THOMAS i Thirteenth Edition
THOMAS' CALCULUS
Thirteenth Edition
Based on the original work by
George B. Thomas, Jr. Massachusetts Institute of Technology
as revised by
Maurice D. Weir Naval Postgraduate School
Joel Hass University of California, Davis
with the assistance of
Christopher Heil Georgia Institute of Technology
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Library of Congress Cataloging-in-Publication Data
Weir, Maurice D.
Thomas' calculus / based on the original work by George B. Thomas, Jr., Massachusetts Institute of Technology; as revised by Maurice D. Weir, Naval Postgraduate School; Joel Hass, University of California, Davis. — Thirteenth edition.
pages cm
Updated edition of: Thomas’ calculus : early transcendentals / as revised by Maurice D. Weir, Joel Hass. c2010.
ISBN 0-321-87896-5 (hardcover)
1. Calculus— Textbooks. 2. Geometry, Analytic— Textbooks. I. Hass, Joel. II. Weir, Maurice D.
QA303.2.W45 2013 515-dc23 2013023097
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Contents
1.1 1.2 1.3 1.4
2.1 2.2 2.3 2.4 2.5 2.6
4.1 4.2 4.3 4.4
Preface ix
Functions 1
Functions and Their Graphs 1
Combining Functions; Shifting and Scaling Graphs 14
Trigonometric Functions 21
Graphing with Software 29
Questions to Guide Your Review 36 Practice Exercises 36
Additional and Advanced Exercises 38
Limits and Continuity — 41
Rates of Change and Tangents to Curves 41 Limit of a Function and Limit Laws | 48
The Precise Definition of a Limit 59 One-Sided Limits 68
Continuity 75
Limits Involving Infinity; Asymptotes of Graphs Questions to Guide Your Review 99
Practice Exercises 100
Additional and Advanced Exercises 102
Derivatives 105
Tangents and the Derivative ata Point 105 The Derivative as a Function 110 Differentiation Rules 118
The Derivative as a Rate of Change 127 Derivatives of Trigonometric Functions 137 The Chain Rule 144
Implicit Differentiation 151
Related Rates 156
Linearization and Differentials 165 Questions to Guide Your Review 177 Practice Exercises 177
Additional and Advanced Exercises 182
Applications of Derivatives 185
Extreme Values of Functions 185
The Mean Value Theorem 193
Monotonic Functions and the First Derivative Test Concavity and Curve Sketching 204
86
199
iv Contents
4.5 Applied Optimization 215
4.6 Newton's Method 227
4.7 Antiderivatives 232 Questions to Guide Your Review 242 Practice Exercises 243 Additional and Advanced Exercises 245
5 Integrals 249
5.1 Area and Estimating with Finite Sums 249
5.2 Sigma Notation and Limits of Finite Sums 259
5.3 The Definite Integral 266
5.4 The Fundamental Theorem of Calculus 278
5.5 Indefinite Integrals and the Substitution Method 289
5.6 Definite Integral Substitutions and the Area Between Curves 296 Questions to Guide Your Review 306 Practice Exercises 306 Additional and Advanced Exercises 309
6 Applications of Definite Integrals 313
6.1 Volumes Using Cross-Sections 313
6.2 Volumes Using Cylindrical Shells 324
6.3 Arc Length 331
6.4 Areas of Surfaces of Revolution 337
6.5 Work and Fluid Forces 342
6.6 Moments and Centers of Mass 351 Questions to Guide Your Review 362 Practice Exercises 362 Additional and Advanced Exercises 364
1 Transcendental Functions 366
7.] Inverse Functions and Their Derivatives 366 7.2 Natural Logarithms 374 7.3 Exponential Functions 382 7.4 Exponential Change and Separable Differential Equations 393 7.5 Indeterminate Forms and L'Hópital's Rule 403 7.6 Inverse Trigonometric Functions 411 7.7 Hyperbolic Functions 424 7.8 Relative Rates of Growth 433
Questions to Guide Your Review 438
Practice Exercises 439
Additional and Advanced Exercises 442
8 Techniques of Integration — 444
8.1 Using Basic Integration Formulas 444 8.2 Integration by Parts 449
10
11
8.3 8.4 8.5 8.6 8.7 8.8 8.9
9.1 9.2 9.3 9.4 9.5
10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9
Trigonometric Integrals 457
Trigonometric Substitutions 463
Integration of Rational Functions by Partial Fractions 468 Integral Tables and Computer Algebra Systems 477 Numerical Integration 482
Improper Integrals 492
Probability 503
Questions to Guide Your Review 516
Practice Exercises 517
Additional and Advanced Exercises 519
First-Order Differential Equations 524
Solutions, Slope Fields, and Euler’s Method 524 First-Order Linear Equations 532
Applications 538
Graphical Solutions of Autonomous Equations 544 Systems of Equations and Phase Planes 551 Questions to Guide Your Review 557
Practice Exercises 557
Additional and Advanced Exercises 558
Infinite Sequences and Series 560
Sequences 560
Infinite Series 572
The Integral Test 581
Comparison Tests 588
Absolute Convergence; The Ratio and Root Tests 592 Alternating Series and Conditional Convergence 598 Power Series 604
Taylor and Maclaurin Series 614
Convergence of Taylor Series 619
10.10 The Binomial Series and Applications of Taylor Series 626
11.1 11:2 11:3 11.4 11.5 11.6 11.7
Questions to Guide Your Review 635 Practice Exercises 636 Additional and Advanced Exercises 638
Parametric Equations and Polar Coordinates
Parametrizations of Plane Curves 641 Calculus with Parametric Curves 649
Polar Coordinates 659
Graphing Polar Coordinate Equations 663 Areas and Lengths in Polar Coordinates 667 Conic Sections 671
Conics in Polar Coordinates 680
Questions to Guide Your Review 687 Practice Exercises 687
Additional and Advanced Exercises 689
641
Contents
vi
Contents
12
12.1 122 12.3 12.4 12.5 12.6
13
13.1 13:2 13.3 13.4 13.5 13.6
14
14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9
Vectors and the Geometry of Space 692
Three-Dimensional Coordinate Systems 692 Vectors 697
The Dot Product 706
The Cross Product 714
Lines and Planes in Space 720
Cylinders and Quadric Surfaces 728 Questions to Guide Your Review 733 Practice Exercises 734
Additional and Advanced Exercises 736
Vector-Valued Functions and Motion in Space
Curves in Space and Their Tangents 739
Integrals of Vector Functions; Projectile Motion 747 Arc Length in Space 756
Curvature and Normal Vectors of aCurve 760 Tangential and Normal Components of Acceleration 766 Velocity and Acceleration in Polar Coordinates 772 Questions to Guide Your Review 776
Practice Exercises 776
Additional and Advanced Exercises 778
Partial Derivatives — 781
Functions of Several Variables 781
Limits and Continuity in Higher Dimensions 789 Partial Derivatives 798
The Chain Rule 809
Directional Derivatives and Gradient Vectors 818 Tangent Planes and Differentials 827
Extreme Values and Saddle Points 836
Lagrange Multipliers 845
Taylor's Formula for Two Variables 854
14.10 Partial Derivatives with Constrained Variables 858
15
15.1 15.2 15.3 15.4 15.5 15.6 15:7 15.8
Questions to Guide Your Review 863 Practice Exercises 864 Additional and Advanced Exercises 867
Multiple Integrals — 870
Double and Iterated Integrals over Rectangles 870
Double Integrals over General Regions 875
Area by Double Integration 884
Double Integrals in Polar Form 888
Triple Integrals in Rectangular Coordinates 894
Moments and Centers of Mass 903
Triple Integrals in Cylindrical and Spherical Coordinates 910 Substitutions in Multiple Integrals 922
739
16
17
16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8
17.1 17:2 17.3 17.4 17.5
A.1 A-2 A3 AA A.5 A.6 A.7 A.8 Ad
Contents
Questions to Guide Your Review 932 Practice Exercises 932 Additional and Advanced Exercises 935
Integrals and Vector Fields 938
Line Integrals 938
Vector Fields and Line Integrals: Work, Circulation, and Flux 945 Path Independence, Conservative Fields, and Potential Functions 957 Green's Theorem in the Plane 968
Surfaces and Area 980
Surface Integrals 991
Stokes’ Theorem 1002
The Divergence Theorem and a Unified Theory 1015
Questions to Guide Your Review 1027
Practice Exercises 1028
Additional and Advanced Exercises 1030
Second-Order Differential Equations online
Second-Order Linear Equations Nonhomogeneous Linear Equations Applications
Euler Equations
Power Series Solutions
Appendices — AP-1
Real Numbers and the Real Line AP-1
Mathematical Induction AP-6
Lines, Circles, and Parabolas AP-10
Proofs of Limit Theorems AP-19
Commonly Occurring Limits AP-22
Theory of the Real Numbers | AP-23
Complex Numbers | AP-26
The Distributive Law for Vector Cross Products AP-35
The Mixed Derivative Theorem and the Increment Theorem AP-36
Answers to Odd-Numbered Exercises A-1 Credits c-1
Index r1
A Brief Table of Integrals = 1-1
vii
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New to this Edition
Preface
Thomas' Calculus, 'Thirteenth Edition, provides a modern introduction to calculus that focuses on conceptual understanding in developing the essential elements of a traditional course. This material supports a three-semester or four-quarter calculus sequence typically taken by students in mathematics, engineering, and the natural sciences. Precise explana- tions, thoughtfully chosen examples, superior figures, and time-tested exercise sets are the foundation of this text. We continue to improve this text in keeping with shifts in both the preparation and the ambitions of today's students, and the applications of calculus to a changing world.
Many of today's students have been exposed to the terminology and computational methods of calculus in high school. Despite this familiarity, their acquired algebra and trigonometry skills sometimes limit their ability to master calculus at the college level. In this text, we seek to balance students’ prior experience in calculus with the algebraic skill development they may still need, without slowing their progress through calculus itself. We have taken care to provide enough review material (in the text and appendices), detailed solutions, and variety of examples and exercises, to support a complete understanding of calculus for students at varying levels. We present the material in a way to encourage stu- dent thinking, going beyond memorizing formulas and routine procedures, and we show students how to generalize key concepts once they are introduced. References are made throughout which tie a new concept to a related one that was studied earlier, or to a gen- eralization they will see later on. After studying calculus from Thomas, students will have developed problem solving and reasoning abilities that will serve them well in many im- portant aspects of their lives. Mastering this beautiful and creative subject, with its many practical applications across so many fields of endeavor, is its own reward. But the real gift of studying calculus is acquiring the ability to think logically and factually, and learning how to generalize conceptually. We intend this book to encourage and support those goals.
In this new edition we further blend conceptual thinking with the overall logic and struc- ture of single and multivariable calculus. We continue to improve clarity and precision, taking into account helpful suggestions from readers and users of our previous texts. While keeping a careful eye on length, we have created additional examples throughout the text. Numerous new exercises have been added at all levels of difficulty, but the focus in this revision has been on the mid-level exercises. A number of figures have been reworked and new ones added to improve visualization. We have written a new section on probability, which provides an important application of integration to the life sciences.
We have maintained the basic structure of the Table of Contents, and retained im- provements from the twelfth edition. In keeping with this process, we have added more improvements throughout, which we detail here:
X Preface
Continuing Features
* Functions In discussing the use of software for graphing purposes, we added a brief subsection on least squares curve fitting, which allows students to take advantage of this widely used and available application. Prerequisite material continues to be re- viewed in Appendices 1-3.
* Continuity We clarified the continuity definitions by confining the term "endpoints" to intervals instead of more general domains, and we moved the subsection on continuous extension of a function to the end of the continuity section.
e Derivatives We included a brief geometric insight justifying l' Hópital's Rule. We also enhanced and clarified the meaning of differentiability for functions of several vari- ables, and added a result on the Chain Rule for functions defined along a path.
* Integrals We wrote a new section reviewing basic integration formulas and the Sub- stitution Rule, using them in combination with algebraic and trigonometric identities, before presenting other techniques of integration.
* Probability We created a new section applying improper integrals to some commonly used probability distributions, including the exponential and normal distributions. Many examples and exercises apply to the life sciences.
* Series We now present the idea of absolute convergence before giving the Ratio and Root Tests, and then state these tests in their stronger form. Conditional convergence is introduced later on with the Alternating Series Test.
* Multivariable and Vector Calculus We give more geometric insight into the idea of multiple integrals, and we enhance the meaning of the Jacobian in using substitutions to evaluate them. The idea of surface integrals of vector fields now parallels the notion for line integrals of vector fields. We have improved our discussion of the divergence and curl of a vector field.
* Exercises and Examples Strong exercise sets are traditional with Thomas' Calculus, and we continue to strengthen them with each new edition. Here, we have updated, changed, and added many new exercises and examples, with particular attention to in- cluding more applications to the life science areas and to contemporary problems. For instance, we updated an exercise on the growth of the U.S. GNP and added new exer- cises addressing drug concentrations and dosages, estimating the spill rate of a ruptured oil pipeline, and predicting rising costs for college tuition.
RIGOR The level of rigor is consistent with that of earlier editions. We continue to distin- guish between formal and informal discussions and to point out their differences. We think starting with a more intuitive, less formal, approach helps students understand a new or dif- ficult concept so they can then appreciate its full mathematical precision and outcomes. We pay attention to defining ideas carefully and to proving theorems appropriate for calculus students, while mentioning deeper or subtler issues they would study in a more advanced course. Our organization and distinctions between informal and formal discussions give the instructor a degree of flexibility in the amount and depth of coverage of the various top- ics. For example, while we do not prove the Intermediate Value Theorem or the Extreme Value Theorem for continuous functions on a = x = b, we do state these theorems precisely, illustrate their meanings in numerous examples, and use them to prove other important re- sults. Furthermore, for those instructors who desire greater depth of coverage, in Appendix 6 we discuss the reliance of the validity of these theorems on the completeness of the real numbers.
Additional Resources
Preface xi
WRITING EXERCISES Writing exercises placed throughout the text ask students to ex- plore and explain a variety of calculus concepts and applications. In addition, the end of each chapter contains a list of questions for students to review and summarize what they have learned. Many of these exercises make good writing assignments.
END-OF-CHAPTER REVIEWS AND PROJECTS In addition to problems appearing after each section, each chapter culminates with review questions, practice exercises covering the entire chapter, and a series of Additional and Advanced Exercises serving to include more challenging or synthesizing problems. Most chapters also include descriptions of several Technology Application Projects that can be worked by individual students or groups of students over a longer period of time. These projects require the use of a com- puter running Mathematica or Maple and additional material that is available over the Internet at www.pearsonhighered.com/thomas and in MyMathLab.
WRITING AND APPLICATIONS As always, this text continues to be easy to read, conversa- tional, and mathematically rich. Each new topic is motivated by clear, easy-to-understand examples and is then reinforced by its application to real-world problems of immediate interest to students. A hallmark of this book has been the application of calculus to science and engineering. These applied problems have been updated, improved, and extended con- tinually over the last several editions.
TECHNOLOGY In a course using the text, technology can be incorporated according to the taste of the instructor. Each section contains exercises requiring the use of technology; these are marked with a |T | if suitable for calculator or computer use, or they are labeled Computer Explorations if a computer algebra system (CAS, such as Maple or Math- ematica) is required.
INSTRUCTOR'S SOLUTIONS MANUAL
Single Variable Calculus (Chapters 1-11), ISBN 0-321-87898-1 | 978-0-321-87898-4 Multivariable Calculus (Chapters 10-16), ISBN 0-321-87901-5 | 978-0-321-87901-1 The /nstructor's Solutions Manual contains complete worked-out solutions to all of the exercises in Thomas’ Calculus.
STUDENT'S SOLUTIONS MANUAL
Single Variable Calculus (Chapters 1-11), ISBN 0-321-95500-5 | 978-0-321-95500-5 Multivariable Calculus (Chapters 10-16), ISBN 0-321-87897-3 | 978-0-321-87897-7 The Student's Solutions Manual is designed for the student and contains carefully worked-out solutions to all the odd-numbered exercises in Thomas’ Calculus.
JUST-IN-TIME ALGEBRA AND TRIGONOMETRY FOR
CALCULUS, Fourth Edition
ISBN 0-321-67104-X | 978-0-321-67104-2
Sharp algebra and trigonometry skills are critical to mastering calculus, and Just-in-Time Algebra and Trigonometry for Calculus by Guntram Mueller and Ronald I. Brent is de- signed to bolster these skills while students study calculus. As students make their way through calculus, this text is with them every step of the way, showing them the necessary algebra or trigonometry topics and pointing out potential problem spots. The easy-to-use table of contents has algebra and trigonometry topics arranged in the order in which stu- dents will need them as they study calculus.
xii
Preface
Technology Resource Manuals
Maple Manual by Marie Vanisko, Carroll College
Mathematica Manual by Marie Vanisko, Carroll College
TI-Graphing Calculator Manual by Elaine McDonald-Newman, Sonoma State University These manuals cover Maple 17, Mathematica 8, and the TI-83 Plus/TI-84 Plus and TI-89, respectively. Each manual provides detailed guidance for integrating a specific software package or graphing calculator throughout the course, including syntax and commands. These manuals are available to qualified instructors through the Thomas’ Calculus Web site, www.pearsonhighered.com/thomas, and MyMathLab.
WEB SITE www.pearsonhighered.com/thomas
The Thomas' Calculus Web site contains the chapter on Second-Order Differential Equa- tions, including odd-numbered answers, and provides the expanded historical biographies and essays referenced in the text. The Technology Resource Manuals and the Technology Application Projects, which can be used as projects by individual students or groups of students, are also available.
MyMathLab? Online Course (access code required)
MyMathLab from Pearson is the world's leading online resource in mathematics, integrat-
ing interactive homework, assessment, and media in a flexible, easy-to-use format. MyMathLab delivers proven results in helping individual students succeed.
e MyMathLab has a consistently positive impact on the quality of learning in higher education math instruction. MyMathLab can be successfully implemented in any environment—lab-based, hybrid, fully online, traditional —and demonstrates the quan- tifiable difference that integrated usage makes in regard to student retention, subse- quent success, and overall achievement.
e MyMathLab's comprehensive online gradebook automatically tracks your students’ re- sults on tests, quizzes, homework, and in the study plan. You can use the gradebook to quickly intervene if your students have trouble, or to provide positive feedback on a job well done. The data within MyMathLab are easily exported to a variety of spreadsheet programs, such as Microsoft Excel. You can determine which points of data you want to export, and then analyze the results to determine success.
MyMathLab provides engaging experiences that personalize, stimulate, and measure learning for each student.
e “Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with.
e Exercises: The homework and practice exercises in MyMathLab are correlated to the exercises in the textbook, and they regenerate algorithmically to give students unlim- ited opportunity for practice and mastery. The software offers immediate, helpful feed- back when students enter incorrect answers.
* Multimedia Learning Aids: Exercises include guided solutions, sample problems, animations, Java™ applets, videos, and eText access for extra help at point-of-use.
e Expert Tutoring: Although many students describe the whole of MyMathLab as “like having your own personal tutor," students using MyMathLab do have access to live tutoring from Pearson, from qualified math and statistics instructors.
Preface xiii
And, MyMathLab comes from an experienced partner with educational expertise and an eye on the future.
* Knowing that you are using a Pearson product means knowing that you are using qual- ity content. It means that our eTexts are accurate and our assessment tools work. It also means we are committed to making MyMathLab as accessible as possible.
* Whether you are just getting started with MyMathLab, or have a question along the way, we're here to help you learn about our technologies and how to incorporate them into your course.
To learn more about how MyMathLab combines proven learning applications with power- ful assessment, visit www.mymathlab.com or contact your Pearson representative.
Video Lectures with Optional Captioning
The Video Lectures with Optional Captioning feature an engaging team of mathemat- ics instructors who present comprehensive coverage of topics in the text. The lecturers' presentations include examples and exercises from the text and support an approach that emphasizes visualization and problem solving. Available only through MyMathLab and MathXL.
MathXL? Online Course (access code required) MathXL? is the homework and assessment engine that runs MyMathLab. (MyMathLab is MathXL plus a learning management system.)
With MathXL, instructors can:
* Create, edit, and assign online homework and tests using algorithmically generated ex- ercises correlated at the objective level to the textbook.
* Create and assign their own online exercises and import TestGen tests for added flexibility. * Maintain records of all student work tracked in MathXL’s online gradebook. With Math XL, students can:
* Take chapter tests in MathXL and receive personalized study plans and/or personalized homework assignments based on their test results.
* Use the study plan and/or the homework to link directly to tutorial exercises for the objectives they need to study.
* Access supplemental animations and video clips directly from selected exercises.
MathXL is available to qualified adopters. For more information, visit our website at www.mathxl.com, or contact your Pearson representative.
TestGen?
TestGen? (www.pearsoned.com/testgen) enables instructors to build, edit, print, and ad- minister tests using a computerized bank of questions developed to cover all the objec- tives of the text. TestGen is algorithmically based, allowing instructors to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modify test bank questions or add new questions. The software and test bank are available for download from Pearson Education's online catalog.
PowerPoint? Lecture Slides
These classroom presentation slides are geared specifically to the sequence and philosophy of the Thomas' Calculus series. Key graphics from the book are included to help bring the concepts alive in the classroom.These files are available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhighered/irc, and MyMathLab.
xiv Preface
Acknowledgments
We would like to express our thanks to the people who made many valuable contributions to this edition as it developed through its various stages:
Accuracy Checkers Lisa Collette
Patricia Nelson
Tom Wegleitner
Reviewers for Recent Editions
Meighan Dillon, Southern Polytechnic State University Anne Dougherty, University of Colorado
Said Fariabi, San Antonio College
Klaus Fischer, George Mason University
Tim Flood, Pittsburg State University
Rick Ford, California State University—Chico
Robert Gardner, East Tennessee State University Christopher Heil, Georgia Institute of Technology
Joshua Brandon Holden, Rose-Hulman Institute of Technology Alexander Hulpke, Colorado State University
Jacqueline Jensen, Sam Houston State University
Jennifer M. Johnson, Princeton University
Hideaki Kaneko, Old Dominion University
Przemo Kranz, University of Mississippi
Xin Li, University of Central Florida
Maura Mast, University of Massachusetts—Boston
Val Mohanakumar, Hillsborough Community College—Dale Mabry Campus Aaron Montgomery, Central Washington University Christopher M. Pavone, California State University at Chico Cynthia Piez, University of Idaho
Brooke Quinlan, Hillsborough Community College—Dale Mabry Campus Rebecca A. Segal, Virginia Commonwealth University Andrew V. Sills, Georgia Southern University
Alex Smith, University of Wisconsin—Eau Claire
Mark A. Smith, Miami University
Donald Solomon, University of Wisconsin—Milwaukee
John Sullivan, Black Hawk College
Maria Terrell, Cornell University
Blake Thornton, Washington University in St. Louis
David Walnut, George Mason University
Adrian Wilson, University of Montevallo
Bobby Winters, Pittsburg State University
Dennis Wortman, University of Massachusetts—Boston
Functions
OVERVIEW Functions are fundamental to the study of calculus. In this chapter we review what functions are and how they are pictured as graphs, how they are combined and trans- formed, and ways they can be classified. We review the trigonometric functions, and we discuss misrepresentations that can occur when using calculators and computers to obtain a function's graph. The real number system, Cartesian coordinates, straight lines, circles, parabolas, and ellipses are reviewed in the Appendices.
l " l Functions and Their Graphs
Functions are a tool for describing the real world in mathematical terms. A function can be represented by an equation, a graph, a numerical table, or a verbal description; we will use all four representations throughout this book. This section reviews these function ideas.
Functions; Domain and Range
The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the length of time the investment is held. The area of a circle depends on the radius of the circle. The dis- tance an object travels at constant speed along a straight-line path depends on the elapsed time.
In each case, the value of one variable quantity, say y, depends on the value of another variable quantity, which we might call x. We say that “y is a function of x” and write this symbolically as
y = f(x) (“y equals f of x").
In this notation, the symbol f represents the function, the letter x is the independent variable representing the input value of f, and y is the dependent variable or output value of f at x.
DEFINITION A function f from a set D to a set Y is a rule that assigns a unique (single) element f(x) e Y to each element x e D.
The set D of all possible input values is called the domain of the function. The set of all output values of f(x) as x varies throughout D is called the range of the function. The range may not include every element in the set Y. The domain and range of a function can be any sets of objects, but often in calculus they are sets of real numbers interpreted as points of a coordinate line. (In Chapters 13—16, we will encounter functions for which the elements of the sets are points in the coordinate plane or in space.)
2 Chapter 1: Functions
x —— f — f(x) Input Output
(domain) (range)
FIGURE 1.1 A diagram showing a function as a kind of machine.
" Cu uu» = f (a) f)
D = domain set Y — set containing the range FIGURE 1.2 A function from a set D to a set Y assigns a unique element of Y to each element in D.
Often a function is given by a formula that describes how to calculate the output value from the input variable. For instance, the equation A = mr? is a rule that calculates the area A of a circle from its radius r (so r, interpreted as a length, can only be positive in this formula). When we define a function y — f(x) with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real x-values for which the formula gives real y-values, which is called the natural domain. If we want to restrict the domain in some way, we must say so. The domain of y — x? is the entire set of real numbers. To restrict the domain of the function to, say, positive values of x, we would write “y = x?, x > 0.”
Changing the domain to which we apply a formula usually changes the range as well. The range of y = x? is [0, 00). The range of y = x?, x = 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation (see Appendix 1), the range is {x?|x = 2} or (y|y = 4} or [4, 00).
When the range of a function is a set of real numbers, the function is said to be real- valued. The domains and ranges of most real-valued functions of a real variable we con- sider are intervals or combinations of intervals. The intervals may be open, closed, or half open, and may be finite or infinite. Sometimes the range of a function is not easy to find.
A function f is like a machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (Figure 1.1). The function keys on a calculator give an example of a function as a machine. For instance, the Vx key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the Vx key.
A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element of the domain D with a unique or single element in the set Y. In Figure 1.2, the arrows indicate that f(a) is associated with a, f(x) is associated with x, and so on. Notice that a function can have the same value at two different input elements in the domain (as occurs with f(a) in Figure 1.2), but each input element x is assigned a single output value f(x).
