of t, the time elapsed since noon;
that is, find t so that d − tstd.
(c)Find f 8 t. What does this function represent?
58.An airplane is flying at a speed of 350 miyh at an altitude of
one mile and passes directly over a radar station at time t − 0.
(a)Express the horizontal distance d (in miles) that the plane
has flown as a function of t.
(b)Express the distance s between the plane and the radar
station as a function of d.
(c) Use composition to express s as a function of t.
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Section 1.4 The Tangent and Velocity Problems
59. The Heaviside function H is defined by
Hstd −
H
45
(c)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 7 seconds and the voltage
is gradually increased to 100 volts over a period of 25 seconds. Write a formula for Vstd in terms of Hstd for t < 32.
0 if t , 0
1 if t > 0
It is used in the study of electric circuits to represent the
sudden surge of electric current, or voltage, when a switch is
instantaneously turned on.
(a) Sketch the graph of the Heaviside function.
(b)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 0 and 120 volts are
applied instantaneously to the circuit. Write a formula
for Vstd in terms of Hstd.
(c)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 5 seconds and 240 volts
are applied instantaneously to the circuit. Write a formula
for Vstd in terms of Hstd. (Note that starting at t − 5
corre­sponds to a translation.)
60.The Heaviside function defined in Exercise 59 can also be
used to define the ramp function y − ctHstd, which represents a gradual increase in voltage or current in a circuit.
(a) Sketch the graph of the ramp function y − tHstd.
(b)Sketch the graph of the voltage Vstd in a circuit if the
switch is turned on at time t − 0 and the voltage is gradually increased to 120 volts over a 60-second time interval.
Write a formula for Vstd in terms of Hstd for t < 60.
61.Let f and t be linear functions with equations f sxd − m1 x 1 b1
and tsxd − m 2 x 1 b 2. Is f 8 t also a linear function? If so,
what is the slope of its graph?
62.If you invest x dollars at 4% interest compounded annually,
then the amount Asxd of the investment after one year is
Asxd − 1.04x. Find A 8 A, A 8 A 8 A, and A 8 A 8 A 8 A. What
do these compositions represent? Find a formula for the composition of n copies of A.
63. (a)If tsxd − 2x 1 1 and hsxd − 4x 2 1 4x 1 7, find a function f such that f 8 t − h. (Think about what operations
you would have to perform on the formula for t to end up
with the formula for h.)
(b)If f sxd − 3x 1 5 and hsxd − 3x 2 1 3x 1 2, find a function t such that f 8 t − h.
64.If f sxd − x 1 4 and hsxd − 4x 2 1, find a function t such
that t 8 f − h.
65.Suppose t is an even function and let h − f 8 t. Is h always an
even function?
66. Suppose t is an odd function and let h − f 8 t. Is h always an
odd function? What if f is odd? What if f is even?
In this section we see how limits arise when we attempt to find the tangent to a curve or
the velocity of an object.
The Tangent Problem
t
(a)
P
t
Example 1 Find an equation of the tangent line to the parabola y − x 2 at the
l
point Ps1, 1d.
(b)
FIGURE 1 C
The word tangent is derived from the Latin word tangens, which means “touching.” Thus
a tangent to a curve is a line that touches the curve. In other words, a tangent line should
have the same direction as the curve at the point of contact. How can this idea be made
precise?
For a circle we could simply follow Euclid and say that a tangent is a line that
intersects the circle once and only once, as in Figure 1(a). For more complicated curves
this definition is inadequate. Figure l(b) shows two lines l and t passing through a point
P on a curve C. The line l intersects C only once, but it certainly does not look like what
we think of as a tangent. The line t, on the other hand, looks like a tangent but it intersects
C twice.
To be specific, let’s look at the problem of trying to find a tangent line t to the parabola
y − x 2 in the following example.
SOLUTION We will be able to find an equation of the tangent line t as soon as we know
its slope m. The difficulty is that we know only one point, P, on t, whereas we need two
points to compute the slope. But observe that we can compute an approximation to m
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
46
Chapter 1 Functions and Limits
y
Q {x, ≈}
y=≈
by choosing a nearby point Qsx, x 2 d on the parabola (as in Figure 2) and computing the
slope mPQ of the secant line PQ. [A secant line, from the Latin word secans, meaning
cutting, is a line that cuts (intersects) a curve more than once.]
We choose x ± 1 so that Q ± P. Then
t
P (1, 1)
mPQ −
x
0
x2 2 1
x21
For instance, for the point Qs1.5, 2.25d we have
FIGURE 2
mPQ −
x
mPQ
2
1.5
1.1
1.01
1.001
3
2.5
2.1
2.01
2.001
x
mPQ
0
0.5
0.9
0.99
0.999
1
1.5
1.9
1.99
1.999
2.25 2 1
1.25
−
− 2.5
1.5 2 1
0.5
The tables in the margin show the values of mPQ for several values of x close to 1. The
closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mPQ is
to 2. This suggests that the slope of the tangent line t should be m − 2.
We say that the slope of the tangent line is the limit of the slopes of the secant lines,
and we express this symbolically by writing
lim mPQ − m and lim
Q lP
y
Q
xl1
x2 2 1
−2
x21
Assuming that the slope of the tangent line is indeed 2, we use the point-slope form
of the equation of a line [y 2 y1 − msx 2 x 1d, see Appendix B] to write the equation of
the tangent line through s1, 1d as
y 2 1 − 2sx 2 1d or y − 2x 2 1
Figure 3 illustrates the limiting process that occurs in this example. As Q approaches
P along the parabola, the corresponding secant lines rotate about P and approach the
tangent line t.
y
y
t
t
t
Q
P
P
0
x
P
0
x
Q
0
x
Q approaches P from the right
y
y
y
t
Q
t
P
0
Q
x
t
P
0
x
0
Q
P
x
Q approaches P from the left
FIGURE 3 ■
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Section 1.4 The Tangent and Velocity Problems
TEC In Visual 1.4 you can see how
the process in Figure 3 works for
additional functions.
t
Q
0.00
0.02
0.04
0.06
0.08
0.10
100.00
81.87
67.03
54.88
44.93
36.76
47
Many functions that occur in science are not described by explicit equations; they are
defined by experimental data. The next example shows how to estimate the slope of the
tangent line to the graph of such a function.
Example 2 The flash unit on a camera operates by storing charge on a capacitor and
releasing it suddenly when the flash is set off. The data in the table describe the charge
Q remaining on the capacitor (measured in microcoulombs) at time t (measured in
seconds after the flash goes off). Use the data to draw the graph of this function and
estimate the slope of the tangent line at the point where t − 0.04. [Note: The slope of
the tangent line represents the electric current flowing from the capacitor to the flash
bulb (measured in microamperes).]
SOLUTION In Figure 4 we plot the given data and use them to sketch a curve that
approximates the graph of the function.
Q (microcoulombs)
100
90
80
70
60
50
FIGURE 4 0
0.02
0.04
0.06
0.08
0.1
t (seconds)
Given the points Ps0.04, 67.03d and Rs0.00, 100.00d on the graph, we find that the
slope of the secant line PR is
mPR −
R
mPR
(0.00, 100.00)
(0.02, 81.87)
(0.06, 54.88)
(0.08, 44.93)
(0.10, 36.76)
2824.25
2742.00
2607.50
2552.50
2504.50
100.00 2 67.03
− 2824.25
0.00 2 0.04
The table at the left shows the results of similar calculations for the slopes of other
secant lines. From this table we would expect the slope of the tangent line at t − 0.04
to lie somewhere between 2742 and 2607.5. In fact, the average of the slopes of the
two closest secant lines is
1
2 s2742
2 607.5d − 2674.75
So, by this method, we estimate the slope of the tangent line to be about 2675.
Another method is to draw an approximation to the tangent line at P and measure
the sides of the triangle ABC, as in Figure 5.
Q (microcoulombs)
100
90
80
A
P
70
60
50
FIGURE 5 0
B
0.02
C
0.04
0.06
0.08
0.1
t (seconds)
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
48
Chapter 1 Functions and Limits
This gives an estimate of the slope of the tangent line as
he physical meaning of the answer
T
in Example 2 is that the electric current flowing from the capacitor to
the flash bulb after 0.04 seconds is
about 2670 microamperes.
2
| AB | < 2 80.4 2 53.6 − 2670
0.06 2 0.02
| BC |
■
The Velocity Problem
If you watch the speedometer of a car as you travel in city traffic, you see that the
speed doesn’t stay the same for very long; that is, the velocity of the car is not constant.
We assume from watching the speedometer that the car has a definite velocity at each
moment, but how is the “instantaneous” velocity defined? Let’s investigate the example
of a falling ball.
Example 3 Suppose that a ball is dropped from the upper observation deck of
the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after
5 seconds.
SOLUTION Through experiments carried out four centuries ago, Galileo discovered that
the distance fallen by any freely falling body is proportional to the square of the time
it has been falling. (This model for free fall neglects air resistance.) If the distance
fallen after t seconds is denoted by sstd and measured in meters, then Galileo’s law is
expressed by the equation
Steve Allen / Stockbyte / Getty Images
sstd − 4.9t 2
The difficulty in finding the velocity after 5 seconds is that we are dealing with a
single instant of time st − 5d, so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval
of a tenth of a second from t − 5 to t − 5.1:
average velocity −
The CN Tower in Toronto was the
tallest freestanding building in the
world for 32 years.
change in position
time elapsed
−
ss5.1d 2 ss5d
0.1
−
4.9s5.1d2 2 4.9s5d2
− 49.49 mys
0.1
The following table shows the results of similar calculations of the average velocity
over successively smaller time periods.
Time interval
Average velocity smysd
5<t<6
53.9
5 < t < 5.1
49.49
5 < t < 5.05
49.245
5 < t < 5.01
5 < t < 5.001
49.049
49.0049
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49
Section 1.4 The Tangent and Velocity Problems
It appears that as we shorten the time period, the average velocity is becoming closer to
49 mys. The instantaneous velocity when t − 5 is defined to be the limiting value of
these average velocities over shorter and shorter time periods that start at t − 5. Thus it
appears that the (instantaneous) velocity after 5 seconds is
s
s=4.9t @
Q
v − 49 mys
slope of secant line
average velocity
P
a
0
a+h
t
s
s=4.9t @
You may have the feeling that the calculations used in solving this problem are very
similar to those used earlier in this section to find tangents. In fact, there is a close
connection between the tangent problem and the problem of finding velocities. If we
draw the graph of the distance function of the ball (as in Figure 6) and we consider the
points Psa, 4.9a 2 d and Qsa 1 h, 4.9sa 1 hd2 d on the graph, then the slope of the secant
line PQ is
mPQ −
slope of tangent line
instantaneous velocity
P
0
a
t
FIGURE 6 1. 4
■
4.9sa 1 hd2 2 4.9a 2
sa 1 hd 2 a
which is the same as the average velocity over the time interval fa, a 1 hg. Therefore
the velocity at time t − a (the limit of these average velocities as h approaches 0) must
be equal to the slope of the tangent line at P (the limit of the slopes of the secant lines).
Examples 1 and 3 show that in order to solve tangent and velocity problems we must
be able to find limits. After studying methods for computing limits in the next four sections, we will return to the problems of finding tangents and velocities in Chapter 2.
Exercises
1.A tank holds 1000 gallons of water, which drains from the
bottom of the tank in half an hour. The values in the table
show the volume V of water remaining in the tank (in gallons)
after t minutes.
t smind
5
10
15
20
25
30
V sgald
694
444
250
111
28
0
(a)If P is the point s15, 250d on the graph of V, find the
slopes of the secant lines PQ when Q is the point on the
graph with t − 5, 10, 20, 25, and 30.
(b)Estimate the slope of the tangent line at P by averaging
the slopes of two secant lines.
(c)Use a graph of the function to estimate the slope of the
tangent line at P. (This slope represents the rate at which
the water is flowing from the tank after 15 minutes.)
2.A cardiac monitor is used to measure the heart rate of a patient
after surgery. It compiles the number of heartbeats after t min­
utes. When the data in the table are graphed, the slope of the
tangent line represents the heart rate in beats per minute.
t smind
Heartbeats
36
38
40
42
44
2530
2661
2806
2948
3080
The monitor estimates this value by calculating the slope of
a secant line. Use the data to estimate the patient’s heart rate
after 42 minutes using the secant line between the points with
the given values of t.
(a) t − 36 and t − 42
(b) t − 38 and t − 42
(c) t − 40 and t − 42
(d) t − 42 and t − 44
What are your conclusions?
3. The point Ps2, 21d lies on the curve y − 1ys1 2 xd.
(a)If Q is the point sx, 1ys1 2 xdd, use your calculator to find
the slope of the secant line PQ (correct to six decimal
places) for the following values of x:
(i) 1.5
(ii) 1.9
(iii) 1.99
(iv) 1.999
(v) 2.5
(vi) 2.1
(vii) 2.01 (viii) 2.001
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50
Chapter 1 Functions and Limits
(b)Using the results of part (a), guess the value of the slope
of the tangent line to the curve at Ps2, 21d.
(c)Using the slope from part (b), find an equation of the
tangent line to the curve at Ps2, 21d.
7.The table shows the position of a motorcyclist after acceler­
ating from rest.
4.The point Ps0.5, 0d lies on the curve y − cos x.
(a)If Q is the point s x, cos xd, use your calculator to find
the slope of the secant line PQ (correct to six decimal
places) for the following values of x:
(i) 0
(ii) 0.4
(iii) 0.49
(iv) 0.499
(v) 1
(vi) 0.6
(vii) 0.51
(viii) 0.501
(b)Using the results of part (a), guess the value of the slope
of the tangent line to the curve at Ps0.5, 0d.
(c)Using the slope from part (b), find an equation of the
tangent line to the curve at Ps0.5, 0d.
(d)Sketch the curve, two of the secant lines, and the
tangent line.
t ssecondsd
0
s (feet)
0
1
2
3
4
5
6
4.9 20.6 46.5 79.2 124.8 176.7
(a)Find the average velocity for each time period:
(i) f2, 4g
(ii) f3, 4g
(iii) f4, 5g
(iv) f4, 6g
(b)Use the graph of s as a function of t to estimate the
instantaneous velocity when t − 3.
8.The displacement (in centimeters) of a particle moving back
and forth along a straight line is given by the equation of
motion s − 2 sin t 1 3 cos t, where t is measured in
seconds.
(a)Find the average velocity during each time period:
(i) [1, 2]
(ii) [1, 1.1]
(iii) [1, 1.01]
(iv) [1, 1.001]
(b)Estimate the instantaneous velocity of the particle
when t − 1.
5. I f a ball is thrown into the air with a velocity of 40 ftys, its
height in feet t seconds later is given by y − 40t 2 16t 2.
(a)Find the average velocity for the time period beginning
when t − 2 and lasting
(i) 0.5 seconds
(ii) 0.1 seconds
(iii) 0.05 seconds
(iv) 0.01 seconds
(b) Estimate the instantaneous velocity when t − 2.
9. The point Ps1, 0d lies on the curve y − sins10yxd.
(a)If Q is the point sx, sins10yxdd, find the slope of the
secant line PQ (correct to four decimal places) for
x − 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9.
Do the slopes appear to be approaching a limit?
(b)Use a graph of the curve to explain why the slopes of
;
the secant lines in part (a) are not close to the slope of
the tangent line at P.
(c)By choosing appropriate secant lines, estimate the slope
of the tangent line at P.
6.If a rock is thrown upward on the planet Mars with a
velocity of 10 mys, its height in meters t seconds later is
given by y − 10t 2 1.86t 2.
(a) Find the average velocity over the given time intervals:
(i) [1, 2]
(ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
(b) Estimate the instantaneous velocity when t − 1.
Having seen in the preceding section how limits arise when we want to find the tangent
to a curve or the velocity of an object, we now turn our attention to limits in general and
numerical and graphical methods for computing them.
Let’s investigate the behavior of the function f defined by f sxd − x 2 2 x 1 2 for
values of x near 2. The following table gives values of f sxd for values of x close to 2 but
not equal to 2.
x
1.0
1.5
1.8
1.9
1.95
1.99
1.995
1.999
f sxd
2.000000
2.750000
3.440000
3.710000
3.852500
3.970100
3.985025
3.997001
x
3.0
2.5
2.2
2.1
2.05
2.01
2.005
2.001
f sxd
8.000000
5.750000
4.640000
4.310000
4.152500
4.030100
4.015025
4.003001
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51
Section 1.5 The Limit of a Function
y
ƒ
approaches
4.
y=≈-x+2
4
From the table and the graph of f (a parabola) shown in Figure 1 we see that the closer
x is to 2 (on either side of 2), the closer f sxd is to 4. In fact, it appears that we can make the
values of f sxd as close as we like to 4 by taking x sufficiently close to 2. We express this by
saying “the limit of the function f sxd − x 2 2 x 1 2 as x approaches 2 is equal to 4.” The
notation for this is
lim sx 2 2 x 1 2d − 4
x l2
In general, we use the following notation.
0
x
2
As x approaches 2,
1 Intuitive Definition of a Limit Suppose f sxd is defined when x is near the
number a. (This means that f is defined on some open interval that contains a,
except possibly at a itself.) Then we write
fiGure 1
lim f sxd − L
xla
“the limit of f sxd, as x approaches a, equals L”
and say
if we can make the values of f sxd arbitrarily close to L (as close to L as we like) by
restricting x to be sufficiently close to a (on either side of a) but not equal to a.
Roughly speaking, this says that the values of f sxd approach L as x approaches a. In
other words, the values of f sxd tend to get closer and closer to the number L as x gets
closer and closer to the number a (from either side of a) but x ± a. (A more precise definition will be given in Section 1.7.)
An alternative notation for
lim f sxd − L
xla
f sxd l L as x l a
is
which is usually read “ f sxd approaches L as x approaches a.”
Notice the phrase “but x ± a” in the definition of limit. This means that in find­ing the
limit of f sxd as x approaches a, we never consider x − a. In fact, f sxd need not even be
defined when x − a. The only thing that matters is how f is defined near a.
Figure 2 shows the graphs of three functions. Note that in part (c), f sad is not defined
and in part (b), f sad ± L. But in each case, regardless of what happens at a, it is true
that lim x l a f sxd − L.
y
y
y
L
L
L
0
a
x
(a)
0
a
(b)
x
0
a
(c)
figure 2 lim f sxd − L in all three cases
xla
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x
52
Chapter 1 Functions and Limits
Example 1 Guess the value of lim
x l1
x,1
f sxd
0.5
0.9
0.99
0.999
0.9999
0.666667
0.526316
0.502513
0.500250
0.500025
x.1
f sxd
1.5
1.1
1.01
1.001
1.0001
0.400000
0.476190
0.497512
0.499750
0.499975
x21
.
x2 2 1
SOLUTION Notice that the function f sxd − sx 2 1dysx 2 2 1d is not defined when x − 1,
but that doesn’t matter because the definition of lim x l a f sxd says that we consider
values of x that are close to a but not equal to a.
The tables at the left give values of f sxd (correct to six decimal places) for values of
x that approach 1 (but are not equal to 1). On the basis of the values in the tables, we
make the guess that
x21
lim
− 0.5
■
xl1 x2 2 1
Example 1 is illustrated by the graph of f in Figure 3. Now let’s change f slightly by
giving it the value 2 when x − 1 and calling the resulting function t:
H
tsxd −
x21
x2 2 1
if x ± 1
2
if x − 1
This new function t still has the same limit as x approaches 1. (See Figure 4.)
1
0.5
y
y
2
y=
x-1
≈-1
y=©
0.5
0
0.5
1
x
0
figure 3
1
x
Figure 4
st 2 1 9 2 3
.
tl0
t2
SOLUTION The table lists values of the function for several values of t near 0.
Example 2 Estimate the value of lim
st 2 1 9 2 3
t2
t
61.0
0.162277 . . .
60.5
0.165525 . . .
60.1
0.166620 . . .
60.05
60.01
0.166655 . . .
0.166666 . . .
As t approaches 0, the values of the function seem to approach 0.1666666 . . . and so
we guess that
lim
tl0
1
st 2 1 9 2 3
−
2
t
6
■
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Section 1.5 The Limit of a Function
t
60.001
60.0001
60.00001
60.000001
st 2 1 9 2 3
t2
0.166667
0.166670
0.167000
0.000000
www.stewartcalculus.com
For a further explanation of why
calculators sometimes give false
values, click on Lies My Calculator
and Computer Told Me. In particular, see the section called The Perils
of Subtraction.
53
In Example 2 what would have happened if we had taken even smaller values of t?
The table in the margin shows the results from one calculator; you can see that something
strange seems to be happening.
If you try these calculations on your own calculator you might get different values,
but eventually you will get the value 0 if you make t sufficiently small. Does this mean
that the answer is really 0 instead of 16 ? No, the value of the limit is 16, as we will show in
the next section. The problem is that the calculator gave false values because st 2 1 9 is
very close to 3 when t is small. (In fact, when t is sufficiently small, a calculator’s value
for st 2 1 9 is 3.000. . . to as many digits as the calculator is capable of carrying.)
Something similar happens when we try to graph the function
f std −
st 2 1 9 2 3
t2
of Example 2 on a graphing calculator or computer. Parts (a) and (b) of Figure 5 show
quite accurate graphs of f , and when we use the trace mode (if available) we can estimate
easily that the limit is about 16. But if we zoom in too much, as in parts (c) and (d), then we
get inaccurate graphs, again because of rounding errors from the subtraction.
0.2
0.2
0.1
0.1
sad 25 < t < 5
sbd 20.1 < t < 0.1
sdd 21027 < t < 1027
scd 21026 < t < 1026
FIGURE 5 Example 3 Guess the value of lim
xl0
x
sin x
x
61.0
60.5
60.4
60.3
60.2
60.1
60.05
60.01
60.005
60.001
0.84147098
0.95885108
0.97354586
0.98506736
0.99334665
0.99833417
0.99958339
0.99998333
0.99999583
0.99999983
sin x
.
x
SOLUTION The function f sxd − ssin xdyx is not defined when x − 0. Using a calculator (and remembering that, if x [ R, sin x means the sine of the angle whose radian
measure is x), we construct a table of values correct to eight decimal places. From the
table at the left and the graph in Figure 6 we guess that
lim
xl0
sin x
−1
x
This guess is in fact correct, as will be proved in Chapter 2 using a geometric argument.
y
_1
figure 6
Example 4 Investigate lim sin
xl0
1
y=
0
1
sinƒx
x
x
.
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
■
54
Chapter 1 Functions and Limits
Computer Algebra Systems
Computer algebra systems (CAS)
have commands that compute limits.
In order to avoid the types of pitfalls
demonstrated in Examples 2, 4, and
5, they don’t find limits by numerical
experimentation. Instead, they use more
sophisticated techniques such as computing infinite series. If you have access
to a CAS, use the limit command to
compute the limits in the examples of
this section and to check your answers
in the exercises of this chapter.
SOLUTION Again the function f sxd − sinsyxd is undefined at 0. Evaluating the
function for some small values of x, we get
f ( 12 ) − sin 2 − 0
f ( 13) − sin 3 − 0 f ( 14) − sin 4 − 0
f s1d − sin − 0
f s0.1d − sin 10 − 0 f s0.01d − sin 100 − 0
Similarly, f s0.001d − f s0.0001d − 0. On the basis of this information we might be
tempted to guess that
lim sin
−0
xl0
x
but this time our guess is wrong. Note that although f s1ynd − sin n − 0 for any
integer n, it is also true that f sxd − 1 for infinitely many values of x (such as 2y5 or
2y101) that approach 0. You can see this from the graph of f shown in Figure 7.
y
y=sin(π/x)
1
_1
x
1
_1
figure 7
The dashed lines near the y-axis indicate that the values of sinsyxd oscillate
between 1 and 21 infinitely often as x approaches 0. (See Exercise 45.)
Since the values of f sxd do not approach a fixed number as x approaches 0,
lim sin
xl0
x
cos 5x
x 1
10,000
1
0.5
0.1
0.05
0.01
1.000028
0.124920
0.001088
0.000222
0.000101
x
0.005
0.001
3
x3 1
cos 5x
10,000
0.00010009
0.00010000
S
Example 5 Find lim x 3 1
xl0
does not exist
x
■
D
cos 5x
.
10,000
SOLUTION As before, we construct a table of values. From the first table in the margin
it appears that
lim
xl0
S
x3 1
cos 5x
10,000
D
−0
But if we persevere with smaller values of x, the second table suggests that
lim
xl0
S
x3 1
cos 5x
10,000
D
− 0.000100 −
1
10,000
In Section 1.8 we will see that lim x l 0 cos 5x − 1; then it follows that the limit is
0.0001.
■
Examples 4 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is
easy to guess the wrong value if we use inappropriate values of x, but it is difficult to
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
55
Section 1.5 The Limit of a Function
know when to stop calculating values. And, as the discussion after Example 2 shows,
sometimes calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits.
One-Sided Limits
Example 6 The Heaviside function H is defined by
Hstd −
y
1
0
t
FIGURE 8 The Heaviside function
H
0 if t , 0
1 if t > 0
[This function is named after the electrical engineer Oliver Heaviside (1850–1925) and
can be used to describe an electric current that is switched on at time t − 0.] Its graph
is shown in Figure 8.
As t approaches 0 from the left, Hstd approaches 0. As t approaches 0 from the right,
Hstd approaches 1. There is no single number that Hstd approaches as t approaches 0.
Therefore lim t l 0 Hstd does not exist.
■
We noticed in Example 6 that Hstd approaches 0 as t approaches 0 from the left and
Hstd approaches 1 as t approaches 0 from the right. We indicate this situation symbolically by writing
lim Hstd − 0 and lim1 Hstd − 1
t l 02
tl0
The notation t l 0 2 indicates that we consider only values of t that are less than 0. Likewise, t l 0 1 indicates that we consider only values of t that are greater than 0.
2 Definition of One-Sided Limits We write
lim f sxd − L
x la2
and say the left-hand limit of f sxd as x approaches a [or the limit of f sxd as
x approaches a from the left] is equal to L if we can make the values of f sxd
arbitrarily close to L by taking x to be sufficiently close to a with x less than a.
Notice that Definition 2 differs from Definition 1 only in that we require x to be less
than a. Similarly, if we require that x be greater than a, we get “the right-hand limit of
f sxd as x approaches a is equal to L” and we write
lim f sxd − L
x l a1
Thus the notation x l a1 means that we consider only x greater than a. These definitions are illustrated in Figure 9.
y
y
L
ƒ
0
FIGURE 9 x
(a) lim ƒ=L
x a_
a
ƒ
L
x
0
a
x
x
(b) lim ƒ=L
x a+
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
56
Chapter 1 Functions and Limits
By comparing Definition l with the definitions of one-sided limits, we see that the
following is true.
3 lim f sxd − L if and only if lim f sxd − L and lim f sxd − L
xla
x l a2
x l a1
y
Example 7 The graph of a function t is shown in Figure 10. Use it to state the values
(if they exist) of the following:
4
3
(a) lim2 tsxd (b) lim1 tsxd (c) lim tsxd
y=©
xl2
xl2
xl2
(d) lim2 tsxd (e) lim1 tsxd (f) lim tsxd
1
xl5
0
1
2
3
4
5
x
xl5
xl5
SOLUTION From the graph we see that the values of tsxd approach 3 as x approaches 2
from the left, but they approach 1 as x approaches 2 from the right. Therefore
(a) lim2 tsxd − 3 and (b) lim1 tsxd − 1
FIGURE 10 xl2
xl2
(c) Since the left and right limits are different, we conclude from (3) that limx l 2 tsxd
does not exist.
The graph also shows that
(d) lim2 tsxd − 2 and (e) lim1 tsxd − 2
xl5
xl5
(f) This time the left and right limits are the same and so, by (3), we have
lim tsxd − 2
xl5
Despite this fact, notice that ts5d ± 2.
Infinite Limits
x
1
x2
61
60.5
60.2
60.1
60.05
60.01
60.001
1
4
25
100
400
10,000
1,000,000
1
if it exists.
xl0 x2
SOLUTION As x becomes close to 0, x 2 also becomes close to 0, and 1yx 2 becomes very
large. (See the table in the margin.) In fact, it appears from the graph of the function
f sxd − 1yx 2 shown in Figure 11 that the values of f sxd can be made arbitrarily large
by taking x close enough to 0. Thus the values of f sxd do not approach a number, so
lim x l 0 s1yx 2 d does not exist.
■
Example 8 Find lim
To indicate the kind of behavior exhibited in Example 8, we use the notation
y
lim
xl0
y=
0
figure 11 ■
1
≈
x
1
−`
x2
This does not mean that we are regarding ` as a number. Nor does it mean that the limit
exists. It simply expresses the particular way in which the limit does not exist: 1yx 2 can
be made as large as we like by taking x close enough to 0.
In general, we write symbolically
lim f sxd − `
xla
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.5 The Limit of a Function
57
to indicate that the values of f sxd tend to become larger and larger (or “increase without
bound”) as x becomes closer and closer to a.
4 Intuitive Definition of an Infinite Limit Let f be a function defined on both
sides of a, except possibly at a itself. Then
lim f sxd − `
xla
means that the values of f sxd can be made arbitrarily large (as large as we please)
by taking x sufficiently close to a, but not equal to a.
Another notation for lim x l a f sxd − ` is
y
y=ƒ
f sxd l ` as x l a
Again, the symbol ` is not a number, but the expression lim x l a f sxd − ` is often read as
0
a
x
“the limit of f sxd, as x approaches a, is infinity”
x=a
figure 12 lim f sxd − `
or
“ f sxd becomes infinite as x approaches a”
or
“ f sxd increases without bound as x approaches a”
xla
When we say a number is “large negative,” we mean that it is negative but its
magnitude (absolute value) is large.
This definition is illustrated graphically in Figure 12.
A similar sort of limit, for functions that become large negative as x gets close to a, is
defined in Definition 5 and is illustrated in Figure 13.
5 Definition Let f be a function defined on both sides of a, except possibly at
a itself. Then
y
lim f sxd − 2`
x=a
0
xla
a
x
means that the values of f sxd can be made arbitrarily large negative by taking x
sufficiently close to a, but not equal to a.
y=ƒ
figure 13
lim f sxd − 2`
The symbol lim x l a f sxd − 2` can be read as “the limit of f sxd, as x approaches a, is
negative infinity” or “ f sxd decreases without bound as x approaches a.” As an example
we have
xla
S D
lim 2
x l0
1
x2
− 2`
Similar definitions can be given for the one-sided infinite limits
lim f sxd − `
x l a2
lim f sxd − 2`
x l a2
lim f sxd − `
x l a1
lim f sxd − 2`
x l a1
remembering that x l a2 means that we consider only values of x that are less than a,
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58
Chapter 1 Functions and Limits
and similarly x l a1 means that we consider only x . a. Illustrations of these four cases
are given in Figure 14.
y
y
a
0
x
(a) lim ƒ=`
x
y
a
0
x
a
0
x
(b) lim ƒ=`
a_
y
(c) lim ƒ=_`
a+
x
a
0
x
x
(d) lim ƒ=_`
a_
x
a+
FIGURE 14 6 Definition The vertical line x − a is called a vertical asymptote of the
curve y − f sxd if at least one of the following statements is true:
lim f sxd − `
x la
lim f sxd − 2`
x la
lim f sxd − `
x l a2
lim f sxd − 2`
x l a2
lim f sxd − `
x l a1
lim f sxd − 2`
x l a1
For instance, the y-axis is a vertical asymptote of the curve y − 1yx 2 because
lim x l 0 s1yx 2 d − `. In Figure 14 the line x − a is a vertical asymptote in each of
the four cases shown. In general, knowledge of vertical asymptotes is very useful in
sketching graphs.
Example 9 Find lim1
x l3
2x
2x
and lim2
.
x
l
3
x23
x23
SOLUTION If x is close to 3 but larger than 3, then the denominator x 2 3 is a small
positive number and 2x is close to 6. So the quotient 2xysx 2 3d is a large positive
number. [For instance, if x − 3.01 then 2xysx 2 3d − 6.02y0.01 − 602.] Thus, intuitively, we see that
lim
y
x l 31
2x
y= x-3
5
x
0
x=3
Likewise, if x is close to 3 but smaller than 3, then x 2 3 is a small negative number
but 2x is still a positive number (close to 6). So 2xysx 2 3d is a numerically large
negative number. Thus
lim2
x l3
Figure 15
2x
−`
x23
2x
− 2`
x23
The graph of the curve y − 2xysx 2 3d is given in Figure 15. The line x − 3 is a
vertical asymptote.
■
Example 10 Find the vertical asymptotes of f sxd − tan x.
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59
Section 1.5 The Limit of a Function
SOLUTION Because
y
there are potential vertical asymptotes where cos x − 0. In fact, since cos x l 01 as
x l sy2d2 and cos x l 02 as x l sy2d1, whereas sin x is positive (near 1) when x
is near y2, we have
1
3π _π
_ 2
_
π
2
sin x
cos x
tan x −
0
π
2
π
3π
2
x
lim
x l sy2d2
tan x − ` and lim
x l sy2d1
tan x − 2`
This shows that the line x − y2 is a vertical asymptote. Similar reasoning shows
that the lines x − y2 1 n, where n is an integer, are all vertical asymptotes of
f sxd − tan x. The graph in Figure 16 confirms this.
Figure 16
y − tan x
1. Explain in your own words what is meant by the equation
(d) lim f sxd
(e) f s3d
xl3
lim f sxd − 5
xl2
y
Is it possible for this statement to be true and yet f s2d − 3?
Explain.
4
2. Explain what it means to say that
2
lim f sxd − 3 and lim1 f sxd − 7
x l 12
x l1
In this situation is it possible that lim x l 1 f sxd exists?
Explain.
3.Explain the meaning of each of the following.
(a) lim f sxd − `
(b) lim1 f sxd − 2`
x l 23
xl4
4.Use the given graph of f to state the value of each quantity,
if it exists. If it does not exist, explain why.
(a) lim2 f sxd
(b) lim1 f sxd
(c) lim f sxd
x l2
xl2
(d) f s2d
xl2
(e) lim f sxd
(f) f s4d
xl4
0
4
x
6.For the function h whose graph is given, state the value of
each quantity, if it exists. If it does not exist, explain why.
(a)
lim hsxd
(b)
x l 232
lim hsxd
(c) lim hsxd
x l 231
x l 23
(d) hs23d
(e) lim2 hsxd
(f) lim1 hsxd
(g) lim hsxd
(h) hs0d
(i) lim hsxd
(j) hs2d
(k) lim1 hsxd
(l)
x l0
xl 0
xl0
y
2
xl2
x l5
lim hsxd
x l 52
4
y
2
0
2
4
x
5.For the function f whose graph is given, state the value of
each quantity, if it exists. If it does not exist, explain why.
(a) lim f sxd
(b) lim2 f sxd
(c) lim1 f sxd
xl1
xl3
_4
_2
0
2
4
6
x
xl3
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■
60
chapter 1 Functions and Limits
7.For the function t whose graph is given, state the value of each
quantity, if it exists. If it does not exist, explain why.
(a) lim2 tstd
(b) lim1 tstd
(c) lim tstd
tl0
tl0
tl0
(d) lim2 tstd
(e) lim1 tstd
(f ) lim tstd
(g) ts2d
(h) lim tstd
tl2
tl2
and explain the significance of these one-sided limits.
f(t)
300
tl2
tl4
150
y
4
0
2
2
8
12
16
t
11–12 Sketch the graph of the function and use it to determine the
values of a for which lim x l a f sxd exists.
t
4
4
H
H
1 1 x if x , 21
11. f sxd − x 2
if 21 < x , 1
2 2 x if x > 1
8.For the function A whose graph is shown, state the following.
(a) lim Asxd
(b) lim2 Asxd
x l23
x l2
(c) lim1 Asxd
(d) lim Asxd
x l2
x l21
1 1 sin x if x , 0
12. f sxd − cos x
if 0 < x < sin x
if x . (e) The equations of the vertical asymptotes
; 13–14 Use the graph of the function f to state the value of each
limit, if it exists. If it does not exist, explain why.
(a) lim2 f sxd (b) lim1 f sxd (c) lim f sxd
y
xl0
0
_3
2
x
5
xl0
xl0
1
x2 1 x
13. f sxd −
1yx 14. f sxd −
112
sx 3 1 x 2
15–18 Sketch the graph of an example of a function f that
satisfies all of the given conditions.
9.For the function f whose graph is shown, state the following.
(a) lim f sxd
(b) lim f sxd
(c) lim f sxd
x l27
x l23
(d) lim2 f sxd
(e) lim1 f sxd
xl6
xl0
15. lim2 f sxd − 21, lim1 f sxd − 2, f s0d − 1
xl0
16. lim f sxd − 1, lim2 f sxd − 22, lim1 f sxd − 2,
xl0
xl6
(f ) The equations of the vertical asymptotes.
xl0
xl3
xl3
f s0d − 21, f s3d − 1
17. lim1 f sxd − 4, lim2 f sxd − 2, lim f sxd − 2,
xl3
y
xl3
x l 22
f s3d − 3, f s22d − 1
18. lim2 f sxd − 2, lim1 f sxd − 0, lim2 f sxd − 3,
xl0
_7
_3
0
6
x
xl0
xl4
lim1 f sxd − 0, f s0d − 2, f s4d − 1
xl4
19–22 Guess the value of the limit (if it exists) by evaluating the
function at the given numbers (correct to six decimal places).
10. A
patient receives a 150-mg injection of a drug every 4 hours.
The graph shows the amount f std of the drug in the blood­
stream after t hours. Find
lim f std and lim1 f std
tl 122
tl 12
19.lim
x l3
x 2 2 3x
, x2 2 9
x − 3.1, 3.05, 3.01, 3.001, 3.0001,
2.9, 2.95, 2.99, 2.999, 2.9999
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Section 1.5 The Limit of a Function
20. lim
x l 23
x 2 2 3x
,
x2 2 9
x − 22.5, 22.9, 22.95, 22.99, 22.999, 22.9999,
23.5, 23.1, 23.05, 23.01, 23.001, 23.0001
sin x
, x − 61, 60.5, 60.2, 60.1, 60.05, 60.01
x 1 tan x
21. lim
xl0
40. (a) Find the vertical asymptotes of the function
x2 1 1
y−
3x 2 2x 2
(b)Confirm your answer to part (a) by graphing the
;
function.
hl 0
h − 60.5, 60.1, 60.01, 60.001, 60.0001
x l1
;
23–26 Use a table of values to estimate the value of the limit.
If you have a graphing device, use it to confirm your result
graphically.
sin 3
1 1 p9
23. lim
24. lim
l 0 tan 2
p l 21 1 1 p 15
25. lim1 x x 26. lim
x l0
tl0
5t 2 1
t
1
1
and lim1 3
x
l
1
x 21
x 21
(a)by evaluating f sxd − 1ysx 3 2 1d for values of x that
approach 1 from the left and from the right,
(b) by reasoning as in Example 9, and
(c) from a graph of f.
41. Determine lim2
s2 1 hd5 2 32
, h
22. lim
3
; 42. (a)By graphing the function f sxd − stan 4xdyx and zooming in toward the point where the graph crosses the
y-axis, estimate the value of lim x l 0 f sxd.
(b)Check your answer in part (a) by evaluating f sxd for
values of x that approach 0.
43. (a)Evaluate the function f sxd − x 2 2 s2 xy1000d for x − 1,
0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of
lim
xl0
2
; 27. (a)By graphing the function f sxd − scos 2x 2 cos xdyx
and zooming in toward the point where the graph
crosses the y-axis, estimate the value of lim x l 0 f sxd.
(b)Check your answer in part (a) by evaluating f sxd for
values of x that approach 0.
; 28. (a)Estimate the value of
lim
xl0
sin x
sin x
by graphing the function f sxd − ssin xdyssin xd. State
your answer correct to two decimal places.
(b)Check your answer in part (a) by evaluating f sxd for
values of x that approach 0.
29–39 Determine the infinite limit.
x11
x11
29. lim1
30. lim2
x l5 x 2 5
x l5 x 2 5
31. lim
x l1
33.
35.
22x
sx
32. lim2
x l3 sx 2 3d 5
sx 2 1d2
lim
x l 221
x21
x21
34. lim 2
x l 0 x sx 1 2d
x 2sx 1 2d
lim
xlsy2d1
1
sec x36. lim2 cot x
x l
x
37. lim 2 x csc x38. lim2
x l2
xl2
2
39. lim1
x l2
x 2 2x 2 8
x 2 2 5x 1 6
x 2 2 2x
x 2 4x 1 4
2
61
S
x2 2
2x
1000
D
(b)Evaluate f sxd for x − 0.04, 0.02, 0.01, 0.005, 0.003,
and 0.001. Guess again.
44. (a)Evaluate hsxd − stan x 2 xdyx 3 for x − 1, 0.5, 0.1,
0.05, 0.01, and 0.005.
tan x 2 x
(b) Guess the value of lim
.
xl0
x3
(c)Evaluate hsxd for successively smaller values of x until
you finally reach a value of 0 for hsxd. Are you still
confident that your guess in part (b) is correct? Explain
why you eventually obtained values of 0 for hsxd. (In
Section 6.8 a method for evaluating this limit will be
explained.)
(d)Graph the function h in the viewing rectangle f21, 1g
;
by f0, 1g. Then zoom in toward the point where the
graph crosses the y-axis to estimate the limit of hsxd as
x approaches 0. Continue to zoom in until you observe
distortions in the graph of h. Compare with the results
of part (c).
raph the function f sxd − sinsyxd of Example 4 in the
; 45. G
viewing rectangle f21, 1g by f21, 1g. Then zoom in toward
the origin several times. Comment on the behavior of this
function.
1
46. Consider the function f sxd − tan .
x
1 1
1
,
,...
(a)Show that f sxd − 0 for x − ,
2 3
4 4
4
,
,...
(b)Show that f sxd − 1 for x − ,
5 9
1
(c) What can you conclude about lim1 tan ?
xl0
x
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62
Chapter 1 Functions and Limits
se a graph to estimate the equations of all the vertical
; 47. U
asymptotes of the curve
y − tans2 sin xd
where m 0 is the mass of the particle at rest and c is the
speed of light. What happens as v l c2?
; 49. (a)Use numerical and graphical evidence to guess the
value of the limit
2 < x < Then find the exact equations of these asymptotes.
48. I n the theory of relativity, the mass of a particle with velocity v is
m0
m−
s1 2 v 2yc 2
lim
xl1
x3 2 1
sx 2 1
(b)How close to 1 does x have to be to ensure that the fun­ction in part (a) is within a distance 0.5 of its limit?
In Section 1.5 we used calculators and graphs to guess the values of limits, but we saw
that such methods don’t always lead to the correct answer. In this section we use the following properties of limits, called the Limit Laws, to calculate limits.
Limit Laws Suppose that c is a constant and the limits
lim f sxd and lim tsxd
xla
xla
exist. Then
1. lim f f sxd 1 tsxdg − lim f sxd 1 lim tsxd
xla
xla
xla
2. lim f f sxd 2 tsxdg − lim f sxd 2 lim tsxd
xla
xla
xla
3. lim fcf sxdg − c lim f sxd
xla
xla
4. lim f f sxd tsxdg − lim f sxd lim tsxd
xla
xla
5. lim
lim f sxd
f sxd
xla
−
tsxd
lim tsxd
xla
Sum Law
Difference Law
Constant Multiple Law
Product Law
Quotient Law
xla
xla
if lim tsxd ± 0
xla
These five laws can be stated verbally as follows:
1. The limit of a sum is the sum of the limits.
2. The limit of a difference is the difference of the limits.
3.The limit of a constant times a function is the constant times the limit of the
function.
4. The limit of a product is the product of the limits.
5.The limit of a quotient is the quotient of the limits (provided that the limit of
the denominator is not 0).
It is easy to believe that these properties are true. For instance, if f sxd is close to L
and tsxd is close to M, it is reasonable to conclude that f sxd 1 tsxd is close to L 1 M.
This gives us an intuitive basis for believing that Law 1 is true. In Section 1.7 we give a
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Section 1.6 Calculating Limits Using the Limit Laws
63
precise definition of a limit and use it to prove this law. The proofs of the remaining laws
are given in Appendix F.
y
f
1
0
g
1
x
Example 1 Use the Limit Laws and the graphs of f and t in Figure 1 to evaluate the
following limits, if they exist.
f sxd
(a) lim f f sxd 1 5tsxdg (b) lim f f sxdtsxdg (c) lim
x l 22
xl1
x l 2 tsxd
SOLUTION (a) From the graphs of f and t we see that
lim f sxd − 1 and lim tsxd − 21
FIGURE 1 x l 22
x l 22
Therefore we have
lim f f sxd 1 5tsxdg − lim f sxd 1 lim f5tsxdg (by Limit Law 1)
x l 22
x l 22
x l 22
− lim f sxd 1 5 lim tsxd (by Limit Law 3)
x l 22
x l 22
− 1 1 5s21d − 24
(b) We see that lim x l 1 f sxd − 2. But lim x l 1 tsxd does not exist because the left and
right limits are different:
lim tsxd − 22 lim1 tsxd − 21
x l 12
xl1
So we can’t use Law 4 for the desired limit. But we can use Law 4 for the one-sided
limits:
lim f f sxdtsxdg − lim2 f sxd lim2 tsxd − 2 s22d − 24
x l 12
x l1
x l1
lim f f sxdtsxdg − lim1 f sxd lim1 tsxd − 2 s21d − 22
x l 11
x l1
x l1
The left and right limits aren’t equal, so lim x l 1 f f sxdtsxdg does not exist.
(c) The graphs show that
lim f sxd < 1.4 and lim tsxd − 0
xl2
xl2
Because the limit of the denominator is 0, we can’t use Law 5. The given limit does not
exist because the denominator approaches 0 while the numerator approaches a nonzero
number.
■
If we use the Product Law repeatedly with tsxd − f sxd, we obtain the following law.
Power Law
fx l a
g
6. lim f f sxdg n − lim f sxd n where n is a positive integer
x la
In applying these six limit laws, we need to use two special limits:
7.
lim c − c
xla
8. lim x − a
xla
These limits are obvious from an intuitive point of view (state them in words or draw
graphs of y − c and y − x), but proofs based on the precise definition are requested in
the exercises for Section 1.7.
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64
Chapter 1 Functions and Limits
If we now put f sxd − x in Law 6 and use Law 8, we get another useful special limit.
9. lim x n − a n where n is a positive integer
xla
A similar limit holds for roots as follows. (For square roots the proof is outlined in
Exercise 1.7.37.)
n
n
10. lim s
x −s
a where n is a positive integer
xla
(If n is even, we assume that a . 0.)
More generally, we have the following law, which is proved in Section 1.8 as a consequence of Law 10.
n
n lim f sxd
f sxd − s
11. lim s
where n is a positive integer
x la
Root Law
x la
f
g
f sxd . 0.
If n is even, we assume that xlim
la
Example 2 Evaluate the following limits and justify each step.
(a) lim s2x 2 2 3x 1 4d
(b) lim
x l5
x l 22
x 3 1 2x 2 2 1
5 2 3x
SOLUTION
(a)
lim s2x 2 2 3x 1 4d − lim s2x 2 d 2 lim s3xd 1 lim 4 (by Laws 2 and 1)
x l5
x l5
x l5
x l5
− 2 lim x 2 2 3 lim x 1 lim 4 (by 3)
x l5
x l5
x l5
− 2s5 2 d 2 3s5d 1 4
(by 9, 8, and 7)
− 39
(b) We start by using Law 5, but its use is fully justified only at the final stage when we
see that the limits of the numerator and denominator exist and the limit of the denominator is not 0.
x 3 1 2x 2 2 1
lim
−
x l 22
5 2 3x
lim sx 3 1 2x 2 2 1d
x l 22
(by Law 5)
lim s5 2 3xd
x l 22
lim x 3 1 2 lim x 2 2 lim 1
−
x l 22
x l 22
x l 22
−
x l 22
lim 5 2 3 lim x
x l 22
s22d3 1 2s22d2 2 1
5 2 3s22d
−2
1
11
(by 1, 2, and 3)
(by 9, 8, and 7)
■
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Section 1.6 Calculating Limits Using the Limit Laws
Newton and Limits
Isaac Newton was born on Christmas
Day in 1642, the year of Galileo’s death.
When he entered Cambridge University
in 1661 Newton didn’t know much
mathematics, but he learned quickly
by reading Euclid and Descartes and by
attending the lectures of Isaac Barrow.
Cam­bridge was closed because of the
plague in 1665 and 1666, and Newton
returned home to reflect on what he
had learned. Those two years were
amazingly productive for at that time
he made four of his major discoveries:
(1) his repre­senta­tion of functions as
sums of infinite series, including the
binomial theorem; (2) his work on differential and integral calculus; (3) his laws
of motion and law of universal gravitation; and (4) his prism experi­ments on
the nature of light and color. Because of
a fear of controversy and criticism, he
was reluctant to publish his discoveries
and it wasn’t until 1687, at the urging
of the astronomer Halley, that Newton
published Principia Mathematica. In
this work, the great­est scientific treatise
ever written, Newton set forth his version of calculus and used it to investigate mechanics, fluid dynamics, and
wave motion, and to explain the motion
of planets and comets.
The beginnings of calculus are
found in the calculations of areas and
volumes by ancient Greek scholars such
as Eudoxus and Archimedes. Although
aspects of the idea of a limit are implicit
in their “method of exhaustion,” Eudoxus
and Archimedes never explicitly formulated the concept of a limit. Like­wise,
mathematicians such as Cavalieri, Fer­mat, and Barrow, the immediate precursors of Newton in the development of
calculus, did not actually use limits. It
was Isaac Newton who was the first to
talk explicitly about limits. He explained
that the main idea behind limits is that
quantities “approach nearer than by
any given difference.” Newton stated
that the limit was the basic concept in
calculus, but it was left to later mathe­
maticians like Cauchy to clarify his ideas
about limits.
65
NOTE If we let f sxd − 2x 2 2 3x 1 4, then f s5d − 39. In other words, we would
have gotten the correct answer in Example 2(a) by substituting 5 for x. Similarly, direct
substitution provides the correct answer in part (b). The functions in Example 2 are
a polynomial and a rational function, respectively, and similar use of the Limit Laws
proves that direct substitution always works for such functions (see Exercises 57 and 58).
We state this fact as follows.
Direct Substitution Property If f is a polynomial or a rational function and a is
in the domain of f, then
lim f sxd − f sad
x la
Functions with the Direct Substitution Property are called continuous at a and will be
studied in Section 1.8. However, not all limits can be evaluated by direct substitution, as
the following examples show.
Example 3 Find lim
xl1
x2 2 1
.
x21
SOLUTION Let f sxd − sx 2 2 1dysx 2 1d. We can’t find the limit by substituting x − 1
because f s1d isn’t defined. Nor can we apply the Quotient Law, because the limit of
the denominator is 0. Instead, we need to do some preliminary algebra. We factor the
numerator as a difference of squares:
x2 2 1
sx 2 1dsx 1 1d
−
x21
x21
The numerator and denominator have a common factor of x 2 1. When we take the
limit as x approaches 1, we have x ± 1 and so x 2 1 ± 0. Therefore we can cancel the
common factor and then compute the limit by direct substitution as follows:
lim
xl1
x2 2 1
sx 2 1dsx 1 1d
− lim
xl1
x21
x21
− lim sx 1 1d
xl1
−111−2
The limit in this example arose in Example 1.4.1 when we were trying to find the
tangent to the parabola y − x 2 at the point s1, 1d.
■
NOTE 1 In Example 3 we do not have an infinite limit even though the denominator
approaches 0 as x l 1. When both numerator and denominator approach 0, the limit
may be infinite or it may be some finite value.
NOTE 2 In Example 3 we were able to compute the limit by replacing the given
function f sxd − sx 2 2 1dysx 2 1d by a simpler function, tsxd − x 1 1, with the same
limit. This is valid because f sxd − tsxd except when x − 1, and in computing a limit as
x approaches 1 we don’t consider what happens when x is actually equal to 1. In general,
we have the following useful fact.
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66
Chapter 1 Functions and Limits
If f sxd − tsxd when x ± a, then lim f sxd − lim tsxd, provided the limits exist.
x la
xla
Example 4 Find lim tsxd where
x l1
tsxd −
H
x 1 1 if x ± 1
if x − 1
SOLUTION Here t is defined at x − 1 and ts1d − , but the value of a limit as x
approaches 1 does not depend on the value of the function at 1. Since tsxd − x 1 1 for
x ± 1, we have
lim tsxd − lim sx 1 1d − 2
xl1
■
xl1
Note that the values of the functions in Examples 3 and 4 are identical except when
x − 1 (see Figure 2) and so they have the same limit as x approaches 1.
y
y
y=ƒ
3
2
2
FIGURE 2 1
1
The graphs of the functions f (from
Example 3) and t (from Example 4)
0
1
2
Example 5 Evaluate lim
hl0
3
y=©
3
x
0
1
2
3
x
s3 1 hd2 2 9
.
h
SOLUTION If we define
Fshd −
s3 1 hd2 2 9
h
then, as in Example 3, we can’t compute lim h l 0 Fshd by letting h − 0 since Fs0d is
undefined. But if we simplify Fshd algebraically, we find that
Fshd −
s9 1 6h 1 h 2 d 2 9
6h 1 h 2
hs6 1 hd
−
−
−61h
h
h
h
(Recall that we consider only h ± 0 when letting h approach 0.) Thus
lim
hl0
Example 6 Find lim
tl0
s3 1 hd2 2 9
− lim s6 1 hd − 6
hl0
h
■
st 2 1 9 2 3
.
t2
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Section 1.6 Calculating Limits Using the Limit Laws
67
SOLUTION We can’t apply the Quotient Law immediately, since the limit of the
denominator is 0. Here the preliminary algebra consists of rationalizing the numerator:
lim
tl0
st 2 1 9 2 3
st 2 1 9 2 3 st 2 1 9 1 3
− lim
2
t
tl0
t2
st 2 1 9 1 3
− lim
st 2 1 9d 2 9
t 2 (st 2 1 9 1 3)
− lim
t2
t (st 2 1 9 1 3)
− lim
1
st 1 9 1 3
tl0
2
tl0
2
tl0
−
1
2 1 9d 1 3
st
slim
t l0
Here we use several properties of
limits (5, 1, 10, 7, 9).
−
1
1
−
313
6
This calculation confirms the guess that we made in Example 1.5.2.
■
Some limits are best calculated by first finding the left- and right-hand limits. The
following theorem is a reminder of what we discovered in Section 1.5. It says that a twosided limit exists if and only if both of the one-sided limits exist and are equal.
1 Theorem lim f sxd − L if and only if lim f sxd − L − lim f sxd
2
1
xla
x la
x la
When computing one-sided limits, we use the fact that the Limit Laws also hold for
one-sided limits.
Example 7 Show that lim | x | − 0.
xl0
SOLUTION Recall that
|x| −
The result of Example 7 looks plausible
from Figure 3.
y
| |
H
x
if x > 0
2x if x , 0
Since x − x for x . 0, we have
| |
lim x − lim1 x − 0
x l 01
y=| x|
x l0
| |
For x , 0 we have x − 2x and so
| |
lim x − lim2 s2xd − 0
x l 02
0
FIGURE 3 x
Therefore, by Theorem 1,
x l0
| |
lim x − 0
xl0
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■
68
Chapter 1 Functions and Limits
Example 8 Prove that lim
xl0
| x | does not exist.
x
| |
| |
SOLUTION Using the facts that x − x when x . 0 and x − 2x when x , 0, we
have
y
| x|
y= x
lim
|x| −
lim2
|x| −
x l 01
1
0
x
_1
x l0
x
x
lim
x
− lim1 1 − 1
x l0
x
lim2
2x
− lim2 s21d − 21
x l0
x
x l 01
x l0
Since the right- and left-hand limits are different, it follows from Theorem 1 that
lim x l 0 x yx does not exist. The graph of the function f sxd − x yx is shown in
Figure 4 and supports the one-sided limits that we found.
| |
FIGURE 4 | |
Example 9 If
f sxd −
H
sx 2 4
8 2 2x
■
if x . 4
if x , 4
determine whether lim x l 4 f sxd exists.
SOLUTION Since f sxd − sx 2 4 for x . 4, we have
It is shown in Example 1.7.3 that
lim x l 01 sx − 0.
lim f sxd − lim1 s x 2 4 − s4 2 4 − 0
x l 41
x l4
Since f sxd − 8 2 2x for x , 4, we have
y
lim f sxd − lim2 s8 2 2xd − 8 2 2 4 − 0
x l 42
x l4
The right- and left-hand limits are equal. Thus the limit exists and
0
x
4
lim f sxd − 0
xl4
FIGURE 5 The graph of f is shown in Figure 5.
■
Other notations for v x b are fxg and :x;. Example 10 The greatest integer function is defined by v x b − the largest integer
The greatest integer function is somethat is less than or equal to x. (For instance, v4 b − 4, v4.8b − 4, v b − 3, vs2 b − 1,
times called the floor function.
1
v22 b − 21.) Show that lim x l3 v x b does not exist.
y
SOLUTION The graph of the greatest integer function is shown in Figure 6. Since
v x b − 3 for 3 < x , 4, we have
4
3
lim v x b − lim1 3 − 3
y=[ x]
2
x l 31
Since v x b − 2 for 2 < x , 3, we have
1
0
x l3
1
2
3
4
5
x
lim v x b − lim2 2 − 2
x l 32
x l3
Because these one-sided limits are not equal, lim xl3 v x b does not exist by Theorem 1. ■
FIGURE 6
Greatest integer function
The next two theorems give two additional properties of limits. Their proofs can be
found in Appendix F.
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Section 1.6 Calculating Limits Using the Limit Laws
69
2 Theorem If f sxd < tsxd when x is near a (except possibly at a) and the limits
of f and t both exist as x approaches a, then
lim f sxd < lim tsxd
xla
xla
3 The Squeeze Theorem If f sxd < tsxd < hsxd when x is near a (except
possibly at a) and
lim f sxd − lim hsxd − L
y
xla
h
g
L
f
0
x
a
FIGURE 7 xla
lim tsxd − L
then
xla
The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the
Pinching Theorem, is illustrated by Figure 7. It says that if tsxd is squeezed between
f sxd and hsxd near a, and if f and h have the same limit L at a, then t is forced to have
the same limit L at a.
1
− 0.
x
SOLUTION First note that we cannot use
Example 11 Show that lim x 2 sin
xl0
lim x 2 sin
xl0
1
1
− lim x 2 lim sin
x
l
0
x
l
0
x
x
because lim x l 0 sins1yxd does not exist (see Example 1.5.4).
Instead we apply the Squeeze Theorem, and so we need to find a function f smaller
than tsxd − x 2 sins1yxd and a function h bigger than t such that both f sxd and hsxd
approach 0. To do this we use our knowledge of the sine function. Because the sine of
any number lies between 21 and 1, we can write.
4
21 < sin
1
<1
x
Any inequality remains true when multiplied by a positive number. We know that
x 2 > 0 for all x and so, multiplying each side of the inequalities in (4) by x 2, we get
y
2x 2 < x 2 sin
y=≈
1
< x2
x
as illustrated by Figure 8. We know that
x
0
y=_≈
FIGURE 8 y − x 2 sins1yxd
lim x 2 − 0 and lim s2x 2 d − 0
xl0
xl0
Taking f sxd − 2x 2, tsxd − x 2 sins1yxd, and hsxd − x 2 in the Squeeze Theorem, we
obtain
1
lim x 2 sin − 0
xl0
x
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■
70
Chapter 1 Functions and Limits
1.Given that
11–32 Evaluate the limit, if it exists.
lim f sxd − 4 lim tsxd − 22 lim hsxd − 0
xl2
xl 2
xl2
x 2 2 5x 1 6
x 2 1 3x
14. lim 2
x l 4 x 2 x 2 12
x25
x l5
3f sxd
(d) lim
x l 2 tsxd
xl2
x l2
13. lim
xl2
(c) lim sf sxd
(e) lim
x 2 2 6x 1 5
x 2 1 3x
12. lim 2
x l 23 x 2 x 2 12
x25
x l5
find the limits that exist. If the limit does not exist, explain
why.
(a) lim f f sxd 1 5tsxdg
(b) lim f tsxdg 3
xl2
11. lim
tsxd
hsxd
(f ) lim
xl2
t l 23
tsxdhsxd
f sxd
hl0
xl0
f sxd
tsxd
(d) lim
(e) lim fx f sxdg
(f ) f s21d 1 lim tsxd
x l3
2
x l2
x l 21
y
y
y=ƒ
1
x l 22
x
1
0
1
1
2
x
3
s3 1 hd21 2 3 21
23. lim
24. lim
x l3 x 2 3
hl0
h
y=©
25. lim
x
27. lim
1
hl0
s9 1 h 2 3
s4u 1 1 2 3
22. lim
ul
2
h
u22
21. lim
tl0
1
0
s2 1 hd3 2 8
h
18. lim
x12
t4 2 1
20. lim 3
3
tl1 t 2 1
x 18
19. lim
hl0
(c) lim f f sxd tsxdg
x l 21
s25 1 hd2 2 25
h
17. lim
2.The graphs of f and t are given. Use them to evaluate each
limit, if it exists. If the limit does not exist, explain why.
(a) lim f f sxd 1 tsxdg
(b) lim f f sxd 2 tsxdg
x l2
t2 2 9
2x 2 1 3x 1 1
16.
lim
x l 21 x 2 2 2x 2 3
2t 2 1 7t 1 3
15. lim
s1 1 t 2 s1 2 t
26. lim
tl0
t
x l 16
S
1
1
2 2
t
t 1t
D
4 2 sx
x 2 2 4x 1 4
2 28. xlim
l 2 x 4 2 3x 2 2 4
16x 2 x
S
1
1
D
sx 2 1 9 2 5
x14
3–9 Evaluate the limit and justify each step by indicating the
appropriate Limit Law(s).
29. lim
3. lim s5x 3 2 3x 2 1 x 2 6d
1
1
2 2
sx 1 hd3 2 x 3
sx 1 hd2
x
31. lim
32. lim
hl0
hl0
h
h
tl0
x l3
4. lim sx 4 2 3xdsx 2 1 5x 1 3d
xl 21
5. lim
t l 22
t4 2 2
2t 2 3t 1 2
2
3
7. lim s1 1 s
x ds2 2 6x 2 1 x 3 d
xl8
9.lim
xl2
Î
6. lim su 4 1 3u 1 6
ul 22
8.
lim
tl2
S
t2 2 2
t 2 3t 1 5
3
2x 2 1 1
3x 2 2
10. (a) What is wrong with the following equation?
x2 1 x 2 6
−x13
x22
D
x l2
is correct.
x2 1 x 2 6
− lim sx 1 3d
x l2
x22
t
30. lim
xl24
; 33. (a) Estimate the value of
2
lim
x l0
x
s1 1 3x 2 1
by graphing the function f sxd − xyss1 1 3x 2 1d.
(b)Make a table of values of f sxd for x close to 0 and guess
the value of the limit.
(c)Use the Limit Laws to prove that your guess is correct.
; 34. (a) Use a graph of
f sxd −
(b) In view of part (a), explain why the equation
lim
t s1 1 t
2
s3 1 x 2 s3
x
to estimate the value of lim x l 0 f sxd to two decimal places.
(b)Use a table of values of f sxd to estimate the limit to four
decimal places.
(c)Use the Limit Laws to find the exact value of the limit.
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71
Section 1.6 Calculating Limits Using the Limit Laws
se the Squeeze Theorem to show that
; 35. U
lim x l 0 sx 2 cos 20xd − 0. Illustrate by graphing the functions f sxd − 2x 2, tsxd − x 2 cos 20x, and hsxd − x 2 on
the same screen.
(b)Does lim x l 2 tsxd exist?
(c) Sketch the graph of t.
50. Let
; 36. Use the Squeeze Theorem to show that
−0
lim sx 3 1 x 2 sin
x l0
x
Illustrate by graphing the functions f, t, and h (in the notation of the Squeeze Theorem) on the same screen.
37. If 4x 2 9 < f sxd < x 2 2 4x 1 7 for x > 0, find lim f sxd.
51. Let
Bstd −
38. If 2x < tsxd < x 2 x 1 2 for all x, evaluate lim tsxd.
2
xl1
39. Prove that lim x 4 cos
x l0
2
− 0.
x
41–46 Find the limit, if it exists. If the limit does not exist,
explain why.
2x 1 12
41. lim s2x 1 x 2 3 d
42. lim
xl3
x l 26 x 1 6
x l 0.5
45. lim2
x l0
|
S
|
2x 2 1
2x 3 2 x 2
|
44. lim
|
x l 22
| |D
1
1
2
x
x
46. lim1
x l0
|
| |
S
tsxd −
| |
D
|
xl0
(iii) lim sgn x
(iv) lim sgn x
xl0
(b) Sketch the graph of t.
x l0
x l0
(v)
x l
lim2 tsxd
x l
|
x2 1 x 2 6
.
x22
|
(a)Find
(i) lim1 tsxd
x l2
|
(vi)
lim tsxd
xl2
x l 22
x l 22.4
(b)If n is an integer, evaluate
(i) lim2 v x b
(ii) lim1 v x b
xln
54. L
et f sxd − v cos x b , 2 < x < .
(a) Sketch the graph of f.
(b) Evaluate each limit, if it exists.
(i)
lim f sxd
(ii)
lim 2 f sxd
x l sy2d
xl0
(iii)
lim 1 f sxd
x l sy2d
(iv)
lim f sxd
x l y2
(c) For what values of a does lim x l a f sxd exist?
x l0
(vi) lim tsxd
x l
(b)For which values of a does lim x l a tsxd not exist?
(c) Sketch a graph of t.
49. Let tsxd −
xl2
53. (a)If the symbol v b denotes the greatest integer function
defined in Example 10, evaluate
(i) lim 1 v x b
(ii) lim v x b
(iii) lim v x b
x ln
48. Let tsxd − sgnssin xd .
(a)Find each of the following limits or explain why it does
not exist.
(i) lim1 tsxd
(ii) lim2 tsxd (iii) lim tsxd
(iv) lim1 tsxd
(v)
lim1 tsxd
(c) For what values of a does lim x l a v x b exist?
(a) Sketch the graph of this function.
(b)Find each of the following limits or explain why it does
not exist.
(i) lim1 sgn x
(ii) lim2 sgn x
x l0
x,1
x−1
1,x<2
x.2
xl1
x l2
1
1
2
x
x
if
if
if
if
(a) Evaluate each of the following, if it exists.
(i)
lim2 tsxd
(ii)
lim tsxd
(iii)
ts1d
(iv)
lim2 tsxd
21 if x , 0
0 if x − 0
1 if x . 0
x l0
if t > 2
x
3
2 2 x2
x23
x l 22
H
if t , 2
st 1 c
x l1
22 x
21x
47. The signum (or sign) function, denoted by sgn, is defined by
sgn x −
4 2 12 t
tl2
52. Let
x l0
43. lim 2
H
Find the value of c so that lim Bstd exists.
40. Prove that lim1 sx f1 1 sin 2s2yxdg − 0.
|
x2 1 1
if x , 1
2
sx 2 2d if x > 1
(a)Find lim x l12 f sxd and lim x l11 f sxd.
(b)Does lim x l1 f sxd exist?
(c) Sketch the graph of f.
xl4
4
H
f sxd −
55. If f sxd − v x b 1 v 2x b , show that lim x l 2 f sxd exists but is
not equal to f s2d.
56. In the theory of relativity, the Lorentz contraction formula
L − L 0 s1 2 v 2yc 2
expresses the length L of an object as a function of its velocity v with respect to an observer, where L 0 is the length of
the object at rest and c is the speed of light. Find lim v l c2 L
and interpret the result. Why is a left-hand limit necessary?
57. If p is a polynomial, show that lim xl a psxd − psad.
(ii) lim2 tsxd
x l2
58. I f r is a rational function, use Exercise 57 to show that
lim x l a rsxd − rsad for every number a in the domain of r.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
72
chapter 1 Functions and Limits
59. If lim
xl1
f sxd 2 8
− 10, find lim f sxd.
xl1
x21
65. Is there a number a such that
lim
x l 22
f sxd
60. If lim 2 − 5, find the following limits.
xl0 x
f sxd
lim f sxd
(b)
lim
(a)
xl0
xl0
x
61. If
f sxd −
H
exists? If so, find the value of a and the value of the limit.
66. The figure shows a fixed circle C1 with equation
sx 2 1d2 1 y 2 − 1 and a shrinking circle C2 with radius r and
center the origin. P is the point s0, rd, Q is the upper point of
intersection of the two circles, and R is the point of intersection
of the line PQ and the x-axis. What happens to R as C2 shrinks,
that is, as r l 0 1?
x 2 if x is rational
0 if x is irrational
y
prove that lim x l 0 f sxd − 0.
P
62. S
how by means of an example that lim x l a f f sxd 1 tsxdg may
exist even though neither lim x l a f sxd nor lim x l a tsxd exists.
C™
63. S
how by means of an example that lim x l a f f sxd tsxdg may
exist even though neither lim x l a f sxd nor lim x l a tsxd exists.
64. Evaluate lim
xl2
3x 2 1 ax 1 a 1 3
x2 1 x 2 2
Q
0
s6 2 x 2 2
.
s3 2 x 2 1
R
C¡
x
The intuitive definition of a limit given in Section 1.5 is inadequate for some purposes
because such phrases as “x is close to 2” and “ f sxd gets closer and closer to L” are vague.
In order to be able to prove conclusively that
lim
xl0
S
x3 1
cos 5x
10,000
D
− 0.0001 or lim
xl0
sin x
−1
x
we must make the definition of a limit precise.
To motivate the precise definition of a limit, let’s consider the function
f sxd −
H
2x 2 1 if x ± 3
6
if x − 3
Intuitively, it is clear that when x is close to 3 but x ± 3, then f sxd is close to 5, and so
lim x l3 f sxd − 5.
To obtain more detailed information about how f sxd varies when x is close to 3, we
ask the following question:
How close to 3 does x have to be so that f sxd differs from 5 by less than 0.l?
It is traditional to use the Greek letter
(delta) in this situation.
|
|
|
|
The distance from x to 3 is x 2 3 and the distance from f sxd to 5 is f sxd 2 5 , so our
problem is to find a number such that
| f sxd 2 5 | , 0.1 if | x 2 3 | , but x ± 3
|
|
If x 2 3 . 0, then x ± 3, so an equivalent formulation of our problem is to find a number such that
| f sxd 2 5 | , 0.1 if 0 , | x 2 3 | , Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.7 The Precise Definition of a Limit
|
73
|
Notice that if 0 , x 2 3 , s0.1dy2 − 0.05, then
| f sxd 2 5 | − | s2x 2 1d 2 5 | − | 2x 2 6 | − 2| x 2 3 | , 2s0.05d − 0.1
| f sxd 2 5 | , 0.1 if 0 , | x 2 3 | , 0.05
that is,
Thus an answer to the problem is given by − 0.05; that is, if x is within a distance of
0.05 from 3, then f sxd will be within a distance of 0.1 from 5.
If we change the number 0.l in our problem to the smaller number 0.01, then by using
the same method we find that f sxd will differ from 5 by less than 0.01 provided that x
differs from 3 by less than (0.01)y2 − 0.005:
| f sxd 2 5 | , 0.01 if 0 , | x 2 3 | , 0.005
Similarly,
| f sxd 2 5 | , 0.001 if 0 , | x 2 3 | , 0.0005
The numbers 0.1, 0.01, and 0.001 that we have considered are error tolerances that we
might allow. For 5 to be the precise limit of f sxd as x approaches 3, we must not only be
able to bring the difference between f sxd and 5 below each of these three numbers; we
must be able to bring it below any positive number. And, by the same reasoning, we can!
If we write « (the Greek letter epsilon) for an arbitrary positive number, then we find as
before that
1 «
| f sxd 2 5 | , « if 0 , | x 2 3 | , − 2
This is a precise way of saying that f sxd is close to 5 when x is close to 3 because (1) says
that we can make the values of f sxd within an arbitrary distance « from 5 by restricting
the val­ues of x to be within a distance «y2 from 3 (but x ± 3).
Note that (1) can be rewritten as follows:
y
ƒ
is in
here
if 3 2 , x , 3 1 sx ± 3d then 5 2 « , f sxd , 5 1 «
5+∑
5
and this is illustrated in Figure 1. By taking the values of x (± 3) to lie in the interval
s3 2 , 3 1 d we can make the values of f sxd lie in the interval s5 2 «, 5 1 «d.
Using (1) as a model, we give a precise definition of a limit.
5-∑
0
x
3
3-∂
3+∂
when x is in here
(x≠3)
2 Precise Definition of a Limit Let f be a function defined on some open
interval that contains the number a, except possibly at a itself. Then we say that
the limit of f sxd as x approaches a is L, and we write
lim f sxd − L
xla
FIGURE 1 if for every number « . 0 there is a number . 0 such that
|
|
|
|
if 0 , x 2 a , then f sxd 2 L , «
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74
Chapter 1 Functions and Limits
|
|
|
|
Since x 2 a is the distance from x to a and f sxd 2 L is the distance from f sxd to
L, and since « can be arbitrarily small, the definition of a limit can be expressed in words
as follows:
lim x l a f sxd 5 L means that the distance between f sxd and L can be made arbitrarily small
by requiring that the distance from x to a be sufficiently small (but not 0).
Alternatively,
lim x l a f sxd 5 L means that the values of f sxd can be made as close as we please to L
by requiring x to be close enough to a (but not equal to a).
We can also reformulate Definition 2 in terms of intervals by observing that the
inequality x 2 a , is equivalent to 2 , x 2 a , , which in turn can be written as a 2 , x , a 1 . Also 0 , x 2 a is true if and only if x 2 a ± 0, that is,
x ± a. Similarly, the inequality f sxd 2 L , « is equivalent to the pair of inequalities
L 2 « , f sxd , L 1 «. Therefore, in terms of intervals, Definition 2 can be stated as
follows:
|
|
|
|
|
|
lim x l a f sxd 5 L means that for every « . 0 (no matter how small « is) we can find
. 0 such that if x lies in the open interval sa 2 , a 1 d and x ± a, then f sxd lies in
the open interval sL 2 «, L 1 «d.
We interpret this statement geometrically by representing a function by an arrow diagram as in Figure 2, where f maps a subset of R onto another subset of R.
f
FIGURE 2 x
a
f(a)
ƒ
The definition of limit says that if any small interval sL 2 «, L 1 «d is given around L,
then we can find an interval sa 2 , a 1 d around a such that f maps all the points in
sa 2 , a 1 d (except possibly a) into the interval sL 2 «, L 1 «d. (See Figure 3.)
f
x
FIGURE 3 a-∂
ƒ
a
a+∂
L-∑
L
L+∑
Another geometric interpretation of limits can be given in terms of the graph of a
function. If « . 0 is given, then we draw the horizontal lines y 5 L 1 « and y 5 L 2 «
and the graph of f. (See Figure 4.) If lim x l a f sxd 5 L, then we can find a number . 0
such that if we restrict x to lie in the interval sa 2 , a 1 d and take x ± a, then the
curve y 5 f sxd lies between the lines y 5 L 2 « and y 5 L 1 «. (See Figure 5.) You can
see that if such a has been found, then any smaller will also work.
It is important to realize that the process illustrated in Figures 4 and 5 must work
for every positive number «, no matter how small it is chosen. Figure 6 shows that if a
smaller « is chosen, then a smaller may be required.
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75
Section 1.7 The Precise Definition of a Limit
y=ƒ
y
y
y=L+∑
ƒ
is in
here
∑
L
∑
y
y=L+∑
L+∑
∑
L
∑
y=L-∑
0
x
a
y=L-∑
L-∑
y=L-∑
0
y=L+∑
a
a-∂
0
x
a+∂
a-∂
a
x
a+∂
when x is in here
(x≠a)
FIGURE 4
FIGURE 5
FIGURE 6
Example 1 Since f sxd − x 3 2 5x 1 6 is a polynomial, we know from the Direct
Substitution Property that lim x l1 f sxd − f s1d − 13 2 5s1d 1 6 − 2. Use a graph to
find a number such that if x is within of 1, then y is within 0.2 of 2, that is,
|
|
|
|
if x 2 1 , then sx 3 2 5x 1 6d 2 2 , 0.2
In other words, find a number that corresponds to « 5 0.2 in the definition of a limit
for the function f sxd 5 x 3 2 5x 1 6 with a 5 1 and L 5 2.
SOLUTION A graph of f is shown in Figure 7; we are interested in the region near the
point s1, 2d. Notice that we can rewrite the inequality
15
| sx
_3
3
_5
2.3
y=˛-5x+6
(1, 2)
y=1.8
FIGURE 8 1.8 , x 3 2 5x 1 6 , 2.2
So we need to determine the values of x for which the curve y 5 x 3 2 5x 1 6 lies
between the horizontal lines y 5 1.8 and y 5 2.2. Therefore we graph the curves
y 5 x 3 2 5x 1 6, y 5 1.8, and y 5 2.2 near the point s1, 2d in Figure 8. Then we use
the cursor to estimate that the x-coordinate of the point of intersection of the line
y 5 2.2 and the curve y 5 x 3 2 5x 1 6 is about 0.911. Similarly, y 5 x 3 2 5x 1 6
intersects the line y 5 1.8 when x < 1.124. So, rounding toward 1 to be safe, we can
say that
y=2.2
0.8
1.7
|
2 5x 1 6d 2 2 , 0.2
20.2 , sx 3 2 5x 1 6d 2 2 , 0.2
as
or equivalently
FIGURE 7 3
1.2
if 0.92 , x , 1.12 then 1.8 , x 3 2 5x 1 6 , 2.2
This interval s0.92, 1.12d is not symmetric about x 5 1. The distance from x 5 1 to the
left endpoint is 1 2 0.92 5 0.08 and the distance to the right endpoint is 0.12. We can
choose to be the smaller of these numbers, that is, 5 0.08. Then we can rewrite our
inequalities in terms of distances as follows:
|
|
|
|
if x 2 1 , 0.08 then sx 3 2 5x 1 6d 2 2 , 0.2
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76
Chapter 1 Functions and Limits
This just says that by keeping x within 0.08 of 1, we are able to keep f sxd within 0.2
of 2.
Although we chose 5 0.08, any smaller positive value of would also have
worked.
■
TEC In Module 1.7/3.4 you can
explore the precise definition of a limit
both graphically and numerically.
The graphical procedure in Example 1 gives an illustration of the definition for
« 5 0.2, but it does not prove that the limit is equal to 2. A proof has to provide a for
every «.
In proving limit statements it may be helpful to think of the definition of limit as a
challenge. First it challenges you with a number «. Then you must be able to produce a
suitable . You have to be able to do this for every « . 0, not just a particular «.
Imagine a contest between two people, A and B, and imagine yourself to be B. Person
A stipulates that the fixed number L should be approximated by the values of f sxd to within
a degree of accuracy « (say, 0.01). Person B then responds by finding a number such
that if 0 , x 2 a , , then f sxd 2 L , «. Then A may become more exacting and
challenge B with a smaller value of « (say, 0.0001). Again B has to respond by finding a
corresponding . Usually the smaller the value of «, the smaller the corresponding value
of must be. If B always wins, no matter how small A makes «, then lim x l a f sxd 5 L.
|
|
|
|
Example 2 Prove that lim s4x 2 5d − 7.
x l3
SOLUTION 1. Preliminary analysis of the problem (guessing a value for ). Let « be a given
positive number. We want to find a number such that
|
|
|
|
if 0 , x 2 3 , then s4x 2 5d 2 7 , «
|
| |
| |
|
|
|
|
|
But s4x 2 5d 2 7 5 4x 2 12 5 4sx 2 3d 5 4 x 2 3 . Therefore we want such that
|
|
|
|
if 0 , x 2 3 , then 4 x 2 3 , «
that is,
|
|
if 0 , x 2 3 , then x 2 3 ,
«
4
This suggests that we should choose 5 «y4.
2. Proof (showing that this works). Given « . 0, choose 5 «y4. If
0 , x 2 3 , , then
|
|
SD
| s4x 2 5d 2 7 | − | 4x 2 12 | − 4| x 2 3 | , 4 − 4
«
4
−«
Thus
|
|
|
|
if 0 , x 2 3 , then s4x 2 5d 2 7 , «
Therefore, by the definition of a limit,
lim s4x 2 5d − 7
x l3
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Section 1.7 The Precise Definition of a Limit
Cauchy and Limits
After the invention of calculus in the
17th century, there followed a period
of free development of the subject in
the 18th century. Mathematicians like
the Bernoulli brothers and Euler were
eager to exploit the power of calculus
and boldly explored the consequences
of this new and wonderful mathematical theory without worrying too much
about whether their proofs were completely correct.
The 19th century, by contrast, was the
Age of Rigor in mathematics. There was
a movement to go back to the foundations of the subject—to provide careful
definitions and rigorous proofs. At the
forefront of this movement was the
French mathematician Augustin-Louis
Cauchy (1789–1857), who started out as
a military engineer before becoming a
mathematics professor in Paris. Cauchy
took Newton’s idea of a limit, which was
kept alive in the 18th century by the
French mathematician Jean d’Alembert,
and made it more precise. His definition
of a limit reads as follows: “When the
successive values attributed to a variable approach indefinitely a fixed value
so as to end by differing from it by as
little as one wishes, this last is called the
limit of all the others.” But when Cauchy
used this definition in examples and
proofs, he often employed delta-epsilon
inequalities similar to the ones in this
section. A typical Cauchy proof starts
with: “Designate by and « two very
small numbers; . . .” He used « because
of the correspondence between epsilon and the French word erreur and because delta corresponds to différence.
Later, the German mathematician Karl
Weierstrass (1815–1897) stated the
definition of a limit exactly as in our
Definition 2.
77
This example is illustrated by Figure 9.
y
y=4x-5
7+∑
7
7-∑
0
figure 9
x
3
3-∂
3+∂
■
Note that in the solution of Example 2 there were two stages—guessing and proving.
We made a preliminary analysis that enabled us to guess a value for . But then in the
second stage we had to go back and prove in a careful, logical fashion that we had made
a correct guess. This procedure is typical of much of mathematics. Sometimes it is necessary to first make an intelligent guess about the answer to a problem and then prove that
the guess is correct.
The intuitive definitions of one-sided limits that were given in Section 1.5 can be pre­
cisely reformulated as follows.
3 Definition of Left-Hand Limit lim f sxd − L
x l a2
if for every number « . 0 there is a number . 0 such that
|
|
if a 2 , x , a then f sxd 2 L , «
4 Definition of Right-Hand Limit lim f sxd − L
x la1
if for every number « . 0 there is a number . 0 such that
|
|
if a , x , a 1 then f sxd 2 L , «
Notice that Definition 3 is the same as Definition 2 except that x is restricted to lie in
the left half sa 2 , ad of the interval sa 2 , a 1 d. In Definition 4, x is restricted to lie
in the right half sa, a 1 d of the interval sa 2 , a 1 d.
Example 3 Use Definition 4 to prove that lim1 sx − 0.
xl0
SOLUTION 1. Guessing a value for . Let « be a given positive number. Here a 5 0 and L 5 0,
so we want to find a number such that
|
|
if 0 , x , then sx 2 0 , «
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78
Chapter 1 Functions and Limits
that is,
if 0 , x , then sx , «
or, squaring both sides of the inequality sx , «, we get
if 0 , x , then x , « 2
This suggests that we should choose 5 « 2.
2. Showing that this works. Given « . 0, let 5 « 2. If 0 , x , , then
sx , s 5 s« 2 5 «
| sx 2 0 | , «
so
According to Definition 4, this shows that lim x l 01 sx − 0.
■
Example 4 Prove that lim x 2 5 9.
xl3
SOLUTION 1. Guessing a value for . Let « . 0 be given. We have to find a number . 0
such that
|
|
|
|
if 0 , x 2 3 , then x 2 2 9 , «
|
|
|
|
|
| |
|
To connect x 2 2 9 with x 2 3 we write x 2 2 9 5 sx 1 3dsx 2 3d . Then
we want
|
|
|
||
|
if 0 , x 2 3 , then x 1 3 x 2 3 , «
|
|
Notice that if we can find a positive constant C such that x 1 3 , C, then
| x 1 3 || x 2 3 | , C | x 2 3 |
|
|
|
|
and we can make C x 2 3 , « by taking x 2 3 , «yC, so we could choose
− «yC.
We can find such a number C if we restrict x to lie in some interval centered at 3.
In fact, since we are interested only in values of x that are close to 3, it is reasonable
to assume that x is within a distance l from 3, that is, x 2 3 , 1. Then 2 , x , 4,
so 5 , x 1 3 , 7. Thus we have x 1 3 , 7, and so C 5 7 is a suitable choice for
the constant.
But now there are two restrictions on x 2 3 , namely
|
|
|
|
|
|
«
«
| x 2 3 | , 1 and | x 2 3 | , C 5 7
To make sure that both of these inequalities are satisfied, we take to be the smaller of
the two numbers 1 and «y7. The notation for this is 5 minh1, «y7j.
2. Showing that this works. Given « . 0, let 5 minh1, «y7j. If
0 , x 2 3 , , then x 2 3 , 1 ? 2 , x , 4 ? x 1 3 , 7 (as in part l).
|
|
|
|
|
|
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
79
Section 1.7 The Precise Definition of a Limit
|
|
We also have x 2 3 , «y7, so
|x
2
| |
||
|
29 − x13 x23 ,7
«
−«
7
This shows that lim x l3 x 2 5 9.
■
As Example 4 shows, it is not always easy to prove that limit statements are true
using the «, definition. In fact, if we had been given a more complicated function such
as f sxd 5 s6x 2 2 8x 1 9dys2x 2 2 1d, a proof would require a great deal of ingenuity.
Fortunately this is unnecessary because the Limit Laws stated in Section 1.6 can be
proved using Definition 2, and then the limits of complicated functions can be found
rigorously from the Limit Laws without resorting to the definition directly.
For instance, we prove the Sum Law: If lim x l a f sxd 5 L and lim x l a tsxd 5 M both
exist, then
lim f f sxd 1 tsxdg − L 1 M
xla
The remaining laws are proved in the exercises and in Appendix F.
Proof of the Sum Law Let « . 0 be given. We must find . 0 such that
|
|
|
|
if 0 , x 2 a , then f sxd 1 tsxd 2 sL 1 Md , «
Triangle Inequality:
|a 1 b| < |a| 1 |b|
(See Appendix A).
Using the Triangle Inequality we can write
| f sxd 1 tsxd 2 sL 1 Md | 5 | s f sxd 2 Ld 1 stsxd 2 Md |
< | f sxd 2 L | 1 | tsxd 2 M |
We make | f sxd 1 tsxd 2 sL 1 Md | less than « by making each of the terms | f sxd 2 L |
and | tsxd 2 M | less than «y2.
5
Since «y2 . 0 and lim x l a f sxd 5 L, there exists a number 1 . 0 such that
|
|
|
«
2
|
if 0 , x 2 a , 1 then f sxd 2 L ,
Similarly, since lim x l a tsxd − M, there exists a number 2 . 0 such that
|
|
|
«
2
|
if 0 , x 2 a , 2 then tsxd 2 M ,
Let − minh1, 2j, the smaller of the numbers 1 and 2. Notice that
|
|
|
|
|
|
if 0 , x 2 a , then 0 , x 2 a , 1 and 0 , x 2 a , 2
«
«
| f sxd 2 L | , 2 and | tsxd 2 M | , 2
and so
Therefore, by (5),
| f sxd 1 tsxd 2 sL 1 Md | < | f sxd 2 L | 1 | tsxd 2 M |
,
«
«
1 5«
2
2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
80
Chapter 1 Functions and Limits
To summarize,
|
|
|
|
if 0 , x 2 a , then f sxd 1 tsxd 2 sL 1 Md , «
Thus, by the definition of a limit,
lim f f sxd 1 tsxdg − L 1 M
■
xla
Infinite Limits
Infinite limits can also be defined in a precise way. The following is a precise version of
Definition 1.5.4.
6 Precise Definition of an Infinite Limit Let f be a function defined on some
open interval that contains the number a, except possibly at a itself. Then
lim f sxd − `
xla
means that for every positive number M there is a positive number such that
|
|
if 0 , x 2 a , then f sxd . M
y
y=M
M
0
a-∂
a
x
a+∂
figure 10
This says that the values of f sxd can be made arbitrarily large (larger than any given
number M) by requiring x to be close enough to a (within a distance , where depends
on M, but with x ± a). A geometric illustration is shown in Figure 10.
Given any horizontal line y 5 M, we can find a number . 0 such that if we restrict
x to lie in the interval sa 2 , a 1 d but x ± a, then the curve y 5 f sxd lies above the
line y − M. You can see that if a larger M is chosen, then a smaller may be required.
1
− `.
x2
SOLUTION Let M be a given positive number. We want to find a number such that
Example 5 Use Definition 6 to prove that lim
xl0
|
|
if 0 , x 2 0 , then 1yx 2 . M
But
1
1
. M &? x 2 , &? sx 2 ,
x2
M
Î
1
1
&? x ,
M
sM
| |
| |
So if we choose − 1ysM and 0 , x , − 1ysM , then 1yx 2 . M. This shows
that 1yx 2 l ` as x l 0.
n
Similarly, the following is a precise version of Definition 1.5.5. It is illustrated by
Figure 11.
y
a-∂
a+∂
a
0
x
7 Definition Let f be a function defined on some open interval that contains
the number a, except possibly at a itself. Then
lim f sxd − 2`
y=N
N
FIGURE 11 xla
means that for every negative number N there is a positive number such that
|
|
if 0 , x 2 a , then f sxd , N
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
81
Section 1.7 The Precise Definition of a Limit
1.Use the given graph of f to find a number such that
|
|
|
|
if x 2 1 , then f sxd 2 1 , 0.2
; 5.Use a graph to find a number such that
Z
if x 2
y
1.2
1
0.8
4
Z
|
|
, then tan x 2 1 , 0.2
; 6.Use a graph to find a number such that
|
Z
|
if x 2 1 , then 0
0.7
; 7.For the limit
x
1 1.1
Z
2x
2 0.4 , 0.1
x2 1 4
lim sx 3 2 3x 1 4d 5 6
xl2
2.Use the given graph of f to find a number such that
|
|
|
|
if 0 , x 2 3 , then f s xd 2 2 , 0.5
illustrate Definition 2 by finding values of that correspond
to « 5 0.2 and « 5 0.1.
; 8.For the limit
y
lim
2.5
x l2
2
4x 1 1
− 4.5
3x 2 4
illustrate Definition 2 by finding values of that correspond
to « 5 0.5 and « 5 0.1.
1.5
; 9.(a) Use a graph to find a number such that
0
2.6 3
x
3.8
3.Use the given graph of f sxd − sx to find a number such
that
|
|
|
|
if x 2 4 , then sx 2 2 , 0.4
y
y=œ„
x
2.4
2
1.6
0
4
?
?
x
4.Use the given graph of f sxd 5 x 2 to find a number such that
|
|
|
|
if x 2 1 , then x 2 2 1 , 12
y
y=≈
1.5
x2 1 4
sx 2 4
. 100
(b) What limit does part (a) suggest is true?
iven that lim x l csc2 x − `, illustrate Definition 6 by
; 10. G
finding values of that correspond to (a) M − 500 and
(b) M − 1000.
11.A machinist is required to manufacture a circular metal disk
with area 1000 cm2.
(a)What radius produces such a disk?
(b)If the machinist is allowed an error tolerance of 65 cm 2
in the area of the disk, how close to the ideal radius in
part (a) must the machinist control the radius?
(c)In terms of the «, definition of lim x l a f sxd 5 L, what
is x? What is f sxd? What is a? What is L? What value of
« is given? What is the corresponding value of ?
crystal growth furnace is used in research to determine how
; 12. A
best to manufacture crystals used in electronic components for
the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input
power. Suppose the relationship is given by
T swd − 0.1w 2 1 2.155w 1 20
1
0.5
0
if 4 , x , 4 1 then ?
1
?
x
where T is the temperature in degrees Celsius and w is the
power input in watts.
(a)How much power is needed to maintain the temperature
at 200°C?
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82
chapter 1 Functions and Limits
(b)If the temperature is allowed to vary from 200°C by
up to 61°C, what range of wattage is allowed for the
input power?
(c)In terms of the «, definition of lim x l a f sxd 5 L, what
is x? What is f sxd? What is a? What is L? What value
of « is given? What is the corresponding value of ?
|
|
13. (a)Find a number such that if x 2 2 , , then
4x 2 8 , «, where « 5 0.1.
(b) Repeat part (a) with « 5 0.01.
|
|
14.Given that limx l 2 s5x 2 7d 5 3, illustrate Definition 2 by
finding values of that correspond to « 5 0.1, « 5 0.05,
and « 5 0.01.
15–18 Prove the statement using the «, definition of a limit
and illustrate with a diagram like Figure 9.
15. lim s1 1 13 xd − 2
16. lim s2x 2 5d 5 3
17. lim s1 2 4xd − 13
18. lim s3x 1 5d − 21
xl3
xl23
xl4
xl22
CAS
35. (a)For the limit lim x l 1 sx 3 1 x 1 1d 5 3, use a graph to
find a value of that corresponds to « 5 0.4.