EXAMPLE 1 Let's verify the natural domains and associated ranges of some simple functions. The domains in each case are the values of x for which the formula makes sense.
Function Domain (x) Range (y) y=x (—09, 00) [ 0, 00) y = 1/x (~œ, 0) U (0, oo) (—oo, 0) U (0, co) y- Vx [0, o9) [ 0, o9) y=V4-x (—09, 4] [ 0, 00) y-V1l1-x [71.1] [0,1]
Solution The formula y = x? gives a real y-value for any real number x, so the domain is (—00, 00). The range of y = x? is [ 0, 00) because the square of any real number is non- negative and every nonnegative number y is the square of its own square root, y — ( vyy for y = 0.
The formula y — 1/x gives a real y-value for every x except x — 0. For consistency in the rules of arithmetic, we cannot divide any number by zero. The range of y = 1/x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = 1/(1/y). That is, for y # 0 the number x = 1/y is the input assigned to the output value y.
The formula y — Vx gives a real y-value only if x = 0. The range of y = Vx is [ 0, 00) because every nonnegative number is some number's square root (namely, it is the square root of its own square).
In y = V4 — x, the quantity 4 — x cannot be negative. That is, 4 — x = 0, or x X 4. The formula gives real y-values for all x =< 4. The range of V4 — x is [0,09), the set of all nonnegative numbers.
x y=x? —2 4 -1 1
0 0
1 1
3 9
2 4
2 4
y
FIGURE 1.5 Graph of the function in Example 2.
1.1 Functions and Their Graphs 3
The formula y = V1 — x? gives a real y-value for every x in the closed interval from —1 to 1. Outside this domain, 1 — x? is negative and its square root is not a real number. The values of 1 — x? vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of V1 — x° is [0,1]. E
Graphs of Functions
If f is a function with domain D, its graph consists of the points in the Cartesian plane whose coordinates are the input-output pairs for f. In set notation, the graph is
{(x, f(x)) |xeD].
The graph of the function f(x) = x + 2 is the set of points with coordinates (x, y) for which y = x + 2. Its graph is the straight line sketched in Figure 1.3.
The graph of a function f is a useful picture of its behavior. If (x, y) is a point on the graph, then y = f(x) is the height of the graph above (or below) the point x. The height may be positive or negative, depending on the sign of f(x) (Figure 1.4).
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FIGURE 1.3 The graph of f(x) = x + 2 FIGURE 1.4 If (x, y) lies on the graph of
is the set of points (x, y) for which y has the f, then the value y = f(x) is the height of
value x + 2. the graph above the point x (or below x if f(x) is negative).
EXAMPLE 2 Graph the function y = x? over the interval [—2, 2].
Solution Make a table of xy-pairs that satisfy the equation y = x?. Plot the points (x, y) whose coordinates appear in the table, and draw a smooth curve (labeled with its equation) through the plotted points (see Figure 1.5). I
How do we know that the graph of y — x? doesn't look like one of these curves?
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Chapter 1: Functions
To find out, we could plot more points. But how would we then connect them? The basic question still remains: How do we know for sure what the graph looks like between the points we plot? Calculus answers this question, as we will see in Chapter 4. Meanwhile, we will have to settle for plotting points and connecting them as best we can.
Representing a Function Numerically
We have seen how a function may be represented algebraically by a formula (the area function) and visually by a graph (Example 2). Another way to represent a function is numerically, through a table of values. Numerical representations are often used by engi- neers and experimental scientists. From an appropriate table of values, a graph of the func- tion can be obtained using the method illustrated in Example 2, possibly with the aid of a computer. The graph consisting of only the points in the table is called a scatterplot.
EXAMPLE 3 Musical notes are pressure waves in the air. The data associated with Figure 1.6 give recorded pressure displacement versus time in seconds of a musical note produced by a tuning fork. The table provides a representation of the pressure function over time. If we first make a scatterplot and then connect approximately the data points (t, p) from the table, we obtain the graph shown in the figure.
p (pressure)
Time Pressure Time Pressure T 0.00091 —0.080 0.00362 0217 i üi 0.00108 0.200 0.00379 0.480 0.4 0.00125 0.480 0.00398 0.681 0.2 sið 0.00144 0.693 0.00416 0.810 —0.2 0.00162 0.816 0.00435 0.827 e 0.00180 0.844 0.00453 0.749 0.00198 0.771 0.00471 0.581 FIGURE 1.6 A smooth curve through the plotted points 0.00216 0.603 0.00489 0.346 gives a graph of the pressure function represented by the 0.00234 0.368 0.00507 0.077 accompanying tabled data (Example 3). 0.00253 0.099 0.00525 —0.164 0.00271 —0.141 0.00543 —0.320 0.00289 —0.309 0.00562 —0.354 0.00307 —0.348 0.00579 —0.248 0.00325 —0.248 0.00598 —0.035 0.00344 —0.041
The Vertical Line Test for a Function
Not every curve in the coordinate plane can be the graph of a function. A function f can have only one value f(x) for each x in its domain, so no vertical line can intersect the graph of a function more than once. If a is in the domain of the function f, then the vertical line x = a will intersect the graph of f at the single point (a, f(a)).
A circle cannot be the graph of a function, since some vertical lines intersect the circle twice. The circle graphed in Figure 1.7a, however, does contain the graphs of functions of x, such as the upper semicircle defined by the function f(x) = V1 — x? and the lower semicircle defined by the function g(x) = — V1 — x? (Figures 1.7b and 1.7c).
FIGURE 1.8 The absolute value function has domain (—09, 0o) and range [ 0, oo).
FIGURE 1.9 To graph the function y — f(x) shown here, we apply different formulas to different parts of its domain (Example 4).
y ^ ^ B P didis 3} e—o T a P4 2r Fm * d = ib 7 y-|xJ 4 l | | 1 =) -1 77 P 33 7 Se Pd oo —2r 4 4
FIGURE 1.10 The graph of the greatest integer function y = |x | lies on or below the line y = x, so it provides an integer floor for x (Example 5).
1.1 Functions and Their Graphs 5
>< mes ><
>x -X X
(a) x? +y? =1 (b y= V1 — x? (c) y=—-V1—- x?
FIGURE 1.7 (a) The circle is not the graph of a function; it fails the vertical line test. (b) The upper semicircle is the graph of a function f(x) = V1 — x?. (c) The lower semicircle is the graph
of a function g(x) = — V1 — x’.
Piecewise-Defined Functions
Sometimes a function is described in pieces by using different formulas on different parts of its domain. One example is the absolute value function
x: xao |x| = =X; x « 0, whose graph is given in Figure 1.8. The right-hand side of the equation means that the
function equals x if x = 0, and equals —x if x < 0. Piecewise-defined functions often arise when real-world data are modeled. Here are some other examples.
First formula
Second formula
EXAMPLE 4 The function =X; x <0 First formula f(x) = ie 0zxzxl Second formula l. x1 Third formula
is defined on the entire real line but has values given by different formulas, depending on the position of x. The values of f are given by y = —x when x < 0, y = x? when 0 =x = 1, and y = 1 when x > 1. The function, however, is just one function whose domain is the entire set of real numbers (Figure 1.9). E
EXAMPLE 5 The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer floor function. It is denoted | x |. Figure 1.10 shows the graph. Observe that
EXAMPLE 6 The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function or the integer ceiling func- tion. It is denoted | x |. Figure 1.11 shows the graph. For positive values of x, this function might represent, for example, the cost of parking x hours in a parking lot that charges $1 for each hour or part of an hour. ii
6 Chapter 1: Functions
4 4^ "yx 3n — 4 2+ oe” ^ y-[x] 10—9* Pd | och d i i y =2 -l 1 2 3 o—$.-1L "d ee 2p 4
FIGURE 1.11 The graph
of the least integer function
y = [x] lies on or above the line y — x, so it provides an integer ceiling for x (Example 6).
(b)
FIGURE 1.12 (a) The graph of y = x? (an even function) is symmetric about the y-axis. (b) The graph of y — x? (an odd function) is symmetric about the origin.
Increasing and Decreasing Functions
If the graph of a function climbs or rises as you move from left to right, we say that the function is increasing. If the graph descends or falls as you move from left to right, the function is decreasing.
DEFINITIONS Let f be a function defined on an interval J and let x, and x; be any two points in J.
1. If fœ) > f(x) whenever x, < x, then f is said to be increasing on /. 2. If f(x) < f(x) whenever x; < x», then f is said to be decreasing on 7.
It is important to realize that the definitions of increasing and decreasing functions must be satisfied for every pair of points x; and x; in J with x, < x». Because we use the inequality < to compare the function values, instead of <, it is sometimes said that f is strictly increasing or decreasing on 7. The interval J may be finite (also called bounded) or infinite (unbounded) and by definition never consists of a single point (Appendix 1).
EXAMPLE 7 The function graphed in Figure 1.9 is decreasing on (—o9, 0 | and increas- ing on [0,1]. The function is neither increasing nor decreasing on the interval [ 1, 00) because of the strict inequalities used to compare the function values in the definitions. Bl
Even Functions and Odd Functions: Symmetry
The graphs of even and odd functions have characteristic symmetry properties.
DEFINITIONS A function y = f(x) is an
even function of x if f(—x) = f(x), odd function of x if f(—x) = —f(x),
for every x in the function's domain.
The names even and odd come from powers of x. If y is an even power of x, as in y =x’ or y = x^ it is an even function of x because (~x) = x? and (—9)* = x*. If yis an odd power of x, as in y = x or y = x°, it is an odd function of x because (~x)! = —x and =x} = =x.
The graph of an even function is symmetric about the y-axis. Since f(—x) = f(x), a point (x, y) lies on the graph if and only if the point (~x, y) lies on the graph (Figure 1.12a). A reflection across the y-axis leaves the graph unchanged.
The graph of an odd function is symmetric about the origin. Since f(—x) = —f(x), a point (x, y) lies on the graph if and only if the point (~x, —y) lies on the graph (Figure 1.12b). Equivalently, a graph is symmetric about the origin if a rotation of 180? about the origin leaves the graph unchanged. Notice that the definitions imply that both x and —x must be in the domain of f.
EXAMPLE 8 Here are several functions illustrating the definition. f(x) = x Even function: (—x)* = x? for all x; symmetry about y-axis.
f@m=ar+1 Even function: (~x) + 1 = x? + 1 for all x; symmetry about y-axis (Figure 1.13a).
f@ =x Odd function: (—x) = —x for all x; symmetry about the origin. f@m=xt+1 Not odd: f(—x) = —x + 1, but —f(x) = —x — 1. The two are not equal.
Not even: (~x) + 1 # x + 1 forall x # 0 (Figure 1.13b). L|
1.1 Functions and Their Graphs 7
(a) (b)
FIGURE 1.13 (a) When we add the constant term 1 to the function
y = x’, the resulting function y = x? + 1 is still even and its graph is still symmetric about the y-axis. (b) When we add the constant term 1 to the function y = x, the resulting function y = x + 1 is no longer odd, since the symmetry about the origin is lost. The function y = x + 1 is also not even (Example 8).
Common Functions
A variety of important types of functions are frequently encountered in calculus. We iden- tify and briefly describe them here.
Linear Functions A function of the form f(x) = mx + b, for constants m and b, is called a linear function. Figure 1.14a shows an array of lines f(x) = mx where b = 0, so these lines pass through the origin. The function f(x) = x where m = 1 and b = Q is called the identity function. Constant functions result when the slope m = O (Figure 1.14b).
A linear function with positive slope whose graph passes through the origin is called a proportionality relationship.
-3 2P 953 TK l L ï i yx 0 1 2 (b)
FIGURE 1.14 (a) Lines through the origin with slope m. (b) A constant func- tion with slope m = 0.
DEFINITION Two variables y and x are proportional (to one another) if one is always a constant multiple of the other; that is, if y — kx for some nonzero constant k.
If the variable y is proportional to the reciprocal 1/x, then sometimes it is said that y is inversely proportional to x (because 1/x is the multiplicative inverse of x).
Power Functions A function f(x) = x^, where a is a constant, is called a power function. There are several important cases to consider.
8
Chapter 1: Functions
(a) a = n, a positive integer.
The graphs of f(x) — x", for n — 1, 2, 3, 4, 5, are displayed in Figure 1.15. These func- tions are defined for all real values of x. Notice that as the power n gets larger, the curves tend to flatten toward the x-axis on the interval (—1, 1), and to rise more steeply for |x| > 1. Each curve passes through the point (1, 1) and through the origin. The graphs of functions with even powers are symmetric about the y-axis; those with odd powers are symmetric about the origin. The even-powered functions are decreasing on the interval (~œ, 0] and increasing on [0, 09); the odd-powered functions are increasing over the entire real line (—09, 00),
i y=x? Z ya 1r 1r l l I 1 =o E -= FIGURE 1.15 Graphs of f(x) = x", n = 1, 2, 3, 4, 5, defined for —oo < x < oo. (b)a=-1 or a=-2. The graphs of the functions f(x) = x! = 1/x and g(x) = x? = 1/x? are shown in
Figure 1.16. Both functions are defined for all x # 0 (you can never divide by zero). The graph of y = 1/x is the hyperbola xy = 1, which approaches the coordinate axes far from the origin. The graph of y — 1/x? also approaches the coordinate axes. The graph of the function f is symmetric about the origin; f is decreasing on the intervals (—09, 0) and (0, co). The graph of the function g is symmetric about the y-axis; g is increasing on (—090, 0) and decreasing on (0, co).
Ne
Domain: x # 0 Range: y #0 0
Domain: x # 0 Range: y>0O
(a) (b)
FIGURE 1.16 Graphs of the power functions f(x) = x“ for part (a) a = —1 and for part (b) a = —2.
WIN
-113 (c) a= 23 and
The functions f(x) = x? = Vx and g(x) = x! = Wx are the square root and cube root functions, respectively. The domain of the square root function is [0, °°), but the cube root function is defined for all real x. Their graphs are displayed in Figure 1.17, along with the graphs of y = x? and y = x?^?. (Recall that x°/? = (x!/2? and x?? = (x!/3)?,)
Polynomials A function p is a polynomial if P(X) = a,x" + a, 4x | + +++ ax + ag
where n is a nonnegative integer and the numbers do, 41, 45, ..., a, are real constants (called the coefficients of the polynomial). All polynomials have domain (—©9, co). If the
1.1 Functions and Their Graphs 9
y ^ y y = x3/2 1 - L >x >x i >x o 1 l 0 Domain: 0 x x < o Domain: =% < x < % Domain: 0 = x < % Domain: —9 < x < c Range: O<y<a Range: —%< y<% Range: 0Sy<% Range: 02 y<% FIGURE 1.17 Graphs of the power functions f(x) = x“ for a = L L, 3, and z.
leading coefficient a, # 0 and n > 0, then n is called the degree of the polynomial. Lin- ear functions with m 7^ 0 are polynomials of degree 1. Polynomials of degree 2, usually written as p(x) = ax? + bx + c, are called quadratic functions. Likewise, cubic functions are polynomials p(x) = ax? + bx? + cx + d of degree 3. Figure 1.18 shows the graphs of three polynomials. Techniques to graph polynomials are studied in Chapter 4.
1 y= (x — 24 + 19x - 1)
(a) (b) (c) FIGURE 1.18 Graphs of three polynomial functions. Rational Functions A rational function is a quotient or ratio f(x) = p(x)/q(x), where
p and q are polynomials. The domain of a rational function is the set of all real x for which q(x) # 0. The graphs of several rational functions are shown in Figure 1.19.
NOT TO SCALE
(a) (b) (c)
FIGURE 1.19 Graphs of three rational functions. The straight red lines approached by the graphs are called asymptotes and are not part of the graphs. We discuss asymptotes in Section 2.6.
10
Chapter 1: Functions
Algebraic Functions Any function constructed from polynomials using algebraic oper- ations (addition, subtraction, multiplication, division, and taking roots) lies within the class of algebraic functions. All rational functions are algebraic, but also included are more complicated functions (such as those satisfying an equation like y? — 9xy + x? = 0, studied in Section 3.7). Figure 1.20 displays the graphs of three algebraic functions.
= 1/3 x-4 y y=x (x ) y y =x(1 -x/5
(a) (b)
FIGURE 1.20 Graphs of three algebraic functions.
Trigonometric Functions The six basic trigonometric functions are reviewed in Section 1.3. The graphs of the sine and cosine functions are shown in Figure 1.21.
(a) f(x) = sin x (b) f(x) = cos x FIGURE 1.21 Graphs of the sine and cosine functions.
Exponential Functions Functions of the form f(x) = a", where the base a > 0 is a positive constant and a # 1, are called exponential functions. All exponential functions have domain (—°%, œ) and range (0, °°), so an exponential function never assumes the value 0. We develop exponential functions in Section 7.3. The graphs of some exponential functions are shown in Figure 1.22.
FIGURE 1.22 Graphs of exponential functions.
1.1 Functions and Their Graphs 11
Logarithmic Functions These are the functions f(x) = log,x, where the base a # 1 is a positive constant. They are the inverse functions of the exponential functions, and we define these functions in Section 7.2. Figure 1.23 shows the graphs of four logarith- mic functions with various bases. In each case the domain is (0,00) and the range is (—09, 0).
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FIGURE 1.23 Graphs of four logarithmic FIGURE 1.24 Graph of a catenary or functions. hanging cable. (The Latin word catena means “chain.”)
Transcendental Functions These are functions that are not algebraic. They include the trigonometric, inverse trigonometric, exponential, and logarithmic functions, and many other functions as well. A particular example of a transcendental function is a catenary. Its graph has the shape of a cable, like a telephone line or electric cable, strung from one support to another and hanging freely under its own weight (Figure 1.24). The function defining the graph is discussed in Section 7.7.
Exercises 1.1
Functions 8. a. y b. » In Exercises 1—6, find the domain and range of each function. 1. f@=14++ 2. f) =1- Vx 3. F(x) = V 5x + 10 4. g(x) = Vx? — 3x _ 4 2 5. f(t) = 3-1 6. G(t) — Ec qe In Exercises 7 and 8, which of the graphs are graphs of functions of x, >x >x and which are not? Give reasons for your answers. d i 7. a. y b. » ^ ^
Finding Formulas for Functions 9. Express the area and perimeter of an equilateral triangle as a function of the triangle's side length x.
10. Express the side length of a square as a function of the length d of the square's diagonal. Then express the area as a function of the diagonal length.
>x >x 11. Express the edge length of a cube as a function of the cube’s diagonal length d. Then express the surface area and volume of the cube as a function of the diagonal length.
12 Chapter 1: Functions
12. A point P in the first quadrant lies on the graph of the function fœ) = Vx. Express the coordinates of P as functions of the slope of the line joining P to the origin.
13. Consider the point (x,y) lying on the graph of the line 2x + 4y = 5. Let L be the distance from the point (x, y) to the origin (0, 0). Write L as a function of x.
14. Consider the point (x, y) lying on the graph of y = Vx — 3. Let L be the distance between the points (x, y) and (4, 0). Write L as a function of y.
Functions and Graphs Find the natural domain and graph the functions in Exercises 15-20.
15. f(x) = 5— 2x 16. f(x) = 1 — 2x — x
17. g(x) = V |x| 18. g(x) = V-x 19. F(t) = t/ t| 20. GO = 1/]t| 21. Find the domain of y — os a 4-Vx-9 22. Find th =f. . Find t =2+ . ind the range of y dad
23. Graph the following equations and explain why they are not graphs of functions of x. a. |y| =x
24. Graph the following equations and explain why they are not graphs of functions of x.
a. |x| + |y| = 1 b. |x + y| =1
Piecewise-Defined Functions Graph the functions in Exercises 25—28.
X 0zxzxi 25. —-4. dd M 12229 | —3; Osxs1 26. = 8&0) cs je 4 — x, xcm 27. Fx) ^ 4 , xt 2x. xcd 1/x, x«0 28. G(x) = @) o O=x Find a formula for each function graphed in Exercises 29-32. 29. ay b. y L1 1 pu 2 *—9 mE Lg 3 >X >t 0 2 of 1 2 3 4 30. a. y b y ^ ; (2, 1)
| -X
31. a. 3 b. y
CLD | ap 2
Ww M a
| 0 | ar >t x —A i5 ij 0 T T 2 The Greatest and Least Integer Functions 33. For what values of x is a. |x| = 0? b. [x] = 0?
34. What real numbers x satisfy the equation |x| = [x]? 35. Does [—x] = —|x] for all real x? Give reasons for your answer.
36. Graph the function
ENN aen
Why is f(x) called the integer part of x?
Increasing and Decreasing Functions
Graph the functions in Exercises 37—46. What symmetries, if any, do the graphs have? Specify the intervals over which the function is increasing and the intervals where it is decreasing.
37. y 2 —? 38. y--L X 39. y- -1 4. y- 77 P. 41. y= Vc 42. y= V—x 43. y = x°/8 44. y = -AVx 45. y = =x? 46. y = (-x?
Even and Odd Functions In Exercises 47—58, say whether the function is even, odd, or neither. Give reasons for your answer.
47. f(x) 23 48. fx) = x^
49. f) 2321 50. fi) =x +x
51. gx) = x) +x 52. ex») = x^ - 32 — 1 _ 1 Lo x
53. g(x) = 2-1] 54. g(x) 2-1
55. h() = —> 56. h(t) = |P |
57. A(t) = 2r+ 1 58. h(t) = 2|t| + 1
Theory and Examples 59. The variable s is proportional to t, and s = 25 when t = 75. Determine t when s = 60.
60. Kinetic energy The kinetic energy K of a mass is proportional to the square of its velocity v. If K — 12,960 joules when v = 18 m/sec, what is K when v = 10 m/sec?
61. The variables r and s are inversely proportional, and r — 6 when s — 4. Determine s when r — 10.
62. Boyle's Law Boyle's Law says that the volume V of a gas at constant temperature increases whenever the pressure P decreases, so that V and P are inversely proportional. If P = 14.7 Ib/in? when V = 1000 inô, then what is V when P = 23.4 Ib/in??
63. A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 14 in. by 22 in. by cutting out equal squares of side x at each corner and then folding up the sides as in the figure. Express the volume V of the box as a func- tion of x.
| |
64. The accompanying figure shows a rectangle inscribed in an isos- celes right triangle whose hypotenuse is 2 units long.
a. Express the y-coordinate of P in terms of x. (You might start by writing an equation for the line AB.)
b. Express the area of the rectangle in terms of x.
y ^
In Exercises 65 and 66, match each equation with its graph. Do not use a graphing device, and give reasons for your answer.
65. a. y = xt b. y = x’ e y = x
66.
67.
68.
69.
70.
71.
72.
1.1 Functions and Their Graphs 13
a. y — 5x b. y= 5 e y-» y ^ g h 0 x f a. Graph the functions f(x) = x/2 and g(x) = 1 + (4/x) to-
gether to identify the values of x for which x d sp b. Confirm your findings in part (a) algebraically. a. Graph the functions f(x) = 3/(x — 1) and g(x) = 2/(x + 1) together to identify the values of x for which 3 2 x= Í = xc A b. Confirm your findings in part (a) algebraically.
For a curve to be symmetric about the x-axis, the point (x, y) must lie on the curve if and only if the point (x, —y) lies on the curve. Explain why a curve that is symmetric about the x-axis is not the graph of a function, unless the function is y = 0.
Three hundred books sell for $40 each, resulting in a revenue of (300)($40) — $12,000. For each $5 increase in the price, 25 fewer books are sold. Write the revenue R as a function of the number x of $5 increases.
A pen in the shape of an isosceles right triangle with legs of length x ft and hypotenuse of length h ft is to be built. If fencing costs $5/ft for the legs and $10/ft for the hypotenuse, write the total cost C of construction as a function of h.
Industrial costs A power plant sits next to a river where the river is 800 ft wide. To lay a new cable from the plant to a loca- tion in the city 2 mi downstream on the opposite side costs $180 per foot across the river and $100 per foot along the land.
|< 2 mi >| P x Q City
| l 800 ft | l l
Power plant NOT TO SCALE
a. Suppose that the cable goes from the plant to a point Q on the opposite side that is x ft from the point P directly opposite the plant. Write a function C(x) that gives the cost of laying the cable in terms of the distance x.
b. Generate a table of values to determine if the least expensive location for point Q is less than 2000 ft or greater than 2000 ft from point P.
14
Chapter 1: Functions
l . 2 Combining Functions; Shifting and Scaling Graphs
In this section we look at the main ways functions are combined or transformed to form new functions.
Sums, Differences, Products, and Quotients
Like numbers, functions can be added, subtracted, multiplied, and divided (except where the denominator is zero) to produce new functions. If f and g are functions, then for every x that belongs to the domains of both f and g (that is, for x e D(f) N D(g)), we define functions f + g, f — g, and fg by the formulas
(gx) = f(x) + gw) F — g(x) = f(x) — go) (fg) = fg). Notice that the + sign on the left-hand side of the first equation represents the operation of addition of functions, whereas the + on the right-hand side of the equation means addition of the real numbers f(x) and g(x).
At any point of D(f) M D(g) at which g(x) # 0, we can also define the function f/g by the formula
(Do = e (where g(x) # 0).
Functions can also be multiplied by constants: If c is a real number, then the function cf is defined for all x in the domain of f by
(cf)x) = cf).
EXAMPLE 1 The functions defined by the formulas fc) Vx and gx) V1- x
have domains D(f) = [0,00) and D(g) = (~œ, 1]. The points common to these domains are the points
[0,œ)N (799, 1] = (0, 1].
The following table summarizes the formulas and domains for the various algebraic com- binations of the two functions. We also write f + g for the product function fg.
Function Formula Domain
fte EHDA = Vxt* VI =x 0,1] = D) D(g)
f-2 (f— 9) = Vx- V1-x [0, 1]
ge e- fl = V1-x- Vx [0,1]
f'g f° 90) = fw = Va — x) [0,1]
f/g Fo = a = alg x 3 [ 0, D (x = 1 excluded)
s/f HE - px = ,/ : x : (0, 1] (x = O excluded) E
The graph of the function f + g is obtained from the graphs of f and g by adding the corresponding y-coordinates f(x) and g(x) at each point x e D(f) M D(g), as in Figure 1.25. The graphs of f + g and f - g from Example | are shown in Figure 1.26.