(b)By using a computer algebra system to solve the cubic
equation x 3 1 x 1 1 5 3 1 «, find the largest possible
value of that works for any given « . 0.
(c)Put « 5 0.4 in your answer to part (b) and compare
with your answer to part (a).
36. Prove that lim
x l2
1
1
− .
x
2
37.Prove that lim sx − sa if a . 0.
F
xla
|
Hint: Use sx 2 sa
|
−
|x 2 a|
sx 1 sa
.
F
38.If H is the Heaviside function defined in Example 1.5.6,
prove, using Definition 2, that lim t l 0 Hstd does not exist.
[Hint: Use an indirect proof as follows. Suppose that the
limit is L. Take « 5 12 in the definition of a limit and try to
arrive at a contradiction.]
19–32 Prove the statement using the «, definition of a limit.
2 1 4x
19. lim
52
20. lim s3 2 45 xd − 25
x l1
x l 10
3
x 2 2 2x 2 8
9 2 4x 2
−6
22. lim
−6
21. lim
xl4
x l21.5 3 1 2x
x24
39. If the function f is defined by
23. lim x − a
24. lim c − c
40.By comparing Definitions 2, 3, and 4, prove Theorem 1.6.1.
25. lim x 2 − 0
26. lim x 3 − 0
xla
xl0
| |
xla
f sxd −
H
0 if x is rational
1 if x is irrational
prove that lim x l 0 f sxd does not exist.
41. How close to 23 do we have to take x so that
1
. 10,000
sx 1 3d4
xl0
27. lim x − 0
28. lim1 s6 1 x − 0
29. lim sx 2 4x 1 5d 5 1
30. lim sx 2 1 2x 2 7d 5 1
42. Prove, using Definition 6, that lim
31. lim sx 2 1d 5 3
32. lim x 3 5 8
43. Prove that lim 2
xl0
2
xl2
2
x l 22
8
xl 26
x l23
xl2
xl2
33.Verify that another possible choice of for showing that
lim x l3 x 2 5 9 in Example 4 is 5 min h2, «y8j.
34.Verify, by a geometric argument, that the largest possible choice of for showing that lim x l3 x 2 − 9 is
− s 9 1 « 2 3.
x l21
1
− `.
sx 1 3d4
5
− 2`.
sx 1 1d 3
44.Suppose that lim x l a f sxd 5 ` and lim x l a tsxd 5 c, where
c is a real number. Prove each statement.
(a) lim f f sxd 1 tsxdg − `
xla
(b) lim f f sxd tsxdg 5 ` if c . 0
xla
(c) lim f f sxd tsxdg 5 2` if c , 0
xl a
We noticed in Section 1.6 that the limit of a function as x approaches a can often be
found simply by calculating the value of the function at a. Functions with this property
are called continuous at a. We will see that the mathematical definition of continuity corresponds closely with the meaning of the word continuity in everyday language. (A continuous process is one that takes place gradually, without interruption or abrupt change.)
1 Definition A function f is continuous at a number a if
lim f sxd − f sad
xl a
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Section 1.8 Continuity
As illustrated in Figure 1, if f is continuous, then the points sx, f sxdd on
the graph of f approach the point
sa, f sadd on the graph. So there is no
gap in the curve.
Notice that Definition l implicitly requires three things if f is continuous at a:
1. f sad is defined (that is, a is in the domain of f )
2. lim f sxd exists
x la
3. lim f sxd 5 f sad
y
ƒ
approaches
f(a).
83
x la
y=ƒ
f(a)
0
x
a
As x approaches a,
figure 1
y
The definition says that f is continuous at a if f sxd approaches f sad as x approaches
a. Thus a continuous function f has the property that a small change in x produces only
a small change in f sxd. In fact, the change in f sxd can be kept as small as we please by
keeping the change in x sufficiently small.
If f is defined near a (in other words, f is defined on an open interval containing a,
except perhaps at a), we say that f is discontinuous at a (or f has a discontinuity at a)
if f is not continuous at a.
Physical phenomena are usually continuous. For instance, the displacement or velocity of a vehicle varies continuously with time, as does a person’s height. But discontinuities do occur in such situations as electric currents. [See Example 1.5.6, where the
Heaviside function is discontinuous at 0 because lim t l 0 Hstd does not exist.]
Geometrically, you can think of a function that is continuous at every number in an
interval as a function whose graph has no break in it: the graph can be drawn without
removing your pen from the paper.
Example 1 Figure 2 shows the graph of a function f. At which numbers is f discontinuous? Why?
0
1
figure 2
2
3
4
5
x
SOLUTION It looks as if there is a discontinuity when a − 1 because the graph has a
break there. The official reason that f is discontinuous at 1 is that f s1d is not defined.
The graph also has a break when a 5 3, but the reason for the discontinuity is
different. Here, f s3d is defined, but lim x l3 f sxd does not exist (because the left and
right limits are different). So f is discontinuous at 3.
What about a 5 5? Here, f s5d is defined and lim x l5 f sxd exists (because the left
and right limits are the same). But
lim f sxd ± f s5d
xl5
So f is discontinuous at 5.
n
Now let’s see how to detect discontinuities when a function is defined by a formula.
Example 2 Where are each of the following functions discontinuous?
(a) f sxd 5
(c) f sxd −
x2 2 x 2 2
x22
H
x 2 2 x 2 2 if x ± 2
x22
1
if x − 2
(b) f sxd −
H
1
if x ± 0
x2
1
if x − 0
(d) f sxd − v x b
SOLUTION (a) Notice that f s2d is not defined, so f is discontinuous at 2. Later we’ll see why f is
continuous at all other numbers.
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84
Chapter 1 Functions and Limits
(b) Here f s0d 5 1 is defined but
lim f sxd − lim
xl0
xl0
1
x2
does not exist. (See Example 1.5.8.) So f is discontinuous at 0.
(c) Here f s2d 5 1 is defined and
lim f sxd − lim
x l2
x l2
x2 2 x 2 2
sx 2 2dsx 1 1d
− lim
− lim sx 1 1d − 3
x
l2
x l2
x22
x22
exists. But
lim f sxd ± f s2d
x l2
so f is not continuous at 2.
(d) The greatest integer function f sxd − v x b has discontinuities at all of the inte­gers because lim x ln v x b does not exist if n is an integer. (See Example 1.6.10 and
Exercise 1.6.53.)
n
Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t be
drawn without lifting the pen from the paper because a hole or break or jump occurs in the
graph. The kind of discontinuity illustrated in parts (a) and (c) is called removable because
we could remove the discontinuity by redefining f at just the single number 2. [The
func­tion tsxd − x 1 1 is continuous.] The discontinuity in part (b) is called an infinite
discontinuity. The discontinuities in part (d) are called jump discontinuities because
the function “jumps” from one value to another.
y
y
y
y
1
1
1
1
0
(a) ƒ=
1
2
x
≈-x-2
x-2
0
1
if x≠0
(b) ƒ= ≈
1
if x=0
0
x
(c) ƒ=
1
2
x
≈-x-2
if x≠2
x-2
1
if x=2
0
1
2
3
x
(d) ƒ=[ x ]
FIGURE 3 Graphs of the functions in Example 2
2 Definition A function f is continuous from the right at a number a if
lim f sxd − f sad
x l a1
and f is continuous from the left at a if
lim f sxd − f sad
x l a2
Example 3 At each integer n, the function f sxd − v x b [see Figure 3(d)] is continuous from the right but discontinuous from the left because
lim f sxd − lim1 v xb − n − f snd
x l n1
x ln
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Section 1.8 Continuity
85
lim f sxd − lim2 v x b − n 2 1 ± f snd
but
x l n2
x ln
n
3 Definition A function f is continuous on an interval if it is continuous at
every number in the interval. (If f is defined only on one side of an endpoint of the
interval, we understand continuous at the endpoint to mean continuous from the
right or continuous from the left.)
Example 4 Show that the function f sxd − 1 2 s1 2 x 2 is continuous on the
interval f21, 1g.
SOLUTION If 21 , a , 1, then using the Limit Laws, we have
lim f sxd − lim (1 2 s1 2 x 2 )
xla
xla
− 1 2 lim s1 2 x 2
(by Laws 2 and 7)
xla
− 1 2 s lim s1 2 x 2 d (by 11)
xla
− 1 2 s1 2 a 2
(by 2, 7, and 9)
− f sad
y
1
-1
0
Thus, by Definition l, f is continuous at a if 21 , a , 1. Similar calculations show that
ƒ=1-œ„„„„„
1-≈
lim f sxd − 1 − f s21d and lim2 f sxd − 1 − f s1d
x l 211
1
x
x l1
so f is continuous from the right at 21 and continuous from the left at 1. Therefore,
according to Definition 3, f is continuous on f21, 1g.
The graph of f is sketched in Figure 4. It is the lower half of the circle
FIGURE 4 x 2 1 sy 2 1d2 − 1
n
Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function as
we did in Example 4, it is often convenient to use the next theorem, which shows how to
build up complicated continuous functions from simple ones.
4 Theorem If f and t are continuous at a and if c is a constant, then the
following functions are also continuous at a:
1. f 1 t
2. f 2 t
4. ft
5. 3. cf
f
if tsad ± 0
t
Proof Each of the five parts of this theorem follows from the corresponding Limit
Law in Section 1.6. For instance, we give the proof of part 1. Since f and t are continuous at a, we have
lim f sxd − f sad and lim tsxd − tsad
xla
xla
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86
Chapter 1 Functions and Limits
Therefore
lim s f 1 tdsxd − lim f f sxd 1 tsxdg
xla
xla
− lim f sxd 1 lim tsxd (by Law 1)
xla
xla
− f sad 1 tsad
− s f 1 tdsad
This shows that f 1 t is continuous at a.
n
It follows from Theorem 4 and Definition 3 that if f and t are continuous on an interval, then so are the functions f 1 t, f 2 t, cf, ft, and (if t is never 0) fyt. The following
theorem was stated in Section 1.6 as the Direct Substitution Property.
5 Theorem (a)Any polynomial is continuous everywhere; that is, it is continuous on
R − s2`, `d.
(b)Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.
Proof
(a) A polynomial is a function of the form
Psxd − cn x n 1 cn21 x n21 1 ∙ ∙ ∙ 1 c1 x 1 c0
where c0 , c1, . . . , cn are constants. We know that
lim c0 − c0 (by Law 7)
xla
and
lim x m − a m m − 1, 2, . . . , n (by 9)
xla
This equation is precisely the statement that the function f sxd − x m is a continuous
function. Thus, by part 3 of Theorem 4, the function tsxd − cx m is continuous. Since P
is a sum of functions of this form and a constant function, it follows from part 1 of
Theorem 4 that P is continuous.
(b) A rational function is a function of the form
f sxd −
Psxd
Qsxd
|
where P and Q are polynomials. The domain of f is D − hx [ R Qsxd ± 0j. We
know from part (a) that P and Q are continuous everywhere. Thus, by part 5 of Theorem 4, f is continuous at every number in D.
n
As an illustration of Theorem 5, observe that the volume of a sphere varies continuously with its radius because the formula Vsrd − 43 r 3 shows that V is a polynomial function of r. Likewise, if a ball is thrown vertically into the air with a velocity
of 50 ftys, then the height of the ball in feet t seconds later is given by the formula
h − 50t 2 16t 2. Again this is a polynomial function, so the height is a continuous function of the elapsed time, as we might expect.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.8 Continuity
87
Knowledge of which functions are continuous enables us to evaluate some limits very
quickly, as the following example shows. Compare it with Example 1.6.2(b).
Example 5 Find lim
x l 22
x 3 1 2x 2 2 1
.
5 2 3x
SOLUTION The function
f sxd −
x 3 1 2x 2 2 1
5 2 3x
|
is rational, so by Theorem 5 it is continuous on its domain, which is h x x ± 53 j.
Therefore
lim
x l22
x 3 1 2x 2 2 1
− lim f sxd − f s22d
x l22
5 2 3x
−
y
P(cos ¨, sin ¨)
1
0
¨
(1, 0)
x
FIGURE 5 n
It turns out that most of the familiar functions are continuous at every number in their
domains. For instance, Limit Law 10 (page 64) is exactly the statement that root functions are continuous.
From the appearance of the graphs of the sine and cosine functions (Figure 1.2.18),
we would certainly guess that they are continuous. We know from the definitions of sin and cos that the coordinates of the point P in Figure 5 are scos , sin d. As l 0, we
see that P approaches the point s1, 0d and so cos l 1 and sin l 0. Thus
6 Another way to establish the limits in
(6) is to use the Squeeze Theorem with
the inequality sin , (for . 0),
which is proved in Section 2.4.
s22d3 1 2s22d2 2 1
1
−2
5 2 3s22d
11
lim cos − 1
lim sin − 0
l0
l0
Since cos 0 − 1 and sin 0 − 0, the equations in (6) assert that the cosine and sine functions are continuous at 0. The addition formulas for cosine and sine can then be used to
deduce that these functions are continuous everywhere (see Exercises 64 and 65).
It follows from part 5 of Theorem 4 that
tan x −
sin x
cos x
is continuous except where cos x − 0. This happens when x is an odd integer multiple
of y2, so y − tan x has infinite discontinuities when x − 6y2, 63y2, 65y2, and
so on (see Figure 6).
y
1
3π _π
_ 2
_
π
2
0
π
2
π
3π
2
x
FIGURE 6 y − tan x
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88
Chapter 1 Functions and Limits
7 Theorem The following types of functions are continuous at every number in
their domains:
•
•
•
•
polynomials
root functions
rational functions
trigonometric functions
Example 6 On what intervals is each function continuous?
(b) tsxd −
(a) f sxd − x 100 2 2x 37 1 75
(c) hsxd − sx 1
x 2 1 2x 1 17
x2 2 1
x11
x11
2 2
x21
x 11
SOLUTION
(a) f is a polynomial, so it is continuous on s2`, `d by Theorem 5(a).
(b) t is a rational function, so by Theorem 5(b), it is continuous on its domain,
which is D − hx x 2 2 1 ± 0j − hx x ± 61j. Thus t is continuous on the intervals
s2`, 21d, s21, 1d, and s1, `d.
(c) We can write hsxd − Fsxd 1 Gsxd 2 Hsxd, where
|
|
Fsxd − sx Gsxd −
x11
x11
Hsxd − 2
x21
x 11
F is continuous on [0, `d by Theorem 7. G is a rational function, so it is continuous
everywhere except when x 2 1 − 0, that is, x − 1. H is also a rational function, but its
denominator is never 0, so H is continuous everywhere. Thus, by parts 1 and 2 of
Theorem 4, h is continuous on the intervals [0, 1d and s1, `d.
■
sin x
.
2 1 cos x
SOLUTION Theorem 7 tells us that y − sin x is continuous. The function in the
denomi­nator, y − 2 1 cos x, is the sum of two continuous functions and is therefore
continuous. Notice that this function is never 0 because cos x > 21 for all x and so
2 1 cos x . 0 everywhere. Thus the ratio
sin x
f sxd −
2 1 cos x
Example 7 Evaluate lim
x l
is continuous everywhere. Hence, by the definition of a continuous function,
lim
x l
sin x
sin 0
− lim f sxd − f sd −
−
−0
x
l
2 1 cos x
2 1 cos 221
n
Another way of combining continuous functions f and t to get a new continuous
function is to form the composite function f 8 t. This fact is a consequence of the following theorem.
This theorem says that a limit symbol
can be moved through a function symbol if the function is continuous and the
limit exists. In other words, the order of
these two symbols can be reversed.
8 Theorem If f is continuous at b and lim tsxd − b, then lim f stsxdd − f sbd.
x la
x la
In other words,
lim f stsxdd − f lim tsxd
xla
S
xl a
D
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.8 Continuity
89
Intuitively, Theorem 8 is reasonable because if x is close to a, then tsxd is close to b,
and since f is continuous at b, if tsxd is close to b, then fstsxdd is close to f sbd. A proof
of Theorem 8 is given in Appendix F.
n
Let’s now apply Theorem 8 in the special case where f sxd − s
x , with n being a positive integer. Then
n
tsxd
f stsxdd − s
and
S
D
f lim tsxd −
xla
tsxd
s xlim
la
n
If we put these expressions into Theorem 8, we get
n
n lim tsxd
lim s
tsxd − s
xla
xla
and so Limit Law 11 has now been proved. (We assume that the roots exist.)
9 Theorem If t is continuous at a and f is continuous at tsad, then the composite function f 8 t given by s f 8 tds xd − f stsxdd is continuous at a.
This theorem is often expressed informally by saying “a continuous function of a continuous function is a continuous function.”
Proof Since t is continuous at a, we have
lim tsxd − tsad
xla
Since f is continuous at b − tsad, we can apply Theorem 8 to obtain
lim f stsxdd − f stsadd
xla
which is precisely the statement that the function hsxd − f s tsxdd is continuous at a;
that is, f 8 t is continuous at a.
n
Example 8 Where are the following functions continuous?
(a) hsxd − sinsx 2 d
(b) Fsxd −
1
sx 1 7 2 4
2
SOLUTION (a) We have hsxd − f s tsxdd, where
tsxd − x 2 and f sxd − sin x
Now t is continuous on R since it is a polynomial, and f is also continuous everywhere.
Thus h − f 8 t is continuous on R by Theorem 9.
(b) Notice that F can be broken up as the composition of four continuous functions:
F − f + t + h + k or Fsxd − f stshsksxdddd
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90
Chapter 1 Functions and Limits
f sxd −
where
1
tsxd − x 2 4 hsxd − sx ksxd − x 2 1 7
x
We know that each of these functions is continuous on its domain (by Theorems 5 and
7), so by Theorem 9, F is continuous on its domain, which is
h x [ R | sx 2 1 7
|
± 4 j − hx x ± 63j − s2`, 23d ø s23, 3d ø s3, `d
n
An important property of continuous functions is expressed by the following theorem,
whose proof is found in more advanced books on calculus.
10 The Intermediate Value Theorem Suppose that f is continuous on the
closed interval fa, bg and let N be any number between f sad and f sbd, where
f sad ± f sbd. Then there exists a number c in sa, bd such that f scd − N.
The Intermediate Value Theorem states that a continuous function takes on every
intermediate value between the function values f sad and f sbd. It is illustrated by Figure
7. Note that the value N can be taken on once [as in part (a)] or more than once [as in
part (b)].
y
y
f(b)
f(b)
N
y=ƒ
f(a)
0
a
FIGURE 7
y
f(a)
y=ƒ
N
y=N
f(b)
0
a
FIGURE 8 b
x
y=ƒ
N
f(a)
c b
x
0
a c¡
(a)
c™
c£
b
x
(b)
If we think of a continuous function as a function whose graph has no hole or break,
then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms
it says that if any horizontal line y − N is given between y − f sad and y − f sbd as in Figure 8, then the graph of f can’t jump over the line. It must intersect y − N somewhere.
It is important that the function f in Theorem 10 be continuous. The Intermediate
Value Theorem is not true in general for discontinuous functions (see Exercise 50).
One use of the Intermediate Value Theorem is in locating roots of equations as in the
following example.
Example 9 Show that there is a root of the equation
4x 3 2 6x 2 1 3x 2 2 − 0
between 1 and 2.
SOLUTION Let f sxd − 4x 3 2 6x 2 1 3x 2 2. We are looking for a solution of the given
equation, that is, a number c between 1 and 2 such that f scd − 0. Therefore we take
a − 1, b − 2, and N − 0 in Theorem 10. We have
f s1d − 4 2 6 1 3 2 2 − 21 , 0
and
f s2d − 32 2 24 1 6 2 2 − 12 . 0
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
91
Section 1.8 Continuity
Thus f s1d , 0 , f s2d; that is, N − 0 is a number between f s1d and f s2d. Now f is
continuous since it is a polynomial, so the Intermediate Value Theorem says there
is a number c between 1 and 2 such that f scd − 0. In other words, the equation
4x 3 2 6x 2 1 3x 2 2 − 0 has at least one root c in the interval s1, 2d.
In fact, we can locate a root more precisely by using the Intermediate Value Theorem
again. Since
0.2
3
f s1.2d − 20.128 , 0 and f s1.3d − 0.548 . 0
3
_1
f s1.22d − 20.007008 , 0 and f s1.23d − 0.056068 . 0
_3
_0.2
FIGURE 9
so a root lies in the interval s1.22, 1.23d.
0.2
3
1.3
1.2
a root must lie between 1.2 and 1.3. A calculator gives, by trial and error,
1.3
1.2
_0.2
FIGURE 10
n
We can use a graphing calculator or computer to illustrate the use of the Intermediate
Value Theorem in Example 9. Figure 9 shows the graph of f in the viewing rectangle
f21, 3g by f23, 3g and you can see that the graph crosses the x-axis between 1 and 2. Fig­
ure 10 shows the result of zooming in to the viewing rectangle f1.2, 1.3g by f20.2, 0.2g.
In fact, the Intermediate Value Theorem plays a role in the very way these graphing
devices work. A computer calculates a finite number of points on the graph and turns on
the pixels that contain these calculated points. It assumes that the function is continuous
and takes on all the intermediate values between two consecutive points. The computer
therefore “connects the dots” by turning on the intermediate pixels.
1.Write an equation that expresses the fact that a function f is
continuous at the number 4.
4.From the graph of t, state the intervals on which t is
continuous.
2.If f is continuous on s2`, `d, what can you say about its
graph?
y
3. (a)From the graph of f , state the numbers at which f is discontinuous and explain why.
(b)For each of the numbers stated in part (a), determine
whether f is continuous from the right, or from the left,
or neither.
_3
_2
0
1
2
3
x
y
5 – 8 Sketch the graph of a function f that is continuous except for
the stated discontinuity.
5.Discontinuous at 2, but continuous from the right there
_4
_2
0
2
4
6
x
6. Discontinuities at 21 and 4, but continuous from the left at 21
and from the right at 4
7.Removable discontinuity at 3, jump discontinuity at 5
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92
chapter 1 Functions and Limits
8.Neither left nor right continuous at 22, continuous only
from the left at 2
9. The toll T charged for driving on a certain stretch of a toll
road is $5 except during rush hours (between 7 am and
10 am and between 4 pm and 7 pm) when the toll is $7.
(a)Sketch a graph of T as a function of the time t, measured in hours past midnight.
(b)Discuss the discontinuities of this function and their
significance to someone who uses the road.
10. Explain why each function is continuous or discontinuous.
(a)The temperature at a specific location as a function
of time
(b)The temperature at a specific time as a function of the
distance due west from New York City
(c)The altitude above sea level as a function of the distance due west from New York City
(d)The cost of a taxi ride as a function of the distance
traveled
(e)The current in the circuit for the lights in a room as a
function of time
11–14 Use the definition of continuity and the properties
of limits to show that the function is continuous at the given
number a.
t 2 1 5t
, a − 2
2t 1 1
a−0
2x 2 2 5x 2 3
22. f sxd −
x23
6
a−3
if x − 3
x2 2 x 2 2
x22
23. f sxd −
2x 2 2 x 2 1
x2 1 1
25.Fsxd −
3
x22
s
x3 2 2
Î
11
1
x
33. y −
1
1 1 sin x
x l2
17– 22 Explain why the function is discontinuous at the given
number a. Sketch the graph of the function.
18. f sxd −
1
x12
H
H
1
x12
1
a − 22
if x ± 22
a − 22
2
37. lim x tan x
x ly4
28. hsxd −
sin x
x11
30. Bsxd −
tan x
s4 2 x 2
32. Fsxd − sinscosssin xdd
34.
y − tan sx
39. f sxd −
40. f sxd −
a−1
36. lim sinsx 1 sin xd
x l
38. lim x 3ysx 2 1 x 2 2
xl2
39 – 40 Show that f is continuous on s2`, `d.
if x − 22
1 2 x 2 if x , 1
19. f sxd −
1yx
if x > 1
x2 1 1
2x 2 x 2 1
2
35 – 38 Use continuity to evaluate the limit.
35. lim x s20 2 x 2
17. f sxd −
26. Gsxd −
; 33 –34 Locate the discontinuities of the function and illustrate
by graphing.
15. f sxd − x 1 sx 2 4 , f4, `d
x21
, s2`, 22d
3x 1 6
x3 2 8
24.
f sxd − 2
x 24
25 – 32 Explain, using Theorems 4, 5, 7, and 9, why the function
is continuous at every number in its domain. State the domain.
2
15 –16 Use the definition of continuity and the properties of
limits to show that the function is continuous on the given
interval.
16. tsxd −
if x ± 3
23 –24 How would you “remove the discontinuity” of f ?
In other words, how would you define f s2d in order to make
f continuous at 2?
31. Msxd −
14. f sxd − 3x 2 5x 1 sx 1 4 , a − 2
3
a−1
if x − 1
29. hsxd − coss1 2 x 2 d
13. psvd − 2s3v 2 1 1 , a − 1
4
if x ± 1
cos x
if x , 0
21. f sxd − 0
if x − 0 1 2 x 2 if x . 0
27. Qsxd −
11. f sxd − sx 1 2x 3 d4, a − 21
12. t std −
H
H
H
x2 2 x
20. f sxd − x 2 2 1
1
H
H
1 2 x 2 if x < 1
sx 2 1 if x . 1
sin x if x , y4
cos x if x > y4
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 1.8 Continuity
41– 43 Find the numbers at which f is discontinuous. At which
of these numbers is f continuous from the right, from the left,
or neither? Sketch the graph of f .
H
H
H
x 2 if x , 21
if 21 < x , 1
41. f sxd − x
1yx if x > 1
x 2 1 1 if x < 1
42. f sxd − 3 2 x if 1 , x < 4
if x . 4
sx
x 1 2 if x , 0
43. f sxd − 2x 2
if 0 < x < 1
2 2 x if x . 1
44. T
he gravitational force exerted by the planet Earth on a unit
mass at a distance r from the center of the planet is
Fsrd −
GMr
if r , R
R3
GM
if r > R
r2
where M is the mass of Earth, R is its radius, and G is the
gravitational constant. Is F a continuous function of r?
45. For what value of the constant c is the function f continuous
on s2`, `d?
f sxd −
H
cx 2 1 2x if x , 2
x 3 2 cx
if x > 2
46. Find the values of a and b that make f continuous everywhere.
f sxd −
x2 2 4
if x , 2
x22
ax 2 2 bx 1 3 if 2 < x , 3
2x 2 a 1 b
if x > 3
47. Suppose f and t are continuous functions such that ts2d − 6
and lim x l2 f3 f sxd 1 f sxd tsxdg − 36. Find f s2d.
48. L
et f sxd − 1yx and tsxd − 1yx 2.
(a) Find s f + tds xd.
(b) Is f + t continuous everywhere? Explain.
49. W
hich of the following functions f has a removable discon­
tinuity at a? If the discontinuity is removable, find a function t
that agrees with f for x ± a and is continuous at a.
x4 2 1
(a) f sxd −
, a − 1
x21
(b) f sxd −
x 3 2 x 2 2 2x
, a − 2
x22
(c) f sxd − v sin x b , a − 93
50. S
uppose that a function f is continuous on [0, 1] except at
0.25 and that f s0d − 1 and f s1d − 3. Let N − 2. Sketch two
pos­sible graphs of f, one showing that f might not satisfy the
conclusion of the Intermediate Value Theorem and one showing that f might still satisfy the conclusion of the Intermediate
Value Theorem (even though it doesn’t satisfy the hypothesis).
51. If f sxd − x 2 1 10 sin x, show that there is a number c such
that f scd − 1000.
52. Suppose f is continuous on f1, 5g and the only solutions of the
equation f sxd − 6 are x − 1 and x − 4. If f s2d − 8, explain
why f s3d . 6.
53– 56 Use the Intermediate Value Theorem to show that there is a
root of the given equation in the specified interval.
53. x 4 1 x 2 3 − 0, s1, 2d
54. 2yx − x 2 sx , s2, 3d
55. cos x − x, s0, 1d
56. sin x − x 2 2 x, s1, 2d
57 – 58 (a) Prove that the equation has at least one real root.
(b) Use your calculator to find an interval of length 0.01 that contains a root.
57. cos x − x 3
58. x 5 2 x 2 1 2x 1 3 − 0
; 59– 60 (a) Prove that the equation has at least one real root.
(b) Use your graphing device to find the root correct to three
decimal places.
1
59. x 5 2 x 2 2 4 − 0
60. sx 2 5 −
x13
61– 62 Prove, without graphing, that the graph of the function has
at least two x-intercepts in the specified interval.
61. y − sin x 3, s1, 2d
62. y − x 2 2 3 1 1yx, s0, 2d
63. Prove that f is continuous at a if and only if
lim f sa 1 hd − f sad
hl0
64. T
o prove that sine is continuous, we need to show that
lim x l a sin x − sin a for every real number a. By Exercise 63
an equivalent statement is that
lim sinsa 1 hd − sin a
hl0
Use (6) to show that this is true.
65. Prove that cosine is a continuous function.
66. (a) Prove Theorem 4, part 3.
(b) Prove Theorem 4, part 5.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
94
chapter 1 Functions and Limits
67. For what values of x is f continuous?
H
0 if x is rational
f sxd −
1 if x is irrational
68. For what values of x is t continuous?
tsxd −
H
0 if x is rational
x if x is irrational
69. Is there a number that is exactly 1 more than its cube?
70. If a and b are positive numbers, prove that the equation
a
b
1 3
−0
x 3 1 2x 2 2 1
x 1x22
has at least one solution in the interval s21, 1d.
1
71. Show that the function
f sxd −
H
x 4 sins1yxd
0
if x ± 0
if x − 0
is continuous on s2`, `d.
| |
72. (a)Show that the absolute value function Fsxd − x is continuous everywhere.
(b)Prove that if f is a continuous function on an interval, then
so is f .
(c)Is the converse of the statement in part (b) also true? In
other words, if f is continuous, does it follow that f is
continuous? If so, prove it. If not, find a counterexample.
| |
| |
73.A Tibetan monk leaves the monastery at 7:00 am and takes his
usual path to the top of the mountain, arriving at 7:00 pm. The
following morning, he starts at 7:00 am at the top and takes
the same path back, arriving at the monastery at 7:00 pm. Use
the Intermediate Value Theorem to show that there is a point
on the path that the monk will cross at exactly the same time
of day on both days.
Review
CONCEPT CHECK
Answers to the Concept Check can be found on the back endpapers.
1.(a) What is a function? What are its domain and range?
(b) What is the graph of a function?
(c)How can you tell whether a given curve is the graph of
a function?
2.Discuss four ways of representing a function. Illustrate your
discussion with examples.
3.(a)What is an even function? How can you tell if a function
is even by looking at its graph? Give three examples of an
even function.
(b)What is an odd function? How can you tell if a function is
odd by looking at its graph? Give three examples of an
odd function.
4.What is an increasing function?
5.What is a mathematical model?
6.Give an example of each type of function.
(a) Linear function
(b) Power function
(c) Exponential function
(d) Quadratic function
(e) Polynomial of degree 5
(f ) Rational function
7.Sketch by hand, on the same axes, the graphs of the following
functions.
(a) f sxd − x
(b) tsxd − x 2
(c) hsxd − x 3
(d) jsxd − x 4
8.Draw, by hand, a rough sketch of the graph of each function.
(a) y − sin x
(b) y − cos x
(c) y − tan x
(d) y − 1yx
(e) y − x
(f) y − sx
| |
9.Suppose that f has domain A and t has domain B.
(a) What is the domain of f 1 t?
(b) What is the domain of f t?
(c) What is the domain of fyt?
10. H
ow is the composite function f 8 t defined? What is its
domain?
11. S
uppose the graph of f is given. Write an equation for each of
the graphs that are obtained from the graph of f as follows.
(a) Shift 2 units upward.
(b) Shift 2 units downward.
(c) Shift 2 units to the right.
(d) Shift 2 units to the left.
(e) Reflect about the x-axis.
(f) Reflect about the y-axis.
(g) Stretch vertically by a factor of 2.
(h) Shrink vertically by a factor of 2.
(i) Stretch horizontally by a factor of 2.
(j) Shrink horizontally by a factor of 2.
12.Explain what each of the following means and illustrate with a
sketch.
(a) lim f sxd − L
(b) lim1 f sxd − L (c) lim2 f sxd − L
x la
(d) lim f sxd − `
x la
x la
x la
(e) lim f sxd − 2`
x la
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
chapter 1 Review
13.Describe several ways in which a limit can fail to exist. Illustrate with sketches.
14.What does it mean to say that the line x − a is a vertical
asymptote of the curve y − f sxd? Draw curves to illustrate the
various possibilities.
15. State the following Limit Laws.
(a) Sum Law
(b) Difference Law
(c) Constant Multiple Law
(d) Product Law
(e) Quotient Law
(f ) Power Law
(g) Root Law
95
16. What does the Squeeze Theorem say?
17. (a) What does it mean for f to be continuous at a?
(b)What does it mean for f to be continuous on the interval
s2`, `d? What can you say about the graph of such a
function?
18. (a)Give examples of functions that are continuous on f21, 1g.
(b)Give an example of a function that is not continuous
on f0, 1g.
19. What does the Intermediate Value Theorem say?
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true,
explain why. If it is false, explain why or give an example that
disproves the statement.
1.If f is a function, then f ss 1 td − f ssd 1 f std.
2.If f ssd − f std, then s − t.
4.If x 1 , x 2 and f is a decreasing function, then f sx 1 d . f sx 2 d.
5.A vertical line intersects the graph of a function at most once.
6.If x is any real number, then sx 2 − x.
x l4
8. lim
x l1
S
8
2x
2
x24
x24
D
− lim
x l4
2x
8
2 lim
x l4 x 2 4
x24
lim sx 2 1 6x 2 7d
x 2 1 6x 2 7
x l1
−
lim sx 2 1 5x 2 6d
x 2 1 5x 2 6
x l1
lim sx 2 3d
x23
xl1
−
9. lim 2
x l 1 x 1 2x 2 4
lim sx 2 1 2x 2 4d
xl1
2
10.
x 29
−x13
x23
11. lim
xl3
15.If lim x l a f sxd exists but lim x l a tsxd does not exist, then
lim x l a f f sxd 1 tsxdg does not exist.
16. If lim x l 6 f f sxd tsxdg exists, then the limit must be f s6d ts6d.
3.If f is a function, then f s3xd − 3 f sxd.
7. lim
14.If neither lim x l a f sxd nor lim x l a tsxd exists, then
lim x l a f f sxd 1 tsxdg does not exist.
x2 2 9
− lim sx 1 3d
xl3
x23
12.If lim x l 5 f sxd − 2 and lim x l 5 tsxd − 0, then
limx l 5 f f sxdytsxdg does not exist.
13.If lim x l5 f sxd − 0 and lim x l 5 tsxd − 0, then
lim x l 5 f f sxdytsxdg does not exist.
17. If p is a polynomial, then lim x l b psxd − psbd.
18.If lim x l 0 f sxd − ` and lim x l 0 tsxd − `, then
lim x l 0 f f sxd 2 tsxdg − 0.
19.If the line x − 1 is a vertical asymptote of y − f sxd, then f is
not defined at 1.
20.If f s1d . 0 and f s3d , 0, then there exists a number c between
1 and 3 such that f scd − 0.
21.If f is continuous at 5 and f s5d − 2 and f s4d − 3, then
lim x l 2 f s4x 2 2 11d − 2.
22.If f is continuous on f21, 1g and f s21d − 4 and f s1d − 3, then
there exists a number r such that r , 1 and f srd − .
| |
23.Let f be a function such that lim x l 0 f sxd − 6. Then there
exists a positive number such that if 0 , x , , then
f sxd 2 6 , 1.
|
| |
|
24.If f sxd . 1 for all x and lim x l 0 f sxd exists, then
lim x l 0 f sxd . 1.
25.The equation x 10 2 10x 2 1 5 − 0 has a root in the
interval s0, 2d.
| |
26.If f is continuous at a, so is f .
| |
27. If f is continuous at a, so is f .
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
96
chapter 1 Functions and Limits
EXERCISES
1. Let f be the function whose graph is given.
(d) y − 12 f sxd 2 1
(c) y − 2 2 f sxd
y
y
f
1
1
x
1
0
1
x
11–16 Use transformations to sketch the graph of the function.
(a)
(b)
(c)
(d)
(e)
(f)
Estimate the value of f s2d.
Estimate the values of x such that f sxd − 3.
State the domain of f.
State the range of f.
On what interval is f increasing?
Is f even, odd, or neither even nor odd? Explain.
2.Determine whether each curve is the graph of a function of x.
If it is, state the domain and range of the function.
y
y
2
2
0
1
0
x
1
x
3.If f sxd − x 2 2 2x 1 3, evaluate the difference quotient
f sa 1 hd 2 f sad
h
4.Sketch a rough graph of the yield of a crop as a function of the
amount of fertilizer used.
5–8 Find the domain and range of the function. Write your answer
in interval notation.
5. f sxd − 2ys3x 2 1d
6. tsxd − s16 2 x 4
7. y − 1 1 sin x
8. Fstd − 3 1 cos 2t
9.Suppose that the graph of f is given. Describe how the graphs
of the following functions can be obtained from the graph of f.
(a) y − f sxd 1 8
(b) y − f sx 1 8d
(c) y − 1 1 2 f sxd
(d) y − f sx 2 2d 2 2
(e) y − 2f sxd
(f ) y − 3 2 f sxd
10.The graph of f is given. Draw the graphs of the following
functions.
(a) y − f sx 2 8d
(b) y − 2f sxd
11. y − sx 2 2d3
12. y − 2 sx
13. y − x 2 2 2 x 1 2
14. y −
15. f sxd − 2cos 2x
16. f sxd −
1
x21
H
1 1 x if x , 0
1 1 x 2 if x > 0
17. Determine whether f is even, odd, or neither even nor odd.
(a) f sxd − 2x 5 2 3x 2 1 2
(b) f sxd − x 3 2 x 7
(c) f sxd − cossx 2 d
(d) f sxd − 1 1 sin x
18.Find an expression for the function whose graph consists of the
line segment from the point s22, 2d to the point s21, 0d together
with the top half of the circle with center the origin and radius 1.
19.If f sxd − sx and tsxd − sin x, find the functions (a) f 8 t,
(b) t 8 f , (c) f 8 f , (d) t 8 t, and their domains.
20.Express the function Fsxd − 1ysx 1 sx as a composition of
three functions.
21.Life expectancy improved dramatically in the 20th century. The
table gives the life expectancy at birth (in years) of males born
in the United States. Use a scatter plot to choose an appropriate
type of model. Use your model to predict the life span of a male
born in the year 2010.
Birth year Life expectancy
1900
1910
1920
1930
1940
1950
48.3
51.1
55.2
57.4
62.5
65.6
Birth year Life expectancy
1960
1970
1980
1990
2000
66.6
67.1
70.0
71.8
73.0
22.A small-appliance manufacturer finds that it costs $9000 to
produce 1000 toaster ovens a week and $12,000 to produce
1500 toaster ovens a week.
(a)Express the cost as a function of the number of toaster
ovens produced, assuming that it is linear. Then sketch the
graph.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
chapter 1 Review
(b)What is the slope of the graph and what does it represent?
(c)What is the y-intercept of the graph and what does it
represent?
39. If 2x 2 1 < f sxd < x 2 for 0 , x , 3, find lim x l1 f sxd.
40. Prove that lim x l 0 x 2 coss1yx 2 d − 0.
41–44 Prove the statement using the precise definition of a limit.
23. The graph of f is given.
3
42.
lim s
x −0
41. lim s14 2 5xd − 4
xl2
y
xl0
2
44.
lim
−`
x l 41 sx 2 4
43. lim sx 2 2 3xd − 22
xl2
0
x l 23
(v) lim f sxd
x l4
x l0
(vi) lim2 f sxd
x l2
24. S
ketch the graph of an example of a function f that satisfies all
of the following conditions:
lim1 f sxd − 22, lim2 f sxd − 1, f s0d − 21,
x l0
lim2 f sxd − `, lim1 f sxd − 2`
x l2
x2 2 9
26. lim 2
x l 3 x 1 2x 2 3
25. lim cossx 1 sin xd
x l0
x2 2 9
x 1 2x 2 3
27. lim
x2 2 9
28.
lim1 2
x l 1 x 1 2x 2 3
2
x l 23
sh 2 1d3 1 1
29. lim
h l0
h
r l9
t2 2 4
30.
lim 3
t l2 t 2 8
sr
sr 2 9d4
42v
32.
lim
v l 41 4 2 v
|
|
33. lim
u 21
u 3 1 5u 2 2 6u
sx 1 6 2 x
34.
lim
xl3
x 3 2 3x 2
35. lim
4 2 ss
s 2 16
v 2 1 2v 2 8
36.
lim
v l2
v 4 2 16
37. lim
1 2 s1 2 x 2
x
4
ul1
sl16
xl0
38. lim
xl1
(i) lim1 f sxd
(ii) lim2 f sxd
(iii) lim f sxd
(iv) lim2 f sxd
(v) lim1 f sxd
(vi) lim f sxd
x l0
x l0
x l3
x l3
S
1
1
1 2
x21
x 2 3x 1 2
x l3
46. Let
tsxd −
2x 2 x 2
22x
x24
if
if
if
if
0<x<2
2,x<3
3,x,4
x>4
(a)For each of the numbers 2, 3, and 4, discover whether t
is continuous from the left, continuous from the right, or
continuous at the number.
(b) Sketch the graph of t.
47–48 Show that the function is continuous on its domain. State
the domain.
sx 2 2 9
4
47. hsxd − s
x 1 x 3 cos x
48.
tsxd − 2
x 22
49–50 Use the Intermediate Value Theorem to show that there is a
root of the equation in the given interval.
49. x 5 2 x 3 1 3x 2 5 − 0, s1, 2d
50. 2 sin x − 3 2 2x, s0, 1d
|
D
x l0
(b) Where is f discontinuous?
(c) Sketch the graph of f.
x l2
25–38 Find the limit.
31. lim
(a) Evaluate each limit, if it exists.
x l 23
(b) State the equations of the vertical asymptotes.
(c) At what numbers is f discontinuous? Explain.
x l0
if x , 0
s2x
f sxd − 3 2 x
if 0 < x , 3
sx 2 3d2 if x . 3
x
1
(a) Find each limit, or explain why it does not exist.
(i) lim1 f sxd
(ii) lim 1 f sxd
(iii) lim f sxd
(iv) lim f sxd
H
45. Let
1
x l2
97
|
51. Suppose that f sxd < tsxd for all x, where lim x l a tsxd − 0.
Find lim x l a f sxd.
52. Let f sxd − v x b 1 v 2x b .
(a) For what values of a does lim x l a f sxd exist?
(b) At what numbers is f discontinuous?
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Principles of
Problem Solving
1 Understand the Problem
There are no hard and fast rules that will ensure success in solving problems. However,
it is possible to outline some general steps in the problem-solving process and to give
some principles that may be useful in the solution of certain problems. These steps and
principles are just common sense made explicit. They have been adapted from George
Polya’s book How To Solve It.
The first step is to read the problem and make sure that you understand it clearly. Ask
yourself the following questions:
What is the unknown?
What are the given quantities?
What are the given conditions?
For many problems it is useful to
draw a diagram
and identify the given and required quantities on the diagram.
Usually it is necessary to
introduce suitable notation
In choosing symbols for the unknown quantities we often use letters such as a, b, c, m, n,
x, and y, but in some cases it helps to use initials as suggestive symbols; for instance, V
for volume or t for time.
2 think of a plan
Find a connection between the given information and the unknown that will enable you
to calculate the unknown. It often helps to ask yourself explicitly: “How can I relate the
given to the unknown?” If you don’t see a connection immediately, the following ideas
may be helpful in devising a plan.
Try to Recognize Something Familiar Relate the given situation to previous knowledge.
Look at the unknown and try to recall a more familiar problem that has a similar unknown.
Try to Recognize Patterns Some problems are solved by recognizing that some kind of
pattern is occurring. The pattern could be geometric, or numerical, or algebraic. If you
can see regularity or repetition in a problem, you might be able to guess what the continuing pattern is and then prove it.
Use Analogy Try to think of an analogous problem, that is, a similar problem, a related
problem, but one that is easier than the original problem. If you can solve the similar,
simpler problem, then it might give you the clues you need to solve the original, more
difficult problem. For instance, if a problem involves very large numbers, you could first
try a similar problem with smaller numbers. Or if the problem involves three-dimensional
geometry, you could look for a similar problem in two-dimensional geometry. Or if the
problem you start with is a general one, you could first try a special case.
Introduce Something Extra It may sometimes be necessary to introduce something new,
an auxiliary aid, to help make the connection between the given and the unknown. For
instance, in a problem where a diagram is useful the auxiliary aid could be a new line
drawn in a diagram. In a more algebraic problem it could be a new unknown that is
related to the original unknown.
98
Take Cases We may sometimes have to split a problem into several cases and give a
different argument for each of the cases. For instance, we often have to use this strategy
in dealing with absolute value.
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Work Backward Sometimes it is useful to imagine that your problem is solved and
work backward, step by step, until you arrive at the given data. Then you may be able
to reverse your steps and thereby construct a solution to the original problem. This procedure is commonly used in solving equations. For instance, in solving the equation
3x 2 5 − 7, we suppose that x is a number that satisfies 3x 2 5 − 7 and work backward. We add 5 to each side of the equation and then divide each side by 3 to get x − 4.
Since each of these steps can be reversed, we have solved the problem.
Establish Subgoals In a complex problem it is often useful to set subgoals (in which the
desired situation is only partially fulfilled). If we can first reach these subgoals, then we
may be able to build on them to reach our final goal.
Indirect Reasoning Sometimes it is appropriate to attack a problem indirectly. In using
proof by contradiction to prove that P implies Q, we assume that P is true and Q is false
and try to see why this can’t happen. Somehow we have to use this information and arrive
at a contradiction to what we absolutely know is true.
Mathematical Induction In proving statements that involve a positive integer n, it is
frequently helpful to use the following principle.
Principle of Mathematical Induction Let Sn be a statement about the positive
integer n. Suppose that
1. S1 is true.
2. Sk11 is true whenever Sk is true.
Then Sn is true for all positive integers n.
This is reasonable because, since S1 is true, it follows from condition 2 swith k − 1d
that S2 is true. Then, using condition 2 with k − 2, we see that S3 is true. Again using
condition 2, this time with k − 3, we have that S4 is true. This procedure can be followed
indefinitely.
3 Carry Out the Plan
In Step 2 a plan was devised. In carrying out that plan we have to check each stage of the
plan and write the details that prove that each stage is correct.
4 Look Back
Having completed our solution, it is wise to look back over it, partly to see if we have
made errors in the solution and partly to see if we can think of an easier way to solve the
problem. Another reason for looking back is that it will familiarize us with the method
of solution and this may be useful for solving a future problem. Descartes said, “Every
problem that I solved became a rule which served afterwards to solve other problems.”
These principles of problem solving are illustrated in the following examples. Before
you look at the solutions, try to solve these problems yourself, referring to these Principles
of Problem Solving if you get stuck. You may find it useful to refer to this section from
time to time as you solve the exercises in the remaining chapters of this book.
As the first example illustrates, it is often necessary to use the problem-solving prin­
ciple of taking cases when dealing with absolute values.
Example 1
|
| |
|
Solve the inequality x 2 3 1 x 1 2 , 11.
Solution Recall the definition of absolute value:
|x| −
H
x
if x > 0
2x if x , 0
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99
It follows that
|x 2 3| −
−
Similarly
|x 1 2| −
−
PS Take cases
H
H
H
H
x23
2sx 2 3d
if x 2 3 > 0
if x 2 3 , 0
x23
2x 1 3
if x > 3
if x , 3
x12
2sx 1 2d
if x 1 2 > 0
if x 1 2 , 0
x12
2x 2 2
if x > 22
if x , 22
These expressions show that we must consider three cases:
x , 22 22 < x , 3 x > 3
Case I If x , 22, we have
| x 2 3 | 1 | x 1 2 | , 11
2x 1 3 2 x 2 2 , 11
22x , 10
x . 25
Case II If 22 < x , 3, the given inequality becomes
2x 1 3 1 x 1 2 , 11
5 , 11 (always true)
Case iii If x > 3, the inequality becomes
x 2 3 1 x 1 2 , 11
2x , 12
x,6
Combining cases I, II, and III, we see that the inequality is satisfied when 25 , x , 6.
So the solution is the interval s25, 6d.
■
In the following example we first guess the answer by looking at special cases and
recognizing a pattern. Then we prove our conjecture by mathematical induction.
In using the Principle of Mathematical Induction, we follow three steps:
Step 1 Prove that Sn is true when n − 1.
Step 2 Assume that Sn is true when n − k and deduce that Sn is true when n − k 1 1.
Step 3 Conclude that Sn is true for all n by the Principle of Mathematical Induction.
100
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Example 2
for fnsxd.
PS Analogy: Try a similar, simpler
problem
If f0sxd − xysx 1 1d and fn11 − f0 8 fn for n − 0, 1, 2, . . . , find a formula
Solution We start by finding formulas for fnsxd for the special cases n − 1, 2, and 3.
S D
x
f1sxd − s f0 8 f0dsxd − f0( f0sxd) − f0
x11
x
x
x11
x11
x
−
−
−
x
2x 1 1
2x 1 1
11
x11
x11
S
x
f2sxd − s f0 8 f1 dsxd − f0( f1sxd) − f0
2x 1 1
D
x
x
2x 1 1
2x 1 1
x
−
−
−
x
3x 1 1
3x 1 1
11
2x 1 1
2x 1 1
S
x
f3sxd − s f0 8 f2 dsxd − f0( f2sxd) − f0
3x 1 1
D
x
x
3x 1 1
3x 1 1
x
−
−
−
x
4x 1 1
4x 1 1
11
3x 1 1
3x 1 1
PS Look for a pattern
We notice a pattern: The coefficient of x in the denominator of fnsxd is n 1 1 in the
three cases we have computed. So we make the guess that, in general,
1 fnsxd −
x
sn 1 1dx 1 1
To prove this, we use the Principle of Mathematical Induction. We have already verified
that (1) is true for n − 1. Assume that it is true for n − k, that is,
fksxd −
Then
x
sk 1 1dx 1 1
S
x
fk11sxd − s f0 8 fk dsxd − f0( fksxd) − f0
sk 1 1dx 1 1
D
x
x
sk 1 1dx 1 1
sk 1 1dx 1 1
x
−
−
−
x
sk 1 2dx 1 1
sk 1 2dx 1 1
11
sk 1 1dx 1 1
sk 1 1dx 1 1
This expression shows that (1) is true for n − k 1 1. Therefore, by mathematical
induction, it is true for all positive integers n.
■
101
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In the following example we show how the problem-solving strategy of introducing
something extra is sometimes useful when we evaluate limits. The idea is to change the
variable—to introduce a new variable that is related to the original variable—in such a
way as to make the problem simpler. Later, in Section 4.5, we will make more extensive
use of this general idea.
3
1 1 cx 2 1
s
, where c is a constant.
xl0
x
Solution As it stands, this limit looks challenging. In Section 1.6 we evaluated several limits in which both numerator and denominator approached 0. There our strategy
was to perform some sort of algebraic manipulation that led to a simplifying cancellation, but here it’s not clear what kind of algebra is necessary.
So we introduce a new variable t by the equation
Example 3 Evaluate lim
3
t−s
1 1 cx
We also need to express x in terms of t, so we solve this equation:
t 3 − 1 1 cx x −
t3 2 1
sif c ± 0d
c
Notice that x l 0 is equivalent to t l 1. This allows us to convert the given limit into
one involving the variable t:
3
1 1 cx 2 1
t21
s
− lim 3
t l1 st 2 1dyc
xl0
x
lim
− lim
t l1
cst 2 1d
t3 2 1
The change of variable allowed us to replace a relatively complicated limit by a simpler
one of a type that we have seen before. Factoring the denominator as a difference of
cubes, we get
lim
t l1
cst 2 1d
cst 2 1d
− lim
t l1 st 2 1dst 2 1 t 1 1d
t3 2 1
− lim
t l1
c
c
−
t2 1 t 1 1
3
In making the change of variable we had to rule out the case c − 0. But if c − 0, the
function is 0 for all nonzero x and so its limit is 0. Therefore, in all cases, the limit
is cy3.
n
The following problems are meant to test and challenge your problem-solving skills.
Some of them require a considerable amount of time to think through, so don’t be discouraged if you can’t solve them right away. If you get stuck, you might find it helpful to
refer to the discussion of the principles of problem solving.
Problems
|
| |
|
2.Solve the inequality | x 2 1 | 2 | x 2 3 | > 5.
1.Solve the equation 2x 2 1 2 x 1 5 − 3.
|
| |
|
3.Sketch the graph of the function f sxd − x 2 2 4 x 1 3 .
102
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|
| |
|
4.Sketch the graph of the function tsxd − x 2 2 1 2 x 2 2 4 .
| |
| |
5.Draw the graph of the equation x 1 x − y 1 y .
6.Sketch the region in the plane consisting of all points sx, yd such that
|x 2 y| 1 |x| 2 |y| < 2
7.The notation maxha, b, . . .j means the largest of the numbers a, b, . . . . Sketch the graph of
each function.
(a)
f sxd − maxhx, 1yxj
(b) f sxd − maxhsin x, cos xj
(c) f sxd − maxhx 2, 2 1 x, 2 2 xj
8.Sketch the region in the plane defined by each of the following equations or inequalities.
(a)
maxhx, 2yj − 1
(b) 21 < maxhx, 2yj < 1
(c) maxhx, y 2 j − 1
9.A driver sets out on a journey. For the first half of the distance she drives at the leisurely
pace of 30 miyh; she drives the second half at 60 miyh. What is her average speed on
this trip?
10. Is it true that f 8 s t 1 hd − f 8 t 1 f 8 h?
11. Prove that if n is a positive integer, then 7 n 2 1 is divisible by 6.
12. Prove that 1 1 3 1 5 1 ∙ ∙ ∙ 1 s2n 2 1d − n 2.
13. If f0sxd − x 2 and fn11sxd − f0s fnsxdd for n − 0, 1, 2, . . . , find a formula for fnsxd.
1
and fn11 − f0 8 fn for n − 0, 1, 2, . . . , find an expression for fnsxd
22x
and use mathematical induction to prove it.
(b)Graph f0 , f1, f2 , f3 on the same screen and describe the effects of repeated composition.
14. (a)If f0sxd −
; 15. Evaluate lim
x l1
3
x 21
s
.
sx 2 1
16. Find numbers a and b such that lim
x l0
17. Evaluate lim
xl0
y
Q
P
FIGURE FOR PROBLEM 18
| 2x 2 1 | 2 | 2x 1 1 | .
x
18.
The figure shows a point P on the parabola y − x 2 and the point Q where the perpendicular
bisector of OP intersects the y-axis. As P approaches the origin along the parabola, what
happens to Q? Does it have a limiting position? If so, find it.
y=≈
0
sax 1 b 2 2
− 1.
x
x
19. Evaluate the following limits, if they exist, where v x b denotes the greatest integer function.
v xb
(a) lim
(b) lim x v 1yx b
xl0
xl0
x
20. Sketch the region in the plane defined by each of the following equations.
(a) v xb 2 1 v yb 2 − 1
(b) v xb 2 2 v yb 2 − 3
2
(c) v x 1 yb − 1
(d) v xb 1 v yb − 1
103
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21. Find all values of a such that f is continuous on R:
f sxd −
H
x 1 1 if x < a
x2
if x . a
22.A fixed point of a function f is a number c in its domain such that f scd − c. (The function
doesn’t move c; it stays fixed.)
(a)Sketch the graph of a continuous function with domain f0, 1g whose range also lies
in f0, 1g. Locate a fixed point of f .
(b)Try to draw the graph of a continuous function with domain f0, 1g and range in f0, 1g
that does not have a fixed point. What is the obstacle?
(c)Use the Intermediate Value Theorem to prove that any continuous function with
domain f0, 1g and range in f0, 1g must have a fixed point.
23.If lim x l a f f sxd 1 tsxdg − 2 and lim x l a f f sxd 2 tsxdg − 1, find lim x l a f f sxd tsxdg.
24. (a)The figure shows an isosceles triangle ABC with /B − /C. The bisector of angle B
intersects the side AC at the point P. Suppose that the base BC remains fixed but the
altitude AM of the triangle approaches 0, so A approaches the midpoint M of BC.
What happens to P during this process? Does it have a limiting position? If so, find it.
|
|
A
P
B
M
C
(b)Try to sketch the path traced out by P during this process. Then find an equation of this
curve and use this equation to sketch the curve.
25. (a)If we start from 0° latitude and proceed in a westerly direction, we can let Tsxd denote
the temperature at the point x at any given time. Assuming that T is a continuous
function of x, show that at any fixed time there are at least two diametrically opposite
points on the equator that have exactly the same temperature.
(b)Does the result in part (a) hold for points lying on any circle on the earth’s surface?
(c)Does the result in part (a) hold for barometric pressure and for altitude above sea level?
104
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2
Derivatives
The maximum sustainable swimming speed S of
salmon depends on the water
temperature T. Exercise 58 in
Section 2.1 asks you to analyze
how S varies as T changes by
estimating the derivative of S
with respect to T.
© Jody Ann / Shutterstock.com
In this chapter we begin our study of differential calculus, which is concerned with how one
quantity changes in relation to another quantity. The central concept of differential calculus is the
derivative, which is an outgrowth of the velocities and slopes of tangents that we considered in
Chapter 1. After learning how to calculate derivatives, we use them to solve problems involving
rates of change and the approximation of functions.
105
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106
Chapter 2 Derivatives
The problem of finding the tangent line to a curve and the problem of finding the velocity
of an object both involve finding the same type of limit, as we saw in Section 1.4. This
special type of limit is called a derivative and we will see that it can be interpreted as a
rate of change in any of the natural or social sciences or engineering.
Tangents
y
Q{ x, ƒ }
ƒ-f(a)
P { a, f(a)}
If a curve C has equation y − f sxd and we want to find the tangent line to C at the point
Psa, f sadd, then we consider a nearby point Qsx, f sxdd, where x ± a, and compute the
slope of the secant line PQ:
mPQ −
x-a
0
a
y
x
x
t
Q
Then we let Q approach P along the curve C by letting x approach a. If mPQ approaches
a number m, then we define the tangent t to be the line through P with slope m. (This
amounts to saying that the tangent line is the limiting position of the secant line PQ as Q
approaches P. See Figure 1.)
1 Definition The tangent line to the curve y − f sxd at the point Psa, f sadd is
the line through P with slope
Q
P
f sxd 2 f sad
x2a
m − lim
Q
xla
f sxd 2 f sad
x2a
provided that this limit exists.
x
0
FIGURE 1
In our first example we confirm the guess we made in Example 1.4.1.
Example 1 Find an equation of the tangent line to the parabola y − x 2 at the
point Ps1, 1d.
SOLUTION Here we have a − 1 and f sxd − x 2, so the slope is
m − lim
x l1
− lim
x l1
f sxd 2 f s1d
x2 2 1
− lim
x l1 x 2 1
x21
sx 2 1dsx 1 1d
x21
− lim sx 1 1d − 1 1 1 − 2
x l1
Point-slope form for a line through the
point sx1 , y1 d with slope m:
Using the point-slope form of the equation of a line, we find that an equation of the
tangent line at s1, 1d is
y 2 y1 − msx 2 x 1 d
y 2 1 − 2sx 2 1d or y − 2x 2 1
n
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107
Section 2.1 Derivatives and Rates of Change
We sometimes refer to the slope of the tangent line to a curve at a point as the slope
of the curve at the point. The idea is that if we zoom in far enough toward the point, the
curve looks almost like a straight line. Figure 2 illustrates this procedure for the curve
y − x 2 in Example 1. The more we zoom in, the more the parabola looks like a line. In
other words, the curve becomes almost indistinguishable from its tangent line.
TEC Visual 2.1 shows an animation
of Figure 2.
2
1.5
(1, 1)
1.1
(1, 1)
2
0
(1, 1)
1.5
0.5
0.9
1.1
FIGURE 2 Zooming in toward the point (1, 1) on the parabola y − x 2
Q { a+h, f(a+h)}
y
t
There is another expression for the slope of a tangent line that is sometimes easier to
use. If h − x 2 a, then x − a 1 h and so the slope of the secant line PQ is
mPQ −
P { a, f(a)}
f(a+h)-f(a)
h
0
a
a+h
x
f sa 1 hd 2 f sad
h
(See Figure 3 where the case h . 0 is illustrated and Q is to the right of P. If it happened
that h , 0, however, Q would be to the left of P.)
Notice that as x approaches a, h approaches 0 (because h − x 2 a) and so the expression for the slope of the tangent line in Definition 1 becomes
FIGURE 3 2
m − lim
hl0
f sa 1 hd 2 f sad
h
Example 2 Find an equation of the tangent line to the hyperbola y − 3yx at the
point s3, 1d.
SOLUTION Let f sxd − 3yx. Then, by Equation 2, the slope of the tangent at s3, 1d is
m − lim
hl0
f s3 1 hd 2 f s3d
h
3
3 2 s3 1 hd
21
31h
31h
− lim
− lim
hl0
hl0
h
h
y
− lim
3
y=
x
x+3y-6=0
hl0
Therefore an equation of the tangent at the point s3, 1d is
(3, 1)
0
y 2 1 − 213 sx 2 3d
x
which simplifies to
FIGURE 4 2h
1
1
− lim 2
−2
hl0
hs3 1 hd
31h
3
x 1 3y 2 6 − 0
The hyperbola and its tangent are shown in Figure 4.
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n
108
Chapter 2 Derivatives
position at
time t=a
0
position at
time t=a+h
s
f(a+h)-f(a)
f(a)
f(a+h)
FIGURE 5 s
Velocities
In Section 1.4 we investigated the motion of a ball dropped from the CN Tower and
defined its velocity to be the limiting value of average velocities over shorter and shorter
time periods.
In general, suppose an object moves along a straight line according to an equation of
motion s − f std, where s is the displacement (directed distance) of the object from the
origin at time t. The function f that describes the motion is called the position function of the object. In the time interval from t − a to t − a 1 h the change in position is
f sa 1 hd 2 f sad. (See Figure 5.)
The average velocity over this time interval is
Q { a+h, f(a+h)}
P { a, f(a)}
h
0
a
mPQ=
a+h
t
f(a+h)-f(a)
h
average velocity
average velocity −
displacement
f sa 1 hd 2 f sad
−
time
h
which is the same as the slope of the secant line PQ in Figure 6.
Now suppose we compute the average velocities over shorter and shorter time intervals fa, a 1 hg. In other words, we let h approach 0. As in the example of the falling ball,
we define the velocity (or instantaneous velocity) vsad at time t − a to be the limit of
these average velocities:
3
vsad − lim
hl0
FIGURE 6
f sa 1 hd 2 f sad
h
This means that the velocity at time t − a is equal to the slope of the tangent line at P
(compare Equations 2 and 3).
Now that we know how to compute limits, let’s reconsider the problem of the falling ball.
Example 3 Suppose that a ball is dropped from the upper observation deck of the
CN Tower, 450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
(b) How fast is the ball traveling when it hits the ground?
Recall from Section 1.4: The dis­tance
(in meters) fallen after t seconds is
4.9t 2.
SOLUTION We will need to find the velocity both when t − 5 and when the ball hits
the ground, so it’s efficient to start by finding the velocity at a general time t. Using the
equation of motion s − f std − 4.9t 2, we have
v std − lim
hl0
f st 1 hd 2 f std
4.9st 1 hd2 2 4.9t 2
− lim
hl0
h
h
− lim
4.9st 2 1 2th 1 h 2 2 t 2 d
4.9s2th 1 h 2 d
− lim
hl0
h
h
− lim
4.9hs2t 1 hd
− lim 4.9s2t 1 hd − 9.8t
hl0
h
hl0
hl0
(a) The velocity after 5 seconds is vs5d − s9.8ds5d − 49 mys.
(b) Since the observation deck is 450 m above the ground, the ball will hit the ground
at the time t when sstd − 450, that is,
4.9t 2 − 450
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Section 2.1 Derivatives and Rates of Change
This gives
t2 −
450
and t −
4.9
Î
109
450
< 9.6 s
4.9
The velocity of the ball as it hits the ground is therefore
v
SÎ D Î
450
4.9
− 9.8
450
< 94 mys
4.9
n
Derivatives
We have seen that the same type of limit arises in finding the slope of a tangent line
(Equation 2) or the velocity of an object (Equation 3). In fact, limits of the form
lim
h l0
f sa 1 hd 2 f sad
h
arise whenever we calculate a rate of change in any of the sciences or engineering, such
as a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit
occurs so widely, it is given a special name and notation.
f 9sad is read “ f prime of a.”
4 Definition The derivative of a function f at a number a, denoted by
f 9sad, is
f sa 1 hd 2 f sad
f 9sad − lim
h l0
h
if this limit exists.
If we write x − a 1 h, then we have h − x 2 a and h approaches 0 if and only if x
approaches a. Therefore an equivalent way of stating the definition of the derivative, as
we saw in finding tangent lines, is
5
f 9sad − lim
xla
f sxd 2 f sad
x2a
Example 4 Find the derivative of the function f sxd − x 2 2 8x 1 9 at the number a.
SOLUTION From Definition 4 we have
Definitions 4 and 5 are equivalent, so
we can use either one to compute the
derivative. In practice, Definition 4
often leads to simpler computations.
f 9sad − lim
h l0
f sa 1 hd 2 f sad
h
− lim
fsa 1 hd2 2 8sa 1 hd 1 9g 2 fa 2 2 8a 1 9g
h
− lim
a 2 1 2ah 1 h 2 2 8a 2 8h 1 9 2 a 2 1 8a 2 9
h
− lim
2ah 1 h 2 2 8h
− lim s2a 1 h 2 8d
h l0
h
h l0
h l0
h l0
− 2a 2 8
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n
110
Chapter 2 Derivatives
We defined the tangent line to the curve y − f sxd at the point Psa, f sadd to be the line
that passes through P and has slope m given by Equation 1 or 2. Since, by Defini­tion 4,
this is the same as the derivative f 9sad, we can now say the following.
The tangent line to y − f sxd at sa, f sadd is the line through sa, f sadd whose slope is
equal to f 9sad, the derivative of f at a.
If we use the point-slope form of the equation of a line, we can write an equation of
the tangent line to the curve y − f sxd at the point sa, f sadd:
y
y=≈-8x+9
y 2 f sad − f 9sadsx 2 ad
Example 5 Find an equation of the tangent line to the parabola y − x 2 2 8x 1 9 at
the point s3, 26d.
x
0
SOLUTION From Example 4 we know that the derivative of f sxd − x 2 2 8x 1 9 at
(3, _6)
the number a is f 9sad − 2a 2 8. Therefore the slope of the tangent line at s3, 26d is
f 9s3d − 2s3d 2 8 − 22. Thus an equation of the tangent line, shown in Figure 7, is
y=_2x
y 2 s26d − s22dsx 2 3d or y − 22x
FIGURE 7 n
Rates of Change
y
Q { ¤, ‡}
P {⁄, fl}
and the corresponding change in y is
Îy
Dy − f sx 2d 2 f sx 1d
Îx
0
⁄
The difference quotient
¤
average rate
rateof
ofchange
change−mPQ
mPQ
average
Dy
f sx 2d 2 f sx 1d
−
Dx
x2 2 x1
x
instantaneous
rateofofchange
change−
instantaneous rate
slope
slopeofoftangent
tangentatatPP
FIGURE 8 Suppose y is a quantity that depends on another quantity x. Thus y is a function of x and
we write y − f sxd. If x changes from x 1 to x 2, then the change in x (also called the increment of x) is
Dx − x 2 2 x 1
is called the average rate of change of y with respect to x over the interval fx 1, x 2g and
can be interpreted as the slope of the secant line PQ in Figure 8.
By analogy with velocity, we consider the average rate of change over smaller and
smaller intervals by letting x 2 approach x 1 and therefore letting Dx approach 0. The limit
of these average rates of change is called the (instantaneous) rate of change of y with
respect to x at x − x1, which (as in the case of velocity) is interpreted as the slope of the
tangent to the curve y − f sxd at Psx 1, f sx 1dd:
6 instantaneous rate of change − lim
Dx l 0
Dy
f sx2 d 2 f sx1d
− lim
x lx
Dx
x2 2 x1
2
1
We recognize this limit as being the derivative f 9sx 1d.
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Section 2.1 Derivatives and Rates of Change
111
We know that one interpretation of the derivative f 9sad is as the slope of the tangent
line to the curve y − f sxd when x − a. We now have a second interpretation:
y
The derivative f 9sad is the instantaneous rate of change of y − f sxd with respect
to x when x − a.
Q
P
x
FIGURE 9 The y-values are changing rapidly
at P and slowly at Q.
The connection with the first interpretation is that if we sketch the curve y − f sxd,
then the instantaneous rate of change is the slope of the tangent to this curve at the point
where x − a. This means that when the derivative is large (and therefore the curve is
steep, as at the point P in Figure 9), the y-values change rapidly. When the derivative is
small, the curve is relatively flat (as at point Q) and the y-values change slowly.
In particular, if s − f std is the position function of a particle that moves along a
straight line, then f 9sad is the rate of change of the displacement s with respect to the
time t. In other words, f 9sad is the velocity of the particle at time t − a. The speed of the
particle is the absolute value of the velocity, that is, f 9sad .
In the next example we discuss the meaning of the derivative of a function that is
defined verbally.
|
|
Example 6 A manufacturer produces bolts of a fabric with a fixed width. The cost of
producing x yards of this fabric is C − f sxd dollars.
(a) What is the meaning of the derivative f 9sxd? What are its units?
(b) In practical terms, what does it mean to say that f 9s1000d − 9?
(c) Which do you think is greater, f 9s50d or f 9s500d? What about f 9s5000d?
SOLUTION (a) The derivative f 9sxd is the instantaneous rate of change of C with respect to x; that
is, f 9sxd means the rate of change of the production cost with respect to the number of
yards produced. (Economists call this rate of change the marginal cost. This idea is
discussed in more detail in Sections 2.7 and 3.7.)
Because
f 9sxd − lim
Dx l 0
Here we are assuming that the cost
function is well behaved; in other
words, Csxd doesn’t oscillate rapidly
near x − 1000.
DC
Dx
the units for f 9sxd are the same as the units for the difference quotient DCyDx. Since
DC is measured in dollars and Dx in yards, it follows that the units for f 9sxd are dollars
per yard.
(b) The statement that f 9s1000d − 9 means that, after 1000 yards of fabric have been
manufactured, the rate at which the production cost is increasing is $9yyard. (When
x − 1000, C is increasing 9 times as fast as x.)
Since Dx − 1 is small compared with x − 1000, we could use the approximation
f 9s1000d <
DC
DC
−
− DC
Dx
1
and say that the cost of manufacturing the 1000th yard (or the 1001st) is about $9.
(c) The rate at which the production cost is increasing (per yard) is probably lower
when x − 500 than when x − 50 (the cost of making the 500th yard is less than the
cost of the 50th yard) because of economies of scale. (The manufacturer makes more
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112
Chapter 2 Derivatives
efficient use of the fixed costs of production.) So
f 9s50d . f 9s500d
But, as production expands, the resulting large-scale operation might become inefficient and there might be overtime costs. Thus it is possible that the rate of increase of
costs will eventually start to rise. So it may happen that
f 9s5000d . f 9s500d
n
In the following example we estimate the rate of change of the national debt with
respect to time. Here the function is defined not by a formula but by a table of values.
t
Dstd
1985
1990
1995
2000
2005
2010
1945.9
3364.8
4988.7
5662.2
8170.4
14,025.2
Example 7 Let Dstd be the US national debt at time t. The table in the margin gives
approximate values of this function by providing end of year estimates, in billions of
dollars, from 1985 to 2010. Interpret and estimate the value of D9s2000d.
SOLUTION The derivative D9s2000d means the rate of change of D with respect to t
when t − 2000, that is, the rate of increase of the national debt in 2000.
According to Equation 5,
Source: US Dept. of the Treasury
D9s2000d − lim
t l 2000
Dstd 2 Ds2000d
t 2 2000
So we compute and tabulate values of the difference quotient (the average rates of
change) as follows.
A Note on Units
The units for the average rate of change
D DyDt are the units for D D divided by
the units for Dt, namely, billions of dollars per year. The instantaneous rate of
change is the limit of the average rates
of change, so it is measured in the same
units: billions of dollars per year.
t
Time interval
Average rate of change −
1985
1990
1995
2005
2010
[1985, 2000]
[1990, 2000]
[1995, 2000]
[2000, 2005]
[2000, 2010]
247.75
229.74
134.70
501.64
836.30
Dstd 2 Ds2000d
t 2 2000
From this table we see that D9s2000d lies somewhere between 134.70 and 501.64 billion
dollars per year. [Here we are making the reasonable assumption that the debt didn’t
fluctuate wildly between 1995 and 2005.] We estimate that the rate of increase of the
national debt of the United States in 2000 was the average of these two numbers, namely,
D9s2000d < 318 billion dollars per year
Another method would be to plot the debt function and estimate the slope of the
tangent line when t − 2000.
n
In Examples 3, 6, and 7 we saw three specific examples of rates of change: the velocity of an object is the rate of change of displacement with respect to time; marginal cost
is the rate of change of production cost with respect to the number of items produced; the
rate of change of the debt with respect to time is of interest in economics. Here is a small
sample of other rates of change: In physics, the rate of change of work with respect to
time is called power. Chemists who study a chemical reaction are interested in the rate of
change in the concentration of a reactant with respect to time (called the rate of reaction).
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.1 Derivatives and Rates of Change
113
A biologist is interested in the rate of change of the population of a colony of bacteria
with respect to time. In fact, the computation of rates of change is important in all of the
natural sciences, in engineering, and even in the social sciences. Further examples will
be given in Section 2.7.
All these rates of change are derivatives and can therefore be interpreted as slopes of
tangents. This gives added significance to the solution of the tangent problem. Whenever
we solve a problem involving tangent lines, we are not just solving a problem in geometry. We are also implicitly solving a great variety of problems involving rates of change
in science and engineering.
1.A curve has equation y − f sxd.
(a)Write an expression for the slope of the secant line
through the points Ps3, f s3dd and Qsx, f sxdd.
(b)Write an expression for the slope of the tangent line at P.
; 2.Graph the curve y − sin x in the viewing rectangles
f22, 2g by f22, 2g, f21, 1g by f21, 1g, and f20.5, 0.5g
by f20.5, 0.5g. What do you notice about the curve as you
zoom in toward the origin?
3.(a)Find the slope of the tangent line to the parabola
y − 4x 2 x 2 at the point s1, 3d
(i) using Definition 1
(ii) using Equation 2
(b) Find an equation of the tangent line in part (a).
(c)Graph the parabola and the tangent line. As a check on
;
your work, zoom in toward the point s1, 3d until the
parabola and the tangent line are indistinguishable.
4.(a)Find the slope of the tangent line to the curve
y − x 2 x 3 at the point s1, 0d
(i) using Definition 1
(ii) using Equation 2
(b) Find an equation of the tangent line in part (a).
(c)Graph the curve and the tangent line in successively
;
smaller viewing rectangles centered at s1, 0d until the
curve and the line appear to coincide.
5–8 Find an equation of the tangent line to the curve at the
given point.
5.y − 4x 2 3x 2, s2, 24d
6. y − x 3 2 3x 1 1, s2, 3d
7. y − sx , s1, 1d
2x 1 1
8. y −
, s1, 1d
x12
9. (a)Find the slope of the tangent to the curve
y − 3 1 4x 2 2 2x 3 at the point where x − a.
(b)Find equations of the tangent lines at the points s1, 5d
and s2, 3d.
(c)Graph the curve and both tangents on a common
;
screen.
10. (a)Find the slope of the tangent to the curve y − 1ysx at
the point where x − a.
;
(b)Find equations of the tangent lines at the points s1, 1d
and (4, 12 ).
(c)Graph the curve and both tangents on a common
screen.
11. (a)A particle starts by moving to the right along a horizontal line; the graph of its position function is shown
in the figure. When is the particle moving to the right?
Moving to the left? Standing still?
(b)Draw a graph of the velocity function.
s (meters)
4
2
0
2
4
6 t (seconds)
12.Shown are graphs of the position functions of two runners,
A and B, who run a 100-meter race and finish in a tie.
s (meters)
80
A
40
0
B
4
8
12
t (seconds)
(a)Describe and compare how the runners run the race.
(b)At what time is the distance between the runners the
greatest?
(c)At what time do they have the same velocity?
13.If a ball is thrown into the air with a velocity of 40 ftys, its
height (in feet) after t seconds is given by y − 40t 2 16t 2.
Find the velocity when t − 2.
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114
chapter 2 Derivatives
14.If a rock is thrown upward on the planet Mars with a velocity
of 10 mys, its height (in meters) after t seconds is given by
H − 10t 2 1.86t 2.
(a) Find the velocity of the rock after one second.
(b) Find the velocity of the rock when t − a.
(c) When will the rock hit the surface?
(d) With what velocity will the rock hit the surface?
15.The displacement (in meters) of a particle moving in a
straight line is given by the equation of motion s − 1yt 2,
where t is measured in seconds. Find the velocity of the
par­ticle at times t − a, t − 1, t − 2, and t − 3.
19.For the function f graphed in Exercise 18:
(a) Estimate the value of f 9s50d.
(b) Is f 9s10d . f 9s30d?
f s80d 2 f s40d
(c) Is f 9s60d .
? Explain.
80 2 40
20.Find an equation of the tangent line to the graph of y − tsxd
at x − 5 if ts5d − 23 and t9s5d − 4.
21.If an equation of the tangent line to the curve y − f sxd at the
point where a − 2 is y − 4x 2 5, find f s2d and f 9s2d.
22.If the tangent line to y − f sxd at (4, 3) passes through the
point (0, 2), find f s4d and f 9s4d.
16.The displacement (in feet) of a particle moving in a straight
line is given by s − 12 t 2 2 6t 1 23, where t is measured in
seconds.
(a)Find the average velocity over each time interval:
(i) f4, 8g
(ii) f6, 8g
(iii) f8, 10g
(iv) f8, 12g
(b) Find the instantaneous velocity when t − 8.
(c)Draw the graph of s as a function of t and draw the secant
lines whose slopes are the average velocities in part (a).
Then draw the tangent line whose slope is the instantaneous velocity in part (b).
25.Sketch the graph of a function t that is continuous on its
domain s25, 5d and where ts0d − 1, t9s0d − 1, t9s22d − 0,
lim x l 251 tsxd − `, and lim x l52 tsxd − 3.
17.For the function t whose graph is given, arrange the
following numbers in increasing order and explain your
reasoning:
26.Sketch the graph of a function f where the domain is s22, 2d,
f 9s0d − 22, lim xl2 f sxd − `, f is continuous at all numbers in its domain except 61, and f is odd.
t9s22d
0
t9s0d
t9s2d
y
_1
1
2
3
4
x
f s40d 2 f s10d
. What does this value repre40 2 10
sent geometrically?
(d)Compute
y
800
29. (a)If Fsxd − 5xys1 1 x 2 d, find F9s2d and use it to find an
equation of the tangent line to the curve y − 5xys1 1 x 2 d
at the point s2, 2d.
(b)Illustrate part (a) by graphing the curve and the tangent
; line on the same screen.
30. (a)If Gsxd − 4x 2 2 x 3, find G9sad and use it to find equations of the tangent lines to the curve y − 4x 2 2 x 3 at
the points s2, 8d and s3, 9d.
(b)Illustrate part (a) by graphing the curve and the tangent
; lines on the same screen.
31–36 Find f 9sad.
31. f sxd − 3x 2 2 4x 1 132. f std − 2t 3 1 t
33. f std −
2t 1 1
34. f sxd − x 22
t13
35. f sxd − s1 2 2x 36. f sxd −
4
s1 2 x
37–42 Each limit represents the derivative of some function f at
some number a. State such an f and a in each case.
400
0
2
28.If tsxd − x 4 2 2, find t9s1d and use it to find an equation of
the tangent line to the curve y − x 4 2 2 at the point s1, 21d.
18.The graph of a function f is shown.
(a)Find the average rate of change of f on the interval
f20, 60g.
(b)Identify an interval on which the average rate of change
of f is 0.
(c)Which interval gives a larger average rate of change,
f40, 60g or f40, 70g?
24.Sketch the graph of a function t for which
ts0d − ts2d − ts4d − 0, t9s1d − t9s3d − 0,
t9s0d − t9s4d − 1, t9s2d − 21, lim x l52 tsxd − `, and
lim x l 211 tsxd − 2`.
27.If f sxd − 3x 2 2 x 3, find f 9s1d and use it to find an equation of
the tangent line to the curve y − 3x 2 2 x 3 at the point s1, 2d.
t9s4d
y=©
0
23.Sketch the graph of a function f for which f s0d − 0,
f 9s0d − 3, f 9s1d − 0, and f 9s2d − 21.
20
40
60
x
37. lim
h l0
2 31h 2 8
s9 1 h 2 3
38. lim
h l0
h
h
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
115
Section 2.1 Derivatives and Rates of Change
1
24
x 6 2 64
x
39. lim
40. lim
x l2 x 2 2
x l 1y4 x 2 1
4
48.The number N of locations of a popular coffeehouse chain is
given in the table. (The numbers of locations as of October 1
are given.)
1
41. lim
h l0
coss 1 hd 1 1
sin 2 2
42. lim
l y6 2 y6
h
43–44 A particle moves along a straight line with equation of
motion s − f std, where s is measured in meters and t in seconds.
Find the velocity and the speed when t − 4.
43. f std − 80t 2 6t 244. f std − 10 1
45
t11
45.A warm can of soda is placed in a cold refrigerator. Sketch
the graph of the temperature of the soda as a function of time.
Is the initial rate of change of temperature greater or less than
the rate of change after an hour?
46.A roast turkey is taken from an oven when its temperature
has reached 185°F and is placed on a table in a room where
the temperature is 75°F. The graph shows how the temperature of the turkey decreases and eventually approaches room
temperature. By measuring the slope of the tangent, estimate
the rate of change of the temperature after an hour.
Year
2004
2006
2008
2010
2012
N
8569
12,440
16,680
16,858
18,066
(a)Find the average rate of growth
(i) from 2006 to 2008
(ii) from 2008 to 2010
In each case, include the units. What can you conclude?
(b)Estimate the instantaneous rate of growth in 2010 by
taking the average of two average rates of change.
What are its units?
(c)Estimate the instantaneous rate of growth in 2010 by
measuring the slope of a tangent.
49.The table shows world average daily oil consumption from
1985 to 2010 measured in thousands of barrels per day.
(a)Compute and interpret the average rate of change from
1990 to 2005. What are the units?
(b)Estimate the instantaneous rate of change in 2000 by
taking the average of two average rates of change.
What are its units?
T (°F)
200
Years
since 1985
Thousands of barrels
of oil per day
0
5
10
15
20
25
60,083
66,533
70,099
76,784
84,077
87,302
P
100
0
30
60
90
120 150
t (min)
47.Researchers measured the average blood alcohol concen­tration Cstd of eight men starting one hour after consumption
of 30 mL of ethanol (corresponding to two alcoholic drinks).
t (hours)
1.0
1.5
2.0
2.5
3.0
Cstd sgydLd
0.033
0.024
0.018
0.012
0.007
(a)Find the average rate of change of C with respect to t
over each time interval:
(i) f1.0, 2.0g
(ii) f1.5, 2.0g
(iii) f2.0, 2.5g
(iv) f2.0, 3.0g
In each case, include the units.
(b)Estimate the instantaneous rate of change at t − 2 and
interpret your result. What are the units?
Source: Adapted from P. Wilkinson et al., “Pharmacokinetics of Ethanol after
Oral Administration in the Fasting State,” Journal of Pharmacokinetics and
Biopharmaceutics 5 (1977): 207–24.
Source: US Energy Information Administration
50. T
he table shows values of the viral load Vstd in HIV patient
303, measured in RNA copiesymL, t days after ABT-538
treatment was begun.
t
4
8
11
15
22
Vstd
53
18
9.4
5.2
3.6
(a)Find the average rate of change of V with respect to t
over each time interval:
(i) f4, 11g
(ii) f8, 11g
(iii) f11, 15g
(iv) f11, 22g
What are the units?
(b) Estimate and interpret the value of the derivative V9s11d.
Source: Adapted from D. Ho et al., “Rapid Turnover of Plasma Virions and
CD4 Lymphocytes in Hiv-1 Infection,” Nature 373 (1995): 123–26.
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116
chapter 2 Derivatives
51.The cost (in dollars) of producing x units of a certain commodity is Csxd − 5000 1 10x 1 0.05x 2.
(a)Find the average rate of change of C with respect to x when
the production level is changed
(i) from x − 100 to x − 105
(ii) from x − 100 to x − 101
(b)Find the instantaneous rate of change of C with respect
to x when x − 100. (This is called the marginal cost. Its
significance will be explained in Section 2.7.)
the oxygen content of water.) The graph shows how oxygen
solubility S varies as a function of the water temperature T.
(a)What is the meaning of the derivative S9sT d? What are its
units?
(b)Estimate the value of S9s16d and interpret it.
S (mg / L)
16
12
52.If a cylindrical tank holds 100,000 gallons of water, which
can be drained from the bottom of the tank in an hour, then
Torricelli’s Law gives the volume V of water remaining in
the tank after t minutes as
8
4
0
1 2
Vstd − 100,000 (1 2 60
t) 0 < t < 60
ind the rate at which the water is flowing out of the tank
F
(the instantaneous rate of change of V with respect to t) as a
function of t. What are its units? For times t − 0, 10, 20, 30,
40, 50, and 60 min, find the flow rate and the amount of water
remaining in the tank. Summarize your findings in a sentence
or two. At what time is the flow rate the greatest? The least?
53.The cost of producing x ounces of gold from a new gold mine
is C − f sxd dollars.
(a)What is the meaning of the derivative f 9sxd? What are its
units?
(b) What does the statement f 9s800d − 17 mean?
(c)Do you think the values of f 9sxd will increase or decrease
in the short term? What about the long term? Explain.
8
16
24
32
40
T (°C)
Source: C. Kupchella et al., Environmental Science: Living Within
the System of Nature, 2d ed. (Boston: Allyn and Bacon, 1989).
58.The graph shows the influence of the temperature T on the
maximum sustainable swimming speed S of Coho salmon.
(a)What is the meaning of the derivative S9sT d? What are its
units?
(b)Estimate the values of S9s15d and S9s25d and interpret them.
S (cm/s)
20
54.The number of bacteria after t hours in a controlled laboratory
experiment is n − f std.
(a)What is the meaning of the derivative f 9s5d? What are its
units?
(b)Suppose there is an unlimited amount of space and
nutrients for the bacteria. Which do you think is larger,
f 9s5d or f 9s10d? If the supply of nutrients is limited,
would that affect your conclusion? Explain.
59–60 Determine whether f 9s0d exists.
55.Let H std be the daily cost (in dollars) to heat an office building when the outside temperature is t degrees Fahrenheit.
(a) What is the meaning of H9s58d? What are its units?
(b)Would you expect H9s58d to be positive or negative?
Explain.
60. f sxd −
0
59. f sxd −
H
H
x sin
10
1
x
0
x 2 sin
0
20
T (°C)
if x ± 0
if x − 0
1
x
if x ± 0
if x − 0
1
; 61. (a)Graph the function f sxd − sin x 2 1000 sins1000xd in the
viewing rectangle f22, 2g by f24, 4g. What slope does
the graph appear to have at the origin?
(b)Zoom in to the viewing window f20.4, 0.4g by
f20.25, 0.25g and estimate the value of f 9s0d. Does this
agree with your answer from part (a)?
(c)Now zoom in to the viewing window f20.008, 0.008g by
57.The quantity of oxygen that can dissolve in water depends on
f20.005, 0.005g. Do you wish to revise your estimate for
the temperature of the water. (So thermal pollution influences
f 9s0d?
56.The quantity (in pounds) of a gourmet ground coffee that is
sold by a coffee company at a price of p dollars per pound
is Q − f s pd.
(a)What is the meaning of the derivative f 9s8d? What are its
units?
(b) Is f 9s8d positive or negative? Explain.
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Section 2.2 The Derivative as a Function
writing Project
117
early methods for finding tangents
The first person to formulate explicitly the ideas of limits and derivatives was Sir Isaac
Newton in the 1660s. But Newton acknowledged that “If I have seen further than other men,
it is because I have stood on the shoulders of giants.” Two of those giants were Pierre Fermat
(1601–1665) and Newton’s mentor at Cambridge, Isaac Barrow (1630–1677). Newton was
familiar with the methods that these men used to find tangent lines, and their methods played a
role in Newton’s eventual formulation of calculus.
The following references contain explanations of these methods. Read one or more of the
references and write a report comparing the methods of either Fermat or Barrow to modern
methods. In particular, use the method of Section 2.1 to find an equation of the tangent line to
the curve y − x 3 1 2x at the point (1, 3) and show how either Fermat or Barrow would have
solved the same problem. Although you used derivatives and they did not, point out similarities between the methods.
1.Carl Boyer and Uta Merzbach, A History of Mathematics (New York: Wiley, 1989),
pp. 389, 432.
2.C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag,
1979), pp. 124, 132.
3.Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Saunders,
1990), pp. 391, 395.
4.Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford
University Press, 1972), pp. 344, 346.
In the preceding section we considered the derivative of a function f at a fixed number a:
1
f 9sad − lim
hl0
f sa 1 hd 2 f sad
h
Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable x, we obtain
2
f 9sxd − lim
hl0
f sx 1 hd 2 f sxd
h
Given any number x for which this limit exists, we assign to x the number f 9sxd. So we
can regard f 9 as a new function, called the derivative of f and defined by Equation 2.
We know that the value of f 9 at x, f 9sxd, can be interpreted geometrically as the slope of
the tangent line to the graph of f at the point sx, f sxdd.
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118
Chapter 2 Derivatives
The function f 9 is called the derivative of f because it has been “derived” from f by
the limiting operation in Equation 2. The domain of f 9 is the set hx f 9sxd existsj and
may be smaller than the domain of f.
|
Example 1 The graph of a function f is given in Figure 1. Use it to sketch the graph
of the derivative f 9.
y
y=ƒ
1
0
x
1
FIGURE 1 SOLUTION We can estimate the value of the derivative at any value of x by drawing the
tangent at the point sx, f sxdd and estimating its slope. For instance, for x − 5 we draw
the tangent at P in Figure 2(a) and estimate its slope to be about 32, so f 9s5d < 1.5. This
allows us to plot the point P9s5, 1.5d on the graph of f 9 directly beneath P. (The slope
of the graph of f becomes the y-value on the graph of f 9.) Repeating this procedure at
several points, we get the graph shown in Figure 2(b). Notice that the tangents at A, B,
and C are horizontal, so the derivative is 0 there and the graph of f 9 crosses the x-axis
(where y − 0) at the points A9, B9, and C9, directly beneath A, B, and C. Between A and
B the tangents have positive slope, so f 9sxd is positive there. (The graph is above the
x-axis.) But between B and C the tangents have negative slope, so f 9sxd is negative
there.
y
B
1
m=0
m=0
y=ƒ
A
0
1
P
m=0
3
mÅ2
5
x
C
TEC Visual 2.2 shows an animation
of Figure 2 for several functions.
(a)
y
y=fª(x)
1
0
FIGURE 2
(b)
Bª
Aª
1
Pª (5, 1.5)
Cª
5
x
n
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Section 2.2 The Derivative as a Function
119
Example 2
(a) If f sxd − x 3 2 x, find a formula for f 9sxd.
(b) Illustrate this formula by comparing the graphs of f and f 9.
2
SOLUTION
f
_2
2
(a) When using Equation 2 to compute a derivative, we must remember that the variable
is h and that x is temporarily regarded as a constant during the calculation of the limit.
f 9sxd − lim
hl0
_2
2
− lim
x 3 1 3x 2h 1 3xh 2 1 h 3 2 x 2 h 2 x 3 1 x
h
− lim
3x 2h 1 3xh 2 1 h 3 2 h
h
hl0
fª
hl0
_2
2
f sx 1 hd 2 f sxd
fsx 1 hd3 2 sx 1 hdg 2 fx 3 2 xg
− lim
hl0
h
h
− lim s3x 2 1 3xh 1 h 2 2 1d − 3x 2 2 1
hl0
(b) We use a graphing device to graph f and f 9 in Figure 3. Notice that f 9sxd − 0
when f has horizontal tangents and f 9sxd is positive when the tangents have positive
slope. So these graphs serve as a check on our work in part (a).
n
_2
FIGURE 3 Example 3 If f sxd − sx , find the derivative of f. State the domain of f 9.
SOLUTION f 9sxd − lim
h l0
− lim
h l0
y
− lim
h l0
1
0
1
x
(a) ƒ=œ„
x
sx 1 h 2 sx
h
S
D
sx 1 h 2 sx sx 1 h 1 sx
?