1.2 Combining Functions; Shifting and Scaling Graphs 15
FIGURE 1.25 Graphical addition of two FIGURE 1.26 The domain of the function f + g
functions. is the intersection of the domains of f and g, the interval [0, 1] on the x-axis where these domains overlap. This interval is also the domain of the function f + g (Example 1).
Composite Functions
Composition is another method for combining functions.
DEFINITION If f and g are functions, the composite function f ° g (“f com- posed with g”) is defined by
(f° g(x) = few). The domain of f ° g consists of the numbers x in the domain of g for which g(x) lies in the domain of f.
The definition implies that f ° g can be formed when the range of g lies in the domain of f. To find (f ° g)(x), first find g(x) and second find f(g(x)). Figure 1.27 pictures f ° g as a machine diagram, and Figure 1.28 shows the composite as an arrow diagram.
fog e f(g(x)) x x—> g > f > fw) 80) FIGURE 1.27 A composite function f ° g uses FIGURE 1.28 Arrow diagram for f » g. If x lies in the the output g(x) of the first function g as the input domain of g and g(x) lies in the domain of f, then the for the second function f. functions f and g can be composed to form (f ° g)(x).
To evaluate the composite function g ° f (when defined), we find f(x) first and then g(f(x)). The domain of g » f is the set of numbers x in the domain of f such that f(x) lies in the domain of g.
The functions f ° g and go f are usually quite different.
16 Chapter 1: Functions
FIGURE 1.29 To shift the graph of f(x) = x? up (or down), we add positive (or negative) constants
to the formula for f (Examples 3a and b).
EXAMPLE 2 If f(x) = Vx and g(x) = x + 1, find (a) (f° gx) (b) (g° H (c) (f » Px) (d) (g° g)().
Solution Composite Domain (a) (f° gy) = few) = Vea) = Vx * 1 [—1, 00) (b) BAHA = g(f@) = f@) +1= Vx +1 [ 0, 00) (c) FAW = fF@) = VfQ) = V vx = x^ [ 0, 00) (d) (g»g)x) = gea) = gt 17x * Dt 17x42 (706, 00) To see why the domain of f ° g is [ —1, cO), notice that g(x) = x + 1 is defined for all real x but belongs to the domain of f only if x + 1 = O0, that is to say, when x = — 1. a
Notice that if f(x) = x? and g(x) = Vx, then (f ° g)G) = (Vx)? = x. However, the domain of f ° g is | 0,00), not (~00, co), since Vx requires x = 0.
Shifting a Graph of a Function
A common way to obtain a new function from an existing one is by adding a constant to each output of the existing function, or to its input variable. The graph of the new function is the graph of the original function shifted vertically or horizontally, as follows.
Shift Formulas
Vertical Shifts
y=fwtk Shifts the graph of f up k units if k > 0 Shifts it down |k| units if k — 0
Horizontal Shifts
y= f(x + h) Shifts the graph of f left h units if h > 0 Shifts it right |h| units if h < 0
EXAMPLE 3
(a) Adding 1 to the right-hand side of the formula y = x? to get y = x? + 1 shifts the graph up 1 unit (Figure 1.29).
(b) Adding —2 to the right-hand side of the formula y — x? to get y — x? — 2 shifts the graph down 2 units (Figure 1.29).
(c) Adding 3 to x in y = x? to get y = (x + 3)? shifts the graph 3 units to the left, while adding —2 shifts the graph 2 units to the right (Figure 1.30).
(d) Adding —2 toxin y = E , and then adding — 1 to the result, gives y = x = 2| = ll and shifts the graph 2 units to the right and 1 unit down (Figure 1.31). E
Scaling and Reflecting a Graph of a Function
To scale the graph of a function y = f(x) is to stretch or compress it, vertically or hori- zontally. This is accomplished by multiplying the function f, or the independent variable x, by an appropriate constant c. Reflections across the coordinate axes are special cases where c = — 1.
FN oU 4 tA
=I 00
FIGURE 1.32 Vertically stretching and compressing the graph y — Vx bya factor of 3 (Example 4a).
1.2 Combining Functions; Shifting and Scaling Graphs 17
Add a positive Add a negative constant to x. y constant to x.
FIGURE 1.30 To shift the graph of y — x? to FIGURE 1.31 The graph of y = |x|
the left, we add a positive constant to x (Example 3c). To shift the graph to the right, we add a nega-
shifted 2 units to the right and 1 unit down (Example 3d).
tive constant to x.
Vertical and Horizontal Scaling and Reflecting Formulas
For c > 1, the graph is scaled:
y = cf(x) Stretches the graph of f vertically by a factor of c.
y- l f(x) Compresses the graph of f vertically by a factor of c.
y = f(cx) Compresses the graph of f horizontally by a factor of c. y = f/d Stretches the graph of f horizontally by a factor of c. Forc = —1, the graph is reflected:
y =—-f@) Reflects the graph of f across the x-axis.
y = f(-x) Reflects the graph of f across the y-axis.
EXAMPLE 4 Here we scale and reflect the graph of y = Vx.
(a)
(b)
(c)
Vertical: Multiplying the right-hand side of y — Vx by 3 to get y = 3 V/x stretches the graph vertically by a factor of 3, whereas multiplying by 1/3 compresses the graph by a factor of 3 (Figure 1.32).
Horizontal: The graph of y — Vx is a horizontal compression of the graph of y= Vx by a factor of 3, and y = Vx/3 is a horizontal stretching by a factor of 3 (Figure 1.33). Note that y — V3x = V3 Vx so a horizontal compression may cor- respond to a vertical stretching by a different scaling factor. Likewise, a horizontal stretching may correspond to a vertical compression by a different scaling factor.
Reflection: The graph of y — — V/x is a reflection of y- Vx across the x-axis, and y = V-—xis a reflection across the y-axis (Figure 1.34). il
><
A FSV X
= y=—-Vx FIGURE 1.33 Horizontally stretching and FIGURE 1.34 Reflections of the graph compressing the graph y = Vx by a factor of y = Vx across the coordinate axes
3 (Example 4b). (Example 4c).
18 Chapter 1: Functions
EXAMPLE 5 Given the function f(x) = x* — 4x? + 10 (Figure 1.352), find formulas to
(a) compress the graph horizontally by a factor of 2 followed by a reflection across the y-axis (Figure 1.35b).
(b) compress the graph vertically by a factor of 2 followed by a reflection across the x-axis (Figure 1.35c).
y y = 16xf + 32x? + 10 7 fo) = x^ — 4x? +10
><
y = 3x4 + 2x3 — 5
>x
*X
(a) (b)
FIGURE 1.35 (a) The original graph of f. (b) The horizontal compression of y — f(x) in part (a) by a factor of 2, followed by
a reflection across the y-axis. (c) The vertical compression of y — f(x) in part (a) by a factor of 2, followed by a reflection across the x-axis (Example 5).
Solution (a) We multiply x by 2 to get the horizontal compression, and by —1 to give reflection
across the y-axis. The formula is obtained by substituting —2x for x in the right-hand side of the equation for f:
y = f(-2x) = (-2x)* — 4(-2xy + 10 = l6x^ + 32x? + 10. (b) The formula is
y--ifo--ié-20-5 = Exercises A Algebraic Combinations e. f(f(-5)) f. (g2) In Exercises 1 and 2, find the domains and ranges of f, g. f + g, and g. FFE) h. e(g(x))
fig
6. If f(x) = x — Land g(x) = 1/(x + 1), find the following. l. fMm=x, gxa=Vx-1
a. f(g(1/2)) b. &(f(1/2)) 2. f) - Vxl, ga) = Vx-1 c. f(g) d. g(fG)) In Exercises 3 and 4, find the domains and ranges of f, g, f/g, and e. f(f(2)) f. e(g(2)) lf. g. FFW) h. (g(a) Se, Be aed In Exercises 7-10, write a formula f h dyad. ees EX um n Exercises 7-10, write a formula for f ° g ° h.
7. fi) " x -* 1, gx) =3x, h(x)24-x
Composites of Functions 8. f) 2 3x 44, g6)-2x— 1, ho) 2 x
5. If f(x) = x + 5 and g(x) = x? — 3, find the following.
1 1 a. f(g(0)) b. g(f(0) % f= Vat, gol tO c. f(gG)) d. g(fG)) 10. (9-3 $ 2 * 3t - Wo) = Vica
Let fx) =x-3, g(x) = Vx, h(x) = x, and j(x) = 2x. Express each of the functions in Exercises 11 and 12 as a composite involving one or more of f, g, h, and j.
ll. a. y ^ Vx - 3 b. y = 2Vx
e y = x d. y = 4x
e y= Vx-3) f y = (2x — 6 12.a. y 2x - 3 b. y = x3?
e y=xX? d. y=x-6
e y=2Vx-3 fi y=Vx - 3 13. Copy and complete the following table.
go) fœ (f° g)Q)
a x-7 Vx ?
b. x +2 3x ?
ORE Vx-5 vx -5
d: = 1 T 1 $
e. ? Ie i x
f. T ? x
14. Copy and complete the following table.
B(x) feo Fega) 1 a c1 |x] ? k=l X Ws x xctl Ce? Vx |x| d. Vx ? |x| 15. Evaluate each expression using the given table of values: x =2 -1 0 1 2 F(x) 1 0 | -3 1 | 2 g(x) 2 1 0| eu |0 a. f(gC- D) b. &(f(0) c. FFCD) d. g(gQ)) e. &(f(-2) f. f(g(1)) 16. Evaluate each expression using the functions T E c X —2=x<0 NO Re BO Eq. Gsen a. f(g(0) b. g(f(3)) c. g(gC- 1) d. f(f2) e. g(f(0)) f. f(g(1/2)
In Exercises 17 and 18, (a) write formulas for f ° g and g » f and find the (b) domain and (c) range of each.
17. f(x) = Vx + 1, gx) =1 18. f(x) = x, g(x) 2 1— Vx
1.2 Combining Functions; Shifting and Scaling Graphs 19
x—2 (F° g)(x) = x.
20. Let f(x) = 2x3 — 4. Find a function y = g(x) so that (fog)G) =x + 2.
Shifting Graphs 21. The accompanying figure shows the graph of y = —x? shifted to two new positions. Write equations for the new graphs.
19. Let f(x) — F * —. Find a function y — g(x) so that
Position (b)
22. The accompanying figure shows the graph of y — x? shifted to two new positions. Write equations for the new graphs. y
^ Position (a)
Position (b)
23. Match the equations listed in parts (a)-(d) to the graphs in the accompanying figure.
a. y= (x= 1-4 b. y-2(x-2» +2 c y-(x-2)» 4*2 d.y-(x-3»-2
y ^
Position 2 Position 1
1 2 Position 4
d, —4)
20 Chapter 1: Functions
24. The accompanying figure shows the graph of y = —x? shifted to four new positions. Write an equation for each new graph.
y ^
(1, 4)
Exercises 25-34 tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation.
25. x? + y! = 49 Down 3, left 2 26. x? + y = 25 Up3,left4 27. y=x° Left 1, down 1
28. y = x° Right 1, down 1 29. y= Vx Left 0.81 30. y - — Vx Right 3 31. y -2x - 7. Up7
32. y — Hx +1)+5 Down5S, right 1
33. y= 1/x Up Ll, right 1 34. y = 1/x? Left 2, down 1
Graph the functions in Exercises 35-54.
55. The accompanying figure shows the graph of a function f(x) with domain [0,2] and range [ 0, 1]. Find the domains and ranges of the following functions, and sketch their graphs.
lr y -f0) 0 » c a. f(x) +2 b. f(x) — 1 c. 2f(x) d. —f(x) e. f(x + 2) f. fx — 1) g f(—x) h. —fix+1)+1
56. The accompanying figure shows the graph of a function g(t) with domain [—4,0] and range [—3,0]. Find the domains and ranges of the following functions, and sketch their graphs.
y ^
—3r- a. g(t) b. —g(t) c. g(t) +3 d. 1 — g(t) e. g(t + 2) f. g(t — 2) g &g(1 — 1) h. —g(t — 4)
Vertical and Horizontal Scaling
Exercises 57—66 tell by what factor and direction the graphs of the given functions are to be stretched or compressed. Give an equation for the stretched or compressed graph.
35. y= Vx * 4 36. y= V9- x 37. y= |x- 2] 38. y= |1-x| - 1 39. y 21 Vx- I1 40. y21— Vx 4l. y = (x + 15 42. y= (x — 8y^ 43. y 2 1 — x? 44. y+ 4 = xi? 45. y= Vx — 1-1 46. y = (x + 292 + 1 Lll zd es 47. y= 48. y- 1-2 zd ao 49. y- (t2 50.y— 549 1 1 51. y= 52.y—-—-1 d m TU 53 y-L41 54. y =- —|— ida. UPC e TP
57. y = x2 — 1, stretched vertically by a factor of 3 58. y = x? — 1, compressed horizontally by a factor of 2
E 1 : 59. y - 1 6 E compressed vertically by a factor of 2 60. y= 14 i. stretched horizontally by a factor of 3
x
61. y= Vx + 1, compressed horizontally by a factor of 4 62. y= Vx + 1, stretched vertically by a factor of 3 63. y — V4 — x?, stretched horizontally by a factor of 2 64. y = V4 — x*, compressed vertically by a factor of 3 65. y = 1 — x°, compressed horizontally by a factor of 3 66. y = 1 — x°, stretched horizontally by a factor of 2
Graphing
In Exercises 67—74, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14—1.17 and applying an appropriate transformation.
67. y - - 2x € 1 68. y= 41-75
69. y (x — 1? +2 7. y=(1-x +2
1.3 Trigonometric Functions 21
Combining Functions
77. Assume that f is an even function, g is an odd function, and both f and g are defined on the entire real line (—00, oo). Which of the following (where defined) are even? odd?
a. fg b. f/g c. g/f d. P = ff e. g = gg f. fog g. gef h. fof i g°g
1
2
7. y—-53-1 fingens gl 73. y =—Wx 74. y = (2x5 75. Graph the function y = |x? — 1|.
76.
Graph the function y = V/|x|.
78.
79.
80.
Can a function be both even and odd? Give reasons for your answer.
(Continuation of Example 1.) Graph the functions f(x) — Vx and g(x) = V1 — x together with their (a) sum, (b) product, (c) two differences, (d) two quotients.
Let f(x) = x — 7 and g(x) = x?. Graph f and g together with
f^gandgof.
Ji A Trigonometric Functions
Cire le of 0”
FIGURE 1.36 The radian measure of the central angle A'CB' is the num- ber 0 = s/r. For a unit circle of radius r = 1, 0 is the length of arc AB that central angle ACB cuts from the unit circle.
This section reviews radian measure and the basic trigonometric functions.
Angles
Angles are measured in degrees or radians. The number of radians in the central angle A'CB' within a circle of radius r is defined as the number of “radius units" contained in the arc s subtended by that central angle. If we denote this central angle by 0 when mea- sured in radians, this means that 0 = s/r (Figure 1.36), or
s — r0 (0 in radians). (1)
If the circle is a unit circle having radius r — 1, then from Figure 1.36 and Equation (1), we see that the central angle 0 measured in radians is just the length of the arc that the angle cuts from the unit circle. Since one complete revolution of the unit circle is 360? or 2a radians, we have
T radians = 180° (2) and
180 BM
l radian = ^5 (~ 57.3) degrees or ] degree = 180 (70.017) radians.
Table 1.1 shows the equivalence between degree and radian measures for some basic angles.
TABLE 1.1 Angles measured in degrees and radians
—90 -45 0 30 45 60 90 120 135 150 180 270 360
=T T 0
Degrees —180 -—135 0 (radians) -T ZÓ
7 2 4 6
22 Chapter 1: Functions
hypotenuse 0 adjacent
h sin 0 = OPP csc 0 = YP hyp opp
dj h cos 0 = = sec 9 = x di tan 9 = SPP cot 0 = 2d adj opp
opposite
FIGURE 1.39 Trigonometric
ratios of an acute angle.
FIGURE 1.40 The trigonometric functions of a general angle 0 are
defined in terms of x, y, and r.
An angle in the xy-plane is said to be in standard position if its vertex lies at the ori- gin and its initial ray lies along the positive x-axis (Figure 1.37). Angles measured counter- clockwise from the positive x-axis are assigned positive measures; angles measured clock- wise are assigned negative measures.
y y ^
Terminal ray
Initial ray
>x
Positive Initial ray
Negative i measure
Terminal measure
ray
> X
FIGURE 1.37 Angles in standard position in the xy-plane.
Angles describing counterclockwise rotations can go arbitrarily far beyond 27 radi- ans or 360°. Similarly, angles describing clockwise rotations can have negative measures of all sizes (Figure 1.38).
y
y y ) 3m 5m 2 x >x >x >x 23m oa 4 4
FIGURE 1.38 Nonzero radian measures can be positive or negative and can go beyond 27.
Angle Convention: Use Radians From now on, in this book it is assumed that all angles are measured in radians unless degrees or some other unit is stated explicitly. When we talk about the angle 77/3, we mean 7/3 radians (which is 60°), not 7/3 degrees. We use radians because it simplifies many of the operations in calculus, and some results we will obtain involving the trigonometric functions are not true when angles are measured in degrees.
The Six Basic Trigonometric Functions
You are probably familiar with defining the trigonometric functions of an acute angle in terms of the sides of a right triangle (Figure 1.39). We extend this definition to obtuse and negative angles by first placing the angle in standard position in a circle of radius r. We then define the trigonometric functions in terms of the coordinates of the point P(x, y) where the angle's terminal ray intersects the circle (Figure 1.40).
5 . y r sine: sinô = 7 cosecant: cscÓ = y
. X r cosine: cos0 = 7 secant: sec@ = x . = y e = x tangent: tan? =% cotangent: cot 6 = y
These extended definitions agree with the right-triangle definitions when the angle is acute. Notice also that whenever the quotients are defined,
. sind |. ol tan 0 = cos 0 cot 0 = an sec 0 = l aoo end
os 0 sin 0
v2 x p
1 1
FIGURE 1.41 Radian angles and side lengths of two common triangles.
S A sin pos all pos >x T € tan pos cos pos
FIGURE 1.42 The CAST rule, remembered by the statement “Calculus Activates Student Thinking,” tells which trigonometric functions
are positive in each quadrant.
1.3 Trigonometric Functions 23
As you can see, tan 0 and sec 0 are not defined if x = cos 0 = 0. This means they are not defined if 0 is 7/2, £37/2,.... Similarly, cot 0 and csc 0 are not defined for values of 0 for which y = 0, namely 0 = 0, tz, t2m,....
The exact values of these trigonometric ratios for some angles can be read from the triangles in Figure 1.41. For instance,
du =- sin 7 = 4 LEE 4 v2 6 2 3 2
LR M cos Z = V3 cage! 4 v2 6 2 3 2 To m... m es
tan 47 1 tan 6 A tan 3 V3
The CAST rule (Figure 1.42) is useful for remembering when the basic trigonometric func- tions are positive or negative. For instance, from the triangle in Figure 1.43, we see that
sin 27 = V3 cos 22. = — 5, tan 22 = — V3.
3 2
FIGURE 1.43 The triangle for calculating the sine and cosine of 27/3 radians. The side lengths come from the geometry of right triangles.
Using a similar method we determined the values of sin 0, cos 0, and tan 0 shown in Table 1.2.
TABLE 1.2 Values of sin 0, cos 6, and tan 0 for selected values of 0
Degrees —180 -135 -90 -45 0 30 45 60 90 120 135 150 180 270 360 0 (radians | —7 — e FS 0 k a 3 2 i — pe 7 — 27 sin 0 0 -— n o | io m 1 uS l a 6 cos 0 -1 = 0 1 NS M Io -1 eS - 0 1 tan 0 0 1 -1 0 32 L WE -V3 -1 — 0 0
24 Chapter 1: Functions
Periodicity and Graphs of the Trigonometric Functions
When an angle of measure 0 and an angle of measure 0 + 277 are in standard position, their terminal rays coincide. The two angles therefore have the same trigonometric func- tion values: sin(0 + 27) = sinO, tan(0 + 27) = tan0, and so on. Similarly, cos(0 — 27) = cos 0, sin(@ — 27) = sin 0, and so on. We describe this repeating behav- ior by saying that the six basic trigonometric functions are periodic.
Periods of Trigonometric Functions
Period 7: tan(x + 7) = tanx cot(x + T) = cot x DEFINITION A function f(x) is periodic if there is a positive number p such that
f(x + p) = f(x) for every value of x. The smallest such value of p is the period of f.
Period 27: sin(x + 27) = sinx cos(x + 2v) = cosx sec(x + 2m) = sec x csc(x + 2m) = esc x
When we graph trigonometric functions in the coordinate plane, we usually denote the independent variable by x instead of 0. Figure 1.44 shows that the tangent and cotangent functions have period p = 7, and the other four functions have period 277. Also, the sym- metries in these graphs reveal that the cosine and secant functions are even and the other four functions are odd (although this does not prove those results).
y
y-tanx Even l y=sinx cos(—x) = cos x >x 3pm fm 0 m 3
sec (~x) = sec x
Domain: ~% < x < c Domain: ~% < x < c Domain: x ES mai a,
: = ] a jx
Odd Range: layl Range Tayal Range: -w<y<o
Period: 27 Period: 27 Period:
i (a) (b) T (o sin (~x) = —sin x " y tan (~x) = —tan x E y=cscx y=cotx csc (~x) = —csc x U 1 1 cot(—x) = —cot x - Sž -a T| T 37 —id T a 2 (^ 2 2 Domain: x #25 zE [| us Domain: x # 0, +7, +27,... Domain: x # 0, Em, +27,... Range: y -loryzl Range: —9 «y«o
Range: y= k a 1
in lge; Period: 27 Period: 2m Period: 7
(d) (e) (f)
FIGURE 1.44 Graphs of the six basic trigonometric functions using radian measure. The shading for each trigonometric function indicates its periodicity.
PEO a Trigonometric Identities
i The coordinates of any point P(x, y) in the plane can be expressed in terms of the point's
distance r from the origin and the angle 0 that ray OP makes with the positive x-axis (Fig- i 6| ure 1.40). Since x/r — cos 0 and y/r — sin 0, we have [cos 6| A 1 >x x = rcos 6, y = rsin 0. When r = 1 we can apply the Pythagorean theorem to the reference right triangle in Figure 1.45 and obtain the equation FIGURE 1.45 The reference cos? 0 + sin? 0 = 1. (3)
triangle for a general angle 0.
1.3 Trigonometric Functions 25
This equation, true for all values of 0, is the most frequently used identity in trigonometry. Dividing this identity in turn by cos? 0 and sin? 0 gives
1 + tan? 0 = sec? 0 1 + cot? 0 = csc” 0
The following formulas hold for all angles A and B (Exercise 58).
Addition Formulas
cos(A + B) = cos Acos B — sin Asin B
4 sin(A + B) = sin Acos B + cos Asin B a
There are similar formulas for cos(A — B) and sin(A — B) (Exercises 35 and 36). All the trigonometric identities needed in this book derive from Equations (3) and (4). For example, substituting 0 for both A and B in the addition formulas gives
Double-Angle Formulas
cos 20 = cos? 0 — sin? 0
. Ee (5) sin 20 — 2sin 0 cos 0
Additional formulas come from combining the equations cos? 0 + sin? 0 = 1, cos? 0 — sin? 0 = cos 20.
We add the two equations to get 2cos? 0 = 1 + cos 26 and subtract the second from the first to get 2sin?@ = 1 — cos 20. This results in the following identities, which are useful in integral calculus.
Half-Angle Formulas cos? — a (6) sin? = 1— 50826 U
The Law of Cosines
If a, b, and c are sides of a triangle ABC and if 0 is the angle opposite c, then
c = @ + b — 2abcos 0. (8)
This equation is called the law of cosines.
26 Chapter 1: Functions
y
B(a cos 0, a sin 0)
FIGURE 1.46 The square of the distance between A and B gives the law of cosines.
><
cos 0
FIGURE 1.47 From the geometry of this figure, drawn for 0 > 0, we get the inequality sin? 0 + (1 — cos 0 x 8°.
We can see why the law holds if we introduce coordinate axes with the origin at C and the positive x-axis along one side of the triangle, as in Figure 1.46. The coordinates of A are (b, 0); the coordinates of B are (acos 0, asin 0). The square of the distance between A and B is therefore c? = (acos 0 — by + (asin 0 a’ (cos? 0 + sin? 0) + D? — 2ab cos 0 i — € —À
1 =a’ + D? — 2abcos 0.
The law of cosines generalizes the Pythagorean theorem. If 0 — 7/2, then cos 0 — 0 and c? = a? + b.
Two Special Inequalities
For any angle 0 measured in radians, the sine and cosine functions satisfy
—|6| = sino - |ü] and -|0| = 1 — cos = |o].
To establish these inequalities, we picture 0 as a nonzero angle in standard position (Figure 1.47). The circle in the figure is a unit circle, so lel equals the length of the circular arc AP. The length of line segment AP is therefore less than |60].
Triangle APQ is a right triangle with sides of length
QP = |sinó, | AQ = 1 — cos 6. From the Pythagorean theorem and the fact that AP < [7 , we get sin? 0 + (1 — cos 0 = (AP) x 0°. (9)
The terms on the left-hand side of Equation (9) are both positive, so each is smaller than their sum and hence is less than or equal to 67:
sinü < 9? and (1 — cos 0 x 0°. By taking square roots, this is equivalent to saying that [sin 0| = |6| and |1 — cos 0| = |6
^
SO —|6| < sino - |ð] and -|0| <1 — cos = |o].
These inequalities will be useful in the next chapter.
Transformations of Trigonometric Graphs
The rules for shifting, stretching, compressing, and reflecting the graph of a function sum- marized in the following diagram apply to the trigonometric functions we have discussed in this section.
Vertical stretch or compression; Vertical shift reflection about y = d if negative P y 7 af(b(x + c) +d
Horizontal stretch or compression; Horizontal shift reflection about x = —c if negative
27
1.3 Trigonometric Functions
The transformation rules applied to the sine function give the general sine function
or sinusoid formula
where |A| is the amplitude,
f(x) = A sin (22 (x — c) + D,
B| is the period, C is the horizontal shift, and D is the vertical
shift. A graphical interpretation of the various terms is given below.
shift (C)
Horizontal
|
Vertical shift (D)
y=AsnZa-c)+D
Amplitude (A wee This axis is the
line y = D.
I——This distance is —>|
the period (B).
Exercises
Radians and Degrees
1.
2.
On a circle of radius 10 m, how long is an arc that subtends a cen- tral angle of (a) 47/5 radians? (b) 110°?
A central angle in a circle of radius 8 is subtended by an arc of length 107. Find the angle’s radian and degree measures.
. You want to make an 80? angle by marking an arc on the perime-
ter of a 12-in.-diameter disk and drawing lines from the ends of the arc to the disk's center. To the nearest tenth of an inch, how long should the arc be?
. If you roll a 1-m-diameter wheel forward 30 cm over level
ground, through what angle will the wheel turn? Answer in radi- ans (to the nearest tenth) and degrees (to the nearest degree).
Evaluating Trigonometric Functions
5:
Copy and complete the following table of function values. If the function is undefined at a given angle, enter “UND.” Do not use a calculator or tables.
0 -r — 27/3 0 a /2 32/4
sin 0 cos 0 tan 0 cot 0 sec 0 csc 0
. Copy and complete the following table of function values. If the
function is undefined at a given angle, enter “UND.” Do not use a calculator or tables.
0 -37/2 —m[3 —7a/6 «[A | 5m[6
In Exercises 7-12, one of sin x, cos x, and tan x is given. Find the other two if x lies in the specified interval.
í a2 T = T 7. sinx = 5° re|Za] 8. tanx = 2, re] 0.3 | 9 zd 0 10 = . COSX= 3, XE PE . COSX=— 73, X€| 5,7 EA! 3 F =. p 3T 11. tanx = y E 12. sinx = > velm 2E]
Graphing Trigonometric Functions Graph the functions in Exercises 13—22. What is the period of each function?
13. sin 2x 14. sin(x/2) 15. cos zx 16. cos Es 17. —sin $ 18. —cos 2zx
19.
20. sin (: + z)
28 Chapter 1: Functions
21. sin(x = z) t1 22. cos(x + 22) = 2,
Graph the functions in Exercises 23—26 in the ts-plane (t-axis horizon- tal, s-axis vertical). What is the period of each function? What sym- metries do the graphs have?
23. s — cot2t 24. s — —tan mt
= Tt É 25. s = «( 2) 26. s — csc (5)
27. a. Graph y = cosx and y = secx together for —37/2 = x = 37/2. Comment on the behavior of sec x in relation to the signs and values of cos x.
ll ll
b. Graph y = sin x and y = csc x together for -m = x = 277. Comment on the behavior of csc x in relation to the signs and values of sin x.
28. Graph y = tan x and y = cot x together for —7 = x = 7. Com- ment on the behavior of cot x in relation to the signs and values of tan x.
29. Graph y = sin x and y = | sin x] together. What are the domain and range of | sin x |?
30. Graph y = sin x and y = [sin x] together. What are the domain and range of | sin x]?
Using the Addition Formulas Use the addition formulas to derive the identities in Exercises 31—36.
31. cos (: = z) = sinx 32. cos (: + z) = —sinx
2 33. sin (: + z) = cos x 34. sin (: — z) = —cos x
35. cos(A — B) = cos Acos B + sin Asin B (Exercise 57 provides a different derivation.)
36. sin(A — B) = sin Acos B — cos Asin B
37. What happens if you take B — A in the trigonometric identity cos(A — B) = cos Acos B + sin Asin B? Does the result agree with something you already know?
38. What happens if you take B = 27 in the addition formulas? Do the results agree with something you already know?
In Exercises 39-42, express the given quantity in terms of sin x and cos x.
39. cos(m + x) 40. sin(2z — x)
. [3T 3 41. sin (i — x) 42. cos (z + x)
. Tm [mom 43. Evaluate sin 12 as sin E + z),
liz T , 27 44. Evaluate cos 12 3s cos( tS )
T . ST 45. Evaluate cos 17 46. Evaluate sin 12
Using the Half-Angle Formulas Find the function values in Exercises 47—50.
27 257 47. cos 8 48. cos 12 49. sin? = 50. sin? 37
12 8
Solving Trigonometric Equations For Exercises 51-54, solve for the angle 0, where 0 = 0 < 277.
51. sin? 0 = i 52. sin? 0 = cos? 0
53. sin20 — cos0 = 0 54. cos 20 + cos0 = 0
Theory and Examples 55. The tangent sum formula The standard formula for the tan- gent of the sum of two angles is
_ tanA + tan B AUCKES ] — tan Atan B' Derive the formula. 56. (Continuation of Exercise 55.) Derive a formula for tan(A — B).
57. Apply the law of cosines to the triangle in the accompanying fig- ure to derive the formula for cos(A — B).
>x
58. a. Apply the formula for cos(A — B) to the identity sin 0 =
cos (5 = o) to obtain the addition formula for sin (A + B).
b. Derive the formula for cos (A + B) by substituting —B for B in the formula for cos (A — B) from Exercise 35.
59. A triangle has sides a = 2 and b = 3 and angle C = 60°. Find the length of side c.
60. A triangle has sides a = 2 and b = 3 and angle C = 40°. Find the length of side c.
61. The law of sines The law of sines says that if a, b, and c are the sides opposite the angles A, B, and C in a triangle, then
snA sinB sinC
a b c
Use the accompanying figures and the identity sin(z — 0) = sin 0, if required, to derive the law.
A A
B C B c
62. A triangle has sides a — 2 and b — 3 and angle C — 60? (as in Exercise 59). Find the sine of angle B using the law of sines.
63. A triangle has side c = 2 and angles A = 7/4 and B = 7/3. Find the length a of the side opposite A.
64. The approximation sin x ~ x It is often useful to know that, when x is measured in radians, sin x ~ x for numerically small val- ues of x. In Section 3.11, we will see why the approximation holds. The approximation error is less than 1 in 5000 if |x| < 0.1.
a. With your grapher in radian mode, graph y = sin x and y = x together in a viewing window about the origin. What do you see happening as x nears the origin?
b. With your grapher in degree mode, graph y = sin x and y = x together about the origin again. How is the picture dif- ferent from the one obtained with radian mode?
General Sine Curves For
f(x) - asin( 22 (x o) +- D,
identify A, B, C, and D for the sine functions in Exercises 65-68 and sketch their graphs.
65. y = 2sin(x + 7) - 1 66. y — sin (mx T) + : ES an ma. = Ly ant 67. y 2sin( Zr) DE. 68. y — PURA L0
COMPUTER EXPLORATIONS In Exercises 69—72, you will explore graphically the general sine function
f(x) - Asin( 22 (x o) + D
as you change the values of the constants A, B, C, and D. Use a CAS or computer grapher to perform the steps in the exercises.
Ji A Graphing with Software
1.4 Graphing with Software 29
69. The period B Set the constants A = 3,C = D — 0.
a. Plot f(x) for the values B = 1, 3, 277, 57 over the interval —4m = x = 4m. Describe what happens to the graph of the general sine function as the period increases.
b. What happens to the graph for negative values of B? Try it with B = —3 and B = —2m. 70. The horizontal shiftC Settheconstants A = 3, B = 6, D = 0.
a. Plot f(x) for the values C = 0, 1, and 2 over the interval —4m = x X 4r. Describe what happens to the graph of the general sine function as C increases through positive values.
b. What happens to the graph for negative values of C?
c. What smallest positive value should be assigned to C so the graph exhibits no horizontal shift? Confirm your answer with a plot.
71. The vertical shift D Set the constants A = 3, B = 6, C = 0.
a. Plot f(x) for the values D = 0, 1, and 3 over the interval —4m = x = 4m. Describe what happens to the graph of the general sine function as D increases through positive values.
b. What happens to the graph for negative values of D? 72. The amplitude A Set the constants B = 6, C = D = 0.
a. Describe what happens to the graph of the general sine func- tion as A increases through positive values. Confirm your answer by plotting f(x) for the values A = 1, 5, and 9.
b. What happens to the graph for negative values of A?
Today a number of hardware devices, including computers, calculators, and smartphones, have graphing applications based on software that enables us to graph very complicated functions with high precision. Many of these functions could not otherwise be easily graphed. However, some care must be taken when using such graphing software, and in this section we address some of the issues that may be involved. In Chapter 4 we will see how calculus helps us determine that we are accurately viewing all the important features of a function's graph.
Graphing Windows
When using software for graphing, a portion of the graph is displayed in a display or viewing window. Depending on the software, the default window may give an incomplete or mislead- ing picture of the graph. We use the term square window when the units or scales used on both axes are the same. This term does not mean that the display window itself is square (usually it is rectangular), but instead it means that the x-unit is the same length as the y-unit.
When a graph is displayed in the default mode, the x-unit may differ from the y-unit of scaling in order to capture essential features of the graph. This difference in scaling can cause visual distortions that may lead to erroneous interpretations of the function's behavior.
30
Chapter 1: Functions
Some graphing software allows us to set the viewing window by specifying one or both of the intervals, a = x = b and c = y S d, and it may allow for equalizing the scales used for the axes as well. The software selects equally spaced x-values in [ a, b] and then plots the points (x, f(x)). A point is plotted if and only if x lies in the domain of the function and f(x) lies within the interval [ c, d]. A short line segment is then drawn between each plotted point and its next neighboring point. We now give illustrative examples of some common problems that may occur with this procedure.
EXAMPLE 1 Graph the function f(x) = x? — 7x? + 28 in each of the following display or viewing windows:
(a) [710,10] by [-10, 10]. (b) [-4,4] by [-50, 10] (©) [-4, 10] by [—60, 60]
Solution
(a) We select a = —10, b = 10, c = —10, and d = 10 to specify the interval of x-values and the range of y-values for the window. The resulting graph is shown in Figure 1.48a. It appears that the window is cutting off the bottom part of the graph and that the interval of x-values is too large. Let's try the next window.
10
"m A NU ' 10 Aili
(a)
—50 (b)
FIGURE 1.48 The graph of f(x) = x? — 7x? + 28 in different viewing windows. Selecting a window that gives a clear
picture of a graph is often a trial-and-error process (Example 1). The default window used by the software may automatically
display the graph in (c).
(b) We see some new features of the graph (Figure 1.48b), but the top is missing and we need to view more to the right of x — 4 as well. The next window should help.
(c) Figure 1.48c shows the graph in this new viewing window. Observe that we get a more complete picture of the graph in this window, and it is a reasonable graph of a third-degree polynomial. a
EXAMPLE 2 When a graph is displayed, the x-unit may differ from the y-unit, as in the graphs shown in Figures 1.48b and 1.48c. The result is distortion in the picture, which may be misleading. The display window can be made square by compressing or stretching the units on one axis to match the scale on the other, giving the true graph. Many software systems have built-in options to make the window “square.” If yours does not, you may have to bring to your viewing some foreknowledge of the true picture.
Figure 1.49a shows the graphs of the perpendicular lines y = x and y = —x + 3V2, together with the semicircle y = V9 — x’, in a nonsquare [—4,4] by [—6, 8] display window. Notice the distortion. The lines do not appear to be perpendicular, and the semi- circle appears to be elliptical in shape.
Figure 1.49b shows the graphs of the same functions in a square window in which the x-units are scaled to be the same as the y-units. Notice that the scaling on the x-axis for Figure 1.49a has been compressed in Figure 1.49b to make the window square. Figure 1.49c gives an enlarged view of Figure 1.49b with a square [ —3, 3] by [0,4] window. B
1.4 Graphing with Software 31
—6 (a)
FIGURE 1.49 Graphs of the perpendicular lines y = x and y = —x + 3 V2 and of the semicircle y = V9 — x? appear distorted (a) in a nonsquare window, but clear (b) and (c) in square windows (Example 2).
0 (c)
Some software may not provide options for the views in (b) or (c).
If the denominator of a rational function is zero at some x-value within the viewing window, graphing software may produce a steep near-vertical line segment from the top to the bottom of the window. Example 3 illustrates steep line segments.
Sometimes the graph of a trigonometric function oscillates very rapidly. When graph- ing software plots the points of the graph and connects them, many of the maximum and minimum points are actually missed. The resulting graph is then very misleading.
EXAMPLE 3 Graph the function f(x) = sin 100x.
Solution Figure 1.50a shows the graph of f in the viewing window [—12, 12] by [—1, 1]. We see that the graph looks very strange because the sine curve should oscillate periodically between —1 and 1. This behavior is not exhibited in Figure 1.50a. We might experiment with a smaller viewing window, say [—6,6] by [—1, 1 ], but the graph is not better (Figure 1.50b). The difficulty is that the period of the trigonometric function y = sin 100x is very small (277/100 ~ 0.063). If we choose the much smaller viewing window [—0.1,0.1] by [-1.1] we get the graph shown in Figure 1.50c. This graph reveals the expected oscillations of a sine curve. El
|
CUM | <A A TOIT
=] =] (a) (b)
FIGURE 1.50 Graphs of the function y = sin 100x in three viewing windows. Because the period is 277/100 ~ 0.063, the smaller window in (c) best displays the true aspects of this rapidly oscillating function (Example 3).
EXAMPLE 4 Graph the function y = cos x + saosin 200x.
Solution In the viewing window [—6,6] by [—1, 1] the graph appears much like the cosine function with some very small sharp wiggles on it (Figure 1.51a). We get a better look when we significantly reduce the window to [—0.2, 0.2] by [ 0.97, 1.01], obtaining the graph in Figure 1.51b. We now see the small but rapid oscillations of the second term, (1/200) sin 200x, added to the comparatively larger values of the cosine curve. |
32
Chapter 1: Functions
(b)
FIGURE 1.51 In (b) we see a close-up view of the function
y = cosx + son 200x graphed in (a). The term cos x clearly dominates
the second term sin 200x, which produces the rapid oscillations along the
ES ' 200 cosine curve. Both views are needed for a clear idea of the graph (Example 4).
Obtaining a Complete Graph
Some graphing software will not display the portion of a graph for f(x) when x « 0. Usu- ally that happens because of the algorithm the software is using to calculate the function values. Sometimes we can obtain the complete graph by defining the formula for the func- tion in a different way, as illustrated in the next example.
EXAMPLE 5 Graph the function y = x3.
Solution Some graphing software displays the graph shown in Figure 1.52a. When we compare it with the graph of y = x!? = Wx in Figure 1.17, we see that the left branch for x < 0 is missing. The reason the graphs differ is that the software algorithm calculates x!/? as e/h, Since the logarithmic function is not defined for negative values of x, the software can produce only the right branch, where x > 0. (Logarithmic and exponential functions are discussed in Chapter 7.)
(a) (b)
FIGURE 1.52 The graph of y = x!? is missing the left branch in (a). In (b) we
graph the function f(x) — c. [x|!, obtaining both branches. (See Example 5.)
Ix To obtain the full picture showing both branches, we can graph the function
Fx) = i |x.
|x| This function equals x!/? except at x = 0 (where f is undefined, although 0'? = 0). A graph of f is displayed in Figure 1.52b. Oo Capturing the Trend of Collected Data
We have pointed out that applied scientists and analysts often collect data to study a par- ticular issue or phenomenon of interest. If there is no known principle or physical law
TABLE 1.3 Tuition and fees at the University of California
Year,x Cost, y 1990 1,820 1995 4,166 2000 3,964 2005 6,802 2010 11,287 2011 13,218
1.4 Graphing with Software 33
relating the independent and dependent variables, the data can be plotted in a scatterplot to help find a curve that captures the overall trend of the data points. This process is called regression analysis, and the curve is called a regression curve.
Many graphing utilities have software that finds the regression curve for a particular type of curve (such as a straight line, a quadratic or other polynomial, or a power curve) and then superimposes the graph of the found curve over the scatterplot. This procedure results in a useful graphical visualization, and often the formula produced for the regression curve can be used to make reasonable estimates or to help explain the issue of interest.
One common method, known as least squares, finds the desired regression curve by minimizing the sum of the squares of the vertical distances between the data points and the curve. The least squares method is an optimization problem. (In Section 14.7 exercises, we discuss how the regression curve is calculated when fitting a straight line to the data.) Here we present a few examples illustrating the technique by using available software to find the curve. Keep in mind that different software packages may have different ways of enter- ing the data points, and different output features as well.
EXAMPLE 6 Table 1.3 shows the annual cost of tuition and fees for a full-time stu- dent attending the University of California for the years 1990—2011. The data in the list cite the beginning of the academic year when the corresponding cost was in effect. Use the table to find a regression line capturing the trend of the data points, and use the line to estimate the cost for academic year 2018-19.
Solution We use regression software that allows for fitting a straight line, and we enter the data from the table to obtain the formula
y = 506.25x — 1.0066 - 106,
where x represents the year and y the cost that took effect that year. Figure 1.53 displays the scatterplot of the data together with the graph of this regression line. From the equation of the line, we find that for x — 2018,
y = 506.25(2018) — 1.0066 - 106 = 15,013
is the estimated cost (rounded to the nearest dollar) for the academic year 2018-19. The last two data points rise above the trend line in the figure, so this estimate may turn out to be low. a
y
14,000 12,000 10,000
8,000
6,000
4,000
2,000 *
0 1985 1990 1995 2000 2005 2010 2015 on
FIGURE 1.53 Scatterplot and regression line for the University of California tuition and fees from Table 1.3 (Example 6).
EXAMPLE 7 The Centers for Disease Control and Prevention recorded the deaths from tuberculosis in the United States for 1970-2006. We list the data in Table 1.4 for 5-year intervals. Find linear and quadratic regression curves capturing the trend of the data points. Which curve might be the better predictor?
34
Chapter 1: Functions
TABLE 1.4 U.S. deaths from
tuberculosis
Year, x Deaths, y 1970 5,217 1975 3:333 1980 1,978 1985 1,752 1990 1,810 1995 1,336 2000 TI6 2005 648
Exercises 1.4
Choosing a Viewing Window In Exercises 1—4, use graphing software to determine which of the given viewing windows displays the most appropriate graph of the
specified function. 1. f(x) = xt — 7€
+ 6x
a. [-1,1] by [-1,1]
c. [—10, 10] by [—10, 10] 2. f(x) = 8 — 4x? — 4x + 16
a. [-1, 1] by [-5,5]
c. [75,5] by [-10,20] 3. f) =5 + 12x — o
a. [-1, 1] by [-1, 1]
c. [74,4] by [-20,20] 4 fx) = V5St4Ax- x
a. [72,2] by [-2.2]
c. [73,7] by [0,10]
Finding a Viewing Window
e
Solution Using regression software that allows us to fit a straight line as well as a qua- dratic curve, we enter the data to obtain the formulas
y = 2.2279 + 10? — 111.04x, line fit and 1451 , 3,483,953 464,757,147 "T y= 359% 2310 2 ab 28 : quadratic fit
where x represents the year and y represents the number of deaths that occurred. A scat- terplot of the data, together with the two trend curves, is displayed in Figure 1.54. In look- ing at the figure, it would appear that the quadratic curve most closely captures the trend of the data, except for the years 1990 and 1995, and would make the better predictor. How- ever, the quadratic seems to have a minimum value near the year 2000, rising upward thereafter, so it would probably not be a useful tool for making good estimates in the years beyond 2010. This example illustrates the danger of using a regression curve to predict values beyond the range of the data used to construct the curve. a
6,000 5,000 4,000 3,000 2,000 1,000
9 1970 1980 1990 2000 2010 da FIGURE 1.54 Scatterplot with the regression line and quadratic curves for tuberculosis deaths in the United States, based on Table 1.4 (Example 7).
should give a picture of the overall behavior of the function. There is more than one choice, but incorrect choices can miss important aspects of the function.
3 2 5. f(x) = x* — 40 + 15 aiam -. me
In Exercises 5—30, find an appropriate graphing software viewing win- dow for the given function and use it to display its graph. The window
2 [—2, 2] by [75.5] 7. f(x) = — 5x* + 10 8. f(x) = 43 — x* [—5, 5] by [—25, 15] 9. f(x) = x V0—- x 10. f(x) = x6 — x?) 1l. y = 2x — 3x73 12. y = xP — 8) [73,3] by [-10, 10] 13. y = 5375 — 2x 14. y = x5 — x) [-20, 20] by [—100, 100] 15. y= |e 1| 16. y = |x’? — x| +3 PEE, [-5,5] by [-10, 10] MEE MEN IE d s 19. fœ = £2 20. fœ = É—1 [74,5] by [715,25] Fu : ed g . g Anl 8 [72,6] by [71,4] 21. f(x) = 2x6 22. f = a9 [-10, 10] by [—10, 10] , d _ 6x 15x + 6 a ud. aket o ?A$9-1— 25. y = sin 250x 26. y = 3 cos 60x
= X c aal 27. y= (45) 28. y ssin( 5) B 1. E 1 29. y 2- x * ToS” 30x 30. y= x + 50995 100x
Use graphing software to graph the functions specified in Exercises 31—36.
Select a viewing window that reveals the key features of the function.
31. Graph the lower half of the circle defined by the equation x? + 2x =4+4 4y — y.
32. Graph the upper branch of the hyperbola y? — 16x? = 1
33. Graph four periods of the function f(x) = — tan 2x.
X
atl.
34. Graph two periods of the function f(x) = 3 cot
35. Graph the function f(x) = sin 2x + cos 3x. 36. Graph the function f(x) = sin’ x.
Regression Lines or Quadratic Curve Fits Use a graphing utility to find the regression curves specified in Exer- cises 37-42.
37. Weight of males The table shows the average weight for men of medium frame based on height as reported by the Metropolitan Life Insurance Company (1983).
Height (in. Weight (Ib) | Height (in.) Weight (Ib)
62 136 70 157
63 138 71 160
64 141 72 163.5
65 141.5 73 167
66 145 74 171
67 148 75 174.5
68 151 76 179
69 154
a. Make a scatterplot of the data.
b. Find and plot a regression line, and superimpose the line on the scatterplot.
€. Does the regression line reasonably capture the trend of the data? What weight would you predict for a male of height 6'7"?
38. Federal minimum wage The federal minimum hourly wage rates have increased over the years. The table shows the rates at the year in which they first took effect, as reported by the U.S. Department of Labor.
Year Wage ($) Year Wage ($) 1978 2.65 1996 4.75 1979 2.90 1997 5.15 1980 3.10 2007 5.85 1981 3:35 2008 6.55 1990 3.80 2009 7.25 1991 4.25
a. Make a scatterplot of the data.
b. Find and plot a regression line, and superimpose the line on the scatterplot.
c. What do you estimate as the minimum wage for the year 2018?
1.4 Graphing with Software 35
39. Median home price The median price of single-family homes in the United States increased quite consistently during the years 1976-2000. Then a housing “bubble” occurred for the years 2001-2010, in which prices first rose dramatically for 6 years and then dropped in a steep "crash" over the next 4 years, causing considerable turmoil in the U.S. economy. The table shows some of the data as reported by the National Association of Realtors.
Year Price ($) Year Price ($) 1976 37400 2000 122600 1980 56250 2002 150000 1984 66500 2004 187500 1988 87500 2006 247500 1992 95800 2008 183300 1996 104200 2010 162500
a. Make a scatterplot of the data.
b. Find and plot the regression line for the years 1976-2002, and superimpose the line on the scatterplot in part (a).
c. How would you interpret the meaning of a data point in the housing "bubble"?
40. Average energy prices The table shows the average residential and transportation prices for energy consumption in the United States for the years 2000-2008, as reported by the U.S. Depart- ment of Energy. The prices are given as dollars paid for one mil- lion BTU (British thermal units) of consumption.
Year Residential ($) ^ Transportation ($)
2000 15 10 2001 16 10 2002 15 9 2003 16 11 2004 18 13 2005 19 16 2006 21 19 2007 21 20 2008 23 25
a. Make a scatterplot of the data sets.
b. Find and plot a regression line for each set of data points, and superimpose the lines on their scatterplots.
c. What do you estimate as the average energy price for resi- dential and transportation use for a million BTU in year 2017?
d. In looking at the trend lines, what do you conclude about the rising costs of energy across the two sectors of usage?
41. Global annual mean surface air temperature A NASA God- dard Institute for Space Studies report gives the annual global mean land-ocean temperature index for the years 1880 to the present. The index number is the difference between the mean temperature over the base years 1951—1980 and the actual tem- perature for the year recorded. For the recorded year, a positive index is the number of degrees Celsius above the base; a negative index is the number below the base. The table lists the index for the years 1940-2010 in 5-year intervals, reported in the NASA data set.
36
Chapter 1: Functions
Year Index (^C) Year Index (^C) 1940 0.04 1980 0.20 1945 0.06 1985 0.05 1950 —0.16 1990 0.36 1955 —0.11 1995 0.39 1960 —0.01 2000 0.35 1965 —0.12 2005 0.62 1970 0.03 2010 0.63 1975 —0.04
a. Make a scatterplot of the data.
b. Find and plot a regression line, and superimpose the line on the scatterplot.
c. Find and plot a quadratic curve that captures the trend of the data, and superimpose the curve on the scatterplot.
42.
Growth of yeast cells The table shows the amount of yeast cells (measured as biomass) growing over a 7-hour period in a nutrient, as recorded by R. Pearl (1927) during a well-known bio- logical experiment.
Hour 0 1 2 3 4 5 6 7
Biomass 9.6 18.3 29.0 47.2 71.1 119.1 174.6 257.3
a. Make a scatterplot of the data.
b. Find and plot a regression quadratic, and superimpose the quadratic curve on the scatterplot.
c. What do you estimate as the biomass of yeast in the nutrient after 11 hours?
d. Do you think the quadratic curve would provide a good estimate of the biomass after 18 hours? Give reasons for your answer.
Chapter du Questions to Guide Your Review
1.
What is a function? What is its domain? Its range? What is an arrow diagram for a function? Give examples.
. What is the graph of a real-valued function of a real variable?
What is the vertical line test?
. What is a piecewise-defined function? Give examples.
. What are the important types of functions frequently encountered
in calculus? Give an example of each type.