(Rationalize the numerator.)
h
sx 1 h 1 sx
− lim
sx 1 hd 2 x
h
− lim
h l 0 h (sx 1 h 1 sx )
h (sx 1 h 1 sx )
− lim
1
1
1
−
−
2sx
sx 1 h 1 sx
sx 1 sx
h l0
h l0
y
f sx 1 hd 2 f sxd
h
We see that f 9sxd exists if x . 0, so the domain of f 9 is s0, `d. This is slightly smaller
than the domain of f , which is f0, `d.
n
1
0
1
1
(b) f ª (x)=
x
2œ„
FIGURE 4
x
Let’s check to see that the result of Example 3 is reasonable by looking at the graphs
of f and f 9 in Figure 4. When x is close to 0, sx is also close to 0, so f 9sxd − 1y(2 sx )
is very large and this corresponds to the steep tangent lines near s0, 0d in Figure 4(a) and
the large values of f 9sxd just to the right of 0 in Figure 4(b). When x is large, f 9sxd is very
small and this corresponds to the flatter tangent lines at the far right of the graph of f and
the horizontal asymptote of the graph of f 9.
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120
Chapter 2 Derivatives
Example 4 Find f 9 if f sxd −
12x
.
21x
SOLUTION
f 9sxd − lim
hl0
a
c
2
b
d
ad 2 bc 1
−
?
e
bd
e
1 2 sx 1 hd
12x
2
2 1 sx 1 hd
21x
− lim
hl0
h
− lim
s1 2 x 2 hds2 1 xd 2 s1 2 xds2 1 x 1 hd
hs2 1 x 1 hds2 1 xd
− lim
s2 2 x 2 2h 2 x 2 2 xhd 2 s2 2 x 1 h 2 x 2 2 xhd
hs2 1 x 1 hds2 1 xd
− lim
23h
23
3
− lim
−2
h l 0 s2 1 x 1 hds2 1 xd
hs2 1 x 1 hds2 1 xd
s2 1 xd2
hl0
hl0
Leibniz
Gottfried Wilhelm Leibniz was born in
Leipzig in 1646 and studied law, theology, philosophy, and mathematics at
the university there, graduating with
a bachelor’s degree at age 17. After
earning his doctorate in law at age 20,
Leibniz entered the diplomatic service
and spent most of his life traveling to
the capitals of Europe on political missions. In particular, he worked to avert a
French military threat against Ger­many
and attempted to reconcile the Catholic
and Protestant churches.
His serious study of mathematics did
not begin until 1672 while he was on
a diplomatic mission in Paris. There he
built a calculating machine and met
scientists, like Huygens, who directed his
attention to the latest develop­ments in
mathematics and science. Leibniz sought
to develop a symbolic logic and system
of notation that would simplify logical
reasoning. In particular, the version
of calculus that he published in 1684
established the notation and the rules for
finding derivatives that we use today.
Unfortunately, a dreadful priority
dispute arose in the 1690s between the
followers of Newton and those of Leibniz
as to who had invented calculus first.
Leibniz was even accused of plagiarism
by members of the Royal Society in
England. The truth is that each man
invented calculus independently. Newton
arrived at his version of calculus first but,
because of his fear of controversy, did not
publish it immediately. So Leibniz’s 1684
account of calculus was the first to be
published.
f sx 1 hd 2 f sxd
h
hl0
n
Other Notations
If we use the traditional notation y − f sxd to indicate that the independent variable is x
and the dependent variable is y, then some common alternative notations for the derivative are as follows:
f 9sxd − y9 −
dy
df
d
−
−
f sxd − D f sxd − Dx f sxd
dx
dx
dx
The symbols D and dydx are called differentiation operators because they indicate the
operation of differentiation, which is the process of calculating a derivative.
The symbol dyydx, which was introduced by Leibniz, should not be regarded as a
ratio (for the time being); it is simply a synonym for f 9sxd. Nonetheless, it is a very useful
and suggestive notation, especially when used in conjunction with increment notation.
Referring to Equation 2.1.6, we can rewrite the definition of derivative in Leibniz notation in the form
dy
Dy
− lim
Dx l 0 Dx
dx
If we want to indicate the value of a derivative dyydx in Leibniz notation at a specific
number a, we use the notation
dy
dx
Z
or x−a
dy
dx
G
x−a
which is a synonym for f 9sad. The vertical bar means “evaluate at.”
3 Definition A function f is differentiable at a if f 9sad exists. It is differentiable on an open interval sa, bd [or sa, `d or s2`, ad or s2`, `d] if it is differentiable at every number in the interval.
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Section 2.2 The Derivative as a Function
121
Example 5 Where is the function f sxd − | x | differentiable?
| |
SOLUTION If x . 0, then x − x and we can choose h small enough that x 1 h . 0
and hence x 1 h − x 1 h. Therefore, for x . 0, we have
|
|
f 9sxd − lim
hl0
− lim
hl0
| x 1 h | 2 | x | − lim
h
hl0
sx 1 hd 2 x
h
h
− lim 1 − 1
hl0
h
and so f is differentiable for any x . 0.
Similarly, for x , 0 we have x − 2x and h can be chosen small enough that
x 1 h , 0 and so x 1 h − 2sx 1 hd. Therefore, for x , 0,
|
| |
|
f 9sxd − lim
hl0
− lim
hl0
| x 1 h | 2 | x | − lim
h
hl0
2sx 1 hd 2 s2xd
h
2h
− lim s21d − 21
hl0
h
and so f is differentiable for any x , 0.
For x − 0 we have to investigate
f s0 1 hd 2 f s0d
h
f 9s0d − lim
hl0
y
− lim
hl0
| 0 1 h | 2 | 0 | − lim | h | h
hl0
h
(if it exists)
Let’s compute the left and right limits separately:
0
and
y
|h| −
h l0
h
h
− lim1 1 − 1
h l0
h
lim2
2h
− lim2 s21d − 21
h l0
h
h l0
h l0
x
_1
(b) y=fª(x)
FIGURE 5 lim2
h
lim1
Since these limits are different, f 9s0d does not exist. Thus f is differentiable at all x
except 0.
A formula for f 9 is given by
1
0
|h| −
h l0
x
(a) y=ƒ=| x |
lim1
f 9sxd −
H
1
if x . 0
21 if x , 0
and its graph is shown in Figure 5(b). The fact that f 9s0d does not exist is reflected
geometrically in the fact that the curve y − x does not have a tangent line at s0, 0d.
[See Figure 5(a).]
n
| |
Both continuity and differentiability are desirable properties for a function to have.
The following theorem shows how these properties are related.
4 Theorem If f is differentiable at a, then f is continuous at a.
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122
Chapter 2 Derivatives
Proof To prove that f is continuous at a, we have to show that lim x l a f sxd − f sad.
We do this by showing that the difference f sxd 2 f sad approaches 0.
The given information is that f is differentiable at a, that is,
f 9sad − lim
xla
PS An important aspect of problem
solving is trying to find a connection
between the given and the unknown.
See Step 2 (Think of a Plan) in Principles
of Problem Solving on page 98.
f sxd 2 f sad
x2a
exists (see Equation 2.1.5). To connect the given and the unknown, we divide and
multiply f sxd 2 f sad by x 2 a (which we can do when x ± a):
f sxd 2 f sad −
f sxd 2 f sad
sx 2 ad
x2a
Thus, using the Product Law and Equation 2.1.5, we can write
lim f f sxd 2 f sadg − lim
xla
xla
− lim
xla
f sxd 2 f sad
sx 2 ad
x2a
f sxd 2 f sad
? lim sx 2 ad
xla
x2a
− f 9sad ? 0 − 0
To use what we have just proved, we start with f sxd and add and subtract f sad:
lim f sxd − lim f f sad 1 s f sxd 2 f saddg
xla
xla
− lim f sad 1 lim f f sxd 2 f sadg
xla
xla
− f sad 1 0 − f sad
Therefore f is continuous at a.
n
NOTE The converse of Theorem 4 is false; that is, there are functions that are continuous but not differentiable. For instance, the function f sxd − x is continuous at 0
because
lim f sxd − lim x − 0 − f s0d
| |
xl0
xl0
| |
(See Example 1.6.7.) But in Example 5 we showed that f is not differentiable at 0.
How Can a Function Fail To Be Differentiable?
| |
We saw that the function y − x in Example 5 is not differentiable at 0 and Figure 5(a)
shows that its graph changes direction abruptly when x − 0. In general, if the graph of a
function f has a “corner” or “kink” in it, then the graph of f has no tangent at this point
and f is not differentiable there. [In trying to compute f 9sad, we find that the left and
right limits are different.]
Theorem 4 gives another way for a function not to have a derivative. It says that if f is
not continuous at a, then f is not differentiable at a. So at any discontinuity (for instance,
a jump discontinuity) f fails to be differentiable.
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123
Section 2.2 The Derivative as a Function
y
A third possibility is that the curve has a vertical tangent line when x − a; that is, f
is continuous at a and
vertical tangent
line
|
|
lim f 9sxd − `
xla
This means that the tangent lines become steeper and steeper as x l a. Figure 6 shows
one way that this can happen; Figure 7(c) shows another. Figure 7 illustrates the three
possibilities that we have discussed.
0
a
x
y
y
y
FIGURE 6
0
a
0
x
a
x
0
a
x
FIGURE 7 Three ways for f not to be
differentiable at a
(a) A corner
(b) A discontinuity
(c) A vertical tangent
A graphing calculator or computer provides another way of looking at differen­tiability. If f is differentiable at a, then when we zoom in toward the point sa, f sadd the graph
straightens out and appears more and more like a line. (See Figure 8. We saw a specific
example of this in Figure 2.1.2.) But no matter how much we zoom in toward a point like
the ones in Figures 6 and 7(a), we can’t eliminate the sharp point or corner (see Figure 9).
y
0
y
a
0
x
FIGURE 8 a
x
FIGURE 9
f is differentiable at a.
f is not differentiable at a.
Higher Derivatives
If f is a differentiable function, then its derivative f 9 is also a function, so f 9 may have
a derivative of its own, denoted by s f 9d9 − f 0. This new function f 0 is called the second
derivative of f because it is the derivative of the derivative of f. Using Leibniz notation,
we write the second derivative of y − f sxd as
d
dx
S D
derivative
of
first
derivative
dy
dx
−
d 2y
dx 2
second
derivative
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124
Chapter 2 Derivatives
Example 6 If f sxd − x 3 2 x, find and interpret f 0sxd.
SOLUTION In Example 2 we found that the first derivative is f 9sxd − 3x 2 2 1. So the
second derivative is
f 99sxd − s f 9d9sxd − lim
h l0
2
f·
_1.5
− lim
f3sx 1 hd2 2 1g 2 f3x 2 2 1g
h
− lim
3x 2 1 6xh 1 3h 2 2 1 2 3x 2 1 1
h
h l0
fª
f
1.5
f 9sx 1 hd 2 f 9sxd
h
h l0
− lim s6x 1 3hd − 6x
h l0
_2
FIGURE 10 TEC In Module 2.2 you can see how
changing the coefficients of a polynomial f affects the appearance of the
graphs of f , f 9, and f 99.
The graphs of f , f 9, and f 0 are shown in Figure 10.
We can interpret f 0sxd as the slope of the curve y − f 9sxd at the point sx, f 9sxdd. In
other words, it is the rate of change of the slope of the original curve y − f sxd.
Notice from Figure 10 that f 0sxd is negative when y − f 9sxd has negative slope
and positive when y − f 9sxd has positive slope. So the graphs serve as a check on our
calculations.
n
In general, we can interpret a second derivative as a rate of change of a rate of change.
The most familiar example of this is acceleration, which we define as follows.
If s − sstd is the position function of an object that moves in a straight line, we know
that its first derivative represents the velocity v std of the object as a function of time:
v std − s9std −
ds
dt
The instantaneous rate of change of velocity with respect to time is called the acceleration astd of the object. Thus the acceleration function is the derivative of the velocity
function and is therefore the second derivative of the position function:
astd − v9std − s0std
or, in Leibniz notation,
a−
dv
d 2s
− 2
dt
dt
Acceleration is the change in velocity you would feel when your car is speeding up or
slowing down.
The third derivative f - is the derivative of the second derivative: f -− s f 0d9. So
f -sxd can be interpreted as the slope of the curve y − f 0sxd or as the rate of change of
f 0sxd. If y − f sxd, then alternative notations for the third derivative are
y- − f -sxd −
d
dx
S D
d2y
dx 2
−
d 3y
dx 3
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125
Section 2.2 The Derivative as a Function
We can also interpret the third derivative physically in the case where the function
is the position function s − sstd of an object that moves along a straight line. Because
s-− ss0d9 − a9, the third derivative of the position function is the derivative of the acceleration function and is called the jerk:
j−
da
d 3s
− 3
dt
dt
Thus the jerk j is the rate of change of acceleration. It is aptly named because a large jerk
means a sudden change in acceleration, which causes an abrupt movement in a vehicle.
The differentiation process can be continued. The fourth derivative f + is usually
denoted by f s4d. In general, the nth derivative of f is denoted by f snd and is obtained from
f by differentiating n times. If y − f sxd, we write
y snd − f sndsxd −
dny
dx n
Example 7 If f sxd − x 3 2 x, find f -sxd and f s4dsxd.
SOLUTION In Example 6 we found that f 0sxd − 6x. The graph of the second derivative
has equation y − 6x and so it is a straight line with slope 6. Since the derivative f -sxd
is the slope of f 0sxd, we have
f -sxd − 6
for all values of x. So f - is a constant function and its graph is a horizontal line. Therefore, for all values of x,
f s4d sxd − 0
n
We have seen that one application of second and third derivatives occurs in analyzing
the motion of objects using acceleration and jerk. We will investigate another application of second derivatives in Section 3.3, where we show how knowledge of f 0 gives us
information about the shape of the graph of f. In Chapter 11 we will see how second and
higher derivatives enable us to represent functions as sums of infinite series.
1–2 Use the given graph to estimate the value of each derivative.
Then sketch the graph of f 9.
1.(a) f 9s23d
(e) f 9s1d
(b) f 9s22d
(f ) f 9s2d
(c) f 9s21d
(g) f 9s3d
(b) f 9s1d
(f) f 9s5d
2.(a) f 9s0d
(e) f 9s4d
(c) f 9s2d
(g) f 9s6d
(d) f 9s3d
(h) f 9s7d
(d) f 9s0d
y
1
y
0
1
1
1
x
x
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126
chapter 2 Derivatives
3.Match the graph of each function in (a)–(d) with the graph of
its derivative in I–IV. Give reasons for your choices.
y
(a)
y
8.
y
(b)
0
0
0
x
y
(c)
x
x
10.
y
x
0
y
x
0
y
III
x
0
x
y
II
0
x
y
11.
0
I
0
x
y
(d)
0
y
9.
x
12. S
hown is the graph of the population function Pstd for yeast
cells in a laboratory culture. Use the method of Example 1 to
graph the derivative P9std. What does the graph of P9 tell us
about the yeast population?
y
IV
P (yeast cells)
0
x
0
x
500
4–11 Trace or copy the graph of the given function f. (Assume
that the axes have equal scales.) Then use the method of Example 1
to sketch the graph of f 9 below it.
4.
y
0
5.
y
0
x
0
5
15 t (hours)
10
13.A rechargeable battery is plugged into a charger. The graph
shows Cstd, the percentage of full capacity that the battery
reaches as a function of time t elapsed (in hours).
(a)What is the meaning of the derivative C9std?
(b)Sketch the graph of C9std. What does the graph
tell you?
x
C
6.
100
y
7.
y
80
Percentage
of full charge
0
x
0
x
60
40
20
0
2
4
6
8
10 12
t (hours)
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.2 The Derivative as a Function
(b)Use symmetry to deduce the values of f 9(221 ), f 9s21d,
f 9s22d, and f 9s23d.
(c) Use the values from parts (a) and (b) to graph f 9.
(d) Guess a formula for f 9sxd.
(e)Use the definition of derivative to prove that your guess
in part (d) is correct.
14. T
he graph (from the US Department of Energy) shows how
driving speed affects gas mileage. Fuel economy F is measured in miles per gallon and speed v is measured in miles
per hour.
(a)What is the meaning of the derivative F9svd?
(b)Sketch the graph of F9svd.
(c)At what speed should you drive if you want to save
on gas?
F
30
19–29 Find the derivative of the function using the definition of
derivative. State the domain of the function and the domain of its
derivative.
(mi/gal)
19. f sxd − 3x 2 8
20. f sxd − mx 1 b
20
21. f std − 2.5t 2 1 6t
22. f sxd − 4 1 8x 2 5x 2
10
23. f sxd − x 2 2 2x 3
24. tstd −
25. tsxd − s9 2 x
26. f sxd −
0
10
20 30 40 50 60 70
127
√ (mi/ h)
27. Gstd −
15.The graph shows how the average age M of first marriage
of Japanese men varied in the last half of the 20th century. Sketch the graph of the derivative function M9std.
During which years was the derivative negative?
1 2 2t
31t
1
st
x2 2 1
2x 2 3
28. f sxd − x 3y2
29. f sxd − x 4
30. (a)Sketch the graph of f sxd − s6 2 x by starting with
the graph of y − s x and using the transformations of
Sec­tion 1.3.
(b) Use the graph from part (a) to sketch the graph of f 9.
(c)Use the definition of a derivative to find f 9sxd. What are
the domains of f and f 9?
(d)Use a graphing device to graph f 9 and compare with
;
your sketch in part (b).
M
27
25
1960
1970
1980
1990
2000 t
16.Make a careful sketch of the graph of the sine function and
below it sketch the graph of its derivative in the same manner as in Example 1. Can you guess what the derivative of
the sine function is from its graph?
2
; 17. Let f sxd − x .
(a)Estimate the values of f 9s0d, f 9( 12 ), f 9s1d, and f 9s2d by
using a graphing device to zoom in on the graph of f.
(b)Use symmetry to deduce the values of f 9(221 ), f 9s21d,
and f 9s22d.
(c)Use the results from parts (a) and (b) to guess a formula
for f 9sxd.
(d)Use the definition of derivative to prove that your guess
in part (c) is correct.
3
; 18. Let f sxd − x .
(a)Estimate the values of f 9s0d, f 9( 12 ), f 9s1d, f 9s2d, and
f 9s3d by using a graphing device to zoom in on the
graph of f.
31. (a) If f sxd − x 4 1 2x, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
;
by comparing the graphs of f and f 9.
;
32. (a) If f sxd − x 1 1yx, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
by comparing the graphs of f and f 9.
33. T
he unemployment rate Ustd varies with time. The table
gives the percentage of unemployed in the US labor force
from 2003 to 2012.
(a) What is the meaning of U9std? What are its units?
(b) Construct a table of estimated values for U9std.
t
Ustd
t
Ustd
2003
2004
2005
2006
2007
6.0
5.5
5.1
4.6
4.6
2008
2009
2010
2011
2012
5.8
9.3
9.6
8.9
8.1
Source: US Bureau of Labor Statistics
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128
chapter 2 Derivatives
34. T
he table gives the number Nstd, measured in thousands, of
minimally invasive cosmetic surgery procedures performed
in the United States for various years t.
t
Nstd (thousands)
2000
2002
2004
2006
2008
2010
2012
5,500
4,897
7,470
9,138
10,897
11,561
13,035
38. Suppose N is the number of people in the United States who
travel by car to another state for a vacation this year when
the average price of gasoline is p dollars per gallon. Do you
expect dNydp to be positive or negative? Explain.
39–42 The graph of f is given. State, with reasons, the numbers
at which f is not differentiable.
y
39.
40.
0
_2
2
y
_2
x
2
4
2
4 x
x
Source: American Society of Plastic Surgeons
(a) What is the meaning of N9std? What are its units?
(b) Construct a table of estimated values for N9std.
(c) Graph N and N9.
(d)How would it be possible to get more accurate values
for N9std?
41.
42.
y
0
2
4
6
x
y
_2
0
35. T
he table gives the height as time passes of a typical pine
tree grown for lumber at a managed site.
Tree age (years)
14
21
28
35
42
49
Height (feet)
41
54
64
72
78
83
Source: Arkansas Forestry Commission
If Hstd is the height of the tree after t years, construct a table
of estimated values for H9 and sketch its graph.
36. W
ater temperature affects the growth rate of brook trout.
The table shows the amount of weight gained by brook trout
after 24 days in various water temperatures.
Temperature (°C)
15.5
17.7
20.0
22.4
24.4
Weight gained (g)
37.2
31.0
19.8
9.7
29.8
| |
raph the function f sxd − x 1 s x . Zoom in repeatedly,
; 43. G
first toward the point (21, 0) and then toward the origin.
What is different about the behavior of f in the vicinity of
these two points? What do you conclude about the differentiability of f ?
oom in toward the points (1, 0), (0, 1), and (21, 0) on
; 44. Z
the graph of the function tsxd − sx 2 2 1d2y3. What do you
notice? Account for what you see in terms of the differentiability of t.
45–46 The graphs of a function f and its derivative f 9 are
shown. Which is bigger, f 9s21d or f 99s1d?
45.
If Wsxd is the weight gain at temperature x, construct a table
of estimated values for W9 and sketch its graph. What are
the units for W9sxd?
y
0
1
x
Source: Adapted from J. Chadwick Jr., “Temperature Effects on Growth
and Stress Physiology of Brook Trout: Implications for Climate Change
Impacts on an Iconic Cold-Water Fish.” Masters Theses. Paper 897. 2012.
scholarworks.umass.edu/theses/897.
37. L
et P represent the percentage of a city’s electrical power
that is produced by solar panels t years after January 1, 2000.
(a) What does dPydt represent in this context?
(b) Interpret the statement
dP
dt
Z
46.
y
1
x
− 3.5
t −2
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129
Section 2.2 The Derivative as a Function
47.The figure shows the graphs of f , f 9, and f 0. Identify each
curve, and explain your choices.
y
a
; 51–52 Use the definition of a derivative to find f 9sxd and f 0sxd.
Then graph f , f 9, and f 0 on a common screen and check to see
if your answers are reasonable.
51. f sxd − 3x 2 1 2x 1 1
b
2
3
s4d
; 53. If f sxd − 2x 2 x , find f 9sxd, f 0sxd, f -sxd, and f sxd.
Graph f , f 9, f 0, and f - on a common screen. Are the
graphs consistent with the geometric interpretations of these
derivatives?
x
c
48.The figure shows graphs of f, f 9, f 0, and f -. Identify each
curve, and explain your choices.
54. (a)The graph of a position function of a car is shown,
where s is measured in feet and t in seconds. Use it to
graph the velocity and acceleration of the car. What is
the acceleration at t − 10 seconds?
a b c d
y
s
x
100
0
49. T
he figure shows the graphs of three functions. One is the
position function of a car, one is the velocity of the car, and
one is its acceleration. Identify each curve, and explain your
choices.
y
a
b
52. f sxd − x 3 2 3x
t
0
20
t
(b)Use the acceleration curve from part (a) to estimate the
jerk at t − 10 seconds. What are the units for jerk?
3
55. Let f sxd − s
x.
(a) If a ± 0, use Equation 2.1.5 to find f 9sad.
(b) Show that f 9s0d does not exist.
3
(c)Show that y − s
x has a vertical tangent line at s0, 0d.
(Recall the shape of the graph of f . See Figure 1.2.13.)
56. (a)
(b)
(c)
(d)
;
c
10
If tsxd − x 2y3, show that t9s0d does not exist.
If a ± 0, find t9sad.
Show that y − x 2y3 has a vertical tangent line at s0, 0d.
Illustrate part (c) by graphing y − x 2y3.
|
|
57.Show that the function f sxd − x 2 6 is not differentiable
at 6. Find a formula for f 9 and sketch its graph.
50.The figure shows the graphs of four functions. One is the
position function of a car, one is the velocity of the car, one
is its acceleration, and one is its jerk. Identify each curve,
and explain your choices.
y
a
0
8et0208x52
08/29/13
| |
59. (a) Sketch the graph of the function f sxd − x x .
(b) For what values of x is f differentiable?
(c) Find a formula for f 9.
| |
60. (a) Sketch the graph of the function tsxd − x 1 x .
(b) For what values of x is t differentiable?
(c) Find a formula for t9.
d
b
58.Where is the greatest integer function f sxd − v x b not differentiable? Find a formula for f 9 and sketch its graph.
c
t
61.Recall that a function f is called even if f s2xd − f sxd
for all x in its domain and odd if f s2xd − 2f sxd for all
such x. Prove each of the following.
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function.
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130
chapter 2 Derivatives
62.The left-hand and right-hand derivatives of f at a are
defined by
f sa 1 hd 2 f sad
f 92 sad − lim2
h l0
h
f 91 sad − lim1
and
h l0
f sa 1 hd 2 f sad
h
if these limits exist. Then f 9sad exists if and only if these
one-sided derivatives exist and are equal.
(a)Find f 92s4d and f 91s4d for the function
f sxd −
0
52x
if x < 0
if 0 , x , 4
1
52x
if x > 4
y=c
c
slope=0
x
0
64.When you turn on a hot-water faucet, the temperature T of the
water depends on how long the water has been running.
(a)Sketch a possible graph of T as a function of the time t that
has elapsed since the faucet was turned on.
(b)Describe how the rate of change of T with respect to t
varies as t increases.
(c) Sketch a graph of the derivative of T.
65.Let be the tangent line to the parabola y − x 2 at the point
s1, 1d. The angle of inclination of is the angle that makes
with the positive direction of the x-axis. Calculate correct to
the nearest degree.
(b) Sketch the graph of f.
(c) Where is f discontinuous?
(d) Where is f not differentiable?
y
63.Nick starts jogging and runs faster and faster for 3 mintues, then
he walks for 5 minutes. He stops at an intersection for 2 minutes,
runs fairly quickly for 5 minutes, then walks for 4 minutes.
(a)Sketch a possible graph of the distance s Nick has covered
after t minutes.
(b) Sketch a graph of dsydt.
If it were always necessary to compute derivatives directly from the definition, as we did
in the preceding section, such computations would be tedious and the evaluation of some
limits would require ingenuity. Fortunately, several rules have been developed for finding
derivatives without having to use the definition directly. These formulas greatly simplify
the task of differentiation.
Let’s start with the simplest of all functions, the constant function f sxd − c. The graph
of this function is the horizontal line y − c, which has slope 0, so we must have f 9sxd − 0.
(See Figure 1.) A formal proof, from the definition of a derivative, is also easy:
f 9sxd − lim
FIGURE 1
hl0
The graph of f sxd − c is the line
y − c, so f 9sxd − 0.
f sx 1 hd 2 f sxd
c2c
− lim
− lim 0 − 0
h
l
0
hl0
h
h
In Leibniz notation, we write this rule as follows.
Derivative of a Constant Function d
scd − 0
dx
y
Power Functions
y=x
We next look at the functions f sxd − x n, where n is a positive integer. If n − 1, the
graph of f sxd − x is the line y − x, which has slope 1. (See Figure 2.) So
slope=1
0
x
FIGURE 2
The graph of f sxd − x is the line
y − x, so f 9sxd − 1.
1
d
sxd − 1
dx
(You can also verify Equation 1 from the definition of a derivative.) We have already
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Section 2.3 Differentiation Formulas
131
investigated the cases n − 2 and n − 3. In fact, in Section 2.2 (Exercises 17 and 18) we
found that
d
d
sx 2 d − 2x sx 3 d − 3x 2
dx
dx
2 For n − 4 we find the derivative of f sxd − x 4 as follows:
f 9sxd − lim
f sx 1 hd 2 f sxd
sx 1 hd4 2 x 4
− lim
hl0
h
h
− lim
x 4 1 4x 3h 1 6x 2h 2 1 4xh 3 1 h 4 2 x 4
h
− lim
4x 3h 1 6x 2h 2 1 4xh 3 1 h 4
h
hl0
hl0
hl0
− lim s4x 3 1 6x 2h 1 4xh 2 1 h 3 d − 4x 3
hl0
Thus
d
sx 4 d − 4x 3
dx
3 Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a
rea­sonable guess that, when n is a positive integer, sdydxdsx n d − nx n21. This turns out to
be true. We prove it in two ways; the second proof uses the Binomial Theorem.
The Power Rule If n is a positive integer, then
d
sx n d − nx n21
dx
First Proof The formula
x n 2 a n − sx 2 adsx n21 1 x n22a 1 ∙ ∙ ∙ 1 xa n22 1 a n21 d
can be verified simply by multiplying out the right-hand side (or by summing the sec-
ond factor as a geometric series). If f sxd − x n, we can use Equation 2.1.5 for f 9sad and
the equation above to write
f sxd 2 f sad
xn 2 an
− lim
xla x 2 a
x2a
f 9sad − lim
xla
− lim sx n21 1 x n22a 1 ∙ ∙ ∙ 1 xa n22 1 a n21 d
xla
− a n21 1 a n22a 1 ∙ ∙ ∙ 1 aa n22 1 a n21
− na n21
second Proof
f 9sxd − lim
hl0
f sx 1 hd 2 f sxd
sx 1 hdn 2 x n
− lim
hl0
h
h
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132
Chapter 2 Derivatives
The Binomial Theorem is given on
Reference Page 1.
In finding the derivative of x 4 we had to expand sx 1 hd4. Here we need to expand
sx 1 hdn and we use the Binomial Theorem to do so:
f 9sxd − lim
F
x n 1 nx n21h 1
G
nsn 2 1d n22 2
x h 1 ∙ ∙ ∙ 1 nxh n21 1 h n 2 x n
2
h
hl0
nsn 2 1d n22 2
x h 1 ∙ ∙ ∙ 1 nxh n21 1 h n
2
h
nx n21h 1
− lim
hl0
− lim
hl0
F
nsn 2 1d n22
x h 1 ∙ ∙ ∙ 1 nxh n22 1 h n21
2
nx n21 1
G
− nx n21
because every term except the first has h as a factor and therefore approaches 0.
■
We illustrate the Power Rule using various notations in Example 1.
Example 1
(a) If f sxd − x 6, then f 9sxd − 6x 5.
(c) If y − t 4, then
dy
− 4t 3.
dt
(b) If y − x 1000, then y9 − 1000x 999.
(d)
d 3
sr d − 3r 2
dr
■
New Derivatives from Old
When new functions are formed from old functions by addition, subtraction, or multiplica­
tion by a constant, their derivatives can be calculated in terms of derivatives of the old
func­tions. In particular, the following formula says that the derivative of a constant times
a function is the constant times the derivative of the function.
The Constant Multiple Rule If c is a constant and f is a differentiable function,
then
d
d
fcf sxdg − c
f sxd
dx
dx
Geometric Interpretation
of the Constant Multiple Rule
y
y=2ƒ
y=ƒ
0
Proof Let tsxd − cf sxd. Then
x
t9sxd − lim
hl0
Multiplying by c − 2 stretches the
graph vertically by a factor of 2. All
the rises have been doubled but the
runs stay the same. So the slopes are
doubled too.
tsx 1 hd 2 tsxd
cf sx 1 hd 2 cf sxd
− lim
hl0
h
h
F
− lim c
hl0
− c lim
hl0
− cf 9sxd
f sx 1 hd 2 f sxd
h
G
f sx 1 hd 2 f sxd
(by Limit Law 3)
h
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.3 Differentiation Formulas
133
Example 2
(a)
d
d
s3x 4 d − 3
sx 4 d − 3s4x 3 d − 12x 3
dx
dx
(b)
d
d
d
s2xd −
fs21dxg − s21d
sxd − 21s1d − 21
dx
dx
dx
■
The next rule tells us that the derivative of a sum of functions is the sum of the
derivatives.
Using prime notation, we can write the
Sum Rule as
The Sum Rule If f and t are both differentiable, then
d
d
d
f f sxd 1 tsxdg −
f sxd 1
tsxd
dx
dx
dx
s f 1 td9 − f 9 1 t9
Proof Let Fsxd − f sxd 1 tsxd. Then
F9sxd − lim
hl0
− lim
hl0
− lim
hl0
− lim
hl0
Fsx 1 hd 2 Fsxd
h
f f sx 1 hd 1 tsx 1 hdg 2 f f sxd 1 tsxdg
h
F
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
1
h
h
G
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
1 lim
(by Limit Law 1)
hl 0
h
h
− f 9sxd 1 t9sxd
■
The Sum Rule can be extended to the sum of any number of functions. For instance,
using this theorem twice, we get
s f 1 t 1 hd9 − fs f 1 td 1 hg9 − s f 1 td9 1 h9 − f 9 1 t9 1 h9
By writing f 2 t as f 1 s21dt and applying the Sum Rule and the Constant Multiple
Rule, we get the following formula.
The Difference Rule If f and t are both differentiable, then
d
d
d
f f sxd 2 tsxdg −
f sxd 2
tsxd
dx
dx
dx
The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined with the Power Rule to differentiate any polynomial, as the following examples
demonstrate.
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134
Chapter 2 Derivatives
Example 3
d
sx 8 1 12x 5 2 4x 4 1 10x 3 2 6x 1 5d
dx
d
d
d
d
d
d
−
sx 8 d 1 12
sx 5 d 2 4
sx 4 d 1 10
sx 3 d 2 6
sxd 1
s5d
dx
dx
dx
dx
dx
dx
− 8x 7 1 12s5x 4 d 2 4s4x 3 d 1 10s3x 2 d 2 6s1d 1 0
− 8x 7 1 60x 4 2 16x 3 1 30x 2 2 6
y
■
Example 4 Find the points on the curve y − x 4 2 6x 2 1 4 where the tangent line is
horizontal.
(0, 4)
SOLUTION Horizontal tangents occur where the derivative is zero. We have
0
x
{œ„
3, _5}
{_ œ„
3, _5}
FIGURE 3
4
2
The curve y − x 2 6x 1 4 and its
horizontal tangents
dy
d
d
d
−
sx 4 d 2 6
sx 2 d 1
s4d
dx
dx
dx
dx
− 4x 3 2 12x 1 0 − 4xsx 2 2 3d
Thus dyydx − 0 if x − 0 or x 2 2 3 − 0, that is, x − 6s3 . So the given curve has
horizontal tangents when x − 0, s3 , and 2s3 . The corresponding points are s0, 4d,
ss3 , 25d, and s2s3 , 25d. (See Figure 3.)
■
Example 5 The equation of motion of a particle is s − 2t 3 2 5t 2 1 3t 1 4, where s
is measured in centimeters and ­t in seconds. Find the acceleration as a function of time.
What is the acceleration after 2 seconds?
SOLUTION The velocity and acceleration are
vstd −
ds
− 6t 2 2 10t 1 3
dt
astd −
dv
− 12t 2 10
dt
The acceleration after 2 s is as2d − 12s2d 2 10 − 14 cmys2.
■
Next we need a formula for the derivative of a product of two functions. By analogy
with the Sum and Difference Rules, one might be tempted to guess, as Leibniz did three
centuries ago, that the derivative of a product is the product of the derivatives. We can
see, however, that this guess is wrong by looking at a particular example. Let f sxd − x
and tsxd − x 2. Then the Power Rule gives f 9sxd − 1 and t9sxd − 2x. But s ftdsxd − x 3,
so s ftd9sxd − 3x 2. Thus s ftd9 ± f 9t9. The correct formula was discovered by Leibniz
(soon after his false start) and is called the Product Rule.
In prime notation:
s ftd9 − ft9 1 t f 9
The Product Rule If f and t are both differentiable, then
d
d
d
f f sxdtsxdg − f sxd
ftsxdg 1 tsxd
f f sxdg
dx
dx
dx
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.3 Differentiation Formulas
135
Proof Let Fsxd − f sxdtsxd. Then
F9sxd − lim
hl0
− lim
hl0
Fsx 1 hd 2 Fsxd
h
f sx 1 hdtsx 1 hd 2 f sxdtsxd
h
In order to evaluate this limit, we would like to separate the functions f and t as in
the proof of the Sum Rule. We can achieve this separation by subtracting and adding
the term f sx 1 hd tsxd in the numerator:
F9sxd − lim
hl0
− lim
hl0
f sx 1 hdtsx 1 hd 2 f sx 1 hdtsxd 1 f sx 1 hd tsxd 2 f sxdtsxd
h
F
f sx 1 hd
tsx 1 hd 2 tsxd
f sx 1 hd 2 f sxd
1 tsxd
h
h
− lim f sx 1 hd lim
hl0
hl0
G
tsx 1 hd 2 tsxd
f sx 1 hd 2 f sxd
1 lim tsxd lim
hl0
hl 0
h
h
− f sxdt9sxd 1 tsxd f 9sxd
Note that lim h l 0 tsxd − tsxd because tsxd is a constant with respect to the variable h.
Also, since f is differentiable at x, it is continuous at x by Theorem 2.2.4, and so
lim h l 0 f sx 1 hd − f sxd. (See Exercise 1.8.63.)
■
In words, the Product Rule says that the derivative of a product of two functions is the
first function times the derivative of the second function plus the second function times
the derivative of the first function.
Example 6 Find F9sxd if Fsxd − s6x 3 ds7x 4 d.
SOLUTION By the Product Rule, we have
F9sxd − s6x 3 d
d
d
s7x 4 d 1 s7x 4 d
s6x 3 d
dx
dx
− s6x 3 ds28x 3 d 1 s7x 4 ds18x 2 d
− 168x 6 1 126x 6 − 294x 6
■
Notice that we could verify the answer to Example 6 directly by first multiplying the
factors:
Fsxd − s6x 3 ds7x 4 d − 42x 7 ? F9sxd − 42s7x 6 d − 294x 6
But later we will meet functions, such as y − x 2 sin x, for which the Product Rule is the
only possible method.
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136
Chapter 2 Derivatives
Example 7 If hsxd − xtsxd and it is known that ts3d − 5 and t9s3d − 2, find h9s3d.
SOLUTION Applying the Product Rule, we get
h9sxd −
d
d
d
fxtsxdg − x
ftsxdg 1 tsxd
fxg
dx
dx
dx
− x t9sxd 1 tsxd s1d
h9s3d − 3t9s3d 1 ts3d − 3 2 1 5 − 11
Therefore
In prime notation:
■
The Quotient Rule If f and t are differentiable, then
SD
f 9
t f 9 2 ft9
−
t
t2
d
dx
F G
f sxd
tsxd
tsxd
−
d
d
f f sxdg 2 f sxd
ftsxdg
dx
dx
ftsxdg 2
proof Let Fsxd − f sxdytsxd. Then
f sx 1 hd
f sxd
2
Fsx 1 hd 2 Fsxd
tsx 1 hd
tsxd
F9sxd − lim
− lim
h l0
hl 0
h
h
− lim
h l0
f sx 1 hdtsxd 2 f sxdtsx 1 hd
htsx 1 hdtsxd
We can separate f and t in this expression by subtracting and adding the term f sxdtsxd
in the numerator:
F9sxd − lim
hl0
− lim
hl0
f sx 1 hdtsxd 2 f sxdtsxd 1 f sxdtsxd 2 f sxdtsx 1 hd
htsx 1 hdtsxd
tsxd
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
2 f sxd
h
h
tsx 1 hdtsxd
lim tsxd lim
−
hl0
hl0
f sx 1 hd 2 f sxd
tsx 1 hd 2 tsxd
2 lim f sxd lim
hl0
hl0
h
h
lim tsx 1 hd lim tsxd
hl0
−
hl0
tsxd f 9sxd 2 f sxdt9sxd
ftsxdg 2
Again t is continuous by Theorem 2.2.4, so lim h l 0 tsx 1 hd − tsxd.
■
In words, the Quotient Rule says that the derivative of a quotient is the denominator
times the derivative of the numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.
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Section 2.3 Differentiation Formulas
137
The theorems of this section show that any polynomial is differentiable on R and any
rational function is differentiable on its domain. Furthermore, the Quotient Rule and the
other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates.
We can use a graphing device to
check that the answer to Example 8
is plausible. Figure 4 shows the graphs
of the function of Example 8 and its
derivative. Notice that when y grows
rapidly (near 22), y9 is large. And
when y grows slowly, y9 is near 0.
Example 8 Let y −
sx 3 1 6d
y9 −
1.5
yª
_4
y
FIGURE 4
d
d
sx 2 1 x 2 2d 2 sx 2 1 x 2 2d
sx 3 1 6d
dx
dx
sx 3 1 6d2
−
sx 3 1 6ds2x 1 1d 2 sx 2 1 x 2 2ds3x 2 d
sx 3 1 6d2
−
s2x 4 1 x 3 1 12x 1 6d 2 s3x 4 1 3x 3 2 6x 2 d
sx 3 1 6d2
−
2x 4 2 2x 3 1 6x 2 1 12x 1 6
sx 3 1 6d2
4
_1.5
x2 1 x 2 2
. Then
x3 1 6
■
NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s
easier to first rewrite a quotient to put it in a form that is simpler for the purpose of differentiation. For instance, although it is possible to differentiate the function
Fsxd −
3x 2 1 2sx
x
using the Quotient Rule, it is much easier to perform the division first and write the function as
Fsxd − 3x 1 2x 21y2
before differentiating.
General Power Functions
The Quotient Rule can be used to extend the Power Rule to the case where the exponent
is a negative integer.
If n is a positive integer, then
d 2n
sx d − 2nx 2n21
dx
proof
d
d
sx 2n d −
dx
dx
xn
−
−
SD
1
xn
d
d
s1d 2 1 sx n d
dx
dx
x n 0 2 1 nx n21
−
n 2
sx d
x 2n
2nx n21
− 2nx n2122n − 2nx 2n21
x 2n
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
■
138
Chapter 2 Derivatives
Example 9
(a) If y −
(b)
d
dt
1
dy
d
1
, then
−
sx 21 d − 2x 22 − 2 2
x
dx
dx
x
SD
6
t3
−6
d 23
18
st d − 6s23dt 24 − 2 4
dt
t
■
So far we know that the Power Rule holds if the exponent n is a positive or negative
integer. If n − 0, then x 0 − 1, which we know has a derivative of 0. Thus the Power
Rule holds for any integer n. What if the exponent is a fraction? In Example 2.2.3 we
found that
d
1
sx −
dx
2 sx
which can be written as
d 1y2
sx d − 12 x21y2
dx
This shows that the Power Rule is true even when n − 12. In fact, it also holds for any real
number n, as we will prove in Chapter 6. (A proof for rational values of n is indicated in
Exercise 2.6.48.) In the meantime we state the general version and use it in the examples
and exercises.
The Power Rule (General Version) If n is any real number, then
d
sx n d − nx n21
dx
Example 10
(a) If f sxd − x , then f 9sxd − x 21.
(b) Let
Then
y−
1
sx 2
3
dy
d 22y3
−
sx d − 223 x2s2y3d21
dx
dx
− 223 x25y3
In Example 11, a and b are constants.
It is customary in mathematics to
use letters near the beginning of the
alphabet to represent constants and
letters near the end of the alphabet to
represent variables.
■
Example 11 Differentiate the function f std − st sa 1 btd.
SOLUTION 1 Using the Product Rule, we have
f 9std − st
d
d
sa 1 btd 1 sa 1 btd
(st )
dt
dt
− st ? b 1 sa 1 btd ? 12 t 21y2
− bst 1
a 1 bt
a 1 3bt
−
2 st
2 st
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.3 Differentiation Formulas
139
SOLUTION 2 If we first use the laws of exponents to rewrite f std, then we can proceed
directly without using the Product Rule.
f std − ast 1 btst − at 1y2 1 bt 3y2
f 9std − 12 at21y2 1 32 bt 1y2
which is equivalent to the answer given in Solution 1.
■
The differentiation rules enable us to find tangent lines without having to resort to the
definition of a derivative. It also enables us to find normal lines. The normal line to a
curve C at a point P is the line through P that is perpendicular to the tangent line at P. (In
the study of optics, one needs to consider the angle between a light ray and the normal
line to a lens.)
Example 12 Find equations of the tangent line and normal line to the curve
y − sx ys1 1 x 2 d at the point s1, 12 d.
SOLUTION According to the Quotient Rule, we have
dy
−
dx
s1 1 x 2 d
d
d
ssx d 2 sx dx
s1 1 x 2 d
dx
s1 1 x 2 d2
1
−
2 sx s2xd
2 sx
s1 1 x 2 d2
−
s1 1 x 2 d 2 4x 2
1 2 3x 2
2 2 −
2 sx s1 1 x d
2 sx s1 1 x 2 d2
s1 1 x 2 d
So the slope of the tangent line at s1, 12 d is
dy
dx
y
y 2 12 − 2 41 sx 2 1d or y − 214 x 1 34
tangent
0
FIGURE 5
x−1
1 2 3 ? 12
1
−2
2 s1s1 1 12 d2
4
We use the point-slope form to write an equation of the tangent line at s1, 12 d:
normal
1
Z
−
2
x
The slope of the normal line at s1, 12 d is the negative reciprocal of 241, namely 4, so
an equation is
y 2 12 − 4sx 2 1d or y − 4x 2 72
The curve and its tangent and normal lines are graphed in Figure 5.
■
Example 13 At what points on the hyperbola xy − 12 is the tangent line parallel to
the line 3x 1 y − 0?
SOLUTION Since xy − 12 can be written as y − 12yx, we have
dy
d
12
− 12
sx 21 d − 12s2x 22 d − 2 2
dx
dx
x
Let the x-coordinate of one of the points in question be a. Then the slope of the tangent
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140
Chapter 2 Derivatives
y
(2, 6)
line at that point is 212ya 2. This tangent line will be parallel to the line 3x 1 y − 0, or
y − 23x, if it has the same slope, that is, 23. Equating slopes, we get
xy=12
0
x
2
12
− 23 or a 2 − 4 or a − 62
a2
Therefore the required points are s2, 6d and s22, 26d. The hyperbola and the tangents
are shown in Figure 6.
■
(_2, _6)
We summarize the differentiation formulas we have learned so far as follows.
3x+y=0
Table of Differentiation Formulas
FIGURE 6
d
scd − 0
dx
d
sx n d − nx n21
dx
scf d9 − cf 9
s f 1 td9 − f 91 t9
s ftd9 − ft9 1 t f 9
s f 2 td9 − f 9 2 t9
SD
f 9
t f 9 2 ft9
−
t
t2
2.3 Exercises
1–22 Differentiate the function.
24. Find the derivative of the function
1. f sxd − 2 2. f sxd − 40
3. f sxd − 5.2x 1 2.3
5. f std − 2t 3 2 3t 2 2 4t
2
4. tsxd − 74 x 2 2 3x 1 12
6. f std − 1.4t 5 2 2.5t 2 1 6.7
7. tsxd − x 2 s1 2 2xd8. Hsud − s3u 2 1dsu 1 2d
9. tstd − 2t 23y4
10. Bs yd − cy26
5
r3
11. Fsrd −
12. y − x 5y3 2 x 2y3
14. y − sx s2 1 xd
15. Rsad − s3a 1 1d2
16. SsRd − 4 R 2
sx
18. y −
sx 1 x
x2
s7
19. Gsqd − s1 1 q d 20. Gstd − s5t 1
t
21 2
21. u −
S
t
2
in two ways: by using the Quotient Rule and by simplifying
first. Show that your answers are equivalent. Which method
do you prefer?
25–44 Differentiate.
25. f sxd − s5x 2 2 2dsx 3 1 3xd
3
x 2 1 4x 1 3
1
x 4 2 5x 3 1 sx
x2
26. Bsud − su 3 1 1ds2u 2 2 4u 2 1d
13. Ss pd − sp 2 p
17. y −
Fsxd −
1
st
D
2
1 1 16t 2
22. Dstd −
s4td 3
23. F
ind the derivative of f sxd − s1 1 2x 2 dsx 2 x 2 d in two
ways: by using the Product Rule and by performing the
multiplication first. Do your answers agree?
27. Fs yd −
S
D
3
1
2 4 s y 1 5y 3 d
y2
y
28. Jsvd − sv 3 2 2 vdsv24 1 v22 d
29. tsxd −
1 1 2x
6t 1 1
30. hstd −
3 2 4x
6t 2 1
31. y −
x2 1 1
1
32. y − 3
x3 2 1
t 1 2t 2 2 1
33. y −
t 3 1 3t
su 1 2d 2
34. y −
t 2 4t 1 3
12u
35. y −
s 2 ss
s2
2
36. y −
sx
21x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.3 Differentiation Formulas
37. f std −
39. Fsxd −
3
t
cx
s
38. y −
t23
1 1 cx
2x 5 1 x 4 2 6x
x3
40. Asvd − v 2y3s2 v 2 1 1 2 v 22 d
41. Gs yd −
43. f sxd −
B
Ay 3 1 B
x
c
x1
x
42. Fstd −
At
Bt 2 1 Ct 3
44. f sxd −
ax 1 b
cx 1 d
45. The general polynomial of degree n has the form
Psxd − a n x n 1 a n21 x n21 1 ∙ ∙ ∙ 1 a 2 x 2 1 a 1 x 1 a 0
141
53. (a)The curve y − 1ys1 1 x 2 d is called a witch of Maria
Agnesi. Find an equation of the tangent line to this
curve at the point s21, 12 d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on the same screen.
54. (a)The curve y − xys1 1 x 2 d is called a serpentine.
Find an equation of the tangent line to this curve at the
point s3, 0.3d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on the same screen.
55–58 Find equations of the tangent line and normal line to the
curve at the given point.
55. y − x 1 sx , s1, 2d
57. y −
56. y 2 − x 3, s1, 1d
3x 1 1
, s1, 2d
x2 1 1
58. y −
sx
, s4, 0.4d
x11
where a n ± 0. Find the derivative of P.
; 46–48 Find f 9sxd. Compare the graphs of f and f 9 and use
them to explain why your answer is reasonable.
46. f sxd − xysx 2 2 1d
47. f sxd − 3x 15 2 5x 3 1 3
1
48. f sxd − x 1
x
; 49. (a)Graph the function
f sxd − x 4 2 3x 3 2 6x 2 1 7x 1 30
in the viewing rectangle f23, 5g by f210, 50g.
(b)Using the graph in part (a) to estimate slopes, make
a rough sketch, by hand, of the graph of f 9. (See
Example 2.2.1.)
(c)Calculate f 9sxd and use this expression, with a graphing device, to graph f 9. Compare with your sketch in
part (b).
2
2
; 50. (a)Graph the function tsxd − x ysx 1 1d in the viewing
rectangle f24, 4g by f21, 1.5g.
(b)Using the graph in part (a) to estimate slopes, make a
rough sketch, by hand, of the graph of t9. (See
Example 2.2.1.)
(c)Calculate t9sxd and use this expression, with a graphing device, to graph t9. Compare with your sketch in
part (b).
51–52 Find an equation of the tangent line to the curve at the
given point.
2x
51. y −
, s1, 1d
x11
52. y − 2x 3 2 x 2 1 2, s1, 3d
59–62 Find the first and second derivatives of the function.
3
60. G srd − sr 1 s
r
59. f sxd − 0.001x 5 2 0.02x 3
61. f sxd −
x2
1 1 2x
62. f sxd −
1
32x
63. T
he equation of motion of a particle is s − t 3 2 3t, where s
is in meters and t is in seconds. Find
(a) the velocity and acceleration as functions of t,
(b) the acceleration after 2 s, and
(c) the acceleration when the velocity is 0.
64.The equation of motion of a particle is
s − t 4 2 2t 3 1 t 2 2 t, where s is in meters and t is in
seconds.
(a)Find the velocity and acceleration as functions of t.
(b)Find the acceleration after 1 s.
(c)Graph the position, velocity, and acceleration functions
;
on the same screen.
65.Biologists have proposed a cubic polynomial to model the
length L of Alaskan rockfish at age A:
L − 0.0155A 3 2 0.372A 2 1 3.95A 1 1.21
where L is measured in inches and A in years. Calculate
dL
dA
Z
A −12
and interpret your answer.
66.The number of tree species S in a given area A in the Pasoh
Forest Reserve in Malaysia has been modeled by the power
function
SsAd − 0.882 A 0.842
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
142
chapter 2 Derivatives
where A is measured in square meters. Find S9s100d and
interpret your answer.
Source: Adapted from K. Kochummen et al., “Floristic Composition of Pasoh
Forest Reserve, A Lowland Rain Forest in Peninsular Malaysia,” Journal of
Tropical Forest Science 3 (1991):1–13.
67.Boyle’s Law states that when a sample of gas is compressed
at a constant temperature, the pressure P of the gas is
inversely proportional to the volume V of the gas.
(a)Suppose that the pressure of a sample of air that occupies 0.106 m 3 at 25°C is 50 kPa. Write V as a function
of P.
(b)Calculate dVydP when P − 50 kPa. What is the meaning of the derivative? What are its units?
; 68.Car tires need to be inflated properly because overinflation
or underinflation can cause premature tread wear. The data
in the table show tire life L (in thousands of miles) for a
certain type of tire at various pressures P (in lbyin2).
P
26
28
31
35
38
42
45
L
50
66
78
81
74
70
59
(a)Use a calculator to model tire life with a quadratic function of the pressure.
(b)Use the model to estimate dLydP when P − 30 and
when P − 40. What is the meaning of the derivative?
What are the units? What is the significance of the signs
of the derivatives?
69.Suppose that f s5d − 1, f 9s5d − 6, ts5d − 23, and
t9s5d − 2. Find the following values.
(a) s ftd9s5d
(b) s fytd9s5d
(c) s tyf d9s5d
70. Suppose that f s4d − 2, ts4d − 5, f 9s4d − 6, and
t9s4d − 23. Find h9s4d.
(a)
hsxd − 3 f sxd 1 8tsxd
(b) hsxd − f sxdtsxd
(c) hsxd −
f sxd
tsxd
(d) hsxd −
tsxd
f sxd 1 tsxd
71. If f sxd − sx tsxd, where ts4d − 8 and t9s4d − 7, find f 9s4d.
72. If hs2d − 4 and h9s2d − 23, find
d
dx
S DZ
hsxd
x
y
F
G
1
0
1
x
75. If t is a differentiable function, find an expression for the
derivative of each of the following functions.
x
(a) y − xtsxd
(b) y −
tsxd
tsxd
(c) y −
x
76. If f is a differentiable function, find an expression for the
derivative of each of the following functions.
f sxd
(a) y − x 2 f sxd
(b) y −
x2
(c) y −
x2
f sxd
(d) y −
1 1 x f sxd
sx
77.Find the points on the curve y − 2x 3 1 3x 2 2 12x 1 1
where the tangent is horizontal.
78.For what values of x does the graph of
f sxd − x 3 1 3x 2 1 x 1 3 have a horizontal tangent?
79. S
how that the curve y − 6x 3 1 5x 2 3 has no tangent line
with slope 4.
80.Find an equation of the tangent line to the curve y − x 4 1 1
that is parallel to the line 32 x 2 y − 15.
81.Find equations of both lines that are tangent to the curve
y − x 3 2 3x 2 1 3x 2 3 and are parallel to the line
3x 2 y − 15.
82. Find equations of the tangent lines to the curve
y−
x−2
73. If f and t are the functions whose graphs are shown, let
usxd − f sxdtsxd and vsxd − f sxdytsxd.
(a) Find u9s1d.
(b) Find v9s5d.
x21
x11
that are parallel to the line x 2 2y − 2.
83.Find an equation of the normal line to the curve y − sx
that is parallel to the line 2x 1 y − 1.
84.Where does the normal line to the parabola y − x 2 2 1 at
the point s21, 0d intersect the parabola a second time?
Illustrate with a sketch.
y
f
1
0
74.Let Psxd − FsxdGsxd and Qsxd − FsxdyGsxd, where F and
G are the functions whose graphs are shown.
(a)Find P9s2d.
(b) Find Q9s7d.
1
g
x
85.Draw a diagram to show that there are two tangent lines to
the parabola y − x 2 that pass through the point s0, 24d.
Find the coordinates of the points where these tangent lines
intersect the parabola.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
143
Section 2.3 Differentiation Formulas
86. (a)Find equations of both lines through the point s2, 23d that
are tangent to the parabola y − x 2 1 x.
(b)Show that there is no line through the point s2, 7d that is
tangent to the parabola. Then draw a diagram to see why.
87. (a)Use the Product Rule twice to prove that if f , t, and h are
differentiable, then s fthd9 − f 9th 1 ft9h 1 fth9.
(b) Taking f − t − h in part (a), show that
d
f f sxdg 3 − 3f f sxdg 2 f 9sxd
dx
(c)Use part (b) to differentiate y − sx 4 1 3x 3 1 17x 1 82d3.
88.Find the nth derivative of each function by calculating the first
few derivatives and observing the pattern that occurs.
(a)f sxd − x n
(b) f sxd − 1yx
89. F
ind a second-degree polynomial P such that Ps2d − 5,
P9s2d − 3, and P99s2d − 2.
90.The equation y99 1 y9 2 2y − x 2 is called a differential
equation because it involves an unknown function y and its
derivatives y9 and y99. Find constants A, B, and C such that the
function y − Ax 2 1 Bx 1 C satisfies this equation. (Differential equations will be studied in detail in Chapter 9.)
91.Find a cubic function y − ax 3 1 bx 2 1 cx 1 d whose graph
has horizontal tangents at the points s22, 6d and s2, 0d.
92. F
ind a parabola with equation y − ax 2 1 bx 1 c that has
slope 4 at x − 1, slope 28 at x − 21, and passes through the
point s2, 15d.
93. I n this exercise we estimate the rate at which the total personal
income is rising in the Richmond-Petersburg, Virginia, metro­politan area. In 1999, the population of this area was 961,400,
and the population was increasing at roughly 9200 people per
year. The average annual income was $30,593 per capita, and
this average was increasing at about $1400 per year (a little
above the national average of about $1225 yearly). Use the
Product Rule and these figures to estimate the rate at which
total personal income was rising in the Richmond-Petersburg
area in 1999. Explain the meaning of each term in the Product
Rule.
94. A
manufacturer produces bolts of a fabric with a fixed width.
The quantity q of this fabric (measured in yards) that is sold is
a function of the selling price p (in dollars per yard), so we can
write q − f s pd. Then the total revenue earned with selling price
p is Rs pd − pf s pd.
(a)What does it mean to say that f s20d − 10,000 and
f 9s20d − 2350?
(b)Assuming the values in part (a), find R9s20d and interpret
your answer.
95.The Michaelis-Menten equation for the enzyme chymotrypsin
is
0.14fSg
v−
0.015 1 fSg
where v is the rate of an enzymatic reaction and [S] is the concentration of a substrate S. Calculate d vyd fSg and interpret it.
96. The biomass Bstd of a fish population is the total mass of the
members of the population at time t. It is the product of the
number of individuals Nstd in the population and the average
mass Mstd of a fish at time t. In the case of guppies, breeding
occurs continually. Suppose that at time t − 4 weeks the population is 820 guppies and is growing at a rate of 50 guppies
per week, while the average mass is 1.2 g and is increasing at
a rate of 0.14 gyweek. At what rate is the biomass increasing
when t − 4?
97.Let
f sxd −
H
x 2 1 1 if x , 1
x 1 1 if x > 1
Is f differentiable at 1? Sketch the graphs of f and f 9.
98.At what numbers is the following function t differentiable?
H
2x
if x < 0
tsxd − 2x 2 x 2 if 0 , x , 2
22x
if x > 2
Give a formula for t9 and sketch the graphs of t and t9.
|
99.(a)For what values of x is the function f sxd − x 2 2 9
differentiable? Find a formula for f 9.
(b) Sketch the graphs of f and f 9.
|
| |
|
|
100.Where is the function hsxd − x 2 1 1 x 1 2 differenti­
able? Give a formula for h9 and sketch the graphs of h and h9.
101.For what values of a and b is the line 2x 1 y − b tangent to
the parabola y − ax 2 when x − 2?
102. (a)If Fsxd − f sxd tsxd, where f and t have derivatives of all
orders, show that F99 − f 99t 1 2 f 9t9 1 f t99.
(b)Find similar formulas for F999 and F s4d.
(c)Guess a formula for F snd.
103.Find the value of c such that the line y − 32 x 1 6 is tangent to
the curve y − csx .
104. Let
f sxd −
H
x2
if x < 2
mx 1 b if x . 2
Find the values of m and b that make f differentiable everywhere.
105. A
n easy proof of the Quotient Rule can be given if we make
the prior assumption that F9sxd exists, where F − fyt. Write
f − Ft; then differentiate using the Product Rule and solve
the resulting equation for F9.
106.A tangent line is drawn to the hyperbola xy − c at a point P.
(a)Show that the midpoint of the line segment cut from this
tangent line by the coordinate axes is P.
(b)Show that the triangle formed by the tangent line and
the coordinate axes always has the same area, no matter
where P is located on the hyperbola.
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144
Chapter 2 Derivatives
107.Evaluate lim
xl1
109. I f c . 12, how many lines through the point s0, cd are normal
lines to the parabola y − x 2 ? What if c < 12 ?
x 1000 2 1
.
x21
108. Draw a diagram showing two perpendicular lines that intersect
on the y-axis and are both tangent to the parabola y − x 2.
Where do these lines intersect?
applied Project
L¡
P
building a better roller coaster
Suppose you are asked to design the first ascent and drop for a new roller coaster. By studying
photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and
the slope of the drop 21.6. You decide to connect these two straight stretches y − L 1sxd and
y − L 2 sxd with part of a parabola y − f sxd − a x 2 1 bx 1 c, where x and f sxd are measured in
feet. For the track to be smooth there can’t be abrupt changes in direction, so you want the linear
segments L 1 and L 2 to be tangent to the parabola at the transition points P and Q. (See the figure.)
To simplify the equations, you decide to place the origin at P.
f
Q
L™
1. (a)Suppose the horizontal distance between P and Q is 100 ft. Write equations in a, b, and c
that will ensure that the track is smooth at the transition points.
(b)Solve the equations in part (a) for a, b, and c to find a formula for f sxd.
(c)Plot L 1, f , and L 2 to verify graphically that the transitions are smooth.
;
(d)Find the difference in elevation between P and Q.
2.
The solution in Problem 1 might look smooth, but it might not feel smooth because the
piecewise defined function [consisting of L 1sxd for x , 0, f sxd for 0 < x < 100, and L 2sxd
for x . 100] doesn’t have a continuous second derivative. So you decide to improve the
design by using a quadratic function qsxd − ax 2 1 bx 1 c only on the interval 10 < x < 90
and connecting it to the linear functions by means of two cubic functions:
© Susana Ortega / Shutterstock.com
7et0301apun01
01/13/10
MasterID: 00344
110.Sketch the parabolas y − x 2 and y − x 2 2 2x 1 2. Do you
think there is a line that is tangent to both curves? If so, find
its equation. If not, why not?
CAS
A review of the trigonometric functions
is given in Appendix D.
tsxd − kx 3 1 lx 2 1 mx 1 n
0 < x , 10
hsxd − px 3 1 qx 2 1 rx 1 s
90 , x < 100
(a)Write a system of equations in 11 unknowns that ensure that the functions and their first
two derivatives agree at the transition points.
(b)Solve the equations in part (a) with a computer algebra system to find formulas for
qsxd, tsxd, and hsxd.
(c)Plot L 1, t, q, h, and L 2, and compare with the plot in Problem 1(c).
Before starting this section, you might need to review the trigonometric functions. In
particular, it is important to remember that when we talk about the function f defined for
all real numbers x by
f sxd − sin x
it is understood that sin x means the sine of the angle whose radian measure is x. A similar convention holds for the other trigonometric functions cos, tan, csc, sec, and cot.
Recall from Section 1.8 that all of the trigonometric functions are continuous at every
number in their domains.
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145
Section 2.4 Derivatives of Trigonometric Functions
If we sketch the graph of the function f sxd − sin x and use the interpretation of f 9sxd
as the slope of the tangent to the sine curve in order to sketch the graph of f 9 (see Exercise 2.2.16), then it looks as if the graph of f 9 may be the same as the cosine curve (see
Figure 1).
y
y=ƒ=sin x
0
TEC Visual 2.4 shows an animation
of Figure 1.
π
2
π
2π
x
y
y=fª(x )
0
π
2
π
x
FIGURE 1
Let’s try to confirm our guess that if f sxd − sin x, then f 9sxd − cos x. From the definition of a derivative, we have
f 9sxd − lim
hl0
We have used the addition formula for
sine. See Appendix D.
− lim
hl0
− lim
hl0
− lim
hl0
1 f sx 1 hd 2 f sxd
sinsx 1 hd 2 sin x
− lim
hl0
h
h
sin x cos h 1 cos x sin h 2 sin x
h
F
F S
sin x
cos h 2 1
h
− lim sin x ? lim
hl0
G
S DG
sin x cos h 2 sin x
cos x sin h
1
h
h
hl0
D
1 cos x
sin h
h
cos h 2 1
sin h
1 lim cos x ? lim
hl0
hl0
h
h
Two of these four limits are easy to evaluate. Since we regard x as a constant when com­
puting a limit as h l 0, we have
lim sin x − sin x and lim cos x − cos x
hl0
hl0
The limit of ssin hdyh is not so obvious. In Example 1.5.3 we made the guess, on the basis
of numerical and graphical evidence, that
2
lim
l0
sin −1
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146
Chapter 2 Derivatives
D
B
We now use a geometric argument to prove Equation 2. Assume first that lies between
0 and y2. Figure 2(a) shows a sector of a circle with center O, central angle , and
radius 1. BC is drawn perpendicular to OA. By the definition of radian measure, we have
arc AB − . Also BC − OB sin − sin . From the diagram we see that
| | |
1
E
Therefore
O
|
| BC | , | AB | , arc AB
sin , so ¨
C
A
(a)
Let the tangent lines at A and B intersect at E. You can see from Figure 2(b) that the
cir­cumference of a circle is smaller than the length of a circumscribed polygon, and so
arc AB , AE 1 EB . Thus
| | | |
B
| | | |
, | AE | 1 | ED |
− | AD | − | OA | tan − arc AB , AE 1 EB
E
A
O
sin ,1
− tan (b)
FIGURE 2
(In Appendix F the inequality < tan is proved directly from the definition of the
length of an arc without resorting to geometric intuition as we did here.) Therefore we
have
sin ,
cos so
cos ,
sin ,1
We know that lim l 0 1 − 1 and lim l 0 cos − 1, so by the Squeeze Theorem, we have
lim
l 01
sin −1
But the function ssin dy is an even function, so its right and left limits must be equal.
Hence, we have
sin lim
−1
l0
so we have proved Equation 2.
We can deduce the value of the remaining limit in (1) as follows:
We multiply numerator and denominator by cos 1 1 in order to put the
function in a form in which we can use
the limits we know.
lim
l0
cos 2 1
− lim
l0
− lim
l0
S
2sin 2
− 2lim
l0
scos 1 1d
− 2lim
l0
− 21 ?
D
S
cos 2 1 cos 1 1
?
cos 1 1
− lim
l0
cos2 2 1
scos 1 1d
sin sin ?
cos 1 1
D
sin sin ? lim
l 0 cos 1 1
S D
0
111
− 0 (by Equation 2)
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Section 2.4 Derivatives of Trigonometric Functions
3
147
cos 2 1
−0
lim
l0
If we now put the limits (2) and (3) in (1), we get
f 9sxd − lim sin x lim
hl0
hl0
cos h 2 1
sin h
1 lim cos x lim
hl0
hl0
h
h
− ssin xd 0 1 scos xd 1 − cos x
So we have proved the formula for the derivative of the sine function:
d
ssin xd − cos x
dx
4 Example 1 Differentiate y − x 2 sin x.
Figure 3 shows the graphs of the function of Example 1 and its deriva­tive.
Notice that y9 − 0 whenever y has a
horizontal tangent.
SOLUTION Using the Product Rule and Formula 4, we have
dy
d
d
− x2
ssin xd 1 sin x
sx 2 d
dx
dx
dx
5
− x 2 cos x 1 2x sin x
yª
_4
y
4
Using the same methods as in the proof of Formula 4, one can prove (see Exercise 20) that
d
scos xd − 2sin x
dx
5
_5
FIGURE 3
■
The tangent function can also be differentiated by using the definition of a derivative,
but it is easier to use the Quotient Rule together with Formulas 4 and 5:
d
d
stan xd −
dx
dx
S D
cos x
−
sin x
cos x
d
d
ssin xd 2 sin x
scos xd
dx
dx
cos2x
−
cos x cos x 2 sin x s2sin xd
cos2x
−
cos2x 1 sin2x
cos2x
−
1
− sec2x
cos2x
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148
Chapter 2 Derivatives
d
stan xd − sec2x
dx
6
The derivatives of the remaining trigonometric functions, csc, sec, and cot, can also
be found easily using the Quotient Rule (see Exercises 17–19). We collect all the differentiation formulas for trigonometric functions in the following table. Remember that
they are valid only when x is measured in radians.
Derivatives of Trigonometric Functions When you memorize this table, it is
helpful to notice that the minus signs
go with the derivatives of the “cofunctions,” that is, cosine, cosecant, and
cotangent.
d
ssin xd − cos x
dx
d
scsc xd − 2csc x cot x
dx
d
scos xd − 2sin x
dx
d
ssec xd − sec x tan x
dx
d
stan xd − sec2x
dx
d
scot xd − 2csc 2x
dx
Example 2 Differentiate f sxd −
of f have a horizontal tangent?
sec x
. For what values of x does the graph
1 1 tan x
SOLUTION The Quotient Rule gives
s1 1 tan xd
f 9sxd −
−
s1 1 tan xd sec x tan x 2 sec x sec2x
s1 1 tan xd2
−
sec x stan x 1 tan2x 2 sec2xd
s1 1 tan xd2
−
sec x stan x 2 1d
s1 1 tan xd2
3
_3
5
_3
FIGURE 4
The horizontal tangents in Example 2
0
4
s
FIGURE 5
d
d
ssec xd 2 sec x
s1 1 tan xd
dx
dx
s1 1 tan xd2
In simplifying the answer we have used the identity tan2x 1 1 − sec2x.
Since sec x is never 0, we see that f 9sxd − 0 when tan x − 1, and this occurs when
x − n 1 y4, where n is an integer (see Figure 4).
■
Trigonometric functions are often used in modeling real-world phenomena. In particular, vibrations, waves, elastic motions, and other quantities that vary in a periodic
manner can be described using trigonometric functions. In the following example we
discuss an instance of simple harmonic motion.
Example 3 An object at the end of a vertical spring is stretched 4 cm beyond its rest
position and released at time t − 0. (See Figure 5 and note that the downward direction
is positive.) Its position at time t is
s − f std − 4 cos t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.4 Derivatives of Trigonometric Functions
149
Find the velocity and acceleration at time t and use them to analyze the motion of the
object.
SOLUTION The velocity and acceleration are
s
v−
ds
d
d
−
s4 cos td − 4
scos td − 24 sin t
dt
dt
dt
a−
dv
d
d
−
s24 sin td − 24
ssin td − 24 cos t
dt
dt
dt
√
a
2
0
π
_2
FIGURE 6
2π t
The object oscillates from the lowest point ss − 4 cmd to the highest point
ss − 24 cmd. The period of the oscillation is 2, the period of cos t.
The speed is v − 4 sin t , which is greatest when sin t − 1, that is, when
cos t − 0. So the object moves fastest as it passes through its equilibrium position
ss − 0d. Its speed is 0 when sin t − 0, that is, at the high and low points.
The acceleration a − 24 cos t − 0 when s − 0. It has greatest magnitude at the
high and low points. See the graphs in Figure 6.
| |
|
|
|
|
■
Example 4 Find the 27th derivative of cos x.
SOLUTION The first few derivatives of f sxd − cos x are as follows:
f 9sxd − 2sin x
PS Look for a pattern.
f 99sxd − 2cos x
f999sxd − sin x
f s4dsxd − cos x
f s5dsxd − 2sin x
We see that the successive derivatives occur in a cycle of length 4 and, in particular,
f sndsxd − cos x whenever n is a multiple of 4. Therefore
f s24dsxd − cos x
and, differentiating three more times, we have
f s27dsxd − sin x
■
Our main use for the limit in Equation 2 has been to prove the differentiation formula
for the sine function. But this limit is also useful in finding certain other trigonometric
limits, as the following two examples show.
Example 5 Find lim
xl0
sin 7x
.
4x
SOLUTION In order to apply Equation 2, we first rewrite the function by multiplying
and dividing by 7:
Note that sin 7x ± 7 sin x.
sin 7x
7
−
4x
4
S D
sin 7x
7x
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150
Chapter 2 Derivatives
If we let − 7x, then l 0 as x l 0, so by Equation 2 we have
lim
xl0
S D
sin 7x
7
sin 7x
− lim
x
l
0
4x
4
7x
−
7
sin 7
7
lim
− 1−
4 l0 4
4
■
Example 6 Calculate lim x cot x.
xl0
SOLUTION Here we divide numerator and denominator by x:
lim x cot x − lim
xl0
xl0
x cos x
sin x
lim cos x
cos x
xl0
−
x l 0 sin x
sin x
lim
xl0
x
x
cos 0
−
(by the continuity of cosine and Equation 2)
1
−1
− lim
1–16 Differentiate.
20. P
rove, using the definition of derivative, that if
f sxd − cos x, then f 9sxd − 2sin x.
1. f sxd − x sin x
2
21–24 Find an equation of the tangent line to the curve at the
given point.
2. f sxd − x cos x 1 2 tan x
3. f sxd − 3 cot x 2 2 cos x4. y − 2 sec x 2 csc x
21. y − sin x 1 cos x, s0, 1d
5. y − sec tan 6. tstd − 4 sec t 1 tan t
22. y − s1 1 xd cos x, s0, 1d
7. y − c cos t 1 t 2 sin t
23. y − cos x 2 sin x, s, 21d
8. y − usa cos u 1 b cot ud
9. y −
13. y −
24. y − x 1 tan x, s, d
x
10. y − sin cos 2 2 tan x
11. f sd −
sin cos x
12. y −
1 1 cos 1 2 sin x
t sin t
sin t
14. y −
11t
1 1 tan t
15. f sd − cos sin 16. y − x sin x tan x
2
17. Prove that
d
scsc xd − 2csc x cot x.
dx
18. Prove that
d
ssec xd − sec x tan x.
dx
19. Prove that
d
scot xd − 2csc 2x.
dx
■
25. (a)Find an equation of the tangent line to the curve
y − 2x sin x at the point sy2, d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on the same screen.
26. (a)Find an equation of the tangent line to the curve
y − 3x 1 6 cos x at the point sy3, 1 3d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on the same screen.
;
27. (a)If f sxd − sec x 2 x, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
by graphing both f and f 9 for x , y2.
| |
28. (a)
If f sxd − sx sin x, find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable
;
by graphing both f and f 9 for 0 < x < 2.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.4 Derivatives of Trigonometric Functions
29.If Hsd − sin , find H9sd and H99sd.
30.If f std − sec t, find f 0sy4d.
31. (a)Use the Quotient Rule to differentiate the function
tan x 2 1
f sxd −
sec x
(b)Simplify the expression for f sxd by writing it in terms
of sin x and cos x, and then find f 9sxd.
(c)Show that your answers to parts (a) and (b) are
equivalent.
32. Suppose f sy3d − 4 and f 9sy3d − 22, and let
tsxd − f sxd sin x and hsxd − scos xdyf sxd. Find
(a) t9sy3d
(b) h9sy3d
33.For what values of x does the graph of f sxd − x 1 2 sin x
have a horizontal tangent?
34.Find the points on the curve y − scos xdys2 1 sin xd at
which the tangent is horizontal.
35.A mass on a spring vibrates horizontally on a smooth
level surface (see the figure). Its equation of motion is
xstd − 8 sin t, where t is in seconds and x in centimeters.
(a) Find the velocity and acceleration at time t.
(b)Find the position, velocity, and acceleration of the mass
at time t − 2y3. In what direction is it moving at that
time?
38. A
n object with weight W is dragged along a horizontal
plane by a force acting along a rope attached to the object.
If the rope makes an angle with the plane, then the magnitude of the force is
F−
39–50 Find the limit.
39. lim
sin 5x
sin x
40. lim
xl0 sin x
3x
41. lim
tan 6t
cos 2 1
42. lim
l
0
sin 2t
sin 43. lim
sin 3x
sin 3x sin 5x
44. lim
xl0
5x 3 2 4x
x2
45. lim
sin 46. lim csc x sinssin xd
xl0
1 tan 47. lim
cos 2 1
sinsx 2 d
48. lim
2
xl0
2
x
xl0
tl0
xl0
l0
49. lim
x l y4
0
x
x
; 36.An elastic band is hung on a hook and a mass is hung on the
lower end of the band. When the mass is pulled downward
and then released, it vibrates vertically. The equation of
motion is s − 2 cos t 1 3 sin t, t > 0, where s is measured
in centi­meters and t in seconds. (Take the positive direction
to be downward.)
(a) Find the velocity and acceleration at time t.
(b) Graph the velocity and acceleration functions.
(c)When does the mass pass through the equilibrium
position for the first time?
(d)How far from its equilibrium position does the mass
travel?
(e)When is the speed the greatest?
37.A ladder 10 ft long rests against a vertical wall. Let be the
angle between the top of the ladder and the wall and let x be
the distance from the bottom of the ladder to the wall. If the
bottom of the ladder slides away from the wall, how fast
does x change with respect to when − y3?
W
sin 1 cos where is a constant called the coefficient of friction.
(a) Find the rate of change of F with respect to .
(b) When is this rate of change equal to 0?
(c)If W − 50 lb and − 0.6, draw the graph of F as
;
a function of and use it to locate the value of for
which dFyd − 0. Is the value consistent with your
answer to part (b)?
l0
equilibrium
position
151
1 2 tan x
sinsx 2 1d
50. lim 2
x
l
1
sin x 2 cos x
x 1x22
51–52 Find the given derivative by finding the first few derivatives and observing the pattern that occurs.
51.
d 99
d 35
sx sin xd
99 ssin xd52.
dx
dx 35
53.Find constants A and B such that the function
y − A sin x 1 B cos x satisfies the differential equation
y99 1 y9 2 2y − sin x.
x sin
; 54. Evaluate xlim
l0
1
and illustrate by graphing
x
y − x sins1yxd.
55.Differentiate each trigonometric identity to obtain a new
(or familiar) identity.
(a) tan x −
sin x
cos x
(c) sin x 1 cos x −
(b) sec x −
1
cos x
1 1 cot x
csc x
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152
chapter 2 Derivatives
56. A
semicircle with diameter PQ sits on an isosceles triangle
PQR to form a region shaped like a two-dimensional
ice-cream cone, as shown in the figure. If Asd is the area of
the semicircle and Bsd is the area of the triangle, find
lim
l 01
57.The figure shows a circular arc of length s and a chord of
length d, both subtended by a central angle . Find
s
lim
l 01 d
Asd
Bsd
d
s
¨
A(¨ )
P
B(¨)
10 cm
Q
; 58.Let f sxd −
10 cm
¨
R
x
.
s1 2 cos 2x
(a)Graph f . What type of discontinuity does it appear to
have at 0?
(b)Calculate the left and right limits of f at 0. Do these
values confirm your answer to part (a)?
Suppose you are asked to differentiate the function
Fsxd − sx 2 1 1
See Section 1.3 for a review of
composite functions.
The differentiation formulas you learned in the previous sections of this chapter do not
enable you to calculate F9sxd.
Observe that F is a composite function. In fact, if we let y − f sud − su and let
u − tsxd − x 2 1 1, then we can write y − Fsxd − f stsxdd, that is, F − f 8 t. We know
how to differentiate both f and t, so it would be useful to have a rule that tells us how to
find the derivative of F − f 8 t in terms of the derivatives of f and t.
It turns out that the derivative of the composite function f 8 t is the product of the
derivatives of f and t. This fact is one of the most important of the differentiation rules and
is called the Chain Rule. It seems plausible if we interpret derivatives as rates of change.
Regard duydx as the rate of change of u with respect to x, dyydu as the rate of change of
y with respect to u, and dyydx as the rate of change of y with respect to x. If u changes
twice as fast as x and y changes three times as fast as u, then it seems reasonable that y
changes six times as fast as x, and so we expect that
dy
dy du
−
dx
du dx
The Chain Rule If t is differentiable at x and f is differentiable at tsxd, then the
composite function F − f 8 t defined by Fsxd − f stsxdd is differentiable at x and
F9 is given by the product
F9sxd − f 9stsxdd ? t9sxd
I n Leibniz notation, if y − f sud and u − tsxd are both differentiable functions,
then
dy
dy du
−
dx
du dx
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Section 2.5 The Chain Rule
James Gregory
The first person to formulate the
Chain Rule was the Scottish mathematician James Gregory (1638–1675),
who also designed the first practical
reflecting telescope. Gregory discovered the basic ideas of calculus at
about the same time as Newton.
He became the first Professor of
Mathematics at the University of
St. Andrews and later held the same
position at the University of Edinburgh. But one year after accepting that position he died at the age
of 36.
153
Comments on the Proof of the Chain Rule Let Du be the change in u correspond-
ing to a change of Dx in x, that is,
Du − tsx 1 Dxd 2 tsxd
Then the corresponding change in y is
Dy − f su 1 Dud 2 f sud
It is tempting to write
dy
Dy
− lim
Dxl 0 Dx
dx
1 − lim
Dy Du
?
Du Dx
− lim
Dy
Du
? lim
Dx
l
0
Du
Dx
− lim
Dy
Du
? lim
Du Dx l 0 Dx
Dx l 0
Dx l 0
Du l 0
−
(Note that Du l 0 as Dx l 0
since t is continuous.)
dy du
du dx
The only flaw in this reasoning is that in (1) it might happen that Du − 0 (even when
Dx ± 0) and, of course, we can’t divide by 0. Nonetheless, this reasoning does at least
suggest that the Chain Rule is true. A full proof of the Chain Rule is given at the end of
this section.
■
The Chain Rule can be written either in the prime notation
s f 8 td9sxd − f 9stsxdd ? t9sxd
2
or, if y − f sud and u − tsxd, in Leibniz notation:
dy
dy du
−
dx
du dx
3
Equation 3 is easy to remember because if dyydu and duydx were quotients, then we
could cancel du. Remember, however, that du has not been defined and duydx should
not be thought of as an actual quotient.
Example 1 Find F9sxd if Fsxd − sx 2 1 1.
SOLUTION 1 (using Equation 2): At the beginning of this section we expressed F as
Fsxd − s f 8 tdsxd − f stsxdd where f sud − su and tsxd − x 2 1 1. Since
f 9sud − 12 u21y2 −
we have
1
2 su
and t9sxd − 2x
F9sxd − f 9stsxdd ? t9sxd
−
1
x
? 2x −
2 sx 2 1 1
sx 2 1 1
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
154
Chapter 2 Derivatives
SOLUTION 2 (using Equation 3): If we let u − x 2 1 1 and y − su , then
dy du
1
1
x
−
s2xd −
s2xd −
2 1 1
du dx
2 su
2 sx 2 1 1
sx
F9sxd −
■
When using Formula 3 we should bear in mind that dyydx refers to the derivative of
y when y is considered as a function of x (called the derivative of y with respect to x),
whereas dyydu refers to the derivative of y when considered as a function of u (the
derivative of y with respect to u). For instance, in Example 1, y can be considered as a
function of x ( y − s x 2 1 1 ) and also as a function of u ( y − su ). Note that
dy
x
dy
1
− F9sxd −
whereas − f 9sud −
2
dx
du
2 su
sx 1 1
NOTE In using the Chain Rule we work from the outside to the inside. Formula 2
says that we differentiate the outer function f [at the inner function tsxd] and then we
multiply by the derivative of the inner function.
d
dx
f
stsxdd
outer
function
evaluated
at inner
function
f9
stsxdd
derivative
of outer
function
evaluated
at inner
function
−
t9sxd
?
derivative
of inner
function
Example 2 Differentiate (a) y − sinsx 2 d and (b) y − sin2x.
SOLUTION 7et0304note1
(a) If y − sinsx 2 d, then the outer function is the sine function and the inner function is
01/13/10
the squaring function,
so the Chain Rule gives
MasterID: 01592
dy
d
−
dx
dx
sin
sx 2 d
outer
function
evaluated
at inner
function
cos
sx 2 d
derivative
of outer
function
evaluated
at inner
function
−
2x
?
derivative
of inner
function
− 2x cossx 2 d
(b) Note that sin2x − ssin
xd2. Here the outer function is the squaring function and the
7et0304note2
inner function is the sine
function. So
01/13/10
01593
dy
dMasterID:
−
ssin xd2
−
2
dx
dx
inner
function
See Reference Page 2 or Appendix D.
?
ssin xd
?
cos x
inner
derivative derivative
evaluated evaluatedderivative derivative
function
of outer of outer
at inner at inner of inner of inner
function function
function function function function
The answer can be left as 2 sin x cos x or written as sin 2x (by a trigonometric identity
7et0304note3
known as the double-angle7et0304note3
formula).
■
01/13/10
01/13/10
MasterID:
MasterID:
0159401594
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.5 The Chain Rule
155
In Example 2(a) we combined the Chain Rule with the rule for differentiating the sine
function. In general, if y − sin u, where u is a differentiable function of x, then, by the
Chain Rule,
dy
dy du
du
−
− cos u
dx
du dx
dx
d
du
ssin ud − cos u
dx
dx
Thus
In a similar fashion, all of the formulas for differentiating trigonometric functions can
be combined with the Chain Rule.
Let’s make explicit the special case of the Chain Rule where the outer function f is
a power function. If y − ftsxdg n, then we can write y − f sud − u n where u − tsxd. By
using the Chain Rule and then the Power Rule, we get
dy
dy du
du
−
− nu n21
− nftsxdg n21 t9sxd
dx
du dx
dx
4 The Power Rule Combined with the Chain Rule If n is any real number
and u − tsxd is differentiable, then
d
du
su n d − nu n21
dx
dx
d
ftsxdg n − nftsxdg n21 t9sxd
dx
Alternatively,
Notice that the derivative in Example 1 could be calculated by taking n − 12 in Rule 4.
Example 3 Differentiate y − sx 3 2 1d100.
SOLUTION Taking u − tsxd − x 3 2 1 and n − 100 in (4), we have
dy
d
d
−
sx 3 2 1d100 − 100sx 3 2 1d99
sx 3 2 1d
dx
dx
dx
− 100sx 3 2 1d99 3x 2 − 300x 2sx 3 2 1d99
■
Example 4 Find f 9sxd if f sxd −
SOLUTION First rewrite f :
Thus
1
.
3
x2 1 x 1 1
s
f sxd − sx 2 1 x 1 1d21y3
f 9sxd − 213 sx 2 1 x 1 1d24y3
d
sx 2 1 x 1 1d
dx
− 213 sx 2 1 x 1 1d24y3s2x 1 1d
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
■
156
Chapter 2 Derivatives
Example 5 Find the derivative of the function
tstd −
S D
t22
2t 1 1
9
SOLUTION Combining the Power Rule, Chain Rule, and Quotient Rule, we get
S D S D
S D
t9std − 9
t22
2t 1 1
8
d
dt
t22
2t 1 1
8
−9
s2t 1 1d 1 2 2st 2 2d
45st 2 2d8
−
s2t 1 1d2
s2t 1 1d10
t22
2t 1 1
■
Example 6 Differentiate y − s2x 1 1d5sx 3 2 x 1 1d4.
SOLUTION In this example we must use the Product Rule before using the Chain Rule:
The graphs of the functions y and y9
in Example 6 are shown in Figure 1.
Notice that y9 is large when y increases
rapidly and y9 − 0 when y has a horizontal tangent. So our answer appears
to be reasonable.
10
dy
d
d
− s2x 1 1d5
sx 3 2 x 1 1d4 1 sx 3 2 x 1 1d4
s2x 1 1d5
dx
dx
dx
− s2x 1 1d5 4sx 3 2 x 1 1d3
d
sx 3 2 x 1 1d
dx
1 sx 3 2 x 1 1d4 5s2x 1 1d4
yª
d
s2x 1 1d
dx
− 4s2x 1 1d5sx 3 2 x 1 1d3s3x 2 2 1d 1 5sx 3 2 x 1 1d4s2x 1 1d4 2
_2
1
y
_10
FIGURE 1
Noticing that each term has the common factor 2s2x 1 1d4sx 3 2 x 1 1d3, we could
factor it out and write the answer as
dy
− 2s2x 1 1d4sx 3 2 x 1 1d3s17x 3 1 6x 2 2 9x 1 3d
dx
■
The reason for the name “Chain Rule” becomes clear when we make a longer chain
by adding another link. Suppose that y − f sud, u − tsxd, and x − hstd, where f , t, and
h are differentiable functions. Then, to compute the derivative of y with respect to t, we
use the Chain Rule twice:
dy
dy dx
dy du dx
−
−
dt
dx dt
du dx dt
Example 7 If f sxd − sinscosstan xdd, then
f 9sxd − cosscosstan xdd
d
cosstan xd
dx
− cosscosstan xddf2sinstan xdg
d
stan xd
dx
− 2cosscosstan xdd sinstan xd sec2x
Notice that we used the Chain Rule twice.
■
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.5 The Chain Rule
157
Example 8 Differentiate y − ssec x 3 .
SOLUTION Here the outer function is the square root function, the middle function is
the secant function, and the inner function is the cubing function. So we have
dy
1
d
−
ssec x 3 d
dx
2 ssec x 3 dx
−
−
1
2 ssec x
sec x 3 tan x 3
3
d
sx 3 d
dx
3x 2 sec x 3 tan x 3
2 ssec x 3
■
How to Prove the Chain Rule
Recall that if y − f sxd and x changes from a to a 1 Dx, we define the increment of y as
Dy − f sa 1 Dxd 2 f sad
According to the definition of a derivative, we have
lim
Dx l 0
Dy
− f 9sad
Dx
So if we denote by « the difference between the difference quotient and the derivative,
we obtain
lim « − lim
Dx l 0
But
«−
Dx l 0
S
D
Dy
2 f 9sad − f 9sad 2 f 9sad − 0
Dx
Dy
2 f 9sad
Dx
?
Dy − f 9sad Dx 1 « Dx
If we define « to be 0 when Dx − 0, then « becomes a continuous function of Dx. Thus,
for a differentiable function f, we can write
5 Dy − f 9sad Dx 1 « Dx
where
« l 0 as Dx l 0
and « is a continuous function of Dx. This property of differentiable functions is what
enables us to prove the Chain Rule.
Proof of the Chain Rule Suppose u − tsxd is differentiable at a and y − f sud is differentiable at b − tsad. If Dx is an increment in x and Du and Dy are the corresponding
increments in u and y, then we can use Equation 5 to write
6 Du − t9sad Dx 1 «1 Dx − ft9sad 1 «1 g Dx
where «1 l 0 as Dx l 0. Similarly
7 Dy − f 9sbd Du 1 «2 Du − f f 9sbd 1 «2 g Du
where «2 l 0 as Du l 0. If we now substitute the expression for Du from Equation 6
into Equation 7, we get
Dy − f f 9sbd 1 «2 gft9sad 1 «1 g Dx
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
158
Chapter 2 Derivatives
Dy
− f f 9sbd 1 «2 gft9sad 1 «1 g
Dx
so
As Dx l 0, Equation 6 shows that Du l 0. So both «1 l 0 and «2 l 0 as Dx l 0.
Therefore
dy
Dy
− lim
− lim f f 9sbd 1 «2 gft9sad 1 «1 g
Dx l 0 Dx
Dx l 0
dx
− f 9sbd t9sad − f 9stsadd t9sad
This proves the Chain Rule.
1–6 Write the composite function in the form f s tsxdd.
[Identify the inner function u − tsxd and the outer function
y − f sud.] Then find the derivative dyydx.
1. y − s1 1 4x
2. y − s2x 1 5d
3. y − tan x
4. y − sinscot xd
3
3
32. Jsd − tan 2 snd
33. y − sins1 1 x 2
34. y − ssins1 1 x 2 d
35. y −
7–46 Find the derivative of the function.
7. Fsxd − s5x 6 1 2x 3 d 48. Fsxd − s1 1 x 1 x 2 d 99
10. tsxd − s2 2 sin xd 3y2
9. f sxd − s5x 1 1
31. y − cosssec 4xd
4
5. y − ssin x 6. y − sin sx
1
11. Astd −
scos t 1 tan td 2
1
12. f sxd − 3 2
sx 2 1
13. f sd − coss 2 d
14. tsd − cos2 3
15. hsvd − v s
1 1 v2
16. f std − t sin t
17. f sxd − s2x 2 3d4 sx 2 1 x 1 1d5
■
S
1 2 cos 2x
1 1 cos 2x
D
39. f std − tanssecscos tdd
40. tsud − fsu 2 2 1d 6 2 3ug 4
41. y − sx 1 sx
42. y − sx 1 sx 1 s x
43. tsxd − s2r sin rx 1 nd p
44. y − cos 4ssin3 xd
45. y − cos ssinstan xd
46. y − fx 1 sx 1 sin2 xd3 g 4
19. hstd − st 1 1d2y3 s2t 2 2 1d3
47. y − cosssin 3d
S
Î
21. tsud −
23. y −
D
8
x
x11
25. hsd − tans 2 sin d
27. y −
cos x
s1 1 sin x
29. Hsrd −
sr 2 2 1d 3
s2r 1 1d 5
S D
S D
Î
1
22. y − x 1
x
5
24. Us yd −
y4 1 1
y2 1 1
26. f std −
t
t2 1 4
28. Fstd −
30. sstd −
t
48. y −
49. y − s1 2 sec t 50. y −
1
s1 1 tan xd 2
4x
sx 1 1
5
51–54 Find an equation of the tangent line to the curve at the
given point.