. What is meant by an increasing function? A decreasing function?
Give an example of each.
. What is an even function? An odd function? What symmetry
properties do the graphs of such functions have? What advantage can we take of this? Give an example of a function that is neither even nor odd.
. If f and g are real-valued functions, how are the domains of
f + g.f-s$ fg, and f/g related to the domains of f and g? Give examples.
. When is it possible to compose one function with another? Give
examples of composites and their values at various points. Does the order in which functions are composed ever matter?
. How do you change the equation y — f(x) to shift its graph verti-
cally up or down by |k| units? Horizontally to the left or right? Give examples.
Chapter ff M Practice Exercises
Functions and Graphs
1,
Express the area and circumference of a circle as functions of the circle’s radius. Then express the area as a function of the circumference.
. Express the radius of a sphere as a function of the sphere’s sur-
face area. Then express the surface area as a function of the volume.
10.
11.
12.
13.
14.
15.
16.
How do you change the equation y = f(x) to compress or stretch the graph by a factor c > 1? Reflect the graph across a coordi- nate axis? Give examples.
What is radian measure? How do you convert from radians to degrees? Degrees to radians?
Graph the six basic trigonometric functions. What symmetries do the graphs have?
What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions?
Starting with the identity sin? 0 + cos? 0 = 1 and the formulas for cos (A + B) and sin (A + B), show how a variety of other trigonometric identities may be derived.
How does the formula for the general sine function f(x) — Asin (Qa /B)(x — C)) + D relate to the shifting, stretching, compressing, and reflection of its graph? Give examples. Graph the general sine curve and identify the constants A, B, C, and D.
Name three issues that arise when functions are graphed using a calculator or computer with graphing software. Give examples.
. A point P in the first quadrant lies on the parabola y = x?.
Express the coordinates of P as functions of the angle of inclina- tion of the line joining P to the origin.
. A hot-air balloon rising straight up from a level field is tracked by
a range finder located 500 ft from the point of liftoff. Express the balloon's height as a function of the angle the line from the range finder to the balloon makes with the ground.
In Exercises 5-8, determine whether the graph of the function is sym- metric about the y-axis, the origin, or neither.
5. y — xl 6. y = x Te y= x -2x-1 8. y= e”
In Exercises 9—16, determine whether the function is even, odd, or neither.
9. y=xr +1 10.y-0-3-x
11. y= 1 — cosx 12. y = sec x tan x A
13, 2 £1 14. y 2 x — sinx x = 2x
15. y 2 x + cosx 16. y = xcosx
17. Suppose that f and g are both odd functions defined on the entire real line. Which of the following (where defined) are even? odd? a. fg b. f£? c. f(sin x) d. g(sec x) e. lel
18. If f(a — x) = f(a + x), show that g(x) = f(x + a) is an even function.
In Exercises 19—28, find the (a) domain and (b) range.
19. y = |x| - 2 20. y=-2 + V1-x
21. y= V16— x? 22. y= 37 *4 1
23. y 2 2e* - 3 24. y = tan (2x — 7) 25. y = 2sin(àx + m) - 1. 26. y= x 27. y= In(x - 3) + I 28. y 2-1 V2- x
29. State whether each function is increasing, decreasing, or neither. a. Volume of a sphere as a function of its radius b. Greatest integer function
c. Height above Earth's sea level as a function of atmospheric pressure (assumed nonzero)
d. Kinetic energy as a function of a particle's velocity
30. Find the largest interval on which the given function is increasing. a. f(x) = |x - 2| +1 b. f) = (œ + 1% c. g(x) = x — 1)' d. Rx) = V2x - 1
Piecewise-Defined Functions In Exercises 31 and 32, find the (a) domain and (b) range.
31 m -4=x=0 ey Vx, O<x<4
=f = 2, —2x-xzx-l 32. y= X; -l<x=1 =% 2. E
In Exercises 33 and 34, write a piecewise formula for the function.
33. » 34. » ^
2 s 25)
>X
0 1 2
>X
Chapter 1 Practice Exercises 37
Composition of Functions In Exercises 35 and 36, find
a. (f° 8)C1). b. (° P). c. (f° f)@). d. (g° 8) (x). 1 1 35. =; ) = —— f= g(x) pu 36 f6)22- x gw=Wet1
In Exercises 37 and 38, (a) write formulas for f ° g and g » f and find the (b) domain and (c) range of each.
37. f(x) 22- x, g(x) = Vx+2
38. f= Vx, ew =VI-x
For Exercises 39 and 40, sketch the graphs of f and f » f. —x—2, —-Azx-z-l1
39. f(x) = 4-1, -l<xsl Xon 1«xz-2
= 1, CH= x= 2
Composition with absolute values In Exercises 41—48, graph f, and f; together. Then describe how applying the absolute value func- tion in f» affects the graph of fı.
fio) fax)
41. x x|
42. x? x|?
43. x x3|
44. x? 4+ x x? + x| 45.4— x? 4 — x!
1 1
46. x ial
47. Vx vial 48. sinx sin |x|
Shifting and Scaling Graphs
49. Suppose the graph of g is given. Write equations for the graphs that are obtained from the graph of g by shifting, scaling, or reflecting, as indicated.
a. Up 4 unit, right 3
b. Down 2 units, left i
c. Reflect about the y-axis d. Reflect about the x-axis e. Stretch vertically by a factor of 5 f. Compress horizontally by a factor of 5 50. Describe how each graph is obtained from the graph of y — f(x).
a. y — f(x — 5) b. y — f(4x) C. y = f(-3x) d. y= fx + 1) e. »- (1-4 f. y--3fQ *i
38 Chapter 1: Functions
In Exercises 51—54, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.15-1.17, and applying an appropriate transformation.
a x a 51. y = 1+ 2 52. y 1 3 53. y= ah +1 54. y = (5x)! 2x
Trigonometry In Exercises 55—58, sketch the graph of the given function. What is the period of the function?
55. y — cos 2x 56. y — sin5 57. y = sin mx 58. y = cos T
59. Sketch the graph y = 2 cos (: = z),
60. Sketch the graph y = 1 + sin (: t z),
In Exercises 61—64, ABC is a right triangle with the right angle at C. The sides opposite angles A, B, and C are a, b, and c, respectively.
61. a. Finda and bif c = 2, B = 7/3. b. Finda and c if b = 2, B = 7/3.
62.
63.
64.
65.
66.
67.
68.
Express a in terms of A and c. Express a in terms of A and b. Express a in terms of B and b. Express c in terms of A and a.
Express sin A in terms of a and c.
TPSDPTIP
Express sin A in terms of b and c.
Height of a pole Two wires stretch from the top T of a vertical pole to points B and C on the ground, where C is 10 m closer to the base of the pole than is B. If wire BT makes an angle of 35? with the horizontal and wire CT makes an angle of 50? with the horizontal, how high is the pole?
Height of a weather balloon Observers at positions A and B 2 km apart simultaneously measure the angle of elevation of a weather balloon to be 40° and 70°, respectively. If the balloon is directly above a point on the line segment between A and B, find the height of the balloon.
Graph the function f(x) = sin x + cos(x/2). What appears to be the period of this function? Confirm your finding in part (b) algebraically. Graph f(x) = sin (1/x).
What are the domain and range of f?
oc POTE
Is f periodic? Give reasons for your answer.
Chapter IS Additional and Advanced Exercises
Functions and Graphs
1. Are there two functions f and g such that f ^g = g» f? Give reasons for your answer.
2. Are there two functions f and g with the following property? The graphs of f and g are not straight lines but the graph of f ° g isa straight line. Give reasons for your answer.
3. If f(x) is odd, can anything be said of g(x) = f(x) — 2? Whatif f is even instead? Give reasons for your answer.
4. If g(x) is an odd function defined for all values of x, can anything be said about g(0)? Give reasons for your answer.
5. Graph the equation |x| + |y| = 1 + x.
6. Graph the equation y + |y| =x + |x|. Derivations and Proofs 7. Prove the following identities.
sin x 1+ cosx
1 — cosx _ sin x
] cos x _ tap? ` 1+ cos x 2
8. Explain the following “proof without words” of the law of cosines. (Source: Kung, Sidney H., “Proof Without Words: The Law of Cosines," Mathematics Magazine, Vol. 63, no. 5, Dec. 1990, p. 342.)
9,
10.
Show that the area of triangle ABC is given by (1/2)absin C = (1/2)bc sin A = (1/2)ca sin B.
C
A z B
Show that the area of triangle ABC is given by Vs(s — als — b\(s — c) where s = (a + b + c)/2 is the semiperimeter of the triangle.
11. Show that if f is both even and odd, then f(x) = 0 for every x in the domain of f.
12. a. Even-odd decompositions Let f be a function whose domain is symmetric about the origin, that is, —x belongs to the domain whenever x does. Show that f is the sum of an even function and an odd function:
fœ = E(x) + Ox), where E is an even function and O is an odd function. (Hint: Let E(x) = (f(x) + f(—x))/2. Show that E(-x) = E(x), so that E is even. Then show that O(x) = f(x) — E(x) is odd.) b. Uniqueness Show that there is only one way to write f as the sum of an even and an odd function. (Hint: One way is given in part (a). If also f(x) = E(x) + O,(x) where E; is even and O, is odd, show that E — E, = O, — O. Then use Exercise 11 to show that E = E, and O = O,.)
Effects of Parameters on Graphs
13. What happens to the graph of y = ax? + bx + c as a. a changes while b and c remain fixed? b. b changes (a and c fixed, a # 0)? c. c changes (a and b fixed, a 0)?
14. What happens to the graph of y = a(x + b? + c as a. a changes while b and c remain fixed? b. b changes (a and c fixed, a # 0)? c. c changes (a and b fixed, a # 0)?
Geometry
15. An object’s center of mass moves at a constant velocity v along a straight line past the origin. The accompanying figure shows the coordinate system and the line of motion. The dots show posi- tions that are 1 sec apart. Why are the areas Aj, A», ..., As in the figure all equal? As in Kepler's equal area law (see Section 13.6),
the line that joins the object's center of mass to the origin sweeps out equal areas in equal times.
y ^
10
Kilometers
Kilometers
Chapter 1 Additional and Advanced Exercises 39
16. a. Find the slope of the line from the origin to the midpoint P of side AB in the triangle in the accompanying figure (a, b — 0).
><
B(0, b)
x O A(a, 0)
b. When is OP perpendicular to AB?
17. Consider the quarter-circle of radius 1 and right triangles ABE and ACD given in the accompanying figure. Use standard area formulas to conclude that
0 .lsin0
LR z5in 0 cos < z < 2 cos 6"
y ^
(0, 1)
I I I L
A E (1,0)
18. Let f(x) = ax + b and g(x) = cx + d. What condition must be satisfied by the constants a, b, c, d in order that (f ° g)(x) = (e ° f)(x) for every value of x?
40 Chapter 1: Functions
Chapter BM Technology Application Projects
An Overview of Mathematica An overview of Mathematica sufficient to complete the Mathematica modules appearing on the Web site. Mathematica/Maple Module:
Modeling Change: Springs, Driving Safety, Radioactivity, Trees, Fish, and Mammals Construct and interpret mathematical models, analyze and improve them, and make predictions using them.
Limits and Continuity
OVERVIEW Mathematicians of the seventeenth century were keenly interested in the study of motion for objects on or near the earth and the motion of planets and stars. This study involved both the speed of the object and its direction of motion at any instant, and they knew the direction at a given instant was along a line tangent to the path of motion. The concept of a limit is fundamental to finding the velocity of a moving object and the tangent to a curve. In this chapter we develop the limit, first intuitively and then formally. We use limits to describe the way a function varies. Some functions vary continuously; small changes in x produce only small changes in f(x). Other functions can have values that jump, vary erratically, or tend to increase or decrease without bound. The notion of limit gives a precise way to distinguish between these behaviors.
P, i l Rates of Change and Tangents to Curves
HISTORICAL BIOGRAPHY * Galileo Galilei
(1564-1642)
Calculus is a tool that helps us understand how a change in one quantity is related to a change in another. How does the speed of a falling object change as a function of time? How does the level of water in a barrel change as a function of the amount of liquid poured into it? We see change occurring in nearly everything we observe in the world and universe, and powerful modern instruments help us see more and more. In this section we introduce the ideas of average and instantaneous rates of change, and show that they are closely related to the slope of a curve at a point P on the curve. We give precise developments of these important concepts in the next chapter, but for now we use an informal approach so you will see how they lead naturally to the main idea of this chapter, the limit. The idea of a limit plays a foundational role throughout calculus.
Average and Instantaneous Speed
In the late sixteenth century, Galileo discovered that a solid object dropped from rest (not moving) near the surface of the earth and allowed to fall freely will fall a distance proportional to the square of the time it has been falling. This type of motion is called free fall. It assumes negligible air resistance to slow the object down, and that gravity is the only force acting on the falling object. If y denotes the distance fallen in feet after t seconds, then Galileo's law is y = 162’,
where 16 is the (approximate) constant of proportionality. (If y is measured in meters, the constant is 4.9.)
A moving object’s average speed during an interval of time is found by dividing the distance covered by the time elapsed. The unit of measure is length per unit time: kilome- ters per hour, feet (or meters) per second, or whatever is appropriate to the problem at hand.
*To learn more about the historical figures mentioned in the text and the development of many major elements and topics of calculus, visit www.aw.com/thomas.
41
42
Chapter 2: Limits and Continuity
EXAMPLE 1 A rock breaks loose from the top of a tall cliff. What is its average speed
(a) during the first 2 sec of fall?
(b) during the 1-sec interval between second 1 and second 2?
Solution The average speed of the rock during a given time interval is the change in distance, Ay, divided by the length of the time interval, Ar. (Increments like Ay and At are reviewed in Appendix 3, and pronounced “delta y" and “delta 1") Measuring distance in feet and time in seconds, we have the following calculations:
Ay 16Qy — 16(0? ft (a) For the first 2 sec: AS 7-0 = 32sec Ay _ 162} 160 m (b) From sec 1 to sec 2: As 2—1 48 SEC [|
We want a way to determine the speed of a falling object at a single instant fj, instead of using its average speed over an interval of time. To do this, we examine what happens when we calculate the average speed over shorter and shorter time intervals starting at fj. The next example illustrates this process. Our discussion is informal here, but it will be made precise in Chapter 3.
EXAMPLE 2 Find the speed of the falling rock in Example 1 at t = 1 and t = 2sec.
Solution We can calculate the average speed of the rock over a time interval [ f, to + A], having length Ar = h, as
Ay 16(% + hy = 164? At h ' (D
We cannot use this formula to calculate the “instantaneous” speed at the exact moment to by simply substituting h = 0, because we cannot divide by zero. But we can use it to cal- culate average speeds over increasingly short time intervals starting at fj = 1 and h = 2. When we do so, by taking smaller and smaller values of h, we see a pattern (Table 2.1).
TABLE 2.1 Average speeds over short time intervals [ fo, to + h]
À d: Ay = 16(% + hy — 164?
verage speed: = ?
Length of Average speed over Average speed over time interval interval of length h interval of length h h starting at , = 1 starting at (, = 2 1 48 80 0.1 33.6 65.6 0.01 32.16 64.16 0.001 32.016 64.016 0.0001 32.0016 64.0016
The average speed on intervals starting at 4 = 1 seems to approach a limiting value of 32 as the length of the interval decreases. This suggests that the rock is falling at a speed of 32 ft/sec at tj = 1 sec. Let's confirm this algebraically.
0
FIGURE 2.1 A secant to the graph y = f(x). Its slope is Ay/ Ax, the average rate of change of f over the interval [x;, x].
FIGURE 2.2 Lis tangent to the circle at P if it passes through P perpendicular to radius OP.
2.1 Rates of Change and Tangents to Curves 43
If we set fj = 1 and then expand the numerator in Equation (1) and simplify, we find that
Ay 16(1 +h)? — 161 16(1 + 2h + I?) — 16 At h ~ h
_ 32h + 1612
n = 32 + 16h.
For values of h different from 0, the expressions on the right and left are equivalent and the average speed is 32 + 16h ft/sec. We can now see why the average speed has the limiting value 32 + 16(0) = 32 ft/sec as h approaches 0.
Similarly, setting t = 2 in Equation (1), the procedure yields
ae ee
At for values of h different from 0. As h gets closer and closer to 0, the average speed has the limiting value 64 ft/sec when fj = 2sec, as suggested by Table 2.1. L|
The average speed of a falling object is an example of a more general idea which we discuss next.
Average Rates of Change and Secant Lines
Given any function y — f(x), we calculate the average rate of change of y with respect to x over the interval [x;, x;] by dividing the change in the value of y, Ay = f(x) — f(x), by the length Ax = x; — x, = h of the interval over which the change occurs. (We use the symbol / for Ax to simplify the notation here and later on.)
DEFINITION The average rate of change of y — f(x) with respect to x over the interval [x;, x5] is
Ay fGo) — f(x) _ f + h — fe)
Ax 3 — X h i
h # 0.
Geometrically, the rate of change of f over [x;, x2] is the slope of the line through the points P(x, f(x1)) and Q(x, f(x;)) (Figure 2.1). In geometry, a line joining two points of a curve is a secant to the curve. Thus, the average rate of change of f from x, to x; is identi- cal with the slope of secant PQ. Let's consider what happens as the point Q approaches the point P along the curve, so the length A of the interval over which the change occurs approaches zero. We will see that this procedure leads to defining the slope of a curve at a point.
Defining the Slope of a Curve
We know what is meant by the slope of a straight line, which tells us the rate at which it rises or falls—its rate of change as a linear function. But what is meant by the slope of a curve at a point P on the curve? If there is a tangent line to the curve at P—a line that just touches the curve like the tangent to a circle—it would be reasonable to identify the slope of the tangent as the slope of the curve at P. So we need a precise meaning for the tangent at a point on a curve.
For circles, tangency is straightforward. A line L is tangent to a circle at a point P if L passes through P perpendicular to the radius at P (Figure 2.2). Such a line just touches the circle. But what does it mean to say that a line L is tangent to some other curve C at a point P?
44 Chapter 2: Limits and Continuity
HISTORICAL BIOGRAPHY Pierre de Fermat
(1601-1665)
To define tangency for general curves, we need an approach that takes into account the behavior of the secants through P and nearby points Q as Q moves toward P along the curve (Figure 2.3). Here is the idea:
1. Start with what we can calculate, namely the slope of the secant PQ.
2. Investigate the limiting value of the secant slope as Q approaches P along the curve. (We clarify the limit idea in the next section.)
3. Ifthe limit exists, take it to be the slope of the curve at P and define the tangent to the curve at P to be the line through P with this slope.
This procedure is what we were doing in the falling-rock problem discussed in Example 2. The next example illustrates the geometric idea for the tangent to a curve.
s NM TW
Secants ^,
Tangent
FIGURE 2.3 The tangent to the curve at P is the line through P whose slope is the limit of the secant slopes as Q — P from either side.
EXAMPLE 3 Find the slope of the parabola y — x? at the point P(2, 4). Write an equation for the tangent to the parabola at this point.
Solution We begin with a secant line through P(2, 4) and Q(2 + h, (2 + h)?) nearby. We then write an expression for the slope of the secant PQ and investigate what happens to the slope as Q approaches P along the curve:
Ay Q-hy-2 P+4n+4-4 Ax h h _ We x 4h h
Secant slope —
=h+ 4.
If h > 0, then Q lies above and to the right of P, as in Figure 2.4. If h < 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope + 4 approaches 4. We take 4 to be the parabola's slope at P.
><
h+4,
= 2 = Yue Secant slope is — =
NOT TO SCALE
FIGURE 2.4 Finding the slope of the parabola y = x? at the point P(2, 4) as the limit of secant slopes (Example 3).
2.1 Rates of Change and Tangents to Curves 45
The tangent to the parabola at P is the line through P with slope 4: y=4+ 4 —- 2) Point-slope equation y=4x-4. E
Instantaneous Rates of Change and Tangent Lines
The rates at which the rock in Example 2 was falling at the instants t = 1 and t = 2 are called instantaneous rates of change. Instantaneous rates and slopes of tangent lines are closely connected, as we see in the following examples.
EXAMPLE 4 Figure 2.5 shows how a population p of fruit flies (Drosophila) grew in a 50-day experiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time f, and the points joined by a smooth curve (colored blue in Figure 2.5). Find the average growth rate from day 23 to day 45.
Solution There were 150 flies on day 23 and 340 flies on day 45. Thus the number of flies increased by 340 — 150 — 190 in 45 — 23 — 22 days. The average rate of change of the population from day 23 to day 45 was
Ap 340 — 150 _ 190 Ar^ 45-23 ^22 ^ 8.6 flies/day.
Average rate of change:
>T
350 300
Q(45, 340)
250 200
Ap-= 190
âp = 8.6 flies/day
At = 22
P(23, 150)
150
Number of flies
100 50
>t
0 10 20 30 40 50 Time (days)
FIGURE 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of change over 22 days is the slope Ap/ At of the secant line (Example 4).
This average is the slope of the secant through the points P and Q on the graph in Figure 2.5. a
The average rate of change from day 23 to day 45 calculated in Example 4 does not tell us how fast the population was changing on day 23 itself. For that we need to examine time intervals closer to the day in question.
EXAMPLE 5 How fast was the number of flies in the population of Example 4 grow- ing on day 23?
Solution To answer this question, we examine the average rates of change over increas- ingly short time intervals starting at day 23. In geometric terms, we find these rates by calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure 2.6).
46 Chapter 2: Limits and Continuity
p Slopeof PO — Ap/At j B(35, 350) — Q (flies / day) 350 pee san Q(45, 340) 340 — 150 _ * (45, 340) 45-23 ^" 8.6 s 250 o 330 — 150 — 2m (40, 330) abeo € M E in 310— 150 — 100 (35, 310) em hae b (30, 265) 263 — 7) = 164 0 ij —20 30 40 39 ^!
A(14, 0) Time (days)
FIGURE 2.6 The positions and slopes of four secants through the point P on the fruit fly graph (Example 5).
Exercises 2.1|
Average Rates of Change
In Exercises 1—6, find the average rate of change of the function over
the given interval or intervals. 1. fé) - x +1
The values in the table show that the secant slopes rise from 8.6 to 16.4 as the t-coordinate of Q decreases from 45 to 30, and we would expect the slopes to rise slightly higher as t continued on toward 23. Geometrically, the secants rotate counterclockwise about P and seem to approach the red tangent line in the figure. Since the line appears to pass through the points (14, 0) and (35, 350), it has slope
350 — 0
35—]4 16.7 flies/day (approximately).
On day 23 the population was increasing at a rate of about 16.7 flies / day. Ei
The instantaneous rates in Example 2 were found to be the values of the average speeds, or average rates of change, as the time interval of length h approached 0. That is, the instantaneous rate is the value the average rate approaches as the length h of the inter- val over which the change occurs approaches zero. The average rate of change corre- sponds to the slope of a secant line; the instantaneous rate corresponds to the slope of the tangent line as the independent variable approaches a fixed value. In Example 2, the inde- pendent variable t approached the values t = 1 and t = 2. In Example 3, the independent variable x approached the value x — 2. So we see that instantaneous rates and slopes of tangent lines are closely connected. We investigate this connection thoroughly in the next chapter, but to do so we need the concept of a limit.
5. R(6) = V40 + 1; [0,2] 6. P(0) = 6? — 48? + 50; [1,2]
Slope of a Curve at a Point In Exercises 7-14, use the method in Example 3 to find (a) the slope
a. [2,3] b. [-1,1] : : : 3 of the curve at the given point P, and (b) an equation of the tangent
2ga) = x! — 2x line at P.
a. [1,3] b. [-2.4] 7. y2x!-5, P2,-1) 3. h(t) = cott 8 y=7- x, PQ,3)
a. [7/4, 37/4] b. [ 7/6, 7/2] 9. y=? — 2 —3, PQ,-3) 4. g(t) = 2 + cost 10, y= —4x, P(l—3)
a. [0,7] b. [77.7] 11. y 2 x, P(2,8)
2.1 Rates of Change and Tangents to Curves 47
12. y 22— x, P(,1) b. What is the average rate of increase of the profits between ? 13. y= xX — 12x, PO,-11) 2012 and 2014? 14. y=- 3x2 +4, P(2,0) c. Use your graph to estimate the rate at which the profits were changing in 2012. Instantaneous Rates of Change 18. Make a table of values for the function F(x) = (x + 2)/(x — 2) 15. Speed of a car The accompanying figure shows the time-to- at the points x = 1.2, x = 11/10, x = 101/100, x = 1001/1000, distance graph for a sports car accelerating from a standstill. x = 10001/10000, and x = 1. 5 a. Find the average rate of change of F(x) over the intervals 650 [1, x] foreach x # 1 in your table. 600 b. Extending the table if necessary, try to determine the rate of 500 change of F(x) at x = 1. a8 19. Let g(x) — Vx for x = 0. — 400 : , 8 a. Find the average rate of change of g(x) with respect to x over S 300 the intervals [1,2], [1, 1.5] and [1,1 + ^]. a 200 b. Make a table of values of the average rate of change of g with respect to x over the interval [ 1, 1 + ^] for some values of h 100 | approaching zero, say h = 0.1, 0.01, 0.001, 0.0001, 0.00001, "i and 0.000001. >t 0 5 10 15 20 c. What does your table indicate is the rate of change of g(x) Elapsed time (sec) with respect to x at x — 1? a. Estimate the slopes of secants PQ), PO», PO, and PQ,, d. Calculate the limit as ^ approaches zero of the average rate of arranging them in order in a table like the one in Figure 2.6. change of g(x) with respect to x over the interval [ 1, 1 A]. What are the appropriate units for these slopes? 20. Let f(t) = 1/t fort # 0 b. Then estimate the car's speed at time t — 20sec. a. Find the average rate of change of f with respect to ¢ over the 16. The accompanying figure shows the plot of distance fallen versus intervals (i) from t = 2 to t = 3, and (ii) from t = 2 to t = T. time for an object that fell from the lunar landing module a dis- b. Make a table of values of the average rate of change of f with
tance 80 m to the surface of the moon. respect to f over the interval [2,7], for some values of T
a. Estimate the slopes of the secants PQ,, PQ;. PQ3, and PQ4,, approaching 2, say T = 2.1, 2.01, 2.001, 2.0001, 2.00001, arranging them in a table like the one in Figure 2.6. and 2.000001. b. About how fast was the object going when it hit the surface? c. What does your table indicate is the rate of change of f with y respect tot at t = 2?
d. Calculate the limit as T approaches 2 of the average rate of change of f with respect to t over the interval from 2 to T. You
g will have to do some algebra before you can substitute T — 2. £ 21. The accompanying graph shows the total distance s traveled by a Ez] . . ps bicyclist after ¢ hours. o 3 a AY A A E Elapsed time (sec) hs o 17. The profits of a small company for each of the first five years of E its operation are given in the following table: S Year Profit in $1000s A ^ > 2010 6 — 2011 2] apsed time (hr) 2012 62 "EE 2013 111 a. Estimate the bicyclist’s average speed over the time intervals
b. Estimate the bicyclist’s instantaneous speed at the times t = L, t = 2, and t = 3.
a. Plot points representing the profit as a function of year, and c. Estimate the bicyclist’s maximum speed and the specific time join them by as smooth a curve as you can. at which it occurs.
48 Chapter 2: Limits and Continuity
22. The accompanying graph shows the total amount of gasoline A in a. Estimate the average rate of gasoline consumption over the the gas tank of an automobile after being driven for ft days. time intervals [0,3], [0,5], and [ 7, 10]. A
Remaining amount (gal)
b. Estimate the instantaneous rate of gasoline consumption at the times t = 1,1 = 4, and t = 8.
c. Estimate the maximum rate of gasoline consumption and the specific time at which it occurs.
>t
Ü 1253 4 5 6 7 8 9 10
Elapsed time (days)
2.2 Limit of a Function and Limit Laws
HISTORICAL ESSAY Limits
FIGURE 2.7 The graph of f is identical with the line y = x I
except at x — 1, where f is not defined (Example 1).
In Section 2.1 we saw that limits arise when finding the instantaneous rate of change of a function or the tangent to a curve. Here we begin with an informal definition of limit and show how we can calculate the values of limits. A precise definition is presented in the next section.
Limits of Function Values
Frequently when studying a function y — f(x), we find ourselves interested in the func- tion's behavior near a particular point c, but not at c. This might be the case, for instance, if c is an irrational number, like 7 or V2, whose values can only be approximated by "close" rational numbers at which we actually evaluate the function instead. Another situ- ation occurs when trying to evaluate a function at c leads to division by zero, which is undefined. We encountered this last circumstance when seeking the instantaneous rate of change in y by considering the quotient function Ay// for h closer and closer to zero. Here's a specific example in which we explore numerically how a function behaves near a particular point at which we cannot directly evaluate the function.
EXAMPLE 1 How does the function
x—1 x—1
f(x) = behave near x = 1?
Solution The given formula defines f for all real numbers x except x = 1 (we cannot divide by zero). For any x # 1, we can simplify the formula by factoring the numerator and canceling common factors:
@-—DOtD_
x—1
f(x) xt+1 for x Al.
The graph of f is the line y = x + 1 with the point (1, 2) removed. This removed point is shown as a “hole” in Figure 2.7. Even though f(1) is not defined, it is clear that we can make the value of f(x) as close as we want to 2 by choosing x close enough to 1 (Table 2.2). L|
TABLE 2.2 As x gets closer to
1, f(x) gets closer to 2. x2—1
x f(x) = x-1
0.9 1.9
1.1 2.1
0.99 1.99
1.01 2.01
0.999 1.999
1.001 2.001
0.999999 1.999999
1.000001 2.000001
(a) Identity function
k yak m —————É | l l i 7 Z >x
(b) Constant function
FIGURE 2.9 The functions in Example 3 have limits at all points c.
2.2 Limit of a Function and Limit Laws 49
Generalizing the idea illustrated in Example 1, suppose f(x) is defined on an open interval about c, except possibly at c itself. If f(x) is arbitrarily close to the number L (as close to L as we like) for all x sufficiently close to c, we say that f approaches the limit L as x approaches c, and write
lim f(x) = L,
which is read “the limit of f(x) as x approaches c is L.” For instance, in Example 1 we would say that f(x) approaches the limit 2 as x approaches 1, and write
lim f@ = 2, or lim — ; = 2.
Essentially, the definition says that the values of f(x) are close to the number L whenever x is close to c (on either side of c).
Our definition here is “informal” because phrases like arbitrarily close and sufficiently close are imprecise; their meaning depends on the context. (To a machinist manufacturing a piston, close may mean within a few thousandths of an inch. To an astronomer studying distant galaxies, close may mean within a few thousand light-years.) Nevertheless, the definition is clear enough to enable us to recognize and evaluate limits of many specific functions. We will need the precise definition given in Section 2.3, however, when we set out to prove theorems about limits or study complicated functions. Here are several more examples exploring the idea of limits.
EXAMPLE 2 The limit value of a function does not depend on how the function is defined at the point being approached. Consider the three functions in Figure 2.8. The function f has limit 2 as x — 1 even though f is not defined at x = 1. The function g has limit 2 as x — 1 even though 2 z^ g(1). The function h is the only one of the three functions in Figure 2.8 whose limit as x — 1 equals its value at x = 1. For h, we have lim,—, A(x) = A(1). This equality of limit and function value is of special importance, and we return to it in Section 2.5. L
žo e ><
TOM C ESS
(a) f(x) = 3
i o x*1 (b) $5) - 1 *
l, x-l
FIGURE 2.8 The limits of f(x), g(x), and A(x) all equal 2 as x approaches 1. However, only h(x) has the same function value as its limit at x = 1 (Example 2).
EXAMPLE 3 (a) If f is the identity function f(x) = x, then for any value of c (Figure 2.9a),
lim fix) = lim x =c. (b) If f is the constant function f(x) = k (function with the constant value k), then for any value of c (Figure 2.9b),
lim f(x) = lim k = k.
50 Chapter 2: Limits and Continuity
For instances of each of these rules we have
lim x = 3 and lim, (4) — lim (4) — 4.
x3
We prove these rules in Example 3 in Section 2.3. E
A function may not have a limit at a particular point. Some ways that limits can fail to exist are illustrated in Figure 2.10 and described in the next example.
(a) Unit step function U(x)
(b) g(x) (c) f@)
FIGURE 2.10 None of these functions has a limit as x approaches 0 (Example 4).
EXAMPLE 4 Discuss the behavior of the following functions, explaining why they have no limit as x — 0.
x J0, x<0 (a) U(x) = 7 ee i xz0 (b) g(x) = 0, x=0 =0 © f=", j sinx, x20 Solution
(a) It jumps: The unit step function U(x) has no limit as x — 0 because its values jump at x = 0. For negative values of x arbitrarily close to zero, U(x) = 0. For positive values of x arbitrarily close to zero, U(x) — 1. There is no single value L approached by U(x) as x — 0 (Figure 2.102).
(b) It grows too "large" to have a limit: g(x) has no limit as x — 0 because the values of g grow arbitrarily large in absolute value as x — 0 and do not stay close to any fixed real number (Figure 2.10b). We say the function is not bounded.
(c) It oscillates too much to have a limit: f(x) has no limit as x — 0 because the func- tion's values oscillate between +1 and —1 in every open interval containing 0. The values do not stay close to any one number as x — 0 (Figure 2.10c). El
2.2 Limit of a Function and Limit Laws 51
The Limit Laws
To calculate limits of functions that are arithmetic combinations of functions having known limits, we can use several fundamental rules.
THEOREM 1—Limit Laws If L, M, c, and k are real numbers and lim fe9-L and lim g(x) = M, then 1. Sum Rule: lim(fG) + g(x) =L+M 2. Difference Rule: lim(f(x) — gx)) =L-M 3. Constant Multiple Rule: lim(k- fo) S kL 4. Product Rule: lim(f(x) *g(x) = L' M 5. Quotient Rule: lim Es -E M#0 6. Power Rule: lim [ f(x) |" = L", n a positive integer 7. Root Rule: lim Vf) = VL-LV" na positive integer (If n is even, we assume that lim f(x) =L> 0.)
In words, the Sum Rule says that the limit of a sum is the sum of the limits. Similarly, the next rules say that the limit of a difference is the difference of the limits; the limit of a con- stant times a function is the constant times the limit of the function; the limit of a product is the product of the limits; the limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0); the limit of a positive integer power (or root) of a function is the integer power (or root) of the limit (provided that the root of the limit is a real number).
It is reasonable that the properties in Theorem 1 are true (although these intuitive arguments do not constitute proofs). If x is sufficiently close to c, then f(x) is close to L and g(x) is close to M, from our informal definition of a limit. It is then reasonable that f(x) + g(x) is close to L + M; f(x) — g(x) is close to L — M; kf(x) is close to KL; f C99) is close to LM; and f(x)/g(x) is close to L/M if M is not zero. We prove the Sum Rule in Section 2.3, based on a precise definition of limit. Rules 2—5 are proved in Appen- dix 4. Rule 6 is obtained by applying Rule 4 repeatedly. Rule 7 is proved in more advanced texts. The Sum, Difference, and Product Rules can be extended to any number of func- tions, not just two.
EXAMPLE 5 Use the observations lim,_.. k = k and lim, x = c (Example 3) and the fundamental rules of limits to find the following limits.
(a) lim(x? + 4x? — 3)
(b) lim
(c) lim, 4x? — 3
52 Chapter 2: Limits and Continuity
Identifying Common Factors
It can be shown that if Q(x) is a poly- nomial and Q(c) = 0, then (x — c) is a factor of Q(x). Thus, if the numerator and denominator of a rational function of x are both zero at x — c, they have (x — c) as a common factor.
T
Solution (a) lim(x? + 4x? — 3) = lim x? + lim 4x? — lim 3 Sum and Difference Rules xc KAGE P a xX—c = + 4c? — 3 Power and Multiple Rules 4 " lim(x* + x? — 1) wo OE eS 1 ye , (b) lim 2 = : 3 Quotient Rule NRC Ac op lim(x 3b 5) x» lim x* + lim x? — lim 1 (—. —,. — EIE 7 , e 7 ii Sum and Difference Rules lim x^ + lim 5 Pa T 4 2 t — 1 = —— Power or Product Rule CHS (c) lim VAx? — 3 = V lim (Ax? — 3) Root Rule with n — 2 x-2 x-2 = V lim 4x? — lim 3 Difference Rule x-—2 x-—2
= V 4(—2y = Product and Multiple Rules
= V16-3
= V13 = Theorem 1 simplifies the task of calculating limits of polynomials and rational functions. To evaluate the limit of a polynomial function as x approaches c, merely substitute c for x in the formula for the function. To evaluate the limit of a rational function as x approaches
a point c at which the denominator is not zero, substitute c for x in the formula for the function. (See Examples 5a and 5b.) We state these results formally as theorems.
THEOREM 2—Limits of Polynomials If P(x) = a,x" + a, 4x"! + +++ + ag, then
lim P(x) = P(c) = a,c" + a,c"! + +++ + a, x
THEOREM 3—Limits of Rational Functions
If P(x) and Q(x) are polynomials and Q(c) # 0, then ji P(x) P(o im —— — : xc OX) | OAc)
EXAMPLE 6 The following calculation illustrates Theorems 2 and 3:
xdg es CD FEED S30. 5 ü x x45 (-1Y +5 6
Eliminating Common Factors from Zero Denominators
Theorem 3 applies only if the denominator of the rational function is not zero at the limit point c. If the denominator is zero, canceling common factors in the numerator and denominator may reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can find the limit by substitution in the simplified fraction.
FIGURE 2.11 The graph of f(x) = G2 -x-2)/(2 — x) in part (a) is the same as the graph of
g(x) = (x + 2)/x in part (b) except
at x = 1, where f is undefined. The functions have the same limit as x — 1 (Example 7).
2.2 Limit of a Function and Limit Laws 53
EXAMPLE 7
Evaluate
Solution We cannot substitute x = 1 because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = 1. It is, so it has a factor of (x — 1) in com- mon with the denominator. Canceling this common factor gives a simpler fraction with the same values as the original for x # 1:
xó)mmm1X€72 @-D)A@t+2) x72 x?-—»x x(x — 1) Met
ifx ~ 1.
Using the simpler fraction, we find the limit of these values as x — 1 by Theorem 3:
See Figure 2.11. a
Using Calculators and Computers to Estimate Limits
When we cannot use the Quotient Rule in Theorem 1 because the limit of the denominator is Zero, we can try using a calculator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example 1, but calculators and computers can sometimes give false values and misleading impressions for functions that are unde- fined at a point or fail to have a limit there. Usually the problem is associated with round- ing errors, as we now illustrate.
EXAMPLE 8
4/72 — Estimate the value of lim Hc = D 10.
x0 X
Solution Table 2.3 lists values of the function obtained on a calculator for several points approaching x = 0. As x approaches 0 through the points +1, 30.5, +0.10, and + 0.01, the function seems to approach the number 0.05.
As we take even smaller values of x, + 0.0005, 3- 0.0001, + 0.00001, and + 0.000001, the function appears to approach the number 0.
Is the answer 0.05 or 0, or some other value? We resolve this question in the next
example. El XE 100 — TABLE 2.3 Computed values of f(x) = a > IB near x = O x f(x) +] 0.049876 +0.5 0.049969 h .05? +0.1 0,049599. LP Oe +0.01 0.050000 + 0.0005 0.050000 + 0.0001 0.000000 approaches 0? +0.00001 0.000000 ( “PP ' + 0.000001 0.000000
54 Chapter 2: Limits and Continuity
><
l l | l | l | l o 0 c
-X
FIGURE 2.12 The graph of f is sand-
wiched between the graphs of g and h.
Using a computer or calculator may give ambiguous results, as in the last example. The calculator could not keep track of enough digits to avoid rounding errors in computing the values of f(x) when x is very small. We cannot substitute x = 0 in the problem, and the numerator and denominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.
EXAMPLE 9 Evaluate Vx? + 100 — 10
lim 5
x0 X
Solution This is the limit we considered in Example 8. We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression
V/x? + 100 + 10 (obtained by changing the sign after the square root). The preliminary algebra rationalizes the numerator:
Vax? + 100 — 10 _ Vx? + 100 — 10 Vx? 100 + 10 x? x Vx? + 100 + 10
|. x? + 100 — 100 x?( Vx + 100 + 10)
2
x
x? (Vx + 100 + 10) 1
Vx2 + 100 + 10
Common factor x?
Cancel x? for x # 0.
Therefore,
. Vx? + 100 — 10 4 1 lim 5 = lim x0 X x—>0 V x + 100 + 10 1 Denominator not 0 at = x = 0; substitute. V0? + 100 + 10
sda = 3 7 0.05.
This calculation provides the correct answer, in contrast to the ambiguous computer results in Example 8. El
We cannot always algebraically resolve the problem of finding the limit of a quotient where the denominator becomes zero. In some cases the limit might then be found with the aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4), or through methods of calculus (illustrated in Section 7.5). The next theorems give helpful tools by using function comparisons.
The Sandwich Theorem
The following theorem enables us to calculate a variety of limits. It is called the Sandwich Theorem because it refers to a function f whose values are sandwiched between the val- ues of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of f must also approach L (Figure 2.12). You will find a proof in Appendix 4.
FIGURE 2.13 Any function u(x) whose graph lies in the region between y = 1 + G2/2) and y = 1 — (2/4) has limit 1 as x — 0 (Example 10).
FIGURE 2.14 The Sandwich Theorem confirms the limits in Example 11.
2.2 Limit of a Function and Limit Laws 55
THEOREM 4—The Sandwich Theorem Suppose that g(x) = f(x) = h(x) for all x in some open interval containing c, except possibly at x — c itself. Suppose also that
lim g(x) = lim h(x) = L.
Then lim, f(x) = L.
The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem. EXAMPLE 10 Given that
x
2 for all x # 0,
J= A < ux) S1 + find lim,—ọ u(x), no matter how complicated u is. Solution Since lim(1 — (2/4) =1 and lim(1 -(x2/2)21
the Sandwich Theorem implies that lim,.9 u(x) = 1 (Figure 2.13). L|
EXAMPLE 11 The Sandwich Theorem helps us establish several important limit rules: (a limsin0 = 0 (b) lim cos = 1 60 00 (c) For any function f, lim | f(x)| = 0 implies lim f(x) = 0. Solution
(a) In Section 1.3 we established that —|e| < sinf < lel for all 0 (see Figure 2.14a). Since limọ>o(—|0|) = limg—so |0| = 0, we have
lim sin 0 = 0. 6—0 (b) From Section 1.3, 0 € 1 — cos0 < lo] for all 0 (see Figure 2.14b), and we have limy.9(1 — cos 0) = O or lim cos 0 = 1. 0—0 (c) Since —|f(x)| = fŒ = |fGO| and —|f(x)| and |fG)| have limit 0 as x c, it follows that lim,- f(x) = 0. ii
Another important property of limits is given by the next theorem. A proof is given in the next section.
THEOREM 5 If f(x) S g(x) for all x in some open interval containing c, except possibly at x = c itself, and the limits of f and g both exist as x approaches c, then
lim f(x) = lim gQ).
Caution The assertion resulting from replacing the less than or equal to (=) inequality by the strict less than (<) inequality in Theorem 5 is false. Figure 2.14a shows that for 6 # 0, —|6| < sin 6 < |6|. So lim, o sin 6 = 0 = limgs9|6], not limg_sp sin 0 < limọ—o |0|.
56 Chapter 2: Limits and Continuity
Exercises 2.23
Limits from Graphs 1. For the function g(x) graphed here, find the following limits or explain why they do not exist.
a. lim g(x) b. lim g(x) c. lim g(x) d. lim g(x) xc a2. x3 32:5
-X
2. For the function f(t) graphed here, find the following limits or explain why they do not exist.
a. lim f() b. lim f() c limf() d. lim fO
s = fA) jx l ie L— t NC 0 1
3. Which of the following statements about the function y — f(x) graphed here are true, and which are false?
a. lim f(x) exists. b. lim f(x) = 0 lim fe) = 1 , lim fe) 1 e. lim f@) 20
a9
ph
lim f(x) exists at every point c in (~ 1, 1).
xe
g. lim f(x) does not exist.
4. Which of the following statements about the function y = f(x) graphed here are true, and which are false?
a. lim f(x) does not exist. b. lim f(x) =2
c. lim f(x) does not exist. paar
d. lim f(x) exists at every point c in (— 1, 1). xc
e. lim f(x) exists at every point c in (1, 3).
xU
y ^ y —f0)
Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. x
5. lim — 6. lim x0 |x| xix—1
7. Suppose that a function f(x) is defined for all real values of x except x — c. Can anything be said about the existence of lim,—,. f(x)? Give reasons for your answer.
8. Suppose that a function f(x) is defined for all x in [—1, 1]. Can anything be said about the existence of lim, 9 f(x)? Give reasons for your answer.
9. If lim, ,, f(x) = 5, must f be defined at x = 1? If it is, must f(1) = 5? Can we conclude anything about the values of f at x = 1? Explain.
10. If f(1) = 5, must lim,., f(x) exist? If it does, then must lim,—.; f(x) = 5? Can we conclude anything about lim,_,, f(x)? Explain.
Calculating Limits Find the limits in Exercises 11—22.
11. lim, (x? — 13) 12. lim(—x? + 5x — 2) 13. lim 8(t — 5)\(t — 7) 14. lim (x° — 2x? + 4x + 8) = x—— zc BEES ` 15. lms; 3 16. E 3s)(2s — 1) » i yt2 17. lim 4x(3x + 4% 18. lim 5—— —— ale y>2y? + 5y + 6 19. lim (5 — y^^ 20. lim Vz - 10 you 2
3 . V5h t 4-2 lim ———— — 22. lim — —— — — hO0N3R 141 h0 h
Limits of quotients Find the limits in Exercises 23-42.
21.
23. lim ~—> i dcc5 e x>5x — 25 x33? + 4x + 3 . x7 + 3x — 10 . x? — "Ix + 10 23, a x+5 25 im Xx 32 2 ES 2 ED 27. lim’ + — j&. qua E ml P-1 A a —2x = 5y? + 8y? 29. lim 2—4 30. lim. x2-2x) + 2x? y20 3y* — 16y*
31. lim 32. lim ea = 1 x0 x . u—) . v—8s . lim ——— 4. lim ———— 33 vu e : m gt 16 , No 4x — x? 35. lim 19 36. im, a.
37. hini-———— — 38. lim
uu A a Me ie S i 2a Mac ; 4-x Wem 313 e T
Limits with trigonometric functions Find the limits in Exercises 43-50.
43. lim(2sinx — 1) 44. lim sin? x 7-20 x—m/A 45. limsec x 46. lim tanx x0 xm/3 ii 48. lim(x? — 1)2 — cosx) x0 3cosx x0
49. lim Vx + 4cos(x + r) 50. lim V7 + sec?x
x—h—m x0
Using Limit Rules
51. Suppose lim,,, f(x) = 1 and lim,.) g(x) = —5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.
2f(x) — ga) _ MCF) 7 se)
So) 727 Tim (Fe) + 7977 i lim 2f@) — lim g(x) =? E (b) (timo +7) 2 lim f(x) — lim g(x) (c)
m 2/3 (imn + tn 7) (0) = 5) 7
Qm 4
52. Let lim; A(x) = 5, lim, p(x) = 1, and lim,—, r(x) = 2. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.
VShG) lim V 5h(x)
lim Oa rw — lim (pG)4 = rW) (a) A lim Sh(x) > (b) (tim pco (im (4 - ro) A /5lim h(x) - : (c)
(tim pco J (tim 4 — lim ro) V (5)(5) 5
| (04-2 2
2.2 Limit of a Function and Limit Laws 57
53. Suppose lim, f(x) = 5 and lim, g(x) = —2. Find a. lim f(x)g@) b. lim 2fG)gQ) ; fe) . lim z—— ——— d D FQ) — «00 54. Suppose lim,—4 f(x) = 0 and lim,—4 g(x) = —3. Find b. lim xf(x)
€. lim(fG) 3809)
a. lim(gQ) + 3) . 8%)
WIEN
55. Suppose lim,.,; f(x) = 7 and lim,_,, g(x) = —3. Find a. lim (fo) + sg) b. lim FO) * gx) c. lim 4g(x) d. lim 69/80)
56. Suppose that lim, ,.» p(x) = 4, lim,.-» r(x) = 0, and lim,» s(x) = —3. Find
a. lim, (p(x) + r(x) + s(x)) b. lim pœ): rx)" s(x) c. lim (74pQ) + 5r(x))/s(x)
c. lim (g@y°
Limits of Average Rates of Change Because of their connection with secant lines, tangents, and instanta- neous rates, limits of the form
. fat hy — feo) lim ———— ———— h—0 h
occur frequently in calculus. In Exercises 57-62, evaluate this limit for the given value of x and function f.
57. fx) =x, x=1
58. f(x) = xi x=-2
59. f(x) = 3x —4, x=2
60. f(x) = 1/x, x =—2
61. f= Vx, x =7
62. f(x) = Vx + l, x=0
Using the Sandwich Theorem 63. If V5 — 2x? < f(x) € V5 — x for-1 < x < 1, find limo f@). 64. If 2 — x? < g(x) = 2 cos x for all x, find lim, +p g(). 65. a. It can be shown that the inequalities X xsinx
l OET
hold for all values of x close to zero. What, if anything, does this tell you about
lim —* sin x
x02 — 2 cosx
Give reasons for your answer.
T|b. Graph y= 1 — (x?/6), y = (xsinx)/(2 — 2 cos x), and y = 1 together for —2 = x = 2. Comment on the behavior of the graphs as x — 0.
58
66. a.
[T] b.
Chapter 2: Limits and Continuity
Suppose that the inequalities
1 x .1-cox .1 M^ "9
hold for values of x close to zero. (They do, as you will see in Section 9.9.) What, if anything, does this tell you about
Give reasons for your answer.
Graph the equations y = (1/2) — (x?/24),
y = (1 — cos x)/x?, and y = 1/2 together for -2 < x S 2. Comment on the behavior of the graphs as x — 0.
Estimating Limits You will find a graphing calculator useful for Exercises 67—74. 67. Let f(x) = (x? — 9)/(x + 3).
a.
c.
Make a table of the values of f at the points x — —3.1, —3.01, —3.001, and so on as far as your calculator can go. Then estimate lim,_,_3 f(x). What estimate do you arrive at if you evaluate f at x = —2.9, —2.99, —2.999,... instead?
Support your conclusions in part (a) by graphing f near c — —3 and using Zoom and Trace to estimate y-values on the graph as x — —3.
Find lim,—.-3 f(x) algebraically, as in Example 7.
68. Let g(x) = (2 — 2)/(x — V2).
a.
c. 69. Let G(x) = (x 4
a.
c. 70. Let h(x) = (x? — 2x
a.
c.
Make a table of the values of g at the points x = 1.4, 1.41, 1.414, and so on through successive decimal approximations of V2. Estimate lim, >z g(x).
Support your conclusion in part (a) by graphing g near
c = V2 and using Zoom and Trace to estimate y-values on the graph as x > v2.
Find lim, =z g(x) algebraically.
6)/ G2 + 4x — 12).
Make a table of the values of G at x = —5.9, —5.99, —5.999, and so on. Then estimate lim,—.-~ G(x). What estimate do you arrive at if you evaluate G at x — —6.1, —6.01,
—6.001, .. . instead?
Support your conclusions in part (a) by graphing G and using
Zoom and Trace to estimate y-values on the graph as x -—6.
Find lim,—-¢ G(x) algebraically.
3)/G? — 4x + 3).
Make a table of the values of h at x = 2.9, 2.99, 2.999, and so on. Then estimate lim, ,, A(x). What estimate do you
arrive at if you evaluate h at x = 3.1, 3.01, 3.001, . . . instead?
Support your conclusions in part (a) by graphing h near c — 3 and using Zoom and Trace to estimate y-values on the graph as x — 3.
Find lim, 4 A(x) algebraically.