51. y − s3x 2 1d26, s0, 1d52.
y − s1 1 x 3 , s2, 3d
53. y − sinssin xd, s, 0d54.
y − sin 2 x cos x, sy2, 0d
2
st 1 1
3
Î
1
x
38. y − sin ( t 1 cos st )
47–50 Find y9 and y 99.
u3 2 1
u3 1 1
36. y − x sin
37. y − cot 2ssin d
18. tsxd − sx 2 1 1d3 sx 2 1 2d6
20. Fstd − s3t 2 1d4 s2t 1 1d23
4
1 1 sin t
1 1 cos t
55. (a)Find an equation of the tangent line to the curve
y − tansx 2y4d at the point s1, 1d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on the same screen.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.5 The Chain Rule
| |
56. (a)The curve y − x ys2 2 x 2 is called a bullet-nose
curve. Find an equation of the tangent line to this curve
at the point s1, 1d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on the same screen.
66. If f is the function whose graph is shown, let hsxd − f s f sxdd
and tsxd − f sx 2 d. Use the graph of f to estimate the value
of each derivative.
(a) h9s2d (b) t9s2d
y
57. (a)If f sxd − x s2 2 x 2 , find f 9sxd.
(b)Check to see that your answer to part (a) is reasonable by
;
comparing the graphs of f and f 9.
; 58. The function f sxd − sinsx 1 sin 2xd, 0 < x < , arises in
applications to frequency modulation (FM) synthesis.
(a)Use a graph of f produced by a graphing device to make
a rough sketch of the graph of f 9.
(b)Calculate f 9sxd and use this expression, with a calculator,
to graph f 9. Compare with your sketch in part (a).
159
y=ƒ
1
0
1
x
67. If tsxd − sf sxd , where the graph of f is shown, evaluate
t9s3d.
y
59. F
ind all points on the graph of the function
f sxd − 2 sin x 1 sin2x at which the tangent line is horizontal.
60.At what point on the curve y − s1 1 2x is the tangent line
perpendicular to the line 6x 1 2y − 1?
61. If Fsxd − f stsxdd, where f s22d − 8, f 9s22d − 4, f 9s5d − 3,
ts5d − 22, and t9s5d − 6, find F9s5d.
62. If hsxd − s4 1 3f sxd , where f s1d − 7 and f 9s1d − 4,
find h9s1d.
x
f sxd
tsxd
f 9sxd
t9sxd
1
2
3
3
1
7
2
8
2
4
5
7
6
7
9
64. Let f and t be the functions in Exercise 63.
(a) If Fsxd − f s f sxdd, find F9s2d.
(b) If Gsxd − tstsxdd, find G9s3d.
65. If f and t are the functions whose graphs are shown, let
usxd − f s tsxdd, vsxd − ts f sxdd, and w sxd − ts tsxdd. Find
each derivative, if it exists. If it does not exist, explain why.
(a) u9s1d (b) v9s1d (c) w9s1d
y
x
68. Suppose f is differentiable on R and is a real number.
Let Fsxd − f sx d and Gsxd − f f sxdg . Find expressions
for (a) F9sxd and (b) G9sxd.
70. If t is a twice differentiable function and f sxd − x tsx 2 d, find
f 99 in terms of t, t9, and t99.
71.If Fsxd − f s3f s4 f sxddd, where f s0d − 0 and f 9s0d − 2,
find F9s0d.
73–74 Find the given derivative by finding the first few deriva­
tives and observing the pattern that occurs.
73. D103 cos 2x74. D 35 x sin x
75. T
he displacement of a particle on a vibrating string is given
by the equation sstd − 10 1 14 sins10 td where s is measured
in centimeters and t in seconds. Find the velocity of the
particle after t seconds.
76. I f the equation of motion of a particle is given by
s − A cosst 1 d, the particle is said to undergo simple
harmonic motion.
(a) Find the velocity of the particle at time t.
(b) When is the velocity 0?
f
g
1
1
72. If Fsxd − f sx f sx f sxddd, where f s1d − 2, f s2d − 3,
f 9s1d − 4, f 9s2d − 5, and f 9s3d − 6, find F9s1d.
(a) If hsxd − f stsxdd, find h9s1d.
(b) If Hsxd − ts f sxdd, find H9s1d.
0
0
69. Let rsxd − f s tshsxddd, where hs1d − 2, ts2d − 3, h9s1d − 4,
t9s2d − 5, and f 9s3d − 6. Find r9s1d.
63. A table of values for f , t, f 9, and t9 is given.
1
f
1
x
77.A Cepheid variable star is a star whose brightness alternately
increases and decreases. The most easily visible such star is
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
160
chapter 2 Derivatives
Delta Cephei, for which the interval between times of maximum brightness is 5.4 days. The average brightness of this
star is 4.0 and its brightness changes by 60.35. In view of
these data, the brightness of Delta Cephei at time t, where
t is measured in days, has been modeled by the function
S D
Bstd − 4.0 1 0.35 sin
2 t
5.4
(a) Find the rate of change of the brightness after t days.
(b)Find, correct to two decimal places, the rate of increase
after one day.
78. I n Example 1.3.4 we arrived at a model for the length of
daylight (in hours) in Philadelphia on the tth day of the
year:
F
Lstd − 12 1 2.8 sin
2
st 2 80d
365
G
Use this model to compare how the number of hours of
day­light is increasing in Philadelphia on March 21 and
May 21.
79. A
particle moves along a straight line with displacement
sstd, velocity vstd, and acceleration astd. Show that
astd − vstd
dv
ds
CAS
83. Use the Chain Rule to prove the following.
(a)The derivative of an even function is an odd function.
(b)The derivative of an odd function is an even function.
84. U
se the Chain Rule and the Product Rule to give an
alternative proof of the Quotient Rule.
[Hint: Write f sxdytsxd − f sxdf tsxdg 21.]
85. (a) If n is a positive integer, prove that
d
ssinn x cos nxd − n sinn21x cossn 1 1dx
dx
(b)Find a formula for the derivative of y − cosnx cos nx
that is similar to the one in part (a).
86.Suppose y − f sxd is a curve that always lies above the
x-axis and never has a horizontal tangent, where f is dif­
ferentiable everywhere. For what value of y is the rate
of change of y 5 with respect to x eighty times the rate of
change of y with respect to x?
87. U
se the Chain Rule to show that if is measured in degrees,
then
Explain the difference between the meanings of the
derivatives dvydt and dvyds.
80. A
ir is being pumped into a spherical weather balloon. At
any time t, the volume of the balloon is Vstd and its radius
is rstd.
(a)What do the derivatives dVydr and dVydt represent?
(b)Express dVydt in terms of drydt.
CAS
(b) Where does the graph of f have horizontal tangents?
(c)Graph f and f 9 on the same screen. Are the graphs
consistent with your answer to part (b)?
81. C
omputer algebra systems have commands that differentiate
functions, but the form of the answer may not be convenient
and so further commands may be necessary to simplify the
answer.
(a)Use a CAS to find the derivative in Example 5 and
compare with the answer in that example. Then use the
simplify command and compare again.
(b)Use a CAS to find the derivative in Example 6. What
happens if you use the simplify command? What happens if you use the factor command? Which form of the
answer would be best for locating horizontal tangents?
82. (a) Use a CAS to differentiate the function
f sxd −
Î
x4 2 x 1 1
x4 1 x 1 1
and to simplify the result.
d
ssin d −
cos d
180
(This gives one reason for the convention that radian
measure is always used when dealing with trigonometric
functions in calculus: the differentiation formulas would not
be as simple if we used degree measure.)
| |
88. (a) Write x − sx 2 and use the Chain Rule to show that
d
x −
dx
| |
|
x
x
| |
|
(b)If f sxd − sin x , find f 9sxd and sketch the graphs of f
and f 9. Where is f not differentiable?
(c)If tsxd − sin x , find t9sxd and sketch the graphs of t
and t9. Where is t not differentiable?
| |
89. If y − f sud and u − tsxd, where f and t are twice differen­tiable functions, show that
d2y
d2y
2 −
dx
du 2
S D
du
dx
2
1
dy d 2u
du dx 2
90.If y − f sud and u − tsxd, where f and t possess third
derivatives, find a formula for d 3 yydx 3 similar to the one
given in Exercise 89.
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Section 2.6 Implicit Differentiation
applied Project
where should a pilot start descent?
y
An approach path for an aircraft landing is shown in the figure and satisfies the following
conditions:
y=P(x)
0
161
(i)The cruising altitude is h when descent starts at a horizontal distance , from
touchdown at the origin.
(ii) The pilot must maintain a constant horizontal speed v throughout descent.
(iii)The absolute value of the vertical acceleration should not exceed a constant k (which
is much less than the acceleration due to gravity).
h
1. F
ind a cubic polynomial Psxd − ax 3 1 bx 2 1 cx 1 d that satisfies condition (i) by
imposing suitable conditions on Psxd and P9sxd at the start of descent and at touchdown.
x
2. Use conditions (ii) and (iii) to show that
6h v 2
<k
,2
Suppose that an airline decides not to allow vertical acceleration of a plane to exceed
3.
k − 860 miyh2. If the cruising altitude of a plane is 35,000 ft and the speed is 300 miyh,
how far away from the airport should the pilot start descent?
; 4. Graph the approach path if the conditions stated in Problem 3 are satisfied.
The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable—for example,
y − sx 3 1 1 or y − x sin x
or, in general, y − f sxd. Some functions, however, are defined implicitly by a relation
between x and y such as
1 x 2 1 y 2 − 25
2 x 3 1 y 3 − 6xy
or
In some cases it is possible to solve such an equation for y as an explicit function (or
several functions) of x. For instance, if we solve Equation 1 for y, we get y − 6s25 2 x 2 ,
so two of the functions determined by the implicit Equation l are f sxd − s25 2 x 2 and
tsxd − 2s25 2 x 2 . The graphs of f and t are the upper and lower semicircles of the
cir­cle x 2 1 y 2 − 25. (See Figure 1.)
y
0
FIGURE 1
(a) ≈+¥=25
y
x
0
25-≈
(b) ƒ=œ„„„„„„
y
x
0
25-≈
(c) ©=_ œ„„„„„„
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x
162
Chapter 2 Derivatives
It’s not easy to solve Equation 2 for y explicitly as a function of x by hand. (A computer algebra system has no trouble, but the expressions it obtains are very complicated.)
Nonetheless, (2) is the equation of a curve called the folium of Descartes shown in
Figure 2 and it implicitly defines y as several functions of x. The graphs of three such
functions are shown in Figure 3. When we say that f is a function defined implicitly by
Equa­tion 2, we mean that the equation
x 3 1 f f sxdg 3 − 6x f sxd
is true for all values of x in the domain of f .
y
0
y
˛+Á =6xy
x
FIGURE 2 The folium of Descartes
0
y
x
y
0
x
0
x
FIGURE 3 Graphs of three functions defined by the folium of Descartes
Fortunately, we don’t need to solve an equation for y in terms of x in order to find the
derivative of y. Instead we can use the method of implicit differentiation. This consists
of differentiating both sides of the equation with respect to x and then solving the resul­ting equation for y9. In the examples and exercises of this section it is always assumed
that the given equation determines y implicitly as a differentiable function of x so that the
method of implicit differentiation can be applied.
Example 1
dy
.
dx
(b) Find an equation of the tangent to the circle x 2 1 y 2 − 25 at the point s3, 4d.
(a) If x 2 1 y 2 − 25, find
SOLUTION 1
(a) Differentiate both sides of the equation x 2 1 y 2 − 25:
d
d
sx 2 1 y 2 d −
s25d
dx
dx
d
d
sx 2 d 1
sy 2 d − 0
dx
dx
Remembering that y is a function of x and using the Chain Rule, we have
d
d
dy
dy
sy 2 d −
sy 2 d
− 2y
dx
dy
dx
dx
Thus
2x 1 2y
dy
−0
dx
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Section 2.6 Implicit Differentiation
163
Now we solve this equation for dyydx:
dy
x
−2
dx
y
(b) At the point s3, 4d we have x − 3 and y − 4, so
dy
3
−2
dx
4
An equation of the tangent to the circle at s3, 4d is therefore
y 2 4 − 234 sx 2 3d or 3x 1 4y − 25
SOLUTION 2
(b) Solving the equation x 2 1 y 2 − 25 for y, we get y − 6s25 2 x 2 . The point
s3, 4d lies on the upper semicircle y − s25 2 x 2 and so we consider the function
f sxd − s25 2 x 2 . Differentiating f using the Chain Rule, we have
f 9sxd − 12 s25 2 x 2 d21y2
d
s25 2 x 2 d
dx
− 12 s25 2 x 2 d21y2s22xd − 2
Example 1 illustrates that even when
it is possible to solve an equation
explicitly for y in terms of x, it may be
easier to use implicit differentiation.
So
f 9s3d − 2
x
s25 2 x 2
3
3
−2
4
s25 2 3 2
and, as in Solution 1, an equation of the tangent is 3x 1 4y − 25.
■
NOTE 1 The formula dyydx − 2xyy in Solution 1 gives the derivative in terms of
both x and y. It is correct no matter which function y is determined by the given equation.
For instance, for y − f sxd − s25 2 x 2 we have
dy
x
x
−2 −2
dx
y
s25 2 x 2
whereas for y − tsxd − 2s25 2 x 2 we have
dy
x
x
x
−2 −2
−
2
dx
y
2s25 2 x
s25 2 x 2
Example 2
(a) Find y9 if x 3 1 y 3 − 6xy.
(b) Find the tangent to the folium of Descartes x 3 1 y 3 − 6xy at the point s3, 3d.
(c) At what point in the first quadrant is the tangent line horizontal?
SOLUTION
(a) Differentiating both sides of x 3 1 y 3 − 6xy with respect to x, regarding y as a
function of x, and using the Chain Rule on the term y 3 and the Product Rule on the
term 6xy, we get
3x 2 1 3y 2 y9 − 6xy9 1 6y
or
x 2 1 y 2 y9 − 2xy9 1 2y
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164
Chapter 2 Derivatives
y 2 y9 2 2xy9 − 2y 2 x 2
We now solve for y9:
y
sy 2 2 2xdy9 − 2y 2 x 2
(3, 3)
0
y9 −
2y 2 x 2
y 2 2 2x
(b) When x − y − 3,
x
y9 −
2 3 2 32
− 21
32 2 2 3
and a glance at Figure 4 confirms that this is a reasonable value for the slope at s3, 3d.
So an equation of the tangent to the folium at s3, 3d is
FIGURE 4
y 2 3 − 21sx 2 3d or x 1 y − 6
4
(c) The tangent line is horizontal if y9 − 0. Using the expression for y9 from part (a),
we see that y9 − 0 when 2y 2 x 2 − 0 (provided that y 2 2 2x ± 0d. Substituting
y − 12 x 2 in the equation of the curve, we get
x 3 1 ( 12 x 2) − 6x ( 12 x 2)
3
4
0
FIGURE 5
which simplifies to x 6 − 16x 3. Since x ± 0 in the first quadrant, we have x 3 − 16. If
x − 16 1y3 − 2 4y3, then y − 12 s2 8y3 d − 2 5y3. Thus the tangent is horizontal at s2 4y3, 2 5y3 d,
which is approximately (2.5198, 3.1748). Looking at Figure 5, we see that our answer
is reasonable.
■
NOTE 2 There is a formula for the three roots of a cubic equation that is like the
quad­ratic formula but much more complicated. If we use this formula (or a computer
algebra system) to solve the equation x 3 1 y 3 − 6xy for y in terms of x, we get three
functions determined by the equation:
3
3
221 x 3 1 s14 x 6 2 8x 3 1 s
221 x 3 2 s14 x 6 2 8x 3
y − f sxd − s
Abel and Galois
The Norwegian mathematician Niels
Abel proved in 1824 that no general
formula can be given for the roots of
a fifth-degree equation in terms of
radicals. Later the French mathematician Evariste Galois proved that it is
impossible to find a general formula
for the roots of an nth-degree equation (in terms of algebraic operations
on the coefficients) if n is any integer
larger than 4.
and
y−
1
2
f2f sxd 6 s23 (s221 x
3
3
3
1 s14 x 6 2 8x 3 2 s
221 x 3 2 s14 x 6 2 8x 3
)g
(These are the three functions whose graphs are shown in Figure 3.) You can see that the
method of implicit differentiation saves an enormous amount of work in cases such as
this. Moreover, implicit differentiation works just as easily for equations such as
y 5 1 3x 2 y 2 1 5x 4 − 12
for which it is impossible to find a similar expression for y in terms of x.
Example 3 Find y9 if sinsx 1 yd − y 2 cos x.
SOLUTION Differentiating implicitly with respect to x and remembering that y is a
function of x, we get
cossx 1 yd s1 1 y9d − y 2s2sin xd 1 scos xds2yy9d
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Section 2.6 Implicit Differentiation
165
(Note that we have used the Chain Rule on the left side and the Product Rule and Chain
Rule on the right side.) If we collect the terms that involve y9, we get
2
cossx 1 yd 1 y 2 sin x − s2y cos xdy9 2 cossx 1 yd ? y9
_2
2
So
y9 −
y 2 sin x 1 cossx 1 yd
2y cos x 2 cossx 1 yd
Figure 6, drawn with the implicit-plotting command of a computer algebra system,
shows part of the curve sinsx 1 yd − y 2 cos x. As a check on our calculation, notice
that y9 − 21 when x − y − 0 and it appears from the graph that the slope is approximately 21 at the origin.
■
_2
FIGURE 6
Figures 7, 8, and 9 show three more curves produced by a computer algebra system
with an implicit-plotting command. In Exercises 41– 42 you will have an opportunity to
create and examine unusual curves of this nature.
4
_4
15
4
12
_15
15
_4
_12
_15
FIGURE 7
sx 2 2 1dsx 2 2 4dsx 2 2 9d
− y 2 s y 2 2 4ds y 2 2 9d
12
_12
FIGURE 8
cossx 2 sin yd − sins y 2 sin xd
FIGURE 9
sinsxyd − sin x 1 sin y
The following example shows how to find the second derivative of a function that is
defined implicitly.
Example 4 Find y99 if x 4 1 y 4 − 16.
SOLUTION Differentiating the equation implicitly with respect to x, we get
4x 3 1 4y 3 y9 − 0
Solving for y9 gives
3 y9 − 2
x3
y3
To find y99 we differentiate this expression for y9 using the Quotient Rule and remembering that y is a function of x:
y99 −
d
dx
−2
S D
2
x3
y3
−2
y 3 sdydxdsx 3 d 2 x 3 sdydxdsy 3 d
sy 3 d2
y 3 3x 2 2 x 3s3y 2 y9d
y6
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166
Chapter 2 Derivatives
If we now substitute Equation 3 into this expression, we get
S D
3x 2 y 3 2 3x 3 y 2 2
y99 − 2
−2
y
x3
y3
6
3sx 2 y 4 1 x 6 d
3x 2sy 4 1 x 4 d
−
2
y7
y7
But the values of x and y must satisfy the original equation x 4 1 y 4 − 16. So the
answer simplifies to
y99 − 2
y
Figure 10 shows the graph of the curve
x 4 1 y 4 − 16 of Example 4. Notice
that it’s a stretched and flat­tened version
of the circle x 2 1 y 2 − 4. For this reason it’s sometimes called a fat circle.
It starts out very steep on the left but
quickly becomes very flat. This can be
seen from the expression
y9 − 2
SD
x3
x
−2
y3
y
3x 2s16d
x2
− 248 7
7
y
y
■
x $+y$ =16
2
0
2 x
3
FIGURE 10
1–4
(a) Find y9 by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get y9
in terms of x.
(c) Check that your solutions to parts (a) and (b) are consistent by
substituting the expression for y into your solution for part (a).
1. 9x 2 2 y 2 − 1
2. 2x 2 1 x 1 xy − 1
3. sx 1 sy − 1
4.
2
1
2 −4
x
y
13. sx 1 y − x 4 1 y 4
14. y sinsx 2 d − x sins y 2 d
15. tansxyyd − x 1 y
16. xy − sx 2 1 y 2
17. sxy − 1 1 x 2 y
18. x sin y 1 y sin x − 1
19. sinsxyd − cossx 1 yd
20. tansx 2 yd −
y
1 1 x2
21. If f sxd 1 x 2 f f sxdg 3 − 10 and f s1d − 2, find f 9s1d.
22. If tsxd 1 x sin tsxd − x 2, find t9s0d.
5–20 Find dyydx by implicit differentiation.
5. x 2 2 4xy 1 y 2 − 4
4
2
2
3
6. 2x 2 1 xy 2 y 2 − 2
3
2
3
23–24 Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dxydy.
23. x 4y 2 2 x 3y 1 2xy 3 − 0
24. y sec x − x tan y
7. x 1 x y 1 y − 5
8. x 2 xy 1 y − 1
x2
− y2 1 1
x1y
10. y 5 1 x 2 y 3 − 1 1 x 4 y
25–32 Use implicit differentiation to find an equation of the
tangent line to the curve at the given point.
12. cossxyd − 1 1 sin y
25. y sin 2x − x cos 2y, sy2, y4d
9.
11. y cos x − x 2 1 y 2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.6 Implicit Differentiation
35–38 Find y99 by implicit differentiation.
26. sinsx 1 yd − 2x 2 2y, s, d
s2, 1d
27. x 2 2 xy 2 y 2 − 1,
28. x 2 1 2xy 1 4y 2 − 12,
(hyperbola)
s2, 1d
29. x 2 1 y 2 − s2x 2 1 2y 2 2 xd2,
(ellipse)
s0, 12 d
35. x 2 1 4y 2 − 4
36. x 2 1 xy 1 y 2 − 3
37. sin y 1 cos x − 1
38. x 3 2 y 3 − 7
(cardioid)
39. If xy 1 y 3 − 1, find the value of y 0 at the point where
x − 0.
y
40. If x 2 1 xy 1 y 3 − 1, find the value of y999 at the point
where x − 1.
x
30. x 2y3 1 y 2y3 − 4,
167
s23 s3, 1d
CAS
(astroid)
ys y 2 2 1ds y 2 2d − xsx 2 1dsx 2 2d
y
0
31. 2sx 2 1 y 2 d2 − 25sx 2 2 y 2 d,
(3, 1) (lemniscate)
CAS
42. (a) The curve with equation
2y 3 1 y 2 2 y 5 − x 4 2 2x 3 1 x 2
y
0
32. y 2s y 2 2 4d − x 2sx 2 2 5d,
At how many points does this curve have horizontal
tangents? Estimate the x-coordinates of these points.
(b)Find equations of the tangent lines at the points (0, 1)
and (0, 2).
(c)Find the exact x-coordinates of the points in part (a).
(d)Create even more fanciful curves by modifying the
equation in part (a).
x
8
41.Fanciful shapes can be created by using the implicit plotting
capabilities of computer algebra systems.
(a) Graph the curve with equation
(0, 22)
x
has been likened to a bouncing wagon. Use a computer
algebra system to graph this curve and discover why.
(b)At how many points does this curve have horizontal
tangent lines? Find the x-coordinates of these points.
(devil’s curve)
43. F
ind the points on the lemniscate in Exercise 31 where the
tangent is horizontal.
y
x
44.Show by implicit differentiation that the tangent to the
ellipse
y2
x2
−1
2 1
a
b2
at the point sx 0 , y 0 d is
x0 x
y0 y
1
−1
a2
b2
33. (a)The curve with equation y 2 − 5x 4 2 x 2 is called a
kampyle of Eudoxus. Find an equation of the tangent
line to this curve at the point s1, 2d.
(b)Illustrate part (a) by graphing the curve and the tangent
;
line on a common screen. (If your graphing device will
graph implicitly defined curves, then use that capability. If not, you can still graph this curve by graphing its
upper and lower halves separately.)
34. (a)The curve with equation y 2 − x 3 1 3x 2 is called the
Tschirnhausen cubic. Find an equation of the tangent
line to this curve at the point s1, 22d.
(b)At what points does this curve have horizontal tangents?
(c)Illustrate parts (a) and (b) by graphing the curve and the
;
tangent lines on a common screen.
45. Find an equation of the tangent line to the hyperbola
x2
y2
−1
2 2
a
b2
at the point sx 0 , y 0 d.
46. Show that the sum of the x- and y-intercepts of any tangent
line to the curve sx 1 sy − sc is equal to c.
47. S
how, using implicit differentiation, that any tangent line at
a point P to a circle with center O is perpendicular to the
radius OP.
48. T
he Power Rule can be proved using implicit differentiation
for the case where n is a rational number, n − pyq, and
y − f sxd − x n is assumed beforehand to be a differentiable
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
168
chapter 2 Derivatives
function. If y − x pyq, then y q − x p. Use implicit differentiation to show that
p s pyqd21
y9 −
x
q
49–52 Two curves are orthogonal if their tangent lines are
perpendicular at each point of intersection. Show that the given
families of curves are orthogonal trajectories of each other; that
is, every curve in one family is orthogonal to every curve in the
other family. Sketch both families of curves on the same axes.
49. x 2 1 y 2 − r 2, ax 1 by − 0
51. y − cx 2, x 2 1 2y 2 − k
52. y − ax 3, x 2 1 3y 2 − b
53. S
how that the ellipse x 2ya 2 1 y 2yb 2 − 1 and the hyperbola
x 2yA2 2 y 2yB 2 − 1 are orthogonal trajectories if A2 , a 2
and a 2 2 b 2 − A2 1 B 2 (so the ellipse and hyperbola have
the same foci).
54.Find the value of the number a such that the families of
curves y − sx 1 cd21 and y − asx 1 kd1y3 are orthogonal
trajectories.
55. (a)The van der Waals equation for n moles of a gas is
P1
57.The equation x 2 2 xy 1 y 2 − 3 represents a “rotated
ellipse,” that is, an ellipse whose axes are not parallel to the
coordinate axes. Find the points at which this ellipse crosses
the x-axis and show that the tangent lines at these points are
parallel.
58. (a)Where does the normal line to the ellipse
x 2 2 xy 1 y 2 − 3 at the point s21, 1d intersect the
ellipse a second time?
(b)Illustrate part (a) by graphing the ellipse and the normal
;
line.
50. x 2 1 y 2 − ax, x 2 1 y 2 − by
S
(b)Plot the curve in part (a). What do you see? Prove that
what you see is correct.
(c)In view of part (b), what can you say about the expression for y9 that you found in part (a)?
CAS
D
n 2a
sV 2 nbd − nRT
V2
where P is the pressure, V is the volume, and T is the
temperature of the gas. The constant R is the universal
gas constant and a and b are positive constants that are
characteristic of a particular gas. If T remains constant,
use implicit differentiation to find dVydP.
(b)Find the rate of change of volume with respect to
pressure of 1 mole of carbon dioxide at a volume
of V − 10 L and a pressure of P − 2.5 atm. Use
a − 3.592 L2-atmymole 2 and b − 0.04267 Lymole.
59.Find all points on the curve x 2 y 2 1 xy − 2 where the slope
of the tangent line is 21.
60. F
ind equations of both the tangent lines to the ellipse
x 2 1 4y 2 − 36 that pass through the point s12, 3d.
61. The Bessel function of order 0, y − J sxd, satisfies the
differential equation xy99 1 y9 1 xy − 0 for all values of x
and its value at 0 is J s0d − 1.
(a)Find J9s0d.
(b)Use implicit differentiation to find J99s0d.
62.The figure shows a lamp located three units to the right of
the y-axis and a shadow created by the elliptical region
x 2 1 4y 2 < 5. If the point s25, 0d is on the edge of the
shadow, how far above the x-axis is the lamp located?
y
?
_5
56. (a)Use implicit differentiation to find y9 if
x 2 1 xy 1 y 2 1 1 − 0
laboratory Project
CAS
0
≈+4¥=5
3
x
Families of implicit curves
In this project you will explore the changing shapes of implicitly defined curves as you vary the
constants in a family, and determine which features are common to all members of the family.
1. Consider the family of curves
y 2 2 2x 2 sx 1 8d − cfs y 1 1d2 sy 1 9d 2 x 2 g
(a)By graphing the curves with c − 0 and c − 2, determine how many points of intersection there are. (You might have to zoom in to find all of them.)
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Section 2.7 Rates of Change in the Natural and Social Sciences
169
(b)Now add the curves with c − 5 and c − 10 to your graphs in part (a). What do you
notice? What about other values of c?
2. (a)Graph several members of the family of curves
x 2 1 y 2 1 cx 2 y 2 − 1
Describe how the graph changes as you change the value of c.
(b)What happens to the curve when c − 21? Describe what appears on the screen.
Can you prove it algebraically?
(c)Find y9 by implicit differentiation. For the case c − 21, is your expression for y9
consistent with what you discovered in part (b)?
We know that if y − f sxd, then the derivative dyydx can be interpreted as the rate of
change of y with respect to x. In this section we examine some of the applications of this
idea to physics, chemistry, biology, economics, and other sciences.
Let’s recall from Section 2.1 the basic idea behind rates of change. If x changes from
x 1 to x 2, then the change in x is
Dx − x 2 2 x 1
and the corresponding change in y is
Dy − f sx 2 d 2 f sx 1 d
The difference quotient
Dy
f sx 2 d 2 f sx 1 d
−
Dx
x2 2 x1
is the average rate of change of y with respect to x over the interval fx 1, x 2 g and can
be interpreted as the slope of the secant line PQ in Figure 1. Its limit as Dx l 0 is the
derivative f 9sx 1 d, which can therefore be interpreted as the instantaneous rate of change
of y with respect to x or the slope of the tangent line at Psx 1, f sx 1 dd. Using Leibniz notation, we write the process in the form
y
Q { ¤, ‡}
Îy
P { ⁄, fl}
Îx
0
⁄
¤
mPQ average rate of change
m=fª(⁄ )=instantaneous rate
of change
FIGURE 1
dy
Dy
− lim
Dx l 0 Dx
dx
x
Whenever the function y − f sxd has a specific interpretation in one of the sciences, its
derivative will have a specific interpretation as a rate of change. (As we discussed in Sec­
tion 2.1, the units for dyydx are the units for y divided by the units for x.) We now look
at some of these interpretations in the natural and social sciences.
Physics
If s − f std is the position function of a particle that is moving in a straight line, then DsyDt
represents the average velocity over a time period Dt, and v − dsydt represents the instantaneous velocity (the rate of change of displacement with respect to time). The instantaneous rate of change of velocity with respect to time is acceleration: astd − v9std − s99std.
This was discussed in Sections 2.1 and 2.2, but now that we know the differentiation
formulas, we are able to solve problems involving the motion of objects more easily.
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170
Chapter 2 Derivatives
Example 1 The position of a particle is given by the equation
s − f std − t 3 2 6t 2 1 9t
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 2 s? After 4 s?
(c) When is the particle at rest?
(d) When is the particle moving forward (that is, in the positive direction)?
(e) Draw a diagram to represent the motion of the particle.
(f) Find the total distance traveled by the particle during the first five seconds.
(g) Find the acceleration at time t and after 4 s.
(h) Graph the position, velocity, and acceleration functions for 0 < t < 5.
(i) When is the particle speeding up? When is it slowing down?
SOLUTION (a) The velocity function is the derivative of the position function.
s − f std − t 3 2 6t 2 1 9t
vstd −
ds
− 3t 2 2 12t 1 9
dt
(b) The velocity after 2 s means the instantaneous velocity when t − 2, that is,
vs2d −
ds
dt
Z
t−2
− 3s2d2 2 12s2d 1 9 − 23 mys
The velocity after 4 s is
vs4d − 3s4d2 2 12s4d 1 9 − 9 mys
(c) The particle is at rest when vstd − 0, that is,
3t 2 2 12t 1 9 − 3st 2 2 4t 1 3d − 3st 2 1dst 2 3d − 0
and this is true when t − 1 or t − 3. Thus the particle is at rest after 1 s and after 3 s.
(d) The particle moves in the positive direction when vstd . 0, that is,
3t 2 2 12t 1 9 − 3st 2 1dst 2 3d . 0
t=3
s=0
t=0
s=0
t=1
s=4
FIGURE 2
s
This inequality is true when both factors are positive st . 3d or when both factors are
negative st , 1d. Thus the particle moves in the positive direction in the time intervals
t , 1 and t . 3. It moves backward (in the negative direction) when 1 , t , 3.
(e) Using the information from part (d) we make a schematic sketch in Figure 2 of the
motion of the particle back and forth along a line (the s-axis).
(f) Because of what we learned in parts (d) and (e), we need to calculate the distances
traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately.
The distance traveled in the first second is
| f s1d 2 f s0d | − | 4 2 0 | − 4 m
From t − 1 to t − 3 the distance traveled is
| f s3d 2 f s1d | − | 0 2 4 | − 4 m
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Section 2.7 Rates of Change in the Natural and Social Sciences
171
From t − 3 to t − 5 the distance traveled is
| f s5d 2 f s3d | − | 20 2 0 | − 20 m
The total distance is 4 1 4 1 20 − 28 m.
(g) The acceleration is the derivative of the velocity function:
25
√
0
s
astd −
a
d 2s
dv
−
− 6t 2 12
dt 2
dt
as4d − 6s4d 2 12 − 12 mys 2
5
(h) Figure 3 shows the graphs of s, v, and a.
( i) The particle speeds up when the velocity is positive and increasing (v and a are
both positive) and also when the velocity is negative and decreasing (v and a are both
negative). In other words, the particle speeds up when the velocity and acceleration
have the same sign. (The particle is pushed in the same direction it is moving.) From
Figure 3 we see that this happens when 1 , t , 2 and when t . 3. The particle slows
down when v and a have opposite signs, that is, when 0 < t , 1 and when 2 , t , 3.
Figure 4 summarizes the motion of the particle.
-12
FIGURE 3
a
√
TEC In Module 2.7 you can see
an animation of Figure 4 with an
expression for s that you can choose
yourself.
s
5
0
_5
t
1
forward
slows
down
FIGURE 4
forward
backward
speeds
up
slows
down
speeds
up
■
Example 2 If a rod or piece of wire is homogeneous, then its linear density is uniform
and is defined as the mass per unit length s − myld and measured in kilograms per
meter. Suppose, however, that the rod is not homogeneous but that its mass measured
from its left end to a point x is m − f sxd, as shown in Figure 5.
x
FIGURE 5
This part of the rod has mass ƒ.
x¡
x™
The mass of the part of the rod that lies between x − x 1 and x − x 2 is given by
Dm − f sx 2 d 2 f sx 1 d, so the average density of that part of the rod is
average density −
Dm
f sx 2 d 2 f sx 1 d
−
Dx
x2 2 x1
If we now let Dx l 0 (that is, x 2 l x 1), we are computing the average density over
smaller and smaller intervals. The linear density at x 1 is the limit of these average
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172
Chapter 2 Derivatives
densities as Dx l 0; that is, the linear density is the rate of change of mass with
respect to length. Symbolically,
Dm
dm
− lim
−
Dx l 0 Dx
dx
Thus the linear density of the rod is the derivative of mass with respect to length.
For instance, if m − f sxd − sx , where x is measured in meters and m in kilograms,
then the average density of the part of the rod given by 1 < x < 1.2 is
Dm
f s1.2d 2 f s1d
s1.2 2 1
−
−
< 0.48 kgym
Dx
1.2 2 1
0.2
while the density right at x − 1 is
−
FIGURE 6
dm
dx
Z
x−1
−
1
2 sx
Z
x−1
− 0.50 kgym
■
Example 3 A current exists whenever electric charges move. Figure 6 shows part of
a wire and electrons moving through a plane surface, shaded red. If DQ is the net
charge that passes through this surface during a time period Dt, then the average current during this time interval is defined as
average current −
DQ
Q2 2 Q1
−
Dt
t2 2 t1
If we take the limit of this average current over smaller and smaller time intervals,
we get what is called the current I at a given time t1:
I − lim
Dt l 0
DQ
dQ
−
Dt
dt
Thus the current is the rate at which charge flows through a surface. It is measured in
units of charge per unit time (often coulombs per second, called amperes).
■
Velocity, density, and current are not the only rates of change that are important in
physics. Others include power (the rate at which work is done), the rate of heat flow,
temperature gradient (the rate of change of temperature with respect to position), and the
rate of decay of a radioactive substance in nuclear physics.
Chemistry
Example 4 A chemical reaction results in the formation of one or more substances
(called products) from one or more starting materials (called reactants). For instance,
the “equation”
2H2 1 O2 l 2H2 O
indicates that two molecules of hydrogen and one molecule of oxygen form two molecules of water. Let’s consider the reaction
A1BlC
where A and B are the reactants and C is the product. The concentration of a reactant A is the number of moles (1 mole − 6.022 3 1023 molecules) per liter and is
denoted by fAg. The concentration varies during a reaction, so fAg, fBg, and fCg are all
functions of time std. The average rate of reaction of the product C over a time interval
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Section 2.7 Rates of Change in the Natural and Social Sciences
t1 < t < t2 is
173
DfCg
fCgst2 d 2 fCgst1 d
−
Dt
t2 2 t1
But chemists are more interested in the instantaneous rate of reaction, which is
obtained by taking the limit of the average rate of reaction as the time interval Dt
approaches 0:
DfCg
dfCg
rate of reaction − lim
−
Dt l 0 Dt
dt
Since the concentration of the product increases as the reaction proceeds, the derivative
dfCgydt will be positive, and so the rate of reaction of C is positive. The concentrations
of the reactants, however, decrease during the reaction, so, to make the rates of reaction
of A and B positive numbers, we put minus signs in front of the derivatives dfAgydt
and dfBgydt. Since fAg and fBg each decrease at the same rate that fCg increases, we
have
dfCg
dfAg
dfBg
rate of reaction −
−2
−2
dt
dt
dt
More generally, it turns out that for a reaction of the form
aA 1 bB l cC 1 dD
we have
2
1 dfAg
1 dfBg
1 dfCg
1 dfDg
−2
−
−
a dt
b dt
c dt
d dt
The rate of reaction can be determined from data and graphical methods. In some cases
there are explicit formulas for the concentrations as functions of time, which enable us
to compute the rate of reaction (see Exercise 24).
■
Example 5 One of the quantities of interest in thermodynamics is compressibility. If
a given substance is kept at a constant temperature, then its volume V depends on its
pressure P. We can consider the rate of change of volume with respect to pressure—
namely, the derivative dVydP. As P increases, V decreases, so dVydP , 0. The compressibility is defined by introducing a minus sign and dividing this derivative by the
volume V:
1 dV
isothermal compressibility − − 2
V dP
Thus measures how fast, per unit volume, the volume of a substance decreases as the
pressure on it increases at constant temperature.
For instance, the volume V (in cubic meters) of a sample of air at 25 8C was found to
be related to the pressure P (in kilopascals) by the equation
V−
5.3
P
The rate of change of V with respect to P when P − 50 kPa is
dV
dP
Z
P−50
Z
−2
5.3
P2
−2
5.3
− 20.00212 m 3ykPa
2500
P−50
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174
Chapter 2 Derivatives
The compressibility at that pressure is
−2
1 dV
V dP
Z
P−50
−
0.00212
− 0.02 sm 3ykPadym 3
5.3
50
■
Biology
Example 6 Let n − f std be the number of individuals in an animal or plant popu­lation at time t. The change in the population size between the times t − t1 and t − t2 is
Dn − f st2 d 2 f st1 d, and so the average rate of growth during the time period t1 < t < t2
is
Dn
f st2 d 2 f st1 d
average rate of growth −
−
Dt
t2 2 t1
The instantaneous rate of growth is obtained from this average rate of growth by
letting the time period Dt approach 0:
growth rate − lim
Dt l 0
Dn
dn
−
Dt
dt
Strictly speaking, this is not quite accurate because the actual graph of a population function n − f std would be a step function that is discontinuous whenever a birth or death
occurs and therefore not differentiable. However, for a large animal or plant population,
we can replace the graph by a smooth approximating curve as in Figure 7.
n
FIGURE 7
A
smooth curve approximating
a growth function
0
t
To be more specific, consider a population of bacteria in a homogeneous nutrient
medium. Suppose that by sampling the population at certain intervals it is determined
that the population doubles every hour. If the initial population is n0 and the time t is
measured in hours, then
f s1d − 2f s0d − 2n0
f s2d − 2f s1d − 2 2n0
f s3d − 2f s2d − 2 3n0
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Section 2.7 Rates of Change in the Natural and Social Sciences
175
Eye of Science / Science Source
and, in general,
f std − 2 t n0
The population function is n − n0 2 t.
This is an example of an exponential function. In Chapter 6 we will discuss exponential functions in general; at that time we will be able to compute their derivatives
■
E. coli bacteria are about 2 micrometers and thereby determine the rate of growth of the bacteria population.
(mm) long and 0.75 mm wide. The
image was produced with a scanning
electron microscope.
Example 7 When we consider the flow of blood through a blood vessel, such as a
vein or artery, we can model the shape of the blood vessel by a cylindrical tube with
radius R and length l as illustrated in Figure 8.
R
FIGURE 8 Blood flow in an artery
r
l
Because of friction at the walls of the tube, the velocity v of the blood is greatest
along the central axis of the tube. The velocity decreases as the distance r from the axis
increases, until v becomes 0 at the wall. The relationship between v and r is given by
the law of laminar flow discovered by the French physician Jean-Louis-Marie
Poiseuille in 1840. This law states that
P
1 v−
sR 2 2 r 2 d
4l
For more detailed information, see
W. Nichols and M. O’Rourke (eds.),
McDonald’s Blood Flow in Arteries:
Theoretical, Experimental, and Clinical
Principles, 5th ed. (New York, 2005).
where is the viscosity of the blood and P is the pressure difference between the ends
of the tube. If P and l are constant, then v is a function of r with domain f0, Rg.
The average rate of change of the velocity as we move from r − r1 outward to
r − r2 is given by
Dv
vsr2 d 2 vsr1 d
−
Dr
r2 2 r1
and if we let Dr l 0, we obtain the velocity gradient, that is, the instantaneous rate of
change of velocity with respect to r:
velocity gradient − lim
Dr l 0
Dv
dv
−
Dr
dr
Using Equation 1, we obtain
dv
P
Pr
−
s0 2 2rd − 2
dr
4l
2l
For one of the smaller human arteries we can take − 0.027, R − 0.008 cm,
l − 2 cm, and P − 4000 dynesycm2, which gives
v−
4000
s0.000064 2 r 2 d
4s0.027d2
< 1.85 3 10 4s6.4 3 10 25 2 r 2 d
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176
Chapter 2 Derivatives
At r − 0.002 cm the blood is flowing at a speed of
vs0.002d < 1.85 3 10 4s64 3 1026 2 4 3 10 26 d
− 1.11 cmys
and the velocity gradient at that point is
dv
dr
Z
r−0.002
−2
4000s0.002d
< 274 scmysdycm
2s0.027d2
To get a feeling for what this statement means, let’s change our units from centi­meters to micrometers (1 cm − 10,000 mm). Then the radius of the artery is 80 mm.
The velocity at the central axis is 11,850 mmys, which decreases to 11,110 mmys at
a distance of r − 20 mm. The fact that dvydr − 274 (mmys)ymm means that, when
r − 20 mm, the velocity is decreasing at a rate of about 74 mmys for each micrometer
that we proceed away from the center.
■
Economics
Example 8 Suppose Csxd is the total cost that a company incurs in producing
x units of a certain commodity. The function C is called a cost function. If the
number of items produced is increased from x 1 to x 2, then the additional cost is
DC − Csx 2 d 2 Csx 1 d, and the average rate of change of the cost is
DC
Csx 2 d 2 Csx 1 d
Csx 1 1 Dxd 2 Csx 1 d
−
−
Dx
x2 2 x1
Dx
The limit of this quantity as Dx l 0, that is, the instantaneous rate of change of
cost with respect to the number of items produced, is called the marginal cost by
economists:
DC
dC
marginal cost − lim
−
Dx l 0 Dx
dx
[Since x often takes on only integer values, it may not make literal sense to let Dx
approach 0, but we can always replace Csxd by a smooth approximating function as in
Example 6.]
Taking Dx − 1 and n large (so that Dx is small compared to n), we have
C9snd < Csn 1 1d 2 Csnd
Thus the marginal cost of producing n units is approximately equal to the cost of producing one more unit [the sn 1 1dst unit].
It is often appropriate to represent a total cost function by a polynomial
Csxd − a 1 bx 1 cx 2 1 dx 3
where a represents the overhead cost (rent, heat, maintenance) and the other terms
represent the cost of raw materials, labor, and so on. (The cost of raw materials may be
proportional to x, but labor costs might depend partly on higher powers of x because of
overtime costs and inefficiencies involved in large-scale operations.)
For instance, suppose a company has estimated that the cost (in dollars) of producing x items is
Csxd − 10,000 1 5x 1 0.01x 2
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Section 2.7 Rates of Change in the Natural and Social Sciences
177
Then the marginal cost function is
C9sxd − 5 1 0.02x
The marginal cost at the production level of 500 items is
C9s500d − 5 1 0.02s500d − $15yitem
This gives the rate at which costs are increasing with respect to the production level
when x − 500 and predicts the cost of the 501st item.
The actual cost of producing the 501st item is
Cs501d 2 Cs500d − f10,000 1 5s501d 1 0.01s501d2 g
2 f10,000 1 5s500d 1 0.01s500d2 g
− $15.01
Notice that C9s500d < Cs501d 2 Cs500d.
■
Economists also study marginal demand, marginal revenue, and marginal profit,
which are the derivatives of the demand, revenue, and profit functions. These will be
considered in Chapter 3 after we have developed techniques for finding the maximum
and minimum values of functions.
Other Sciences
Rates of change occur in all the sciences. A geologist is interested in knowing the rate
at which an intruded body of molten rock cools by conduction of heat into surrounding
rocks. An engineer wants to know the rate at which water flows into or out of a reservoir.
An urban geographer is interested in the rate of change of the population density in a city
as the distance from the city center increases. A meteorologist is concerned with the rate
of change of atmospheric pressure with respect to height (see Exercise 6.5.19).
In psychology, those interested in learning theory study the so-called learning curve,
which graphs the performance Pstd of someone learning a skill as a function of the
training time t. Of particular interest is the rate at which performance improves as time
passes, that is, dPydt.
In sociology, differential calculus is used in analyzing the spread of rumors (or innovations or fads or fashions). If pstd denotes the proportion of a population that knows a rumor
by time t, then the derivative dpydt represents the rate of spread of the rumor (see Exer­cise 6.2.65).
A Single Idea, Many Interpretations
Velocity, density, current, power, and temperature gradient in physics; rate of reaction
and compressibility in chemistry; rate of growth and blood velocity gradient in biology;
marginal cost and marginal profit in economics; rate of heat flow in geology; rate of
improvement of performance in psychology; rate of spread of a rumor in sociology—
these are all special cases of a single mathematical concept, the derivative.
This is an illustration of the fact that part of the power of mathematics lies in its
abstractness. A single abstract mathematical concept (such as the derivative) can have different interpretations in each of the sciences. When we develop the properties of the
mathematical concept once and for all, we can then turn around and apply these results to
all of the sciences. This is much more efficient than developing properties of special concepts in each separate science. The French mathematician Joseph Fourier (1768–1830)
put it succinctly: “Mathematics compares the most diverse phenomena and discovers the
secret analogies that unite them.”
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178
Chapter 2 Derivatives
1–4 A particle moves according to a law of motion s − f std,
t > 0, where t is measured in seconds and s in feet.
(a) Find the velocity at time t.
(b) What is the velocity after 1 second?
(c) When is the particle at rest?
(d) When is the particle moving in the positive direction?
(e) Find the total distance traveled during the first 6 seconds.
(f)Draw a diagram like Figure 2 to illustrate the motion of the
particle.
(g) Find the acceleration at time t and after 1 second.
; (h)Graph the position, velocity, and acceleration functions
for 0 < t < 6.
(i)When is the particle speeding up? When is it slowing down?
1. f std − t 3 2 9t 2 1 24t
2. f std − 0.01t 4 2 0.04t 3
3. f std − sinsty2d
4. f std −
9t
t2 1 9
5.Graphs of the velocity functions of two particles are shown,
where t is measured in seconds. When is each particle
speeding up? When is it slowing down? Explain.
(a)√ √
(b)√ √
0 0
1 1
t t
0 0 1 1
t t
6.Graphs of the position functions of two particles are shown,
where t is measured in seconds. When is each particle
speeding up? When is it slowing down? Explain.
(a)s s
(b)s s
0 0 1 1
t t
0 0 1 1
t t
7.The height (in meters) of a projectile shot vertically upward
from a point 2 m above ground level with an initial velocity
of 24.5 mys is h − 2 1 24.5t 2 4.9t 2 after t seconds.
(a) Find the velocity after 2 s and after 4 s.
(b) When does the projectile reach its maximum height?
(c) What is the maximum height?
(d) When does it hit the ground?
(e) With what velocity does it hit the ground?
8.If a ball is thrown vertically upward with a velocity of
80 ftys, then its height after t seconds is s − 80t 2 16t 2.
(a) What is the maximum height reached by the ball?
(b)What is the velocity of the ball when it is 96 ft above
the ground on its way up? On its way down?
9.If a rock is thrown vertically upward from the surface of
Mars with velocity 15 mys, its height after t seconds is
h − 15t 2 1.86t 2.
(a) What is the velocity of the rock after 2 s?
(b)What is the velocity of the rock when its height is
25 m on its way up? On its way down?
10. A particle moves with position function
s − t 4 2 4t 3 2 20t 2 1 20t t > 0
(a)At what time does the particle have a velocity of
20 mys?
(b)At what time is the acceleration 0? What is the significance of this value of t?
11. (a)A company makes computer chips from square wafers
of silicon. It wants to keep the side length of a wafer
very close to 15 mm and it wants to know how the area
Asxd of a wafer changes when the side length x changes.
Find A9s15d and explain its meaning in this situation.
(b)Show that the rate of change of the area of a square
with respect to its side length is half its perimeter. Try
to explain geometrically why this is true by drawing a
square whose side length x is increased by an amount
Dx. How can you approximate the resulting change in
area DA if Dx is small?
12. (a)Sodium chlorate crystals are easy to grow in the shape
of cubes by allowing a solution of water and sodium
chlorate to evaporate slowly. If V is the volume of
such a cube with side length x, calculate dVydx when
x − 3 mm and explain its meaning.
(b)Show that the rate of change of the volume of a cube
with respect to its edge length is equal to half the surface area of the cube. Explain geometrically why this
result is true by arguing by analogy with Exercise 11(b).
13. (a)Find the average rate of change of the area of a circle
with respect to its radius r as r changes from
(i) 2 to 3
(ii) 2 to 2.5
(iii) 2 to 2.1
(b) Find the instantaneous rate of change when r − 2.
(c)Show that the rate of change of the area of a circle with
respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically why this
is true by drawing a circle whose radius is increased by
an amount Dr. How can you approximate the resulting
change in area DA if Dr is small?
14.A stone is dropped into a lake, creating a circular ripple that
travels outward at a speed of 60 cmys. Find the rate at
which the area within the circle is increasing after (a) 1 s,
(b) 3 s, and (c) 5 s. What can you conclude?
15. A
spherical balloon is being inflated. Find the rate of
increase of the surface area sS − 4r 2 d with respect to the
radius r when r is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can you make?
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Section 2.7 Rates of Change in the Natural and Social Sciences
16. (a)The volume of a growing spherical cell is V − 43 r 3,
where the radius r is measured in micrometers
(1 μm − 1026 m). Find the average rate of change of V
with respect to r when r changes from
(i) 5 to 8 μm (ii) 5 to 6 μm
(iii) 5 to 5.1 μm
(b)Find the instantaneous rate of change of V with respect to
r when r − 5 μm.
(c)Show that the rate of change of the volume of a sphere
with respect to its radius is equal to its surface area.
Explain geometrically why this result is true. Argue by
analogy with Exercise 13(c).
17.The mass of the part of a metal rod that lies between its left
end and a point x meters to the right is 3x 2 kg. Find the linear
density (see Example 2) when x is (a) 1 m, (b) 2 m, and
(c) 3 m. Where is the density the highest? The lowest?
18. I f a tank holds 5000 gallons of water, which drains from the
bottom of the tank in 40 minutes, then Torricelli’s Law gives
the volume V of water remaining in the tank after t minutes as
1
td 0 < t < 40
V − 5000 s1 2 40
2
Find the rate at which water is draining from the tank after
(a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what
time is the water flowing out the fastest? The slowest?
Summarize your findings.
19.The quantity of charge Q in coulombs (C) that has passed
through a point in a wire up to time t (measured in seconds) is
given by Qstd − t 3 2 2t 2 1 6t 1 2. Find the current when
(a) t − 0.5 s and (b) t − 1 s. [See Example 3. The unit of
current is an ampere (1 A − 1 Cys).] At what time is the
current lowest?
20. N
ewton’s Law of Gravitation says that the magnitude F of the
force exerted by a body of mass m on a body of mass M is
F−
21.The force F acting on a body with mass m and velocity v is
the rate of change of momentum: F − sdydtdsmvd. If m is
constant, this becomes F − ma, where a − dvydt is the
acceleration. But in the theory of relativity the mass of a
particle varies with v as follows: m − m 0 ys1 2 v 2yc 2 , where
m 0 is the mass of the particle at rest and c is the speed of
light. Show that
F−
22.Some of the highest tides in the world occur in the Bay of
Fundy on the Atlantic Coast of Canada. At Hopewell Cape
the water depth at low tide is about 2.0 m and at high tide it is
about 12.0 m. The natural period of oscillation is a little more
than 12 hours and on June 30, 2009, high tide occurred at
6:45 am. This helps explain the following model for the water
depth D (in meters) as a function of the time t (in hours after
midnight) on that day:
Dstd − 7 1 5 cosf0.503st 2 6.75dg
How fast was the tide rising (or falling) at the following
times?
(a) 3:00 am
(b) 6:00 am
(c) 9:00 am
(d) Noon
23.Boyle’s Law states that when a sample of gas is compressed
at a constant temperature, the product of the pressure and the
volume remains constant: PV − C.
(a)Find the rate of change of volume with respect to
pressure.
(b)A sample of gas is in a container at low pressure and
is steadily compressed at constant temperature for
10 minutes. Is the volume decreasing more rapidly at
the beginning or the end of the 10 minutes? Explain.
(c)Prove that the isothermal compressibility (see
Example 5) is given by − 1yP.
24. I f, in Example 4, one molecule of the product C is formed
from one molecule of the reactant A and one molecule of the
reactant B, and the initial concentrations of A and B have a
common value fAg − fBg − a molesyL, then
fCg − a 2ktysakt 1 1d
where k is a constant.
(a) Find the rate of reaction at time t.
(b) Show that if x − fCg, then
GmM
r2
where G is the gravitational constant and r is the distance
between the bodies.
(a)Find dFydr and explain its meaning. What does the minus
sign indicate?
(b)Suppose it is known that the earth attracts an object with
a force that decreases at the rate of 2 Nykm when
r − 20,000 km. How fast does this force change when
r − 10,000 km?
m0a
s1 2 v 2yc 2 d3y2
179
dx
− ksa 2 xd2
dt
; 25.The table gives the population of the world Pstd, in millions,
where t is measured in years and t − 0 corresponds to the
year 1900.
t
Population
(millions)
t
Population
(millions)
0
10
20
30
40
50
1650
1750
1860
2070
2300
2560
60
70
80
90
100
110
3040
3710
4450
5280
6080
6870
(a)Estimate the rate of population growth in 1920 and in
1980 by averaging the slopes of two secant lines.
(b)Use a graphing device to find a cubic function (a thirddegree polynomial) that models the data.
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180
chapter 2 Derivatives
(c)Use your model in part (b) to find a model for the rate
of population growth.
(d)Use part (c) to estimate the rates of growth in 1920 and
1980. Compare with your estimates in part (a).
(e) Estimate the rate of growth in 1985.
; 26.The table shows how the average age of first marriage of
Japanese women has varied since 1950.
t
Astd
t
Astd
1950
1955
1960
1965
1970
1975
1980
23.0
23.8
24.4
24.5
24.2
24.7
25.2
1985
1990
1995
2000
2005
2010
25.5
25.9
26.3
27.0
28.0
28.8
(a)Use a graphing calculator or computer to model these
data with a fourth-degree polynomial.
(b)Use part (a) to find a model for A9std.
(c)Estimate the rate of change of marriage age for women
in 1990.
(d)Graph the data points and the models for A and A9.
27.Refer to the law of laminar flow given in Example 7. Consider a blood vessel with radius 0.01 cm, length 3 cm,
pressure difference 3000 dynesycm2, and viscosity
− 0.027.
(a)Find the velocity of the blood along the centerline
r − 0, at radius r − 0.005 cm, and at the wall
r − R − 0.01 cm.
(b)Find the velocity gradient at r − 0, r − 0.005,
and r − 0.01.
(c)Where is the velocity the greatest? Where is the velocity
changing most?
28. T
he frequency of vibrations of a vibrating violin string is
given by
f−
1
2L
Î
T
where L is the length of the string, T is its tension, and is
its linear density. [See Chapter 11 in D. E. Hall, Musical
Acoustics, 3rd ed. (Pacific Grove, CA: Brooks/Cole, 2002).]
(a) Find the rate of change of the frequency with respect to
(i) the length (when T and are constant),
(ii) the tension (when L and are constant), and
(iii) the linear density (when L and T are constant).
(b)The pitch of a note (how high or low the note sounds)
is determined by the frequency f . (The higher the frequency, the higher the pitch.) Use the signs of the
derivatives in part (a) to determine what happens to the
pitch of a note
(i)when the effective length of a string is decreased
by placing a finger on the string so a shorter portion of the string vibrates,
(ii)when the tension is increased by turning a tuning
peg,
(iii)when the linear density is increased by switching
to another string.
29.Suppose that the cost (in dollars) for a company to produce
x pairs of a new line of jeans is
Csxd − 2000 1 3x 1 0.01x 2 1 0.0002x 3
(a)Find the marginal cost function.
(b)Find C9s100d and explain its meaning. What does it
predict?
(c)Compare C9s100d with the cost of manufacturing the
101st pair of jeans.
30.The cost function for a certain commodity is
Csqd − 84 1 0.16q 2 0.0006q 2 1 0.000003q 3
(a) Find and interpret C9s100d.
(b)Compare C9s100d with the cost of producing the
101st item.
31. If psxd is the total value of the production when there are
x workers in a plant, then the average productivity of the
workforce at the plant is
psxd
Asxd −
x
(a)Find A9sxd. Why does the company want to hire more
workers if A9sxd . 0?
(b)Show that A9sxd . 0 if p9sxd is greater than the average
productivity.
32. I f R denotes the reaction of the body to some stimulus
of strength x, the sensitivity S is defined to be the rate
of change of the reaction with respect to x. A particular
example is that when the brightness x of a light source is
increased, the eye reacts by decreasing the area R of the
pupil. The experimental formula
R−
40 1 24x 0.4
1 1 4x 0.4
h as been used to model the dependence of R on x when R is
measured in square millimeters and x is measured in appropriate units of brightness.
(a) Find the sensitivity.
(b)Illustrate part (a) by graphing both R and S as functions
;
of x. Comment on the values of R and S at low levels of
brightness. Is this what you would expect?
33. T
he gas law for an ideal gas at absolute temperature T (in
kelvins), pressure ­P (in atmospheres, atm), and volume V
(in liters) is PV − nRT, where n is the number of moles of
the gas and R − 0.0821 is the gas constant. Suppose that, at
a certain instant, P − 8.0 atm and is increasing at a rate of
0.10 atmymin and V − 10 L and is decreasing at a rate of
0.15 Lymin. Find the rate of change of T with respect to
time at that instant if n − 10 moles.
34.Invasive species often display a wave of advance as they
colonize new areas. Mathematical models based on ran-
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Section 2.8 Related Rates
dom dispersal and reproduction have demonstrated that the
speed with which such waves move is given by the function
f srd − 2 sDr , where r is the reproductive rate of individuals and D is a parameter quantifying dispersal. Calculate the
derivative of the wave speed with respect to the reproductive
rate r and explain its meaning.
35. I n the study of ecosystems, predator-prey models are often
used to study the interaction between species. Consider populations of tundra wolves, given by Wstd, and caribou, given by
Cstd, in northern Canada. The interaction has been modeled by
the equations
dC
dW
− aC 2 bCW − 2cW 1 dCW
dt
dt
(a)What values of dCydt and dWydt correspond to stable
populations?
(b)How would the statement “The caribou go extinct” be
represented mathematically?
181
(c)Suppose that a − 0.05, b − 0.001, c − 0.05, and
d − 0.0001. Find all population pairs sC, W d that lead to
stable populations. According to this model, is it possible
for the two species to live in balance or will one or both
species become extinct?
36.In a fish farm, a population of fish is introduced into a pond and
harvested regularly. A model for the rate of change of the fish
population is given by the equation
S
D
dP
Pstd
Pstd 2 Pstd
− r0 1 2
dt
Pc
where r0 is the birth rate of the fish, Pc is the maximum population that the pond can sustain (called the carrying capacity),
and is the percentage of the population that is harvested.
(a)What value of dPydt corresponds to a stable population?
(b)If the pond can sustain 10,000 fish, the birth rate is 5%, and
the harvesting rate is 4%, find the stable population level.
(c) What happens if is raised to 5%?
If we are pumping air into a balloon, both the volume and the radius of the balloon are
increasing and their rates of increase are related to each other. But it is much easier to
measure directly the rate of increase of the volume than the rate of increase of the radius.
In a related rates problem the idea is to compute the rate of change of one quantity in
terms of the rate of change of another quantity (which may be more easily measured).
The procedure is to find an equation that relates the two quantities and then use the Chain
Rule to differentiate both sides with respect to time.
Example 1 Air is being pumped into a spherical balloon so that its volume increases
at a rate of 100 cm3ys. How fast is the radius of the balloon increasing when the diameter is 50 cm?
PS According to the Principles of Problem Solving discussed on page 98, the
first step is to understand the problem.
This includes reading the problem
carefully, identifying the given and the
unknown, and introducing suitable
notation.
SOLUTION We start by identifying two things:
the given information:
the rate of increase of the volume of air is 100 cm3ys
and the unknown:
the rate of increase of the radius when the diameter is 50 cm
In order to express these quantities mathematically, we introduce some suggestive
notation:
Let V be the volume of the balloon and let r be its radius.
The key thing to remember is that rates of change are derivatives. In this problem, the
volume and the radius are both functions of the time t. The rate of increase of the volume with respect to time is the derivative dVydt, and the rate of increase of the radius
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182
Chapter 2 Derivatives
is drydt. We can therefore restate the given and the unknown as follows:
PS The second stage of problem solving is to think of a plan for connecting
the given and the unknown.
Given:
dV
− 100 cm3ys
dt
Unknown:
dr
dt
when r − 25 cm
In order to connect dVydt and drydt, we first relate V and r by the formula for the
volume of a sphere:
V − 43 r 3
In order to use the given information, we differentiate each side of this equation with
respect to t. To differentiate the right side, we need to use the Chain Rule:
dV
dV dr
dr
−
− 4r 2
dt
dr dt
dt
Now we solve for the unknown quantity:
dr
1 dV
−
dt
4r 2 dt
Notice that, although dVydt is constant,
drydt is not constant.
If we put r − 25 and dVydt − 100 in this equation, we obtain
dr
1
1
−
100 −
dt
4s25d2
25
wall
The radius of the balloon is increasing at the rate of 1ys25d < 0.0127 cmys.
Example 2 A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ftys, how fast is the top of the ladder sliding
down the wall when the bottom of the ladder is 6 ft from the wall?
10
y
x
ground
FIGURE 1
dy
dt
■
=?
SOLUTION We first draw a diagram and label it as in Figure 1. Let x feet be the distance from the bottom of the ladder to the wall and y feet the distance from the top of
the ladder to the ground. Note that x and y are both functions of t (time, measured in
seconds).
We are given that dxydt − 1 ftys and we are asked to find dyydt when x − 6 ft (see
Figure 2). In this problem, the relationship between x and y is given by the Pythagorean
Theorem:
x 2 1 y 2 − 100
Differentiating each side with respect to t using the Chain Rule, we have
y
2x
x
and solving this equation for the desired rate, we obtain
dx
dt
FIGURE 2
dx
dy
1 2y
−0
dt
dt
=1
dy
x dx
−2
dt
y dt
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Section 2.8 Related Rates
183
When x − 6, the Pythagorean Theorem gives y − 8 and so, substituting these values
and dxydt − 1, we have
dy
6
3
− 2 s1d − 2 ftys
dt
8
4
The fact that dyydt is negative means that the distance from the top of the ladder to
the ground is decreasing at a rate of 34 ftys. In other words, the top of the ladder is sliding
down the wall at a rate of 34 ftys.
■
Example 3 A water tank has the shape of an inverted circular cone with base radius
2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m 3ymin, find
the rate at which the water level is rising when the water is 3 m deep.
2
r
4
SOLUTION We first sketch the cone and label it as in Figure 3. Let V, r, and h be the
volume of the water, the radius of the surface, and the height of the water at time t,
where t is measured in minutes.
We are given that dVydt − 2 m 3ymin and we are asked to find dhydt when h is 3 m.
The quantities V and h are related by the equation
V − 13 r 2h
h
FIGURE 3
but it is very useful to express V as a function of h alone. In order to eliminate r, we use
the similar triangles in Figure 3 to write
r
2
h
− r −
h
4
2
and the expression for V becomes
V−
SD
1
h
3
2
2
h−
3
h
12
Now we can differentiate each side with respect to t:
dV
2 dh
−
h
dt
4
dt
so
dh
4 dV
−
dt
h 2 dt
Substituting h − 3 m and dVydt − 2 m 3ymin, we have
dh
4
8
−
2 ? 2 −
dt
s3d
9
The water level is rising at a rate of 8ys9d < 0.28 mymin.
PS Look back: What have we learned
from Examples 1–3 that will help us
solve future problems?
Problem Solving Strategy It is useful to recall some of the problem-solving principles from page 98 and adapt them to related rates in light of our experience in
Examples 1–3:
1.
2.
3.
4.
Read the problem carefully.
Draw a diagram if possible.
Introduce notation. Assign symbols to all quantities that are functions of time.
Express the given information and the required rate in terms of derivatives.
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■
184
Chapter 2 Derivatives
WARNING A common error is
to substitute the given numerical
information (for quantities that vary
with time) too early. This should be
done only after the differentiation.
(Step 7 follows Step 6.) For instance,
in Example 3 we dealt with general
values of h until we finally substituted
h − 3 at the last stage. (If we had put
h − 3 earlier, we would have gotten
dVydt − 0, which is clearly wrong.)
x
C
y
z
B
A
5.
Write an equation that relates the various quantities of the problem. If necessary,
use the geometry of the situation to eliminate one of the variables by substitution
(as in Example 3).
6. Use the Chain Rule to differentiate both sides of the equation with respect to t.
7.
Substitute the given information into the resulting equation and solve for the
unknown rate.
The following examples are further illustrations of the strategy.
Example 4 Car A is traveling west at 50 miyh and car B is traveling north at
60 miyh. Both are headed for the intersection of the two roads. At what rate are
the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the
intersection?
SOLUTION We draw Figure 4, where C is the intersection of the roads. At a given time
t, let x be the distance from car A to C, let y be the distance from car B to C, and let z
be the distance between the cars, where x, y, and z are measured in miles.
We are given that dxydt − 250 miyh and dyydt − 260 miyh. (The derivatives are
negative because x and y are decreasing as t increases.) We are asked to find dzydt. The
equation that relates x, y, and z is given by the Pythagorean Theorem:
z2 − x 2 1 y 2
FIGURE 4
Differentiating each side with respect to t, we have
2z
dz
dx
dy
− 2x
1 2y
dt
dt
dt
dz
1
−
dt
z
S
x
dx
dy
1y
dt
dt
D
When x − 0.3 mi and y − 0.4 mi, the Pythagorean Theorem gives z − 0.5 mi, so
dz
1
−
f0.3s250d 1 0.4s260dg
dt
0.5
− 278 miyh
The cars are approaching each other at a rate of 78 miyh.
■
Example 5 A man walks along a straight path at a speed of 4 ftys. A searchlight is
located on the ground 20 ft from the path and is kept focused on the man. At what rate
is the searchlight rotating when the man is 15 ft from the point on the path closest to
the searchlight?
¨
SOLUTION We draw Figure 5 and let x be the distance from the man to the point on the
path closest to the searchlight. We let be the angle between the beam of the searchlight and the perpendicular to the path.
We are given that dxydt − 4 ftys and are asked to find dydt when x − 15. The
equation that relates x and can be written from Figure 5:
FIGURE 5
x
− tan x − 20 tan 20
x
20
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Section 2.8 Related Rates
185
Differentiating each side with respect to t, we get
dx
d
− 20 sec2
dt
dt
so
d
1
dx
−
cos2
dt
20
dt
−
1
1
cos2 s4d − cos2
20
5
When x − 15, the length of the beam is 25, so cos − 45 and
d
1
−
dt
5
SD
4
5
2
−
16
− 0.128
125
The searchlight is rotating at a rate of 0.128 radys.
1.If V is the volume of a cube with edge length x and the cube
expands as time passes, find dVydt in terms of dxydt.
2.(a)If A is the area of a circle with radius r and the circle
expands as time passes, find dAydt in terms of drydt.
(b)Suppose oil spills from a ruptured tanker and spreads in
a circular pattern. If the radius of the oil spill increases at
a constant rate of 1 mys, how fast is the area of the spill
increasing when the radius is 30 m?
3.Each side of a square is increasing at a rate of 6 cmys. At what
rate is the area of the square increasing when the area of the
square is 16 cm2?
4.The length of a rectangle is increasing at a rate of 8 cmys and
its width is increasing at a rate of 3 cmys. When the length is
20 cm and the width is 10 cm, how fast is the area of the
rectangle increasing?
5.A cylindrical tank with radius 5 m is being filled with water
at a rate of 3 m3ymin. How fast is the height of the water
increasing?
6.The radius of a sphere is increasing at a rate of 4 mmys. How
fast is the volume increasing when the diameter is 80 mm?
7.The radius of a spherical ball is increasing at a rate of
2 cmymin. At what rate is the surface area of the ball
increasing when the radius is 8 cm?
8.The area of a triangle with sides of lengths a and b and
contained angle is
A − 12 ab sin (a)If a − 2 cm, b − 3 cm, and increases at a rate of
0.2 radymin, how fast is the area increasing when
− y3?
■
(b)If a − 2 cm, b increases at a rate of 1.5 cmymin, and increases at a rate of 0.2 radymin, how fast is the area
increasing when b − 3 cm and − y3?
(c)If a increases at a rate of 2.5 cmymin, b increases at a rate
of 1.5 cmymin, and increases at a rate of 0.2 radymin,
how fast is the area increasing when a − 2 cm, b − 3 cm,
and − y3?
9.Suppose y − s2x 1 1 , where x and y are functions of t.
(a)If dxydt − 3, find dyydt when x − 4.
(b)If dyydt − 5, find dxydt when x − 12.
10.Suppose 4x 2 1 9y 2 − 36, where x and y are functions of t.
(a)If dyydt − 13, find dxydt when x − 2 and y − 23 s5 .
(b)If dxydt − 3, find dy ydt when x − 22 and y − 23 s5 .
11.If x 2 1 y 2 1 z 2 − 9, dxydt − 5, and dyydt − 4, find dzydt
when sx, y, zd − s2, 2, 1d.
12. A
particle is moving along a hyperbola xy − 8. As it reaches
the point s4, 2d, the y-coordinate is decreasing at a rate of
3 cmys. How fast is the x-coordinate of the point changing at
that instant?
13–16
(a) What quantities are given in the problem?
(b) What is the unknown?
(c) Draw a picture of the situation for any time t.
(d) Write an equation that relates the quantities.
(e) Finish solving the problem.
13.A plane flying horizontally at an altitude of 1 mi and a speed of
500 miyh passes directly over a radar station. Find the rate at
which the distance from the plane to the station is increasing
when it is 2 mi away from the station.
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186
chapter 2 Derivatives
14. If
a snowball melts so that its surface area decreases at a
rate of 1 cm2ymin, find the rate at which the diameter
decreases when the diameter is 10 cm.
15. A
street light is mounted at the top of a 15-ft-tall pole.
A man 6 ft tall walks away from the pole with a speed
of 5 ftys along a straight path. How fast is the tip of his
shadow moving when he is 40 ft from the pole?
16. At
noon, ship A is 150 km west of ship B. Ship A is sailing
east at 35 kmyh and ship B is sailing north at 25 kmyh.
How fast is the distance between the ships changing at
4:00 pm?
17. T
wo cars start moving from the same point. One travels
south at 60 miyh and the other travels west at 25 miyh. At
what rate is the distance between the cars increasing two
hours later?
18.A spotlight on the ground shines on a wall 12 m away. If a
man 2 m tall walks from the spotlight toward the building at
a speed of 1.6 mys, how fast is the length of his shadow on
the building decreasing when he is 4 m from the building?
19. A
man starts walking north at 4 ftys from a point P. Five
minutes later a woman starts walking south at 5 ftys from a
point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?
20. A
baseball diamond is a square with side 90 ft. A batter hits
the ball and runs toward first base with a speed of 24 ftys.
(a)At what rate is his distance from second base decreasing when he is halfway to first base?
(b)At what rate is his distance from third base increasing at
the same moment?
23. A
t noon, ship A is 100 km west of ship B. Ship A is sailing
south at 35 kmyh and ship B is sailing north at 25 kmyh.
How fast is the distance between the ships changing at
4:00 pm?
24. A
particle moves along the curve y − 2 sinsxy2d. As the
particle passes through the point ( 13 , 1), its x-coordinate
increases at a rate of s10 cmys. How fast is the distance
from the particle to the origin changing at this instant?
25. W
ater is leaking out of an inverted conical tank at a rate
of 10,000 cm 3ymin at the same time that water is being
pumped into the tank at a constant rate. The tank has height
6 m and the diameter at the top is 4 m. If the water level is
rising at a rate of 20 cmymin when the height of the water
is 2 m, find the rate at which water is being pumped into the
tank.
26. A
trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a
height of 1 ft. If the trough is being filled with water at a
rate of 12 ft 3ymin, how fast is the water level rising when
the water is 6 inches deep?
27. A
water trough is 10 m long and a cross-section has the
shape of an isosceles trapezoid that is 30 cm wide at the
bottom, 80 cm wide at the top, and has height 50 cm. If the
trough is being filled with water at the rate of 0.2 m 3ymin,
how fast is the water level rising when the water is 30 cm
deep?
28. A
swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the
shallow end, and 9 ft deep at its deepest point. A crosssection is shown in the figure. If the pool is being filled at a
rate of 0.8 ft 3ymin, how fast is the water level rising when
the depth at the deepest point is 5 ft?
3
6
6
90 ft
21. T
he altitude of a triangle is increasing at a rate of 1 cmymin
while the area of the triangle is increasing at a rate of
2 cm 2ymin. At what rate is the base of the triangle changing
when the altitude is 10 cm and the area is 100 cm2 ?
12
16
6
29.Gravel is being dumped from a conveyor belt at a rate of
30 ft 3ymin, and its coarseness is such that it forms a pile in
the shape of a cone whose base diameter and height are
always equal. How fast is the height of the pile increasing
when the pile is 10 ft high?
22.A boat is pulled into a dock by a rope attached to the bow
of the boat and passing through a pulley on the dock that is
1 m higher than the bow of the boat. If the rope is pulled in
at a rate of 1 mys, how fast is the boat approaching the dock
when it is 8 m from the dock?
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Section 2.8 Related Rates
30. A
kite 100 ft above the ground moves horizontally at a
speed of 8 ftys. At what rate is the angle between the string
and the horizontal decreasing when 200 ft of string has been
let out?
31.The sides of an equilateral triangle are increasing at a rate
of 10 cmymin. At what rate is the area of the triangle
increasing when the sides are 30 cm long?
32. H
ow fast is the angle between the ladder and the ground
changing in Example 2 when the bottom of the ladder is 6 ft
from the wall?
33.The top of a ladder slides down a vertical wall at a rate of
0.15 mys. At the moment when the bottom of the ladder is
3 m from the wall, it slides away from the wall at a rate of
0.2 mys. How long is the ladder?
187
40.Brain weight B as a function of body weight W in fish has
been modeled by the power function B − 0.007W 2y3, where
B and W are measured in grams. A model for body weight
as a function of body length L (measured in centimeters) is
W − 0.12L2.53. If, over 10 million years, the average length
of a certain species of fish evolved from 15 cm to 20 cm at
a constant rate, how fast was this species’ brain growing
when the average length was 18 cm?
41. T
wo sides of a triangle have lengths 12 m and 15 m. The
angle between them is increasing at a rate of 2 8ymin. How
fast is the length of the third side increasing when the angle
between the sides of fixed length is 60°?
34.According to the model we used to solve Example 2, what
happens as the top of the ladder approaches the ground? Is
the model appropriate for small values of y?
42.Two carts, A and B, are connected by a rope 39 ft long that
passes over a pulley P (see the figure). The point Q is on the
floor 12 ft directly beneath P and between the carts. Cart A
is being pulled away from Q at a speed of 2 ftys. How fast
is cart B moving toward Q at the instant when cart A is 5 ft
from Q?
35.If the minute hand of a clock has length r (in centimeters),
find the rate at which it sweeps out area as a function of r.
P
; 36.A faucet is filling a hemispherical basin of diameter 60 cm
with water at a rate of 2 Lymin. Find the rate at which the
water is rising in the basin when it is half full. [Use the
following facts: 1 L is 1000 cm3. The volume of the portion
of a sphere with radius r from the bottom to a height h is
V − (rh 2 2 13 h 3), as we will show in Chapter 5.]
37. B
oyle’s Law states that when a sample of gas is compressed
at a constant temperature, the pressure P and volume V
satisfy the equation PV − C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the
pressure is 150 kPa, and the pressure is increasing at a rate
of 20 kPaymin. At what rate is the volume decreasing at this
instant?
38.When air expands adiabatically (without gaining or losing
heat), its pressure P and volume V are related by the equation PV 1.4 − C, where C is a constant. Suppose that at
a certain instant the volume is 400 cm3 and the pressure is
80 kPa and is decreasing at a rate of 10 kPaymin. At what
rate is the volume increasing at this instant?
39.If two resistors with resistances R1 and R2 are connected in
parallel, as in the figure, then the total resistance R, measured in ohms (V), is given by
1
1
1
−
1
R
R1
R2
If R1 and R2 are increasing at rates of 0.3 Vys and 0.2 Vys,
respectively, how fast is R changing when R1 − 80 V and
R2 − 100 V?
R¡
R™
12 ft
A
B
Q
43.A television camera is positioned 4000 ft from the base of a
rocket launching pad. The angle of elevation of the camera
has to change at the correct rate in order to keep the rocket
in sight. Also, the mechanism for focusing the camera has to
take into account the increasing distance from the camera to
the rising rocket. Let’s assume the rocket rises vertically and
its speed is 600 ftys when it has risen 3000 ft.
(a)How fast is the distance from the television camera to
the rocket changing at that moment?
(b)If the television camera is always kept aimed at the
rocket, how fast is the camera’s angle of elevation
changing at that same moment?
44.A lighthouse is located on a small island 3 km away from
the nearest point P on a straight shoreline and its light
makes four revolutions per minute. How fast is the beam of
light moving along the shoreline when it is 1 km from P?
45. A
plane flies horizontally at an altitude of 5 km and passes
directly over a tracking telescope on the ground. When the
angle of elevation is y3, this angle is decreasing at a rate
of y6 radians per minute. How fast is the plane traveling at
that time?
46.A Ferris wheel with a radius of 10 m is rotating at a rate of
one revolution every 2 minutes. How fast is a rider rising
when his seat is 16 m above ground level?
47. A
plane flying with a constant speed of 300 kmyh passes
over a ground radar station at an altitude of 1 km and climbs
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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188
chapter 2 Derivatives
at an angle of 308. At what rate is the distance from the plane to
the radar station increasing a minute later?
48. T
wo people start from the same point. One walks east at
3 miyh and the other walks northeast at 2 miyh. How fast is
the distance between the people changing after 15 minutes?
49. A
runner sprints around a circular track of radius 100 m at
a constant speed of 7 mys. The runner’s friend is standing
50. T
he minute hand on a watch is 8 mm long and the hour hand is
4 mm long. How fast is the distance between the tips of the
hands changing at one o’clock?
We have seen that a curve lies very close to its tangent line near the point of tangency. In
fact, by zooming in toward a point on the graph of a differentiable function, we noticed
that the graph looks more and more like its tangent line. (See Figure 2.1.2.) This observation is the basis for a method of finding approximate values of functions.
The idea is that it might be easy to calculate a value f sad of a function, but difficult
(or even impossible) to compute nearby values of f . So we settle for the easily computed
values of the linear function L whose graph is the tangent line of f at sa, f sadd. (See
Figure 1.)
In other words, we use the tangent line at sa, f sadd as an approximation to the curve
y − f sxd when x is near a. An equation of this tangent line is
y
y=ƒ
{a, f(a)}
at a distance 200 m from the center of the track. How fast is
the distance between the friends changing when the distance
between them is 200 m?
y=L(x)
y − f sad 1 f 9sadsx 2 ad
0
x
and the approximation
FIGURE 1
1 f sxd < f sad 1 f 9sadsx 2 ad
is called the linear approximation or tangent line approximation of f at a. The linear
function whose graph is this tangent line, that is,
2 Lsxd − f sad 1 f 9sadsx 2 ad
is called the linearization of f at a.
Example 1 Find the linearization of the function f sxd − sx 1 3 at a − 1 and use
it to approximate the numbers s3.98 and s4.05 . Are these approximations overestimates or underestimates?
SOLUTION The derivative of f sxd − sx 1 3d1y2 is
f 9sxd − 12 sx 1 3d21y2 −
1
2sx 1 3
and so we have f s1d − 2 and f 9s1d − 14. Putting these values into Equation 2, we see
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.9 Linear Approximations and Differentials
189
that the linearization is
Lsxd − f s1d 1 f 9s1dsx 2 1d − 2 1 14 sx 2 1d −
7
x
1
4
4
The corresponding linear approximation (1) is
sx 1 3 <
7
x
1 (when x is near 1)
4
4
In particular, we have
y
7
1.05
7
s3.98 < 74 1 0.98
4 − 1.995 and s4.05 < 4 1 4 − 2.0125
x
y= 4 + 4
_3
FIGURE 2
(1, 2)
0
1
y= œ„„„„
x+3
x
The linear approximation is illustrated in Figure 2. We see that, indeed, the tangent
line approximation is a good approximation to the given function when x is near l. We
also see that our approximations are overestimates because the tangent line lies above
the curve.
Of course, a calculator could give us approximations for s3.98 and s4.05 , but the
linear approximation gives an approximation over an entire interval.
■
In the following table we compare the estimates from the linear approximation in
Example 1 with the true values. Notice from this table, and also from Figure 2, that the tangent line approximation gives good estimates when x is close to 1 but the accuracy of the
approximation deteriorates when x is farther away from 1.
x
From Lsxd
Actual value
s3.9
0.9
1.975
1.97484176. . .
s3.98
0.98
1.995
1.99499373. . .
s4
1
2
2.00000000. . .
s4.05
1.05
2.0125
2.01246117. . .
s4.1
1.1
2.025
2.02484567. . .
s5
s6
2
2.25
2.23606797. . .
3
2.5
2.44948974. . .
How good is the approximation that we obtained in Example 1? The next example
shows that by using a graphing calculator or computer we can determine an interval
throughout which a linear approximation provides a specified accuracy.
Example 2 For what values of x is the linear approximation
sx 1 3 <
7
x
1
4
4
accurate to within 0.5? What about accuracy to within 0.1?
SOLUTION Accuracy to within 0.5 means that the functions should differ by less
than 0.5:
Z
sx 1 3 2
S DZ
7
x
1
4
4
, 0.5
Equivalently, we could write
sx 1 3 2 0.5 ,
7
x
1 , sx 1 3 1 0.5
4
4
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190
Chapter 2 Derivatives
4.3
y= œ„„„„
x+3+0.5
L(x)
P
Q
y= œ„„„„
x+3-0.5
_4
10
_1
FIGURE 3
3
is accurate to within 0.5 when 22.6 , x , 8.6. (We have rounded 22.66 up and 8.66
down to be safe.)
Similarly, from Figure 4 we see that the approximation is accurate to within 0.1
when 21.1 , x , 3.9.
■
Q
y= œ„„„„
x+3+0.1
Applications to Physics
y= œ„„„„
x+3-0.1
P
_2
1
This says that the linear approximation should lie between the curves obtained by shifting the curve y − sx 1 3 upward and downward by an amount 0.5. Figure 3 shows
the tangent line y − s7 1 xdy4 intersecting the upper curve y − sx 1 3 1 0.5 at P
and Q. Zooming in and using the cursor, we estimate that the x-coordinate of P is about
22.66 and the x-coordinate of Q is about 8.66. Thus we see from the graph that the
approximation
7
x
sx 1 3 < 1
4
4
5
FIGURE 4
Linear approximations are often used in physics. In analyzing the consequences of an
equation, a physicist sometimes needs to simplify a function by replacing it with its linear
approximation. For instance, in deriving a formula for the period of a pendulum, physics textbooks obtain the expression a T − 2t sin for tangential acceleration and then
replace sin by with the remark that sin is very close to if is not too large. [See,
for exam­ple, Physics: Calculus, 2d ed., by Eugene Hecht (Pacific Grove, CA: Brooks/
Cole, 2000), p. 431.] You can verify that the linearization of the function f sxd − sin x at
a − 0 is Lsxd − x and so the lin­ear approximation at 0 is
sin x < x
(see Exercise 40). So, in effect, the derivation of the formula for the period of a pendulum
uses the tangent line approximation for the sine function.
Another example occurs in the theory of optics, where light rays that arrive at shallow
angles relative to the optical axis are called paraxial rays. In paraxial (or Gaussian) optics,
both sin and cos are replaced by their linearizations. In other words, the linear
approximations
sin < and cos < 1
are used because is close to 0. The results of calculations made with these approximations became the basic theoretical tool used to design lenses. [See Optics, 4th ed., by
Eugene Hecht (San Francisco, 2002), p. 154.]
In Section 11.11 we will present several other applications of the idea of linear approximations to physics and engineering.
Differentials
If dx ± 0, we can divide both sides of
Equation 3 by dx to obtain
dy
− f 9sxd
dx
We have seen similar equations before,
but now the left side can genuinely be
interpreted as a ratio of differentials.
The ideas behind linear approximations are sometimes formulated in the terminology and
notation of differentials. If y − f sxd, where f is a differentiable function, then the differential dx is an independent variable; that is, dx can be given the value of any real number.
The differential dy is then defined in terms of dx by the equation
3 dy − f 9sxd dx
So dy is a dependent variable; it depends on the values of x and dx. If dx is given a specific value and x is taken to be some specific number in the domain of f , then the numerical value of dy is determined.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.9 Linear Approximations and Differentials
y
Q
Îy
dx=Îx
0
x
The geometric meaning of differentials is shown in Figure 5. Let Psx, f sxdd and
Qsx 1 Dx, f sx 1 Dxdd be points on the graph of f and let dx − Dx. The corresponding
change in y is
Dy − f sx 1 Dxd 2 f sxd
R
P
dy
S
x+Î x
191
x
y=ƒ
FIGURE 5
The slope of the tangent line PR is the derivative f 9sxd. Thus the directed distance from
S to R is f 9sxd dx − dy. Therefore dy represents the amount that the tangent line rises or
falls (the change in the linearization) when x changes by an amount dx, whereas Dy represents the amount that the curve y − f sxd rises or falls when x changes by an amount
dx − Dx.
Example 3 Compare the values of Dy and dy if y − f sxd − x 3 1 x 2 2 2x 1 1 and
x changes (a) from 2 to 2.05 and (b) from 2 to 2.01.
SOLUTION (a) We have
f s2d − 2 3 1 2 2 2 2s2d 1 1 − 9
f s2.05d − s2.05d3 1 s2.05d2 2 2s2.05d 1 1 − 9.717625
Dy − f s2.05d 2 f s2d − 0.717625
Figure 6 shows the function in Example 3 and a comparison of dy and Dy
when a − 2. The viewing rectangle is
f1.8, 2.5g by f6, 18g.
dy − f 9sxd dx − s3x 2 1 2x 2 2d dx
In general,
When x − 2 and dx − Dx − 0.05, this becomes
dy − f3s2d2 1 2s2d 2 2g0.05 − 0.7
y=˛+≈-2x+1
(b)
dy
Îy
f s2.01d − s2.01d3 1 s2.01d2 2 2s2.01d 1 1 − 9.140701
Dy − f s2.01d 2 f s2d − 0.140701
(2, 9)
FIGURE 6
When dx − Dx − 0.01,
dy − f3s2d2 1 2s2d 2 2g0.01 − 0.14
■
Notice that the approximation Dy < dy becomes better as Dx becomes smaller in
Example 3. Notice also that dy was easier to compute than Dy. For more complicated
functions it may be impossible to compute Dy exactly. In such cases the approximation
by differentials is especially useful.
In the notation of differentials, the linear approximation (1) can be written as
f sa 1 dxd < f sad 1 dy
For instance, for the function f sxd − sx 1 3 in Example 1, we have
dy − f 9sxd dx −
dx
2 sx 1 3
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192
Chapter 2 Derivatives
If a − 1 and dx − Dx − 0.05, then
dy −
and
0.05
− 0.0125
2 s1 1 3
s4.05 − f s1.05d < f s1d 1 dy − 2.0125
just as we found in Example 1.
Our final example illustrates the use of differentials in estimating the errors that occur
because of approximate measurements.
Example 4 The radius of a sphere was measured and found to be 21 cm with a pos­sible error in measurement of at most 0.05 cm. What is the maximum error in using this
value of the radius to compute the volume of the sphere?
4
SOLUTION If the radius of the sphere is r, then its volume is V − 3 r 3. If the error
in the measured value of r is denoted by dr − Dr, then the corresponding error in the
calculated value of V is DV, which can be approximated by the differential
dV − 4r 2 dr
When r − 21 and dr − 0.05, this becomes
dV − 4s21d2 0.05 < 277
The maximum error in the calculated volume is about 277 cm3.
■
NOTE Although the possible error in Example 4 may appear to be rather large, a
better picture of the error is given by the relative error, which is computed by dividing
the error by the total volume:
DV
dV
4r 2 dr
dr
<
− 4 3 −3
V
V
r
3 r
Thus the relative error in the volume is about three times the relative error in the radius.
In Example 4 the relative error in the radius is approximately dryr − 0.05y21 < 0.0024
and it produces a relative error of about 0.007 in the volume. The errors could also be
expressed as percentage errors of 0.24% in the radius and 0.7% in the volume.
1–4 Find the linearization Lsxd of the function at a.
1. f sxd − x 2 x 1 3, a − 22
3
2
2.f sxd − sin x, a − y6
3. f sxd − sx ,
a−4
4. f sxd − 2ysx 2 2 5 , a − 3
; 5.Find the linear approximation of the function
f sxd − s1 2 x at a − 0 and use it to approximate the
numbers s0.9 and s0.99 . Illustrate by graphing f and the
tangent line.
; 6.Find the linear approximation of the function
3
tsxd − s
1 1 x at a − 0 and use it to approximate the
3
3
numbers s
0.95 and s
1.1 . Illustrate by graphing t and the
tangent line.
; 7–10 Verify the given linear approximation at a − 0. Then
determine the values of x for which the linear approximation is
accurate to within 0.1.
4
7. s
1 1 2x < 1 1 12 x
9. 1ys1 1 2xd4 < 1 2 8x
8. s1 1 xd23 < 1 2 3x
10. tan x < x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 2.9 Linear Approximations and Differentials
11–14 Find the differential dy of each function.
11. (a) y − sx 2 2 3d 22
12. (a) y −
1 1 2u
1 1 3u
(b) y − s1 2 t 4
(b) y − 2 sin 2
1 2 v2
1 1 v2
13. (a) y − tan st
(b) y −
14. (a) y − st 2 cos t
1
(b) y − sin x
x
15–18 (a) Find the differential dy and (b) evaluate dy for the given
values of x and dx.
15. y − tan x, x − y4, dx − 20.1
16. y − cos x, x − 13, dx − 20.02
17. y − s3 1 x 2 , x − 1, dx − 20.1
18. y −
x11
, x − 2, dx − 0.05
x21
19–22 Compute Dy and dy for the given values of x and dx − Dx.
Then sketch a diagram like Figure 5 showing the line segments with
lengths dx, dy, and Dy.
19. y − x 2 2 4x,
x − 3, Dx − 0.5
20. y − x 2 x 3,
x − 0, Dx − 20.3
21. y − sx 2 2 ,
x − 3, Dx − 0.8
3
22. y − x , x − 1, Dx − 0.5
23–28 Use a linear approximation (or differentials) to estimate the
given number.
23. s1.999d4
24. 1y4.002
3
25. s
1001
26. s100.5
27. tan 2°
28. cos 29°
29–30 Explain, in terms of linear approximations or differentials,
why the approximation is reasonable.
29. sec 0.08 < 1
30. s4.02 < 2.005
31. T
he edge of a cube was found to be 30 cm with a possible error
in measurement of 0.1 cm. Use differentials to estimate the
maximum possible error, relative error, and percentage error
in computing (a) the volume of the cube and (b) the sur­face
area of the cube.