71. Let f(x) = (x? — 1)/(|x| — 1).
a. Make tables of the values of f at values of x that approach c = —] from above and below. Then estimate lim,_,_; f(x). b. Support your conclusion in part (a) by graphing f near c — —] and using Zoom and Trace to estimate y-values on the graph as x — —1. c. Find lim,- f(x) algebraically. 72. Let F(x) = (x? + 3x + 2)/(2 — |x|). a. Make tables of values of F at values of x that approach c = —2 from above and below. Then estimate lim,_._, F(x). b. Support your conclusion in part (a) by graphing F near c = —2 and using Zoom and Trace to estimate y-values on the graph as x — —2. c. Find lim,._, F(x) algebraically.
73. Let g(0) = (sin 0)/0.
a.
b.
Make a table of the values of g at values of 0 that approach 0, = 0 from above and below. Then estimate lim, ,o (0).
Support your conclusion in part (a) by graphing g near 0, = 0.
74. Let GA) = (1 — cos t/f.
a.
Make tables of values of G at values of t that approach f, = 0 from above and below. Then estimate lim; ,9 G(f).
Support your conclusion in part (a) by graphing G near lo = 0.
Theory and Examples
75. If x* € f(x) € x? for x in [-1,1] and x? < f(x) € x* for x < —] and x > 1, at what points c do you automatically know lim,—.. f(x)? What can you say about the value of the limit at these points?
76. Suppose that g(x) = f(x) = h(x) for all x # 2 and suppose that
lim g(x) = lim A(x) = —5.
Can we conclude anything about the values of f, g, and h at x = 2? Could f(2) = 0? Could lim,_,, f(x) = 0? Give reasons for your answers.
77. If lim — = 1, find lim f(o : xod do CONUM pay TO): 78. If lim m = |, find x>-2 X a. lim fœ) X b. dis D 79. a. If lim = 3, find li . 8. im e 7 in lim f(x). itn” 3 mun i =p x 2 CDS HDI lim pod
80. If lim 2? — ], find
[T] 81.
B
on the origin as necessary.
b. Confirm your estimate in part (a) with a proof.
[T] 82.
in on the origin as necessary.
Graph g(x) = xsin(1/x) to estimate lim, ,9 g(x), zooming in
a. Graph h(x) = x?cos(1/x?) to estimate lim,9 A(x), zooming — 87, Jim
2.3 The Precise Definition of a Limit 59
83. lim ————
d e c VL M a. in oe x-1 (x s 1)
Witx-1
85. lim x
x0
86 Mer Lc CM 47-4 ] — cosx xo0 xsinx
b. Confirm your estimate in part (a) with a proof. 88. lim 2x?
COMPUTER EXPLORATIONS
Graphical Estimates of Limits
x03 — 3cosx
In Exercises 83-88, use a CAS to perform the following steps:
a. Plot the function near the point c being approached.
b. From your plot guess the value of the limit.
2 The Precise Definition of a Limit
Upper bound: y-9
To satisfy this
Lower bound: y=s
í l l | Ly 34 ——* Restrict
to this
FIGURE 2.15 Keeping x within 1 unit
of x = 4 will keep y within 2 units of y = 7 (Example 1).
urn eae v =
We now turn our attention to the precise definition of a limit. We replace vague phrases like “gets arbitrarily close to” in the informal definition with specific conditions that can be applied to any particular example. With a precise definition, we can avoid misunder- standings, prove the limit properties given in the preceding section, and establish many important limits.
To show that the limit of f(x) as x — c equals the number L, we need to show that the gap between f(x) and L can be made “as small as we choose" if x is kept “close enough" to c. Let us see what this would require if we specified the size of the gap between f(x) and L.
EXAMPLE 1 Consider the function y = 2x — 1 near x = 4. Intuitively it appears that y is close to 7 when x is close to 4, so lim,_,4(2x — 1) = 7. However, how close to x = 4 does x have to be so that y = 2x — 1 differs from 7 by, say, less than 2 units?
Solution We are asked: For what values of x is |y — 7| « 2? To find the answer we first express |y — 7| in terms of x:
ly - 7| = |Qx - 1) - 7| = [2x - 8].
The question then becomes: what values of x satisfy the inequality |2x — 8| < 2? To find out, we solve the inequality:
2g — 8| «2 =) 2x 80952 6 < 2x « 10 3<x<5 Solve for x.
=l<x7=4 <1: Solve for x — 4.
Keeping x within | unit of x = 4 will keep y within 2 units of y = 7 (Figure 2.15). Oo
60 Chapter 2: Limits and Continuity
y
10 feo . f(x) lies L in here 1 L~i0 for all x zc in here "CNN - ox 0 c—6 c cô
FIGURE 2.16 How should we define ô > 0 so that keeping x within the interval (c — 0, c + 8) will keep f(x) within the
, 1 1 — — |? interval (u 10 L+ 5 y ^ Lt e? f(x) lies L 4 fe) in here L-—e«bp for all x zc in here ô ô —— ——— x 0 {—e o - >x c—óà c ct ó
FIGURE 2.17 The relation of ô and € in the definition of limit.
In the previous example we determined how close x must be to a particular value c to ensure that the outputs f(x) of some function lie within a prescribed interval about a limit value L. To show that the limit of f(x) as x — c actually equals L, we must be able to show that the gap between f(x) and L can be made less than amy prescribed error, no matter how small, by holding x close enough to c.
Definition of Limit
Suppose we are watching the values of a function f(x) as x approaches c (without taking on the value of c itself). Certainly we want to be able to say that f(x) stays within one- tenth of a unit from L as soon as x stays within some distance 6 of c (Figure 2.16). But that in itself is not enough, because as x continues on its course toward c, what is to prevent f(x) from jittering about within the interval from L — (1/10) to L + (1/10) without tending toward L?
We can be told that the error can be no more than 1/100 or 1/1000 or 1/100,000. Each time, we find a new 6-interval about c so that keeping x within that interval satisfies the new error tolerance. And each time the possibility exists that f(x) jitters away from L at some stage.
The figures on the next page illustrate the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents e-challenges to prove that the limit does not exist or, more precisely, that there is room for doubt. The scholar answers every challenge with a 6-interval around c that keeps the function values within e of L.
How do we stop this seemingly endless series of challenges and responses? We can do so by proving that for every error tolerance e that the challenger can produce, we can present a matching distance 6 that keeps x “close enough" to c to keep f(x) within that e-tolerance of L (Figure 2.17). This leads us to the precise definition of a limit.
DEFINITION Let f(x) be defined on an open interval about c, except possibly at c itself. We say that the limit of f(x) as x approaches c is the number L, and write lim f(x) = L, if, for every number e > 0, there exists a corresponding number 6 > 0 such
that for all x,
0O«|x-e «8 => |f@-L| <e.
One way to think about the definition is to suppose we are machining a generator shaft to a close tolerance. We may try for diameter L, but since nothing is perfect, we must be satisfied with a diameter f(x) somewhere between L — e and L + e. The 6 is the mea- sure of how accurate our control setting for x must be to guarantee this degree of accuracy in the diameter of the shaft. Notice that as the tolerance for error becomes stricter, we may have to adjust 6. That is, the value of 6, how tight our control setting must be, depends on the value of e, the error tolerance.
Examples: Testing the Definition
The formal definition of limit does not tell how to find the limit of a function, but it enables us to verify that a conjectured limit value is correct. The following examples show how the definition can be used to verify limit statements for specific functions. However, the real purpose of the definition is not to do calculations like this, but rather to prove gen- eral theorems so that the calculation of specific limits can be simplified, such as the theo- rems stated in the previous section.
2.3 The Precise Definition of a Limit 61
y y y =f) 1 L+ 109 L 1 | L- To mm | | | | i i | = l | | o| s U UNGE e P | c dic | uw e va € — dino c+ Ojo C — 81/100 c + 8in00 The challenge: Response: New challenge: Response: Make |f(x) = L| <€ =} |x — c| < 8o (a number) Make|fG) - L| <€ =i Ix — e| < ioo y y ^ y =f) LF L | | l >x c New challenge: Response: =! € = 1000 |x= e| <8ino00 x ^ T y y —fQ) y =f@) y =f) CRIS 1 L + 790,000 L + 169,900 EM 1 L L l I I 1 d i E A | | | Lae | L = 100,000 100,000 E | ! a | ! on l 5 7 >x 7 | 2 | >x o I >x New challenge: Response: New challenge: _ 1 € = 100,000 [x — e| < 81/100,000 Ens
EXAMPLE 2 Show that lim (5x — 3) = 2. Solution Set c = 1, f(x) = 5x — 3, and L = 2 in the definition of limit. For any given
€ > 0, we have to find a suitable 6 > O so that if x ~ 1 and x is within distance ô of c = 1, that is, whenever
0< |x= 1| <8, it is true that f(x) is within distance e of L = 2, so |f(x) — 2| < e.
62 Chapter 2: Limits and Continuity
y We find 6 by working backward from the e-inequality: -5x-3 p lx — 3 2| e ses] Se Six- 1| «e |x — 1| < e/5.
Thus, we can take 6 = €/5 (Figure 2.18). If 0 < |x — 1| < 8 = e/5, then l5x — 3) — 2| = |5x — 5| = 5|x — 1| < 5(e/5 = e,
which proves that lim,-.,(5x — 3) = 2. The value of à = €/5 is not the only value that will make 0 < |x — 1| < 8 imply [5x — 5| < e. Any smaller positive 5 will do as well. The definition does not ask for a
“best” positive 6, just one that will work. E -3 EXAMPLE 3 Prove the following results presented graphically in Section 2.2. NOT TO SCALE (a) lim x=c
FIGURE 2.18 If f(x) = 5x — 3, then (b) lim k =k (kconstant) 0 « |x — 1| < e/5 guarantees that xc | f(x) - 2| < e (Example 2). Solution (a) Let e > 0 be given. We must find 6 > 0 such that for all x
0< |x-c| «6 implies Ix- e| < e. 4 B The implication will hold if 6 equals e or any smaller positive number (Figure 2.19). F T This proves that lim, =, x = c. [^ er
(b) Let e > 0 be given. We must find 6 > 0 such that for all x Oc«|x-c| «8 implies — |k — k| <e.
Since k — k = 0, we can use any positive number for 6 and the implication will hold (Figure 2.20). This proves that lim, ,. k = k. ig
| | | | | | | 1 C
Finding Deltas Algebraically for Given Epsilons
FIGURE 2.19 For the function In Examples 2 and 3, the interval of values about c for which |f(x) — L| was less than e f(x) = x, we find that 0 < |x — c| < ô was symmetric about c and we could take 6 to be half the length of that interval. When will guarantee | f(x) — c| < e whenever such symmetry is absent, as it usually is, we can take 6 to be the distance from c to the ô < e (Example 3a). interval's nearer endpoint.
EXAMPLE 4 For the limit lim,.,5 Vx — 1 = 2, find a ô > 0 that works for e = 1. That is, find a 6 > O such that for all x
ü«lx-5|€8 = [Vx 192 1,
e Solution We organize the search into two steps. k—e
1. Solve the inequality | Vx — 1 — 2| « 1 to find an interval containing x — 5 on which the inequality holds for all x # 5.
[We = 2) al mE ix -1 < Ve 1=2 21
1< Vx-1<3 FIGURE 2.20 For the function x
f(x) = k, we find that | f(x) — k| < € Leg dms for any positive ô (Example 3b). 2<x< 10
><
l l l I l l | 0 c—ó c
2.3 The Precise Definition of a Limit 63
3 3 The inequality holds for all x in the open interval (2, 10), so it holds for all x # 5 in this interval as well.
2. Find a value of 6 > 0 to place the centered interval 5 — 6 < x < 5 + 6 (centered at x — 5) inside the interval (2, 10). The distance from 5 to the nearer endpoint of (2, 10) is 3 (Figure 2.21). If we take 6 = 3 or any smaller positive number, then the inequality 0 < |x — 5| < 6 will automatically place x between 2 and 10 to make
| Vx — 1 — 2| « 1 (Figure 2.22):
FIGURE 2.21 An open interval of ra- dius 3 about x — 5 will lie inside the open interval (2, 10).
0< x-5|*«3 = [Wee T3 <i. El
How to Find Algebraically a ô for a Given f, L, c, and £ > 0 The process of finding a 6 > 0 such that for all x
0< |x-c| <6 => lf) - L| «e can be accomplished in two steps.
1. Solve the inequality | f(x) — L| < e to find an open interval (a, b) contain- ing c on which the inequality holds for all x # c.
NOT TO SCALE 2. Find a value of 5 > 0 that places the open interval (c — 8, c + ô) centered at c inside the interval (a, b). The inequality | f(x) — L| < e will hold for all
FIGURE 2.22 The function and inter- sc oundüsSantervil
vals in Example 4.
EXAMPLE 5 Prove that lim,_., f(x) = 4 if
x^, x #2 fe I
Solution Our task is to show that given e > 0 there exists a 6 > 0 such that for all x 0«|x-2| «6 => |fG) — 4| « e. 1. Solve the inequality |f(x) — 4| < e to find an open interval containing x = 2 on which the inequality holds for all x # 2. For x # c = 2, we have f(x) = x’, and the inequality to solve is |x? — 4| < e: ie - 4| «e -e<w-4<e 4-e<xv<4+.e
IQ. D |
l
. 4-e< |x] = 4+€ Assumes € < 4; see below.
l :
A t l about x = 2 0 2N E Vcg SM VA e. aoii vV4-e Wate The inequalit X) — 4| < e holds for all x # 2 in the open interval (V4 — e, FIGURE 2.23 An interval containing q y H ) | P (
V4 + €) (Figure 2.23).
x = 2 so that the function in Example 5
satisfies | f(a) — 4| < e. 2. Find a value of 6 > 0 that places the centered interval (2 — 6,2 + 6) inside the
interval ( V4 — e, V4 * e).
Take ô to be the distance from x = 2 to the nearer endpoint of (V4 — e, V4 + e). In other words, take 6 = min {2 — V4—eV4-e-— 2}, the minimum (the
64
Chapter 2: Limits and Continuity
smaller) of the two numbers 2 — V4 — e and V4 + e — 2. If ô has this or any smaller positive value, the inequality 0 < |x — 2| < 6 will automatically place x
between V4 — e and V4 + e to make | f(x) — 4| < e. For all x, P= <8 = f@ -4| < e.
This completes the proof for e < 4. If € = 4, then we take 6 to be the distance from x = 2 to the nearer endpoint of
the interval (0, V4 + €). In other words, take 6 = min 18 V4 t e-— a). (See Figure 2.23.) a
Using the Definition to Prove Theorems
We do not usually rely on the formal definition of limit to verify specific limits such as those in the preceding examples. Rather, we appeal to general theorems about limits, in particular the theorems of Section 2.2. The definition is used to prove these theorems (Appendix 5). As an example, we prove part 1 of Theorem 1, the Sum Rule.
EXAMPLE 6 Given that lim,_,. f(x) = L and lim, g(x) = M, prove that lim (f(x) + g(x)) = L + M. Solution Let e > 0 be given. We want to find a positive number 6 such that for all x 0< |x-c| <6 => FŒ + g(x) — (L + M)| < e. Regrouping terms, we get
FŒ + 6) — (L + M)|
[6 (x) — L) + (g(x) — M)| Triangle Inequality: Ifc) — L| + |gG) — MI. la + b| = |a| + |p|
IA
Since lim, ,, f(x) = L, there exists a number 6, > 0 such that for all x
0< |x- c| < ô => Ife) — L| < «/2. Similarly, since lim, ,. g(x) = M, there exists a number 6, > 0 such that for all x
0 < |x — c| < ê — eG) — M| < €/2. Let ô = min {6,, ô}, the smaller of 5, and ô. If 0 < |x — c| < 6 then |x — c| < ô, so | f(x) — L| < e/2, and |x — c| < ô, so |g(x) — M| < e/2. Therefore
fe) + s - Cc M| «5-57 €
This shows that lim, , (f(x) + g0)0) ^ L + M. Oo
Next we prove Theorem 5 of Section 2.2.
EXAMPLE 7 Given that lim,_,. f(x) = L and lim,_.. g(x) = M, and that f(x) = g(x) for all x in an open interval containing c (except possibly c itself), prove that L = M.
Solution We use the method of proof by contradiction. Suppose, on the contrary, that L > M. Then by the limit of a difference property in Theorem 1,
lim(gG) — f(9) = M — L.
2.3 The Precise Definition of a Limit 65
Therefore, for any € > 0, there exists 6 > 0 such that
(gx) — fo) -M -D| «e
whenever 0 < |x — c| < 6.
Since L — M > 0 by hypothesis, we take e = L — M in particular and we have a num-
ber 6 > O such that
(ig) — f) -(M- D| <L-M
whenever 0 < |x — c| < 6.
Since a < |a| for any number a, we have
(gx) — fo) -(M- D«-L-M
which simplifies to
g(x) < FQ)
whenever 0 < |x —c| <6
whenever 0 < |x — c| < ô.
But this contradicts f(x) = g(x). Thus the inequality L > M must be false. Therefore
L x M.
Exercises
Centering Intervals About a Point
In Exercises 1—6, sketch the interval (a, b) on the x-axis with the point c inside. Then find a value of 6 — O such that for all x0«lix-c «8 > a<x<b.
la=1, b-7, c=5
2 a=1, b=7, c=2
= —7/2, b=—-1/2, c — —3 = —7/2, b - 1/2, c= —3/2 = 4/9, b=4/7, c= M2
= 2.7591, b= 32391, c=3
D u e w
a a a a
Finding Deltas Graphically In Exercises 7-14, use the graphs to find a 6 > 0 such that for all x 0< |x- e| «6 = |f@ - L| < e.
7 8. y y=2x-4 6.2 f(x) =2x-4 6r----- l c=5 oN AF | L=6 I e€ = 0.2 l 0 Z $5 p 4.9 5.1 NOT TO SCALE
NOT TO SCALE
><
f(x) =2Vx4+1
c=
AIU = AIU
l | l l l | | -10 2.61 3 341
NOT TO SCALE
11. 12.
NOT TO SCALE
NOT TO SCALE
66 Chapter 2: Limits and Continuity
13. 14. y ^ fo-1 ixl crt 2 2.01 fe € = 0.01 2 Ii onera eis 1.99 ----4- l I ! I I I I I I | I I I ae I I | | d. 0 7 1 S *x zoi ? T99
NOT TO SCALE
Finding Deltas Algebraically
Each of Exercises 15-30 gives a function f(x) and numbers L, c, and € — 0. Ineach case, find an open interval about c on which the inequal- ity |fG) — L| < e holds. Then give a value for 6 > 0 such that for all x satisfying 0 < |x — c| < & the inequality |f(x) — L| < e holds.
15. fi x lj L=5, c-4, e-001
16. f) =2x-2, L=-6 c=-2, e-00 17. f@=Vxt+1, L=1, c=0, e-01 18. {H= Vx, LS1À c=1/4 xe 19. fo) VI9-xX L3, c=10, e-1 20. f) 2 Vx - 2, L=4 c=23, e-21 24. j(0 = ie L1/4 c4 e-005
22. f0 3i) L3, c=V3, e-01
23. f(x) = x, L=4, c=-2, e — 05
24. fo =1/x, L=-l, c=-l, e-01
25. fi) = x2— 5, L=11, c=4, e=1 26. f(x) = 120/x, L=5, c = 24, ELE 27. f(x) = m, m>O0, L= 2m, c=2, e= 0.03
28. f(x) = mx, m > 0, L = 3m, c=3, e=c>0 29. f(x) = mx + b, m > 0, L = (m/2) + b,
c = 1/2, e=c>0 30. f(x) = mx + b, m > 0, L=m+ b, e 1,
e = 0.05
Using the Formal Definition
Each of Exercises 31—36 gives a function f(x), a point c, and a posi- tive number e. Find L = lim f(x). Then find a number 6 > 0 such that for all x dis
0<|x-cl <6 => lf) — L| < e. 31. f(x) = 3 — 2x, c = 3, e = 0.02 32. f(x) = —3x - 2, c — —l, € — 0.03
x -4 B _ 33. f(x) = x2 c—2, € — 0.05
x + 6x45 B
34. fx) = BS, €-0905 35. f(x) = V1 — 5x, c= —3, e = 0.5 36. f(x) = 4/x, c= 2, e= 0.4
Prove the limit statements in Exercises 37—50. 37. lim (9 —x)-5 38. lim 3x —7)22 Kn x3
39. lim Vx -5-2 40. lim V4 — x = : . x? xzl 41. lim f() = 1 if fœ) = { x1 2, x= 1 2 z—2 42. lim f®) =4 if fœ = {i lá x2 UP X-—-—2 "M C 43. lim xcd 44. im; 273 "P 2-9 2 B 45. im xq 6 46. lim z 2 47. lim fœ@=2 if fey = [ios x«l "ue P : B 6x — 4, x21 2x x«0 48. li =0 if = i lim f@) =0 if fa) l Ee
49. lim x sin + = 0 x0
50. lim x? sinl =0 x0
Theory and Examples 51. Define what it means to say that lim g(x) = k.
52. Prove that lim f(x) = L if and only if lim f(h * c) 7» L. x- ^c (nd
53. A wrong statement about limits Show by example that the following statement is wrong.
The number L is the limit of f(x) as x approaches c if f(x) gets closer to L as x approaches c.
Explain why the function in your example does not have the given value of L as a limit as x — c. 54. Another wrong statement about limits Show by example that
the following statement is wrong.
The number L is the limit of f(x) as x approaches c if, given any € > 0, there exists a value of x for which | f(x) — L| < e.
Explain why the function in your example does not have the given value of L as a limit as x > c.
55. Grinding engine cylinders Before contracting to grind engine
cylinders to a cross-sectional area of 9 in?, you need to know how much deviation from the ideal cylinder diameter of c — 3.385 in. you can allow and still have the area come within 0.01 in? of the required 9 in?. To find out, you let A = 7(x/2)? and look for the interval in which you must hold x to make |A — 9| < 0.01. What interval do you find?
56. Manufacturing electrical resistors Ohm’s law for electrical circuits like the one shown in the accompanying figure states that V = RI. In this equation, V is a constant voltage, / is the current in amperes, and R is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which V will be 120 volts and J is to be 5 + 0.1 amp. In what interval does R have to lie for / to be within 0.1 amp of the value J) = 5?
fay}, tT O
2.3 The Precise Definition of a Limit 67
When Is a Number L Not the Limit of f(x) as x— c? Showing L is not a limit We can prove that lim,_,. f(x) # L by providing an e > 0 such that no possible 6 > 0 satisfies the condition
forallx, 0 < |x— c| «6 => |f(x) — L| < e.
We accomplish this for our candidate e by showing that for each ô > 0 there exists a value of x such that
0clx-c|«6 and fe) — L| = e.
» =f@)
fœ)
|
|
|
|
|
i >x 0| c-6\| c c+ô
a value of x for which 0<|x-c| <6 and| f) - L| ze
xs
x, 57. Let fœ) = et feo ~ x> 1.
><
y =f@)
>x
a. Lete = 1/2. Show that no possible 6 > 0 satisfies the fol- lowing condition:
Forallx, O«|x-1|«8 = |f(@ — 2| < 1/2.
That is, for each 6 > 0 show that there is a value of x such that
0c|x-1|«8 an |fG)-2| = 1/2.
This will show that lim,—, f(x) # 2.
b. Show that lim,.,, f(x) 7 1. €. Show that lim,—, f(x) # 1.5.
68 Chapter 2: Limits and Continuity
x] x«2 60. a. For the function graphed here, show that lim,._, g(x) # 2. 58. Leth(x) = 43, x= b. Does lim,—.-; g(x) appear to exist? If so, what is the value of 2. x2. the limit? If not, why not? y A y ^ y = h(x) y=2 oo >x COMPUTER EXPLORATIONS Show that In Exercises 61-66, you will further explore finding deltas graphi- a dimh sd cally. Use a CAS to perform the following steps: T2 a. Plot the function y — f(x) near the point c being approached. b. lim h(x) ¥ 3 ae A x2 b. Guess the value of the limit L and then evaluate the limit sym- c. lim h(x) # 2 bolically to see if you guessed correctly. x2
c. Using the value e = 0.2, graph the banding lines y; = L — €
59. For the function graphed here, explain why and y; = L + e together with the function f near c.
= lim Tet d. From your graph in part (c), estimate a 6 > 0 such that for all x B: lim f) # 4.8 0<|x-cl <6 => fco — L| < e.
c. lim f(x) 73 Test your estimate by plotting f, yı, and y» over the interval
uic 0 < |x — e| < 8. For your viewing window use c — 28 =
x Sc + 268 and L — 2e S y < L + 2e. If any function val-
i ues lie outside the interval [L — €, L + €], your choice of 6 was too large. Try again with a smaller estimate. i e. Repeat parts (c) and (d) successively for e — 0.1, 0.05, and 0.001. 48r 4 34 9,2 Aa ‘ 6. fo) =F co3 62 f = 35S. c= 0 y =f) fi m 3r _ sin 2x i _ xl — cos x _ iN 63. f(x) = EE c=0 64. f(x) care f 0 3 — 65. f(x) = x c=1 3x2 — (Ix + D Vx +5 0 3 Tue 66. f(x) = cd a c=l
2 " 4 One-Sided Limits
In this section we extend the limit concept to one-sided limits, which are limits as x approaches the number c from the left-hand side (where x < c) or the right-hand side (x > c) only.
Approaching a Limit from One Side
To have a limit L as x approaches c, a function f must be defined on both sides of c and its values f(x) must approach L as x approaches c from either side. That is, f must be defined in some open interval about c, but not necessarily at c. Because of this, ordinary limits are called two-sided.
><
> X
FIGURE 2.24 Different right-hand and left-hand limits at the origin.
>X
-2 0 2
FIGURE 2.26 The function
f(x) = V4 — x? has right-hand limit 0 at x = —2 and left-hand limit 0 at x = 2 (Example 1).
2.4 One-Sided Limits 69
If f fails to have a two-sided limit at c, it may still have a one-sided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a right-hand limit. From the left, it is a left-hand limit.
The function f(x) = x/ |x] (Figure 2.24) has limit 1 as x approaches 0 from the right, and limit — 1 as x approaches 0 from the left. Since these one-sided limit values are not the same, there is no single number that f(x) approaches as x approaches 0. So f(x) does not have a (two-sided) limit at 0.