32. T
he radius of a circular disk is given as 24 cm with a maxi­mum
error in measurement of 0.2 cm.
(a)Use differentials to estimate the maximum error in the
calculated area of the disk.
(b) What is the relative error? What is the percentage error?
193
33. T
he circumference of a sphere was measured to be 84 cm with
a possible error of 0.5 cm.
(a)Use differentials to estimate the maximum error in the
calculated surface area. What is the relative error?
(b)Use differentials to estimate the maximum error in the
calculated volume. What is the relative error?
34. U
se differentials to estimate the amount of paint needed to
apply a coat of paint 0.05 cm thick to a hemispherical dome
with diameter 50 m.
35. (a)Use differentials to find a formula for the approximate
volume of a thin cylindrical shell with height h, inner
radius r, and thickness Dr.
(b)What is the error involved in using the formula from
part (a)?
36.One side of a right triangle is known to be 20 cm long and
the opposite angle is measured as 30°, with a possible error
of 61°.
(a)Use differentials to estimate the error in computing the
length of the hypotenuse.
(b)What is the percentage error?
37.If a current I passes through a resistor with resistance R, Ohm’s
Law states that the voltage drop is V − RI. If V is constant and
R is measured with a certain error, use differentials to show that
the relative error in calculating I is approximately the same (in
magnitude) as the relative error in R.
38.When blood flows along a blood vessel, the flux F (the volume
of blood per unit time that flows past a given point) is proportional to the fourth power of the radius R of the blood vessel:
F − kR 4
(This is known as Poiseuille’s Law; we will show why it
is true in Section 8.4.) A partially clogged artery can be
expanded by an operation called angioplasty, in which a
balloon-tipped catheter is inflated inside the artery in order
to widen it and restore the normal blood flow.
Show that the relative change in F is about four times the
relative change in R. How will a 5% increase in the radius
affect the flow of blood?
39. E
stablish the following rules for working with differentials
(where c denotes a constant and u and v are functions of x).
(a) dc − 0
(b) dscud − c du
(c) dsu 1 vd − du 1 dv
(d) dsuvd − u dv 1 v du
SD
(e) d
u
v
−
v du 2 u dv
v2
(f) dsx n d − nx n21 dx
40. O
n page 431 of Physics: Calculus, 2d ed., by Eugene Hecht
(Pacific Grove, CA: Brooks/Cole, 2000), in the course of
deriving the formula T − 2 sLyt for the period of a
pendulum of length L, the author obtains the equation
a T − 2t sin for the tangential acceleration of the bob of the
pendulum. He then says, “for small angles, the value of in
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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194
chapter 2 Derivatives
(b)Are your estimates in part (a) too large or too small?
Explain.
radians is very nearly the value of sin ; they differ by less
than 2% out to about 20°.”
(a)Verify the linear approximation at 0 for the sine
function:
y
sin x < x
;
y=fª(x)
(b)Use a graphing device to determine the values of x for
which sin x and x differ by less than 2%. Then verify
Hecht’s statement by converting from radians to
degrees.
1
0
;
x
42.Suppose that we don’t have a formula for tsxd but we know
that ts2d − 24 and t9sxd − sx 2 1 5 for all x.
(a)Use a linear approximation to estimate ts1.95d
and ts2.05d.
(b)Are your estimates in part (a) too large or too small?
Explain.
41.Suppose that the only information we have about a function
f is that f s1d − 5 and the graph of its derivative is as
shown.
(a)Use a linear approximation to estimate f s0.9d and
f s1.1d.
laboratory Project
1
Taylor Polynomials
The tangent line approximation Lsxd is the best first-degree (linear) approximation to f sxd near
x − a because f sxd and Lsxd have the same rate of change (derivative) at a. For a better approximation than a linear one, let’s try a second-degree (quadratic) approximation Psxd. In other
words, we approximate a curve by a parabola instead of by a straight line. To make sure that the
approximation is a good one, we stipulate the following:
(i) Psad − f sad
(ii) P9sad − f 9sad
(iii) P99sad − f 99sad
(P and f should have the same value at a.)
(P and f should have the same rate of change at a.)
(The slopes of P and f should change at the same rate at a.)
1. F
ind the quadratic approximation Psxd − A 1 Bx 1 Cx 2 to the function f sxd − cos x that
satisfies conditions (i), (ii), and (iii) with a − 0. Graph P, f , and the linear approximation
Lsxd − 1 on a common screen. Comment on how well the functions P and L approximate f.
2.
Determine the values of x for which the quadratic approximation f sxd < Psxd in Problem 1 is
accurate to within 0.1. [Hint: Graph y − Psxd, y − cos x 2 0.1, and y − cos x 1 0.1 on a
common screen.]
3.
To approximate a function f by a quadratic function P near a number a, it is best to write P in
the form
Psxd − A 1 Bsx 2 ad 1 Csx 2 ad2
Show that the quadratic function that satisfies conditions (i), (ii), and (iii) is
Psxd − f sad 1 f 9sadsx 2 ad 1 12 f 99sadsx 2 ad2
4.
Find the quadratic approximation to f sxd − sx 1 3 near a − 1. Graph f , the quadratic
approximation, and the linear approximation from Example 2.9.2 on a common screen.
What do you conclude?
5. I nstead of being satisfied with a linear or quadratic approximation to f sxd near x − a,
let’s try to find better approximations with higher-degree polynomials. We look for an
nth-degree polynomial
Tnsxd − c0 1 c1 sx 2 ad 1 c2 sx 2 ad2 1 c3 sx 2 ad3 1 ∙ ∙ ∙ 1 cn sx 2 adn
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chapter 2 Review
such that Tn and its first n derivatives have the same values at x − a as f and its first n
derivatives. By differentiating repeatedly and setting x − a, show that these conditions are
satisfied if c0 − f sad, c1 − f 9sad, c2 − 12 f 99 sad, and in general
ck −
f skdsad
k!
where k! − 1 ? 2 ? 3 ? 4 ? ∙ ∙ ∙ ? k. The resulting polynomial
Tn sxd − f sad 1 f 9sadsx 2 ad 1
195
f 99sad
f sndsad
sx 2 ad2 1 ∙ ∙ ∙ 1
sx 2 adn
2!
n!
is called the nth-degree Taylor polynomial of f centered at a.
6. F
ind the 8th-degree Taylor polynomial centered at a − 0 for the function f sxd − cos x.
Graph f together with the Taylor polynomials T2 , T4 , T6 , T8 in the viewing rectangle
f25, 5g by f21.4, 1.4g and comment on how well they approximate f .
2
Review
CONCEPT CHECK
1.Write an expression for the slope of the tangent line to the
curve y − f sxd at the point sa, f sadd.
2.Suppose an object moves along a straight line with position
f std at time t. Write an expression for the instantaneous velocity of the object at time t − a. How can you interpret this
velocity in terms of the graph of f ?
3.If y − f sxd and x changes from x 1 to x 2, write expressions for
the following.
(a)The average rate of change of y with respect to x over the
interval fx 1, x 2 g.
(b)The instantaneous rate of change of y with respect to x
at x − x 1.
4.Define the derivative f 9sad. Discuss two ways of interpreting
this number.
5.(a)What does it mean for f to be differentiable at a?
(b)What is the relation between the differentiability and
continuity of a function?
(c)Sketch the graph of a function that is continuous but not
differentiable at a − 2.
6.Describe several ways in which a function can fail to be
differentiable. Illustrate with sketches.
Answers to the Concept Check can be found on the back endpapers.
7.What are the second and third derivatives of a function f ?
If f is the position function of an object, how can you interpret f 0 and f -?
8.State each differentiation rule both in symbols and in words.
(a) The Power Rule
(b) The Constant Multiple Rule
(c) The Sum Rule
(d) The Difference Rule
(e) The Product Rule
(f) The Quotient Rule
(g) The Chain Rule
9.State the derivative of each function.
(a) y − x n
(b) y − sin x
(d) y − tan x
(e) y − csc x
(g) y − cot x
(c) y − cos x
(f) y − sec x
10. Explain how implicit differentiation works.
11. G
ive several examples of how the derivative can be interpreted
as a rate of change in physics, chemistry, biology, economics,
or other sciences.
12. (a)Write an expression for the linearization of f at a.
(b)If y − f sxd, write an expression for the differential dy.
(c)If dx − Dx, draw a picture showing the geometric meanings of Dy and dy.
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196
chapter 2 Derivatives
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true, explain
why. If it is false, explain why or give an example that disproves the
statement.
1.If f is continuous at a, then f is differentiable at a.
2.If f and t are differentiable, then
d
f f sxd 1 tsxdg − f 9sxd 1 t9sxd
dx
|
| |
|
8.If f 9srd exists, then lim x l r f sxd − f srd.
10.
d 2y
−
dx 2
S D
dy
dx
xl2
tsxd 2 ts2d
− 80
x22
2
11. A
n equation of the tangent line to the parabola y − x 2
at s22, 4d is y 2 4 − 2xsx 1 2d.
4.If f and t are differentiable, then
d
f s tsxdd − f 9s tsxdd t9sxd
dx
g
12.
d
f 9sxd
.
sf sxd −
dx
2 sf sxd
5.If f is differentiable, then
d
x 2 1 x − 2x 1 1
dx
9.If tsxd − x 5, then lim
3.If f and t are differentiable, then
d
f f sxd tsxdg − f 9sxd t9sxd
dx
f
7.
d
d
stan2xd −
ssec 2xd
dx
dx
13. The derivative of a polynomial is a polynomial.
14. The derivative of a rational function is a rational function.
d
f 9sxd
f ssx d −
.
6.If f is differentiable, then
dx
2 sx
15. If f sxd − sx 6 2 x 4 d 5, then f s31dsxd − 0.
EXERCISES
1.The displacement (in meters) of an object moving in a straight
line is given by s − 1 1 2t 1 14 t 2, where t is measured in
seconds.
(a)Find the average velocity over each time period.
(i) f1, 3g (ii) f1, 2g (iii) f1, 1.5g (iv) f1, 1.1g
(b) Find the instantaneous velocity when t − 1.
5.The figure shows the graphs of f , f 9, and f 0. Identify each
curve, and explain your choices.
y
a
b
2.The graph of f is shown. State, with reasons, the numbers
at which f is not differentiable.
0
y
_1 0
2
4
6.Find a function f and a number a such that
x
6
lim
h l0
3–4 Trace or copy the graph of the function. Then sketch a graph of
its derivative directly beneath.
y
4.
3.
y
0
x
0
x
c
x
s2 1 hd6 2 64
− f 9sad
h
7.The total cost of repaying a student loan at an interest rate of
r % per year is C − f srd.
(a)What is the meaning of the derivative f 9srd? What are its
units?
(b) What does the statement f 9s10d − 1200 mean?
(c) Is f 9srd always positive or does it change sign?
8.The total fertility rate at time t, denoted by Fstd, is an estimate of the average number of children born to each woman
(assuming that current birth rates remain constant). The graph
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
197
chapter 2 Review
of the total fertility rate in the United States shows
the fluctuations from 1940 to 2010.
(a)Estimate the values of F9s1950d, F9s1965d, and
F9s1987d.
(b) What are the meanings of these derivatives?
(c)Can you suggest reasons for the values of these
derivatives?
y
3.0
2.0
19. y −
baby
boom
3.5
2.5
17. y − x 2 sin x
baby
boomlet
y=F(t)
1.5
1940
1950
1960
1970
1980
1990
2000
2010 t
9.Let Pstd be the percentage of Americans under the age of 18
at time t. The table gives values of this function in census
years from 1950 to 2010.
t
Pstd
t
Pstd
1950
1960
1970
1980
31.1
35.7
34.0
28.0
1990
2000
2010
25.7
25.7
24.0
(a) What is the meaning of P9std? What are its units?
(b) Construct a table of estimated values for P9std.
(c) Graph P and P9.
(d)How would it be possible to get more accurate values
for P9std?
10–11 Find f 9sxd from first principles, that is, directly from the
def­inition of a derivative.
10. f sxd −
42x
31x
11. f sxd − x 3 1 5x 1 4
12. (a)If f sxd − s3 2 5x , use the definition of a derivative
to find f 9sxd.
(b) Find the domains of f and f 9.
(c)Graph f and f 9 on a common screen. Compare the
;
graphs to see whether your answer to part (a) is
reasonable.
13–40 Calculate y9.
13. y − sx 1 x d
1
1
14. y −
2 5 3
sx
sx
x2 2 x 1 2
15. y −
sx
tan x
16. y −
1 1 cos x
2
3 4
1
x2
s7
20. y − sinscos xd
1
sinsx 2 sin xd
21. y − tan s1 2 x
22. y −
23. xy 4 1 x 2 y − x 1 3y
24. y − secs1 1 x 2 d
25. y −
baby
bust
t4 2 1
t4 1 1
S D
18. y − x 1
sec 2
1 1 tan 2
26. x 2 cos y 1 sin 2y − xy
27. y − s1 2 x 21 d21
3
28. y − 1ys
x 1 sx
29. sinsxyd − x 2 2 y
30. y − ssin s x
31. y − cots3x 2 1 5d
32. y −
sx 1 d4
x 4 1 4
33. y − sx cos sx
34. y −
sin mx
x
35. y − tan2ssin d
36. x tan y − y 2 1
5
37. y − s
x tan x
38. y −
39. y − sinstan s1 1 x 3 d
40. y − sin2 scosssin x d
sx 2 1dsx 2 4d
sx 2 2dsx 2 3d
41. If f std − s4t 1 1, find f 99s2d.
42. If tsd − sin , find t99sy6d.
43. Find y99 if x 6 1 y 6 − 1.
44. Find f sndsxd if f sxd − 1ys2 2 xd.
45–46 Find the limit.
sec x
45. lim
x l 0 1 2 sin x
46. lim
tl0
t3
tan3 2t
47–48 Find an equation of the tangent to the curve at the given
point.
47. y − 4 sin2 x, sy6, 1d
48. y −
x2 2 1
, s0, 21d
x2 1 1
49–50 Find equations of the tangent line and normal line to the
curve at the given point.
49. y − s1 1 4 sin x , s0, 1d
50. x 2 1 4xy 1 y 2 − 13, s2, 1d
51. (a) If f sxd − x s5 2 x , find f 9sxd.
(b)Find equations of the tangent lines to the curve
y − x s5 2 x at the points s1, 2d and s4, 4d.
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198
;
;
chapter 2 Derivatives
(c)Illustrate part (b) by graphing the curve and tangent
lines on the same screen.
(d)Check to see that your answer to part (a) is reasonable
by comparing the graphs of f and f 9.
52. (a)If f sxd − 4x 2 tan x, 2y2 , x , y2, find f 9
and f 0.
(b)Check to see that your answers to part (a) are reason;
able by comparing the graphs of f , f 9, and f 99.
61–68 Find f 9 in terms of t9.
61. f sxd − x 2tsxd
62. f sxd − tsx 2 d
63. f sxd − f tsxdg 2
64. f sxd − x a tsx b d
65. f sxd − ts tsxdd
66. f sxd − sins tsxdd
67. f sxd − tssin xd
68. f sxd − tstan sx d
53.At what points on the curve y − sin x 1 cos x,
0 < x < 2, is the tangent line horizontal?
69–71 Find h9 in terms of f 9 and t9.
54. F
ind the points on the ellipse x 2 1 2y 2 − 1 where the
tangent line has slope 1.
69. hsxd −
55. F
ind a parabola y − ax 2 1 bx 1 c that passes through the
point s1, 4d and whose tangent lines at x − 21 and x − 5
have slopes 6 and 22, respectively.
71. hsxd − f s tssin 4xdd
56. H
ow many tangent lines to the curve y − xysx 1 1) pass
through the point s1, 2d? At which points do these tangent
lines touch the curve?
57. If f sxd − sx 2 adsx 2 bdsx 2 cd, show that
f 9sxd
1
1
1
−
1
1
f sxd
x2a
x2b
x2c
58. (a) By differentiating the double-angle formula
cos 2x − cos2x 2 sin2x
obtain the double-angle formula for the sine function.
(b) By differentiating the addition formula
sinsx 1 ad − sin x cos a 1 cos x sin a
obtain the addition formula for the cosine function.
59.Suppose that
f s1d − 2
ts1d − 3
(a)
(b)
(c)
(d)
f 9s1d − 3
t9s1d − 1
f s2d − 1
ts2d − 1
f 9s2d − 2
t9s2d − 4
If Ssxd − f sxd 1 tsxd, find S9s1d.
If Psxd − f sxd tsxd, find P9s2d.
If Qsxd − f sxdytsxd, find Q9s1d.
If Csxd − f stsxdd, find C9s2d.
60.If f and t are the functions whose graphs are shown, let
Psxd − f sxd tsxd, Qsxd − f sxdytsxd, and Csxd − f s tsxdd.
Find (a) P9s2d, (b) Q9s2d, and (c) C9s2d.
y
f
Î
f sxd
tsxd
72.A particle moves along a horizontal line so that its coordinate at time t is x − sb 2 1 c 2 t 2 , t > 0, where b and c
are positive constants.
(a) Find the velocity and acceleration functions.
(b)Show that the particle always moves in the positive
direction.
73.A particle moves on a vertical line so that its coordinate at
time t is y − t 3 2 12t 1 3, t > 0.
(a) Find the velocity and acceleration functions.
(b)When is the particle moving upward and when is it
moving downward?
(c)Find the distance that the particle travels in the time
interval 0 < t < 3.
(d)Graph the position, velocity, and acceleration functions
;
for 0 < t < 3.
(e)When is the particle speeding up? When is it slowing
down?
74. T
he volume of a right circular cone is V − 13 r 2h, where
r is the radius of the base and h is the height.
(a)Find the rate of change of the volume with respect to
the height if the radius is constant.
(b)Find the rate of change of the volume with respect to
the radius if the height is constant.
75.The mass of part of a wire is x s1 1 sx d kilograms, where
x is measured in meters from one end of the wire. Find the
linear density of the wire when x − 4 m.
Csxd − 920 1 2x 2 0.02x 2 1 0.00007x 3
1
1
70. hsxd −
76.The cost, in dollars, of producing x units of a certain commodity is
g
0
f sxd tsxd
f sxd 1 tsxd
x
(a) Find the marginal cost function.
(b) Find C9s100d and explain its meaning.
(c)Compare C9s100d with the cost of producing the
101st item.
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chapter 2 Review
77.The volume of a cube is increasing at a rate of 10 cm3ymin.
How fast is the surface area increasing when the length of
an edge is 30 cm?
78.A paper cup has the shape of a cone with height 10 cm and
radius 3 cm (at the top). If water is poured into the cup at a
rate of 2 cm3ys, how fast is the water level rising when the
water is 5 cm deep?
79.A balloon is rising at a constant speed of 5 ftys. A boy is
cycling along a straight road at a speed of 15 ftys. When he
passes under the balloon, it is 45 ft above him. How fast is
the distance between the boy and the balloon increasing
3 s later?
80. A
waterskier skis over the ramp shown in the figure at a
speed of 30 ftys. How fast is she rising as she leaves the
ramp?
3
83.(a)Find the linearization of f sxd − s
1 1 3x at a − 0.
State the corresponding linear approximation and use
3
it to give an approximate value for s
1.03 .
(b)Determine the values of x for which the linear approxi;
mation given in part (a) is accurate to within 0.1.
84.Evaluate dy if y − x 3 2 2x 2 1 1, x − 2, and dx − 0.2.
85.A window has the shape of a square surmounted by a
semi­circle. The base of the window is measured as having
width 60 cm with a possible error in measurement of
0.1 cm. Use differentials to estimate the maximum error
possible in computing the area of the window.
86–88 Express the limit as a derivative and evaluate.
86. lim
x l1
88. lim
x 17 2 1
x21
l y3
89.Evaluate lim
xl0
81.The angle of elevation of the sun is decreasing at a rate of
0.25 radyh. How fast is the shadow cast by a 400-ft-tall
building increasing when the angle of elevation of the sun
is y6?
; 82.(a)Find the linear approximation to f sxd − s25 2 x 2
near 3.
(b)Illustrate part (a) by graphing f and the linear
approximation.
(c)For what values of x is the linear approximation
accurate to within 0.1?
87. lim
hl0
4
16 1 h 2 2
s
h
cos 2 0.5
2 y3
4 ft
15 ft
199
s1 1 tan x 2 s1 1 sin x
.
x3
90.Suppose f is a differentiable function such that
f s tsxdd − x and f 9sxd − 1 1 f f sxdg 2. Show that
t9sxd − 1ys1 1 x 2 d.
91.Find f 9sxd if it is known that
d
f f s2xdg − x 2
dx
92.Show that the length of the portion of any tangent line to
the astroid x 2y3 1 y 2y3 − a 2y3 cut off by the coordinate
axes is constant.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Problems Plus
Before you look at the example, cover up the solution and try it yourself first.
Example How many lines are tangent to both of the parabolas y − 21 2 x 2 and
y − 1 1 x 2 ? Find the coordinates of the points at which these tangents touch the
parabolas.
SOLUTION To gain insight into this problem, it is essential to draw a diagram. So we
sketch the parabolas y − 1 1 x 2 (which is the standard parabola y − x 2 shifted 1 unit
upward) and y − 21 2 x 2 (which is obtained by reflecting the first parabola about the
x-axis). If we try to draw a line tangent to both parabolas, we soon discover that there
are only two possibilities, as illustrated in Figure 1.
Let P be a point at which one of these tangents touches the upper parabola and let a
be its x-coordinate. (The choice of notation for the unknown is important. Of course we
could have used b or c or x 0 or x1 instead of a. However, it’s not advisable to use x in
place of a because that x could be confused with the variable x in the equation of the
parabola.) Then, since P lies on the parabola y − 1 1 x 2, its y-coordinate must be
1 1 a 2. Because of the symmetry shown in Figure 1, the coordinates of the point Q
where the tangent touches the lower parabola must be s2a, 2s1 1 a 2 dd.
To use the given information that the line is a tangent, we equate the slope of the
line PQ to the slope of the tangent line at P. We have
y
P
1
x
Q
_1
FIGURE 1
mPQ −
1 1 a 2 2 s21 2 a 2 d
1 1 a2
−
a 2 s2ad
a
If f sxd − 1 1 x 2, then the slope of the tangent line at P is f 9sad − 2a. Thus the condition that we need to use is that
1 1 a2
− 2a
a
Solving this equation, we get 1 1 a 2 − 2a 2, so a 2 − 1 and a − 61. Therefore the
points are (1, 2) and s21, 22d. By symmetry, the two remaining points are s21, 2d
and s1, 22d.
Problems
1.Find points P and Q on the parabola y − 1 2 x 2 so that the triangle ABC formed by the
x-axis and the tangent lines at P and Q is an equilateral triangle. (See the figure.)
y
;
A
P
B
C
2.
Find the point where the curves y − x 3 2 3x 1 4 and y − 3sx 2 2 xd are tangent to each
other, that is, have a common tangent line. Illustrate by sketching both curves and the
common tangent.
3.Show that the tangent lines to the parabola y − ax 2 1 bx 1 c at any two points with
x-coordinates p and q must intersect at a point whose x-coordinate is halfway between p
and q.
Q
0
■
x
4.
Show that
d
dx
FIGURE FOR PROBLEM 1
5.
If f sxd − lim
tlx
S
sin2 x
cos2 x
1
1 1 cot x
1 1 tan x
D
− 2cos 2x
sec t 2 sec x
, find the value of f 9sy4d.
t2x
6.Find the values of the constants a and b such that
lim
xl0
3
ax 1 b 2 2
5
s
−
x
12
200
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
7.Prove that
dn
ssin4 x 1 cos4 xd − 4n21 coss4x 1 ny2d.
dx n
8.If f is differentiable at a, where a . 0, evaluate the following limit in terms of f 9sad:
lim
y
xla
y=≈
1
10.Find all values of c such that the parabolas y − 4x 2 and x − c 1 2y 2 intersect each other at
right angles.
1
11.How many lines are tangent to both of the circles x 2 1 y 2 − 4 and x 2 1 sy 2 3d 2 − 1?
At what points do these tangent lines touch the circles?
x
x 46 1 x 45 1 2
, calculate f s46ds3d. Express your answer using factorial notation:
11x
n! − 1 ? 2 ? 3 ? ∙ ∙ ∙ ? sn 2 1d ? n.
12.If f sxd −
FIGURE FOR PROBLEM 9
y
13. The
figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length
1.2 m. The pin P slides back and forth along the x-axis as the wheel rotates counter­
clockwise at a rate of 360 revolutions per minute.
(a)Find the angular velocity of the connecting rod, dydt, in radians per second,
when − y3.
(b) Express the distance x − OP in terms of .
(c) Find an expression for the velocity of the pin P in terms of .
A
å
¨
sx 2 sa
9.
The figure shows a circle with radius 1 inscribed in the parabola y − x 2. Find the center of
the circle.
0
O
f sxd 2 f sad
|
P (x, 0) x
|
14. T
angent lines T1 and T2 are drawn at two points P1 and P2 on the parabola y − x 2 and they
intersect at a point P. Another tangent line T is drawn at a point between P1 and P2 ; it intersects T1 at Q1 and T2 at Q2. Show that
FIGURE FOR PROBLEM 13
| PQ | 1 | PQ | − 1
| PP | | PP |
y
yT
0
yN
P
xN
N
2
2
15. L
et T and N be the tangent and normal lines to the ellipse x 2y9 1 y 2y4 − 1 at any point P
on the ellipse in the first quadrant. Let x T and yT be the x- and y-intercepts of T and x N and
yN be the intercepts of N. As P moves along the ellipse in the first quadrant (but not on the
axes), what values can x T , yT , x N, and yN take on? First try to guess the answers just by looking at the figure. Then use calculus to solve the problem and see how good your intuition is.
T
2
1
1
3
FIGURE FOR PROBLEM 15
xT
x
16. Evaluate lim
xl0
sins3 1 xd2 2 sin 9
.
x
17. (a)Use the identity for tansx 2 yd (see Equation 14b in Appendix D) to show that if two
lines L 1 and L 2 intersect at an angle , then
tan −
m 2 2 m1
1 1 m1 m 2
where m1 and m 2 are the slopes of L 1 and L 2, respectively.
(b)The angle between the curves C1 and C2 at a point of intersection P is defined to be
the angle between the tangent lines to C1 and C2 at P (if these tangent lines exist). Use
part (a) to find, correct to the nearest degree, the angle between each pair of curves at
each point of intersection.
(i) y − x 2 and y − sx 2 2d2
(ii) x 2 2 y 2 − 3 and x 2 2 4x 1 y 2 1 3 − 0
201
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y
0
y=›
∫
P(⁄, ›)
å
18.Let Psx 1, y1d be a point on the parabola y 2 − 4px with focus Fs p, 0d. Let be the angle
between the parabola and the line segment FP, and let be the angle between the horizontal
line y − y1 and the parabola as in the figure. Prove that − . (Thus, by a prin­ciple of geometrical optics, light from a source placed at F will be reflected along a line parallel to the
x-axis. This explains why paraboloids, the surfaces obtained by rotating parabolas about
their axes, are used as the shape of some automobile headlights and mirrors for
telescopes.)
x
F(p, 0)
¥=4px
19.
Suppose that we replace the parabolic mirror of Problem 18 by a spherical mirror. Although
the mirror has no focus, we can show the existence of an approximate focus. In the figure, C
is a semicircle with center O. A ray of light coming in toward the mirror parallel to the axis
along the line PQ will be reflected to the point R on the axis so that /PQO − /OQR (the
angle of incidence is equal to the angle of reflection). What happens to the point R as P is
taken closer and closer to the axis?
FIGURE FOR PROBLEM 18
Q
¨
A
R
P
¨
20. If f and t are differentiable functions with f s0d − ts0d − 0 and t9s0d ± 0, show that
lim
O
xl0
C
21. Evaluate lim
xl0
f sxd
f 9s0d
−
tsxd
t9s0d
sinsa 1 2xd 2 2 sinsa 1 xd 1 sin a
.
x2
22. G
iven an ellipse x 2ya 2 1 y 2yb 2 − 1, where a ± b, find the equation of the set of all points
from which there are two tangents to the curve whose slopes are (a) reciprocals and
(b) negative reciprocals.
FIGURE FOR PROBLEM 19
23. Find the two points on the curve y − x 4 2 2x 2 2 x that have a common tangent line.
24. S
uppose that three points on the parabola y − x 2 have the property that their normal lines
intersect at a common point. Show that the sum of their x-coordinates is 0.
25.A lattice point in the plane is a point with integer coordinates. Suppose that circles with
radius r are drawn using all lattice points as centers. Find the smallest value of r such that
any line with slope 25 intersects some of these circles.
26. A
cone of radius r centimeters and height h centimeters is lowered point first at a rate of
1 cmys into a tall cylinder of radius R centimeters that is partially filled with water. How fast
is the water level rising at the instant the cone is completely submerged?
27.A container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top. It
is partially filled with a liquid that oozes through the sides at a rate proportional to the area
of the container that is in contact with the liquid. (The surface area of a cone is rl, where
r is the radius and l is the slant height.) If we pour the liquid into the container at a rate of
2 cm3ymin, then the height of the liquid decreases at a rate of 0.3 cmymin when the height
is 10 cm. If our goal is to keep the liquid at a constant height of 10 cm, at what rate should
we pour the liquid into the container?
CAS
28. (a)The cubic function f sxd − xsx 2 2dsx 2 6d has three distinct zeros: 0, 2, and 6. Graph
f and its tangent lines at the average of each pair of zeros. What do you notice?
(b)Suppose the cubic function f sxd − sx 2 adsx 2 bdsx 2 cd has three distinct zeros:
a, b, and c. Prove, with the help of a computer algebra system, that a tangent line
drawn at the average of the zeros a and b intersects the graph of f at the third zero.
202
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3
Applications of Differentiation
When flying, some small
birds—like the finch pictured
here—alternate between
flapping their wings and
keeping them folded while
gliding. In the project on
page 271, we will investigate
how frequently a bird should
flap its wings in order to
minimize the energy required.
© Targn Pleiades / Shutterstock.com
We have already investigated some of the applications of derivatives, but now that we
know the differen­tiation rules we are in a better position to pursue the applications of differentiation in greater depth. Here we learn how derivatives affect the shape of a graph of a function and,
in particular, how they help us locate maximum and minimum values of functions. Many practical problems require us to minimize a cost or maximize an area or somehow find the best possible
outcome of a situation. In particular, we will be able to investigate the optimal shape of a can and
to explain the location of rainbows in the sky.
203
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204
Chapter 3 Applications of Differentiation
Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Here
are examples of such problems that we will solve in this chapter:
•
•
hat is the shape of a can that minimizes manufacturing costs?
W
What is the maximum acceleration of a space shuttle? (This is an important
question to the astronauts who have to withstand the effects of acceleration.)
What is the radius of a contracted windpipe that expels air most rapidly during
a cough?
At what angle should blood vessels branch so as to minimize the energy
expended by the heart in pumping blood?
•
•
These problems can be reduced to finding the maximum or minimum values of a function. Let’s first explain exactly what we mean by maximum and minimum values.
We see that the highest point on the graph of the function f shown in Figure 1 is the
point s3, 5d. In other words, the largest value of f is f s3d − 5. Likewise, the smallest
value is f s6d − 2. We say that f s3d − 5 is the absolute maximum of f and f s6d − 2 is
the absolute minimum. In general, we use the following definition.
y
4
2
0
4
2
x
6
1 Definition Let c be a number in the domain D of a function f. Then f scd is
the
• absolute maximum value of f on D if f scd > f sxd for all x in D.
Figure 1
•
y
f(d)
f(a)
a
0
b
c
d
e
x
Figure 2
Abs min f sad, abs max f sdd,
loc min f scd, f sed, loc max f sbd, f sdd
6
4
loc
min
2
0
Figure 3
loc
max
loc
and
abs
min
I
J
K
4
8
12
An absolute maximum or minimum is sometimes called a global maximum or minimum. The maximum and minimum values of f are called extreme values of f .
Figure 2 shows the graph of a function f with absolute maximum at d and absolute
minimum at a. Note that sd, f sddd is the highest point on the graph and sa, f sadd is the
lowest point. In Figure 2, if we consider only values of x near b [for instance, if we
restrict our attention to the interval sa, cd], then f sbd is the largest of those values of f sxd
and is called a local maximum value of f. Likewise, f scd is called a local minimum value
of f because f scd < f sxd for x near c [in the interval sb, dd, for instance]. The function f
also has a local minimum at e. In general, we have the following definition.
2 Definition The number f scd is a
•
•
y
x
absolute minimum value of f on D if f scd < f sxd for all x in D.
local maximum value of f if f scd > f sxd when x is near c.
local minimum value of f if f scd < f sxd when x is near c.
In Definition 2 (and elsewhere), if we say that something is true near c, we mean that it
is true on some open interval containing c. For instance, in Figure 3 we see that f s4d − 5
is a local minimum because it’s the smallest value of f on the interval I. It’s not the absolute minimum because f sxd takes smaller values when x is near 12 (in the interval K,
for instance). In fact f s12d − 3 is both a local minimum and the absolute minimum.
Similarly, f s8d − 7 is a local maximum, but not the absolute maximum because f takes
larger values near 1.
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Section 3.1 Maximum and Minimum Values
205
Example 1 The function f sxd − cos x takes on its (local and absolute) maximum value of 1 infinitely many times, since cos 2n − 1 for any integer n and
21 < cos x < 1 for all x. (See Figure 4.) Likewise, coss2n 1 1d − 21 is its minimum value, where n is any integer.
y
Local and absolute maximum
0
π
FIGURE 4
y=≈
0
3π
Local and absolute minimum
y − cosx
y
2π
x
n
Example 2 If f sxd − x 2, then f sxd > f s0d because x 2 > 0 for all x. Therefore
x
FIGURE 5 Mimimum value 0, no maximum
y
f s0d − 0 is the absolute (and local) minimum value of f. This corresponds to the fact
that the origin is the lowest point on the parabola y − x 2. (See Figure 5.) However,
there is no highest point on the parabola and so this function has no maximum value. n
Example 3 From the graph of the function f sxd − x 3, shown in Figure 6, we see that
this function has neither an absolute maximum value nor an absolute minimum value.
In fact, it has no local extreme values either.
n
Example 4 The graph of the function
y=˛
f sxd − 3x 4 2 16x 3 1 18x 2 21 < x < 4
0
x
FIGURE 6 No mimimum, no maximum
is shown in Figure 7. You can see that f s1d − 5 is a local maximum, whereas the
absolute maximum is f s21d − 37. (This absolute maximum is not a local maximum
because it occurs at an endpoint.) Also, f s0d − 0 is a local minimum and f s3d − 227
is both a local and an absolute minimum. Note that f has neither a local nor an absolute
maximum at x − 4.
y
(_1, 37)
y=3x$-16˛+18≈
(1, 5)
_1
FIGURE 7 1
2
3
4
5
x
(3, _27)
n
We have seen that some functions have extreme values, whereas others do not. The
following theorem gives conditions under which a function is guaranteed to possess
extreme values.
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206
Chapter 3 Applications of Differentiation
3 The Extreme Value Theorem If f is continuous on a closed interval fa, bg,
then f attains an absolute maximum value f scd and an absolute minimum value
f sdd at some numbers c and d in fa, bg.
The Extreme Value Theorem is illustrated in Figure 8. Note that a function can attain
an extreme value more than once. Although the Extreme Value Theorem is intuitively
very plausible, it is difficult to prove and so we omit the proof.
y
y
y
FIGURE 8 Functions continuous on a closed
interval always attain extreme values.
0
a
c
d b
0
x
a
c
d=b
x
0
a c¡
d
c™ b
x
Figures 9 and 10 show that a function need not possess extreme values if either
hypothe­sis (continuity or closed interval) is omitted from the Extreme Value Theorem.
y
y
3
1
0
1
2
x
FIGURE 9
This function has minimum value
f(2)=0, but no maximum value.
y
{c, f(c)}
{d, f (d )}
0
c
FIGURE 11
d
x
0
2
x
FIGURE 10
This continuous function g has
no maximum or minimum.
The function f whose graph is shown in Figure 9 is defined on the closed interval
[0, 2] but has no maximum value. (Notice that the range of f is [0, 3). The function
takes on val­ues arbitrarily close to 3, but never actually attains the value 3.) This does
not contradict the Extreme Value Theorem because f is not continuous. [Nonetheless, a
discontinuous function could have maximum and minimum values. See Exercise 13(b).]
The function t shown in Figure 10 is continuous on the open interval (0, 2) but has
neither a maximum nor a minimum value. [The range of t is s1, `d. The function takes
on arbitrarily large values.] This does not contradict the Extreme Value Theorem because
the interval (0, 2) is not closed.
The Extreme Value Theorem says that a continuous function on a closed interval has a
maximum value and a minimum value, but it does not tell us how to find these extreme
values. Notice in Figure 8 that the absolute maximum and minimum values that are
between a and b occur at local maximum or minimum values, so we start by looking for
local extreme values.
Figure 11 shows the graph of a function f with a local maximum at c and a local
minimum at d. It appears that at the maximum and minimum points the tangent lines are
hor­izontal and therefore each has slope 0. We know that the derivative is the slope of the
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.1 Maximum and Minimum Values
207
tan­gent line, so it appears that f 9scd − 0 and f 9sdd − 0. The following theorem says that
this is always true for differentiable functions.
Fermat
Fermat’s Theorem is named after
Pierre Fermat (1601–1665), a French
lawyer who took up mathematics as
a hobby. Despite his amateur status,
Fermat was one of the two inventors
of analytic geometry (Descartes was
the other). His methods for finding
tangents to curves and maximum and
minimum values (before the invention
of limits and derivatives) made him a
forerunner of Newton in the creation
of differ­ential calculus.
4 Fermat’s Theorem If f has a local maximum or minimum at c, and if f 9scd
exists, then f 9scd − 0.
Proof Suppose, for the sake of definiteness, that f has a local maximum at c. Then,
according to Definition 2, f scd > f sxd if x is sufficiently close to c. This implies that if h
is sufficiently close to 0, with h being positive or negative, then
f scd > f sc 1 hd
and therefore
f sc 1 hd 2 f scd < 0
5
We can divide both sides of an inequality by a positive number. Thus, if h . 0 and h is
sufficiently small, we have
f sc 1 hd 2 f scd
<0
h
Taking the right-hand limit of both sides of this inequality (using Theorem 1.6.2), we get
lim
hl 01
f sc 1 hd 2 f scd
< lim1 0 − 0
h l0
h
But since f 9scd exists, we have
f 9scd − lim
hl0
f sc 1 hd 2 f scd
f sc 1 hd 2 f scd
− lim1
h l0
h
h
and so we have shown that f 9scd < 0.
If h , 0, then the direction of the inequality (5) is reversed when we divide by h:
f sc 1 hd 2 f scd
> 0 h , 0
h
So, taking the left-hand limit, we have
f 9scd − lim
hl0
f sc 1 hd 2 f scd
f sc 1 hd 2 f scd
− lim2
>0
h
l
0
h
h
We have shown that f 9scd > 0 and also that f 9scd < 0. Since both of these inequalities
must be true, the only possibility is that f 9scd − 0.
We have proved Fermat’s Theorem for the case of a local maximum. The case of
a local minimum can be proved in a similar manner, or we could use Exercise 70 to
n
deduce it from the case we have just proved (see Exercise 71).
The following examples caution us against reading too much into Fermat’s Theorem:
We can’t expect to locate extreme values simply by setting f 9sxd − 0 and solving for x.
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208
Chapter 3 Applications of Differentiation
y
Example 5 If f sxd − x 3, then f 9sxd − 3x 2, so f 9s0d − 0. But f has no maximum
or minimum at 0, as you can see from its graph in Figure 12. (Or observe that x 3 . 0
for x . 0 but x 3 , 0 for x , 0.) The fact that f 9s0d − 0 simply means that the curve
y − x 3 has a horizontal tangent at s0, 0d. Instead of having a maximum or minimum at
s0, 0d, the curve crosses its horizontal tangent there.
n
y=˛
0
x
| |
Example 6 The function f sxd − x has its (local and absolute) minimum value
at 0, but that value can’t be found by setting f 9sxd − 0 because, as was shown in
Example 2.2.5, f 9s0d does not exist. (See Figure 13.)
Figure 12
If f sxd − x 3, then f 9s0d − 0, but f
has no maximum or minimum.
warning Examples 5 and 6 show that we must be careful when using Fermat’s Theorem. Example 5 demonstrates that even when f 9scd − 0 there need not be a maximum
or minimum at c. (In other words, the converse of Fermat’s Theorem is false in general.) Fur­thermore, there may be an extreme value even when f 9scd does not exist (as in
Example 6).
Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the numbers c where f 9scd − 0 or where f 9scd does not exist. Such numbers
are given a special name.
y
y=| x|
0
n
x
Figure 13
6 Definition A critical number of a function f is a number c in the domain of
f such that either f 9scd − 0 or f 9scd does not exist.
| |
If f sxd − x , then f s0d − 0 is
a minimum value, but f 9s0d does
not exist.
Example 7 Find the critical numbers of f sxd − x 3y5s4 2 xd.
SOLUTION The Product Rule gives
Figure 14 shows a graph of the function
f in Example 7. It supports our answer
because there is a horizontal tangent
when x − 1.5 fwhere f 9sxd − 0g and
a vertical tangent when x − 0 fwhere
f 9sxd is undefinedg.
3.5
_0.5
5
f 9sxd − x 3y5s21d 1 s4 2 xd(53 x22y5) − 2x 3y5 1
−
3s4 2 xd
5x 2 y5
25x 1 3s4 2 xd
12 2 8x
−
5x 2y5
5x 2y5
[The same result could be obtained by first writing f sxd − 4x 3y5 2 x 8y5.] Therefore
f 9sxd − 0 if 12 2 8x − 0, that is, x − 32, and f 9sxd does not exist when x − 0. Thus the
critical numbers are 32 and 0.
n
In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare
Definition 6 with Theorem 4):
_2
Figure 14
7 If f has a local maximum or minimum at c, then c is a critical number of f.
To find an absolute maximum or minimum of a continuous function on a closed
interval, we note that either it is local [in which case it occurs at a critical number by (7)]
or it occurs at an endpoint of the interval, as we see from the examples in Figure 8. Thus
the following three-step procedure always works.
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Section 3.1 Maximum and Minimum Values
209
The Closed Interval Method To find the absolute maximum and minimum
values of a continuous function f on a closed interval fa, bg:
1. Find the values of f at the critical numbers of f in sa, bd.
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum value;
the smallest of these values is the absolute minimum value.
Example 8 Find the absolute maximum and minimum values of the function
f sxd − x 3 2 3x 2 1 1 212 < x < 4
f
g
SOLUTION Since f is continuous on 2 12 , 4 , we can use the Closed Interval Method:
f sxd − x 3 2 3x 2 1 1
f 9sxd − 3x 2 2 6x − 3xsx 2 2d
Since f 9sxd exists for all x, the only critical numbers of f occur when f 9sxd − 0, that is,
x − 0 or x − 2. Notice that each of these critical numbers lies in the interval s212 , 4d.
The values of f at these critical numbers are
y
20
y=˛-3≈+1
(4, 17)
15
10
5
_1 0
_5
f s0d − 1 f s2d − 23
The values of f at the endpoints of the interval are
1
f s212 d − 18 f s4d − 17
2
(2, _3)
3
x
4
FIGURE 15 Comparing these four numbers, we see that the absolute maximum value is f s4d − 17
and the absolute minimum value is f s2d − 23.
Note that in this example the absolute maximum occurs at an endpoint, whereas the
absolute minimum occurs at a critical number. The graph of f is sketched in Figure 15.
n
If you have a graphing calculator or a computer with graphing software, it is possible
to estimate maximum and minimum values very easily. But, as the next example shows,
calculus is needed to find the exact values.
Example 9
(a) Use a graphing device to estimate the absolute minimum and maximum values of
the function f sxd − x 2 2 sin x, 0 < x < 2.
(b) Use calculus to find the exact minimum and maximum values.
8
0
_1
FIGURE 16 2π
SOLUTION (a) Figure 16 shows a graph of f in the viewing rectangle f0, 2g by f21, 8g. By
moving the cursor close to the maximum point, we see that the y-coordinates don’t
change very much in the vicinity of the maximum. The absolute maximum value is
about 6.97 and it occurs when x < 5.2. Similarly, by moving the cursor close to the
minimum point, we see that the absolute minimum value is about 20.68 and it occurs
when x < 1.0. It is possible to get more accurate estimates by zooming in toward the
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210
Chapter 3 Applications of Differentiation
maximum and minimum points (or using a built-in maximum or minimum feature),
but instead let’s use calculus.
(b) The function f sxd − x 2 2 sin x is continuous on f0, 2g. Since
f 9sxd − 1 2 2 cos x, we have f 9sxd − 0 when cos x − 12 and this occurs when
x − y3 or 5y3. The values of f at these critical numbers are
f sy3d −
and
f s5y3d −
2 2 sin
−
2 s3 < 20.684853
3
3
3
5
5
5
2 2 sin
−
1 s3 < 6.968039
3
3
3
The values of f at the endpoints are
f s0d − 0 and f s2d − 2 < 6.28
Comparing these four numbers and using the Closed Interval Method, we see that the
absolute minimum value is f sy3d − y3 2 s3 and the absolute maximum value is
f s5y3d − 5y3 1 s3 . The values from part (a) serve as a check on our work.
n
Example 10 The Hubble Space Telescope was deployed on April 24, 1990, by the
space shuttle Discovery. A model for the velocity of the shuttle during this mission,
from liftoff at t − 0 until the solid rocket boosters were jettisoned at t − 126 seconds,
is given by
vstd − 0.001302t 3 2 0.09029t 2 1 23.61t 2 3.083
NASA
(in feet per second). Using this model, estimate the absolute maximum and minimum
values of the acceleration of the shuttle between liftoff and the jettisoning of the
boosters.
SOLUTION We are asked for the extreme values not of the given velocity function,
but rather of the acceleration function. So we first need to differentiate to find the
acceleration:
astd − v9std −
d
s0.001302t 3 2 0.09029t 2 1 23.61t 2 3.083d
dt
− 0.003906t 2 2 0.18058t 1 23.61
We now apply the Closed Interval Method to the continuous function a on the interval
0 < t < 126. Its derivative is
a9std − 0.007812t 2 0.18058
The only critical number occurs when a9std − 0:
t1 −
0.18058
< 23.12
0.007812
Evaluating astd at the critical number and at the endpoints, we have
as0d − 23.61 ast1 d < 21.52 as126d < 62.87
So the maximum acceleration is about 62.87 ftys2 and the minimum acceleration is
about 21.52 ftys2.
n
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211
Section 3.1 Maximum and Minimum Values
1.Explain the difference between an absolute minimum and a
local minimum.
(c)Sketch the graph of a function that has a local maximum
at 2 and is not continuous at 2.
2.Suppose f is a continuous function defined on a closed
interval fa, bg.
(a)What theorem guarantees the existence of an absolute
max­imum value and an absolute minimum value for f ?
(b)What steps would you take to find those maximum and
minimum values?
12. (a)Sketch the graph of a function on [21, 2] that has an
absolute maximum but no local maximum.
(b)Sketch the graph of a function on [21, 2] that has a local
maximum but no absolute maximum.
3–4 For each of the numbers a, b, c, d, r, and s, state whether the
function whose graph is shown has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum
nor a minimum.
3.
4.
y
0 a b
c d
r
y
0
s x
a
b
c d
r
s x
5–6 Use the graph to state the absolute and local maximum and
minimum values of the function.
5.
y
6.
y
0
15–28 Sketch the graph of f by hand and use your sketch to
find the absolute and local maximum and minimum values of f.
(Use the graphs and transformations of Sections 1.2 and 1.3.)
15. f sxd − 12 s3x 2 1d, x < 3
16. f sxd − 2 2 13 x, x > 22
17. f sxd − 1yx, x > 1
19. f sxd − sin x, 0 < x , y2
20. f sxd − sin x, 0 , x < y2
y=ƒ
1
14. (a)Sketch the graph of a function that has two local maxima, one local minimum, and no absolute minimum.
(b)Sketch the graph of a function that has three local minima, two local maxima, and seven critical numbers.
18. f sxd − 1yx, 1 , x , 3
y=©
1
13. (a)Sketch the graph of a function on [21, 2] that has an
absolute maximum but no absolute minimum.
(b)Sketch the graph of a function on [21, 2] that is discontinuous but has both an absolute maximum and an
absolute minimum.
21. f sxd − sin x, 2y2 < x < y2
1
x
0
1
x
22. f std − cos t, 23y2 < t < 3y2
23. f sxd − 1 1 sx 1 1d 2, 22 < x , 5
7–10 Sketch the graph of a function f that is continuous on [1, 5]
and has the given properties.
7.Absolute maximum at 5, absolute minimum at 2,
local maximum at 3, local minima at 2 and 4
8.Absolute maximum at 4, absolute minimum at 5,
local maximum at 2, local minimum at 3
9.Absolute minimum at 3, absolute maximum at 4,
local maximum at 2
10.Absolute maximum at 2, absolute minimum at 5, 4 is a
critical number but there is no local maximum or minimum
there.
11. (a)Sketch the graph of a function that has a local maximum
at 2 and is differentiable at 2.
(b)Sketch the graph of a function that has a local maximum
at 2 and is continuous but not differentiable at 2.
| |
24. f sxd − x
25. f sxd − 1 2 sx
26. f sxd − 1 2 x 3
27. f sxd −
28. f sxd −
H
H
x2
if 21 < x < 0
2 2 3x if 0 , x < 1
2x 1 1 if 0 < x , 1
4 2 2x if 1 < x < 3
29–42 Find the critical numbers of the function.
29. f sxd − 4 1 13 x 2 12 x 2
30. f sxd − x 3 1 6x 2 2 15x
31. f sxd − 2x 3 2 3x 2 2 36x
32. f sxd − 2x 3 1 x 2 1 2x
33. tstd − t 4 1 t 3 1 t 2 1 1
34. tstd − 3t 2 4
|
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
|
212
Chapter 3 Applications of Differentiation
35. tsyd −
y21
y2 2 y 1 1
36. hs pd −
p21
p2 1 4
60. f sxd − x 4 2 3x 3 1 3x 2 2 x, 0 < x < 2
61. f sxd − x sx 2 x 2
3
38. tsxd − s
4 2 x2
37. hstd − t 3y4 2 2 t 1y4
62. f sxd − x 2 2 cos x, 22 < x < 0
39. Fsxd − x 4y5sx 2 4d 2 40. tsd − 4 2 tan 41. f sd − 2 cos 1 sin242. tsxd − s1 2 x 2
63.Between 0°C and 30°C, the volume V (in cubic centimeters)
of 1 kg of water at a temperature T is given approximately by
the formula
; 43–44 A formula for the derivative of a function f is given. How
many critical numbers does f have?
43. f 9sxd − 1 1
V − 999.87 2 0.06426T 1 0.0085043T 2 2 0.0000679T 3
210 sin x
x 2 2 6x 1 10
Find the temperature at which water has its maximum density.
100 cos 2 x
21
10 1 x 2
44. f 9sxd −
45–56 Find the absolute maximum and absolute minimum values
of f on the given interval.
where is a positive constant called the coefficient of friction
and where 0 < < y2. Show that F is minimized when
tan − .
45. f sxd − 12 1 4x 2 x 2, f0, 5g
46. f sxd − 5 1 54x 2 2x 3, f0, 4g
47. f sxd − 2x 3 2 3x 2 2 12x 1 1, f22, 3g
65.The water level, measured in feet above mean sea level, of
Lake Lanier in Georgia, USA, during 2012 can be modeled by
the function
48. f sxd − x 2 6x 1 5, f23, 5g
3
2
49. f sxd − 3x 4 2 4x 3 2 12x 2 1 1, f22, 3g
Lstd − 0.01441t 3 2 0.4177t 2 1 2.703t 1 1060.1
50. f std − st 2 2 4d 3, f22, 3g
1
, f0.2, 4g
x
x
52. f sxd − 2
, f0, 3g
x 2x11
where t is measured in months since January 1, 2012. Estimate
when the water level was highest during 2012.
51. f sxd − x 1
3
53. f std − t 2 s
t , f21, 4g
54. f std −
st
, f0, 2g
1 1 t2
55. f std − 2 cos t 1 sin 2t, f0, y2g
56. f std − t 1 cot sty2d, fy4, 7y4g
; 66.On May 7, 1992, the space shuttle Endeavour was launched
on mission STS-49, the purpose of which was to install a new
perigee kick motor in an Intelsat communications satellite.
The table gives the velocity data for the shuttle between liftoff
and the jettisoning of the solid rocket boosters.
(a)Use a graphing calculator or computer to find the cubic
polynomial that best models the velocity of the shuttle for
the time interval t [ f0, 125g. Then graph this polynomial.
(b)Find a model for the acceleration of the shuttle and use it
to estimate the maximum and minimum values of the
acceleration during the first 125 seconds.
57.If a and b are positive numbers, find the maximum value
of f sxd − x as1 2 xd b, 0 < x < 1.
; 58.Use a graph to estimate the critical numbers of
f sxd − 1 1 5x 2 x 3 correct to one decimal place.
|
|
; 59–62
(a) Use a graph to estimate the absolute maximum and
minimum values of the function to two decimal places.
(b) Use calculus to find the exact maximum and minimum values.
59. f sxd − x 5 2 x 3 1 2, 21 < x < 1
64.An object with weight W is dragged along a horizontal plane
by a force acting along a rope attached to the object. If the
rope makes an angle with the plane, then the magnitude of
the force is
W
F−
sin 1 cos Event
Time (s)
Velocity (ftys)
Launch
Begin roll maneuver
End roll maneuver
Throttle to 89%
Throttle to 67%
Throttle to 104%
Maximum dynamic pressure
Solid rocket booster separation
0
10
15
20
32
59
62
125
0
185
319
447
742
1325
1445
4151
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Applied Project The Calculus of Rainbows
67.When a foreign object lodged in the trachea (windpipe) forces
a person to cough, the diaphragm thrusts upward causing an
increase in pressure in the lungs. This is accompanied by a
contraction of the trachea, making a narrower channel for
the expelled air to flow through. For a given amount of air to
escape in a fixed time, it must move faster through the narrower
channel than the wider one. The greater the velocity of the airstream, the greater the force on the foreign object. X rays show
that the radius of the circular tracheal tube contracts to about
two-thirds of its normal radius during a cough. According to
a mathematical model of coughing, the velocity v of the airstream is related to the radius r of the trachea by the equation
68.Show that 5 is a critical number of the function
tsxd − 2 1 sx 2 5d 3
but t does not have a local extreme value at 5.
69.Prove that the function
f sxd − x 101 1 x 51 1 x 1 1
has neither a local maximum nor a local minimum.
70.If f has a local minimum value at c, show that the function
tsxd − 2f sxd has a local maximum value at c.
vsrd − ksr0 2 rdr 2 12 r0 < r < r0
where k is a constant and r0 is the normal radius of the trachea.
The restriction on r is due to the fact that the tracheal wall
stiffens under pressure and a contraction greater than 12 r0 is
prevented (otherwise the person would suffocate).
(a)Determine the value of r in the interval 12 r0 , r0 at which
v has an absolute maximum. How does this compare with
experimental evidence?
(b)What is the absolute maximum value of v on the
interval?
(c) Sketch the graph of v on the interval f0, r0 g.
f
213
g
71.Prove Fermat’s Theorem for the case in which f has a
local minimum at c.
72.A cubic function is a polynomial of degree 3; that is, it has
the form f sxd − ax 3 1 bx 2 1 cx 1 d, where a ± 0.
(a)Show that a cubic function can have two, one, or no
critical number(s). Give examples and sketches to
illustrate the three possibilities.
(b)How many local extreme values can a cubic function
have?
The calculus of rainbows
APPLIED Project
Rainbows are created when raindrops scatter sunlight. They have fascinated mankind since
ancient times and have inspired attempts at scientific explanation since the time of Aristotle. In
this project we use the ideas of Descartes and Newton to explain the shape, location, and colors
of rainbows.
å A
from
sun
∫
∫
O
B
∫
D(å )
∫
å
to
observer
C
Formation of the primary rainbow
1.The figure shows a ray of sunlight entering a spherical raindrop at A. Some of the light
is reflected, but the line AB shows the path of the part that enters the drop. Notice
that the light is refracted toward the normal line AO and in fact Snell’s Law says that
sin − k sin , where is the angle of incidence, is the angle of refraction, and k < 43
is the index of refraction for water. At B some of the light passes through the drop and is
refracted into the air, but the line BC shows the part that is reflected. (The angle of incidence equals the angle of reflection.) When the ray reaches C, part of it is reflected, but for
the time being we are more interested in the part that leaves the raindrop at C. (Notice that
it is refracted away from the normal line.) The angle of deviation Dsd is the amount of
clockwise rotation that the ray has undergone during this three-stage process. Thus
Dsd − s 2 d 1 s 2 2d 1 s 2 d − 1 2 2 4
Show that the minimum value of the deviation is Dsd < 1388 and occurs when < 59.48.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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214
Chapter 3 Applications of Differentiation
The significance of the minimum deviation is that when < 59.48 we have D9sd < 0,
so DDyD < 0. This means that many rays with < 59.48 become deviated by approximately the same amount. It is the concentration of rays coming from near the direction of
minimum deviation that creates the brightness of the primary rainbow. The figure at the left
shows that the angle of elevation from the observer up to the highest point on the rainbow
is 180 8 2 1388 − 428. (This angle is called the rainbow angle.)
rays from sun
138°
rays from sun
42°
observer
C
laboratory
Project
∫
D
∫
∫
å
to
observer
from
sun
å
∫
∫
∫
A
Formation of the secondary rainbow
2.Problem 1 explains the location of the primary rainbow, but how do we explain the colors?
Sunlight comprises a range of wavelengths, from the red range through orange, yellow,
green, blue, indigo, and violet. As Newton discovered in his prism experiments of 1666, the
index of refraction is different for each color. (The effect is called dispersion.) For red light
the refractive index is k < 1.3318, whereas for violet light it is k < 1.3435. By repeating
the calculation of Problem 1 for these values of k, show that the rainbow angle is about
42.38 for the red bow and 40.68 for the violet bow. So the rainbow really consists of seven
bows corresponding to the seven colors.
Theindividual
Paradox
3.PerhapsFigure
you have
seen a data
fainter
secondary
rainbow above
the primary
bow. That
Bimportant.
1 displays
from
an observational
study that
clearly depicts
thisresults
trend.
from the part of a ray that enters a raindrop and is refracted at A, reflected twice (at B and
C),
andthe
refracted
it leaves
thecorresponds
drop at D (see
theinitial
figureexpectationt.
at the left). This time the devia1.
Draw
causal as
diagram
that
to the
tion angle Dsd is the total amount of counterclockwise rotation that the ray undergoes in
2.
Suppose.
this
four-stage process. Show that
B
3.
Suppose.
Dsd − 2 2 6 1 2
and Dsd has a minimum value when
cos −
Î
k2 2 1
8
Taking k − 43, show that the minimum deviation is about 1298 and so the rainbow angle for
the secondary rainbow is about 518, as shown in the figure at the left.
4.Show that the colors in the secondary rainbow appear in the opposite order from those in
the primary rainbow.
© Pichugin Dmitry / Shutterstock.com
42° 51°
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215
Section 3.2 The Mean Value Theorem
We will see that many of the results of this chapter depend on one central fact, which is
called the Mean Value Theorem. But to arrive at the Mean Value Theorem we first need
the following result.
Rolle’s Theorem Let f be a function that satisfies the following three hypotheses:
1. f is continuous on the closed interval fa, bg.
2. f is differentiable on the open interval sa, bd.
3. f sad − f sbd
Then there is a number c in sa, bd such that f 9scd − 0.
Rolle
Rolle’s Theorem was first published
in 1691 by the French mathematician
Michel Rolle (1652–1719) in a book
entitled Méthode pour resoudre les
Egalitez. He was a vocal critic of the
methods of his day and attacked calculus as being a “collection of ingenious
fallacies.” Later, however, he became
convinced of the essential correctness
of the methods of calculus.
y
0
Before giving the proof let’s take a look at the graphs of some typical functions that
satisfy the three hypotheses. Figure 1 shows the graphs of four such functions. In each
case it appears that there is at least one point sc, f scdd on the graph where the tangent is
hori­zontal and therefore f 9scd − 0. Thus Rolle’s Theorem is plausible.
y
a
c¡
c™ b
(a)
FIGURE 1 x
0
y
y
a
c
b
x
(b)
0
a
c¡
c™
b
x
0
a
(c)
c
b
x
(d)
Proof There are three cases:
PS Take cases
CASE I f sxd − k, a constant
Then f 9sxd − 0, so the number c can be taken to be any number in sa, bd.
CASE II f sxd . f sad for some x in sa, bd [as in Figure 1(b) or (c)]
y the Extreme Value Theorem (which we can apply by hypothesis 1), f has a maxiB
mum value somewhere in fa, bg. Since f sad − f sbd, it must attain this maximum value
at a number c in the open interval sa, bd. Then f has a local maximum at c and, by
hypothesis 2, f is differentiable at c. Therefore f 9scd − 0 by Fermat’s Theorem.
CASE III f sxd , f sad for some x in sa, bd [as in Figure 1(c) or (d)]
By the Extreme Value Theorem, f has a minimum value in fa, bg and, since
f sad − f sbd, it attains this minimum value at a number c in sa, bd. Again f 9scd − 0 by
Fermat’s Theorem.
n
Example 1 Let’s apply Rolle’s Theorem to the position function s − f std of a
moving object. If the object is in the same place at two different instants t − a and
t − b, then f sad − f sbd. Rolle’s Theorem says that there is some instant of time t − c
between a and b when f 9scd − 0; that is, the velocity is 0. (In particular, you can see
that this is true when a ball is thrown directly upward.)
n
Example 2 Prove that the equation x 3 1 x 2 1 − 0 has exactly one real root.
SOLUTION First we use the Intermediate Value Theorem (1.8.10) to show that a root
exists. Let f sxd − x 3 1 x 2 1. Then f s0d − 21 , 0 and f s1d − 1 . 0. Since f is a
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
216
Chapter 3 Applications of Differentiation
Figure 2 shows a graph of the function f sxd − x 3 1 x 2 1 discussed in
Example 2. Rolle’s Theorem shows
that, no matter how much we enlarge
the viewing rectangle, we can never
find a second x-intercept.
3
_2
polynomial, it is continuous, so the Intermediate Value Theorem states that there is a
number c between 0 and 1 such that f scd − 0. Thus the given equation has a root.
To show that the equation has no other real root, we use Rolle’s Theorem and argue
by contradiction. Suppose that it had two roots a and b. Then f sad − 0 − f sbd and,
since f is a polynomial, it is differentiable on sa, bd and continuous on fa, bg. Thus, by
Rolle’s Theorem, there is a number c between a and b such that f 9scd − 0. But
f 9sxd − 3x 2 1 1 > 1 for all x
2
(since x 2 > 0) so f 9sxd can never be 0. This gives a contradiction. Therefore the equation can’t have two real roots.
n
Our main use of Rolle’s Theorem is in proving the following important theorem,
which was first stated by another French mathematician, Joseph-Louis Lagrange.
_3
FIGURE 2
The Mean Value Theorem Let f be a function that satisfies the following
hypotheses:
1. f is continuous on the closed interval fa, bg.
The Mean Value Theorem is an
example of what is called an existence
theorem. Like the Intermediate Value
Theorem, the Extreme Value Theorem,
and Rolle’s Theorem, it guarantees that
there exists a number with a certain
property, but it doesn’t tell us how to
find the number.
2. f is differentiable on the open interval sa, bd.
Then there is a number c in sa, bd such that
1
f 9scd −
or, equivalently,
2 f sbd 2 f sad
b2a
f sbd 2 f sad − f 9scdsb 2 ad
Before proving this theorem, we can see that it is reasonable by interpreting it geomet­
rically. Figures 3 and 4 show the points Asa, f sadd and Bsb, f sbdd on the graphs of two
dif­ferentiable functions. The slope of the secant line AB is
3 mAB −
f sbd 2 f sad
b2a
which is the same expression as on the right side of Equation 1. Since f 9scd is the slope of
the tangent line at the point sc, f scdd, the Mean Value Theorem, in the form given by Equa­
tion 1, says that there is at least one point Psc, f scdd on the graph where the slope of the
tangent line is the same as the slope of the secant line AB. In other words, there is a point
P where the tangent line is parallel to the secant line AB. (Imagine a line far away that
stays parallel to AB while moving toward AB until it touches the graph for the first time.)
y
y
P { c, f(c)}
P¡
B
P™
A
A{ a, f(a)}
B { b, f(b)}
0
FIGURE 3
a
c
b
x
0
a
c¡
c™
b
x
FIGURE 4
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.2 The Mean Value Theorem
y
h(x)
A
ƒ
0
a
y=ƒ
B
x
f(a)+
b
f(b)-f(a)
(x-a)
b-a
x
Proof We apply Rolle’s Theorem to a new function h defined as the difference
between f and the function whose graph is the secant line AB. Using Equation 3
and the point-slope equation of a line, we see that the equation of the line AB can
be written as
f sbd 2 f sad
y 2 f sad −
sx 2 ad
b2a
y − f sad 1
or as
f sbd 2 f sad
sx 2 ad
b2a
So, as shown in Figure 5,
figure 5
4 Lagrange and the
Mean Value Theorem
The Mean Value Theorem was first
formulated by Joseph-Louis Lagrange
(1736–1813), born in Italy of a French
father and an Italian mother. He was a
child prodigy and became a professor in
Turin at the tender age of 19. Lagrange
made great contributions to number
theory, theory of functions, theory of
equations, and analytical and celestial
mechanics. In particular, he applied
calculus to the analysis of the stability
of the solar system. At the invitation
of Frederick the Great, he succeeded
Euler at the Berlin Academy and, when
Frederick died, Lagrange accepted King
Louis XVI’s invitation to Paris, where he
was given apartments in the Louvre
and became a professor at the Ecole
Polytechnique. Despite all the trappings
of luxury and fame, he was a kind and
quiet man, living only for science.
217
hsxd − f sxd 2 f sad 2
f sbd 2 f sad
sx 2 ad
b2a
First we must verify that h satisfies the three hypotheses of Rolle’s Theorem.
1. The function h is continuous on fa, bg because it is the sum of f and a first-degree
polynomial, both of which are continuous.
2.
The function h is differentiable on sa, bd because both f and the first-degree polynomial are differentiable. In fact, we can compute h9 directly from Equation 4:
h9sxd − f 9sxd 2
f sbd 2 f sad
b2a
(Note that f sad and f f sbd 2 f sadgysb 2 ad are constants.)
3.
hsad − f sad 2 f sad 2
f sbd 2 f sad
sa 2 ad − 0
b2a
hsbd − f sbd 2 f sad 2
f sbd 2 f sad
sb 2 ad
b2a
− f sbd 2 f sad 2 f f sbd 2 f sadg − 0
Therefore hsad − hsbd.
Since h satisfies the hypotheses of Rolle’s Theorem, that theorem says there is a number c in sa, bd such that h9scd − 0. Therefore
0 − h9scd − f 9scd 2
and so
f 9scd −
f sbd 2 f sad
b2a
f sbd 2 f sad
b2a
n
Example 3 To illustrate the Mean Value Theorem with a specific function, let’s
consider f sxd − x 3 2 x, a − 0, b − 2. Since f is a polynomial, it is continuous and
differentiable for all x, so it is certainly continuous on f0, 2g and differentiable on s0, 2d.
Therefore, by the Mean Value Theorem, there is a number c in s0, 2d such that
f s2d 2 f s0d − f 9scds2 2 0d
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
218
Chapter 3 Applications of Differentiation
y
y=˛- x
B
Now f s2d − 6, f s0d − 0, and f 9sxd − 3x 2 2 1, so this equation becomes
6 − s3c 2 2 1d2 − 6c 2 2 2
O
c
FIGURE 6 2
x
which gives c 2 − 43, that is, c − 62ys3 . But c must lie in s0, 2d, so c − 2ys3 . Figure 6 illustrates this calculation: The tangent line at this value of c is parallel to the
secant line OB.
n
Example 4 If an object moves in a straight line with position function s − f std, then
the average velocity between t − a and t − b is
f sbd 2 f sad
b2a
and the velocity at t − c is f 9scd. Thus the Mean Value Theorem (in the form of Equation 1) tells us that at some time t − c between a and b the instantaneous velocity f 9scd
is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours, then
the speedometer must have read 90 kmyh at least once.
In general, the Mean Value Theorem can be interpreted as saying that there is a number at which the instantaneous rate of change is equal to the average rate of change
over an interval.
n
The main significance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The next example provides an instance of this principle.
Example 5 Suppose that f s0d − 23 and f 9sxd < 5 for all values of x. How large can
f s2d possibly be?
SOLUTION We are given that f is differentiable (and therefore continuous) everywhere.
In particular, we can apply the Mean Value Theorem on the interval f0, 2g. There exists
a number c such that
f s2d 2 f s0d − f 9scds2 2 0d
so
f s2d − f s0d 1 2f 9scd − 23 1 2f 9scd
We are given that f 9sxd < 5 for all x, so in particular we know that f 9scd < 5. Multiplying both sides of this inequality by 2, we have 2f 9scd < 10, so
f s2d − 23 1 2f 9scd < 23 1 10 − 7
The largest possible value for f s2d is 7.
n
The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following theorem. Others will be found
in the following sections.
5 Theorem If f 9sxd − 0 for all x in an interval sa, bd, then f is constant on sa, bd.
Proof Let x 1 and x 2 be any two numbers in sa, bd with x 1 , x 2. Since f is differentiable on sa, bd, it must be differentiable on sx 1, x 2 d and continuous on fx 1, x 2 g. By
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Section 3.2 The Mean Value Theorem
219
applying the Mean Value Theorem to f on the interval fx 1, x 2 g, we get a number c such
that x 1 , c , x 2 and
f sx 2 d 2 f sx 1d − f 9scdsx 2 2 x 1d
6 Since f 9sxd − 0 for all x, we have f 9scd − 0, and so Equation 6 becomes
f sx 2 d 2 f sx 1 d − 0 or f sx 2 d − f sx 1 d
Therefore f has the same value at any two numbers x 1 and x 2 in sa, bd. This means that
f is constant on sa, bd.
n
Corollary 7 says that if two functions
have the same derivatives on an interval, then their graphs must be vertical
translations of each other there. In
other words, the graphs have the same
shape, but could be shifted up or down.
7 Corollary If f 9sxd − t9sxd for all x in an interval sa, bd, then f 2 t is constant
on sa, bd; that is, f sxd − tsxd 1 c where c is a constant.
Proof Let Fsxd − f sxd 2 tsxd. Then
F9sxd − f 9sxd 2 t9sxd − 0
for all x in sa, bd. Thus, by Theorem 5, F is constant; that is, f 2 t is constant.
n
NOTE Care must be taken in applying Theorem 5. Let
f sxd −
H
x
1
if x . 0
−
x
21 if x , 0
| |
|
The domain of f is D − hx x ± 0j and f 9sxd − 0 for all x in D. But f is obviously not
a constant function. This does not contradict Theorem 5 because D is not an interval.
Notice that f is constant on the interval s0, `d and also on the interval s2`, 0d.
1.The graph of a function f is shown. Verify that f satisfies the
hypotheses of Rolle’s Theorem on the interval f0, 8g. Then
estimate the value(s) of c that satisfy the conclusion of Rolle’s
Theorem on that interval.
y
3.The graph of a function t is shown.
y
y=©
y=ƒ
1
0
1
0
1
x
2.Draw the graph of a function defined on f0, 8g such that
f s0d − f s8d − 3 and the function does not satisfy the
conclusion of Rolle’s Theorem on f0, 8g.
1
x
(a)Verify that t satisfies the hypotheses of the Mean Value
Theorem on the interval f0, 8g.
(b)Estimate the value(s) of c that satisfy the conclusion of the
Mean Value Theorem on the interval f0, 8g.
(c)Estimate the value(s) of c that satisfy the conclusion of the
Mean Value Theorem on the interval f2, 6g.
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220
Chapter 3 Applications of Differentiation
4.Draw the graph of a function that is continuous on f0, 8g
where f s0d − 1 and f s8d − 4 and that does not satisfy the
conclusion of the Mean Value Theorem on f0, 8g.
5–8 Verify that the function satisfies the three hypotheses of
Rolle’s Theorem on the given interval. Then find all numbers c
that satisfy the conclusion of Rolle’s Theorem.
5. f sxd − 2 x 2 2 4 x 1 5, f21, 3g
6. f sxd − x 3 2 2x 2 2 4x 1 2, f22, 2g
g
8. f sxd − x 1 1y x, 12 , 2
9.
Let f sxd − 1 2 x 2y3. Show that f s21d − f s1d but there is
no number c in s21, 1d such that f 9scd − 0. Why does this
not contradict Rolle’s Theorem?
10.Let f sxd − tan x. Show that f s0d − f sd but there is no
number c in s0, d such that f 9scd − 0. Why does this not
contradict Rolle’s Theorem?
11–14 Verify that the function satisfies the hypotheses of the
Mean Value Theorem on the given interval. Then find all num­
bers c that satisfy the conclusion of the Mean Value Theorem.
11. f sxd − 2x 2 2 3x 1 1, f0, 2g
12. f sxd − x 3 2 3x 1 2, f22, 2g
3
13. f sxd − s
x , f0, 1g
24. (a)Suppose that f is differentiable on R and has two roots.
Show that f 9 has at least one root.
(b)Suppose f is twice differentiable on R and has three
roots. Show that f 0 has at least one real root.
(c) Can you generalize parts (a) and (b)?
25.If f s1d − 10 and f 9sxd > 2 for 1 < x < 4, how small can
f s4d possibly be?
7. f sxd − sins xy2d, fy2, 3y2g
f
23. (a)Show that a polynomial of degree 3 has at most three
real roots.
(b)Show that a polynomial of degree n has at most n real
roots.
14. f sxd − 1yx, f1, 3g
; 15–16 Find the number c that satisfies the conclusion of the
Mean Value Theorem on the given interval. Graph the function,
the secant line through the endpoints, and the tangent line at
sc, f scdd. Are the secant line and the tangent line parallel?
26.Suppose that 3 < f 9sxd < 5 for all values of x. Show that
18 < f s8d 2 f s2d < 30.
27.Does there exist a function f such that f s0d − 21,
f s2d − 4, and f 9sxd < 2 for all x?
28.Suppose that f and t are continuous on fa, bg and differenti­able on sa, bd. Suppose also that f sad − tsad and
f 9sxd , t9sxd for a , x , b. Prove that f sbd , tsbd. [Hint:
Apply the Mean Value Theorem to the function h − f 2 t.]
29. Show that sin x , x if 0 , x , 2.
30.Suppose f is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists
a number c in s2b, bd such that f 9scd − f sbdyb.
31. Use the Mean Value Theorem to prove the inequality
| sin a 2 sin b | < | a 2 b | for all a and b
32.If f 9sxd − c (c a constant) for all x, use Corollary 7 to show
that f sxd − cx 1 d for some constant d.
33. Let f sxd − 1yx and
15. f sxd − sx , f0, 4g
16. f sxd − x 3 2 2x, f22, 2g
17.Let f sxd − s x 2 3d22. Show that there is no value of c in
s1, 4d such that f s4d 2 f s1d − f 9scds4 2 1d. Why does this
not contradict the Mean Value Theorem?
|
|
18.Let f sxd − 2 2 2 x 2 1 . Show that there is no value of
c such that f s3d 2 f s0d − f 9scds3 2 0d. Why does this not
contradict the Mean Value Theorem?
19–20 Show that the equation has exactly one real root.
19. 2 x 1 cos x − 020. 2x 2 1 2 sin x − 0
21.Show that the equation x 3 2 15x 1 c − 0 has at most one
root in the interval f22, 2g.
22.Show that the equation x 4 1 4x 1 c − 0 has at most two
real roots.
tsxd −
1
x
1
11
x
if x . 0
if x , 0
Show that f 9sxd − t9sxd for all x in their domains. Can we
conclude from Corollary 7 that f 2 t is constant?
34.At 2:00 pm a car’s speedometer reads 30 miyh. At 2:10 pm it
reads 50 miyh. Show that at some time between 2:00 and
2:10 the acceleration is exactly 120 miyh2.
35.Two runners start a race at the same time and finish in a tie.
Prove that at some time during the race they have the same
speed. [Hint: Consider f std − tstd 2 hstd, where t and h
are the position functions of the two runners.]
36.A number a is called a fixed point of a function f if
f sad − a. Prove that if f 9sxd ± 1 for all real numbers x,
then f has at most one fixed point.
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Section 3.3 How Derivatives Affect the Shape of a Graph
y
221
Many of the applications of calculus depend on our ability to deduce facts about a function f from information concerning its derivatives. Because f 9sxd represents the slope of
the curve y − f sxd at the point sx, f sxdd, it tells us the direction in which the curve proceeds at each point. So it is reasonable to expect that information about f 9sxd will provide
us with information about f sxd.
D
B
What Does f 9 Say About f ?
C
A
x
0
FIGURE 1
To see how the derivative of f can tell us where a function is increasing or decreasing,
look at Figure 1. (Increasing functions and decreasing functions were defined in Section
1.1.) Between A and ­B and between C and D, the tangent lines have positive slope and
so f 9sxd . 0. Between B and C, the tangent lines have negative slope and so f 9sxd , 0.
Thus it appears that f increases when f 9sxd is positive and decreases when f 9sxd is negative. To prove that this is always the case, we use the Mean Value Theorem.
Increasing/Decreasing Test
Let’s abbreviate the name of this test to
the I/D Test.
(a) If f 9sxd . 0 on an interval, then f is increasing on that interval.
(b) If f 9sxd , 0 on an interval, then f is decreasing on that interval.
Proof
(a) Let x 1 and x 2 be any two numbers in the interval with x1 , x2. According to the
definition of an increasing function (page 19), we have to show that f sx1 d , f sx2 d.
Because we are given that f 9sxd . 0, we know that f is differentiable on fx1, x2 g. So,
by the Mean Value Theorem, there is a number c between x1 and x2 such that
1
f sx 2 d 2 f sx 1 d − f 9scdsx 2 2 x 1 d
Now f 9scd . 0 by assumption and x 2 2 x 1 . 0 because x 1 , x 2. Thus the right side of
Equation 1 is positive, and so
f sx 2 d 2 f sx 1 d . 0 or f sx 1 d , f sx 2 d
This shows that f is increasing.
Part (b) is proved similarly.
n
Example 1 Find where the function f sxd − 3x 4 2 4x 3 2 12x 2 1 5 is increasing
and where it is decreasing.
SOLUTION We start by differentiating f :
f 9sxd − 12x 3 2 12x 2 2 24x − 12xsx 2 2dsx 1 1d
_1
0
2
x
To use the IyD Test we have to know where f 9sxd . 0 and where f 9sxd , 0. To
solve these inequalities we first find where f 9sxd − 0, namely, at x − 0, 2, and 21.
These are the critical numbers of f , and they divide the domain into four intervals
(see the number line at the left). Within each interval, f 9sxd must be always positive or always negative. (See Examples 3 and 4 in Appendix A.) We can determine
which is the case for each interval from the signs of the three factors of f 9sxd, namely,
12x, x 2 2, and x 1 1, as shown in the following chart. A plus sign indicates that the
given expression is positive, and a minus sign indicates that it is negative. The last col-
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222
Chapter 3 Applications of Differentiation
umn of the chart gives the conclusion based on the IyD Test. For instance, f 9sxd , 0 for
0 , x , 2, so f is decreasing on (0, 2). (It would also be true to say that f is decreasing on the closed interval f0, 2g.)
20
Interval
_2
3
x22
x11
f 9sxd
f
2
2
2
2
decreasing on (2`, 21)
21 , x , 0
2
2
1
1
increasing on (21, 0)
0,x,2
x.2
1
1
2
1
1
1
2
1
decreasing on (0, 2)
increasing on (2, `)
x , 21
_30
FIGURE 2
12x
The graph of f shown in Figure 2 confirms the information in the chart.
n
Local Extreme Values
Recall from Section 3.1 that if f has a local maximum or minimum at c, then c must be
a critical number of f (by Fermat’s Theorem), but not every critical number gives rise to
a maximum or a minimum. We therefore need a test that will tell us whether or not f has
a local maximum or minimum at a critical number.
You can see from Figure 2 that f s0d − 5 is a local maximum value of f because f
increases on s21, 0d and decreases on s0, 2d. Or, in terms of derivatives, f 9sxd . 0 for
21 , x , 0 and f 9sxd , 0 for 0 , x , 2. In other words, the sign of f 9sxd changes
from positive to negative at 0. This observation is the basis of the following test.
The First Derivative Test Suppose that c is a critical number of a continuous
function f.
(a) If f 9 changes from positive to negative at c, then f has a local maximum at c.
(b) If f 9 changes from negative to positive at c, then f has a local minimum at c.
(c) If f 9 is positive to the left and right of c, or negative to the left and right of c,
then f has no local maximum or minimum at c.
The First Derivative Test is a consequence of the IyD Test. In part (a), for instance,
since the sign of f 9sxd changes from positive to negative at c, f is increasing to the left of
c and decreasing to the right of c. It follows that f has a local maximum at c.
It is easy to remember the First Derivative Test by visualizing diagrams such as those
in Figure 3.
y
y
fª(x)>0
y
fª(x)<0
fª(x)<0
fª(x)>0
fª(x)<0
0
c
(a) Local maximum
y
x
0
fª(x)>0
c
(b) Local minimum
fª(x)<0
fª(x)>0
x
0
c
x
(c) No maximum or minimum
0
c
x
(d) No maximum or minimum
FIGURE 3
Example 2 Find the local minimum and maximum values of the function f in
Example 1.
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223
Section 3.3 How Derivatives Affect the Shape of a Graph
SOLUTION From the chart in the solution to Example 1 we see that f 9sxd changes from
negative to positive at 21, so f s21d − 0 is a local minimum value by the First Derivative Test. Similarly, f 9 changes from negative to positive at 2, so f s2d − 227 is also a
local minimum value. As noted previously, f s0d − 5 is a local maximum value because
f 9sxd changes from positive to negative at 0.
n
Example 3 Find the local maximum and minimum values of the function
tsxd − x 1 2 sin x 0 < x < 2
SOLUTION As in Example 1, we start by finding the critical numbers. The derivative is
t9sxd − 1 1 2 cos x
so t9sxd − 0 when cos x −
The solutions of this equation are 2y3 and 4y3.
Because t is differentiable everywhere, the only critical numbers are 2y3 and 4y3.
We split the domain into intervals according to the critical numbers. Within each
interval, t9sxd is either always positive or always negative and so we analyze t in the
following chart.
212.
The 1 signs in the chart come from the
fact that t9sxd . 0 when cos x . 2 12.
From the graph of y − cos x, this is
true in the indicated intervals.
6
t9sxd − 1 1 2 cos x
Interval
0 , x , 2y3
1
increasing on s0, 2y3d
2y3 , x , 4y3
4y3 , x , 2
2
1
decreasing on s2y3, 4y3d
increasing on s4y3, 2d
Because t9sxd changes from positive to negative at 2y3, the First Derivative Test tells
us that there is a local maximum at 2y3 and the local maximum value is
ts2y3d −
0
t
2π
3
4π
3
2π
−
2
1 s3 < 3.83
3
Likewise, t9sxd changes from negative to positive at 4y3 and so
ts4y3d −
FIGURE 4 S D
2
2
2
s3
1 2 sin
−
12
3
3
3
2
S D
4
4
4
s3
1 2 sin
−
12 2
3
3
3
2
−
4
2 s3 < 2.46
3
is a local minimum value. The graph of t in Figure 4 supports our conclusion.
tsxd − x 1 2 sin x
n
What Does f 99 Say About f ?
Figure 5 shows the graphs of two increasing functions on sa, bd. Both graphs join point
A to point B but they look different because they bend in different directions. How can
we dis­tinguish between these two types of behavior?
B
y
B
y
g
f
A
A
0
FIGURE 5 a
b
(a)
x
0
b
a
(b)
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
x
224
Chapter 3 Applications of Differentiation
B
y
In Figure 6 tangents to these curves have been drawn at several points. In (a) the curve
lies above the tangents and f is called concave upward on sa, bd. In (b) the curve lies
below the tangents and t is called concave downward on sa, bd.
f
Definition If the graph of f lies above all of its tangents on an interval I, then it is
called concave upward on I. If the graph of f lies below all of its tangents on I, it
is called concave downward on I.
A
0
x
Figure 7 shows the graph of a function that is concave upward (abbreviated CU) on
the intervals sb, cd, sd, ed, and se, pd and concave downward (CD) on the intervals sa, bd,
sc, dd, and sp, qd.
(a) Concave upward
B
y
y
g
D
B
P
C
A
0
x
0 a
c
CD
(b) Concave downward
FIGURE 6 b
d
CU
e
CD
CU
p
CU
q
x
CD
FIGURE 7 Let’s see how the second derivative helps determine the intervals of concavity. Look­
ing at Figure 6(a), you can see that, going from left to right, the slope of the tangent
increas­es. This means that the derivative f 9 is an increasing function and therefore its
derivative f 0 is positive. Likewise, in Figure 6(b) the slope of the tangent decreases from
left to right, so f 9 decreases and therefore f 0 is negative. This reasoning can be reversed
and suggests that the following theorem is true. A proof is given in Appendix F with the
help of the Mean Value Theorem.
Concavity Test
(a) If f 0sxd . 0 for all x in I, then the graph of f is concave upward on I.
(b) If f 0sxd , 0 for all x in I, then the graph of f is concave downward on I.
Example 4 Figure 8 shows a population graph for Cyprian honeybees raised in an
apiary. How does the rate of population increase change over time? When is this rate
highest? Over what intervals is P concave upward or concave downward?
P
80
Number of bees
(in thousands)
60
40
20
0
FIGURE 8 3
6
9
12
15
18
t
Time (in weeks)
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Section 3.3 How Derivatives Affect the Shape of a Graph
225
SOLUTION By looking at the slope of the curve as t increases, we see that the rate
of increase of the population is initially very small, then gets larger until it reaches a
maximum at about t − 12 weeks, and decreases as the population begins to level off.
As the population approaches its maximum value of about 75,000 (called the carrying
capacity), the rate of increase, P9std, approaches 0. The curve appears to be concave
upward on (0, 12) and concave downward on (12, 18).
n
In Example 4, the population curve changed from concave upward to concave downward at approximately the point (12, 38,000). This point is called an inflection point of
the curve. The significance of this point is that the rate of population increase has its
maximum value there. In general, an inflection point is a point where a curve changes its
direction of concavity.
Definition A point P on a curve y − f sxd is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward or
from concave downward to concave upward at P.
For instance, in Figure 7, B, C, D, and P are the points of inflection. Notice that if a
curve has a tangent at a point of inflection, then the curve crosses its tangent there.
In view of the Concavity Test, there is a point of inflection at any point where the
second derivative changes sign.
y
(4, 6)
6
(2, 3)
3
0
dec
Example 5 Sketch a possible graph of a function f that satisfies the following
conditions:
(i) f s0d − 0, f s2d − 3, f s4d − 6, f 9s0d − f 9s4d − 0
(ii) f 9sxd . 0 for 0 , x , 4, f 9sxd , 0 for x , 0 and for x . 4
(iii) f 0sxd . 0 for x , 2, f 0sxd , 0 for x . 2
2
x
4
inc
dec
CU
CD
FIGURE 9 y
f
P
f ª(c)=0
0
c
x
FIGURE 10 f 99scd . 0, f is concave upward
Another application of the second derivative is the following test for identifying local
maximum and minimum values. It is a consequence of the Concavity Test and serves as
an alternative to the First Derivative Test.
The Second Derivative Test Suppose f 0 is continuous near c.
(a) If f 9scd − 0 and f 0scd . 0, then f has a local minimum at c.
(b) If f 9scd − 0 and f 0scd , 0, then f has a local maximum at c.
ƒ
f(c)
SOLUTION Condition (i) tells us that the graph has horizontal tangents at the points
s0, 0d and s4, 6d. Condition (ii) says that f is increasing on the interval s0, 4d and
decreasing on the intervals s2`, 0d and s4, `d. It follows from the I/D Test that
f s0d − 0 is a local minimum and f s4d − 6 is a local maximum.
Condition (iii) says that the graph is concave upward on the interval s2`, 2d and
concave downward on s2, `d. Because the curve changes from concave upward to concave downward when x − 2, the point s2, 3d is an inflection point.
We use this information to sketch the graph of f in Figure 9. Notice that we made
the curve bend upward when x , 2 and bend downward when x . 2.
n
x
For instance, part (a) is true because f 0sxd . 0 near c and so f is concave upward
near c. This means that the graph of f lies above its horizontal tangent at c and so f has
a local minimum at c. (See Figure 10.)
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226
Chapter 3 Applications of Differentiation
Example 6 Discuss the curve y − x 4 2 4x 3 with respect to concavity, points of
inflection, and local maxima and minima. Use this information to sketch the curve.
SOLUTION If f sxd − x 4 2 4x 3, then
f 9sxd − 4x 3 2 12x 2 − 4x 2sx 2 3d
f 0sxd − 12x 2 2 24x − 12xsx 2 2d
To find the critical numbers we set f 9sxd − 0 and obtain x − 0 and x − 3. (Note that f 9
is a polynomial and hence defined everywhere.) To use the Second Derivative Test we
evaluate f 0 at these critical numbers:
f 0s0d − 0 f 0s3d − 36 . 0
y
Since f 9s3d − 0 and f 0s3d . 0, f s3d − 227 is a local minimum. [In fact, the expression for f 9sxd shows that f decreases to the left of 3 and increases to the right of 3.]
Since f 0s0d − 0, the Second Derivative Test gives no information about the critical
number 0. But since f 9sxd , 0 for x , 0 and also for 0 , x , 3, the First Derivative
Test tells us that f does not have a local maximum or minimum at 0.
Since f 0sxd − 0 when x − 0 or 2, we divide the real line into intervals with these
numbers as endpoints and complete the following chart.
y=x$-4˛
(0, 0)
2
inflection
points
3
(2, _16)
(3, _27)
FIGURE 11 x
Interval
f 99sxd − 12 x sx 2 2d
s2`, 0d
1
upward
s0, 2d
s2, `d
2
1
downward
upward
Concavity
The point s0, 0d is an inflection point since the curve changes from concave upward
to concave downward there. Also s2, 216d is an inflection point since the curve
changes from concave downward to concave upward there.
Using the local minimum, the intervals of concavity, and the inflection points, we
sketch the curve in Figure 11.
n
Note The Second Derivative Test is inconclusive when f 0scd − 0. In other words,
at such a point there might be a maximum, there might be a minimum, or there might be
neither (as in Example 6). This test also fails when f 0scd does not exist. In such cases the
First Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test is often the easier one to use.
Example 7 Sketch the graph of the function f sxd − x 2y3s6 2 xd1y3.
SOLUTION Calculation of the first two derivatives gives
Use the differentiation rules to check
these calculations.
f 9sxd −
42x
28
f 0sxd − 4y3
x 1y3s6 2 xd2y3
x s6 2 xd5y3
Since f 9sxd − 0 when x − 4 and f 9sxd does not exist when x − 0 or x − 6, the critical
numbers are 0, 4, and 6.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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227
Section 3.3 How Derivatives Affect the Shape of a Graph
Interval
TEC In Module 3.3 you can practice
using information about f 9, f 0, and
asymptotes to determine the shape of
the graph of f.
|
62x
62x
x 1y3
s6 2 xd2y3
f 9sxd
f
x,0
1
2
1
2
decreasing on s2`, 0d
0,x,4
1
1
1
1
increasing on s0, 4d
4,x,6
x.6
2
2
1
1
1
1
2
2
decreasing on s4, 6d
decreasing on s6, `d
To find the local extreme values we use the First Derivative Test. Since f 9 changes
from negative to positive at 0, f s0d − 0 is a local minimum. Since f 9 changes from positive to negative at 4, f s4d − 2 5y3 is a local maximum. The sign of f 9 does not change
at 6, so there is no minimum or maximum there. (The Second Derivative Test could be
used at 4 but not at 0 or 6 since f 0 does not exist at either of these numbers.)
Looking at the expression for f 0sxd and noting that x 4y3 > 0 for all x, we have
f 0sxd , 0 for x , 0 and for 0 , x , 6 and f 0sxd . 0 for x . 6. So f is concave
downward on s2`, 0d and s0, 6d and concave upward on s6, `d, and the only inflection point is s6, 0d. The graph is sketched in Figure 12. Note that the curve has vertical
tangents at s0, 0d and s6, 0d because f 9sxd l ` as x l 0 and as x l 6.
Try reproducing the graph in Fig­ure 12
with a graphing calculator or computer.
Some machines produce the complete
graph, some produce only the portion
to the right of the y-axis, and some
produce only the portion between
x − 0 and x − 6. For an explanation
and cure, see Example 7 in “Graphing Calculators and Computers” at
www.stewartcalculus.com. An equivalent expression that gives the correct
graph is
y − sx 2 d1y3 ?
42x
|
y
4
(4, 2%?# )
3
2
0
6 2 x|
||
1y3
1
0
1
1
x
0
1
3.Suppose you are given a formula for a function f.
(a)How do you determine where f is increasing or
decreasing?
4
7 x
5
n
4. ( a) State the First Derivative Test.
(b)State the Second Derivative Test. Under what circum­
stances is it inconclusive? What do you do if it fails?
2. y
1
3
(b)How do you determine where the graph of f is concave
upward or concave downward?
(c) How do you locate inflection points?
1–2 Use the given graph of f to find the following.
(a) The open intervals on which f is increasing.
(b) The open intervals on which f is decreasing.
(c) The open intervals on which f is concave upward.
(d) The open intervals on which f is concave downward.
(e) The coordinates of the points of inflection.
y
2
y=x @ ?#(6-x)!?#
FIGURE 12 1.
|
x
5–6 The graph of the derivative f 9 of a function f is shown.
(a) On what intervals is f increasing or decreasing?
(b) At what values of x does f have a local maximum or
minimum?
5. y
y
y=fª(x)
0
2
4
y=fª(x)
6
7et0403x05–06
09/10/09
x
0
2
4
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
6
x
228
6.
Chapter 3 Applications of Differentiation
y
18. ( a) Find the critical numbers of f sxd − x 4sx 2 1d3.
(b)What does the Second Derivative Test tell you about the
behavior of f at these critical numbers?
(c) What does the First Derivative Test tell you?
y=fª(x)
0
2
4
6
8
x
19. Suppose f 0 is continuous on s2`, `d.
(a)If f 9s2d − 0 and f 0s2d − 25, what can you say about f ?
(b)If f 9s6d − 0 and f 0s6d − 0, what can you say about f ?
7.In each part state the x-coordinates of the inflection points
of f. Give reasons for your answers.
(a) The curve is the graph of f.
(b) The curve is the graph of f 9.
(c) The curve is the graph of f 0.
y
0
2
4
6
x
8
8.The graph of7et0403x07
the first derivative f 9 of a function f is shown.
(a) On what09/10/09
intervals is f increasing? Explain.
(b)At what values of x does f have a local maximum or
MasterID:
minimum?
Explain. 00490
(c)On what intervals is f concave upward or concave down­
ward? Explain.
(d)What are the x-coordinates of the inflection points of f ?
Why?
y
y=fª(x)
0
2
4
6
8
x
9–14
(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f.
(c) Find the intervals of concavity and the inflection points.
9.f sxd − x 3 2 3x 2 2 9x 1 4
10. f sxd − 2x 3 2 9x 2 1 12x 2 3
x
11. f sxd − x 4 2 2x 2 1 312.
f sxd − 2
x 11
13. f sxd − sin x 1 cos x, 0 < x < 2
14. f sxd − cos2 x 2 2 sin x, 0 < x < 2
15–17 Find the local maximum and minimum values of f using
both the First and Second Derivative Tests. Which method do you
prefer?
x2
15. f sxd − 1 1 3x 2 2 2x 316.
f sxd −
x21
4
17. f sxd − sx 2 s
x
20–27 Sketch the graph of a function that satisfies all of the given
conditions.
20. (a) f 9sxd , 0 and f 0sxd , 0 for all x
(b) f 9sxd . 0 and f 0sxd . 0 for all x
21. (a) f 9sxd . 0 and f 0sxd , 0 for all x
(b) f 9sxd , 0 and f 0sxd . 0 for all x
22. Vertical asymptote x − 0, f 9sxd . 0 if x , 22,
f 9sxd , 0 if x . 22 sx ± 0d,
f 0sxd , 0 if x , 0, f 0sxd . 0 if x . 0
23. f 9s0d − f 9s2d − f 9s4d − 0, f 9sxd . 0 if x , 0 or 2 , x , 4, f 9sxd , 0 if 0 , x , 2 or x . 4, f 0sxd . 0 if 1 , x , 3, f 0sxd , 0 if x , 1 or x . 3
24. f 9sxd . 0 for all x ± 1, vertical asymptote x − 1,
f 99sxd . 0 if x , 1 or x . 3, f 99sxd , 0 if 1 , x , 3
25. f 9s5d − 0, f 9sxd , 0 when x , 5,
f 9sxd . 0 when x . 5, f 99s2d − 0, f 99s8d − 0,
f 99sxd , 0 when x , 2 or x . 8,
f 0sxd . 0 for 2 , x , 8
26. f 9s0d − f 9s4d − 0, f 9sxd − 1 if x , 21,
f 9sxd . 0 if 0 , x , 2,
f 9sxd , 0 if 21 , x , 0 or 2 , x , 4 or x . 4,
lim2 f 9sxd − `, lim1 f 9sxd − 2`,
x l2
x l2
f 99sxd . 0 if 21 , x , 2 or 2 , x , 4, f 0sxd , 0 if x . 4
27. f s0d − f 9s0d − f 9s2d − f 9s4d − f 9s6d − 0,
f 9sxd . 0 if 0 , x , 2 or 4 , x , 6,
f 9sxd , 0 if 2 , x , 4 or x . 6,
f 0sxd . 0 if 0 , x , 1 or 3 , x , 5,
f 0sxd , 0 if 1 , x , 3 or x . 5, f s2xd − f sxd
28. Suppose f s3d − 2, f 9s3d − 12, and f 9sxd . 0 and f 0sxd , 0
for all x.
(a) Sketch a possible graph for f.
(b)How many solutions does the equation f sxd − 0 have?
Why?
(c) Is it possible that f 9s2d − 13? Why?
29. Suppose f is a continuous function where f sxd . 0 for all x,
f s0d − 4, f 9s xd . 0 if x , 0 or x . 2, f 9s xd , 0
if 0 , x , 2, f 99s21d − f 99s1d − 0, f 99s xd . 0 if
x , 21 or x . 1, f 99s xd , 0 if 21 , x , 1.
(a)Can f have an absolute maximum? If so, sketch a possible
graph of f. If not, explain why.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.3 How Derivatives Affect the Shape of a Graph
(b)Can f have an absolute minimum? If so, sketch a possible graph of f. If not, explain why.
(c)Sketch a possible graph for f that does not achieve an
absolute minimum.
229
33–44 (a) Find the intervals of increase or decrease.
(b) Find the local maximum and minimum values.
(c) Find the intervals of concavity and the inflection points.
(d) Use the information from parts (a)–(c) to sketch the graph.
Check your work with a graphing device if you have one.
30. T
he graph of a function y − f sxd is shown. At which
point(s) are the following true?
33. f sxd − x 3 2 12x 1 2
2
dy
d y
and
are both positive.
dx
dx 2
dy
d 2y
are both negative.
(b) and
dx
dx 2
dy
d 2y
(c)
is negative but
is positive.
dx
dx 2
(a)
y
C
A
D
34. f sxd − 36x 1 3x 2 2 2x 3
35. f sxd − 12 x 4 2 4x 2 1 336.
tsxd − 200 1 8x 3 1 x 4
37. hsxd − sx 1 1d5 2 5x 2 238.
hsxd − 5x 3 2 3x 5
39. Fsxd − x s6 2 x 40.
G
­ sxd − 5x 2y3 2 2x 5y3
41. Csxd − x 1y3sx 1 4d42.
f sxd − 2 sx 2 4x 2
E
43. f sd − 2 cos 1 cos 2 , 0 < < 2
44. Ssxd − x 2 sin x, 0 < x < 4
B
45. S
uppose the derivative of a function f is
f 9sxd − sx 1 1d2 sx 2 3d5 sx 2 6d 4. On what interval is f
increasing?
x
0
31–32 The graph of the derivative f 9 of a continuous function
f is shown.
(a) On what intervals is f increasing? Decreasing?
(b) At what values of x does f have a local maximum? Local
minimum?
(c) On what intervals is f concave upward? Concave
downward?
(d) State the x-coordinate(s) of the point(s) of inflection.
(e) Assuming that f s0d − 0, sketch a graph of f.
31. y
y=fª(x)
46. U
se the methods of this section to sketch the curve
y − x 3 2 3a 2x 1 2a 3, where a is a positive constant. What
do the members of this family of curves have in common?
How do they differ from each other?
; 47–48
(a) Use a graph of f to estimate the maximum and minimum
values. Then find the exact values.
(b) Estimate the value of x at which f increases most rapidly.
Then find the exact value.
x11
47. f sxd −
sx 2 1 1
48. f sxd − x 1 2 cos x, 0 < x < 2
2
0
2
4
6
8 x
_2
32.
; 49–50
(a) Use a graph of f to give a rough estimate of the intervals of
concavity and the coordinates of the points of inflection.
(b) Use a graph of f 0 to give better estimates.
49. f sxd − sin 2x 1 sin 4x, 0 < x < y
50. f sxd − sx 2 1d2 sx 1 1d3
y=fª(x)
CAS
2
0
_2
2
4
6
8 x
51–52 Estimate the intervals of concavity to one decimal place
by using a computer algebra system to compute and graph f 0.
51. f sxd −
52. f sxd −
x4 1 x3 1 1
sx 2 1 x 1 1
sx 1 1d3sx 2 1 5d
sx 3 1 1dsx 2 1 4d
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
230
Chapter 3 Applications of Differentiation
53. A
graph of a population of yeast cells in a new laboratory
culture as a function of time is shown.
(a) Describe how the rate of population increase varies.
(b) When is this rate highest?
(c)On what intervals is the population function concave
upward or downward?
(d) Estimate the coordinates of the inflection point.
f sxd − ax 3 1 bx 2 1 cx 1 d
that has a local maximum value of 3 at x − 22 and a local
minimum value of 0 at x − 1.
60. Show that the curve
y−
700
600
500
Number
400
of
yeast cells 300
200
11x
1 1 x2
h as three points of inflection and they all lie on one straight
line.
100
0
59. Find a cubic function
2
4
6
8
10 12 14 16 18
Time (in hours)
54. I n an episode of The Simpsons television show, Homer
reads from a newspaper and announces “Here’s good news!
According to this eye-catching article, SAT scores are
declining at a slower rate.” Interpret Homer’s statement in
terms of a function and its first and second derivatives.
55. T
he president announces that the national deficit is increasing, but at a decreasing rate. Interpret this statement in terms
of a function and its first and second derivatives.
56. L
et f std be the temperature at time t where you live and suppose that at time t − 3 you feel uncomfortably hot. How do
you feel about the given data in each case?
(a) f 9s3d − 2, f 0s3d − 4
(b) f 9s3d − 2, f 0s3d − 24
(c) f 9s3d − 22, f 0s3d − 4
(d) f 9s3d − 22, f 0s3d − 24
57. Let Kstd be a measure of the knowledge you gain by
studying for a test for t hours. Which do you think is larger,
Ks8d 2 Ks7d or Ks3d 2 Ks2d? Is the graph of K concave
upward or concave downward? Why?
58. C
offee is being poured into the mug shown in the figure at a
constant rate (measured in volume per unit time). Sketch a
rough graph of the depth of the coffee in the mug as a function of time. Account for the shape of the graph in terms of
concavity. What is the significance of the inflection point?
61. (a)If the function f sxd − x 3 1 ax 2 1 bx has the local
minimum value 292 s3 at x − 1ys3 , what are the values
of a and b?
(b)Which of the tangent lines to the curve in part (a) has the
smallest slope?
62. F
or what values of a and b is s2, 2.5d an inflection point of the
curve x 2 y 1 ax 1 by − 0? What additional inflection points
does the curve have?
63. S
how that the inflection points of the curve y − x sin x lie on
the curve y 2sx 2 1 4d − 4x 2.
64–66 Assume that all of the functions are twice differentiable
and the second derivatives are never 0.
64. (a)If f and t are concave upward on I, show that f 1 t is
concave upward on I.
(b)If f is positive and concave upward on I, show that the
function tsxd − f f sxdg 2 is concave upward on I.
65. (a)If f and t are positive, increasing, concave upward functions on I, show that the product function f t is concave
upward on I.
(b)Show that part (a) remains true if f and t are both
decreasing.
(c)Suppose f is increasing and t is decreasing. Show, by
giving three examples, that f t may be concave upward,
concave downward, or linear. Why doesn’t the argument
in parts (a) and (b) work in this case?
66. S
uppose f and t are both concave upward on s2`, `d.
Under what condition on f will the composite function
hsxd − f s tsxdd be concave upward?
67. Show that tan x . x for 0 , x , y2. [Hint: Show that
f sxd − tan x 2 x is increasing on s0, y2d.]
68. Prove that, for all x . 1,
2 sx . 3 2
1
x
69. S
how that a cubic function (a third-degree polynomial)
always has exactly one point of inflection. If its graph has
three x-intercepts x 1, x 2, and x 3, show that the x-coordinate of
the inflection point is sx 1 1 x 2 1 x 3 dy3.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
231
Section 3.4 Limits at Infinity; Horizontal Asymptotes
or what values of c does the polynomial
; 70. F
Psxd − x 4 1 cx 3 1 x 2 have two inflection points? One
inflection point? None? Illustrate by graphing P for several
values of c. How does the graph change as c decreases?
71. P
rove that if sc, f scdd is a point of inflection of the graph
of f and f 0 exists in an open interval that contains c, then
f 0scd − 0. [Hint: Apply the First Derivative Test and
Fermat’s Theorem to the function t − f 9.]
72. S
how that if f sxd − x 4, then f 0s0d − 0, but s0, 0d is not an
inflection point of the graph of f .
76. For what values of c is the function
f sxd − cx 1
increasing on s2`, `d?
77. T
he three cases in the First Derivative Test cover the situations
one commonly encounters but do not exhaust all possibilities.
Consider the functions f, t, and h whose values at 0 are all 0
and, for x ± 0,
1
1
f sxd − x 4 sin tsxd − x 4 2 1 sin
x
x
S
| |
73. S
how that the function tsxd − x x has an inflection point at
s0, 0d but t0s0d does not exist.
74. Suppose that f 09 is continuous and f 9scd − f 0scd − 0, but
f -scd . 0. Does f have a local maximum or minimum
at c? Does f have a point of inflection at c?
75. Suppose f is differentiable on an interval I and f 9sxd . 0 for
all numbers x in I except for a single number c. Prove that f is
increasing on the entire interval I.
1
x 13
2
hsxd − x 4 22 1 sin
S
D
D
1
x
(a)Show that 0 is a critical number of all three functions but
their derivatives change sign infinitely often on both sides
of 0.
(b)Show that f has neither a local maximum nor a local min­
imum at 0, t has a local minimum, and h has a local
maximum.
In Sections 1.5 and 1.7 we investigated infinite limits and vertical asymptotes. There we
let x approach a number and the result was that the values of y became arbitrarily large
(positive or negative). In this section we let x become arbitrarily large (positive or negative) and see what happens to y. We will find it very useful to consider this so-called end
behavior when sketching graphs.
Let’s begin by investigating the behavior of the function f defined by
f sxd −
x
0
61
62
63
64
65
610
650
6100
61000
f sxd
21
0
0.600000
0.800000
0.882353
0.923077
0.980198
0.999200
0.999800
0.999998
x2 2 1
x2 1 1
as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 1.
y
0
y=1
1
y=
≈-1
≈+1
x
FIGURE 1
As x grows larger and larger you can see that the values of f sxd get closer and closer
to 1. (The graph of f approaches the horizontal line y − 1 as we look to the right.) In
fact, it seems that we can make the values of f sxd as close as we like to 1 by taking x
sufficiently large. This situation is expressed symbolically by writing
lim
xl`
x2 2 1
−1
x2 1 1
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
232
Chapter 3 Applications of Differentiation
In general, we use the notation
lim f sxd − L
xl`
to indicate that the values of f sxd approach L as x becomes larger and larger.
1 Intuitive Definition of a Limit at Infinity Let f be a function defined on
some interval sa, `d. Then
lim f sxd − L
xl`
means that the values of f sxd can be made arbitrarily close to L by requiring x to
be sufficiently large.
Another notation for lim x l ` f sxd − L is
f sxd l L as x l `
The symbol ` does not represent a number. Nonetheless, the expression lim f sxd − L
x l`
is often read as
“the limit of f sxd, as x approaches infinity, is L”
or
“the limit of f sxd, as x becomes infinite, is L”
or
“the limit of f sxd, as x increases without bound, is L”
The meaning of such phrases is given by Definition 1. A more precise definition, similar
to the «, definition of Section 1.7, is given at the end of this section.
Geometric illustrations of Definition 1 are shown in Figure 2. Notice that there are
many ways for the graph of f to approach the line y − L (which is called a horizontal
asymptote) as we look to the far right of each graph.
y
y
y=L
y
y=ƒ
y=ƒ
0
FIGURE 2 xl`
y=ƒ
y=L
x
Examples illustrating lim f sxd − L
y=L
0
x
0
x
Referring back to Figure 1, we see that for numerically large negative values of x,
the values of f sxd are close to 1. By letting x decrease through negative values without
bound, we can make f sxd as close to 1 as we like. This is expressed by writing
lim
x l2`
x2 2 1
−1
x2 1 1
The general definition is as follows.
2 Definition Let f be a function defined on some interval s2`, ad. Then
lim f sxd − L
x l 2`
means that the values of f sxd can be made arbitrarily close to L by requiring x to
be sufficiently large negative.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.4 Limits at Infinity; Horizontal Asymptotes
y
233
Again, the symbol 2` does not represent a number, but the expression lim f sxd − L
x l 2`
is often read as
y=ƒ
“the limit of f sxd, as x approaches negative infinity, is L”
y=L
0
x
Definition 2 is illustrated in Figure 3. Notice that the graph approaches the line y − L as
we look to the far left of each graph.
y
3 Definition The line y − L is called a horizontal asymptote of the curve
y − f sxd if either
lim f sxd − L or lim f sxd − L
y=ƒ
y=L
x l`
0
x l 2`
x
FIGURE 3 Examples illustrating lim f sxd − L
For instance, the curve illustrated in Figure 1 has the line y − 1 as a horizontal asymp­
tote because
x l 2`
lim
xl`
x2 2 1
−1
x2 1 1
The curve y − f sxd sketched in Figure 4 has both y − 21 and y − 2 as horizontal asymptotes because
lim f sxd − 21
xl`
and
lim f sxd − 2
x l2`
y
2
y=2
0
y=_1
y=ƒ
_1
x
FIGURE 4
y
Example 1 Find the infinite limits, limits at infinity, and asymptotes for the function
f whose graph is shown in Figure 5.
SOLUTION We see that the values of f sxd become large as x l 21 from both sides, so
2
0
2
lim f sxd − `
x l21
x
Notice that f sxd becomes large negative as x approaches 2 from the left, but large posi­
tive as x approaches 2 from the right. So
FIGURE 5
lim f sxd − 2` and lim1 f sxd − `
x l 22
x l2
Thus both of the lines x − 21 and x − 2 are vertical asymptotes.
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234
Chapter 3 Applications of Differentiation
As x becomes large, it appears that f sxd approaches 4. But as x decreases through
negative values, f sxd approaches 2. So
lim f sxd − 4 and lim f sxd − 2
xl`
x l2`
This means that both y − 4 and y − 2 are horizontal asymptotes.
Example 2 Find lim
xl`
n
1
1
and lim .
x l2` x
x
SOLUTION Observe that when x is large, 1yx is small. For instance,
1
1
1
− 0.01 − 0.0001 − 0.000001
100
10,000
1,000,000
y
y=∆
0
FIGURE 6 lim
xl`
1
1
− 0, lim
−0
x l2` x
x
In fact, by taking x large enough, we can make 1yx as close to 0 as we please. There­
fore, according to Definition 1, we have
lim
x
xl`
1
−0
x
Similar reasoning shows that when x is large negative, 1yx is small negative, so we also
have
1
lim
−0
x l2` x
It follows that the line y − 0 (the x-axis) is a horizontal asymptote of the curve
y − 1yx. (This is an equilateral hyperbola; see Figure 6.)
n
Most of the Limit Laws that were given in Section 1.6 also hold for limits at infinity. It
can be proved that the Limit Laws listed in Section 1.6 (with the exception of Laws 9 and
10) are also valid if “x l a” is replaced by “x l `” or “x l 2`.” In particular, if we
combine Laws 6 and 11 with the results of Example 2, we obtain the following important
rule for calculating limits.
4 Theorem If r . 0 is a rational number, then
lim
xl`
1
−0
xr
If r . 0 is a rational number such that x r is defined for all x, then
lim
x l2`
1
−0
xr
Example 3 Evaluate
lim
x l`
3x 2 2 x 2 2
5x 2 1 4x 1 1
and indicate which properties of limits are used at each stage.
SOLUTION As x becomes large, both numerator and denominator become large, so it
isn’t obvious what happens to their ratio. We need to do some preliminary algebra.
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Section 3.4 Limits at Infinity; Horizontal Asymptotes
235
To evaluate the limit at infinity of any rational function, we first divide both the
numerator and denominator by the highest power of x that occurs in the denominator.
(We may assume that x ± 0, since we are interested only in large values of x.) In this
case the highest power of x in the denominator is x 2, so we have
3x 2 2 x 2 2
1
2
32 2 2
3x 2 x 2 2
x2
x
x
lim
− lim
− lim
x l ` 5x 2 1 4x 1 1
x l ` 5x 2 1 4x 1 1
x l`
4
1
51 1 2
2
x
x
x
2
S
S
D
D
1
2
2 2
x
x
−
4
1
lim 5 1 1 2
xl`
x
x
lim 3 2
xl`
1
2 2 lim
x l`
x l` x
x l`
−
1
lim 5 1 4 lim 1 lim
x l`
x l` x
x l`
y
lim 3 2 lim
y=0.6
0
FIGURE 7 2
y−
3x 2 x 2 2
5x 2 1 4x 1 1
1
(by Limit Law 5)
x
−
32020
51010
−
3
5
1
x2
1
x2
(by 1, 2, and 3)
(by 7 and Theorem 4)
A similar calculation shows that the limit as x l 2` is also 35. Figure 7 illustrates the
results of these calculations by showing how the graph of the given rational function
approaches the horizontal asymptote y − 35 − 0.6.
n
Example 4 Find the horizontal and vertical asymptotes of the graph of the function
f sxd −
s2x 2 1 1
3x 2 5
SOLUTION Dividing both numerator and denominator by x and using the properties of
limits, we have
Î
s2x 2 1 1
x
s2x 2 1 1
lim
− lim
− lim
x l ` 3x 2 5
xl`
xl`
3x 2 5
x
lim
−
xl`
lim
xl`
2x 2 1 1
x2
(since sx 2 − x for x . 0)
3x 2 5
x
Î
Î
S D
21
32
5
x
1
x2
1
x2
s2 1 0
s2
−
−
−
325?0
3
1
lim 3 2 5 lim
xl`
xl` x
lim 2 1 lim
xl`
xl`
Therefore the line y − s2 y3 is a horizontal asymptote of the graph of f.
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236
Chapter 3 Applications of Differentiation
In computing the limit as x l 2`, we must remember that for x , 0, we have
sx 2 − x − 2x. So when we divide the numerator by x, for x , 0 we get
| |
Î
s2x 2 1 1
s2x 2 1 1
−
−2
x
2 sx 2
Î
2x 2 1 1
−2
x2
21
1
x2
Therefore
s2x 1 1
lim
− lim
x l2`
x l2`
3x 2 5
2
21
1
x2
5
32
x
Î
2
−
1
s2
x2
−2
1
3
3 2 5 lim
x l2` x
2 1 lim
x l2`
Thus the line y − 2s2 y3 is also a horizontal asymptote.
A vertical asymptote is likely to occur when the denominator, 3x 2 5, is 0, that is,
when x − 53. If x is close to 53 and x . 53, then the denominator is close to 0 and 3x 2 5
is positive. The numerator s2x 2 1 1 is always positive, so f sxd is positive. Therefore
y
œ„2
y= 3
lim
x
x l s5y3d1
œ„2
y=_ 3
x=
s2x 2 1 1
−`
3x 2 5
(Notice that the numerator does not approach 0 as x l 5y3).
If x is close to 53 but x , 53, then 3x 2 5 , 0 and so f sxd is large negative. Thus
5
3
lim
FIGURE 8 y−
Î
2
x l s5y3d2
s2 x 2 1 1
s2x 2 1 1
− 2`
3x 2 5
The vertical asymptote is x − 53. All three asymptotes are shown in Figure 8.
3x 2 5
n
Example 5 Compute lim (sx 2 1 1 2 x).
x l`
SOLUTION Because both sx 2 1 1 and x are large when x is large, it’s difficult to see
We can think of the given function as
having a denominator of 1.
what happens to their difference, so we use algebra to rewrite the function. We first
multiply numerator and denominator by the conjugate radical:
lim (sx 2 1 1 2 x) − lim (sx 2 1 1 2 x) x l`
x l`
− lim
y
x l`
y=œ„„„„„-x
≈+1
FIGURE 9
1
sx 2 1 1d 2 x 2
1
− lim
x l ` sx 2 1 1 1 x
sx 2 1 1 1 x
Notice that the denominator of this last expression (sx 2 1 1 1 x) becomes large as
x l ` (it’s bigger than x). So
1
0
sx 2 1 1 1 x
sx 2 1 1 1 x
x
lim (sx 2 1 1 2 x) − lim
x l`
Figure 9 illustrates this result.
x l`
1
−0
sx 2 1 1 1 x
n
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Section 3.4 Limits at Infinity; Horizontal Asymptotes
237
1
x
Example 6 Evaluate lim sin .
xl`
PS The problem-solving strategy for
Example 6 is introducing something extra (see page 98). Here, the something
extra, the auxiliary aid, is the new
variable t.
SOLUTION If we let t − 1yx, then t l 01 as x l `. Therefore
lim sin
xl`
1
− lim1 sin t − 0
tl0
x
(See Exercise 73.)
n
Example 7 Evaluate lim sin x.
xl`
SOLUTION As x increases, the values of sin x oscillate between 1 and 21 infinitely often
and so they don’t approach any definite number. Thus lim x l` sin x does not exist.
n
Infinite Limits at Infinity
The notation
lim f sxd − `
x l`
is used to indicate that the values of f sxd become large as x becomes large. Similar mean­
ings are attached to the following symbols:
lim f sxd − ` lim f sxd − 2` lim f sxd − 2`
x l 2`
x l`
x l 2`
Example 8 Find lim x 3 and lim x 3.
xl`
x l2`
SOLUTION When x becomes large, x 3 also becomes large. For instance,
y
10 3 − 1000 100 3 − 1,000,000 1000 3 − 1,000,000,000
y=˛
0
x
In fact, we can make x 3 as big as we like by requiring x to be large enough. Therefore
we can write
lim x 3 − `
xl`
Similarly, when x is large negative, so is x 3. Thus
lim x 3 − 2`
FIGURE 10 x l2`
lim x 3 − `, lim x 3 − 2`
xl`
x l2`
These limit statements can also be seen from the graph of y − x 3 in Figure 10.
n
Example 9 Find lim sx 2 2 xd.
x l`
SOLUTION It would be wrong to write
lim sx 2 2 xd − lim x 2 2 lim x − ` 2 `
x l`
x l`
x l`
The Limit Laws can’t be applied to infinite limits because ` is not a number
(` 2 ` can’t be defined). However, we can write
lim sx 2 2 xd − lim xsx 2 1d − `
x l`
x l`
because both x and x 2 1 become arbitrarily large and so their product does too.
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n
238
Chapter 3 Applications of Differentiation
Example 10 Find lim
xl`
x2 1 x
.
32x
SOLUTION As in Example 3, we divide the numerator and denominator by the highest
power of x in the denominator, which is just x:
lim
x l`
x2 1 x
x11
− lim
− 2`
x l` 3
32x
21
x
because x 1 1 l ` and 3yx 2 1 l 0 2 1 − 21 as x l `.
n
The next example shows that by using infinite limits at infinity, together with inter­
cepts, we can get a rough idea of the graph of a polynomial without computing derivatives.
Example 11 Sketch the graph of y − sx 2 2d4sx 1 1d3sx 2 1d by finding its inter­
cepts and its limits as x l ` and as x l 2`.
SOLUTION The y-intercept is f s0d − s22d4s1d3s21d − 216 and the x-intercepts are
found by setting y − 0: x − 2, 21, 1. Notice that since sx 2 2d4 is never negative,
the function doesn’t change sign at 2; thus the graph doesn’t cross the x-axis at 2. The
graph crosses the axis at 21 and 1.
When x is large positive, all three factors are large, so
y
0
_1
1
2
x
lim sx 2 2d4sx 1 1d3sx 2 1d − `
xl`
When x is large negative, the first factor is large positive and the second and third fac­
tors are both large negative, so
_16
lim sx 2 2d4sx 1 1d3sx 2 1d − `
x l2`
FIGURE 11
y − sx 2 2d4 sx 1 1d3 sx 2 1d
Combining this information, we give a rough sketch of the graph in Figure 11.
n
Precise Definitions
Definition 1 can be stated precisely as follows.
5 Precise Definition of a Limit at Infinity Let f be a function defined on some
interval sa, `d. Then
lim f sxd − L
xl`
means that for every « . 0 there is a corresponding number N such that
|
|
if x . N then f sxd 2 L , «
In words, this says that the values of f sxd can be made arbitrarily close to L (within a
distance «, where « is any positive number) by requiring x to be sufficiently large (larger
than N, where N depends on «). Graphically it says that by keeping x large enough
(larger than some number N) we can make the graph of f lie between the given hori­
zontal lines y − L 2 « and y − L 1 « as in Figure 12. This must be true no matter how
small we choose «.
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239
Section 3.4 Limits at Infinity; Horizontal Asymptotes
y
y=ƒ
y=L+∑
∑
L ∑
y=L-∑
ƒ is
in here
0
FIGURE 12
x
N
lim f sxd − L
when x is in here
xl`
Figure 13 shows that if a smaller value of « is chosen, then a larger value of N may
be required.
L
FIGURE 13
y=ƒ
y=L+∑
y=L-∑
0
lim f sxd − L
N
x
xl`
Similarly, a precise version of Definition 2 is given by Definition 6, which is illus­
trated in Figure 14.
6 Definition Let f be a function defined on some interval s2`, ad. Then
lim f sxd − L
x l 2`
means that for every « . 0 there is a corresponding number N such that
|
|
if x , N then f sxd 2 L , «
y
y=ƒ
y=L+∑
L
y=L-∑
FIGURE 14
0
N
x
lim f sxd − L
x l2`
In Example 3 we calculated that
lim
xl`
3x 2 2 x 2 2
3
−
2
5x 1 4x 1 1
5
In the next example we use a graphing device to relate this statement to Definition 5 with
L − 35 − 0.6 and « − 0.1.
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240
Chapter 3 Applications of Differentiation
TEC In Module 1.7y3.4 you can
explore the precise definition of a limit
both graphically and numerically.
Example 12 Use a graph to find a number N such that
if x . N then Z
3x 2 2 x 2 2
2 0.6
5x 2 1 4x 1 1
Z
, 0.1
SOLUTION We rewrite the given inequality as
3x 2 2 x 2 2
, 0.7
5x 2 1 4x 1 1
0.5 ,
1
We need to determine the values of x for which the given curve lies between the hori­
zontal lines y − 0.5 and y − 0.7. So we graph the curve and these lines in Figure 15.
Then we use the cursor to estimate that the curve crosses the line y − 0.5 when
x < 6.7. To the right of this number it seems that the curve stays between the lines
y − 0.5 and y − 0.7. Rounding up to be safe, we can say that
y=0.7
y=0.5
y=
3≈-x-2
5≈+4x+1
if x . 7 then 15
0
FIGURE 15
Z
3x 2 2 x 2 2
2 0.6
5x 2 1 4x 1 1
Z
, 0.1
In other words, for « − 0.1 we can choose N − 7 (or any larger number) in Defini­­tion 5.
1
− 0.
x
Example 13 Use Definition 5 to prove that lim
xl`
SOLUTION Given « . 0, we want to find N such that
if x . N then n
Z
1
20
x
Z
,«
In computing the limit we may assume that x . 0. Then 1yx , « &? x . 1y«. Let’s
choose N − 1y«. So
if x . N −
Z
1
1
then 20
«
x
Z
−
1
,«
x
Therefore, by Definition 5,
lim
xl`
1
−0
x
Figure 16 illustrates the proof by showing some values of « and the corresponding
values of N.
y
y
∑=1
0
FIGURE 16 N=1
x
∑=0.2
0
y
N=5
x
∑=0.1
0
N=10
x
n
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241
Section 3.4 Limits at Infinity; Horizontal Asymptotes
y
M
Finally we note that an infinite limit at infinity can be defined as follows. The geomet­
ric illustration is given in Figure 17.
y=M
0
7 Definition of an Infinite Limit at Infinity Let f be a function defined on
some interval sa, `d. Then
lim f sxd − `
x
N
xl`
FIGURE 17
means that for every positive number M there is a corresponding positive number
N such that
if x . N then f sxd . M
lim f sxd − `
xl`
Similar definitions apply when the symbol ` is replaced by 2`. (See Exercise 74.)
y
1.Explain in your own words the meaning of each of the
following.
(a) lim f sxd − 5
(b) lim f sxd − 3
xl`
x l 2`
1
2. (a)Can the graph of y − f sxd intersect a vertical asymptote?
Can it intersect a horizontal asymptote? Illustrate by
sketching graphs.
(b)How many horizontal asymptotes can the graph of
y − f sxd have? Sketch graphs to illustrate the possibilities.
3.For the function f whose graph is given, state the following.
(a) lim f sxd
(b) lim f sxd
x l`
x l 2`
(c) lim f sxd
(d) lim f sxd
x l1
; 5.Guess the value of the limit
lim
x l`
x l3
y
; 6.(a) Use a graph of
1
x
x l`
x l 2`
(c) lim tsxd
(d) lim2 tsxd
(e) lim1 tsxd
(f ) The equations of the asymptotes
x l2
S D
f sxd − 1 2
4.For the function t whose graph is given, state the following.
(a) lim tsxd
(b) lim tsxd
xl0
x2
2x
by evaluating the function f sxd − x 2y2 x for x − 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f to
support your guess.
(e) The equations of the asymptotes
1
x
1
xl2
2
x
x
to estimate the value of lim x l ` f sxd correct to two
decimal places.
(b)Use a table of values of f sxd to estimate the limit to
four decimal places.
7–8 Evaluate the limit and justify each step by indicating the
appropriate properties of limits.
7. lim
xl`
Î
2x 2 2 7
9x 3 1 8x 2 4
8.
lim
2
xl`
5x 1 x 2 3
3 2 5x 1 x 3
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242
chapter 3 Applications of Differentiation
9–32 Find the limit or show that it does not exist.
9. lim
xl`
2
3x 2 2
2x 1 1
12x
10.
lim
xl` x3 2 x 1 1
x22
11. lim 2
x l 2` x 1 1
35–40 Find the horizontal and vertical asymptotes of each
curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes.
4x 3 1 6x 2 2 2
12.
lim
x l 2` 2x 3 2 4x 1 5
35. y −
5 1 4x
x13
2x 2 1 1
36.
y−
2
3x 1 2x 2 1
13. lim
st 1 t 2
2t 2 t 2
t 2 t st
14.
lim
tl ` 2t 3y2 1 3t 2 5
37. y −
2x 2 1 x 2 1
x2 1 x 2 2
1 1 x4
38.
y− 2
x 2 x4
15. lim
s2x 2 1 1d2
sx 2 1d2sx 2 1 xd
x2
16.
lim
x l ` sx 4 1 1
39. y −
x3 2 x
x 2 6x 1 5
x29
40.
y−
2
s4x 1 3x 1 2
17. lim
s1 1 4x 6
2 2 x3
18. lim
19. lim
sx 1 3x 2
4x 2 1
20. lim
tl`
xl`
xl`
xl`
s1 1 4x 6
2 2 x3
x l 2`
xl`
22. lim (s4x 2 1 3x 1 2 x )
x l2`
23. lim (sx 1 ax 2 sx 1 bx
x l`
f sxd −
3x 3 1 500x 2
x 1 500x 2 1 100x 1 2000
3
by graphing f for 210 < x < 10. Then calculate the equation of the asymptote by evaluating the limit. How do you
explain the discrepancy?
x l`
2
; 41. Estimate the horizontal asymptote of the function
x 1 3x 2
4x 2 1
21. lim (s9x 2 1 x 2 3x)
2
2
; 42. (a) Graph the function
)
f sxd −
24. lim cos x
x l`
x 4 2 3x 2 1 x
25. lim 3
xl` x 2 x 1 2
26. lim sx 1 1
27. lim sx 2 1 2x 7 d
1 1 x6
28.
lim 4
x l 2` x 1 1
29. lim s x 2 sx d
30. lim sx 2 2 x 4 d
x l 2`
xl`
31. lim x sin
xl`
2
x l`
xl`
1
x
32. lim sx sin
xl`
1
x
; 33. (a) Estimate the value of
s2x 2 1 1
3x 2 5
How many horizontal and vertical asymptotes do you
observe? Use the graph to estimate the values of the
limits
lim
x l`
s2x 2 1 1
s2x 2 1 1
and lim
x l 2`
3x 2 5
3x 2 5
(b)By calculating values of f sxd, give numerical estimates
of the limits in part (a).
(c)Calculate the exact values of the limits in part (a). Did
you get the same value or different values for these two
limits? [In view of your answer to part (a), you might
have to check your calculation for the second limit.]
43. Let P and Q be polynomials. Find
lim (sx 2 1 x 1 1 1 x)
x l 2`
lim
by graphing the function f sxd − sx 1 x 1 1 1 x.
xl`
2
(b)Use a table of values of f sxd to guess the value of the
limit.
(c)Prove that your guess is correct.
; 34. (a) Use a graph of
f sxd − s3x 2 1 8x 1 6 2 s3x 2 1 3x 1 1
to estimate the value of lim x l ` f sxd to one decimal
place.
(b)Use a table of values of f sxd to estimate the limit to
four decimal places.
(c)Find the exact value of the limit.
Psxd
Qsxd
if the degree of P is (a) less than the degree of Q and
(b) greater than the degree of Q.
44. M
ake a rough sketch of the curve y − x n (n an integer)
for the following five cases:
(i) n − 0
(ii) n . 0, n odd
(iii) n . 0, n even
(iv) n , 0, n odd
(v) n , 0, n even
Then use these sketches to find the following limits.
(a) lim1 x n
(b) lim2 x n
x l0
(c) lim x n
x l`
x l0
(d) lim x n
x l 2`
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.4 Limits at Infinity; Horizontal Asymptotes
45. F
ind a formula for a function f that satisfies the following
conditions:
lim f sxd − 0, lim f sxd − 2`, f s2d − 0,
x l 6`
x l0
243
60. ts0d − 0, t0sxd , 0 for x ± 0, lim x l2` tsxd − `,
lim x l ` tsxd − 2`, lim x l 02 t9sxd − 2`,
lim x l 01 t9sxd − `
lim2 f sxd − `, lim1 f sxd − 2`
x l3
x l3
46. F
ind a formula for a function that has vertical asymptotes
x − 1 and x − 3 and horizontal asymptote y − 1.
47. A function f is a ratio of quadratic functions and has a
vertical asymptote x − 4 and just one x-intercept, x − 1.
It is known that f has a removable discontinuity at x − 21
and lim x l21 f sxd − 2. Evaluate
(a)f s0d
(b) lim f sxd
xl`
48–51 Find the horizontal asymptotes of the curve and use them,
together with concavity and intervals of increase and decrease, to
sketch the curve.
48. y −
50. y −
1 1 2x 2
1 1 x2
x
sx 2 1 1
49. y −
12x
11x
51. y −
x
x2 1 1
52–56 Find the limits as x l ` and as x l 2`. Use this information, together with intercepts, to give a rough sketch of the
graph as in Example 11.
52. y − 2x 3 2 x 4
53.
y − x 4 2 x6
54. y − x sx 1 2d sx 2 1d
3
2
55. y − s3 2 xds1 1 xd 2s1 2 xd 4 sin x
.
x
(b)Graph f sxd − ssin xdyx. How many times does the graph
cross the asymptote?
61. (a)Use the Squeeze Theorem to evaluate lim
; xl`
; 62. By the end behavior of a function we mean the behavior of
its values as x l ` and as x l 2`.
(a)Describe and compare the end behavior of the functions
Psxd − 3x 5 2 5x 3 1 2x Qsxd − 3x 5
by graphing both functions in the viewing rectangles
f22, 2g by f22, 2g and f210, 10g by f210,000, 10,000g.
(b)Two functions are said to have the same end behavior if
their ratio approaches 1 as x l `. Show that P and Q
have the same end behavior.
63. Find lim x l ` f sxd if
4x 2 1
4x 2 1 3x
, f sxd ,
x
x2
for all x . 5.
64. (a)A tank contains 5000 L of pure water. Brine that contains 30 g of salt per liter of water is pumped into the
tank at a rate of 25 Lymin. Show that the concentration
of salt after t minutes (in grams per liter) is
56. y − x 2sx 2 2 1d 2sx 1 2d
57–60 Sketch the graph of a function that satisfies all of the
given conditions.
57.f 9s2d − 0, f s2d − 21, f s0d − 0,
f 9sxd , 0 if 0 , x , 2, f 9sxd . 0 if x . 2,
f 0sxd , 0 if 0 < x , 1 or if x . 4,
f 0sxd . 0 if 1 , x , 4, lim x l ` f sxd − 1,
f s2xd − f sxd for all x
Cstd −
(b) What happens to the concentration as t l `?
; 65. Use a graph to find a number N such that
Z
if x . N then f 0sxd . 0 if x . 4, lim x l ` f sxd − 0,
f s2xd − 2f sxd for all x
59. f s1d − f 9s1d − 0, lim x l21 f sxd − `, lim x l22 f sxd − 2`,
lim x l 0 f sxd − 2`, lim x l2` f sxd − `, lim x l ` f sxd − 0,
f 0sxd . 0 for x . 2, f 0sxd , 0 for x , 0 and for
0,x,2
Z
3x2 1 1
2 1.5 , 0.05
2x 2 1 x 1 1
; 66. For the limit
lim
xl`
58. f 9s2d − 0, f 9s0d − 1, f 9sxd . 0 if 0 , x , 2,
f 9sxd , 0 if x . 2, f 0sxd , 0 if 0 , x , 4,
30t
200 1 t
1 2 3x
sx 2 1 1
− 23
illustrate Definition 5 by finding values of N that correspond
to « − 0.1 and « − 0.05.
; 67. For the limit
lim
x l2`
1 2 3x
sx 2 1 1
−3
illustrate Definition 6 by finding values of N that correspond
to « − 0.1 and « − 0.05.
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244
chapter 3 Applications of Differentiation
; 68. For the limit
lim
xl`
71. Use Definition 6 to prove that lim
3x
sx 2 3
x l2`
−`
illustrate Definition 7 by finding a value of N that corresponds to M − 100.
69. (a)How large do we have to take x so that
1yx 2 , 0.0001?
(b) Taking r − 2 in Theorem 4, we have the statement
72. Prove, using Definition 7, that lim x 3 − `.
xl`
73. (a) Prove that
lim f sxd − lim1 f s1ytd
tl0
xl`
and
f s1ytd
lim f sxd − t lim
l 02
xl2`
if these limits exist.
(b) Use part (a) and Exercise 61 to find
1
−0
x2
lim
xl`
Prove this directly using Definition 5.
70. (a)How large do we have to take x so that
1ysx , 0.0001?
(b) Taking r − 12 in Theorem 4, we have the statement
lim
xl`
1
sx
lim x sin
x l 01
lim f sxd − 2`
x l2`
−0
y=8˛-21≈+18x+2
_2
4
_10
FIGURE 1
8
0
y=8˛-21≈+18x+2
6
FIGURE 2
2
1
x
74. Formulate a precise definition of
Then use your definition to prove that
Prove this directly using Definition 5.
30
1
− 0.
x
lim s1 1 x 3 d − 2`
x l2`
So far we have been concerned with some particular aspects of curve sketching: domain,
range, symmetry, limits, continuity, and vertical asymptotes in Chapter 1; deriva­tives and
tangents in Chapter 2; and extreme values, intervals of increase and decrease, concavity,
points of inflection, and horizontal asymptotes in this chapter. It is now time to put all of
this information together to sketch graphs that reveal the important features of functions.
You might ask: Why don’t we just use a graphing calculator or computer to graph a
curve? Why do we need to use calculus?
It’s true that current technology is capable of producing very accurate graphs. But
even the best graphing devices have to be used intelligently. It is easy to arrive at a
misleading graph, or to miss important details of a curve, when relying solely on technology. (See “Graphing Calculators and Computers” at www.stewartcalculus.com, especially Examples 1, 3, 4, and 5. See also Section 3.6.) The use of calculus enables us to
discover the most interesting aspects of graphs and in many cases to calculate maximum
and minimum points and inflection points exactly instead of approximately.
For instance, Figure 1 shows the graph of f sxd − 8x 3 2 21x 2 1 18x 1 2. At first
glance it seems reasonable: it has the same shape as cubic curves like y − x 3, and it
appears to have no maximum or minimum point. But if you compute the derivative, you
will see that there is a maximum when x − 0.75 and a minimum when x − 1. Indeed,
if we zoom in to this portion of the graph, we see that behavior exhibited in Figure 2.
Without calculus, we could easily have overlooked it.
In the next section we will graph functions by using the interaction between calculus
and graphing devices. In this section we draw graphs by first considering the following
information. We don’t assume that you have a graphing device, but if you do have one
you should use it as a check on your work.
Guidelines for Sketching a Curve
The following checklist is intended as a guide to sketching a curve y − f sxd by hand. Not
every item is relevant to every function. (For instance, a given curve might not have an
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245
Section 3.5 Summary of Curve Sketching
y
0
x
(a) Even function: reflectional symmetry
y
x
0
(b) Odd function: rotational symmetry
FIGURE 3
asymptote or possess symmetry.) But the guidelines provide all the information you need
to make a sketch that displays the most important aspects of the function.
A. Domain It’s often useful to start by determining the domain D of f , that is, the set
of values of x for which f sxd is defined.
B. Intercepts The y-intercept is f s0d and this tells us where the curve intersects the
y-axis. To find the x-intercepts, we set y − 0 and solve for x. (You can omit this step
if the equa­tion is difficult to solve.)
C. Symmetry
(i ) If f s2xd − f sxd for all x in D, that is, the equation of the curve is unchanged
when x is replaced by 2x, then f is an even function and the curve is symmetric
about the y-axis. This means that our work is cut in half. If we know what the curve
looks like for x > 0, then we need only reflect about the y-axis to obtain the complete curve [see Figure 3(a)]. Here are some examples: y − x 2, y − x 4, y − x , and
y − cos x.
(ii) If f s2xd − 2f sxd for all x in D, then f is an odd function and the curve
is sym­metric about the origin. Again we can obtain the complete curve if we know
what it looks like for x > 0. [Rotate 180° about the origin; see Figure 3(b).] Some
simple examples of odd functions are y − x, y − x 3, y − x 5, and y − sin x.
(iii) If f sx 1 pd − f sxd for all x in D, where p is a positive constant, then f is
called a periodic function and the smallest such number p is called the period. For
instance, y − sin x has period 2 and y − tan x has period . If we know what the
graph looks like in an interval of length p, then we can use translation to sketch the
entire graph (see Figure 4).
| |
y
period p
FIGURE 4
Periodic function:
translational symmetry
a-p
0
a
a+p
a+2p
x
D. Asymptotes
(i) Horizontal Asymptotes. Recall from Section 3.4 that if either lim x l ` f sxd − L
or lim x l2 ` f sxd − L, then the line y − L is a horizontal asymptote of the curve
y − f sxd. If it turns out that lim x l ` f sxd − ` (or 2`), then we do not have an
asymptote to the right, but this fact is still useful information for sketching the curve.
(ii) Vertical Asymptotes. Recall from Section 1.5 that the line x − a is a vertical
asymptote if at least one of the following statements is true:
1
lim f sxd − `
x l a1
lim f sxd − 2`
x l a1
lim f sxd − `
x l a2
lim f sxd − 2`
x l a2
(For rational functions you can locate the vertical asymptotes by equating the denominator to 0 after canceling any common factors. But for other functions this method
does not apply.) Furthermore, in sketching the curve it is very useful to know exactly
which of the statements in (1) is true. If f sad is not defined but a is an endpoint of the
domain of f, then you should compute lim x l a2 f sxd or lim x l a1 f sxd, whether or not
this limit is infinite.
(iii) Slant Asymptotes. These are discussed at the end of this section.
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246
Chapter 3 Applications of Differentiation
E.Intervals of Increase or Decrease Use the I/D Test. Compute f 9sxd and find the
intervals on which f 9sxd is positive ( f is increasing) and the intervals on which f 9sxd
is negative ( f is decreasing).
F.Local Maximum and Minimum Values Find the critical numbers of f [the numbers c where f 9scd − 0 or f 9scd does not exist]. Then use the First Derivative Test.
If f 9 changes from positive to negative at a critical number c, then f scd is a local
maximum. If f 9 changes from negative to positive at c, then f scd is a local minimum.
Although it is usually prefer­able to use the First Derivative Test, you can use the
Second Derivative Test if f 9scd − 0 and f 0scd ± 0. Then f 0scd . 0 implies that f scd
is a local minimum, whereas f 0scd , 0 implies that f scd is a local maximum.
G.Concavity and Points of Inflection Compute f 0sxd and use the Concavity Test. The
curve is concave upward where f 0sxd . 0 and concave downward where f 0sxd , 0.
Inflection points occur where the direction of concavity changes.
H.Sketch the Curve Using the information in items A–G, draw the graph. Sketch the
asymptotes as dashed lines. Plot the intercepts, maximum and minimum points, and
inflection points. Then make the curve pass through these points, rising and falling
according to E, with concavity according to G, and approaching the asymptotes. If
additional accuracy is desired near any point, you can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds.
Example 1 Use the guidelines to sketch the curve y −
A. The domain is
hx
|x
2
2 1 ± 0j − hx
2x 2
.
x 21
2
| x ± 61j − s2`, 21d ø s21, 1d ø s1, `d
B. The x- and y-intercepts are both 0.
C. Since f s2xd − f sxd, the function f is even. The curve is symmetric about the y-axis.
D.
y
lim
x l6`
Therefore the line y − 2 is a horizontal asymptote.
Since the denominator is 0 when x − 61, we compute the following limits:
lim1
x l1
y=2
lim 1
x l 21
0
x=_1
x
x=1
2x 2
−`
x2 2 1
2x 2
− 2`
x 21
2
We have shown the curve approaching
its horizontal asymptote from above
in Figure 5. This is confirmed by the
intervals of increase and decrease.
lim2
2x 2
− 2`
x 21
lim 2
2x 2
−`
x 21
x l1
x l 21
2
2
Therefore the lines x − 1 and x − 21 are vertical asymptotes. This information
about limits and asymptotes enables us to draw the preliminary sketch in Figure 5,
showing the parts of the curve near the asymptotes.
FIGURE 5
Preliminary sketch
2x 2
2
− lim
−2
2
x l6` 1 2 1yx 2
x 21
E.
f 9sxd −
sx 2 2 1ds4xd 2 2x 2 ? 2x
24x
− 2
2
2
sx 2 1d
sx 2 1d2
Since f 9sxd . 0 when x , 0 sx ± 21d and f 9sxd , 0 when x . 0 sx ± 1d, f is
increasing on s2`, 21d and s21, 0d and decreasing on s0, 1d and s1, `d.
F.
The only critical number is x − 0. Since f 9 changes from positive to negative at 0,
f s0d − 0 is a local maximum by the First Derivative Test.
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Section 3.5 Summary of Curve Sketching
y
G.
y=2
f 0sxd −
| |
f 0sxd . 0 &? x 2 2 1 . 0 &? x . 1
x
| |
and f 0sxd , 0 &? x , 1. Thus the curve is concave upward on the intervals
s2`, 21d and s1, `d and concave downward on s21, 1d. It has no point of inflection since 1 and 21 are not in the domain of f.
H. Using the information in E–G, we finish the sketch in Figure 6.
n
x=1
FIGURE 6
Finished sketch of y −
sx 2 2 1d2 s24d 1 4x ? 2sx 2 2 1d2x
12x 2 1 4
− 2
2
4
sx 2 1d
sx 2 1d3
Since 12x 2 1 4 . 0 for all x, we have
0
x=_1
247
2x 2
x 21
2
Example 2 Sketch the graph of f sxd −
|
|
x2
.
sx 1 1
A. Domain − hx x 1 1 . 0j − hx x . 21j − s21, `d
B. The x- and y-intercepts are both 0.
C. Symmetry: None
D. Since
x2
lim
−`
x l ` sx 1 1
there is no horizontal asymptote. Since sx 1 1 l 0 as x l 211 and f sxd is
always positive, we have
x2
lim 1
−`
x l 21 sx 1 1
E.
and so the line x − 21 is a vertical asymptote.
f 9sxd −
3x 2 1 4x
xs3x 1 4d
sx 1 1 s2xd 2 x 2 ? 1y( 2sx 1 1 )
−
−
x11
2sx 1 1d3y2
2sx 1 1d3y2
e see that f 9sxd − 0 when x − 0 (notice that 243 is not in the domain of f ), so the
W
only critical number is 0. Since f 9sxd , 0 when 21 , x , 0 and f 9sxd . 0 when
x . 0, f is decreasing on s21, 0d and increasing on s0, `d.
F.
Since f 9s0d − 0 and f 9 changes from negative to positive at 0, f s0d − 0 is a local
(and absolute) minimum by the First Derivative Test.
y
G.
y=
x=_1
FIGURE 7
0
≈
œ„„„„
x+1
x
f 0sxd −
2sx 1 1d3y2s6x 1 4d 2 s3x 2 1 4xd3sx 1 1d1y2
3x 2 1 8x 1 8
−
3
4sx 1 1d
4sx 1 1d5y2
Note that the denominator is always positive. The numerator is the quadratic
3x 2 1 8x 1 8, which is always positive because its discriminant is
b 2 2 4ac − 232, which is negative, and the coefficient of x 2 is positive. Thus
f 0sxd . 0 for all x in the domain of f, which means that f is concave upward on
s21, `d and there is no point of inflection.
H. The curve is sketched in Figure 7.
n
Example 3 Sketch the graph of f sxd −
cos x
.
2 1 sin x
A. The domain is R.
B.
The y-intercept is f s0d − 12. The x-intercepts occur when cos x − 0, that is,
x − sy2d 1 n, where n is an integer.
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248
Chapter 3 Applications of Differentiation
C.f is neither even nor odd, but f sx 1 2d − f sxd for all x and so f is periodic and
has period 2. Thus, in what follows, we need to consider only 0 < x < 2 and
then extend the curve by translation in part H.
D. Asymptotes: None
E.
f 9sxd −
s2 1 sin xds2sin xd 2 cos x scos xd
2 sin x 1 1
−2
2
s2 1 sin xd
s2 1 sin xd 2
The denominator is always positive, so f 9sxd . 0 when 2 sin x 1 1 , 0 &?
sin x , 221 &? 7y6 , x , 11y6. So f is increasing on s7y6, 11y6d and
decreasing on s0, 7y6d and s11y6, 2d.
F.
From part E and the First Derivative Test, we see that the local minimum value is
f s7y6d − 21ys3 and the local maximum value is f s11y6d − 1ys3 .
G.If we use the Quotient Rule again and simplify, we get
f 0sxd − 2
2 cos x s1 2 sin xd
s2 1 sin xd 3
Because s2 1 sin xd 3 . 0 and 1 2 sin x > 0 for all x, we know that f 0sxd . 0
when cos x , 0, that is, y2 , x , 3y2. So f is concave upward on sy2, 3y2d
and concave downward on s0, y2d and s3y2, 2d. The inflection points are
sy2, 0d and s3y2, 0d.
H.The graph of the function restricted to 0 < x < 2 is shown in Figure 8. Then we
extend it, using periodicity, to the complete graph in Figure 9.
y
”
1
2
11π 1
6 , œ„
3
π
π
2
y
’
3π
2
1
2
2π x
_π
π
2π
3π
x
1
- ’
” 7π
6 , œ„3
FIGURE 8
n
FIGURE 9
Slant Asymptotes
y
Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If
y=ƒ
lim f f sxd 2 smx 1 bdg − 0
xl`
ƒ-(mx+b)
y=mx+b
0
x
FIGURE 10
where m ± 0, then the line y − mx 1 b is called a slant asymptote because the vertical distance between the curve y − f sxd and the line y − mx 1 b approaches 0, as
in Fig­ure 10. (A similar situation exists if we let x l 2`.) For rational functions,
slant asymp­totes occur when the degree of the numerator is one more than the degree of
the denominator. In such a case the equation of the slant asymptote can be found by long
division as in the following example.
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249
Section 3.5 Summary of Curve Sketching
Example 4 Sketch the graph of f sxd −
x3
.
x 11
2
A. The domain is R − s2`, `d.
B. The x- and y-intercepts are both 0.
C. Since f s2xd − 2f sxd, f is odd and its graph is symmetric about the origin.
D.
Since x 2 1 1 is never 0, there is no vertical asymptote. Since f sxd l ` as x l `
and f sxd l 2` as x l 2`, there is no horizontal asymptote. But long division
gives
x3
x
f sxd − 2
−x2 2
x 11
x 11
This equation suggests that y − x is a candidate for a slant asymptote. In fact,
f sxd 2 x − 2
x
−2
x2 1 1
1
x
11
1
x2
l 0 as x l 6`
So the line y − x is a slant asymptote.
E.
f 9sxd −
sx 2 1 1ds3x 2 d 2 x 3 ? 2x
x 2sx 2 1 3d
−
2
2
sx 1 1d
sx 2 1 1d2
Since f 9sxd . 0 for all x (except 0), f is increasing on s2`, `d.
F.
Although f 9s0d − 0, f 9 does not change sign at 0, so there is no local maximum or
minimum.
G.
y
y=
˛
≈+1
f 0sxd −
sx 2 1 1d2 s4x 3 1 6xd 2 sx 4 1 3x 2 d ? 2sx 2 1 1d2x
2xs3 2 x 2 d
−
2
4
sx 1 1d
sx 2 1 1d3
Since f 0sxd − 0 when x − 0 or x − 6s3 , we set up the following chart:
Interval
”œ„3,
0
”_œ„3, _
3œ„
3
’
4
y=x
FIGURE 11
3œ„
3
’
4
x
inflection
points
x , 2s3
2s3 , x , 0
0 , x , s3
x . s3
x
3 2 x2
sx 2 1 1d3
f 99sxd
f
2
2
1
1
CU on (2`, 2s3 )
2
1
1
2
1
1
1
1
1
2
1
2
CD on (2s3 , 0)
CU on ( 0, s3 )
CD on (s3 , `)
The points of inflection are (2s3 , 243 s3 ), s0, 0d, and (s3 , 34 s3 ).
H. The graph of f is sketched in Figure 11.
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n
250
chapter 3 Applications of Differentiation
1–40 Use the guidelines of this section to sketch the curve.
3
1. y − x 1 3x
2
2
2. y − 2 1 3x 2 x
3
3. y − x 4 2 4x
4. y − x 4 2 8x 2 1 8
5. y − xsx 2 4d3
6. y − x 5 2 5x
7. y −
1 5
5x
2
8 3
3x
8. y − s4 2 x d
2 5
1 16x
2
x 1 5x
25 2 x 2
x
x21
10. y −
11. y −
x 2 x2
2 2 3x 1 x 2
12. y − 1 1
13. y −
x
x2 2 4
14. y −
1
x2 2 4
15. y −
x2
x 13
16. y −
sx 2 1d2
x2 1 1
17. y −
x21
x2
18. y −
x
x 21
19. y −
x3
x 11
20. y −
x3
x22
9. y −
2
3
1
1
1 2
x
x
3
21. y − sx 2 3dsx
3
22. y − sx 2 4ds
x
23. y − sx 2 1 x 2 2
24. y − sx 2 1 x 2 x
25. y −
27. y −
x
26. y − x s2 2 x 2
sx 1 1
2
s1 2 x 2
x
28. y −
x
sx 2 2 1
29. y − x 2 3x 1y3
30. y − x 5y3 2 5x 2y3
3
31. y − s
x2 2 1
3
32. y − s
x3 1 1
33. y − sin3 x
34. y − x 1 cos x
35. y − x tan x, 2y2 , x , y2
where m 0 is the rest mass of the particle, m is the mass when
the particle moves with speed v relative to the observer, and c is
the speed of light. Sketch the graph of m as a function of v.
42.In the theory of relativity, the energy of a particle is
E − sm 02 c 4 1 h 2 c 2y2
where m 0 is the rest mass of the particle, is its wave length,
and h is Planck’s constant. Sketch the graph of E as a function
of . What does the graph say about the energy?
43.The figure shows a beam of length L embedded in concrete
walls. If a constant load W is distributed evenly along its
length, the beam takes the shape of the deflection curve
y−2
W 4
WL 3
WL 2 2
x 1
x 2
x
24EI
12EI
24EI
where E and I are positive constants. (E is Young’s modulus of
elasticity and I is the moment of inertia of a cross-section of
the beam.) Sketch the graph of the deflection curve.
y
W
0
L
44.Coulomb’s Law states that the force of attraction between
two charged particles is directly proportional to the product
of the charges and inversely proportional to the square of the
distance between them. The figure shows particles with charge
1 located at positions 0 and 2 on a coordinate line and a particle
with charge 21 at a position x between them. It follows from
Coulomb’s Law that the net force acting on the middle particle
is
k
k
Fsxd − 2 2 1
0,x,2
x
sx 2 2d2
where k is a positive constant. Sketch the graph of the net force
function. What does the graph say about the force?
36. y − 2x 2 tan x, 2y2 , x , y2
37. y − sin x 1 s3 cos x, 22 < x < 2
+1
_1
+1
38. y − csc x 2 2 sin x, 0 , x , 0
x
2
39. y −
sin x
1 1 cos x
40. y −
sin x
2 1 cos x
41.In the theory of relativity, the mass of a particle is
m−
m0
s1 2 yc
v2
2
x
7et0405x60
45–48 Find an equation
of the slant asymptote. Do not sketch the
curve.
09/11/09
45. y −
x 2 1 1MasterID:
x11
00518
46. y −
4x 3 2 10x 2 2 11x 1 1
x 2 2 3x
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251
Section 3.6 Graphing with Calculus and Calculators
47. y −
2x 3 2 5x 2 1 3x
x2 2 x 2 2
48. y −
57.Show that the lines y − sbyadx and y − 2sbyadx are slant
asymptotes of the hyperbola
26x 4 1 2x 3 1 3
2x 3 2 x
x2
y2
2 2 −1
2
a
b
49–54 Use the guidelines of this section to sketch the curve. In
guideline D find an equation of the slant asymptote.
x2
49. y −
x21
58. Let f sxd − sx 3 1 1dyx. Show that
1 1 5x 2 2x 2
50. y −
x22
51. y −
x3 1 4
x2
52. y −
x3
sx 1 1d2
53. y −
2x 3 1 x 2 1 1
x2 1 1
54. y −
sx 1 1d3
sx 2 1d2
lim f f sxd 2 x 2 g − 0
x l 6`
This shows that the graph of f approaches the graph of y − x 2,
and we say that the curve y − f sxd is asymptotic to the
parabola y − x 2. Use this fact to help sketch the graph of f.
59.Discuss the asymptotic behavior of
f sxd −
x4 1 1
x
55.Show that the curve y − s4x 2 1 9 has two slant asymptotes:
y − 2x and y − 22x. Use this fact to help sketch the curve.
in the same manner as in Exercise 58. Then use your results to
help sketch the graph of f.
56.Show that the curve y − sx 2 1 4x has two slant asymptotes:
y − x 1 2 and y − 2x 2 2. Use this fact to help sketch the
curve.
60.Use the asymptotic behavior of f sxd − cos x 1 1yx 2 to sketch
its graph without going through the curve-sketching procedure
of this section.
You may want to read “Graphing
Calculators and Computers” at
www.stewartcalculus.com if you
haven’t already. In particular, it
explains how to avoid some of the pitfalls of graphing devices by choosing
appropriate viewing rectangles.
The method we used to sketch curves in the preceding section was a culmination of much
of our study of differential calculus. The graph was the final object that we produced.
In this section our point of view is completely different. Here we start with a graph
produced by a graphing calculator or computer and then we refine it. We use calculus
to make sure that we reveal all the important aspects of the curve. And with the use of
graphing devices we can tackle curves that would be far too complicated to consider
without technology. The theme is the interaction between calculus and calculators.
Example 1 Graph the polynomial f sxd − 2x 6 1 3x 5 1 3x 3 2 2x 2. Use the graphs
of f 9 and f 0 to estimate all maximum and minimum points and intervals of concavity.
SOLUTION If we specify a domain but not a range, many graphing devices will deduce
a suitable range from the values computed. Figure 1 shows the plot from one such
device if we specify that 25 < x < 5. Although this viewing rectangle is useful for
showing that the asymptotic behavior (or end behavior) is the same as for y − 2x 6, it is
obviously hiding some finer detail. So we change to the viewing rectangle f23, 2g by
f250, 100g shown in Figure 2.
100
41,000
y=ƒ
y=ƒ
_5
FIGURE 1
_1000
_3
5
2
_50
FIGURE 2
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252
Chapter 3 Applications of Differentiation
20
From Figure 2 it appears that there is an absolute minimum value of about 215.33
when x < 21.62 (by using the cursor) and f is decreasing on s2`, 21.62d and
increasing on s21.62, `d. Also there appears to be a horizontal tangent at the origin and
inflection points when x − 0 and when x is somewhere between 22 and 21.
Now let’s try to confirm these impressions using calculus. We differentiate and get
y=fª(x)
_3
2
f 9sxd − 12x 5 1 15x 4 1 9x 2 2 4x
_5
f 0sxd − 60x 4 1 60x 3 1 18x 2 4
FIGURE 3
1
y=ƒ
_1
1
_1
FIGURE 4
10
_3
2
y=f·(x)
n
_30
FIGURE 5
When we graph f 9 in Figure 3 we see that f 9sxd changes from negative to positive when
x < 21.62; this confirms (by the First Derivative Test) the minimum value that we
found earlier. But, perhaps to our surprise, we also notice that f 9sxd changes from positive to negative when x − 0 and from negative to positive when x < 0.35. This means
that f has a local maximum at 0 and a local minimum when x < 0.35, but these were
hidden in Figure 2. Indeed, if we now zoom in toward the origin in Figure 4, we see
what we missed before: a local maximum value of 0 when x − 0 and a local minimum
value of about 20.1 when x < 0.35.
What about concavity and inflection points? From Figures 2 and 4 there appear to
be inflection points when x is a little to the left of 21 and when x is a little to the right
of 0. But it’s difficult to determine inflection points from the graph of f, so we graph
the second derivative f 0 in Figure 5. We see that f 0 changes from positive to negative
when x < 21.23 and from negative to positive when x < 0.19. So, correct to two decimal places, f is concave upward on s2`, 21.23d and s0.19, `d and concave downward
on s21.23, 0.19d. The inflection points are s21.23, 210.18d and s0.19, 20.05d.
We have discovered that no single graph reveals all the important features of this
polynomial. But Figures 2 and 4, when taken together, do provide an accurate picture.
Example 2 Draw the graph of the function
f sxd −
x 2 1 7x 1 3
x2
in a viewing rectangle that contains all the important features of the function. Estimate
the maximum and minimum values and the intervals of concavity. Then use calculus to
find these quantities exactly.
SOLUTION Figure 6, produced by a computer with automatic scaling, is a disaster.
Some graphing calculators use f210, 10g by f210, 10g as the default viewing rectangle, so let’s try it. We get the graph shown in Figure 7; it’s a major improvement.
3 10!*
10
y=ƒ
_10
y=ƒ
_5
FIGURE 6
5
10
_10
FIGURE 7
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Section 3.6 Graphing with Calculus and Calculators
253
The y-axis appears to be a vertical asymptote and indeed it is because
lim
xl0
10
y=ƒ
y=1
_20
20
Figure 7 also allows us to estimate the x-intercepts: about 20.5 and 26.5. The
exact values are obtained by using the quadratic formula to solve the equation
x 2 1 7x 1 3 − 0; we get x − (27 6 s37 )y2.
To get a better look at horizontal asymptotes, we change to the viewing rectangle
f220, 20g by f25, 10g in Figure 8. It appears that y − 1 is the horizontal asymptote
and this is easily confirmed:
_5
FIGURE 8
lim
x l 6`
2
_3
0
x 2 1 7x 1 3
−`
x2
S
x 2 1 7x 1 3
7
3
− lim 1 1 1 2
2
x
l
6`
x
x
x
D
−1
To estimate the minimum value we zoom in to the viewing rectangle f23, 0g by
f24, 2g in Figure 9. The cursor indicates that the absolute minimum value is about
23.1 when x < 20.9, and we see that the function decreases on s2`, 20.9d and s0, `d
and increases on s20.9, 0d. The exact values are obtained by differentiating:
y=ƒ
f 9sxd − 2
_4
7
6
7x 1 6
2 2
3 − 2
x
x
x3
This shows that f 9sxd . 0 when 267 , x , 0 and f 9sxd , 0 when x , 267 and when
37
x . 0. The exact minimum value is f (2 67 ) − 2 12
< 23.08.
Figure 9 also shows that an inflection point occurs somewhere between x − 21 and
x − 22. We could estimate it much more accurately using the graph of the second
deriv­ative, but in this case it’s just as easy to find exact values. Since
FIGURE 9
f 0sxd −
14
18
2s7x 1 9d
3 1
4 −
x
x
x4
we see that f 0sxd . 0 when x . 297 sx ± 0d. So f is concave upward on s297 , 0d and
s0, `d and concave downward on s2`, 297 d. The inflection point is s297 , 271
27 d.
The analysis using the first two derivatives shows that Figure 8 displays all the
major aspects of the curve.
n
Example 3 Graph the function f sxd −
10
_10
y=ƒ
10
_10
FIGURE 10
x 2sx 1 1d3
.
sx 2 2d2sx 2 4d4
SOLUTION Drawing on our experience with a rational function in Example 2, let’s
start by graphing f in the viewing rectangle f210, 10g by f210, 10g. From Figure 10
we have the feeling that we are going to have to zoom in to see some finer detail and
also zoom out to see the larger picture. But, as a guide to intelligent zooming, let’s
first take a close look at the expression for f sxd. Because of the factors sx 2 2d2 and
sx 2 4d4 in the denominator, we expect x − 2 and x − 4 to be the vertical asymptotes. Indeed
lim
x l2
x 2sx 1 1d3
x 2sx 1 1d3
−
` and lim
−`
x l 4 sx 2 2d2sx 2 4d4
sx 2 2d2sx 2 4d4
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254
Chapter 3 Applications of Differentiation
To find the horizontal asymptotes, we divide numerator and denominator by x 6:
x 2 sx 1 1d3
?
x 2sx 1 1d3
x3
x3
−
2
4 −
2
sx 2 2d sx 2 4d
sx 2 2d
sx 2 4d4
?
x2
x4
y
_1
1
2
3
4
S D
S DS D
12
2
x
2
12
4
x
4
This shows that f sxd l 0 as x l 6`, so the x-axis is a horizontal asymptote.
It is also very useful to consider the behavior of the graph near the x-intercepts
using an analysis like that in Example 3.4.11. Since x 2 is positive, f sxd does not
change sign at 0 and so its graph doesn’t cross the x-axis at 0. But, because of the
factor sx 1 1d3, the graph does cross the x-axis at 21 and has a horizontal tangent
there. Putting all this information together, but without using derivatives, we see that
the curve has to look something like the one in Figure 11.
Now that we know what to look for, we zoom in (several times) to produce the
graphs in Figures 12 and 13 and zoom out (several times) to get Figure 14.
x
FIGURE 11
0.05
0.0001
500
y=ƒ
y=ƒ
_100
3
1
1
11
x
x
1
_1.5
0.5
y=ƒ
_0.05
_0.0001
FIGURE 13
FIGURE 12
_1
_10
10
FIGURE 14
We can read from these graphs that the absolute minimum is about 20.02 and
occurs when x < 220. There is also a local maximum <0.00002 when x < 20.3 and
a local minimum <211 when x < 2.5. These graphs also show three inflection points
near 235, 25, and 21 and two between 21 and 0. To estimate the inflection points
closely we would need to graph f 0, but to compute f 0 by hand is an unreasonable
chore. If you have a computer algebra system, then it’s easy to do (see Exercise 13).
We have seen that, for this particular function, three graphs (Figures 12, 13, and 14)
are necessary to convey all the useful information. The only way to display all these
features of the function on a single graph is to draw it by hand. Despite the exaggerations and distortions, Figure 11 does manage to summarize the essential nature of the
function. 1.2
n
1.1
y=ƒ
Example 4 Graph the function f sxd − sinsx 1 sin 2xd. For 0 < x < , estimate
0
π
all maximum
and minimum values, intervals
of increase and decrease, and inflection
0
π
points.
y=fª(x)
SOLUTION We first note that
f is periodic with period 2. Also, f is odd and
1 for all x. So the choice of a viewing rectangle is not a problem for this func| f sxd | <_1.2
_1.1
FIGURE 15
tion: We start with f0, g by f21.1, 1.1g. (See Figure 15.) It appears that there are three
local maximum values and two local minimum values in that window. To confirm
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255
Section 3.6 Graphing with Calculus and Calculators
this and locate them more accurately, we calculate that
The family of functions
f sxd − sinsx 1 sin cxd
where c is a constant, occurs in applications to frequency modulation (FM)
synthesis. A sine wave is modulated by
a wave with a different frequency
ssin cxd. The case where c − 2 is
studied in Example 4. Exercise 19
explores another special case.
f 9sxd − cossx 1 sin 2xd ? s1 1 2 cos 2xd
and graph both f and f 9 in Figure 16.
Using zoom-in and the First Derivative Test, we find the following approximate
values:
Intervals of increase:
s0, 0.6d, s1.0, 1.6d, s2.1, 2.5d
Intervals of decrease:
1.2
Local maximum values: f s0.6d < 1, f s1.6d < 1, f s2.5d < 1
y=ƒ
π
s0.6, 1.0d, s1.6, 2.1d, s2.5, d
0
Local minimum values:
π
f s1.0d < 0.94, f s2.1d < 0.94
The second derivative is
y=f ª(x)
f 0sxd − 2s1 1 2 cos 2xd2 sinsx 1 sin 2xd 2 4 sin 2x cossx 1 sin 2xd
_1.2
Graphing both f and f 0 in Figure 17, we obtain the following approximate values:
FIGURE 16
Concave upward on:
1.2
Concave downward on: s0, 0.8d, s1.3, 1.8d, s2.3, d
f
0
s0.8, 1.3d, s1.8, 2.3d
Inflection points:
π
f·
s0, 0d, s0.8, 0.97d, s1.3, 0.97d, s1.8, 0.97d, s2.3, 0.97d
Having checked that Figure 15 does indeed represent f accurately for 0 < x < ,
we can state that the extended graph in Figure 18 represents f accurately for
22 < x < 2.
n
_1.2
FIGURE 17
Our final example is concerned with families of functions. This means that the functions in the family are related to each other by a formula that contains one or more arbitrary constants. Each value of the constant gives rise to a member of the family and the
idea is to see how the graph of the function changes as the constant changes.
1.2
2π
_2π
_1.2
Example 5 How does the graph of f sxd − 1ysx 2 1 2x 1 cd vary as c varies?
SOLUTION The graphs in Figures 19 and 20 (the special cases c − 2 and c − 22)
show two very different-looking curves.
FIGURE 18
2
_5
y=
2
4
1
≈+2x+2
_5
_2
y=
1
≈+2x-2
4
_2
FIGURE 19
FIGURE 20
c−2
c − 22
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256
Chapter 3 Applications of Differentiation
Before drawing any more graphs, let’s see what members of this family have in
common. Since
1
lim
−0
x l 6` x 2 1 2x 1 c
for any value of c, they all have the x-axis as a horizontal asymptote. A vertical asymptote will occur when x 2 1 2x 1 c − 0. Solving this quadratic equation, we get
x − 21 6 s1 2 c . When c . 1, there is no vertical asymptote (as in Figure 19).
When c − 1, the graph has a single vertical asymptote x − 21 because
lim
x l21
1
1
− lim
−`
x l21 sx 1 1d2
x 2 1 2x 1 1
When c , 1, there are two vertical asymptotes: x − 21 6 s1 2 c (as in Figure 20).
Now we compute the derivative:
f 9sxd − 2
TEC See an animation of Figure 21 in
Visual 3.6.
c=_1
FIGURE 21
The family of functions
f sxd − 1ysx 2 1 2x 1 cd
2x 1 2
sx 1 2x 1 cd2
2
This shows that f 9sxd − 0 when x − 21 (if c ± 1), f 9sxd . 0 when x , 21, and
f 9sxd , 0 when x . 21. For c > 1, this means that f increases on s2`, 21d
and decreases on s21, `d. For c . 1, there is an absolute maximum value
f s21d − 1ysc 2 1d. For c , 1, f s21d − 1ysc 2 1d is a local maximum value and the
intervals of increase and decrease are interrupted at the vertical asymptotes.
Figure 21 is a “slide show” displaying five members of the family, all graphed in the
viewing rectangle f25, 4g by f22, 2g. As predicted, a transition takes place from two
vertical asymptotes to one at c − 1, and then to none for c . 1. As c increases from
1, we see that the maximum point becomes lower; this is explained by the fact that
1ysc 2 1d l 0 as c l `. As c decreases from 1, the vertical asymptotes become more
widely separated because the distance between them is 2 s 1 2 c , which becomes large
as c l 2`. Again, the maximum point approaches the x-axis because 1ysc 2 1d l 0
as c l 2`.
c=0
c=1
c=2
c=3
There is clearly no inflection point when c < 1. For c . 1 we calculate that
f 0sxd −
2s3x 2 1 6x 1 4 2 cd
sx 2 1 2x 1 cd3
and deduce that inflection points occur when x − 21 6 s3sc 2 1dy3. So the inflection
points become more spread out as c increases and this seems plausible from the last
two parts of Figure 21.
n
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Section 3.6 Graphing with Calculus and Calculators
257
;
1–8 Produce graphs of f that reveal all the important aspects of
the curve. In particular, you should use graphs of f 9 and f 0 to
estimate the intervals of increase and decrease, extreme values,
intervals of concavity, and inflection points.
1. f sxd − x 5 2 5x 4 2 x 3 1 28x 2 2 2x
CAS
15–18 Use a computer algebra system to graph f and to find f 9
and f 0. Use graphs of these derivatives to estimate the intervals
of increase and decrease, extreme values, intervals of concavity,
and inflection points of f.
15. f sxd −
x 3 1 5x 2 1 1
x 1 x3 2 x2 1 2
3. f sxd − x 6 2 5x 5 1 25x 3 2 6x 2 2 48x
16. f sxd −
x4 2 x 3 2 8
x2 2 x 2 6
x
5. f sxd − 3
x 1 x2 1 1
x 2y3
1 1 x 1 x4
17. f sxd − sx 1 5 sin x , x < 20
2. f sxd − 22x 6 1 5x 5 1 140x 3 2 110x 2
4. f sxd −
6.f sxd − 6 sin x 2 x , 25 < x < 3
2
7.f sxd − 6 sin x 1 cot x, 2 < x < sin x
8.f sxd −
, 22 < x < 2
x
9–10 Produce graphs of f that reveal all the important aspects
of the curve. Estimate the intervals of increase and decrease and
intervals of concavity, and use calculus to find these intervals
exactly.
9. f sxd − 1 1
10. f sxd −
1
8
1
1 2 1 3
x
x
x
1
2 3 10 8
2
x8
x4
11–12 Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to
producing graphs (with a graphing device) that display the major
features of the curve. Use these graphs to estimate the maximum
and minimum values.
11. f sxd −
sx 1 4dsx 2 3d
x 4sx 2 1d
12. f sxd −
s2 x 1 3d 2 sx 2 2d 5
x 3 sx 2 5d 2
2
18. f sxd −
CAS
13. I f f is the function considered in Example 3, use a com­
puter algebra system to calculate f 9 and then graph it to
confirm that all the maximum and minimum values are as
given in the example. Calculate f 0 and use it to estimate the
intervals of concavity and inflection points.
14. I f f is the function of Exercise 12, find f 9 and f 0 and
use their graphs to estimate the intervals of increase and
decrease and concavity of f.
2x 2 1
4
x4 1 x 1 1
s
19.In Example 4 we considered a member of the family of
functions f sxd − sinsx 1 sin cxd that occur in FM synthesis. Here we investigate the function with c − 3. Start by
graphing f in the viewing rectangle f0, g by f21.2, 1.2g.
How many local maximum points do you see? The graph
has more than are visible to the naked eye. To discover the
hidden maximum and minimum points you will need to
examine the graph of f 9 very carefully. In fact, it helps to
look at the graph of f 0 at the same time. Find all the maximum and minimum values and inflection points. Then graph
f in the viewing rectangle f22, 2g by f21.2, 1.2g and
comment on symmetry.
20–25 Describe how the graph of f varies as c varies. Graph
several members of the family to illustrate the trends that you
discover. In particular, you should investigate how maximum
and minimum points and inflection points move when c changes.
You should also identify any transitional values of c at which the
basic shape of the curve changes.
20. f sxd − x 3 1 cx
21. f sxd − x 2 1 6 x 1 cyx (Trident of Newton)
22. f sxd − x sc 2 2 x 2
24. f sxd −
CAS
4
sin x
c 1 cos x
23. f sxd −
cx
1 1 c 2x 2
25. f sxd − cx 1 sin x
26.Investigate the family of curves given by the equation
f sxd − x 4 1 cx 2 1 x. Start by determining the transitional
value of c at which the number of inflection points changes.
Then graph several members of the family to see what
shapes are possible. There is another transitional value of c
at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered.
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258
chapter 3 Applications of Differentiation
27. (a)Investigate the family of polynomials given by the equa­
tion f sxd − cx 4 2 2 x 2 1 1. For what values of c does the
curve have minimum points?
(b)Show that the minimum and maximum points of every
curve in the family lie on the parabola y − 1 2 x 2. Illustrate by graphing this parabola and several members of the
family.
PS
28. (a)Investigate the family of polynomials given by the equa­tion
f sxd − 2x 3 1 cx 2 1 2 x. For what values of c does the
curve have maximum and minimum points?
(b)Show that the minimum and maximum points of every
curve in the family lie on the curve y − x 2 x 3. Illustrate by
graphing this curve and several members of the family.
The methods we have learned in this chapter for finding extreme values have practical applications in many areas of life. A businessperson wants to minimize costs and
maximize profits. A traveler wants to minimize transportation time. Fermat’s Principle in
optics states that light follows the path that takes the least time. In this section we solve
such problems as maximizing areas, volumes, and profits and minimizing distances,
times, and costs.
In solving such practical problems the greatest challenge is often to convert the word
problem into a mathematical optimization problem by setting up the function that is to
be maximized or minimized. Let’s recall the problem-solving principles discussed on
page 98 and adapt them to this situation:
Steps In Solving Optimization Problems
1.Understand the Problem The first step is to read the problem carefully until it is
clearly understood. Ask yourself: What is the unknown? What are the given quantities? What are the given conditions?
2.Draw a Diagram In most problems it is useful to draw a diagram and identify the
given and required quantities on the diagram.
3.Introduce Notation Assign a symbol to the quantity that is to be maximized or
minimized (let’s call it Q for now). Also select symbols sa, b, c, . . . , x, yd for other
unknown quantities and label the diagram with these symbols. It may help to use
initials as suggestive symbols—for example, A for area, h for height, t for time.
4. Express Q in terms of some of the other symbols from Step 3.
5.
If Q has been expressed as a function of more than one variable in Step 4, use the
given information to find relationships (in the form of equations) among these
variables. Then use these equations to eliminate all but one of the variables in the
expression for Q. Thus Q will be expressed as a function of one variable x, say,
Q − f sxd. Write the domain of this function in the given context.
6.
Use the methods of Sections 3.1 and 3.3 to find the absolute maximum or minimum
value of f. In particular, if the domain of f is a closed interval, then the Closed
Interval Method in Section 3.1 can be used.
Example 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular field
that borders a straight river. He needs no fence along the river. What are the dimensions
of the field that has the largest area?
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Section 3.7 Optimization Problems
SOLUTION In order to get a feeling for what is happening in this problem, let’s experiment with some specific cases. Figure 1 (not to scale) shows three possible ways of
laying out the 2400 ft of fencing.
PS Understand the problem
PS Analogy: Try special cases
PS Draw diagrams
400
1000
2200
100
100
Area=100 · 2200=220,000 ft@
FIGURE 1
y
FIGURE 2
A
700
1000
700
Area=700 · 1000=700,000 ft@
1000
Area=1000 · 400=400,000 ft@
We see that when we try shallow, wide fields or deep, narrow fields, we get relatively small areas. It seems plausible that there is some intermediate configuration that
produces the largest area.
Figure 2 illustrates the general case. We wish to maximize the area A of the rectangle. Let x and y be the depth and width of the rectangle (in feet). Then we express A
in terms of x and y:
A − xy
PS Introduce notation
x
259
x
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the
fencing is 2400 ft. Thus
2x 1 y − 2400
From this equation we have y − 2400 2 2x, which gives
A − xy − xs2400 2 2xd − 2400x 2 2x 2
Note that the largest x can be is 1200 (this uses all the fence for the depth and none for
the width) and x can’t be negative, so the function that we wish to maximize is
Asxd − 2400x 2 2x 2 0 < x < 1200
The derivative is A9sxd − 2400 2 4x, so to find the critical numbers we solve the
equation
2400 2 4x − 0
which gives x − 600. The maximum value of A must occur either at this critical number
or at an endpoint of the interval. Since As0d − 0, As600d − 720,000, and As1200d − 0,
the Closed Interval Method gives the maximum value as As600d − 720,000.
[Alternatively, we could have observed that A0sxd − 24 , 0 for all x, so A is
always concave downward and the local maximum at x − 600 must be an absolute
maximum.]
The corresponding y-value is y − 2400 2 2s600d − 1200, so the rectangular field
should be 600 ft deep and 1200 ft wide.
n
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260
Chapter 3 Applications of Differentiation
Example 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensions
that will minimize the cost of the metal to manufacture the can.
SOLUTION Draw the diagram as in Figure 3, where r is the radius and h the height (both
in centimeters). In order to minimize the cost of the metal, we minimize the total surface
area of the cylinder (top, bottom, and sides). From Figure 4 we see that the sides are
made from a rectangular sheet with dimensions 2r and h. So the surface area is
h
r
A − 2r 2 1 2rh
FIGURE 3
We would like to express A in terms of one variable, r. To eliminate h we use the
fact that the volume is given as 1 L, which is equivalent to 1000 cm3. Thus
2πr
r 2h − 1000
r
which gives h − 1000ysr 2 d. Substitution of this into the expression for A gives
h
A − 2r 2 1 2r
Area 2{πr@}
Area (2πr)h
S D
1000
r 2
− 2r 2 1
2000
r
We know r must be positive, and there are no limitations on how large r can be. Therefore the function that we want to minimize is
FIGURE 4
Asrd − 2r 2 1
2000
r . 0
r
To find the critical numbers, we differentiate:
A9srd − 4r 2
y
y=A(r)
1000
0
10
r
FIGURE 5
In the Applied Project on page 270 we
investigate the most economical shape
for a can by taking into account other
manufacturing costs.
2000
4sr 3 2 500d
−
2
r
r2
3
Then A9srd − 0 when r 3 − 500, so the only critical number is r − s
500y .
Since the domain of A is s0, `d, we can’t use the argument of Example 1 concern3
ing endpoints. But we can observe that A9srd , 0 for r , s
500y and A9srd . 0 for
3
r . s500y , so A is decreasing for all r to the left of the critical number and increas3
ing for all r to the right. Thus r − s
500y must give rise to an absolute minimum.
[Alternatively, we could argue that Asrd l ` as r l 0 1 and Asrd l ` as r l `, so
there must be a minimum value of Asrd, which must occur at the critical number. See
Figure 5.]
3
The value of h corresponding to r − s
500y is
h−
Î
1000
1000
−2
2 −
r
s500yd2y3
3
500
− 2r
3
Thus, to minimize the cost of the can, the radius should be s
500y cm and the height
should be equal to twice the radius, namely, the diameter.
n
Note 1 The argument used in Example 2 to justify the absolute minimum is a variant of the First Derivative Test (which applies only to local maximum or minimum values) and is stated here for future reference.
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Section 3.7 Optimization Problems
261
First Derivative Test for Absolute Extreme Values Suppose that c is a critical
number of a continuous function f defined on an interval.
(a) If f 9sxd . 0 for all x , c and f 9sxd , 0 for all x . c, then f scd is the absolute maximum value of f.
(b) If f 9sxd , 0 for all x , c and f 9sxd . 0 for all x . c, then f scd is the absolute minimum value of f.
TEC Module 3.7 takes you through
six additional optimization problems,
including animations of the physical
situations.
Note 2 An alternative method for solving optimization problems is to use implicit
differentiation. Let’s look at Example 2 again to illustrate the method. We work with the
same equations
A − 2r 2 1 2rh r 2h − 1000
but instead of eliminating h, we differentiate both equations implicitly with respect to r:
A9 − 4r 1 2rh9 1 2h r 2h9 1 2rh − 0
The minimum occurs at a critical number, so we set A9 − 0, simplify, and arrive at the
equations
2r 1 rh9 1 h − 0 rh9 1 2h − 0
Subtraction of these equations gives 2r 2 h − 0, or h − 2r.
y
Example 3 Find the point on the parabola y 2 − 2x that is closest to the point s1, 4d.
(1, 4)
SOLUTION The distance between the point s1, 4d and the point sx, yd is
d − ssx 2 1d2 1 sy 2 4d2
(x, y)
1
0
¥=2x
1 2 3 4
x
(See Figure 6.) But if sx, yd lies on the parabola, then x − 12 y 2, so the expression for d
becomes
d − s(12 y2 2 1)2 1 sy 2 4d2
figure 6
(Alternatively, we could have substituted y − s2x to get d in terms of x alone.)
Instead of minimizing d, we minimize its square:
d 2 − f syd − ( 12 y 2 2 1) 2 1 sy 2 4d2
(You should convince yourself that the minimum of d occurs at the same point as the
minimum of d 2, but d 2 is easier to work with.) Note that there are no restrictions on y,
so the domain is all real numbers. Differentiating, we obtain
f 9syd − 2( 12 y 2 2 1) y 1 2sy 2 4d − y 3 2 8
so f 9syd − 0 when y − 2. Observe that f 9syd , 0 when y , 2 and f 9s yd . 0 when
y . 2, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y − 2. (Or we could simply say that because of the geometric nature
of the problem, it’s obvious that there is a closest point but not a farthest point.) The
corresponding value of x is x − 12 y 2 − 2. Thus the point on y 2 − 2x closest to s1, 4d is
s2, 2d. [The distance between the points is d − sf s2d − s5 .]
n
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262
Chapter 3 Applications of Differentiation
Example 4 A man launches his boat from point A on a bank of a straight river, 3 km
wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickly as
possible (see Figure 7). He could row his boat directly across the river to point C and
then run to B, or he could row directly to B, or he could row to some point D between
C and B and then run to B. If he can row 6 kmyh and run 8 kmyh, where should he
land to reach B as soon as possible? (We assume that the speed of the water is negligible compared with the speed at which the man rows.)
3 km
C
A
x
D
SOLUTION If we let x be the distance from C to D, then the running distance is
| DB | − 8 2 x and the Pythagorean Theorem gives the rowing distance as
| AD | − sx 1 9 . We use the equation
8 km
2
time −
B
distance
rate
Then the rowing time is sx 2 1 9 y6 and the running time is s8 2 xdy8, so the total time
T as a function of x is
FIGURE 7
Tsxd −
82x
sx 2 1 9
1
6
8
The domain of this function T is f0, 8g. Notice that if x − 0, he rows to C and if x − 8,
he rows directly to B. The derivative of T is
T9sxd −
x
6sx 1 9
2
2
1
8
Thus, using the fact that x > 0, we have
T9sxd − 0 &? x
1
− &? 4x − 3sx 2 1 9
8
6sx 2 1 9
&? 16x 2 − 9sx 2 1 9d &? 7x 2 − 81
&? x −
The only critical number is x − 9ys7 . To see whether the minimum occurs at this
critical number or at an endpoint of the domain f0, 8g, we follow the Closed Interval
Method by evaluating T at all three points:
T
y=T(x)
Ts0d − 1.5 T
1
0
9
s7
2
FIGURE 8
4
6
x
S D
9
s7
−11
s7
s73
< 1.33 Ts8d −
< 1.42
8
6
Since the smallest of these values of T occurs when x − 9ys7 , the absolute minimum
value of T must occur there. Figure 8 illustrates this calculation by showing the graph
of T.
Thus the man should land the boat at a point 9ys7 km (<3.4 km) downstream from
his starting point.
n
Example 5 Find the area of the largest rectangle that can be inscribed in a semicircle
of radius r.
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Section 3.7 Optimization Problems
y
SOLUTION 1 Let’s take the semicircle to be the upper half of the circle x 2 1 y 2 − r 2
(x, y)
2x
_r
263
y
r x
0
FIGURE 9
with center the origin. Then the word inscribed means that the rectangle has two
vertices on the semicircle and two vertices on the x-axis as shown in Figure 9.
Let sx, yd be the vertex that lies in the first quadrant. Then the rectangle has sides of
lengths 2x and y, so its area is
A − 2xy
To eliminate y we use the fact that sx, yd lies on the circle x 2 1 y 2 − r 2 and so
y − sr 2 2 x 2 . Thus
A − 2xsr 2 2 x 2
The domain of this function is 0 < x < r. Its derivative is
A9 − 2sr 2 2 x 2 2
2x 2
2sr 2 2 2x 2 d
−
2
sr 2 x
sr 2 2 x 2
2
which is 0 when 2x 2 − r 2, that is, x − rys2 (since x > 0). This value of x gives a
maximum value of A since As0d − 0 and Asrd − 0. Therefore the area of the largest
inscribed rectangle is
S D
A
r
s2
−2
r
s2
Î
r2 2
r2
− r2
2
SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable.
Let be the angle shown in Figure 10. Then the area of the rectangle is
r
¨
r cos ¨
FIGURE 10
r sin ¨
Asd − s2r cos dsr sin d − r 2s2 sin cos d − r 2 sin 2
We know that sin 2 has a maximum value of 1 and it occurs when 2 − y2. So Asd
has a maximum value of r 2 and it occurs when − y4.
Notice that this trigonometric solution doesn’t involve differentiation. In fact, we
didn’t need to use calculus at all.
n
Applications to Business and Economics
In Section 2.7 we introduced the idea of marginal cost. Recall that if Csxd, the cost function, is the cost of producing x units of a certain product, then the marginal cost is the
rate of change of C with respect to x. In other words, the marginal cost function is the
derivative, C9sxd, of the cost function.
Now let’s consider marketing. Let psxd be the price per unit that the company can
charge if it sells x units. Then p is called the demand function (or price function) and
we would expect it to be a decreasing function of x. (More units sold corresponds to a
lower price.) If x units are sold and the price per unit is psxd, then the total revenue is
Rsxd − quantity 3 price − xpsxd
and R is called the revenue function. The derivative R9 of the revenue function is called
the marginal revenue function and is the rate of change of revenue with respect to the
num­ber of units sold.
If x units are sold, then the total profit is
Psxd − Rsxd 2 Csxd
and P is called the profit function. The marginal profit function is P9, the derivative of
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264
Chapter 3 Applications of Differentiation
the profit function. In Exercises 59–63 you are asked to use the marginal cost, revenue,
and profit functions to minimize costs and maximize revenues and profits.
Example 6 A store has been selling 200 flat-screen TVs a week at $350 each. A market survey indicates that for each $10 rebate offered to buyers, the number of TVs sold
will increase by 20 a week. Find the demand function and the revenue function. How
large a rebate should the store offer to maximize its revenue?
SOLUTION If x is the number of TVs sold per week, then the weekly increase in sales is
x 2 200. For each increase of 20 units sold, the price is decreased by $10. So for each
1
additional unit sold, the decrease in price will be 20
3 10 and the demand function is
1
psxd − 350 2 10
20 sx 2 200d − 450 2 2 x
The revenue function is
Rsxd − xpsxd − 450x 2 12 x 2
Since R9sxd − 450 2 x, we see that R9sxd − 0 when x − 450. This value of x gives an
absolute maximum by the First Derivative Test (or simply by observing that the graph
of R is a parabola that opens downward). The corresponding price is
ps450d − 450 2 12 s450d − 225
and the rebate is 350 2 225 − 125. Therefore, to maximize revenue, the store should
offer a rebate of $125.
n
1.Consider the following problem: Find two numbers whose sum
is 23 and whose product is a maximum.
(a)Make a table of values, like the one below, so that the sum
of the numbers in the first two columns is always 23. On
the basis of the evidence in your table, estimate the answer
to the problem.
(b)Use calculus to solve the problem and compare with your
answer to part (a).
First number
Second number
Product
1
2
3
.
.
.
22
21
20
.
.
.
22
42
60
.
.
.
6.What is the minimum vertical distance between the parabolas
y − x 2 1 1 and y − x 2 x 2 ?
7.Find the dimensions of a rectangle with perimeter 100 m whose
area is as large as possible.
8.Find the dimensions of a rectangle with area 1000 m2 whose
perimeter is as small as possible.
9.A model used for the yield Y of an agricultural crop as a function of the nitrogen level N in the soil (measured in appropriate
units) is
kN
Y−
1 1 N2
where k is a positive constant. What nitrogen level gives the
best yield?
3.Find two positive numbers whose product is 100 and whose
sum is a minimum.
10.The rate sin mg carbonym 3yhd at which photosynthesis takes
place for a species of phytoplankton is modeled by the
function
100 I
P− 2
I 1I14
4.The sum of two positive numbers is 16. What is the smallest
possible value of the sum of their squares?
where I is the light intensity (measured in thousands of footcandles). For what light intensity is P a maximum?
2.Find two numbers whose difference is 100 and whose product
is a minimum.
5.What is the maximum vertical distance between the line
y − x 1 2 and the parabola y − x 2 for 21 < x < 2?
11.Consider the following problem: A farmer with 750 ft of
fencing wants to enclose a rectangular area and then divide
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.7 Optimization Problems
it into four pens with fencing parallel to one side of the
rectangle. What is the largest possible total area of the
four pens?
(a)Draw several diagrams illustrating the situation, some
with shallow, wide pens and some with deep, narrow
pens. Find the total areas of these configurations. Does it
appear that there is a maximum area? If so, estimate it.
(b)Draw a diagram illustrating the general situation.
Introduce notation and label the diagram with your
symbols.
(c) Write an expression for the total area.
(d)Use the given information to write an equation that
relates the variables.
(e)Use part (d) to write the total area as a function of one
variable.
(f )Finish solving the problem and compare the answer
with your estimate in part (a).
12.Consider the following problem: A box with an open top
is to be constructed from a square piece of cardboard, 3 ft
wide, by cutting out a square from each of the four corners
and bending up the sides. Find the largest volume that such
a box can have.
(a)Draw several diagrams to illustrate the situation, some
short boxes with large bases and some tall boxes with
small bases. Find the volumes of several such boxes.
Does it appear that there is a maximum volume? If so,
estimate it.
(b)Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols.
(c) Write an expression for the volume.
(d)Use the given information to write an equation that
relates the variables.
(e)Use part (d) to write the volume as a function of one
variable.
(f )Finish solving the problem and compare the answer
with your estimate in part (a).
13.A farmer wants to fence in an area of 1.5 million square feet
in a rectangular field and then divide it in half with a fence
parallel to one of the sides of the rectangle. How can he do
this so as to minimize the cost of the fence?
14.A box with a square base and open top must have a volume
of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.
15.If 1200 cm2 of material is available to make a box with a
square base and an open top, find the largest possible volume of the box.
16.A rectangular storage container with an open top is to have
a volume of 10 m3. The length of its base is twice the width.
Material for the base costs $10 per square meter. Material
for the sides costs $6 per square meter. Find the cost of
materials for the cheapest such container.
17.Do Exercise 16 assuming the container has a lid that is
made from the same material as the sides.
265
18.A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed
along the barn, and the fencing along the west side of the
plot is shared with a neighbor who will split the cost of that
portion of the fence. If the fencing costs $20 per linear foot
to install and the farmer is not willing to spend more than
$5000, find the dimensions for the plot that would enclose
the most area.
19. I f the farmer in Exercise 18 wants to enclose 8000 square
feet of land, what dimensions will minimize the cost of
the fence?
20. (a)Show that of all the rectangles with a given area, the
one with smallest perimeter is a square.
(b)Show that of all the rectangles with a given perimeter,
the one with greatest area is a square.
21.Find the point on the line y − 2x 1 3 that is closest to the
origin.
22.Find the point on the curve y − sx that is closest to the
point s3, 0d.
23.Find the points on the ellipse 4x 2 1 y 2 − 4 that are farthest
away from the point s1, 0d.
; 24.Find, correct to two decimal places, the coordinates of the
point on the curve y − sin x that is closest to the point s4, 2d.
25.Find the dimensions of the rectangle of largest area that can
be inscribed in a circle of radius r.
26.Find the area of the largest rectangle that can be inscribed in
the ellipse x 2ya 2 1 y 2yb 2 − 1.
27.Find the dimensions of the rectangle of largest area that can
be inscribed in an equilateral triangle of side L if one side of
the rectangle lies on the base of the triangle.
28.Find the area of the largest trapezoid that can be inscribed
in a circle of radius 1 and whose base is a diameter of the
circle.
29.Find the dimensions of the isosceles triangle of largest area
that can be inscribed in a circle of radius r.
30.If the two equal sides of an isosceles triangle have length a,
find the length of the third side that maximizes the area of
the triangle.
31.A right circular cylinder is inscribed in a sphere of radius r.
Find the largest possible volume of such a cylinder.
32.A right circular cylinder is inscribed in a cone with height h
and base radius r. Find the largest possible volume of such
a cylinder.
33.A right circular cylinder is inscribed in a sphere of radius r.
Find the largest possible surface area of such a cylinder.
34.A Norman window has the shape of a rectangle surmounted
by a semicircle. (Thus the diameter of the semicircle is
equal to the width of the rectangle. See Exercise 1.1.62.) If
the perimeter of the window is 30 ft, find the dimensions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
266
chapter 3 Applications of Differentiation
of the window so that the greatest possible amount of light is
admitted.
35.The top and bottom margins of a poster are each 6 cm and the
side margins are each 4 cm. If the area of printed material on
the poster is fixed at 384 cm2, find the dimensions of the poster
with the smallest area.
36.A poster is to have an area of 180 in2 with 1-inch margins at the
bottom and sides and a 2-inch margin at the top. What dimen­
sions will give the largest printed area?
37.A piece of wire 10 m long is cut into two pieces. One piece
is bent into a square and the other is bent into an equilateral
triangle. How should the wire be cut so that the total area
enclosed is (a) a maximum? (b) A minimum?
38.Solve Exercise 37 if one piece is bent into a square and the
other into a circle.
39.If you are offered one slice from a round pizza (in other words,
a sector of a circle) and the slice must have a perimeter of
32 inches, what diameter pizza will reward you with the
largest slice?
40.A fence 8 ft tall runs parallel to a tall building at a distance of
4 ft from the building. What is the length of the shortest ladder
that will reach from the ground over the fence to the wall of the
building?
41.A cone-shaped drinking cup is made from a circular piece of
paper of radius R by cutting out a sector and joining the edges
CA and CB. Find the maximum capacity of such a cup.
A
B
R
C
42.A cone-shaped paper drinking cup is to be made to hold 27 cm3
of water. Find the height and radius of the cup that will use the
smallest amount of paper.
43.A cone with height h is inscribed in a larger cone with
height H so that its vertex is at the center of the base of the
larger cone. Show that the inner cone has maximum volume
when h − 13 H.
45.If a resistor of R ohms is connected across a battery of E volts
with internal resistance r ohms, then the power (in watts) in the
external resistor is
E 2R
P−
sR 1 rd 2
If E and r are fixed but R varies, what is the maximum value of
the power?
46.For a fish swimming at a speed v relative to the water, the
energy expenditure per unit time is proportional to v 3. It is
believed that migrating fish try to minimize the total energy
required to swim a fixed distance. If the fish are swimming
against a current u su , vd, then the time required to swim a
distance L is Lysv 2 ud and the total energy E required to swim
the distance is given by
L
Esvd − av 3 v2u
where a is the proportionality constant.
(a) Determine the value of v that minimizes E.
(b) Sketch the graph of E.
Note: This result has been verified experimentally; migrating
fish swim against a current at a speed 50% greater than the
current speed.
47.In a beehive, each cell is a regular hexagonal prism, open at
one end with a trihedral angle at the other end as in the figure.
It is believed that bees form their cells in such a way as to min­
imize the surface area for a given side length and height, thus
using the least amount of wax in cell construction. Examination
of these cells has shown that the measure of the apex angle is
amazingly consistent. Based on the geometry of the cell, it can
be shown that the surface area S is given by
S − 6sh 2 32 s 2 cot 1 (3s 2s3 y2) csc where s, the length of the sides of the hexagon, and h, the
height, are constants.
(a) Calculate dSyd.
(b) What angle should the bees prefer?
(c)Determine the minimum surface area of the cell (in terms
of s and h).
Note: Actual measurements of the angle in beehives have
been made, and the measures of these angles seldom differ
from the calculated value by more than 28.
trihedral
angle ¨
rear
of cell
44.An object with weight W is dragged along a horizontal plane
by a force acting along a rope attached to the object. If the rope
makes an angle with a plane, then the magnitude of the force
is
h
W
F−
sin 1 cos b
where is a constant called the coefficient of friction. For what
value of is F smallest?
s
front
of cell
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.7 Optimization Problems
48.A boat leaves a dock at 2:00 pm and travels due south at a
speed of 20 kmyh. Another boat has been heading due east
at 15 kmyh and reaches the same dock at 3:00 pm. At what
time were the two boats closest together?
49.Solve the problem in Example 4 if the river is 5 km wide
and point B is only 5 km downstream from A.
50.A woman at a point A on the shore of a circular lake with
radius 2 mi wants to arrive at the point C diametrically
opposite A on the other side of the lake in the shortest
possible time (see the figure). She can walk at the rate of
4 miyh and row a boat at 2 miyh. How should she proceed?
B
A
¨
2
2
C
51.An oil refinery is located on the north bank of a straight
river that is 2 km wide. A pipeline is to be constructed from
the refinery to storage tanks located on the south bank of
the river 6 km east of the refinery. The cost of laying pipe
is $400,000ykm over land to a point P on the north bank
and $800,000ykm under the river to the tanks. To minimize
the cost of the pipeline, where should P be located?
; 52.Suppose the refinery in Exercise 51 is located 1 km north
of the river. Where should P be located?
53.The illumination of an object by a light source is directly
proportional to the strength of the source and inversely
proportional to the square of the distance from the source.
If two light sources, one three times as strong as the other,
are placed 10 ft apart, where should an object be placed
on the line between the sources so as to receive the least
illumination?
54.Find an equation of the line through the point s3, 5d that
cuts off the least area from the first quadrant.
55.Let a and b be positive numbers. Find the length of the
shortest line segment that is cut off by the first quadrant and
passes through the point sa, bd.
56.At which points on the curve y − 1 1 40x 3 2 3x 5 does the
tangent line have the largest slope?
57.What is the shortest possible length of the line segment that
is cut off by the first quadrant and is tangent to the curve
y − 3yx at some point?
267
58.What is the smallest possible area of the triangle that is cut
off by the first quadrant and whose hypotenuse is tangent to
the parabola y − 4 2 x 2 at some point?
59. (a)If Csxd is the cost of producing x units of a commodity,
then the average cost per unit is csxd − Csxdyx. Show
that if the average cost is a minimum, then the marginal
cost equals the average cost.
(b)If Csxd − 16,000 1 200x 1 4x 3y2, in dollars, find (i)
the cost, average cost, and marginal cost at a production
level of 1000 units; (ii) the production level that will
minimize the average cost; and (iii) the minimum average cost.
60. (a)Show that if the profit Psxd is a maximum, then the
marginal revenue equals the marginal cost.
(b)If Csxd − 16,000 1 500x 2 1.6x 2 1 0.004x 3 is the
cost function and psxd − 1700 2 7x is the demand
function, find the production level that will maximize
profit.
61.A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at $10, the average attendance had
been 27,000. When ticket prices were lowered to $8, the
average attendance rose to 33,000.
(a) Find the demand function, assuming that it is linear.
(b) How should ticket prices be set to maximize revenue?
62.During the summer months Terry makes and sells necklaces
on the beach. Last summer he sold the necklaces for $10
each and his sales averaged 20 per day. When he increased
the price by $1, he found that the average decreased by two
sales per day.
(a) Find the demand function, assuming that it is linear.
(b)If the material for each necklace costs Terry $6, what
should the selling price be to maximize his profit?
63.A retailer has been selling 1200 tablet computers a week
at $350 each. The marketing department estimates that an
additional 80 tablets will sell each week for every $10 that
the price is lowered.
(a) Find the demand function.
(b)What should the price be set at in order to maximize
revenue?
(c)If the retailer’s weekly cost function is
Csxd − 35,000 1 120x
what price should it choose in order to maximize its
profit?
64.A company operates 16 oil wells in a designated area. Each
pump, on average, extracts 240 barrels of oil daily. The
company can add more wells but every added well reduces
the average daily ouput of each of the wells by 8 barrels.
How many wells should the company add in order to maximize daily production?
65.Show that of all the isosceles triangles with a given perime­
ter, the one with the greatest area is equilateral.
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268
chapter 3 Applications of Differentiation
66.Consider the situation in Exercise 51 if the cost of laying
pipe under the river is considerably higher than the cost of
laying pipe over land ($400,000ykm). You may suspect that
in some instances, the minimum distance possible under the
river should be used, and P should be located 6 km from the
refinery, directly across from the storage tanks. Show that this
is never the case, no matter what the “under river” cost is.
x2
y2
67.Consider the tangent line to the ellipse 2 1 2 − 1
a
b
at a point s p, qd in the first quadrant.
(a)Show that the tangent line has x-intercept a 2yp and
y-intercept b 2yq.
(b)Show that the portion of the tangent line cut off by the
coordinate axes has minimum length a 1 b.
(c)Show that the triangle formed by the tangent line and
the coordinate axes has minimum area ab.
CAS
68.The frame for a kite is to be made from six pieces of wood.
The four exterior pieces have been cut with the lengths
indicated in the figure. To maximize the area of the kite, how
long should the diagonal pieces be?
c
0
20
40
60
71.Let v1 be the velocity of light in air and v2 the velocity of
light in water. According to Fermat’s Principle, a ray of light
will travel from a point A in the air to a point B in the water
by a path ACB that minimizes the time taken. Show that
sin 1
v1
−
sin 2
v2
where 1 (the angle of incidence) and 2 (the angle of refraction) are as shown. This equation is known as Snell’s Law.
A
¨¡
C
b
a
√
¨™
a
B
b
; 69.A point P needs to be located somewhere on the line AD
so that the total length L of cables linking P to the points
A, B. and C is minimized (see the figure). Express L as a
function of x − AP and use the graphs of L and dLydx to
estimate the minimum value of L.
| |
72.Two vertical poles PQ and ST are secured by a rope PRS
going from the top of the first pole to a point R on the
ground between the poles and then to the top of the second
pole as in the figure. Show that the shortest length of such a
rope occurs when 1 − 2 .
P
S
A
P
¨¡
5m
Q
B
2m
D
3m
C
70.The graph shows the fuel consumption c of a car (measured
in gallons per hour) as a function of the speed v of the car. At
very low speeds the engine runs inefficiently, so initially c
decreases as the speed increases. But at high speeds the fuel
consumption increases. You can see that csvd is minimized
for this car when v < 30 miyh. However, for fuel efficiency,
what must be minimized is not the consumption in gallons
per hour but rather the fuel consumption in gallons per mile.
Let’s call this consumption G. Using the graph, estimate the
speed at which G has its minimum value.
¨™
R
T
73.The upper right-hand corner of a piece of paper, 12 in. by
8 in., as in the figure, is folded over to the bottom edge. How
would you fold it so as to minimize the length of the fold? In
other words, how would you choose x to minimize y?
12
y
8
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.7 Optimization Problems
74.A steel pipe is being carried down a hallway 9 ft wide. At
the end of the hall there is a right-angled turn into a narrower
hallway 6 ft wide. What is the length of the longest pipe that
can be carried horizontally around the corner?
269
blood. (Poiseuille established this law experimentally, but it
also follows from Equation 8.4.2.) The figure shows a main
blood vessel with radius r1 branching at an angle into a
smaller vessel with radius r 2.
C
6
r™
¨
b
vascular
branching
9
A
r¡
¨
B
75.An observer stands at a point P, one unit away from a track.
Two runners start at the point S in the figure and run along the
track. One runner runs three times as fast as the other. Find
the maximum value of the observer’s angle of sight between
the runners. [Hint: Maximize tan .]
(a)Use Poiseuille’s Law to show that the total resistance of
the blood along the path ABC is
R−C
P
¨
1
a
S
D
a 2 b cot b csc 1
r14
r24
where a and b are the distances shown in the figure.
(b) Prove that this resistance is minimized when
cos −
S
76.A rain gutter is to be constructed from a metal sheet of width
30 cm by bending up one-third of the sheet on each side
through an angle . How should be chosen so that the gutter
will carry the maximum amount of water?
¨
10 cm
¨
10 cm
10 cm
77.Find the maximum area of a rectangle that can be circum­
scribed about a given rectangle with length L and width W.
[Hint: Express the area as a function of an angle .]
78.The blood vascular system consists of blood vessels (arteries,
arterioles, capillaries, and veins) that convey blood from the
heart to the organs and back to the heart. This system should
work so as to minimize the energy expended by the heart in
pumping the blood. In particular, this energy is reduced when
the resistance of the blood is lowered. One of Poiseuille’s
Laws gives the resistance R of the blood as
R−C
L
r4
where L is the length of the blood vessel, r is the radius, and
C is a positive constant determined by the viscosity of the
r24
r14
(c)Find the optimal branching angle (correct to the nearest
degree) when the radius of the smaller blood vessel is
two-thirds the radius of the larger vessel.
79.Ornithologists have determined that some species of birds
tend to avoid flights over large bodies of water during daylight
hours. It is believed that more energy is required to fly over
water than over land because air generally rises over land and
falls over water during the day. A bird with these tendencies
is released from an island that is 5 km from the nearest point
B on a straight shoreline, flies to a point C on the shoreline,
and then flies along the shoreline to its nesting area D. Assume
that the bird instinctively chooses a path that will minimize its
energy expenditure. Points B and D are 13 km apart.
(a)In general, if it takes 1.4 times as much energy to fly over
water as it does over land, to what point C should the
bird fly in order to minimize the total energy expended in
returning to its nesting area?
(b)Let W and L denote the energy (in joules) per kilometer
flown over water and land, respectively. What would a
large value of the ratio WyL mean in terms of the bird’s
flight? What would a small value mean? Determine the
ratio WyL corresponding to the minimum expenditure of
energy.
(c)What should the value of WyL be in order for the bird to
fly directly to its nesting area D? What should the value
of WyL be for the bird to fly to B and then along the shore
to D?
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270
Chapter 3 Applications of Differentiation
(d)If the ornithologists observe that birds of a certain species reach the shore at a point 4 km from B, how many
times more energy does it take a bird to fly over water
than over land?
island
5 km
C
B
13 km
D
nest
to use the fact that the intensity of illumination for a single
source is directly proportional to the strength of the source
and inversely proportional to the square of the distance from
the source.
(a)Find an expression for the intensity Isxd at the point P.
(b)If d − 5 m, use graphs of Isxd and I9sxd to show that the
intensity is minimized when x − 5 m, that is, when P is
at the midpoint of ,.
(c)If d − 10 m, show that the intensity (perhaps surpris­
ingly) is not minimized at the midpoint.
(d)Somewhere between d − 5 m and d − 10 m there is
a transitional value of d at which the point of minimal
illumination abruptly changes. Estimate this value of d
by graphical methods. Then find the exact value of d.
P
; 80.Two light sources of identical strength are placed 10 m
apart. An object is to be placed at a point P on a line ,,
paral­lel to the line joining the light sources and at a distance
d meters from it (see the figure). We want to locate P on ,
so that the intensity of illumination is minimized. We need
applied Project
h
r
The shape of a can
x
d
10 m
7et0407x78
09/11/09
MasterID: 00567
In this project we investigate the most economical shape for a can. We first interpret this to mean
that the volume V of a cylindrical can is given and we need to find the height h and radius r that
minimize the cost of the metal to make the can (see the figure). If we disregard any waste metal
in the manufacturing process, then the problem is to minimize the surface area of the cylinder.
We solved this problem in Example 3.7.2 and we found that h − 2r; that is, the height should be
the same as the diameter. But if you go to your cupboard or your supermarket with a ruler, you
will discover that the height is usually greater than the diameter and the ratio hyr varies from 2
up to about 3.8. Let’s see if we can explain this phenomenon.
The material for the cans is cut from sheets of metal. The cylindrical sides are formed by
1.
bending rectangles; these rectangles are cut from the sheet with little or no waste. But if the
top and bottom discs are cut from squares of side 2r (as in the second figure), this leaves
considerable waste metal, which may be recycled but has little or no value to the can makers.
If this is the case, show that the amount of metal used is minimized when
h
8
−
< 2.55
r
Discs cut from squares
2.
A more efficient packing of the discs is obtained by dividing the metal sheet into hexagons
and cutting the circular lids and bases from the hexagons (see the last figure). Show that if this
strategy is adopted, then
h
4 s3
−
< 2.21
r
Discs cut from hexagons
3. The values of hyr that we found in Problems 1 and 2 are a little closer to the ones that
actually occur on supermarket shelves, but they still don’t account for everything. If we
look more closely at some real cans, we see that the lid and the base are formed from discs
with radius larger than r that are bent over the ends of the can. If we allow for this we would
increase hyr. More significantly, in addition to the cost of the metal we need to incorporate
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
applied Project Planes and Birds: Minimizing Energy
271
the manufacturing of the can into the cost. Let’s assume that most of the expense is incurred
in joining the sides to the rims of the cans. If we cut the discs from hexagons as in Problem 2,
then the total cost is proportional to
4 s3 r 2 1 2rh 1 ks4r 1 hd
where k is the reciprocal of the length that can be joined for the cost of one unit area of metal.
Show that this expression is minimized when
3
V
s
−
k
Î
3
h
2 2 hyr
r
hyr 2 4s3
3
lot s
V yk as a function of x − hyr and use your graph to argue that when a can is large or
; 4. P
joining is cheap, we should make hyr approximately 2.21 (as in Problem 2). But when the can
is small or joining is costly, hyr should be substantially larger.
5. O
ur analysis shows that large cans should be almost square but small cans should be tall and
thin. Take a look at the relative shapes of the cans in a supermarket. Is our conclusion usually true in practice? Are there exceptions? Can you suggest reasons why small cans are not
always tall and thin?
© Targn Pleiades / Shutterstock.com
applied Project
planes and birds: Minimizing energy
Small birds like finches alternate between flapping their wings and keeping them folded while
gliding (see Figure 1). In this project we analyze this phenomenon and try to determine how
frequently a bird should flap its wings. Some of the principles are the same as for fixed-wing
aircraft and so we begin by considering how required power and energy depend on the speed of
airplanes.1
FIGURE 1
1.
The power needed to propel an airplane forward at velocity v is
P − Av 3 1
BL 2
v
where A and B are positive constants specific to the particular aircraft and L is the lift, the
upward force supporting the weight of the plane. Find the speed that minimizes the required
power.
2. T
he speed found in Problem 1 minimizes power but a faster speed might use less fuel. The
energy needed to propel the airplane a unit distance is E − Pyv. At what speed is energy
minimized?
1. Adapted from R. McNeill Alexander, Optima for Animals (Princeton, NJ: Princeton University
Press, 1996.)
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
272
Chapter 3 Applications of Differentiation
3.
Hows much faster is the speed for minimum energy than the speed for minimum power?
4.
In applying the equation of Problem 1 to bird flight we split the term Av 3 into two parts:
Ab v 3 for the bird’s body and Aw v 3 for its wings. Let x be the fraction of flying time spent in
flapping mode. If m is the bird’s mass and all the lift occurs during flapping, then the lift
is mtyx and so the power needed during flapping is
Pflap − sA b 1 Awdv 3 1
Bsmtyxd2
v
The power while wings are folded is Pfold − Abv 3. Show that the average power over an entire
flight cycle is
P − x Pflap 1 s1 2 xdPfold − Abv 3 1 x Awv 3 1
Bm2t 2
xv
5. F
or what value of x is the average power a minimum? What can you conclude if the bird flies
slowly? What can you conclude if the bird flies faster and faster?
6.
The average energy over a cycle is E − Pyv. What value of x minimizes E ?
Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per
month for five years. You would like to know what monthly interest rate the dealer is, in
effect, charging you. To find the answer, you have to solve the equation
1
0.15
0
0.012
_0.05
FIGURE 1
Try to solve Equation 1 numerically
using your calculator or computer.
Some machines are not able to solve it.
Others are successful but require you to
specify a starting point for the search.
48xs1 1 xd60 2 s1 1 xd60 1 1 − 0
(The details are explained in Exercise 39.) How would you solve such an equation?
For a quadratic equation ax 2 1 bx 1 c − 0 there is a well-known formula for the
solutions. For third- and fourth-degree equations there are also formulas for the solutions, but they are extremely complicated. If f is a polynomial of degree 5 or higher,
there is no such formula (see the note on page 164). Likewise, there is no formula that
will enable us to find the exact roots of a transcendental equation such as cos x − x.
We can find an approximate solution to Equation 1 by plotting the left side of the
equation. Using a graphing device, and after experimenting with viewing rectangles, we
pro­duce the graph in Figure 1.
We see that in addition to the solution x − 0, which doesn’t interest us, there is a
solution between 0.007 and 0.008. Zooming in shows that the root is approximately
0.0076. If we need more accuracy we could zoom in repeatedly, but that becomes tiresome. A faster alternative is to use a calculator or computer algebra system to solve the
equation numerically. If we do so, we find that the root, correct to nine decimal places,
is 0.007628603.
How do these devices solve equations? They use a variety of methods, but most of
them make some use of Newton’s method, also called the Newton-Raphson method.
We will explain how this method works, partly to show what happens inside a calculator
or computer, and partly as an application of the idea of linear approximation.
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Section 3.8 Newton’s Method
y
{x ¡, f(x¡)}
y=ƒ
L
x™ x¡
r
0
x
FIGURE 2
273
The geometry behind Newton’s method is shown in Figure 2. We wish to solve an
equation of the form f sxd − 0, so the roots of the equation correspond to the x-intercepts
of the graph of f. The root that we are trying to find is labeled r in the figure. We start
with a first approximation x 1, which is obtained by guess­ing, or from a rough sketch of
the graph of f , or from a computer-generated graph of f. Consider the tangent line L to
the curve y − f sxd at the point sx 1, f sx 1dd and look at the x-intercept of L, labeled x 2.
The idea behind Newton’s method is that the tangent line is close to the curve and so
its x-intercept, x2, is close to the x-intercept of the curve (namely, the root r that we are
seeking). Because the tangent is a line, we can easily find its x-intercept.
To find a formula for x2 in terms of x1 we use the fact that the slope of L is f 9sx1 d, so
its equation is
y 2 f sx 1 d − f 9sx 1 dsx 2 x 1 d
Since the x-intercept of L is x 2 , we know that the point sx 2 , 0d is on the line, and so
0 2 f sx 1 d − f 9sx 1 dsx 2 2 x 1 d
If f 9sx 1d ± 0, we can solve this equation for x 2 :
x2 − x1 2
We use x 2 as a second approximation to r.
Next we repeat this procedure with x 1 replaced by the second approximation x 2 , using
the tangent line at sx 2 , f sx 2 dd. This gives a third approximation:
y
{x¡, f(x¡)}
x3 − x2 2
{x™, f(x™)}
r
0
x¢
x£
x™ x¡
x
2
Sequences were briefly introduced
in A Preview of Calculus on page 5.
A more thorough discussion starts
in Section 11.1.
FIGURE 4
x n11 − x n 2
f sx n d
f 9sx n d
If the numbers x n become closer and closer to r as n becomes large, then we say that
the sequence converges to r and we write
lim x n − r
y
0
f sx 2 d
f 9sx 2 d
If we keep repeating this process, we obtain a sequence of approximations x 1, x 2, x 3, x 4, . . .
as shown in Figure 3. In general, if the nth approximation is x n and f 9sx n d ± 0, then
the next approximation is given by
FIGURE 3
x£
f sx 1 d
f 9sx 1 d
nl`
x¡
r
x™
x
A
lthough the sequence of successive approximations converges to the desired root for
fun­ctions of the type illustrated in Figure 3, in certain circumstances the sequence may
not converge. For example, consider the situation shown in Figure 4. You can see that
x 2 is a worse approximation than x 1. This is likely to be the case when f 9sx 1d is close to
0. It might even happen that an approximation (such as x 3 in Figure 4) falls outside the
domain of f. Then Newton’s method fails and a better initial approximation x 1 should
be chosen. See Exercises 29–32 for specific examples in which Newton’s method works
very slowly or does not work at all.
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274
Chapter 3 Applications of Differentiation
Example 1 Starting with x 1 − 2, find the third approximation x 3 to the root of the
equation x 3 2 2x 2 5 − 0.
TEC In Module 3.8 you can investigate how Newton’s method works for
several functions and what happens
when you change x 1.
SOLUTION We apply Newton’s method with
f sxd − x 3 2 2x 2 5 and f 9sxd − 3x 2 2 2
Newton himself used this equation to illustrate his method and he chose x 1 − 2 after
some experimentation because f s1d − 26, f s2d − 21, and f s3d − 16. Equation 2
becomes
Figure 5 shows the geometry behind
x™
2.2
y=10x-21
FIGURE 5
f sx1 d
x13 2 2x 1 2 5
− x1 2
f 9sx1d
3x12 2 2
2 3 2 2s2d 2 5
− 2.1
3s2d2 2 2
Then with n − 2 we obtain
x3 − x2 2
_2
x2 − x1 2
−22
1
1.8
f sx nd
x n3 2 2x n 2 5
− xn 2
f 9sx nd
3x n2 2 2
With n − 1 we have
the first step in Newton’s method in
Example 1. Since f 9s2d − 10, the
tangent line to y − x 3 2 2x 2 5 at
s2, 21d has equation y − 10x 2 21
so its x-intercept is x 2 − 2.1.
y=˛-2x-5
x n11 − x n 2
x 23 2 2x 2 2 5
s2.1d3 2 2s2.1d 2 5
− 2.1 2
< 2.0946
2
3x 2 2 2
3s2.1d2 2 2
It turns out that this third approximation x 3 < 2.0946 is accurate to four decimal
places.
n
Suppose that we want to achieve a given accuracy, say to eight decimal places, using
Newton’s method. How do we know when to stop? The rule of thumb that is generally
used is that we can stop when successive approximations x n and x n11 agree to eight decimal places. (A precise statement concerning accuracy in Newton’s method will be given
in Exercise 11.11.39.)
Notice that the procedure in going from n to n 1 1 is the same for all values of n. (It is
called an iterative process.) This means that Newton’s method is particularly convenient
for use with a programmable calculator or a computer.
6
Example 2 Use Newton’s method to find s
2 correct to eight decimal places.
6 2
SOLUTION First we observe that finding s
is equivalent to finding the positive root of
the equation
x6 2 2 − 0
so we take f sxd − x 6 2 2. Then f 9sxd − 6x 5 and Formula 2 (Newton’s method)
becomes
x n11 − x n 2
fsx nd
x n6 2 2
− xn 2
f 9sx nd
6x n5
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Section 3.8 Newton’s Method
275
If we choose x 1 − 1 as the initial approximation, then we obtain
x 2 < 1.16666667
x 3 < 1.12644368
x 4 < 1.12249707
x 5 < 1.12246205
x 6 < 1.12246205
Since x 5 and x 6 agree to eight decimal places, we conclude that
6
2 < 1.12246205
s
to eight decimal places.
n
Example 3 Find, correct to six decimal places, the root of the equation cos x − x.
SOLUTION We first rewrite the equation in standard form:
cos x 2 x − 0
y
Therefore we let f sxd − cos x 2 x. Then f 9sxd − 2sin x 2 1, so Formula 2 becomes
y=x
y=cos x
1
π
2
x
π
x n11 − x n 2
cos x n 2 x n
cos x n 2 x n
− xn 1
2sin x n 2 1
sin x n 1 1
In order to guess a suitable value for x 1 we sketch the graphs of y − cos x and y − x in
Figure 6. It appears that they intersect at a point whose x-coordinate is somewhat less
than 1, so let’s take x 1 − 1 as a convenient first approximation. Then, remembering to
put our calculator in radian mode, we get
FIGURE 6
x 2 < 0.75036387
x 3 < 0.73911289
x 4 < 0.73908513
x 5 < 0.73908513
Since x 4 and x 5 agree to six decimal places (eight, in fact), we conclude that the root of
the equation, correct to six decimal places, is 0.739085.
n
1
y=cos x
Instead of using the rough sketch in Figure 6 to get a starting approximation for
Newton’s method in Example 3, we could have used the more accurate graph that a
calculator or computer provides. Figure 7 suggests that we use x1 − 0.75 as the initial
approximation. Then Newton’s method gives
y=x
0
FIGURE 7
1
x 2 < 0.73911114 x 3 < 0.73908513 x 4 < 0.73908513
and so we obtain the same answer as before, but with one fewer step.
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276
Chapter 3 Applications of Differentiation
1.The figure shows the graph of a function f. Suppose that
Newton’s method is used to approximate the root s of the
equation f sxd − 0 with initial approximation x 1 − 6.
(a)Draw the tangent lines that are used to find x 2 and x 3,
and esti­mate the numerical values of x 2 and x 3.
(b)Would x 1 − 8 be a better first approximation? Explain.
7.
2
2 x 2 1 1 − 0, x 1 − 2
x
8. x 7 1 4 − 0, x1 − 21
; 9.Use Newton’s method with initial approximation x1 − 21
to find x 2, the second approximation to the root of the equation x 3 1 x 1 3 − 0. Explain how the method works by
first graphing the function and its tangent line at s21, 1d.
; 10.Use Newton’s method with initial approximation x1 − 1 to
find x 2, the second approximation to the root of the equation
x 4 2 x 2 1 − 0. Explain how the method works by first
graphing the function and its tangent line at s1, 21d.
11–12 Use Newton’s method to approximate the given number
correct to eight decimal places.
4
11. s
75
2.Follow the instructions for Exercise 1(a) but use x 1 − 1
as the starting approximation for finding the root r.
8
12. s
500
3.Suppose the tangent line to the curve y − f sxd at the point
s2, 5d has the equation y − 9 2 2x. If Newton’s method is
used to locate a root of the equation f sxd − 0 and the initial
approximation is x1 − 2, find the second approximation x 2.
13–14 (a) Explain how we know that the given equation must
have a root in the given interval. (b) Use Newton’s method to
approximate the root correct to six decimal places.
4.For each initial approximation, determine graphically what
happens if Newton’s method is used for the function whose
graph is shown.
(a) x1 − 0
(b) x1 − 1
(c) x1 − 3
(d) x1 − 4(e) x1 − 5
14. 22 x 5 1 9x 4 2 7x 3 2 11x − 0, f3, 4g
13. 3x 4 2 8x 3 1 2 − 0, f2, 3g
15–16 Use Newton’s method to approximate the indicated root
of the equation correct to six decimal places.
15. The positive root of sin x − x 2
y
16. The positive root of 3 sin x − x
0
3
1
17–22 Use Newton’s method to find all solutions of the equation
correct to six decimal places.
x
5
17. 3 cos x − x 1 1
7et0408x04
5.For which of the initial approximations x1 − a, b, c, and d
09/11/09
do you think Newton’s method will work and lead to the
MasterID:
00578
root of the
equation f sxd
− 0?
y
a
0
6.2 x 2 3x 1 2 − 0 , x 1 − 21
1
3
−s
x 21
x
21. x 3 − cos x
20. sx 2 1d 2 − sx
22. sin x − x 2 2 2
; 23–26 Use Newton’s method to find all the solutions of the
equation correct to eight decimal places. Start by drawing a
graph to find initial approximations.
b
c
d
x
6–8 Use
Newton’s method with the specified initial approxima7et0408x05
tion x01/19/10
1 to find x 3, the third approximation to the root of the given
equation. (Give your answer to four decimal places.)
MasterID:
03017
3
2
19.
18. sx 1 1 − x 2 2 x
23. 22x7 2 5x 4 1 9x 3 1 5 − 0
24. x 5 2 3x 4 1 x 3 2 x 2 2 x 1 6 − 0
25.
x
− s1 2 x
x2 1 1
26. cossx 2 2 xd − x 4
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Section 3.8 Newton’s Method
27. (a)Apply Newton’s method to the equation x 2 2 a − 0 to
derive the following square-root algorithm (used by the
ancient Babylonians to compute sa ):
x n11 −
S
1
a
xn 1
2
xn
D
(b)Use part (a) to compute s1000 correct to six decimal
places.
28. (a)Apply Newton’s method to the equation 1yx 2 a − 0 to
derive the following reciprocal algorithm:
x n11 − 2x n 2 ax n2
(This algorithm enables a computer to find reciprocals
without actually dividing.)
(b)Use part (a) to compute 1y1.6984 correct to six decimal
places.
29.Explain why Newton’s method doesn’t work for finding the
root of the equation
(b)Find the absolute minimum value of f correct to four
decimal places.
34.Use Newton’s method to find the absolute maximum value
of the function f sxd − x cos x, 0 < x < , correct to six
decimal places.
35.Use Newton’s method to find the coordinates of the inflection point of the curve y − x 2 sin x, 0 < x < , correct
to six decimal places.
36.Of the infinitely many lines that are tangent to the curve
y − 2sin x and pass through the origin, there is one that has
the largest slope. Use Newton’s method to find the slope of
that line correct to six decimal places.
37.Use Newton’s method to find the coordinates, correct to six
decimal places, of the point on the parabola y − sx 2 1d 2
that is closest to the origin.
38.In the figure, the length of the chord AB is 4 cm and the
length of the arc AB is 5 cm. Find the central angle , in
radians, correct to four decimal places. Then give the
answer to the nearest degree.
x 3 2 3x 1 6 − 0
if the initial approximation is chosen to be x 1 − 1.
30. (a)Use Newton’s method with x 1 − 1 to find the root of
the equation x 3 2 x − 1 correct to six decimal places.
(b)Solve the equation in part (a) using x 1 − 0.6 as the
initial approximation.
(c)Solve the equation in part (a) using x 1 − 0.57. (You
definitely need a programmable calculator for this
part.)
(d)Graph f sxd − x 3 2 x 2 1 and its tangent lines at
;
x1 − 1, 0.6, and 0.57 to explain why Newton’s method
is so sensitive to the value of the initial approximation.
31.Explain why Newton’s method fails when applied to the
3
equation s
x − 0 with any initial approximation x 1 ± 0.
Illustrate your explanation with a sketch.
32. If
f sxd −
H
5 cm
A
4 cm
B
¨
39.A car dealer sells a new car for $18,000. He also offers to
sell the same car for payments of $375 per month for five
years. What monthly interest rate is this dealer charging?
To solve this problem you will need to use the formula
for the present value A of an annuity consisting of n equal
payments of size R with interest rate i per time period:
A−
R
f1 2 s1 1 i d2n g
i
Replacing i by x, show that
sx
2s2x
if x > 0
if x , 0
then the root of the equation f sxd − 0 is x − 0. Explain
why Newton’s method fails to find the root no matter which
initial approximation x 1 ± 0 is used. Illustrate your explanation with a sketch.
33. (a)Use Newton’s method to find the critical numbers of the
function
f sxd − x 6 2 x 4 1 3x 3 2 2x
277
correct to six decimal places.
48xs1 1 xd60 2 s1 1 xd60 1 1 − 0
Use Newton’s method to solve this equation.
40.The figure shows the sun located at the origin and the earth
at the point s1, 0d. (The unit here is the distance between the
centers of the earth and the sun, called an astronomical
unit: 1 AU < 1.496 3 10 8 km.) There are five locations
L 1, L 2, L 3, L 4, and L 5 in this plane of rotation of the earth
about the sun where a satellite remains motionless with
respect to the earth because the forces acting on the
satellite (including the gravitational attractions of the earth
and the sun) balance each other. These locations are called
libration points. (A solar research satellite has been placed
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278
Chapter 3 Applications of Differentiation
at one of these libration points.) If m1 is the mass of the sun,
m 2 is the mass of the earth, and r − m 2ysm1 1 m 2 d, it turns out
that the x-coordinate of L 1 is the unique root of the fifth-degree
equation
Using the value r < 3.04042 3 10 26, find the locations of the
libration points (a) L 1 and (b) L 2.
y
psxd − x 5 2 s2 1 rdx 4 1 s1 1 2rdx 3 2 s1 2 rdx 2
L¢
sun
1 2s1 2 rdx 1 r 2 1 − 0
earth
L∞
and the x-coordinate of L 2 is the root of the equation
L¡
L™
x
L£
psxd 2 2rx 2 − 0
A physicist who knows the velocity of a particle might wish to know its position at a
given time. An engineer who can measure the variable rate at which water is leaking
from a tank wants to know the amount leaked over a certain time period. A biologist who
knows the rate at which a bacteria population is increasing might want to deduce what
the size of the population will be at some future time. In each case, the problem is to
find a function F whose derivative is a known function f. If such a function F exists, it
is called an antiderivative of f.
Definition A function F is called an antiderivative of f on an interval I if
F9sxd − f sxd for all x in I.
For instance, let f sxd − x 2. It isn’t difficult to discover an antiderivative of f if we
keep the Power Rule in mind. In fact, if Fsxd − 13 x 3, then F9sxd − x 2 − f sxd. But the
function Gsxd − 13 x 3 1 100 also satisfies G9sxd − x 2. Therefore both F and G are antiderivatives of f. Indeed, any function of the form Hsxd − 13 x 3 1 C, where C is a constant,
is an antiderivative of f. The question arises: are there any others?
To answer this question, recall that in Section 3.2 we used the Mean Value Theorem to
prove that if two functions have identical derivatives on an interval, then they must differ
by a constant (Corollary 3.2.7). Thus if F and G are any two antiderivatives of f , then
y
˛
y= 3 +3
˛
y= 3 +2
˛
y= 3 +1
0
x
y= ˛
3
˛
y= 3 -1
˛
y= 3 -2
FIGURE 1
Members of the family of
antiderivatives of f sxd − x 2
F9sxd − f sxd − G9sxd
so Gsxd 2 Fsxd − C, where C is a constant. We can write this as Gsxd − Fsxd 1 C, so
we have the following result.
1 Theorem If F is an antiderivative of f on an interval I, then the most general
antiderivative of f on I is
Fsxd 1 C
where C is an arbitrary constant.
Going back to the function f sxd − x 2, we see that the general antiderivative of f is
1 C. By assigning specific values to the constant C, we obtain a family of functions
whose graphs are vertical translates of one another (see Figure 1). This makes sense
because each curve must have the same slope at any given value of x.
1 3
3x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 3.9 Antiderivatives
279
Example 1 Find the most general antiderivative of each of the following functions.
(a) f sxd − sin x (b) f sxd − x n, n > 0 (c) f sxd − x 23
SOLUTION
(a) If Fsxd − 2cos x , then F9sxd − sin x, so an antiderivative of sin x is 2cos x. By
Theorem 1, the most general antiderivative is Gsxd − 2cos x 1 C.
(b) We use the Power Rule to discover an antiderivative of x n:
d
dx
S D
x n11
n11
−
sn 1 1dx n
− xn
n11
Therefore the general antiderivative of f sxd − x n is
Fsxd −
x n11
1C
n11
This is valid for n > 0 because then f sxd − x n is defined on an interval.
(c) If we put n − 23 in the antiderivative formula from part (b), we get the particular
antiderivative Fsxd − x 22ys22d. But notice that f sxd − x 23 is not defined at x − 0.
Thus Theorem 1 tells us only that the general antiderivative of f is x 22ys22d 1 C on
any interval that does not contain 0. So the general antiderivative of f sxd − 1yx 3 is
1
1 C1 if x . 0
2x 2
F sxd −
1
2 2 1 C2 if x , 0
2x
2
■
As in Example 1, every differentiation formula, when read from right to left, gives
rise to an antidifferentiation formula. In Table 2 we list some particular antiderivatives.
Each for­mula in the table is true because the derivative of the function in the right column
appears in the left column. In particular, the first formula says that the antiderivative of
a constant times a function is the constant times the antiderivative of the function. The
second formula says that the antiderivative of a sum is the sum of the antiderivatives. (We
use the notation F9− f , G9 − t.)
able of
2 T
Antidifferentiation
Formulas
To obtain the most general anti­derivative
from the particular ones in Table 2, we
have to add a constant (or constants), as
in Example 1.
Function
Particular antiderivative
Function
Particular antiderivative
c f sxd
cFsxd
cos x
sin x
f sxd 1 tsxd
Fsxd 1 Gsxd
sin x
2cos x
x n sn ± 21d
x n11
n11
sec2x
tan x
sec x tan x
sec x
Example 2 Find all functions t such that
t9sxd − 4 sin x 1
2x 5 2 sx
x
SOLUTION We first rewrite the given function as follows:
t9sxd − 4 sin x 1
2x 5
1
sx
2
− 4 sin x 1 2x 4 2
x
x
sx
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
280
Chapter 3 Applications of Differentiation
Thus we want to find an antiderivative of
t9sxd − 4 sin x 1 2x 4 2 x21y2
Using the formulas in Table 2 together with Theorem 1, we obtain
tsxd − 4s2cos xd 1 2
We often use a capital letter F to represent an antiderivative of a function f.
If we begin with derivative notation, f 9,
an antiderivative is f, of course.
x5
x1y2
2 1 1C
5
2
− 24 cos x 1 25 x 5 2 2sx 1 C
n
In applications of calculus it is very common to have a situation as in Example 2,
where it is required to find a function, given knowledge about its derivatives. An equation
that involves the derivatives of a function is called a differential equation. These will be
studied in some detail in Chapter 9, but for the present we can solve some elementary differential equations. The general solution of a differential equation involves an arbitrary
con­stant (or constants) as in Example 2. However, there may be some extra conditions
given that will determine the constants and therefore uniquely specify the solution.
Example 3 Find f if f 9sxd − xsx and f s1d − 2.
SOLUTION The general antiderivative of
f 9sxd − xsx − x 3y2
f sxd −
is
x 5y2
5
2
1 C − 25 x 5y2 1 C
To determine C we use the fact that f s1d − 2:
f s1d − 25 1 C − 2
Solving for C, we get C − 2 2 25 − 85, so the particular solution is
f sxd −
2x 5y2 1 8
5
n
Example 4 Find f if f 0sxd − 12x 2 1 6x 2 4, f s0d − 4, and f s1d − 1.
SOLUTION The general antiderivative of f 0sxd − 12x 2 1 6x 2 4 is
f 9sxd − 12
x3
x2
16
2 4x 1 C − 4x 3 1 3x 2 2 4x 1 C
3
2
Using the antidifferentiation rules once more, we find that
f sxd − 4
x4
x3
x2
13
24
1 Cx 1 D − x 4 1 x 3 2 2x 2 1 Cx 1 D
4
3
2
To determine C and D we use the given conditions that f s0d − 4 and f s1d − 1. Since
f s0d − 0 1 D − 4, we have D − 4. Since
f s1d − 1 1 1 2 2 1 C 1 4 − 1
we have C − 23. Therefore the required function is
f sxd − x 4 1 x 3 2 2x 2 2 3x 1 4
n
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Section 3.9 Antiderivatives
281
If we are given the graph of a function f, it seems reasonable that we should be able
to sketch the graph of an antiderivative F. Suppose, for instance, that we are given that
Fs0d − 1. Then we have a place to start, the point s0,1d, and the direction in which
we move our pencil is given at each stage by the derivative F9sxd − f sxd. In the next
example we use the principles of this chapter to show how to graph F even when we
don’t have a formula for f. This would be the case, for instance, when f sxd is determined
by experimental data.
Example 5 The graph of a function f is given in Figure 2. Make a rough sketch of
an antiderivative F, given that Fs0d − 2.
y
y=ƒ
0
1
2
3
4
x
FIGURE 2
y
y=F(x)
2
Rectilinear Motion
1
0
SOLUTION We are guided by the fact that the slope of y − Fsxd is f sxd. We start at the
point s0, 2d and draw F as an initially decreasing function since f sxd is negative when
0 , x , 1. Notice that f s1d − f s3d − 0, so F has horizontal tangents when x − 1 and
x − 3. For 1 , x , 3, f sxd is positive and so F is increasing. We see that F has a local
minimum when x − 1 and a local maximum when x − 3. For x . 3, f sxd is negative
and so F is decreasing on s3, `d. Since f sxd l 0 as x l `, the graph of F becomes
flatter as x l `. Also notice that F0sxd − f 9sxd changes from positive to negative at
x − 2 and from negative to positive at x − 4, so F has inflection points when x − 2 and
x − 4. We use this information to sketch the graph of the antiderivative in Figure 3. n
1
FIGURE 3
x
Antidifferentiation is particularly useful in analyzing the motion of an object moving in
a straight line. Recall that if the object has position function s − f std, then the velocity
function is vstd − s9std. This means that the position function is an antiderivative of the
velocity function. Likewise, the acceleration function is astd − v9std, so the velocity function is an antiderivative of the acceleration. If the acceleration and the initial values ss0d
and vs0d are known, then the position function can be found by antidifferentiating twice.
Example 6 A particle moves in a straight line and has acceleration given by
astd − 6t 1 4. Its initial velocity is vs0d − 26 cmys and its initial displacement is
ss0d − 9 cm. Find its position function sstd.
SOLUTION Since v9std − astd − 6t 1 4, antidifferentiation gives
vstd − 6
t2
1 4t 1 C − 3t 2 1 4t 1 C
2
Note that vs0d − C. But we are given that vs0d − 26, so C − 26 and
vstd − 3t 2 1 4t 2 6
Since vstd − s9std, s is the antiderivative of v:
sstd − 3
t3
t2
14
2 6t 1 D − t 3 1 2t 2 2 6t 1 D
3
2
This gives ss0d − D. We are given that ss0d − 9, so D − 9 and the required position
function is
sstd − t 3 1 2t 2 2 6t 1 9
n
An object near the surface of the earth is subject to a gravitational force that produces
a downward acceleration denoted by t. For motion close to the ground we may assume
that t is constant, its value being about 9.8 mys2 (or 32 ftys2).
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
282
Chapter 3 Applications of Differentiation
Example 7 A ball is thrown upward with a speed of 48 ftys from the edge of a cliff
432 ft above the ground. Find its height above the ground t seconds later. When does it
reach its maximum height? When does it hit the ground?
SOLUTION The motion is vertical and we choose the positive direction to be upward.
At time t the distance above the ground is sstd and the velocity vstd is decreasing. Therefore the acceleration must be negative and we have
astd −
dv
− 232
dt
Taking antiderivatives, we have
vstd − 232t 1 C
To determine C we use the given information that vs0d − 48. This gives 48 − 0 1 C, so
vstd − 232t 1 48
The maximum height is reached when vstd − 0, that is, after 1.5 seconds. Since
s9std − vstd, we antidifferentiate again and obtain
Figure 4 shows the position function of
the ball in Example 7. The graph corroborates the con­clusions we reached:
The ball reaches its maximum height
after 1.5 seconds and hits the ground
after about 6.9 seconds.
500
sstd − 216t 2 1 48t 1 D
Using the fact that ss0d − 432, we have 432 − 0 1 D and so
sstd − 216t 2 1 48t 1 432
The expression for sstd is valid until the ball hits the ground. This happens when
sstd − 0, that is, when
216t 2 1 48t 1 432 − 0
or, equivalently,
t 2 2 3t 2 27 − 0
Using the quadratic formula to solve this equation, we get
t−
8
0
We reject the solution with the minus sign since it gives a negative value for t. Therefore
the ball hits the ground after 3(1 1 s13 )y2 < 6.9 seconds.
n
FIGURE 4
1–20 Find the most general antiderivative of the function.
(Check your answer by differentiation.)
1. f sxd − 4x 1 7
2.
f sxd − x 2 2 3x 1 2
3. f sxd − 2x 3 2 23 x 2 1 5x
4.
f sxd − 6x 5 2 8x 4 2 9x 2
5. f sxd − xs12 x 1 8d
6.
f sxd − sx 2 5d 2
8.
f sxd − x 3.4 2 2x s221
7. f sxd − 7x 2y5 1 8x 24y5
9. f sxd − s2
3
11. f sxd − 3sx 2 2 s
x
3 6 3s13
2
10.
f sxd − 2
12.
3
f sxd − s
x 2 1 x sx
13. f sxd −
15. tstd −
10
x9
14. tsxd −
1 1 t 1 t2
st
17. hsd − 2 sin 2 sec 2 5 2 4x 3 1 2x 6
x6
16. f std − 3 cos t 2 4 sin t
18. tsv d − 5 1 3 sec 2 v
19. f std − 8st 2 sec t tan t
20. f sxd − 1 1 2 sin x 1 3ysx
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Section 3.9 Antiderivatives
; 21–22 Find the antiderivative F of f that satisfies the given
condition. Check your answer by comparing the graphs of f
and F.
283
47.The graph of a function is shown in the figure. Make a
rough sketch of an antiderivative F, given that Fs0d − 1.
y
21. f sxd − 5x 4 2 2x 5, Fs0d − 4
22. f sxd − x 1 2 sin x, Fs0d − 26
y=ƒ
0
x
1
23–42 Find f.
48.The graph of the velocity function of a particle is shown
in the figure. Sketch the graph of a position function.
23. f 0sxd − 20x 3 2 12x 2 1 6x
24. f 0sxd − x 6 2 4x 4 1 x 1 1
3
25. f 0sxd − 4 2 s
x
26. f 0sxd − x 2y3 1 x 22y3
27. f -std − 12 1 sin t
28. f -std − st 2 2 cos t
√
0
29. f 9sxd − 1 1 3sx , f s4d − 25
t
30. f 9sxd − 5x 4 2 3x 2 1 4, f s21d − 2
31. f 9sxd − sx s6 1 5xd, f s1d − 10
49.The graph of f 9 is shown in the figure. Sketch the graph
of f if f is continuous on f0, 3g and f s0d − 21.
32. f 9std − t 1 1yt , t . 0, f s1d − 6
3
y
33.f 9std − sec t ssec t 1 tan td, 2y2 , t , y2, f sy4d − 21
2
34. f 9sxd − sx 1 1dysx , f s1d − 5
35. f 0sxd − 22 1 12x 2 12x 2, f s0d − 4,
3
36. f 0sxd − 8x 1 5,
f s1d − 0,
f 9s0d − 12
0
_1
f 9s1d − 8
37. f 0sd − sin 1 cos , f s0d − 3, f 9s0d − 4
38. f 0std − 4 2 6yt 4, f s1d − 6, f 9s2d − 9,
t.0
39. f 0sxd − 4 1 6x 1 24x , f s0d − 3, f s1d − 10
40. f 0sxd − 20x 3 1 12x 2 1 4, f s0d − 8, f s1d − 5
3
41. f 0std − s
t 2 cos t, f s0d − 2, f s1d − 2
42. f -sxd − cos x, f s0d − 1, f 9s0d − 2, f 0s0d − 3
51. f sxd −
44.Find a function f such that f 9sxd − x 3 and the line
x 1 y − 0 is tangent to the graph of f .
f
x
c
53. vstd − sin t 2 cos t, ss0d − 0
a
x
b
sin x
, 22 < x < 2
1 1 x2
53–58 A particle is moving with the given data. Find the position of the particle.
46. y
b
x
52. f sxd − sx 4 2 2 x 2 1 2 2 2, 23 < x < 3
45–46 The graph of a function f is shown. Which graph is
an antiderivative of f and why?
f
2
; 51–52 Draw a graph of f and use it to make a rough sketch of
the antiderivative that passes through the origin.
43.Given that the graph of f passes through the point (2, 5)
and that the slope of its tangent line at sx, f sxdd is 3 2 4x,
find f s1d.
a
1
; 50. (a) Use a graphing device to graph f sxd − 2x 2 3 sx .
(b)Starting with the graph in part (a), sketch a rough graph
of the antiderivative F that satisfies Fs0d − 1.
(c)Use the rules of this section to find an expression
for Fsxd.
(d)Graph F using the expression in part (c). Compare with
your sketch in part (b).
2
45. y
y=fª(x)
1
54. vstd − t 2 2 3 st , ss4d − 8
55. astd − 2t 1 1, ss0d − 3, v s0d − 22
c
56. astd − 3 cos t 2 2 sin t, ss0d − 0, v s0d − 4
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284
chapter 3 Applications of Differentiation
57. astd − 10 sin t 1 3 cos t, ss0d − 0, ss2d − 12
58. astd − t 2 2 4t 1 6, ss0d − 0, ss1d − 20
59.A stone is dropped from the upper observation deck (the Space
Deck) of the CN Tower, 450 m above the ground.
(a)Find the distance of the stone above ground level at time t.
(b) How long does it take the stone to reach the ground?
(c) With what velocity does it strike the ground?
(d)If the stone is thrown downward with a speed of 5 mys,
how long does it take to reach the ground?
60.Show that for motion in a straight line with constant acceleration a, initial velocity v 0, and initial displacement s 0, the dis­
placement after time t is
66.The linear density of a rod of length 1 m is given by
sxd − 1ysx , in grams per centimeter, where x is measured in
centimeters from one end of the rod. Find the mass of the rod.
67.Since raindrops grow as they fall, their surface area increases
and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 mys and its downward acceleration is
a−
H
9 2 0.9t
0
if 0 < t < 10
if t . 10
If the raindrop is initially 500 m above the ground, how long
does it take to fall?
s − 12 at 2 1 v 0 t 1 s 0
68.A car is traveling at 50 miyh when the brakes are fully applied,
producing a constant deceleration of 22 ftys2. What is the distance traveled before the car comes to a stop?
61.An object is projected upward with initial velocity v 0 meters per
second from a point s0 meters above the ground. Show that
69.What constant acceleration is required to increase the speed of
a car from 30 miyh to 50 miyh in 5 seconds?
fvstdg 2 − v02 2 19.6fsstd 2 s0 g
62.Two balls are thrown upward from the edge of the cliff in
Example 7. The first is thrown with a speed of 48 ftys and the
other is thrown a second later with a speed of 24 ftys. Do the
balls ever pass each other?
63.A stone was dropped off a cliff and hit the ground with a speed
of 120 ftys. What is the height of the cliff?
64.If a diver of mass m stands at the end of a diving board with
length L and linear density , then the board takes on the shape
of a curve y − f sxd, where
EI y 0 − mtsL 2 xd 1 12 tsL 2 xd2
E and I are positive constants that depend on the material of the
board and t s, 0d is the acceleration due to gravity.
(a)Find an expression for the shape of the curve.
(b)Use f sLd to estimate the distance below the horizontal at
the end of the board.
y
0
x
65.A company estimates that the marginal cost (in dollars per
item) of producing x items is 1.92 2 0.002x. If the cost of
producing one item is $562, find the cost of producing 100
items.
70.A car braked with a constant deceleration of 16 ftys2, producing skid marks measuring 200 ft before coming to a stop.
How fast was the car traveling when the brakes were first
applied?
71.A car is traveling at 100 kmyh when the driver sees an accident 80 m ahead and slams on the brakes. What constant
deceleration is required to stop the car in time to avoid a
pileup?
72.A model rocket is fired vertically upward from rest. Its
acceler­ation for the first three seconds is astd − 60t, at which
time the fuel is exhausted and it becomes a freely “falling”
body. Fourteen seconds later, the rocket’s parachute opens,
and the (downward) velocity slows linearly to 218 ftys in
5 seconds. The rocket then “floats” to the ground at that rate.
(a)Determine the position function s and the velocity function v (for all times t). Sketch the graphs of s and v.
(b)At what time does the rocket reach its maximum height,
and what is that height?
(c) At what time does the rocket land?
73.A high-speed bullet train accelerates and decelerates at the rate
of 4 ftys2. Its maximum cruising speed is 90 miyh.
(a)What is the maximum distance the train can travel if it
accelerates from rest until it reaches its cruising speed and
then runs at that speed for 15 minutes?
(b)Suppose that the train starts from rest and must come to
a complete stop in 15 minutes. What is the maximum
distance it can travel under these conditions?
(c)Find the minimum time that the train takes to travel
between two consecutive stations that are 45 miles apart.
(d)The trip from one station to the next takes 37.5 minutes.
How far apart are the stations?
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
CHAPTER 3 Review
3
285
Review
CONCEPT CHECK
1.Explain the difference between an absolute maximum and
a local maximum. Illustrate with a sketch.
2.What does the Extreme Value Theorem say?
3.(a) State Fermat’s Theorem.
(b) Define a critical number of f.
4.Explain how the Closed Interval Method works.
5.(a) State Rolle’s Theorem.
(b)State the Mean Value Theorem and give a geometric
interpretation.
6.(a) State the Increasing/Decreasing Test.
(b)What does it mean to say that f is concave upward on an
interval I?
(c) State the Concavity Test.
(d) What are inflection points? How do you find them?
7.(a) State the First Derivative Test.
(b) State the Second Derivative Test.
(c)What are the relative advantages and disadvantages of
these tests?
Answers to the Concept Check can be found on the back endpapers.
8.Explain the meaning of each of the following statements.
(a) lim f sxd − L
(b) lim f sxd − L
(c) lim f sxd − `
xl`
x l 2`
xl`
(d)The curve y − f sxd has the horizontal asymptote y − L.
9.If you have a graphing calculator or computer, why do you need
calculus to graph a function?
10. (a)Given an initial approximation x1 to a root of the equation
f sxd − 0, explain geometrically, with a diagram, how the
second approximation x 2 in Newton’s method is obtained.
(b)Write an expression for x 2 in terms of x1, f sx 1 d,
and f 9sx 1d.
(c)Write an expression for x n11 in terms of x n , f sx n d, and
f 9sx n d.
(d)Under what circumstances is Newton’s method likely to fail
or to work very slowly?
11. (a) What is an antiderivative of a function f ?
(b)Suppose F1 and F2 are both antiderivatives of f on an
interval I. How are F1 and F2 related?
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true,
explain why. If it is false, explain why or give an example that
disproves the statement.
1.If f 9scd − 0, then f has a local maximum or minimum at c.
2.If f has an absolute minimum value at c, then f 9scd − 0.
3.If f is continuous on sa, bd, then f attains an absolute maximum value f scd and an absolute minimum value f sd d at some
numbers c and d in sa, bd.
10. T
here exists a function f such that f sxd , 0, f 9sxd , 0,
and f 0 sxd . 0 for all x.
11. If f and t are increasing on an interval I, then f 1 t is
increasing on I.
12. I f f and t are increasing on an interval I, then f 2 t is
increasing on I.
13. I f f and t are increasing on an interval I, then f t is
increasing on I.
4.If f is differentiable and f s21d − f s1d, then there is a number c such that c , 1 and f 9scd − 0.
14. If f and t are positive increasing functions on an interval I,
then f t is increasing on I.
5.If f 9sxd , 0 for 1 , x , 6, then f is decreasing on (1, 6).
15. If f is increasing and f sxd . 0 on I, then tsxd − 1yf sxd is
decreasing on I.
| |
6.If f 0s2d − 0, then s2, f s2dd is an inflection point of the
curve y − f sxd.
7.If f 9sxd − t9sxd for 0 , x , 1, then f sxd − tsxd for
0 , x , 1.
8.There exists a function f such that f s1d − 22, f s3d − 0,
and f 9sxd . 1 for all x.
9.There exists a function f such that f sxd . 0, f 9sxd , 0,
and f 0 sxd . 0 for all x.
16. If f is even, then f 9 is even.
17. If f is periodic, then f 9 is periodic.
18. The most general antiderivative of f sxd − x 22 is
Fsxd − 2
1
1C
x
19. If f 9sxd exists and is nonzero for all x, then f s1d ± f s0d.
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
286
Chapter 3 Applications of Differentiation
EXERCISES
1–6 Find the local and absolute extreme values of the function on
the given interval.
1.f sxd − x 3 2 9x 2 1 24 x 2 2, f0, 5g
(b)For what values of x does f have a local maximum or
minimum?
(c) Sketch the graph of f 0.
(d) Sketch a possible graph of f.
2.f sxd − x s1 2 x , f21, 1g
3.f sxd −
y
3x 2 4
, f22, 2g
x2 1 1
y=f ª(x)
_2
_1
4.f sxd − sx 1 x 1 1 , f22, 1g
2
0
1
2
3
4
5
6
7
x
5.f sxd − x 1 2 cos x, f2, g
6.f sxd − sin x 1 cos 2 x, f0, g
17–28 Use the guidelines of Section 3.5 to sketch the curve.
17. y − 2 2 2x 2 x 3
7–12 Find the limit.
18. y − 22 x 3 2 3x 2 1 12 x 1 5
3x 4 1 x 2 5
7. lim
x l ` 6x 4 2 2x 2 1 1
8. lim
tl`
9. lim
19. y − 3x 4 2 4x 3 1 2
t3 2 t 1 2
s2t 2 1dst 2 1 t 1 1d
x l 2`
s4x 2 1 1
3x 2 1
10.
11. lim ss4x 2 1 3x 2 2xd
xl`
sin 4 x
12. lim
x l ` sx
13–15 Sketch the graph of a function that satisfies the given
conditions.
x
1 2 x2
1
1
2
x2
sx 2 2d 2
21. y −
1
xsx 2 3d2
22. y −
23. y −
sx 2 1d 3
x2
24. y − s1 2 x 1 s1 1 x
lim sx 2 1 x 3 d
x l 2`
20. y −
25. y − x s2 1 x
26. y − x 2y3sx 2 3d 2
27. y − sin 2 x 2 2 cos x
28. y − 4x 2 tan x, 2y2 , x , y2
13. f s0d − 0, f 9s22d − f 9s1d − f 9s9d − 0,
lim f sxd − 0,
xl`
lim f sxd − 2`,
x l6
f 9sxd , 0 on s2`, 22d, s1, 6d, and s9, `d,
f 9sxd . 0 on s22, 1d and s6, 9d,
f 0sxd . 0 on s2`, 0d and s12, `d,
f 0sxd , 0 on s0, 6d and s6, 12d
14.f s0d − 0, f is continuous and even, f 9sxd − 2x if 0 , x , 1, f 9sxd − 21 if 1 , x , 3,
f 9sxd − 1 if x . 3
; 29–32 Produce graphs of f that reveal all the important aspects
of the curve. Use graphs of f 9 and f 0 to estimate the intervals of
increase and decrease, extreme values, intervals of concavity,
and inflection points. In Exercise 29 use calculus to find these
quantities exactly.
29. f sxd −
x2 2 1
x3
30. f sxd −
x3 1 1
x6 1 1
31. f sxd − 3x 6 2 5x 5 1 x 4 2 5x 3 2 2x 2 1 2
15.f is odd, f 9sxd , 0 for 0 , x , 2, f 9sxd . 0 for x . 2, f 0sxd . 0 for 0 , x , 3,
f 0sxd , 0 for x . 3, lim f sxd − 22
xl`
16.The figure shows the graph of the derivative f 9of a function f.
(a) On what intervals is f increasing or decreasing?
32. f sxd − x 2 1 6.5 sin x, 25 < x < 5
33.Show that the equation 3x 1 2 cos x 1 5 − 0 has exactly one
real root.
34.Suppose that f is continuous on f0, 4g, f s0d − 1, and
2 < f 9sxd < 5 for all x in s0, 4d. Show that 9 < f s4d < 21.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
chapter 3 Review
35.By applying the Mean Value Theorem to the function
f sxd − x 1y5 on the interval f32, 33g, show that
; 48.A manufacturer determines that the cost of making x units
of a commodity is
Csxd − 1800 1 25x 2 0.2x 2 1 0.001x 3
5
2,s
33 , 2.0125
36.For what values of the constants a and b is s1, 3d a point of
inflection of the curve y − ax 3 1 bx 2 ?
37.Let tsxd − f sx 2 d, where f is twice differentiable for all x,
f 9sxd . 0 for all x ± 0, and f is concave downward on
s2`, 0d and concave upward on s0, `d.
(a) At what numbers does t have an extreme value?
(b) Discuss the concavity of t.
38.Find two positive integers such that the sum of the first
number and four times the second number is 1000 and
the product of the numbers is as large as possible.
39.Show that the shortest distance from the point sx 1, y1 d to the
straight line Ax 1 By 1 C − 0 is
| Ax
1
|
1 By1 1 C
sA2 1 B 2
40.Find the point on the hyperbola x y − 8 that is closest to the
point s3, 0d.
41.Find the smallest possible area of an isosceles triangle that
is circumscribed about a circle of radius r.
42.Find the volume of the largest circular cone that can be
inscribed in a sphere of radius r.
|
| |
|
43.In D ABC, D lies on AB, CD AB, AD − BD − 4 cm,
and CD − 5 cm. Where should a point P be chosen on
CD so that the sum PA 1 PB 1 PC is a minimum?
|
|
| | | | | |
44. Solve Exercise 43 when | CD | − 2 cm.
and the demand function is psxd − 48.2 2 0.03x.
(a)Graph the cost and revenue functions and use the
graphs to estimate the production level for maximum
profit.
(b)Use calculus to find the production level for maximum
profit.
(c)Estimate the production level that minimizes the average cost.
49.Use Newton’s method to find the root of the equation
x 5 2 x 4 1 3x 2 2 3x 2 2 − 0 in the interval f1, 2g correct to six decimal places.
50.Use Newton’s method to find all solutions of the equation
sin x − x 2 2 3x 1 1 correct to six decimal places.
51.Use Newton’s method to find the absolute maximum
value of the function f std − cos t 1 t 2 t 2 correct to eight
decimal places.
52.Use the guidelines in Section 3.5 to sketch the curve
y − x sin x, 0 < x < 2. Use Newton’s method when
necessary.
53–54 Find the most general antiderivative of the function.
53. f sxd − 4 sx 2 6x 2 1 3
54. tsxd − cos x 1 2 sec 2 x
55–58 Find f.
55. f 9std − 2t 2 3 sin t, f s0d − 5
45. The velocity of a wave of length L in deep water is
v−K
Î
287
L
C
1
C
L
where K and C are known positive constants. What is the
length of the wave that gives the minimum velocity?
56. f 9sud −
u 2 1 su
, f s1d − 3
u
57. f 0sxd − 1 2 6x 1 48x 2, f s0d − 1, f 9s0d − 2
58. f 0sxd − 5x 3 1 6x 2 1 2, f s0d − 3, f s1d − 22
46.A metal storage tank with volume V is to be constructed
in the shape of a right circular cylinder surmounted by a
hemisphere. What dimensions will require the least amount
of metal?
59–60 A particle is moving according to the given data. Find
the position of the particle.
47.A hockey team plays in an arena with a seating capacity of
15,000 spectators. With the ticket price set at $12, average
attendance at a game has been 11,000. A market survey
indicates that for each dollar the ticket price is lowered,
average attendance will increase by 1000. How should the
owners of the team set the ticket price to maximize their
revenue from ticket sales?
60. astd − sin t 1 3 cos t, ss0d − 0, v s0d − 2
59. vstd − 2t 2 sin t, ss0d − 3
; 61.Use a graphing device to draw a graph of the function
f sxd − x 2 sinsx 2 d, 0 < x < , and use that graph to sketch
the antiderivative F of f that satisfies the initial condition
Fs0d − 0.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
288
chapter 3 Applications of Differentiation
tal, as shown in the figure. Show that the range of the
projectile, measured up the slope, is given by
; 62. Investigate the family of curves given by
f sxd − x 4 1 x 3 1 cx 2
Rsd −
In particular you should determine the transitional value of
c at which the number of critical numbers changes and the
transitional value at which the number of inflection points
changes. Illustrate the various possible shapes with graphs.
(b) Determine so that R is a maximum.
(c)Suppose the plane is at an angle below the horizontal. Determine the range R in this case, and determine
the angle at which the projectile should be fired to
maximize R.
63.A canister is dropped from a helicopter 500 m above the
ground. Its parachute does not open, but the canister has
been designed to withstand an impact velocity of 100 mys.
Will it burst?
y
64.In an automobile race along a straight road, car A passed
car B twice. Prove that at some time during the race their
accelera­tions were equal. State the assumptions that you
make.
¨
65.A rectangular beam will be cut from a cylindrical log of
radius 10 inches.
(a)Show that the beam of maximal cross-sectional area is
a square.
(b)Four rectangular planks will be cut from the four
sections of the log that remain after cutting the square
beam. Determine the dimensions of the planks that will
have maximal cross-sectional area.
depth
x
7et04rx80
(c)Suppose that the strength of a rectangular beam is
proportional to the product of its width and the square
of its depth. Find the dimensions of the strongest beam
that can be cut from the cylindrical log.
¨
h
d
40
66.If a projectile is fired with an initial velocity v at an angle
of inclination from the horizontal, then its trajectory,
neglecting air resistance, is the parabola
(a)Suppose the projectile is fired from the base of a plane
that is inclined at an angle , . 0, from the horizon-
R
67.A light is to be placed atop a pole of height h feet to
illuminate a09/09/09
busy traffic circle, which has a radius of 40 ft.
MasterID:
00594
The intensity
of illumination
I at any point P on the circle
is directly proportional to the cosine of the angle (see
the figure) and inversely proportional to the square of the
distance d from the source.
(a) How tall should the light pole be to maximize I?
(b)Suppose that the light pole is h feet tall and that a
woman is walking away from the base of the pole at
the rate of 4 ftys. At what rate is the intensity of the
light at the point on her back 4 ft above the ground
decreasing when she reaches the outer edge of the
traffic circle?
width
t
x 2 0 , ,
2v 2 cos 2
2
å
0
10
y − stan dx 2
2v 2 cos sins 2 d
t cos2
CAS
68.If f sxd −
P
cos 2 x
, 2 < x < , use the graphs of f,
sx 1 x 1 1
f 9, and f 0 to estimate the x-coordinates of the maximum and
minimum points and inflection points of f.
2
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Problems Plus
One of the most important principles of problem solving is analogy (see page 98). If you
are having trouble getting started on a problem, it is sometimes helpful to start by solving
a similar, but simpler, problem. The following example illustrates the principle. Cover up
the solution and try solving it yourself first.
Example If x, y, and z are positive numbers, prove that
sx 2 1 1dsy 2 1 1dsz 2 1 1d
>8
xyz
SOLUTION It may be difficult to get started on this problem. (Some students have
tackled it by multiplying out the numerator, but that just creates a mess.) Let’s try to
think of a similar, simpler problem. When several variables are involved, it’s often helpful to think of an analogous problem with fewer variables. In the present case we can
reduce the number of variables from three to one and prove the analogous inequality
1 x2 1 1
>2
x
for x . 0
In fact, if we are able to prove (1), then the desired inequality follows because
sx 2 1 1ds y 2 1 1dsz 2 1 1d
−
xyz
S DS DS D
x2 1 1
x
y2 1 1
y
z2 1 1
z
>2?2?2−8
The key to proving (1) is to recognize that it is a disguised version of a minimum problem. If we let
f sxd −
x2 1 1
1
− x 1 x . 0
x
x
then f 9sxd − 1 2 s1yx 2 d, so f 9sxd − 0 when x − 1. Also, f 9sxd , 0 for 0 , x , 1 and
f 9sxd . 0 for x . 1. Therefore the absolute minimum value of f is f s1d − 2. This
means that
x2 1 1
>2
x
PS Look Back
What have we learned from the solution to this example?
● To solve a problem involving several
variables, it might help to solve a
similar problem with just one
variable.
● When trying to prove an inequality,
it might help to think of it as a maximum or minimum problem.
for all positive values of x
and, as previously mentioned, the given inequality follows by multiplication.
The inequality in (1) could also be proved without calculus. In fact, if x . 0, we
have
x2 1 1
> 2 &?
x
x 2 1 1 > 2x &? x 2 2 2x 1 1 > 0
&? sx 2 1d2 > 0
Because the last inequality is obviously true, the first one is true too.
■
289
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|
|
1.
Show that sin x 2 cos x < s2 for all x.
Problems
| |
2.
S
how that x y s4 2 x ds4 2 y 2 d < 16 for all numbers x and y such that x < 2 and
y < 2.
2
2
2
| |
3.
S
how that the inflection points of the curve y − ssin xdyx lie on the curve y 2 sx 4 1 4d − 4.
4.
Find the point on the parabola y − 1 2 x 2 at which the tangent line cuts from the first
quadrant the triangle with the smallest area.
5.
Find the highest and lowest points on the curve x 2 1 x y 1 y 2 − 12.
6.
Water is flowing at a constant rate into a spherical tank. Let Vstd be the volume of water in
the tank and Hstd be the height of the water in the tank at time t.
(a)What are the meanings of V9std and H9std? Are these derivatives positive, negative, or
zero?
(b)Is V 0std positive, negative, or zero? Explain.
(c)Let t1, t 2, and t 3 be the times when the tank is one-quarter full, half full, and threequarters full, respectively. Are the values H 0st1d, H 0st 2 d, and H 0st 3 d positive, negative,
or zero? Why?
7.
Find the absolute maximum value of the function
f sxd −
y
1
1
1
11 x
11 x22
| |
|
|
8.
Find a function f such that f 9s21d − 12 , f 9s0d − 0, and f 0sxd . 0 for all x, or prove that
such a function cannot exist.
Q
9.
I f Psa, a 2 d is any point on the parabola y − x 2, except for the origin, let Q be the point
where the normal line at P intersects the parabola again (see the figure).
(a) Show that the y-coordinate of Q is smallest when a − 1ys2 .
(b) Show that the line segment PQ has the shortest possible length when a − 1ys2 .
P
10.An isosceles triangle is circumscribed about the unit circle so that the equal sides meet
at the point s0, ad on the y-axis (see the figure). Find the value of a that minimizes the
lengths of the equal sides. (You may be surprised that the result does not give an equilateral triangle.).
x
0
FIGURE for PROBLEM 9
y
y=≈
B
11. T
he line y − mx 1 b intersects the parabola y − x 2 in points A and B. (See the figure.) Find the point P on the arc AOB of the parabola that maximizes the area of the
triangle PAB.
A
y=mx+b
O
P
FIGURE for PROBLEM 11
x
| |
12.
Sketch the graph of a function f such that f 9sxd , 0 for all x, f 0sxd . 0 for x . 1,
f 0sxd , 0 for x , 1, and lim x l6` f f sxd 1 xg − 0.
| |
13.
Determine the values of the number a for which the function f has no critical number:
f sxd − sa 2 1 a 2 6d cos 2x 1 sa 2 2dx 1 cos 1
290
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
14.Sketch the region in the plane consisting of all points sx, yd such that
|
|
2xy < x 2 y < x 2 1 y 2
| | |
|
15.
Let ABC be a triangle with /BAC − 1208 and AB AC − 1.
(a) Express the length of the angle bisector AD in terms of x − AB .
(b) Find the largest possible value of AD .
|
| |
|
|
C
|
16.
(a)Let ABC be a triangle with right angle A and hypotenuse a − BC . (See the figure.)
If the inscribed circle touches the hypotenuse at D, show that
D
| CD | − 12 s| BC | 1 | AC | 2 | AB |d
(b)If − 12 /C, express the radius r of the inscribed circle in terms of a and .
(c) If a is fixed and varies, find the maximum value of r.
A
B
FIGURE for PROBLEM 16
17.
A triangle with sides a, b, and c varies with time t, but its area never changes. Let be the
angle opposite the side of length a and suppose always remains acute.
(a) Express dydt in terms of b, c, , dbydt, and dcydt.
(b) Express daydt in terms of the quantities in part (a).
18.ABCD is a square piece of paper with sides of length 1 m. A quarter-circle is drawn from
B to D with center A. The piece of paper is folded along EF, with E on AB and F on AD,
so that A falls on the quarter-circle. Determine the maximum and minimum areas that the
triangle AEF can have.
19.The speeds of sound c1 in an upper layer and c2 in a lower layer of rock and the thick­ness
h of the upper layer can be determined by seismic exploration if the speed of sound in the
lower layer is greater than the speed in the upper layer. A dynamite charge is detonated at
a point P and the transmitted signals are recorded at a point Q, which is a distance D
from P. The first signal to arrive at Q travels along the surface and takes T1 seconds. The
next signal travels from P to a point R, from R to S in the lower layer, and then to Q,
taking T2 seconds. The third signal is reflected off the lower layer at the midpoint O of RS
and takes T3 seconds to reach Q. (See the figure.)
P
Q
D
speed of sound=c¡
h
¨
¨
R
O
S
speed of sound=c™
(a) Express T1, T2, and T3 in terms of D, h, c1, c2, and .
(b)Show that T2 is a minimum when sin − c1yc2.
(c)Suppose that D − 1 km, T1 − 0.26 s, T2 − 0.32 s, and T3 − 0.34 s. Find c1, c2, and h.
N
ote: Geophysicists use this technique when studying the structure of the earth’s crust,
whether searching for oil or examining fault lines.
20.For what values of c is there a straight line that intersects the curve
y − x 4 1 cx 3 1 12x 2 2 5x 1 2
in four distinct points?
291
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d
B
E
x
C
r
F
D
FIGURE for PROBLEM 21
21. O
ne of the problems posed by the Marquis de l’Hospital in his calculus textbook Analyse
des Infiniment Petits concerns a pulley that is attached to the ceiling of a room at a point
C by a rope of length r. At another point B on the ceiling, at a distance d from C (where
d . r), a rope of length , is attached and passed through the pulley at F and connected to
a weight W. The weight is released and comes to rest at its equilibrium position D. (See
the figure.) As l’Hospital argued, this happens when the distance ED is maximized.
Show that when the system reaches equilibrium, the value of x is
|
|
r
(r 1 sr 2 1 8d 2 )
4d
Notice that this expression is independent of both W and ,.
22.Given a sphere with radius r, find the height of a pyramid of minimum volume whose base
is a square and whose base and triangular faces are all tangent to the sphere. What if the
base of the pyramid is a regular n-gon? (A regular n-gon is a polygon with n equal sides and
angles.) (Use the fact that the volume of a pyramid is 13 Ah, where A is the area of the base.)
23.Assume that a snowball melts so that its volume decreases at a rate proportional to its
surface area. If it takes three hours for the snowball to decrease to half its original volume,
how much longer will it take for the snowball to melt completely?
24.A hemispherical bubble is placed on a spherical bubble of radius 1. A smaller hemispherical bubble is then placed on the first one. This process is continued until n chambers,
including the sphere, are formed. (The figure shows the case n − 4.) Use mathematical
induction to prove that the maximum height of any bubble tower with n chambers
is 1 1 sn .
292
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
4
Integrals
The photo shows Lake Lanier,
which is a reservoir
in Georgia, USA. In Exercise 63 in Section 4.4 you will
estimate the amount of water
that flowed into Lake Lanier
during a certain time period.
JRC, Inc. / Alamy
In chapter 2 we used the tangent and velocity problems to introduce the derivative, which is
the central idea in differential calculus. In much the same way, this chapter starts with the area
and distance problems and uses them to formulate the idea of a definite integral, which is the
basic concept of integral calculus. We will see in Chapters 5 and 8 how to use the integral to solve
problems concerning volumes, lengths of curves, population predictions, cardiac output, forces
on a dam, work, consumer surplus, and baseball, among many others.
There is a connection between integral calculus and differential calculus. The Fundamental
Theorem of Calculus relates the integral to the derivative, and we will see in this chapter that it
greatly simplifies the solution of many problems.
293
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294
Chapter 4 Integrals
Now is a good time to read (or reread)
A Preview of Calculus (see page 1). It
discusses the unifying ideas of calculus
and helps put in perspec­tive where we
have been and where we are going.
y
y=ƒ
x=a
S
x=b
a
0
FIGURE 1
x
b
|
In this section we discover that in trying to find the area under a curve or the distance
traveled by a car, we end up with the same special type of limit.
The Area Problem
We begin by attempting to solve the area problem: find the area of the region S that
lies under the curve y − f sxd from a to b. This means that S, illustrated in Figure 1, is
bounded by the graph of a continuous function f [where f sxd > 0], the vertical lines
x − a and x − b, and the x-axis.
In trying to solve the area problem we have to ask ourselves: what is the meaning
of the word area? This question is easy to answer for regions with straight sides. For a
rectangle, the area is defined as the product of the length and the width. The area of a
triangle is half the base times the height. The area of a polygon is found by dividing it
into triangles (as in Figure 2) and adding the areas of the triangles.
S − hsx, yd a < x < b, 0 < y < f sxdj
A™
w
h
A¢
A¡
b
l
FIGURE 2
A£
A= 21 bh
A=lw
A=A¡+A™+A£+A¢
However, it isn’t so easy to find the area of a region with curved sides. We all have an
intuitive idea of what the area of a region is. But part of the area problem is to make this
intuitive idea precise by giving an exact definition of area.
Recall that in defining a tangent we first approximated the slope of the tangent line by
slopes of secant lines and then we took the limit of these approximations. We pursue a
sim­ilar idea for areas. We first approximate the region S by rectangles and then we take
the limit of the areas of these rectangles as we increase the number of rectangles. The
follow­ing example illustrates the procedure.
y
Example 1 Use rectangles to estimate the area under the parabola y − x 2 from 0 to 1
(1, 1)
(the parabolic region S illustrated in Figure 3).
SOLUTION We first notice that the area of S must be somewhere between 0 and 1
because S is contained in a square with side length 1, but we can certainly do better
than that. Suppose we divide S into four strips S1, S2, S3, and S4 by drawing the vertical
lines x − 14, x − 12, and x − 34 as in Figure 4(a).
y=≈
S
0
1
x
y
y
(1, 1)
(1, 1)
y=≈
FIGURE 3
S¡
0
FIGURE 4
S¢
S™
1
4
S£
1
2
(a)
3
4
1
x
0
1
4
1
2
3
4
1
x
(b)
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295
Section 4.1 Areas and Distances
We can approximate each strip by a rectangle that has the same base as the strip and
whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words,
the heights of these rectangles are the values of the function f sxd − x 2 at the right end­-
f g f 41 , 12 g, f 12 , 34 g, and f 34 , 1g.
points of the subintervals 0, 14 ,
Each rectangle has width 14 and the heights are ( 14 ) , ( 12 ) , ( 34 ) , and 12. If we let R 4 be
the sum of the areas of these approximating rectangles, we get
2
2
2
R4 − 14 ? ( 14 ) 1 14 ? ( 12 ) 1 14 ? ( 34 ) 1 14 ? 12 − 15
32 − 0.46875
2
2
2
From Figure 4(b) we see that the area A of S is less than R 4, so
A , 0.46875
y
Instead of using the rectangles in Figure 4(b) we could use the smaller rectangles in
Figure 5 whose heights are the values of f at the left endpoints of the subintervals. (The
leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these
approximating rectangles is
(1, 1)
y=≈
7
L 4 − 14 ? 0 2 1 14 ? ( 14 ) 1 14 ? ( 12 ) 1 14 ? ( 34 ) − 32
− 0.21875
2
2
2
We see that the area of S is larger than L 4, so we have lower and upper estimates for A:
0
1
4
1
2
3
4
1
x
0.21875 , A , 0.46875
We can repeat this procedure with a larger number of strips. Figure 6 shows what
happens when we divide the region S into eight strips of equal width.
FIGURE 5
y
y
(1, 1)
y=≈
0
FIGURE 6
Approximating S with eight rectangles
1
8
1
(a) Using left endpoints
(1, 1)
x
0
1
8
1
x
(b) Using right endpoints
By computing the sum of the areas of the smaller rectangles sL 8 d and the sum of the
areas of the larger rectangles sR 8 d, we obtain better lower and upper estimates for A:
0.2734375 , A , 0.3984375
n
Ln
Rn
10
20
30
50
100
1000
0.2850000
0.3087500
0.3168519
0.3234000
0.3283500
0.3328335
0.3850000
0.3587500
0.3501852
0.3434000
0.3383500
0.3338335
So one possible answer to the question is to say that the true area of S lies somewhere
between 0.2734375 and 0.3984375.
We could obtain better estimates by increasing the number of strips. The table at
the left shows the results of similar calculations (with a computer) using n rectangles
whose heights are found with left endpoints sL n d or right endpoints sR n d. In particular,
we see by using 50 strips that the area lies between 0.3234 and 0.3434. With 1000
strips we narrow it down even more: A lies between 0.3328335 and 0.3338335. A good
estimate is obtained by averaging these numbers: A < 0.3333335.
n
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296
Chapter 4 Integrals
From the values in the table in Example 1, it looks as if R n is approaching
increases. We confirm this in the next example.
1
3
as n
Example 2 For the region S in Example 1, show that the sum of the areas of the
upper approximating rectangles approaches 13, that is,
lim R n − 13
nl`
SOLUTION R n is the sum of the areas of the n rectangles in Figure 7. Each rectangle
has width 1yn and the heights are the values of the function f sxd − x 2 at the points
1yn, 2yn, 3yn, . . . , nyn; that is, the heights are s1ynd2, s2ynd2, s3ynd2, . . . , snynd2. Thus
y
(1, 1)
y=≈
Rn −
0
1
1
n
x
FIGURE 7
1
n
SD SD SD
1
n
2
1
1
n
2
2
n
1
1
n
3
n
2
1 ∙∙∙ 1
−
1 1 2
s1 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 d
n n2
−
1 2
s1 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 d
n3
1
n
SD
n
n
2
Here we need the formula for the sum of the squares of the first n positive integers:
1
12 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 −
nsn 1 1ds2n 1 1d
6
Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E.
Putting Formula 1 into our expression for R n, we get
Rn −
Here we are computing the limit of
the sequence hR n j. Sequences and their
limits were discussed in A Preview of
Calculus and will be studied in detail in
Section 11.1. The idea is very similar to
a limit at infinity (Section 3.4) except
that in writing lim n l ` we restrict n to
be a positive integer. In particular, we
know that
1
lim − 0
nl ` n
When we write lim n l ` Rn − 13 we
mean that we can make Rn as close to 13
as we like by taking n sufficiently large.
1 nsn 1 1ds2n 1 1d
sn 1 1ds2n 1 1d
−
3 n
6
6n 2
Thus we have
lim R n − lim
nl `
nl `
sn 1 1ds2n 1 1d
6n 2
− lim
1
6
− lim
1
6
nl`
nl`
−
S DS D
S DS D
n11
n
11
1
n
2n 1 1
n
21
1
n
1
1
12−
6
3
n
It can be shown that the lower approximating sums also approach 13, that is,
lim L n − 13
nl`
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297
Section 4.1 Areas and Distances
From Figures 8 and 9 it appears that, as n increases, both L n and R n become better and
bet­ter approximations to the area of S. Therefore we define the area A to be the limit of
the sums of the areas of the approximating rectangles, that is,
TEC In Visual 4.1 you can create
pictures like those in Figures 8 and 9
for other values of n.
A − lim R n − lim L n − 13
nl`
nl`
y
y
n=10 R¡¸=0.385
0
y
n=50 R∞¸=0.3434
n=30 R£¸Å0.3502
1
x
0
1
x
0
1
x
1
x
FIGURE 8 Right endpoints produce upper sums because f sxd − x 2 is increasing.
y
y
n=10 L¡¸=0.285
0
y
n=50 L∞¸=0.3234
n=30 L£¸Å0.3169
1
x
0
1
x
0
FIGURE 9 Left endpoints produce lower sums because f sxd − x 2 is increasing.
Let’s apply the idea of Examples 1 and 2 to the more general region S of Figure 1.
We start by subdividing S into n strips S1, S2 , . . . , Sn of equal width as in Figure 10.
y
y=ƒ
S¡
FIGURE 10
0
a
S™
⁄
S£
x2
Si
‹
. . . xi-1
Sn
xi
. . . xn-1
b
x
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
298
Chapter 4 Integrals
The width of the interval fa, bg is b 2 a, so the width of each of the n strips is
Dx −
b2a
n
These strips divide the interval fa, bg into n subintervals
fx 0 , x 1 g, fx 1, x 2 g, fx 2 , x 3 g, . . . ,
fx n21, x n g
where x 0 − a and x n − b. The right endpoints of the subintervals are
x 1 − a 1 Dx,
x 2 − a 1 2 Dx,
x 3 − a 1 3 Dx,
∙
∙
∙
Let’s approximate the ith strip Si by a rectangle with width Dx and height f sx i d, which
is the value of f at the right endpoint (see Figure 11). Then the area of the ith rectangle
is f sx i d Dx. What we think of intuitively as the area of S is approximated by the sum of
the areas of these rectangles, which is
R n − f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dx
y
Îx
f(xi)
0
FIGURE 11
a
⁄
x2
‹
xi-1
b
xi
x
Figure 12 shows this approximation for n − 2, 4, 8, and 12. Notice that this approximation appears to become better and better as the number of strips increases, that is, as
n l `. Therefore we define the area A of the region S in the following way.
y
y
0
a
⁄
(a) n=2
b x
0
y
a
⁄
x2
(b) n=4
‹
b
x
0
y
b
a
(c) n=8
x
0
b
a
x
(d) n=12
FIGURE 12
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
299
Section 4.1 Areas and Distances
2 Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:
A − lim R n − lim f f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dxg
nl`
nl`
It can be proved that the limit in Definition 2 always exists, since we are assuming that
f is continuous. It can also be shown that we get the same value if we use left endpoints:
3
A − lim L n − lim f f sx 0 d Dx 1 f sx 1 d Dx 1 ∙ ∙ ∙ 1 f sx n21 d Dxg
nl`
nl`
In fact, instead of using left endpoints or right endpoints, we could take the height of
the ith rectangle to be the value of f at any number x*i in the ith subinterval fx i21, x i g.
We call the numbers x1*, x2*, . . . , x n* the sample points. Figure 13 shows approximating
rectangles when the sample points are not chosen to be endpoints. So a more general
expression for the area of S is
A − lim f f sx1* d Dx 1 f sx2* d Dx 1 ∙ ∙ ∙ 1 f sx*n d Dxg
4
nl`
y
Îx
f(x *)
i
0
FIGURE 13
a
⁄
x*¡
x2
x™*
‹
x£*
xi-1
xi
b
xn-1
x *i
x
x n*
Note It can be shown that an equivalent definition of area is the following: A is the
unique number that is smaller than all the upper sums and bigger than all the lower sums.
We saw in Examples 1 and 2, for instance, that the area s A − 13 d is trapped between
all the left approximating sums L n and all the right approximating sums Rn. The function
in those examples, f sxd − x 2, happens to be increasing on f0, 1g and so the lower sums
arise from left endpoints and the upper sums from right endpoints. (See Figures 8 and 9.)
In gen­eral, we form lower (and upper) sums by choosing the sample points x*i so that
f sx*i d is the minimum (and maximum) value of f on the ith subinterval. (See Figure 14
and Exercises 7–8.)
y
FIGURE 14
Lower sums (short rectangles) and
upper sums (tall rectangles)
0
a
b
x
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300
Chapter 4 Integrals
This tells us to
end with i=n.
This tells us
to add.
This tells us to
start with i=m.
We often use sigma notation to write sums with many terms more compactly. For
instance,
n
µ
n
o f sx i d Dx − f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dx
i−1
f(xi) Îx
i=m
So the expressions for area in Equations 2, 3, and 4 can be written as follows:
n
o f sx i d Dx
n l ` i−1
If you need practice with sigma notation,
look at the examples and try some of
the exercises in Appendix E.
A − lim
n
o f sx i21 d Dx
n l ` i−1
A − lim
n
o f sxi*d Dx
A − lim
n l ` i−1
We can also rewrite Formula 1 in the following way:
n
o i2 −
i−1
nsn 1 1ds2n 1 1d
6
Example 3 Let A be the area of the region that lies under the graph of f sxd − cos x
between x − 0 and x − b, where 0 < b < y2.
(a) Using right endpoints, find an expression for A as a limit. Do not evaluate the limit.
(b) Estimate the area for the case b − y2 by taking the sample points to be midpoints
and using four subintervals.
SOLUTION
(a) Since a − 0, the width of a subinterval is
Dx −
b20
b
−
n
n
So x 1 − byn, x 2 − 2byn, x 3 − 3byn, x i − ibyn, and x n − nbyn. The sum of the areas of
the approximating rectangles is
R n − f sx 1 d Dx 1 f sx 2 d Dx 1 ∙ ∙ ∙ 1 f sx n d Dx
− scos x 1 d Dx 1 scos x 2 d Dx 1 ∙ ∙ ∙ 1 scos x n d Dx
S D S D
− cos
b
2b
1 cos
n
n
b
n
S D
b
nb
1 ∙ ∙ ∙ 1 cos
n
n
b
n
According to Definition 2, the area is
A − lim R n − lim
nl`
nl`
b
n
S
cos
b
2b
3b
nb
1 cos
1 cos
1 ∙ ∙ ∙ 1 cos
n
n
n
n
D
Using sigma notation we could write
b
nl` n
A − lim
n
o cos
i−1
ib
n
It is very difficult to evaluate this limit directly by hand, but with the aid of a computer
algebra system it isn’t hard (see Exercise 31). In Section 4.3 we will be able to find A
more easily using a different method.
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301
Section 4.1 Areas and Distances
(b) With n − 4 and b − y2 we have Dx − sy2dy4 − y8, so the subintervals are
f0, y8g, fy8, y4g, fy4, 3y8g, and f3y8, y2g. The midpoints of these subinter­
vals are
3
5
7
x1* −
x2* −
x3* −
x4* −
16
16
16
16
y
y=cos x
1
0
π
8
FIGURE 15
π
4
3π
8
π
2
x
and the sum of the areas of the four approximating rectangles (see Figure 15) is
4
M4 −
o f sxi*d Dx
i−1
− f sy16d Dx 1 f s3y16d Dx 1 f s5y16d Dx 1 f s7y16d Dx
S D S D S D S D
S
D
− cos
−
8
16
cos
3
1 cos
8
16
5
1 cos
8
16
7
1 cos
8
16
3
5
7
1 cos
1 cos
1 cos
16
16
16
16
8
< 1.006
So an estimate for the area is
A < 1.006
n
The Distance Problem
Now let’s consider the distance problem: find the distance traveled by an object during a
certain time period if the velocity of the object is known at all times. (In a sense this is the
inverse problem of the velocity problem that we discussed in Section 1.4.) If the velocity
remains constant, then the distance problem is easy to solve by means of the formula
distance − velocity 3 time
But if the velocity varies, it’s not so easy to find the distance traveled. We investigate the
problem in the following example.
Example 4 Suppose the odometer on our car is broken and we want to estimate the
distance driven over a 30-second time interval. We take speedometer readings every
five seconds and record them in the following table:
Time (s)
0
5
10
15
20
25
30
Velocity (miyh)
17
21
24
29
32
31
28
In order to have the time and the velocity in consistent units, let’s convert the velocity
readings to feet per second (1 miyh − 5280y3600 ftys):
Time (s)
0
5
10
15
20
25
30
Velocity (ftys)
25
31
35
43
47
45
41
During the first five seconds the velocity doesn’t change very much, so we can estimate
the distance traveled during that time by assuming that the velocity is constant. If we
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302
Chapter 4 Integrals
take the velocity during that time interval to be the initial velocity (25 ftys), then we
obtain the approximate distance traveled during the first five seconds:
25 ftys 3 5 s − 125 ft
Similarly, during the second time interval the velocity is approximately constant and
we take it to be the velocity when t − 5 s. So our estimate for the distance traveled
from t − 5 s to t − 10 s is
31 ftys 3 5 s − 155 ft
If we add similar estimates for the other time intervals, we obtain an estimate for the
total distance traveled:
s25 3 5d 1 s31 3 5d 1 s35 3 5d 1 s43 3 5d 1 s47 3 5d 1 s45 3 5d − 1130 ft
We could just as well have used the velocity at the end of each time period instead
of the velocity at the beginning as our assumed constant velocity. Then our estimate
becomes
s31 3 5d 1 s35 3 5d 1 s43 3 5d 1 s47 3 5d 1 s45 3 5d 1 s41 3 5d − 1210 ft
If we had wanted a more accurate estimate, we could have taken velocity readings
every two seconds, or even every second.
n
√
40
20
0
10
FIGURE 16
20
30
t
Perhaps the calculations in Example 4 remind you of the sums we used earlier to
estimate areas. The similarity is explained when we sketch a graph of the velocity function of the car in Figure 16 and draw rectangles whose heights are the initial velocities
for each time interval. The area of the first rectangle is 25 3 5 − 125, which is also
our estimate for the dis­tance traveled in the first five seconds. In fact, the area of each
rectangle can be interpreted as a distance because the height represents velocity and the
width represents time. The sum of the areas of the rectangles in Figure 16 is L 6 − 1130,
which is our initial estimate for the total distance traveled.
In general, suppose an object moves with velocity v − f std, where a < t < b and
f std > 0 (so the object always moves in the positive direction). We take velocity readings at times t0 s− ad, t1, t2 , . . . , tn s− bd so that the velocity is approximately constant
on each subinterval. If these times are equally spaced, then the time between consecutive
readings is Dt − sb 2 adyn. During the first time interval the velocity is approximately
f st0 d and so the distance traveled is approximately f st0 d Dt. Similarly, the distance traveled during the second time interval is about f st1 d Dt and the total distance traveled during the time inter­val fa, bg is approximately
f st0 d Dt 1 f st1 d Dt 1 ∙ ∙ ∙ 1 f stn21 d Dt −
n
o f sti21 d Dt
i−1
If we use the velocity at right endpoints instead of left endpoints, our estimate for the
total distance becomes
f st1 d Dt 1 f st2 d Dt 1 ∙ ∙ ∙ 1 f stn d Dt −
n
o f sti d Dt
i−1
The more frequently we measure the velocity, the more accurate our estimates become,
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Section 4.1 Areas and Distances
303
so it seems plausible that the exact distance d traveled is the limit of such expressions:
n
n
lim o f sti d Dt
o f sti21 d Dt − nl`
nl` i−1
i−1
5
d − lim
We will see in Section 4.4 that this is indeed true.
Because Equation 5 has the same form as our expressions for area in Equations 2
and 3, it follows that the distance traveled is equal to the area under the graph of the
velocity func­tion. In Chapter 5 we will see that other quantities of interest in the natural
and social sciences—such as the work done by a variable force or the cardiac output of
the heart—can also be interpreted as the area under a curve. So when we compute areas
in this chapter, bear in mind that they can be interpreted in a variety of practical ways.
1.(a)By reading values from the given graph of f, use five rect­
angles to find a lower estimate and an upper estimate for
the area under the given graph of f from x − 0 to x − 10.
In each case sketch the rectangles that you use.
(b) Find new estimates using ten rectangles in each case.
4.(a)Estimate the area under the graph of f sxd − sin x from
x − 0 to x − y2 using four approximating rect­angles
and right endpoints. Sketch the graph and the rectangles.
Is your estimate an underestimate or an overestimate?
(b)Repeat part (a) using left endpoints.
y
4
y=ƒ
2
0
8
4
x
2. (a)Use six rectangles to find estimates of each type for the
area under the given graph of f from x − 0 to x − 12.
(i)
L 6 (sample points are left endpoints)
(ii)
R 6 (sample points are right endpoints)
(iii)
M6 (sample points are midpoints)
(b)Is L 6 an underestimate or overestimate of the true area?
(c) Is R 6 an underestimate or overestimate of the true area?
(d)Which of the numbers L 6, R 6, or M6 gives the best estimate? Explain.
y
8
y=ƒ
4
5. (a)Estimate the area under the graph of f sxd − 1 1 x 2
from x − 21 to x − 2 using three rectangles and right
end­points. Then improve your estimate by using six
rect­angles. Sketch the curve and the approximating
rectangles.
(b) Repeat part (a) using left endpoints.
(c) Repeat part (a) using midpoints.
(d)From your sketches in parts (a)–(c), which appears to be
the best estimate?
; 6.(a)Graph the function
f sxd − 1ys1 1 x 2 d
22 < x < 2
(b)Estimate the area under the graph of f using four
approximating rectangles and taking the sample points
to be (i) right endpoints and (ii) midpoints. In each case
sketch the curve and the rectangles.
(c)Improve your estimates in part (b) by using eight
rectangles.
7.Evaluate the upper and lower sums for f sxd − 2 1 sin x,
0 < x < , with n − 2, 4, and 8. Illustrate with diagrams
like Figure 14.
4
0
3.(a)Estimate the area under the graph of f sxd − 1yx from
x − 1 to x − 2 using four approximating rectangles
and right endpoints. Sketch the graph and the rectangles.
Is your estimate an underestimate or an overestimate?
(b)Repeat part (a) using left endpoints.
8
12 x
8.Evaluate the upper and lower sums for f sxd − 1 1 x 2,
21 < x < 1, with n − 3 and 4. Illustrate with diagrams like
Figure 14.
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304
Chapter 4 Integrals
9–10 With a programmable calculator (or a computer), it is
pos­sible to evaluate the expressions for the sums of areas of
approximating rectangles, even for large values of n, using
looping. (On a TI use the Is. command or a For-EndFor loop,
on a Casio use Isz, on an HP or in BASIC use a FOR-NEXT
loop.) Compute the sum of the areas of approximating rect­
angles using equal subintervals and right end­points for n − 10,
30, 50, and 100. Then guess the value of the exact area.
9.The region under y − x 4 from 0 to 1
10. The region under y − cos x from 0 to y2
time period using the velocities at the beginning of the
time intervals.
(b)Give another estimate using the velocities at the end of
the time periods.
(c)Are your estimates in parts (a) and (b) upper and lower
estimates? Explain.
15.Oil leaked from a tank at a rate of rstd liters per hour. The rate
decreased as time passed and values of the rate at two-hour
time intervals are shown in the table. Find lower and upper
estimates for the total amount of oil that leaked out.
t (h)
CAS
CAS
11.Some computer algebra systems have commands that will
draw approximating rectangles and evaluate the sums of
their areas, at least if x*i is a left or right endpoint. (For
instance, in Maple use leftbox, rightbox, leftsum, and rightsum.)
(a)If f sxd − 1ysx 2 1 1d, 0 < x < 1, find the left and
right sums for n − 10, 30, and 50.
(b) Illustrate by graphing the rectangles in part (a).
(c)Show that the exact area under f lies between 0.780
and 0.791.
12. (a)If f sxd − xysx 1 2d, 1 < x < 4, use the commands
discussed in Exercise 11 to find the left and right
sums for n − 10, 30, and 50.
(b)Illustrate by graphing the rectangles in part (a).
(c)Show that the exact area under f lies between 1.603
and 1.624.
13.The speed of a runner increased steadily during the first
three seconds of a race. Her speed at half-second intervals
is given in the table. Find lower and upper estimates for
the distance that she traveled during these three seconds.
t (s)
0
0.5
1.0
1.5
2.0
2.5
3.0
v (ftys)
0
6.2
10.8
14.9
18.1
19.4
20.2
14.The table shows speedometer readings at 10-second
intervals during a 1-minute period for a car racing at the
Daytona International Speedway in Florida.
(a)Estimate the distance the race car traveled during this
Time (s)
Velocity (miyh)
0
10
20
30
40
50
60
182.9
168.0
106.6
99.8
124.5
176.1
175.6
rstd (Lyh)
0
2
4
6
8
10
8.7
7.6
6.8
6.2
5.7
5.3
16. W
hen we estimate distances from velocity data, it is sometimes necessary to use times t0 , t1, t2 , t3 , . . . that are not
equally spaced. We can still estimate distances using the time
periods Dt i − t i 2 t i21. For example, on May 7, 1992, the
space shuttle Endeavour was launched on mission STS-49,
the purpose of which was to install a new perigee kick motor
in an Intelsat communications satellite. The table, provided
by NASA, gives the velocity data for the shuttle between
liftoff and the jettisoning of the solid rocket boosters. Use
these data to estimate the height above the earth’s surface of
the Endeavour, 62 seconds after liftoff.
Time ssd Velocity sftysd
Event
Launch
0
0
Begin roll maneuver
10
185
End roll maneuver
15
319
Throttle to 89%
20
447
Throttle to 67%
32
742
Throttle to 104%
59
1325
Maximum dynamic pressure
62
1445
125
4151
Solid rocket booster separation
17. T
he velocity graph of a braking car is shown. Use it to esti­mate
the distance traveled by the car while the brakes are applied.
√
(ft /s)
60
40
20
0
2
4
t
6
(seconds)
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305
Section 4.1 Areas and Distances
18.The velocity graph of a car accelerating from rest to a speed
of 120 kmyh over a period of 30 seconds is shown. Estimate
the distance traveled during this period.
√
(km / h)
Source: A. Gumel et al., “Modelling Strategies for Controlling SARS
Outbreaks,” Proceedings of the Royal Society of London: Series B 271
(2004): 2223–32.
80
21–23 Use Definition 2 to find an expression for the area
under the graph of f as a limit. Do not evaluate the limit.
40
0
Singapore between March 1 and May 24, 2003, using
both left endpoints and right endpoints.
(b)How would you interpret the number of SARS deaths
as an area under a curve?
t
30
(seconds)
20
10
19. I n someone infected with measles, the virus level N (measured in number of infected cells per mL of blood plasma)
reaches a peak density at about t − 12 days (when a rash
appears) and then decreases fairly rapidly as a result of
immune response. The area under the graph of Nstd from
t − 0 to t − 12 (as shown in the figure) is equal to the total
amount of infection needed to develop symptoms (measured
in density of infected cells 3 time). The function N has been
modeled by the function
21. f sxd −
2x
, 1 < x < 3
x2 1 1
22. f sxd − x 2 1 s1 1 2x , 4 < x < 7
23. f sxd − ssin x , 0 < x < 24–25 Determine a region whose area is equal to the given
limit. Do not evaluate the limit.
n
o
n l ` i−1
24. lim
3
n
Î
11
3i
n
n
o
n l ` i−1
25. lim
i
tan
4n
4n
f std − 2tst 2 21dst 1 1d
Use this model with six subintervals and their midpoints
to estimate the total amount of infection needed to develop
symptoms of measles.
N
1000
N=f(t)
0
12
26. (a)Use Definition 2 to find an expression for the area
under the curve y − x 3 from 0 to 1 as a limit.
(b)The following formula for the sum of the cubes of
the first n integers is proved in Appendix E. Use it to
evaluate the limit in part (a).
13 1 2 3 1 3 3 1 ∙ ∙ ∙ 1 n 3 −
21
t (days)
Source: J. M. Heffernan et al., “An In-Host Model of Acute Infection: Measles
as a Case Study,” Theoretical Population Biology 73 (2006): 134– 47.
20. The
table shows the number of people per day who died
from SARS in Singapore at two-week intervals beginning on
March 1, 2003.
Date
Deaths per day
Date
Deaths per day
March 1
March 15
March 29
April 12
0.0079
0.0638
0.1944
0.4435
April 26
May 10
May 24
0.5620
0.4630
0.2897
(a)By using an argument similar to that in Example 4,
estimate the number of people who died of SARS in
F
nsn 1 1d
2
G
2
27. Let A be the area under the graph of an increasing continuous function f from a to b, and let ­L n and Rn be the
approximations to A with n subintervals using left and
right endpoints, respectively.
(a)How are A, L n, and Rn related?
(b) Show that
b2a
f f sbd 2 f sadg
n
Then draw a diagram to illustrate this equation by
showing that the n rectangles representing R n 2 L n
can be reassem­bled to form a single rectangle whose
area is the right side of the equation.
(c) Deduce that
Rn 2 L n −
Rn 2 A ,
b2a
f f sbd 2 f sadg
n
28. If A is the area under the curve y − sin x from 0 to
y2, use Exercise 27 to find a value of n such that
Rn 2 A , 0.0001.
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306
CAS
CAS
CAS
Chapter 4 Integrals
29. (a)Express the area under the curve y − x 5 from 0 to 2
as a limit.
(b)Use a computer algebra system to find the sum in
your expression from part (a).
(c) Evaluate the limit in part (a).
30.(a)Express the area under the curve y − x 4 1 5x 2 1 x
from 2 to 7 as a limit.
(b)Use a computer algebra system to evaluate the sum in
part (a).
(c)Use a computer algebra system to find the exact area
by evaluating the limit of the expression in part (b).
algebra system both to evaluate the sum and compute the
limit.) In particular, what is the area if b − y2?
32. (a)Let A n be the area of a polygon with n equal sides
inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2yn,
show that
A n − 12 nr 2 sin
S D
2
n
(b)Show that lim n l ` A n − r 2. [Hint: Use Equation 2.4.2
on page 145.]
31.Find the exact area under the cosine curve y − cos x from
x − 0 to x − b, where 0 < b < y2. (Use a computer
We saw in Section 4.1 that a limit of the form
n
1
f f sx1*d Dx 1 f sx2*d Dx 1 ∙ ∙ ∙ 1 f sx n*d Dxg
o f sx*i d Dx − nlim
l`
n l ` i−1
lim
arises when we compute an area. We also saw that it arises when we try to find the dis­
tance traveled by an object. It turns out that this same type of limit occurs in a wide vari­
ety of situations even when f is not necessarily a positive function. In Chapters 5 and 8
we will see that limits of the form (1) also arise in finding lengths of curves, volumes of
solids, centers of mass, force due to water pressure, and work, as well as other quantities.
We therefore give this type of limit a special name and notation.
2 Definition of a Definite Integral If f is a function defined for a < x < b,
we divide the interval fa, bg into n subintervals of equal width Dx − sb 2 adyn.
We let x 0 s− ad, x 1, x 2 , . . . , x n s− bd be the endpoints of these subintervals and we
let x1*, x2*, . . . , x n* be any sample points in these subintervals, so x*i lies in the ith
subinterval fx i21, x i g. Then the definite integral of f from a to b is
y
n
o f sx*i d Dx
n l ` i−1
b
f sxd dx − lim
a
p rovided that this limit exists and gives the same value for all possible choices of
sample points. If it does exist, we say that f is integrable on fa, bg.
The precise meaning of the limit that defines the integral is as follows:
For every number « . 0 there is an integer N such that
Zy
b
a
f sxd dx 2
n
o f sx*i d Dx
i−1
Z
,«
for every integer n . N and for every choice of x*i in fx i21, x i g.
Note 1 The symbol y was introduced by Leibniz and is called an integral sign. It
is an elongated S and was chosen because an integral is a limit of sums. In the notation
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307
Section 4.2 The Definite Integral
ya f sxd dx, f sxd is called the integrand and a and b are called the limits of integration;
b
a is the lower limit and b is the upper limit. For now, the symbol dx has no meaning by
b
itself; ya f sxd dx is all one symbol. The dx simply indicates that the independent vari­able
is x. The procedure of calculating an integral is called integration.
Note 2 The definite integral ya f sxd dx is a number; it does not depend on x. In fact,
b
we could use any letter in place of x without changing the value of the integral:
y
b
a
f sxd dx − y f std dt − y f srd dr
b
a
Note 3 The sum
a
n
o f sx*i d Dx
i−1
Riemann
Bernhard Riemann received his Ph.D.
under the direction of the legendary
Gauss at the University of Göttingen and
remained there to teach. Gauss, who
was not in the habit of praising other
mathematicians, spoke of Riemann’s
“creative, active, truly mathematical
mind and gloriously fertile originality.”
The definition (2) of an integral that we
use is due to Riemann. He also made
major contributions to the theory of
functions of a complex variable, mathematical physics, number theory, and
the foundations of geometry. Riemann’s
broad concept of space and geometry
turned out to be the right setting, 50
years later, for Einstein’s general relativity theory. Riemann’s health was poor
throughout his life, and he died of
tuberculosis at the age of 39.
that occurs in Definition 2 is called a Riemann sum after the German mathematician
Bernhard Riemann (1826 –1866). So Definition 2 says that the definite integral of an
integrable function can be approximated to within any desired degree of accuracy by a
Riemann sum.
We know that if f happens to be positive, then the Riemann sum can be interpreted
as a sum of areas of approximating rectangles (see Figure 1). By comparing Definition 2
b
with the definition of area in Section 4.1, we see that the definite integral ya f sxd dx can
be interpreted as the area under the curve y − f sxd from a to b. (See Figure 2.)
y
If f sxd > 0, the Riemann sum o f sx*i d Dx
is the sum of areas of rectangles.
y=ƒ
b
x
FIGURE 3
o f sx*i d Dx is an approximation
to the net area.
Îx
0
a
x *i
y=ƒ
0
x
b
a
FIGURE 2
b
x
If f sxd > 0, the integral ya f sxd dx is the
area under the curve y − f sxd from a to b.
b
If f takes on both positive and negative values, as in Figure 3, then the Riemann sum
is the sum of the areas of the rectangles that lie above the x-axis and the negatives of the
areas of the rectangles that lie below the x-axis (the areas of the blue rectangles minus
the areas of the gold rectangles). When we take the limit of such Riemann sums, we get
the situation illustrated in Figure 4. A definite integral can be interpreted as a net area,
that is, a difference of areas:
b
y f sxd dx − A 1 2 A 2
where A 1 is the area of the region above the x-axis and below the graph of f , and A 2 is
the area of the region below the x-axis and above the graph of f .
y=ƒ
0 a
FIGURE 4
b
y
a
y
a
y
FIGURE 1
0 a
y
b
f sxd dx is the net area.
b x
Note 4 Although we have defined ya f sxd dx by dividing fa, bg into subintervals of
b
equal width, there are situations in which it is advantageous to work with subintervals of
unequal width. For instance, in Exercise 4.1.16, NASA provided velocity data at times
that were not equally spaced, but we were still able to estimate the distance traveled. And
there are methods for numerical integration that take advantage of unequal subintervals.
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308
Chapter 4 Integrals
If the subinterval widths are Dx 1, Dx 2 , . . . , Dx n , we have to ensure that all these
widths approach 0 in the limiting process. This happens if the largest width, max Dx i ,
approaches 0. So in this case the definition of a definite integral becomes
y
b
a
f sxd dx −
n
o f sx*i d Dx i
lim
max Dx i l 0 i−1
Note 5 We have defined the definite integral for an integrable function, but not all
functions are integrable (see Exercises 71–72). The following theorem shows that the
most commonly occurring functions are in fact integrable. The theorem is proved in
more advanced courses.
3 Theorem If f is continuous on fa, bg, or if f has only a finite number of jump
discontinuities, then f is integrable on fa, bg; that is, the definite integral yab f sxd dx
exists.
If f is integrable on fa, bg, then the limit in Definition 2 exists and gives the same
value no matter how we choose the sample points x*i . To simplify the calculation of the
integral we often take the sample points to be right endpoints. Then x*i − x i and the definition of an integral simplifies as follows.
4 Theorem If f is integrable on fa, bg, then
y
b
a
where
Dx −
f sxd dx − lim
n
o f sx id Dx
nl` i−1
b2a
and x i − a 1 i Dx
n
Example 1 Express
n
o sx 3i 1 x i sin x i d Dx
n l ` i−1
lim
as an integral on the interval f0, g.
SOLUTION Comparing the given limit with the limit in Theorem 4, we see that they
will be identical if we choose f sxd − x 3 1 x sin x. We are given that a − 0 and b − .
Therefore, by Theorem 4, we have
n
o sx 3i 1 x i sin x i d Dx − y0 sx 3 1 x sin xd dx
n l ` i−1
lim
n
Later, when we apply the definite integral to physical situations, it will be important to
recognize limits of sums as integrals, as we did in Example 1. When Leibniz chose the
notation for an integral, he chose the ingredients as reminders of the limiting process. In
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Section 4.2 The Definite Integral
309
general, when we write
n
o f sx *i d Dx − ya f sxd dx
n l ` i−1
b
lim
we replace lim o by y, x*i by x, and Dx by dx.
Evaluating Integrals
When we use a limit to evaluate a definite integral, we need to know how to work
with sums. The following three equations give formulas for sums of powers of positive
integers. Equation 5 may be familiar to you from a course in algebra. Equations 6 and 7
were discussed in Section 4.1 and are proved in Appendix E.
n
nsn 1 1d
2
oi−
i−1
5 n
nsn 1 1ds2n 1 1d
6
o i2 −
i−1
6 F
n
o
7 i−1
i3 −
nsn 1 1d
2
G
2
The remaining formulas are simple rules for working with sigma notation:
Formulas 8 –11 are proved by writing
out each side in expanded form. The
left side of Equation 9 is
ca 1 1 ca 2 1 ∙ ∙ ∙ 1 ca n
n
o c − nc
8 i−1
n
o
9 i−1
The right side is
csa 1 1 a 2 1 ∙ ∙ ∙ 1 a n d
These are equal by the distributive
property. The other formulas are
discussed in Appendix E.
n
o
10 i−1
n
o
11 i−1
n
ca i − c
sa i 1 bi d −
sa i 2 bi d −
o ai
i−1
n
o
n
ai 1
i−1
n
o
i−1
o bi
i−1
n
ai 2
o bi
i−1
Example 2
(a) Evaluate the Riemann sum for f sxd − x 3 2 6x, taking the sample points to be right
endpoints and a − 0, b − 3, and n − 6.
(b) Evaluate y sx 3 2 6xd dx.
3
0
SOLUTION
(a) With n − 6 the interval width is
Dx −
b2a
320
1
−
−
n
6
2
and the right endpoints are x 1 − 0.5, x 2 − 1.0, x 3 − 1.5, x 4 − 2.0, x 5 − 2.5, and
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310
Chapter 4 Integrals
x 6 − 3.0. So the Riemann sum is
6
R6 −
o f sx i d Dx
i−1
− f s0.5d Dx 1 f s1.0d Dx 1 f s1.5d Dx 1 f s2.0d Dx 1 f s2.5d Dx 1 f s3.0d Dx
y
− 12 s22.875 2 5 2 5.625 2 4 1 0.625 1 9d
5
y=˛-6x
− 23.9375
0
x
3
FIGURE 5
Notice that f is not a positive function and so the Riemann sum does not represent
a sum of areas of rectangles. But it does represent the sum of the areas of the blue
rectangles (above the x-axis) minus the sum of the areas of the gold rectangles
(below the x-axis) in Figure 5.
(b) With n subintervals we have
Dx −
b2a
3
−
n
n
So x 0 − 0, x 1 − 3yn, x 2 − 6yn, x 3 − 9yn, and, in general, x i − 3iyn. Since we are
using right endpoints, we can use Theorem 4:
y
3
0
In the sum, n is a constant (unlike i),
so we can move 3yn in front of the
o sign.
− lim
3
n
− lim
3
n
nl`
nl`
− lim
nl`
− lim
nl`
y
5
y=˛-6x
− lim
nl`
A¡
0
A™
3
x
FIGURE 6
y
3
0
sx 3 2 6xd dx − A1 2 A2 − 26.75
n
SD
o FS D
S DG
oF
G
n
o f sx i d Dx − nlim
of
n l ` i−1
l ` i−1
sx 3 2 6xd dx − lim
−
F
n
i−1
n
3
n
3i
n
26
(Equation 9 with c − 3yn)
27 3
18
i
3 i 2
n
n
i−1
81
n4
3
3i
n
3i
n
n
o
i3 2
i−1
54
n2
n
G
oi
i−1
(Equations 11 and 9)
H F G
J
F S D S DG
81
n4
nsn 1 1d
2
81
4
11
1
n
2
2
54 nsn 1 1d
n2
2
2
2 27 1 1
(Equations 7 and 5)
1
n
81
27
2 27 − 2
− 26.75
4
4
This integral can’t be interpreted as an area because f takes on both positive and negative values. But it can be interpreted as the difference of areas A 1 2 A 2, where A 1 and
A 2 are shown in Figure 6.
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Section 4.2 The Definite Integral
311
Figure 7 illustrates the calculation by showing the positive and negative terms in
the right Riemann sum R n for n − 40. The values in the table show the Riemann sums
approaching the exact value of the integral, 26.75, as n l `.
y
5
y=˛-6x
0
x
3
FIGURE 7
R40 < 26.3998
n
Rn
40
100
500
1000
5000
26.3998
26.6130
26.7229
26.7365
26.7473
n
A much simpler method for evaluating the integral in Example 2 will be given in
Example 4.4.3.
Example 3
(a) Set up an expression for y25 x 4 dx as a limit of sums.
(b) Use a computer algebra system to evaluate the expression.
Because f sxd − x 4 is positive, the
integral in Example 3 represents the
area shown in Figure 8.
SOLUTION
y
(a) Here we have f sxd − x 4, a − 2, b − 5, and
Dx −
y=x$
300
b2a
3
−
n
n
x0 − 2, x 1 − 2 1 3yn, x 2 − 2 1 6yn, x 3 − 2 1 9yn, and
xi − 2 1
0
2
5
x
3i
n
From Theorem 4, we get
y
FIGURE 8
5
2
n
o
n l ` i −1
x 4 dx − lim
− lim
nl`
3
n
n
of
n l ` i −1
f sx i d Dx − lim
n
o
i−1
S D
21
3i
n
S D
21
3i
n
3
n
4
(b) If we ask a computer algebra system to evaluate the sum and simplify, we obtain
n
o
i−1
S D
21
3i
n
4
−
2062n 4 1 3045n 3 1 1170n 2 2 27
10n 3
Now we ask the computer algebra system to evaluate the limit:
y
5
2
x 4 dx − lim
nl`
−
3
n
n
o
i−1
S D
21
3i
n
4
− lim
nl`
3s2062n 4 1 3045n 3 1 1170n 2 2 27d
10n 4
3s2062d
3093
−
− 618.6
10
5
We will learn a much easier method for the evaluation of integrals in the next section.
n
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312
Chapter 4 Integrals
y
Example 4 Evaluate the following integrals by interpreting each in terms of areas.
y= œ„„„„„
1-≈
or
≈+¥=1
1
0
(a)
1
0
s1 2 x 2 dx
(b)
y s1 2 x
1
FIGURE 9
(3, 2)
y
A¡
1
3
0
3
sx 2 1d dx
2
dx − 14 s1d2 −
4
(In Section 7.3 we will be able to prove that the area of a circle of radius r is r 2.)
(b) The graph of y − x 2 1 is the line with slope 1 shown in Figure 10. We compute
the integral as the difference of the areas of the two triangles:
y=x-1
0 A™
3
0
SOLUTION
0
y
y
(a) Since f sxd − s1 2 x 2 > 0, we can interpret this integral as the area under the
curve y − s1 2 x 2 from 0 to 1. But, since y 2 − 1 2 x 2, we get x 2 1 y 2 − 1, which
shows that the graph of f is the quarter-circle with radius 1 in Figure 9. Therefore
x
1
y
x
_1
FIGURE 10
TEC Module 4.2 / 7.7 shows how the
Midpoint Rule estimates improve as n
increases.
sx 2 1d dx − A 1 2 A 2 − 12 s2 ∙ 2d 2 12 s1 ∙ 1d − 1.5
n
The Midpoint Rule
We often choose the sample point x*i to be the right endpoint of the ith subinterval
because it is convenient for computing the limit. But if the purpose is to find an approximation to an integral, it is usually better to choose x*i to be the midpoint of the interval,
which we denote by x i . Any Riemann sum is an approximation to an integral, but if we
use midpoints we get the following approximation.
Midpoint Rule y
b
a
f sxd dx <
where
Dx −
n
o f sx i d Dx − Dx f f sx 1 d 1 ∙ ∙ ∙ 1 f sx n dg
i−1
b2a
n
x i − 12 sx i21 1 x i d − midpoint of fx i21, x i g
and
Example 5 Use the Midpoint Rule with n − 5 to approximate y
2
1
1
dx.
x
SOLUTION The endpoints of the five subintervals are 1, 1.2, 1.4, 1.6, 1.8, and 2.0,
so the midpoints are 1.1, 1.3, 1.5, 1.7, and 1.9. The width of the subintervals is
Dx − s2 2 1dy5 − 15, so the Midpoint Rule gives
y
2
1
1
dx < Dx f f s1.1d 1 f s1.3d 1 f s1.5d 1 f s1.7d 1 f s1.9dg
x
−
1
5
S
1
1
1
1
1
1
1
1
1
1.1
1.3
1.5
1.7
1.9
D
< 0.691908
Since f sxd − 1yx . 0 for 1 < x < 2, the integral represents an area, and the approxi­­
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Section 4.2 The Definite Integral
y
0
mation given by the Midpoint Rule is the sum of the areas of the rectangles shown in
Figure 11.
n
1
y= x
1
313
2
x
At the moment we don’t know how accurate the approximation in Example 5 is,
but in Section 7.7 we will learn a method for estimating the error involved in using the
Midpoint Rule. At that time we will discuss other methods for approximating definite
integrals.
If we apply the Midpoint Rule to the integral in Example 2, we get the picture in Fig­
ure 12. The approximation M40 < 26.7563 is much closer to the true value 26.75 than
the right endpoint approximation, R 40 < 26.3998, shown in Figure 7.
FIGURE 11
y
TEC In Visual 4.2 you can compare
left, right, and midpoint approximations to the integral in Example 2 for
different values of n.
5
y=˛-6x
0
3
x
FIGURE 12
M40 < 26.7563
Properties of the Definite Integral
b
When we defined the definite integral ya f sxd dx, we implicitly assumed that a , b. But
the definition as a limit of Riemann sums makes sense even if a . b. Notice that if we
reverse a and b, then Dx changes from sb 2 adyn to sa 2 bdyn. Therefore
y
a
b
f sxd dx − 2y f sxd dx
b
a
If a − b, then Dx − 0 and so
y
a
a
f sxd dx − 0
We now develop some basic properties of integrals that will help us to evaluate integrals in a simple manner. We assume that f and t are continuous functions.
Properties of the Integral
1.
y
b
2.
y
b
a
a
c dx − csb 2 ad, where c is any constant
f f sxd 1 tsxdg dx − y f sxd dx 1 y tsxd dx
b
a
b
a
3. y cf sxd dx − c y f sxd dx, where c is any constant
b
a
4.
y
b
a
b
a
f f sxd 2 tsxdg dx − y f sxd dx 2 y tsxd dx
b
a
b
a
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314
Chapter 4 Integrals
Property 1 says that the integral of a constant function f sxd − c is the constant times
the length of the interval. If c . 0 and a , b, this is to be expected because csb 2 ad is
the area of the shaded rectangle in Figure 13.
y
y=c
c
area=c(b-a)
y
FIGURE 13
b
a
a
b
f+g
y
f
b
a
n
o f f sx i d 1 tsx i dg Dx
n l ` i−1
f f sxd 1 tsxdg dx − lim
Fo
n
0
b x
a
− lim
nl`
b
a
f f sxd 1 tsxdg dx −
y
i−1
f sx i d Dx 1
n
b
a
f sxd dx 1
G
o tsx i d Dx
i−1
n
FIGURE 14
y
x
Property 2 says that the integral of a sum is the sum of the integrals. For positive
functions it says that the area under f 1 t is the area under f plus the area under t.
Figure 14 helps us understand why this is true: in view of how graphical addition works,
the corresponding vertical line segments have equal height.
In general, Property 2 follows from Theorem 4 and the fact that the limit of a sum is
the sum of the limits:
y
g
0
c dx − csb 2 ad
n
o f sx i d Dx 1 nlim
o tsx i d Dx
n l ` i−1
l ` i−1
− lim
y tsxd dx
− y f sxd dx 1 y tsxd dx
b
b
a
Property 3 seems intuitively reasonable
because we know that multiplying
a function by a positive number c
stretches or shrinks its graph vertically
by a factor of c. So it stretches or
shrinks each approximating rectangle
by a factor c and therefore it has the
effect of multiplying the area by c.
b
a
a
Property 3 can be proved in a similar manner and says that the integral of a constant
times a function is the constant times the integral of the function. In other words, a constant (but only a constant) can be taken in front of an integral sign. Property 4 is proved
by writing f 2 t − f 1 s2td and using Properties 2 and 3 with c − 21.
Example 6 Use the properties of integrals to evaluate y s4 1 3x 2 d dx.
1
0
SOLUTION Using Properties 2 and 3 of integrals, we have
y
1
0
s4 1 3x 2 d dx − y 4 dx 1 y 3x 2 dx − y 4 dx 1 3 y x 2 dx
1
1
0
1
0
1
0
0
We know from Property 1 that
y
1
0
4 dx − 4s1 2 0d − 4
and we found in Example 4.1.2 that y x 2 dx − 13. So
1
0
y
1
0
s4 1 3x 2 d dx − y 4 dx 1 3 y x 2 dx
1
0
1
0
− 4 1 3 ∙ 13 − 5
n
The next property tells us how to combine integrals of the same function over adjacent
intervals.
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Section 4.2 The Definite Integral
y
y
5. y=ƒ
0
a
c
b
c
a
315
f sxd dx 1 y f sxd dx − y f sxd dx
b
b
c
a
This is not easy to prove in general, but for the case where f sxd > 0 and a , c , b
Property 5 can be seen from the geometric interpretation in Figure 15: the area under
y − f sxd from a to c plus the area from c to b is equal to the total area from a to b.
x
Example 7 If it is known that y010 f sxd dx − 17 and y08 f sxd dx − 12, find y810 f sxd dx.
FIGURE 15
SOLUTION By Property 5, we have
y
8
0
so
y
10
8
f sxd dx 1 y f sxd dx − y f sxd dx
10
10
8
0
f sxd dx − y f sxd dx 2 y f sxd dx − 17 2 12 − 5
10
8
0
0
n
Properties 1–5 are true whether a , b, a − b, or a . b. The following properties, in
which we compare sizes of functions and sizes of integrals, are true only if a < b.
Comparison Properties of the Integral
6. If f sxd > 0 for a < x < b, then y f sxd dx > 0.
b
a
7. If f sxd > tsxd for a < x < b, then y f sxd dx > y tsxd dx.
b
b
a
a
8. If m < f sxd < M for a < x < b, then
msb 2 ad < y f sxd dx < Msb 2 ad
b
a
y
b
M
y=ƒ
m
0
If f sxd > 0, then ya f sxd dx represents the area under the graph of f, so the geometric
interpretation of Property 6 is simply that areas are positive. (It also follows directly from
the definition because all the quantities involved are positive.) Property 7 says that a bigger function has a bigger integral. It follows from Properties 6 and 4 because f 2 t > 0.
Property 8 is illustrated by Figure 16 for the case where f sxd > 0. If f is continuous,
we could take m and M to be the absolute minimum and maximum values of f on the
inter­val fa, bg. In this case Property 8 says that the area under the graph of f is greater
than the area of the rectangle with height m and less than the area of the rectangle with
height M.
a
FIGURE 16
b
x
Proof of Property 8 Since m < f sxd < M, Property 7 gives
y
b
a
m dx < y f sxd dx < y M dx
b
a
b
a
Using Property 1 to evaluate the integrals on the left and right sides, we obtain
msb 2 ad < y f sxd dx < Msb 2 ad
b
a
n
Property 8 is useful when all we want is a rough estimate of the size of an integral
with­out going to the bother of using the Midpoint Rule.
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316
Chapter 4 Integrals
Example 8 Use Property 8 to estimate y sx dx.
4
1
y
y=œ„x
y=2
2
SOLUTION Since f sxd − sx is an increasing function, its absolute minimum on f1, 4g
is m − f s1d − 1 and its absolute maximum on f1, 4g is M − f s4d − s4 − 2. Thus
Property 8 gives
1s4 2 1d < y sx dx < 2s4 2 1d
4
1
1
y=1
0
1
4
FIGURE 17
3 < y sx dx < 6
4
or
x
The result of Example 8 is illustrated in Figure 17. The area under y − sx from 1 to 4
is greater than the area of the lower rectangle and less than the area of the large rectangle.
1.Evaluate the Riemann sum for f sxd − x 2 1, 26 < x < 4,
with five subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann
sum represents.
2.If
n
1
4
6.The graph of t is shown. Estimate y22
tsxd dx with six sub-­
intervals using (a) right endpoints, (b) left endpoints, and
(c) midpoints.
y
f sxd − cos x 0 < x < 3y4
1
evaluate the Riemann sum with n − 6, taking the sample
points to be left endpoints. (Give your answer correct to six
decimal places.) What does the Riemann sum represent?
Illustrate with a diagram.
x
1
3.If f sxd − x 2 2 4, 0 < x < 3, find the Riemann sum with
n − 6, taking the sample points to be midpoints. What does
the Riemann sum represent? Illustrate with a diagram.
4.(a)Find the Riemann sum for f sxd − 1yx, 1 < x < 2, with
four terms, taking the sample points to be right endpoints.
(Give your answer correct to six decimal places.) Explain
what the Riemann sum represents with the aid of a sketch.
(b)Repeat part (a) with midpoints as the sample points.
5.The graph of a function f is given. Estimate y010 f sxd dx using
five subintervals with (a) right endpoints, (b) left endpoints,
and (c) midpoints.
y
1
0
1
x
7.A table of values of an increasing function f is shown. Use
30
the table to find lower and upper estimates for y10 f sxd dx.
x
10
14
18
22
26
30
f sxd
212
26
22
1
3
8
8.The table gives the values of a function obtained from an
9
experiment. Use them to estimate y3 f sxd dx using three
equal subintervals with (a) right endpoints, (b) left end­
points, and (c) midpoints. If the function is known to be an
increasing function, can you say whether your estimates
are less than or greater than the exact value of the integral?
x
f sxd
3
4
5
23.4 22.1 20.6
6
7
8
9
0.3
0.9
1.4
1.8
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317
Section 4.2 The Definite Integral
26.(a)Find an approximation to the integral y04 sx 2 2 3xd dx
using a Riemann sum with right endpoints and n − 8.
(b)Draw a diagram like Figure 3 to illustrate the approximation in part (a).
(c) Use Theorem 4 to evaluate y04 sx 2 2 3xd dx.
(d)Interpret the integral in part (c) as a difference of areas
and illustrate with a diagram like Figure 4.
9–12 Use the Midpoint Rule with the given value of n to
approximate the integral. Round the answer to four decimal
places.
9.
y
8
sin sx dx, n − 4
10.
y
1
0
11.
y
2
x
dx, n − 5
x11
12.
y
0
0
0
sx 3 1 1 dx, n − 5
x sin2 x dx, n − 4
27. Prove that y x dx −
b
CAS
13.If you have a CAS that evaluates midpoint approximations
and graphs the corresponding rectangles (use RiemannSum
or middlesum and middlebox commands in Maple),
check the answer to Exercise 11 and illustrate with a graph.
Then repeat with n − 10 and n − 20.
14.With a programmable calculator or computer (see the
instructions for Exercise 4.1.9), compute the left and right
Riemann sums for the function f sxd − xysx 1 1d on the
interval f0, 2g with n − 100. Explain why these estimates
show that
2
x
0.8946 , y
dx , 0.9081
0 x 1 1
15.Use a calculator or computer to make a table of values of
right Riemann sums R n for the integral y0 sin x dx with
n − 5, 10, 50, and 100. What value do these numbers
appear to be approaching?
16.Use a calculator or computer to make a table of values
of left and right Riemann sums L n and R n for the integral
y02 s1 1 x 4 dx with n − 5, 10, 50, and 100. Between what
two numbers must the value of the integral lie? Can you
2
make a similar statement for the integral y21
s1 1 x 4 dx?
Explain.
a
28. Prove that y x 2 dx −
b
a
18. lim
o x i s1 1 x i3
n l ` i−1
b3 2 a3
.
3
29–30 Express the integral as a limit of Riemann sums. Do not
evaluate the limit.
1
3
5
30. y x 2 1
dx
29. y s4 1 x 2 dx
1
2
x
S
CAS
D
31–32 Express the integral as a limit of sums. Then evaluate,
using a computer algebra system to find both the sum and the
limit.
31.
y
0
sin 5x dx
32.
y
10
2
x 6 dx
33.The graph of f is shown. Evaluate each integral by inter-­
preting it in terms of areas.
(a)
y
2
(c)
y
7
0
5
f sxd dx
(b)
y
5
f sxd dx
(d)
y
9
0
0
f sxd dx
f sxd dx
y
17–20 Express the limit as a definite integral on the given
interval.
n
sin x i
17. lim o
Dx, f0, g
nl` i−1 1 1 x i
n
b2 2 a2
.
2
y=ƒ
2
0
2
4
6
8
x
Dx, f2, 5g
n
o f5sx*i d3 2 4 x*i g Dx, f2, 7]
n l ` i−1
19. lim
n
20. lim
o
n l ` i−1
x *i
Dx, f1, 3g
sx *i d2 1 4
(a)
21–25 Use the form of the definition of the integral given in
Theorem 4 to evaluate the integral.
21.
y
5
23.
y
0
25.
y
1
2
s4 2 2xd dx
22.
y
4
sx 2 1 x d dx
24.
y
2
22
0
sx 3 2 3x 2 d dx
34.The graph of t consists of two straight lines and a semi­circle. Use it to evaluate each integral.
1
0
sx 2 2 4x 1 2 d dx
y
2
0
tsxd dx (b)
y
6
2
tsxd dx (c)
y
7
0
tsxd dx
y
4
2
y=©
s2x 2 x 3 d dx
0
4
7 x
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318
Chapter 4 Integrals
35–40 Evaluate the integral by interpreting it in terms of areas.
35.
y
37.
y (1 1 s9 2 x ) dx
39.
y | 12 x | dx
2
21
s1 2 xd dx
0
2
23
3
24
36.
y ( 13 x 2 2) dx
38.
y ( x 2 s25 2 x ) dx
40.
y | 2x 2 1 | dx
9
0
5
52.If Fsxd − y2x f std dt, where f is the function whose graph is
given, which of the following values is largest?
(A) Fs0d
(B) Fs1d
(C) Fs2d
(D) Fs3d
(E) Fs4d
2
y
25
1
y=f(t)
0
0
41. Evaluate y s1 1 x 4 dx.
1
1
2
3
t
4
1
42.Given that y sin4 x dx − 38 , what is y sin4 d?
0
0
43.In Example 4.1.2 we showed that y0 x 2 dx − 13. Use this fact
1
53.Each of the regions A, B, and C bounded by the graph of f
and the x-axis has area 3. Find the value of
y
and the properties of integrals to evaluate y0 s5 2 6x 2 d dx.
1
44.Use the properties of integrals and the result of Example 3
to evaluate y25 s1 1 3x 4 d dx.
47. Write as a single integral in the form yab f sxd dx:
y
2
22
f sxd dx 1 y f sxd dx 2 y
5
21
2
22
f sxd dx
48. If y f sxd dx − 7.3 and y f sxd dx − 5.9, find y f sxd dx.
8
2
4
2
8
4
49.If y09 f sxd dx − 37 and y09 tsxd dx − 16, find
y
9
0
50. Find y f sxd dx if
5
0
_4
y
_2
A
0
2
C
x
54.Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must y02 f sxd dx
lie? Which property of integrals allows you to make your
conclusion?
55–58 Use the properties of integrals to verify the inequality
without evaluating the integrals.
55.
56.
f2 f sxd 1 3tsxdg dx
f f sxd 1 2x 1 5g dx
B
45.Use the results of Exercises 27 and 28 and the properties of
integrals to evaluate y14 s2x 2 2 3x 1 1d dx.
46.Use the result of Exercise 27 and the fact that
y2
y0 cos x dx − 1 (from Exercise 4.1.31), together with the
properties of integrals, to evaluate y0y2 s2 cos x 2 5xd dx.
2
24
y
4
0
y
1
0
sx 2 2 4x 1 4d dx > 0
s1 1 x 2 dx < y s1 1 x dx
1
0
57. 2 < y s1 1 x 2 dx < 2 s2
1
21
H
3
f sxd −
x
58.
for x , 3
for x > 3
51.For the function f whose graph is shown, list the following
quantities in increasing order, from smallest to largest, and
explain your reasoning.
y3
s3 < y sin x dx <
y6
12
12
59–64 Use Property 8 of integrals to estimate the value of the
integral.
1
x 1 4 dx
59.
y
1
61.
y
y3
y
63.
y
1
2
65–66 Use properties of integrals, together with Exercises 27 and
28, to prove the inequality.
(A) y f sxd dx
(B) y f sxd dx
(D) y f sxd dx
(E) f 9s1d
8
0
8
4
0
3
0
5
(C) y f sxd dx
8
3
x
65.
0
x 3 dx
y4
21
y
3
1
tan x dx
s1 1 x 4 dx
sx 4 1 1 dx >
26
3
60.
y
3
62.
y
2
64.
y
2
66.
0
0
y
sx 3 2 3x 1 3d dx
sx 2 2 sin xd dx
y2
0
x sin x dx <
2
8
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discovery project Area Functions
67. W
hich of the integrals y12 sx dx, y12 s1yx dx, and y12 ssx dx
has the largest value? Why?
0.5
0.5
68. W
hich of the integrals y0 cossx 2 d dx, y0 cos sx dx is larger?
Why?
69. Prove Property 3 of integrals.
uy
b
a
|
u
f sxd dx < y
|
a
| f sxd | dx
|
|
b
[Hint: 2 f sxd < f sxd < f sxd .]
(b) Use the result of part (a) to show that
uy
2
0
u
72.Let f s0d − 0 and f sxd − 1yx if 0 , x < 1. Show that f
is not integrable on f0, 1g. [Hint: Show that the first term
in the Riemann sum, f sx1* d Dx, can be made arbitrarily
large.]
73–74 Express the limit as a definite integral.
n
70. (a)If f is continuous on fa, bg, show that
f sxd sin 2x dx < y
2
0
| f sxd | dx
o
n l ` i−1
73. lim
74. lim
nl`
1
n
i4
[Hint: Consider f sxd − x 4.]
n5
n
o
i−1
1
1 1 siynd2
2
75.Find y1 x 22 dx. Hint: Choose x *i to be the geometric mean
of x i21 and x i (that is, x *i − sx i21 x i ) and use the identity
1
1
1
−
2
msm 1 1d
m
m11
71.Let f sxd − 0 if x is any rational number and f sxd − 1 if x is
any irrational number. Show that f is not integrable on f0, 1g.
discovery Project
319
area functions
1.
(a)Draw the line y − 2t 1 1 and use geometry to find the area under this line, above the
t-axis, and between the vertical lines t − 1 and t − 3.
(b)If x . 1, let Asxd be the area of the region that lies under the line y − 2t 1 1
between t − 1 and t − x. Sketch this region and use geometry to find an expression
for Asxd.
(c) Differentiate the area function Asxd. What do you notice?
2.
(a) If x > 21, let
Asxd − y s1 1 t 2 d dt
x
21
Asxd represents the area of a region. Sketch that region.
(b) Use the result of Exercise 4.2.28 to find an expression for Asxd.
(c) Find A9sxd. What do you notice?
(d)If x > 21 and h is a small positive number, then Asx 1 hd 2 Asxd represents the
area of a region. Describe and sketch the region.
(e)Draw a rectangle that approximates the region in part (d). By comparing the areas of
these two regions, show that
Asx 1 hd 2 Asxd
< 1 1 x2
h
(f) Use part (e) to give an intuitive explanation for the result of part (c).
2
; 3. (a)Draw the graph of the function f sxd − cossx d in the viewing rectangle f0, 2g
by f21.25, 1.25g.
(b) If we define a new function t by
tsxd − y cos st 2 d dt
x
0
then tsxd is the area under the graph of f from 0 to x [until f sxd becomes negative, at
which point tsxd becomes a difference of areas]. Use part (a) to determine the value
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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320
Chapter 4 Integrals
of x at which tsxd starts to decrease. [Unlike the integral in Problem 2, it is impossible
to evaluate the integral defining t to obtain an explicit expression for tsxd.]
(c)Use the integration command on your calculator or computer to estimate ts0.2d,
ts0.4d, ts0.6d, . . . , ts1.8d, ts2d. Then use these values to sketch a graph of t.
(d)Use your graph of t from part (c) to sketch the graph of t9 using the interpretation of
t9sxd as the slope of a tangent line. How does the graph of t9 compare with the graph
of f ?
4.
Suppose f is a continuous function on the interval fa, bg and we define a new function t
by the equation
x
tsxd − y f std dt
a
Based on your results in Problems 1–3, conjecture an expression for t9sxd.
y
The Fundamental Theorem of Calculus is appropriately named because it establishes a
con­nection between the two branches of calculus: differential calculus and integral
calculus. Differential calculus arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, the area problem. Newton’s mentor at
Cambridge, Isaac Barrow (1630 –1677), discovered that these two problems are actually closely related. In fact, he realized that differentiation and integration are inverse
processes. The Fundamental Theorem of Calculus gives the precise inverse relationship
between the derivative and the integral. It was Newton and Leibniz who exploited this
relationship and used it to develop calculus into a systematic mathema­tical method. In
particular, they saw that the Fundamental Theorem enabled them to compute areas and
integrals very easily without having to compute them as limits of sums as we did in Sections 4.1 and 4.2.
The first part of the Fundamental Theorem deals with functions defined by an equa­
tion of the form
y=f(t )
1
area=©
0
a
x
b
t
FIGURE 1
y
2
x
a
where f is a continuous function on fa, bg and x varies between a and b. Observe that t
depends only on x, which appears as the variable upper limit in the integral. If x is a fixed
number, then the integral yax f std dt is a definite number. If we then let x vary, the number
yax f std dt also varies and defines a function of x denoted by tsxd.
If f happens to be a positive function, then tsxd can be interpreted as the area under the
graph of f from a to x, where x can vary from a to b. (Think of t as the “area so far”
function; see Figure 1.)
y=f(t)
1
0
tsxd − y f std dt
1
2
4
t
Example 1 If f is the function whose graph is shown in Figure 2 and
tsxd − y0x f std dt, find the values of ts0d, ts1d, ts2d, ts3d, ts4d, and ts5d. Then sketch a
rough graph of t.
SOLUTION First we notice that ts0d − y0 f std dt − 0. From Figure 3 we see that ts1d is
0
the area of a triangle:
FIGURE 2
ts1d − y f std dt − 12 s1 2d − 1
1
0
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321
Section 4.3 The Fundamental Theorem of Calculus
To find ts2d we add to ts1d the area of a rectangle:
ts2d − y f std dt − y f std dt 1 y f std dt − 1 1 s1 ? 2d − 3
2
1
0
2
0
1
We estimate that the area under f from 2 to 3 is about 1.3, so
ts3d − ts2d 1 y f std dt < 3 1 1.3 − 4.3
3
2
y
2
y
2
y
2
y
2
y
2
1
1
1
1
1
0
t
1
0
1
g(1)=1
2
t
0
g(2)=3
1
2
3
t
0
1
2
4
t
0
1
2
4
t
g(3)Å4.3
g(4)Å3
g(5)Å1.7
FIGURE 3
y
For t . 3, f std is negative and so we start subtracting areas:
4
g
3
ts4d − ts3d 1 y f std dt < 4.3 1 s21.3d − 3.0
4
3
2
ts5d − ts4d 1 y f std dt < 3 1 s21.3d − 1.7
5
4
1
0
1
2
4
3
We use these values to sketch the graph of t in Figure 4. Notice that, because f std
is positive for t , 3, we keep adding area for t , 3 and so t is increasing up to x − 3,
where it attains a maximum value. For x . 3, t decreases because f std is negative. n
5 x
FIGURE 4
tsxd − y f std dt
x
If we take f std − t and a − 0, then, using Exercise 4.2.27, we have
0
tsxd − y t dt −
x
0
Notice that t9sxd − x, that is, t9 − f . In other words, if t is defined as the integral of f by
Equation 1, then t turns out to be an antiderivative of f , at least in this case. And if we
sketch the derivative of the function t shown in Figure 4 by estimating slopes of tangents,
we get a graph like that of f in Figure 2. So we suspect that t9 − f in Example 1 too.
To see why this might be generally true we consider any continuous function f with
x
f sxd > 0. Then tsxd − ya f std dt can be interpreted as the area under the graph of f from
a to x, as in Figure 1.
In order to compute t9sxd from the definition of a derivative we first observe that, for
h . 0, tsx 1 hd 2 tsxd is obtained by subtracting areas, so it is the area under the graph of
f from x to x 1 h (the blue area in Figure 5). For small h you can see from the figure that
this area is approximately equal to the area of the rectangle with height f sxd and width h:
y
h
ƒ
0
a
FIGURE 5
x
x2
2
x+h
b
tsx 1 hd 2 tsxd < hf sxd
t
so
tsx 1 hd 2 tsxd
< f sxd
h
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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322
Chapter 4 Integrals
Intuitively, we therefore expect that
t9sxd − lim
hl0
tsx 1 hd 2 tsxd
− f sxd
h
The fact that this is true, even when f is not necessarily positive, is the first part of the
Fun­damental Theorem of Calculus.
We abbreviate the name of this theorem
as FTC1. In words, it says that the
derivative of a definite integral with
respect to its upper limit is the integrand evaluated at the upper limit.
The Fundamental Theorem of Calculus, Part 1 If f is continuous on fa, bg, then
the function t defined by
tsxd − y f std dt a < x < b
x
a
is continuous on fa, bg and differentiable on sa, bd, and t9sxd − f sxd.
Proof If x and x 1 h are in sa, bd, then
tsx 1 hd 2 tsxd − y
x1h
a
−
Sy
y
x
x1h
x
x
a
a
−y
f std dt 2 y f std dt
f std dt 1 y
x1h
x
D
f std dt 2 y f std dt (by Property 5)
x
a
f std dt
and so, for h ± 0,
y=ƒ
2
m
0
x u
y
x1h
x
f std dt
For now let’s assume that h . 0. Since f is continuous on fx, x 1 hg, the Extreme
Value Theorem says that there are numbers u and v in fx, x 1 hg such that f sud − m
and f svd − M, where m and M are the absolute minimum and maximum values of f on
fx, x 1 hg. (See Figure 6.)
By Property 8 of integrals, we have
M
√=x+h
tsx 1 hd 2 tsxd
1
−
h
h
x
FIGURE 6
mh < y
x1h
f sudh < y
x1h
x
that is,
x
f std dt < Mh
f std dt < f svdh
Since h . 0, we can divide this inequality by h:
f sud <
1
h
y
x1h
x
f std dt < f svd
Now we use Equation 2 to replace the middle part of this inequality:
3
TEC Module 4.3 provides visual
evidence for FTC1.
f sud <
tsx 1 hd 2 tsxd
< f svd
h
Inequality 3 can be proved in a similar manner for the case where h , 0. (See Exercise 69.)
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Section 4.3 The Fundamental Theorem of Calculus
323
Now we let h l 0. Then u l x and v l x, since u and v lie between x and x 1 h.
Therefore
lim f sud − lim f sud − f sxd and lim f svd − lim f svd − f sxd
hl0
ulx
hl0
vlx
because f is continuous at x. We conclude, from (3) and the Squeeze Theorem, that
4
t9sxd − lim
hl0
tsx 1 hd 2 tsxd
− f sxd
h
If x − a or b, then Equation 4 can be interpreted as a one-sided limit. Then Theo­
rem 2.2.4 (modified for one-sided limits) shows that t is continuous on fa, bg.
n
Using Leibniz notation for derivatives, we can write FTC1 as
5
d
dx
y
x
a
f std dt − f sxd
when f is continuous. Roughly speaking, Equation 5 says that if we first integrate f
and then differentiate the result, we get back to the original function f.
Example 2 Find the derivative of the function tsxd − y s1 1 t 2 dt.
x
0
SOLUTION Since f std − s1 1 t 2 is continuous, Part 1 of the Fundamental Theorem
of Calculus gives
t9sxd − s1 1 x 2
y
1
f
0
n
Example 3 Although a formula of the form tsxd − yax f std dt may seem like a strange
S
way of defining a function, books on physics, chemistry, and statistics are full of such
functions. For instance, the Fresnel function
x
1
Ssxd − y sinst 2y2d dt
x
0
is named after the French physicist Augustin Fresnel (1788 –1827), who is famous for
his works in optics. This function first appeared in Fresnel’s theory of the diffraction of
light waves, but more recently it has been applied to the design of highways.
Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function:
FIGURE 7
f sxd − sinsx 2y2d
Ssxd − y sinst 2y2d dt
x
0
S9sxd − sinsx 2y2d
y
0.5
1
FIGURE 8
The Fresnel function
Ssxd − y sinst 2y2d dt
x
0
x
This means that we can apply all the methods of differential calculus to analyze S (see
Exercise 63).
Figure 7 shows the graphs of f sxd − sinsx 2y2d and the Fresnel function
Ssxd − y0x f std dt. A computer was used to graph S by computing the value of this integral for many values of x. It does indeed look as if Ssxd is the area under the graph of f
from 0 to x [until x < 1.4 when Ssxd becomes a difference of areas]. Figure 8 shows a
larger part of the graph of S.
If we now start with the graph of S in Figure 7 and think about what its derivative
should look like, it seems reasonable that S9sxd − f sxd. [For instance, S is increasing
when f sxd . 0 and decreasing when f sxd , 0.] So this gives a visual confirmation of
Part 1 of the Fundamental Theorem of Calculus.
n
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
324
Chapter 4 Integrals
d x4
y sec t dt.
dx 1
SOLUTION Here we have to be careful to use the Chain Rule in conjunction with FTC1.
Let u − x 4. Then
Example 4 Find
d
dx
y
x4
1
sec t dt −
d
dx
−
d
du
y
u
1
sec t dt
Fy
− sec u
u
1
G
sec t dt
du
dx
du
dx
(by the Chain Rule)
(by FTC1)
− secsx 4 d ? 4x 3
n
In Section 4.2 we computed integrals from the definition as a limit of Riemann sums
and we saw that this procedure is sometimes long and difficult. The second part of
the Fun­damental Theorem of Calculus, which follows easily from the first part, provides
us with a much simpler method for the evaluation of integrals.
The Fundamental Theorem of Calculus, Part 2 If f is continuous on fa, bg, then
y
We abbreviate this theorem as FTC2.
b
a
f sxd dx − Fsbd 2 Fsad
where F is any antiderivative of f, that is, a function F such that F9 − f.
Proof Let tsxd − ya f std dt. We know from Part 1 that t9sxd − f sxd; that is, t is an
x
antiderivative of f. If F is any other antiderivative of f on fa, bg, then we know from
Corollary 3.2.7 that F and t differ by a constant:
6
Fsxd − tsxd 1 C
for a , x , b. But both F and t are continuous on fa, bg and so, by taking limits of
both sides of Equation 6 (as x l a1 and x l b2), we see that it also holds when x − a
and x − b. So Fsxd − tsxd 1 C for all x in fa, bg.
If we put x − a in the formula for tsxd, we get
tsad − y f std dt − 0
a
a
So, using Equation 6 with x − b and x − a, we have
Fsbd 2 Fsad − ftsbd 1 Cg 2 ftsad 1 Cg
− tsbd 2 tsad − tsbd − y f std dt
b
a
n
Part 2 of the Fundamental Theorem states that if we know an antiderivative F of f,
then we can evaluate yab f sxd dx simply by subtracting the values of F at the endpoints
of the interval fa, bg. It’s very surprising that yab f sxd dx, which was defined by a complicated pro­cedure involving all of the values of f sxd for a < x < b, can be found by
knowing the val­ues of Fsxd at only two points, a and b.
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Section 4.3 The Fundamental Theorem of Calculus
325
Although the theorem may be surprising at first glance, it becomes plausible if we
interpret it in physical terms. If vstd is the velocity of an object and sstd is its position at
time t, then vstd − s9std, so s is an antiderivative of v. In Section 4.1 we considered an
object that always moves in the positive direction and made the guess that the area under
the velocity curve is equal to the distance traveled. In symbols:
y
b
a
vstd dt − ssbd 2 ssad
That is exactly what FTC2 says in this context.
Example 5 Evaluate the integral y x 3 dx.
1
22
SOLUTION The function f sxd − x 3 is continuous on f22, 1g and we know from Sec­-
tion 3.9 that an antiderivative is Fsxd − 14 x 4, so Part 2 of the Fundamental Theorem
gives
y
1
22
x 3 dx − Fs1d 2 Fs22d − 14 s1d4 2 14 s22d4 − 2 15
4
Notice that FTC2 says we can use any antiderivative F of f. So we may as well use
the simplest one, namely Fsxd − 14 x 4, instead of 14 x 4 1 7 or 14 x 4 1 C.
n
We often use the notation
g
b
Fsxd a − Fsbd 2 Fsad
So the equation of FTC2 can be written as
y
b
a
g
b
f sxd dx − Fsxd a where F9 − f
Other common notations are Fsxd
|
b
a
and fFsxdg ba .
Example 6 Find the area under the parabola y − x 2 from 0 to 1.
SOLUTION An antiderivative of f sxd − x 2 is Fsxd − 13 x 3. The required area A is found
using Part 2 of the Fundamental Theorem:
In applying the Fundamental Theorem
we use a particular antiderivative F
of f . It is not necessary to use the
most general antiderivative.
A − y x 2 dx −
1
0
x3
3
G
1
−
0
13
03
1
2
−
3
3
3
n
If you compare the calculation in Example 6 with the one in Example 4.1.2, you will
see that the Fundamental Theorem gives a much shorter method.
y
1
Example 7 Find the area under the cosine curve from 0 to b, where 0 < b < y2.
y=cos x
SOLUTION Since an antiderivative of f sxd − cos x is Fsxd − sin x, we have
0
FIGURE 9
π
2
x
g
A − y cos x dx − sin x 0 − sin b 2 sin 0 − sin b
b
area=1
0
b
In particular, taking b − y2, we have proved that the area under the cosine curve
from 0 to y2 is sinsy2d − 1. (See Figure 9.)
n
When the French mathematician Gilles de Roberval first found the area under the
sine and cosine curves in 1635, this was a very challenging problem that required a great
deal of ingenuity. If we didn’t have the benefit of the Fundamental Theorem, we would
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326
Chapter 4 Integrals
have to compute a difficult limit of sums using obscure trigonometric identities (or a
computer algebra system as in Exercise 4.1.31). It was even more difficult for Roberval
because the apparatus of limits had not been invented in 1635. But in the 1660s and
1670s, when the Fundamental Theorem was discovered by Barrow and exploited by
Newton and Leibniz, such problems became very easy, as you can see from Example 7.
Example 8 What is wrong with the following calculation?
y
3
21
1
x21
dx
−
x2
21
G
3
21
−2
4
1
21−2
3
3
SOLUTION To start, we notice that this calculation must be wrong because the answer
is negative but f sxd − 1yx 2 > 0 and Property 6 of integrals says that yab f sxd dx > 0
when f > 0. The Fundamental Theorem of Calculus applies to continuous functions. It
can’t be applied here because f sxd − 1yx 2 is not continuous on f21, 3g. In fact, f has
an infinite discontinuity at x − 0, so
y
3
21
1
dx does not exist.
x2
n
Differentiation and Integration as Inverse Processes
We end this section by bringing together the two parts of the Fundamental Theorem.
The Fundamental Theorem of Calculus Suppose f is continuous on fa, bg.
1. If tsxd − ya f std dt, then t9sxd − f sxd.
x
2. ya f sxd dx − Fsbd 2 Fsad, where F is any antiderivative of f , that is, F9− f.
b
We noted that Part 1 can be rewritten as
d
dx
y
x
a
f std dt − f sxd
which says that if f is integrated and then the result is differentiated, we arrive back at
the original function f. Since F9sxd − f sxd, Part 2 can be rewritten as
y
b
a
F9sxd dx − Fsbd 2 Fsad
This version says that if we take a function F, first differentiate it, and then integrate the
result, we arrive back at the original function F, but in the form Fsbd 2 Fsad. Taken
together, the two parts of the Fundamental Theorem of Calculus say that differentiation
and integration are inverse processes. Each undoes what the other does.
The Fundamental Theorem of Calculus is unquestionably the most important theo­
rem in calculus and, indeed, it ranks as one of the great accomplishments of the human
mind. Before it was discovered, from the time of Eudoxus and Archimedes to the time of
Galileo and Fermat, problems of finding areas, volumes, and lengths of curves were so
difficult that only a genius could meet the challenge. But now, armed with the systematic
method that Newton and Leibniz fashioned out of the Funda­mental Theorem, we will
see in the chap­ters to come that these challenging problems are accessible to all of us.
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327
Section 4.3 The Fundamental Theorem of Calculus
1.Explain exactly what is meant by the statement that
“differenti­ation and integration are inverse processes.”
x
2.Let tsxd − y0 f std dt, where f is the function whose graph is
shown.
(a)Evaluate tsxd for x − 0, 1, 2, 3, 4, 5, and 6.
(b)Estimate ts7d.
(c)Where does t have a maximum value? Where does it
have a minimum value?
(d)Sketch a rough graph of t.
y
f
1
0
4
1
t
6
f
1
7. tsxd −
y
9. tssd −
y
x
t
4.Let tsxd − y0x f std dt, where f is the function whose graph is
shown.
(a) Evaluate ts0d and ts6d.
(b) Estimate tsxd for x − 1, 2, 3, 4, and 5.
(c) On what interval is t increasing?
(d) Where does t have a maximum value?
(e) Sketch a rough graph of t.
(f)Use the graph in part (e) to sketch the graph of t9sxd.
Compare with the graph of f.
y
2
f
2
5
t
8. tsxd − y cosst 2 d dt
x
st 1 t 3 dt
0
s
1
10. hsud − y
st 2 t 2 d8 dt
5
u
0
x
st
dt
t11
F
G
Hint: y s1 1 sec t dt − 2y s1 1 sec t dt
0
x
0
x
12. Rs yd − y t 3 sin t dt
2
y
13. hsxd − y
1yx
15. y − y
3x12
17. y − y
y4
1
5
x
0
7–18 Use Part 1 of the Fundamental Theorem of Calculus to find
the derivative of the function.
2
1
6. tsxd − y s2 1 sin td dt
x
1
0
y
0
5. tsxd − y t 2 dt
11. Fsxd − y s1 1 sec t dt
3.Let tsxd − y0x f std dt, where f is the function whose graph is
shown.
(a) Evaluate ts0d, ts1d, ts2d, ts3d, and ts6d.
(b) On what interval is t increasing?
(c) Where does t have a maximum value?
(d) Sketch a rough graph of t.
0
5–6 Sketch the area represented by tsxd. Then find t9sxd in two
ways: (a) by using Part 1 of the Fundamental Theorem and (b) by
evaluating the integral using Part 2 and then differentiating.
sx
14. hsxd − y
sin 4t dt
sx
1
t
dt
1 1 t3
z2
dz
z 11
4
16. y − y cos2 d
x4
0
18. y − ysin x s1 1 t 2 dt
1
tan d 19–38 Evaluate the integral.
19.
y
21.
y ( 45 t
23.
y
9
25.
y
27.
y
1
29.
y
4
31.
y
y2
3
1
sx 2 1 2x 2 4d dx
2
0
1
2 34 t 2 1 25 t ) dt
sx dx
y6
0
3
sin d
su 1 2dsu 2 3d du
2 1 x2
1
y6
sx
dx
csc t cot t dt
20.
y
22.
y (1 2 8v
24.
y
8
26.
y
5
28.
y
4
30.
y
2
32.
y
y3
1
21
x 100 dx
1
0
1
1 16v7) d v
x 22y3 dx
25
0
3
dx
s4 2 td st dt
21
y4
s3u 2 2dsu 1 1d du
csc 2 d
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
328
33.
Chapter 4 Integrals
y
1
0
35.
y
2
37.
y
38.
y
2
s1 1 rd3 dr
5
v 1 3v
0
y
36.
y
6
dv
v4
1
34.
H
H
2
1
54. tsxd − y
s4 1 1
ds
s2
Î
18
1
55. hsxd − y
22
x3
3
dz
z
t sin t dt
56. tsxd − y
cosst 2 d dt
sx
x2
tan x
1
dt
s2 1 t 4
cos t
dt. Find an equation of the tangent line
t
to the curve y − Fsxd at the point with x-coordinate .
57.Let Fsxd − y
sin x if 0 < x , y2
f sxd dx where f sxd −
cos x if y2 < x < f sxd dx where f sxd −
112x
122x
x
x
58.If f sxd − y0 s1 2 t 2 d cos 2 t dt, on what interval is f
increasing?
2
if 22 < x < 0
4 2 x 2 if 0 , x < 2
59.On what interval is the curve
; 39–42 Sketch the region enclosed by the given curves and
calculate its area.
y−y
39. y − sx , y − 0, x − 4
x
60.Let Fsxd − y1 f std dt, where f is the function whose graph
is shown. Where is F concave downward?
41. y − 4 2 x 2, y − 0
y
42. y − 2 x 2 x 2, y − 0
; 43–46 Use a graph to give a rough estimate of the area of
the region that lies beneath the given curve. Then find the exact
area.
3
43. y − s
x , 0 < x < 27
44. y − x 24, 1 < x < 6
45. y − sin x, 0 < x < 46. y − sec 2 x, 0 < x < y3
_1
62.If f sxd − y0
t 0sy6d.
sin x
2
x 3 dx
21
48.
y
2
y6
cos x dx
; 49–52 What is wrong with the equation?
y
1
50.
y
2
51.
y
y
49.
52.
x 24 dx −
22
G
G
1
y3
−2
22
4
2
dx − 2 2
x3
x
21
0
x23
23
3
−
2
21
g
0
g
CAS
y3
F
3x
2x
−0
u2 2 1
du
u2 1 1
G
Hint: y f sud du − y f sud du 1 y f sud du
3x
2x
0
2x
sins t 2y2d dt − 0.2
64. The sine integral function
x
0
53–56 Find the derivative of the function.
53. tsxd − y
x
Sisxd − y
− 23
3x
0
t
63.The Fresnel function S was defined in Example 3 and
graphed in Figures 7 and 8.
(a)At what values of x does this function have local maxi­
mum values?
(b) On what intervals is the function concave upward?
(c)Use a graph to solve the following equation correct to
two decimal places:
0
1
y
s1 1 t 2 dt and ts yd − y3 f sxd dx, find
y
2
sec tan d − sec sec 2 x dx − tan x
CAS
3
8
0
61.If f s1d − 12, f 9 is continuous, and y14 f 9sxd dx − 17, what is
the value of f s4d?
47–48 Evaluate the integral and interpret it as a difference of
areas. Illustrate with a sketch.
y
t2
dt
t 1t12
2
concave downward?
40. y − x 3, y − 0, x − 1
47.
x
0
sin t
dt
t
is important in electrical engineering. [The integrand
f std − ssin tdyt is not defined when t − 0, but we know that
its limit is 1 when t l 0. So we define f s0d − 1 and this
makes f a continuous function everywhere.]
(a) Draw the graph of Si.
(b)At what values of x does this function have local maxi­
mum values?
(c)Find the coordinates of the first inflection point to the
right of the origin.
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Section 4.3 The Fundamental Theorem of Calculus
(d) Does this function have horizontal asymptotes?
(e)Solve the following equation correct to one decimal
place:
y
x
0
sin t
dt − 1
t
x
3
f
1
2
4
6
8
t
y
x2
dx < 0.1
x 1 x2 1 1
4
74. Let
f sxd −
0
x
22x
0
if
if
if
if
x,0
0<x<1
1,x<2
x.2
tsxd − y f std dt
x
and
0
(a) Find an expression for tsxd similar to the one for f sxd.
(b) Sketch the graphs of f and t.
(c) Where is f differentiable? Where is t differentiable?
1
3
5
7
9
t
_0.2
67–68 Evaluate the limit by first recognizing the sum as a
Riemann sum for a function defined on f0, 1g.
n
S D
SÎ Î Î
o
n l ` i−1
67. lim
1
n
f std
dt − 2 sx for all x . 0
t2
76.Suppose h is a function such that hs1d − 22, h9s1d − 2,
h 0s1d − 3, hs2d − 6, h9s2d − 5, h 0s2d − 13, and h 0 is
continuous everywhere. Evaluate y12 h 0sud du.
0.2
0
x
a
f
0.4
nl`
10
5
61y
_2
68. lim
0<y
75. Find a function f and a number a such that
0
_1
66.
73. Show that
by comparing the integrand to a simpler function.
65–66 Let tsxd − y0 f std dt, where f is the function whose
graph is shown.
(a)At what values of x do the local maximum and minimum
values of t occur?
(b) Where does t attain its absolute maximum value?
(c) On what intervals is t concave downward?
(d) Sketch the graph of t.
65.
y
2
329
Cstd −
i4
i
5 1
n
n2
1
1
n
2
1
n
3
1 ∙∙∙ 1
n
ÎD
n
n
69. Justify (3) for the case h , 0.
70.If f is continuous and t and h are differentiable functions,
find a formula for
d hsxd
y f std dt
dx tsxd
71. (a) Show that 1 < s1 1 x 3 < 1 1 x 3 for x > 0.
(b) Show that 1 < y01 s1 1 x 3 dx < 1.25.
72. (a) Show that cossx 2d > cos x for 0 < x < 1.
(b) Deduce that y0
y6
77.A manufacturing company owns a major piece of equip­
ment that depreciates at the (continuous) rate f − f std,
where t is the time measured in months since its last overhaul. Because a fixed cost A is incurred each time the
machine is overhauled, the company wants to determine the
optimal time T (in months) between overhauls.
t
(a)Explain why y0 f ssd ds represents the loss in value of
the machine over the period of time t since the last
overhaul.
(b) Let C − Cstd be given by
cossx 2d dx > 12.
1
t
F
A 1 y f ssd ds
t
0
G
What does C represent and why would the company
want to minimize C?
(c)Show that C has a minimum value at the numbers t − T
where CsT d − f sT d.
78.A high-tech company purchases a new computing system
whose initial value is V. The system will depreciate at the
rate f − f std and will accumulate maintenance costs at the
rate t − tstd, where t is the time measured in months. The
com­pany wants to determine the optimal time to replace the
system.
(a) Let
1 t
Cstd − y f f ssd 1 tssdg ds
t 0
Show that the critical numbers of C occur at the num­
bers t where Cstd − f std 1 tstd.
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330
Chapter 4 Integrals
(b) Suppose that
f std −
and
H
V
V
2
t
15
450
if 0 , t < 30
0
if t . 30
The following exercises are intended only for those who have
already covered Chapter 6.
79–84 Evaluate the integral.
79.
Vt 2
t . 0
tstd −
12,900
81.
Determine the length of time T for the total deprecia­­­tion
Dstd − y0t f ssd ds to equal the initial value V.
(c) Determine the absolute minimum of C on s0, T g.
(d)Sketch the graphs of C and f 1 t in the same coordinate
system, and verify the result in part (a) in this case.
1
dx80.
2x
y
9
y
1ys2
y
1
1
1y2
83.
21
y
1
dx82.
y
1
e u11 du84.
y
3
4
s1 2 x 2
0
0
1
10 x dx
4
dt
t 11
2
y 3 2 2y 2 2 y
dy
y2
We saw in Section 4.3 that the second part of the Fundamental Theorem of Calculus provides a very powerful method for evaluating the definite integral of a function, assuming
that we can find an antiderivative of the function. In this section we introduce a notation
for antiderivatives, review the formulas for antiderivatives, and use them to evaluate definite integrals. We also reformulate FTC2 in a way that makes it easier to apply to science
and engineering problems.
Indefinite Integrals
Both parts of the Fundamental Theorem establish connections between antiderivatives
x
and definite integrals. Part 1 says that if f is continuous, then ya f std dt is an antiderivative
b
of f. Part 2 says that ya f sxd dx can be found by evaluating Fsbd 2 Fsad, where F is an
antiderivative of f.
We need a convenient notation for antiderivatives that makes them easy to work with.
Because of the relation between antiderivatives and integrals given by the Fundamental
Theorem, the notation y f sxd dx is traditionally used for an antiderivative of f and is
called an indefinite integral. Thus
y f sxd dx − Fsxd means F9sxd − f sxd
For example, we can write
yx
2
dx −
x3
1C
3
because
d
dx
S
D
x3
1 C − x2
3
So we can regard an indefinite integral as representing an entire family of functions (one
antiderivative for each value of the constant C ).
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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Section 4.4 Indefinite Integrals and the Net Change Theorem
331
You should distinguish carefully between definite and indefinite integrals. A def­inite
integral yab f sxd dx is a number, whereas an indefinite integral y f sxd dx is a func­tion (or
family of functions). The connection between them is given by Part 2 of the Fundamental
Theorem: if f is continuous on fa, bg, then
y
b
a
g
f sxd dx − y f sxd dx
b
a
The effectiveness of the Fundamental Theorem depends on having a supply of anti­
derivatives of functions. We therefore restate the Table of Antidifferentiation Formulas
from Section 3.9, together with a few others, in the notation of indefinite integrals. Any
formula can be verified by differentiating the function on the right side and ob­tain­ing the
integrand. For instance,
d
y sec x dx − tan x 1 C because dx stan x 1 Cd − sec x
2
1 Table of
Indefinite Integrals
2
y cf sxd dx − c y f sxd dx
y f f sxd 1 tsxdg dx − y f sxd dx 1 y tsxd dx
y k dx − kx 1 C
yx
y sin x dx − 2cos x 1 C
y cos x dx − sin x 1 C
y sec x dx − tan x 1 C
y csc x dx − 2cot x 1 C
y sec x tan x dx − sec x 1 C
y csc x cot x dx − 2csc x 1 C
2
n
dx −
x n11
1 C sn ± 21d
n11
2
Recall from Theorem 3.9.1 that the most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative. We adopt the convention
that when a formula for a general indefinite integral is given, it is valid only on an
interval. Thus we write
y
1
1
dx − 2 1 C
x2
x
with the understanding that it is valid on the interval s0, `d or on the interval s2`, 0d.
This is true despite the fact that the general antiderivative of the function f sxd − 1yx 2,
x ± 0, is
2
1
1 C1
x
if x , 0
2
1
1 C2
x
if x . 0
Fsxd −
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
332
Chapter 4 Integrals
Example 1 Find the general indefinite integral
4
y s10x
_1.5
1.5
4
SOLUTION Using our convention and Table 1, we have
y s10x
_4
4
2 2 sec2xd dx − 10 y x 4 dx 2 2 y sec2x dx
− 10
FIGURE 1
The indefinite integral in Example 1
is graphed in Figure 1 for several
values of C. Here the value of C is
the y-intercept.
2 2 sec 2xd dx
x5
2 2 tan x 1 C
5
− 2x 5 2 2 tan x 1 C
You should check this answer by differentiating it.
n
cos d.
sin2
Example 2 Evaluate y
SOLUTION This indefinite integral isn’t immediately apparent from Table 1, so we use
trigonometric identities to rewrite the function before integrating:
y
cos d − y
sin2
S DS D
1
sin cos sin d
− y csc cot d − 2csc 1 C
n
Example 3 Evaluate y sx 3 2 6xd dx.
3
0
SOLUTION Using FTC2 and Table 1, we have
y
3
sx 3 2 6xd dx −
0
x4
x2
26
4
2
G
3
0
− ( 41 3 4 2 3 3 2 ) 2 ( 41 0 4 2 3 0 2 )
− 81
4 2 27 2 0 1 0 − 26.75
Compare this calculation with Example 4.2.2(b).
n
Example 4 Find y sx 2 12 sin xd dx.
12
0
SOLUTION The Fundamental Theorem gives
y
12
0
x2
sx 2 12 sin xd dx −
2 12s2cos xd
2
G
12
0
− 12 s12d2 1 12scos 12 2 cos 0d
− 72 1 12 cos 12 2 12
− 60 1 12 cos 12
This is the exact value of the integral. If a decimal approximation is desired, we can use
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 4.4 Indefinite Integrals and the Net Change Theorem
Figure 2 shows the graph of the integrand in Example 4. We know from
Section 4.2 that the value of the integral
can be interpreted as a net area: the sum
of the areas labeled with a plus sign
minus the areas labeled with a minus
sign.
y
y=x-12 sin x
a calculator to approximate cos 12. Doing so, we get
y
12
0
sx 2 12 sin xd dx < 70.1262
n
2t 2 1 t 2 st 2 1
dt.
1
t2
SOLUTION First we need to write the integrand in a simpler form by carrying out the
division:
Example 5 Evaluate y
10
y
9
1
0
333
12 x
9
9
2t 2 1 t 2 s t 2 1
dt − y s2 1 t 1y2 2 t22 d dt
2
1
t
− 2t 1
FIGURE 2
t 3y2
3
2
2
t21
21
G
9
− 2t 1 23 t 3y2 1
1
1
t
G
9
1
− (2 9 1 23 9 3y2 1 19 ) 2 (2 1 1 23 13y2 1 11 )
− 18 1 18 1 19 2 2 2 23 2 1 − 32 49
n
Applications
Part 2 of the Fundamental Theorem says that if f is continuous on fa, bg, then
y
b
a
f sxd dx − Fsbd 2 Fsad
where F is any antiderivative of f. This means that F9 − f, so the equation can be rewritten as
y
b
a
F9sxd dx − Fsbd 2 Fsad
We know that F9sxd represents the rate of change of y − Fsxd with respect to x and
Fsbd 2 Fsad is the change in y when x changes from a to b. [Note that y could, for
instance, increase, then decrease, then increase again. Although y might change in both
directions, Fsbd 2 Fsad represents the net change in y.] So we can reformulate FTC2 in
words as follows.
Net Change Theorem The integral of a rate of change is the net change:
y
b
a
F9sxd dx − Fsbd 2 Fsad
This principle can be applied to all of the rates of change in the natural and social sciences that we discussed in Section 2.7. Here are a few instances of this idea:
If Vstd is the volume of water in a reservoir at time t, then its derivative V9std is the
rate at which water flows into the reservoir at time t. So
●
y
t2
t1
V9std dt − Vst2 d 2 Vst1 d
is the change in the amount of water in the reservoir between time t1 and time t2.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
334
Chapter 4 Integrals
If fCgstd is the concentration of the product of a chemical reaction at time t, then
the rate of reaction is the derivative dfCgydt. So
●
y
t2
t1
dfCg
dt − fCgst2 d 2 fCgst1 d
dt
is the change in the concentration of C from time t1 to time t2.
If the mass of a rod measured from the left end to a point x is msxd, then the linear
density is sxd − m9sxd. So
●
y
b
a
sxd dx − msbd 2 msad
is the mass of the segment of the rod that lies between x − a and x − b.
If the rate of growth of a population is dnydt, then
●
y
t2
t1
dn
dt − nst 2 d 2 nst1 d
dt
is the net change in population during the time period from t1 to t2. (The population increases when births happen and decreases when deaths occur. The net
change takes into account both births and deaths.)
If Csxd is the cost of producing x units of a commodity, then the marginal cost is
the derivative C9sxd. So
●
y
x2
x1
C9sxd dx − Csx 2 d 2 Csx 1 d
is the increase in cost when production is increased from x1 units to x2 units.
If an object moves along a straight line with position function sstd, then its velocity is vstd − s9std, so
●
y
2
t2
t1
vstd dt − sst2 d 2 sst1 d
is the net change of position, or displacement, of the particle during the time
period from t1 to t2. In Section 4.1 we guessed that this was true for the case
where the object moves in the positive direction, but now we have proved that it is
always true