Intuitively, if f(x) is defined on an interval (c, b), where c < b, and approaches arbi- trarily close to L as x approaches c from within that interval, then f has right-hand limit L at c. We write
lim f(x) = L.
The symbol “x — c*” means that we consider only values of x greater than c.
Similarly, if f(x) is defined on an interval (a, c), where a < c and approaches arbi- trarily close to M as x approaches c from within that interval, then f has left-hand limit M at c. We write
lim f(x) — M. The symbol “x — c " means that we consider only x-values less than c.
These informal definitions of one-sided limits are illustrated in Figure 2.25. For the function f(x) = x/|x| in Figure 2.24 we have
lim, f@=1 and lim fe») = -1.
y y ^ ^ L feo feo u >x >x 0 Co— X 0 X — C (a lim f()-L (b lim f(x) =M
FIGURE 2.25 (a) Right-hand limit as x approaches c. (b) Left-hand limit as x approaches c.
EXAMPLE 1 The domain of f(x) = V4 — x? is [—2, 2]; its graph is the semicircle in Figure 2.26. We have
lim V4 — x =0 and lim V4 — x? = 0.
2 I> The function does not have a left-hand limit at x = —2 or a right-hand limit at x = 2. It does not have a two-sided limit at either —2 or 2 because each point does not belong to an open interval over which f is defined. a
One-sided limits have all the properties listed in Theorem 1 in Section 2.2. The right-hand limit of the sum of two functions is the sum of their right-hand limits, and so on. The theorems for limits of polynomials and rational functions hold with one-sided limits, as do the Sandwich Theorem and Theorem 5. One-sided limits are related to limits in the following way.
THEOREM 6 A function f(x) has a limit as x approaches c if and only if it has left-hand and right-hand limits there and these one-sided limits are equal:
limfo)-L e lim fœ) =L and — lim fQ) = L.
70
y = f@)
0 1 2 3 4
FIGURE 2.27 Graph of the function in Example 2.
Lote @ f(x) f(x) lies Le in here L-ew for all x £c in here = ô, x e >x 0 c c+6 FIGURE 2.28 Intervals associated with
the definition of right-hand limit.
y ^ L+em @ f) f(x) lies Le in here L-e for all x £c in here — € Hd >x 0 c—óà c
FIGURE 2.29 Intervals associated with the definition of left-hand limit.
Chapter 2: Limits and Continuity
EXAMPLE 2 For the function graphed in Figure 2.27,
At x — 0: lim, .o f(x) = 1, lim, 59- f(x) and lim,_.9 f(x) do not exist. The function is not de- fined to the left of x = 0.
Atx= 1: lim,—.;- f(x) = 0 even though f(1) = 1, lim, f(x) = 1, lim,—.; f(x) does not exist. The right- and left-hand limits are not equal.
At x = 2: lim, f(x) = 1, lim, 55« f(x) = 1, lim,» f(x) = 1 even though f(2) = 2.
Atx = 3 lim,+3- f(x) = lim,..3+ f(x) = lim,.; f(x) = f(3) = 2.
Atx=4 lim,_.4- f(x) = 1 even though f(4) # 1, lim,_.4+ f(x) and lim,—.4 f(x) do not exist. The function is not defined to the right of x = 4.
At every other point c in [0,4], f(x) has limit f(c). El
Precise Definitions of One-Sided Limits
The formal definition of the limit in Section 2.3 is readily modified for one-sided limits.
DEFINITIONS We say that f(x) has right-hand limit L at c, and write
lim, f@Mm=L (see Figure 2.28) if for every number e > 0 there exists a corresponding number 6 > 0 such that for all x
CLIS CEO => fŒ- L| < e,
We say that f has left-hand limit L at c, and write
lim. fœ) =L (see Figure 2.29) if for every number e > 0 there exists a corresponding number 6 > 0 such that for all x
c-8<x<e => IfG) — L| < e.
EXAMPLE 3 Prove that
Solution Let e > 0 be given. Here c = 0 and L = 0, so we want to find a 6 > 0 such that for all x
0cx«à8 =>» |vx-O0|«e Or
Vx < e.
0<x<6 =>
fo) = Vx
FIGURE 2.30 lim, Vx = 0 in Example 3.
2.4 One-Sided Limits 71
Squaring both sides of this last inequality gives x«e if O<x<6.
If we choose ô = €? we have
0<x<ôĝ= = Wee or 0Ocx«e >» |Vx-Ol<e. According to the definition, this shows that lim, sg Vx — 0 (Figure 2.30). Bg
The functions examined so far have had some kind of limit at each point of interest. In general, that need not be the case.
EXAMPLE 4 Show that y — sin(1/x) has no limit as x approaches zero from either side (Figure 2.31).
FIGURE 2.31 The function y = sin(1/x) has neither a right- hand nor a left-hand limit as x approaches zero (Example 4). The graph here omits values very near the y-axis.
Solution As x approaches zero, its reciprocal, 1/x, grows without bound and the values of sin (1/x) cycle repeatedly from —1 to 1. There is no single number L that the function’s values stay increasingly close to as x approaches zero. This is true even if we restrict x to positive values or to negative values. The function has neither a right-hand limit nor a left- hand limit at x = 0. Oo
Limits Involving (sin 0)/0
A central fact about (sin 0)/0 is that in radian measure its limit as 0 — 0 is 1. We can see this in Figure 2.32 and confirm it algebraically using the Sandwich Theorem. You will see the importance of this limit in Section 3.5, where instantaneous rates of change of the trigonometric functions are studied.
NOT TO SCALE
FIGURE 2.32 The graph of f(0) = (sin 0)/0 suggests that the right- and left-hand limits as 0 approaches 0 are both 1.
72 Chapter 2: Limits and Continuity
><
FIGURE 2.33 The figure for the proof of Theorem 7. By definition, TA/OA = tan 6, but OA = 1, so TA = tan 0.
Equation (2) is where radian measure comes in: The area of sector OAP is 0/2
only if 0 is measured in radians.
THEOREM 7—Limit of the Ratio sin 0/0 as 0— 0
lin —5— = 1 (0 in radians) (1)
Proof The plan is to show that the right-hand and left-hand limits are both 1. Then we will know that the two-sided limit is 1 as well.
To show that the right-hand limit is 1, we begin with positive values of @ less than T /2 (Figure 2.33). Notice that
Area AOAP < area sector OAP < area AOAT.
We can express these areas in terms of 0 as follows:
Area AOAP = line X height — T (Din 0) — 7 sin 0
2 Area sector OAP = Ld = l ay =e (2)
2 2 2
|. 1 : _ 1 _ 1
Area AOAT = 3 base X height = 5 (tan 0) — z tan 0. Thus, I. ss 1 1 z sin 0 < 79 < z tan 6.
This last inequality goes the same way if we divide all three terms by the number (1/2) sin@, which is positive, since 0 < 0 < 7/2:
sin@ ` cos @ Taking reciprocals reverses the inequalities:
1 > 356 > cos.
Since limg—.9:cos@ = 1 (Example 11b, Section 2.2), the Sandwich Theorem gives
. sind lim = (0. 0
1.
To consider the left-hand limit, we recall that sin 0 and 0 are both odd functions (Sec- tion 1.1). Therefore, f(0) = (sin 0)/0 is an even function, with a graph symmetric about the y-axis (see Figure 2.32). This symmetry implies that the left-hand limit at 0 exists and has the same value as the right-hand limit:
Yn sinô _ | 2 lim 92 i» 6—0- oo 0 so limọ—o (sin 09)/0 = 1 by Theorem 6. Ei
. cosh—]l _ . Sin2x _ 2 EXAMPLE 5 Show that (a) lim h — 0 and (b) lim 5x 5"
2.4 One-Sided Limits 73
Solution (a) Using the half-angle formula cos h = 1 — 2 sin? (h /2), we calculate
cosh — 1 . 2 sin? (h/2) A UT im — —— ——
I = | h>0 h h=>0 h .. sind. = —]im ——sin 0 Let 0 — h/2. 650 0 mE u Eq. (1) and Example 11a = -QX0) = 0. in Section 2.2
(b) Equation (1) does not apply to the original fraction. We need a 2x in the denominator, not a 5x. We produce it by multiplying numerator and denominator by 2/5: sin2x _ ,. (2/5)* sin 2x
ub x C215 GO
| 2, sin2x Now, Eq. (1) applies 5, ' 2x with 6 = 2x.
EXAMPLE 6 Find lim ee t [—!
Solution From the definition of tan ¢ and sec 2f, we have
tanfsec2t _ |; 1 1 sint 1
l 3t 3 t^ cost cos 2t _ li smr. 1] . 1 3:00 t cost cos2t E 1 Eq. (1) and Example 11b g 3DU) B 3 in Section 2.2 d
Exercises EE
Finding Limits Graphically 1. Which of the following statements about the function y = f(x) graphed here are true, and which are false?
y = f@)
=i t—1" ü " ae, fo) = 1 Pe un a=? a. lim ,f00-71 b. lim f(x) does not exist. ub ci uu i s o^ fx) =2 d. lim. fe922 T 2m PEERS: f. im mca e. ‘im, fa) =1 f. lim f(x) does not exist. b limfo cd i =1 g. pe f(x) = lim f(x) m lim fuo j. pi Te h. im fe») m every c in the open interval (— 1, 1). k. a: TO) Goes MOUSSE l. pn Toc i. lim f(x) exists at every c in the open interval (1, 3). 2 whch of te elongata he moton y= SO — je uoo fi Jo) des tex.
74 Chapter 2: Limits and Continuity
3-x x«2 6. Let g(x) = Vx sin(1/x).
3. Let f(x) = UNT ed, adem 1
a. Find lim, ,,- f(x) and lim, ,,- f(x). b. Does lim, ,; f(x) exist? If so, what is it? If not, why not? 2 c. Find lim, ,4- f(x) and lim, ,4- f(x). d. Does lim, 4 f(x) exist? If so, what is it? If not, why not? a. Does lim, 9 g(x) exist? If so, what is it? If not, why not? b. Does lim, ,$- g(x) exist? If so, what is it? If not, why not? 3X «2 : ] - "ee c. Does lim,—o g(x) exist? If so, what is it? If not, why not? _ J2, x=2 4. Let f(x) = x x, x*1 > xc. 7. a. Graph f(x) = 0. ped: b. Find lim, ,,- f(x) and lim,_,,+ f(x). c. Does lim,_,, f(x) exist? If so, what is it? If not, why not? yo Boo 1-x, x*1 ; 8. a. Graph f(x) = { 2, x=1. j b. Find lim,_.;+ f(x) and lim,—.;- f(x). i : c. Does lim,_.; f(x) exist? If so, what is it? If not, why not? — LLL Loop od =2 0 2 Graph the functions in Exercises 9 and 10. Then answer these questions. a. Find lim,» f(x), lim,—»- f(x), and f(2). a. What are the domain and range of f? b. Does lim, f(x) exist? If so, what is it? If not, why not? b. At what points c, if any, does lim,—,, f(x) exist? c. Find lim,- f(x) and lim, =>- f(x). c. At what points does only the left-hand limit exist? d. Does lim,—_, f(x) exist? If so, what is it? If not, why not? d. At what points does only the right-hand limit exist? V1-x, 0sx<1 0, x=0 5. Let f(x) = i 9. f(x) = 41, l1sx<2 sing, x > 0. 2, x= 2 X, =lsy<0 or O< xs 1 10. fx) = 41, x=0 0, x<-l o x>1 Finding One-Sided Limits Algebraically Find the limits in Exercises 11-18. , +2 : ee 1 Hn zu. x1 12. am x+2 : x 2x F5 13. mG + E + 5) 14. lim 1 HO 3 —X a. Does lim, o f(x) exist? If so, what is it? If not, why not? *aorAxcl x 7 b. Does lim,_.9- f(x) exist? If so, what is it? If not, why not? ji: mm V + 4h +5- V5
€. Does lim,—o f(x) exist? If so, what is it? If not, why not? h0* h
V6 — V5R? + 11h 6
15 xdi h | lx 2] | ET 17. a. dim, 0 + 3) z772 b. lm x +3) PEN V 2. = 1 V 2. = TEM ec ho ea xr [x — I| xr |x — 1|
Use the graph of the greatest integer function y = | x |, Figure 1.10 in Section 1.1, to help you find the limits in Exercises 19 and 20.
le] Le]
19. a. Hm y b. jim gy 20. a. lim(t — |1]) b. lim(t — |£]) 14* tro
; ,. sind Using lim B " 1
2.5 Continuity 75
41. lim AU - 42. lim eod S. 6—0 0? cot 30 60 sin? 0 cot? 20
Theory and Examples
43. Once you know lim,—,+ f(x) and lim,—,- f(x) at an interior point of the domain of f, do you then know lim,_., f(x)? Give reasons for your answer.
44. If you know that lim, ,, f(x) exists, can you find its value by cal- culating lim,—..+ f(x)? Give reasons for your answer.
45. Suppose that f is an odd function of x. Does knowing that lim,—o+ f(x) = 3 tell you anything about lim, ,- f(x)? Give rea- sons for your answer.
46. Suppose that f is an even function of x. Does knowing that lim, .5- f(x) = 7 tell you anything about either lim,—._,- f(x) or lim,—.-5+ f(x)? Give reasons for your answer.
Formal Definitions of One-Sided Limits 47. Given e > 0, find an interval J = (5, 5 + ô), 6 > 0, such that if
. sin V 20 . sinkt x lies in J, then Vx — 5 « e. What limit is being verified and 21. lim ———— 22. lim (k constant) EE 9 60 4/20 10 t what is its value? . sin3y . h 48. Given e > 0, find an interval J = (4 — 8,4), 6 > 0, such that if 23. lim 4y 24. pi sin 3h x lies in J, then V4 — x « e. What limit is being verified and tan 2x 2t what is its value? 25. lim 26. lim —— m" . - x0 0 tan f Use the definitions of right-hand and left-hand limits to prove the limit statements in Exercises 49 and 50. 21. lim £596 24 28. lim 6x?(cot x)(csc 2x) P "er — EN 2 . 49. lim |x] --—] 50. lim ix-2| =] = , x> pot l= 29. lim% S X COS x 30. lim? s sin x X SCR A i 51. Greatest integer function Find (a) lim, 49» [x] and (b) 31. lim 1 — cos 32. lim — = COS lim, ,495- |x |; then use limit definitions to verify your findings. eo sin20 x20 sin? 3x (c) Based on your conclusions in parts (a) and (b), can you say 33. lim sin(1 — cos f) 34. lim sin (sin h) anything about lim,—.199 | x |? Give reasons for your answer. “+0 1-— cost ` 10 sinh 2 sin(1/x) " : i x^sin(l/x, x< sin 0 . sin 5x 52. One-sided limits Let f(x) = { í Ja: gm sin 20 36 m sin 4x Vx, x > 0. 37. lim 0 cos 0 38. lim sin 0 cot 20 Find (a) lim,—ọ f(x) and (b) lim, ,- f(x); then use limit defini- 9*0 9 tions to verify your findings. (c) Based on your conclusions in 39. lim tan 3x 40. ig parts (a) and (b), can you say anything about lim,_.) f(x)? Give x0 sin 8x »20 ycot4y reasons for your answer. 2 ê 5 Continuity y When we plot function values generated in a laboratory or collected in the field, we often
500
375
Distance fallen (m) N nn e
Elapsed time (sec)
FIGURE 2.34 Connecting plotted points by an unbroken curve from experimental
data Qi. Q2, Qs...
. for a falling object.
Continuity at a Point
connect the plotted points with an unbroken curve to show what the function's values are likely to have been at the points we did not measure (Figure 2.34). In doing so, we are assuming that we are working with a continuous function, so its outputs vary regularly and consistently with the inputs, and do not jump abruptly from one value to another without taking on the values in between. Intuitively, any function y — f(x) whose graph can be sketched over its domain in one unbroken motion is an example of a continuous function. Such functions play an important role in the study of calculus and its applications.
To understand continuity, it helps to consider a function like that in Figure 2.35, whose limits we investigated in Example 2 in the last section.
76 Chapter 2: Limits and Continuity
FIGURE 2.35 The function is not continuous at x = l,x = 2, and x = 4 (Example 1).
Continuity Two-sided
from the right ^ continuity Continuity — from the left Ü& OG l s l 1 y=f@ | l l | | l | l L >y a 5 b
FIGURE 2.36 Continuity at points a, b, and c.
EXAMPLE 1 At which numbers does the function f in Figure 2.35 appear to be not continuous? Explain why. What occurs at other numbers in the domain?
Solution First we observe that the domain of the function is the closed interval | 0, 4], so we will be considering the numbers x within that interval. From the figure, we notice right away that there are breaks in the graph at the numbers x = 1, x = 2, and x = 4. The breaks appear as jumps, which we identify later as “jump discontinuities.” These are num- bers for which the function is not continuous, and we discuss each in turn.
Numbers at which the graph of f has breaks:
At x = 1, the function fails to have a limit. It does have both a left-hand limit, lim,_,;- f(x) = 0, as well as a right-hand limit, lim, ,,- f(x) = 1, but the limit values are different, resulting in a jump in the graph. The function is not continuous at x — 1.
At x = 2, the function does have a limit, lim, ,; f(x) = 1, but the value of the func- tion is f(2) — 2. The limit and function values are not the same, so there is a break in the graph and f is not continuous at x — 2.
At x = 4, the function does have a left-hand limit at this right endpoint, lim, ,4- f(x) = 1, but again the value of the function f(4) — i differs from the value of the limit. We see again a break in the graph of the function at this endpoint and the function is not continu- ous from the left.
Numbers at which the graph of f has no breaks:
At x = 0, the function has a right-hand limit at this left endpoint, lim, ,9- f(x) = 1, and the value of the function is the same, f(0) — 1. So no break occurs in the graph of the function at this endpoint, and the function is continuous from the right at x — 0.
At x = 3, the function has a limit, lim, ,4 f(x) = 2. Moreover, the limit is the same value as the function there, f(3) — 2. No break occurs in the graph and the function is continuous at x — 3.
At all other numbers x = c in the domain, which we have not considered, the func- tion has a limit equal to the value of the function at the point, so lim, ,, f(x) = f(c). For example, lim, ,5/5 f(x) = f (3) = 3. No breaks appear in the graph of the function at any of these remaining numbers and the function is continuous at each of them. a
The following definitions capture the continuity ideas we observed in Example 1.
DEFINITIONS Let c be a real number on the x-axis.
The function f is continuous at c if lim f(x) = f(o).
The function f is right-continuous at c (or continuous from the right) if lim f(x) = f(o).
The function f is left-continuous at c (or continuous from the left) if
lim fa) = f(o.
From Theorem 6, it follows immediately that a function f is continuous at an interior point c of its domain if and only if it is both right-continuous and left-continuous at c (Fig- ure 2.36). We say that a function is continuous over a closed interval |a, b] if it is right- continuous at a, left-continuous at b, and continuous at all interior points of the interval.
FIGURE 2.37 A function that is continuous over its domain (Example 2).
FIGURE 2.38 A function that has a jump discontinuity at the origin (Example 3).
4r —- 3r- e—o y=([x] 2} o—o lr e—o l o— l l L— >y =! 1 2 3 4 — —O —2rT
FIGURE 2.39 The greatest integer function is continuous at every noninte- ger point. It is right-continuous, but not left-continuous, at every integer point (Example 4).
2.5 Continuity 77
This definition applies to the infinite closed intervals [ a, oo) and (~œ, b ] as well, but only one endpoint is involved. If a function is not continuous at an interior point c of its domain, we say that f is discontinuous at c, and that c is a point of discontinuity of f. Note that a function f can be continuous, right-continuous, or left-continuous only at a point c for which f(c) is defined.
EXAMPLE 2 The function f(x) = V4 — x? is continuous over its domain [—2, 2] (Figure 2.37). It is right-continuous at x = —2, and left-continuous at x = 2. E
EXAMPLE 3 The unit step function U(x), graphed in Figure 2.38, is right-continuous at x = 0, but is neither left-continuous nor continuous there. It has a jump discontinuity at x=0. Oo
We summarize continuity at an interior point in the form of a test.
Continuity Test
A function f(x) is continuous at a point x = c if and only if it meets the follow- ing three conditions.
1. f(c) exists (c lies in the domain of f). 2. lim,—., f(x) exists (f has a limit as x > c). 3. lim, f(x) = f(c) (the limit equals the function value).
For one-sided continuity and continuity at an endpoint of an interval, the limits in parts 2 and 3 of the test should be replaced by the appropriate one-sided limits.
EXAMPLE 4 The function y = | x | introduced in Section 1.1 is graphed in Figure 2.39. It is discontinuous at every integer because the left-hand and right-hand limits are not equal as x — n: lim|x| 2n— 1 and lim [x] =n.
Since | n | = n, the greatest integer function is right-continuous at every integer n (but not left-continuous).
The greatest integer function is continuous at every real number other than the inte- gers. For example,
lim. lx] md — ES]. In general, if n — 1 < c < n, n an integer, then lim|x| 2n — 1 7 |c]. a
xc
Figure 2.40 displays several common types of discontinuities. The function in Figure 2.40a is continuous at x = 0. The function in Figure 2.40b would be continuous if it had f(0) — 1. The function in Figure 2.40c would be continuous if f(0) were 1 instead of 2. The discontinuity in Figure 2.40c is removable. The function has a limit as x — 0, and we can remove the discontinuity by setting f(0) equal to this limit.
The discontinuities in Figure 2.40d through f are more serious: lim,—.9 f(x) does not exist, and there is no way to improve the situation by changing f at 0. The step function in Figure 2.40d has a jump discontinuity: The one-sided limits exist but have different val- ues. The function f(x) = 1/x? in Figure 2.40e has an infinite discontinuity. The function in Figure 2.40f has an oscillating discontinuity: It oscillates too much to have a limit as x 0.
78 Chapter 2: Limits and Continuity
y =f) ú
FIGURE 2.40 The function in (a) is continuous at x = 0; the functions in (b) through (f) are not.
Continuous Functions
Generally, we want to describe the continuity behavior of a function throughout its entire domain, not only at a single point. We know how to do that if the domain is a closed interval. In the same way, we define a continuous function as one that is continuous at every point in its domain. This is a property of the function. A function always has a specified domain, so if we change the domain, we change the function, and this may change its continuity property as well. If a function is discontinuous at one or more points of its domain, we say it is a discontinuous function.
EXAMPLE 5
(a) The function y = 1/x (Figure 2.41) is a continuous function because it is continuous at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = 0.
(b) The identity function f(x) = x and constant functions are continuous everywhere by Example 3, Section 2.3. a
Algebraic combinations of continuous functions are continuous wherever they are defined.
THEOREM 8—Properties of Continuous Functions If the functions f and g are continuous at x — c, then the following algebraic combinations are continuous at x — c, 1. Sums: JOE
FIGURE 2.41 The function y = 1/x 2. Differences: f-g
» conMonOUS bi its hatufal domain: 1 3. Constant multiples: k* f, for any number k
has a point of discontinuity at the origin,
so it is discontinuous on any interval 4. Products: Pug
containing x = 0 (Example 5). 5. Quotients: f/g, provided g(c) # 0 6. Powers: f", na positive integer 7. Roots: Vf. provided it is defined on an open interval
containing c, where n is a positive integer
2.5 Continuity 79
Most of the results in Theorem 8 follow from the limit rules in Theorem 1, Section 2.2. For instance, to prove the sum property we have
lim + g)69 = limf) + gG)
= lim f@ + lim g(x) Sum Rule, Theorem 1 = f(c) + g(c) Continuity of f, g atc = (f + gXo). This shows that f + g is continuous. EXAMPLE 6 (a) Every polynomial P(x) = a,x” + a, 4x" ! + +++ + ag is continuous because
lim P(x) = P(c) by Theorem 2, Section 2.2.
(b) If P(x) and Q(x) are polynomials, then the rational function P(x)/ Q(x) is continuous
wherever it is defined (Q(c) # 0) by Theorem 3, Section 2.2. L| EXAMPLE 7 The function f(x) = |x| is continuous. If x > 0, we have f(x) = x, a polynomial. If x < 0, we have f(x) = —x, another polynomial. Finally, at the origin, lim,. |x| = 0 = Jo]. L|
The functions y = sin x and y = cos x are continuous at x = 0 by Example 11 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 70). It fol- lows from Theorem 8 that all six trigonometric functions are then continuous wherever they are defined. For example, y = tanx is continuous on ::: U (—7/2, 7/2) U (7/2, 30/2) U +>.
Composites
All composites of continuous functions are continuous. The idea is that if f(x) is continuous at x = c and g(x) is continuous at x = f(c), then go f is continuous at x = c (Figure 2.42). In this case, the limit as x — c is g(f(c)).
Continuous * at f(c) ^ c foe gCf (o)
Continuous atc
FIGURE 2.42 Composites of continuous functions are continuous.
THEOREM 9—Composite of Continuous Functions If f is continuous at c and g is continuous at f(c), then the composite g ° f is continuous at c.
Intuitively, Theorem 9 is reasonable because if x is close to c, then f(x) is close to f (c), and since g is continuous at f(c), it follows that g(f(x)) is close to g(f(c)).
The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of a proof of Theorem 9, see Exercise 6 in Appendix 4.
80 Chapter 2: Limits and Continuity
FIGURE 2.43 The graph suggests that
T 2T
y = |G sinx)/Q2 4 (Example 8d).
- 2)| is continuous
EXAMPLE 8 Show that the following functions are continuous on their natural
domains. 2/3
ee E m EP:
(a) y X 2x — 5 (b) y TE xm» x sin x
= d = | (c) y E (d) y PONE Solution (a) The square root function is continuous on [ 0, oo) because it is a root of the continu-
(b)
(c)
(d)
ous identity function f(x) = x (Part 7, Theorem 8). The given function is then the composite of the polynomial f(x) = x? — 2x — 5 with the square root function g(t) = Vt, and is continuous on its natural domain.
The numerator is the cube root of the identity function squared; the denominator is an everywhere-positive polynomial. Therefore, the quotient is continuous.
The quotient (x — 2)/(x? — 2) is continuous for all x # + V2, and the function is the composition of this quotient with the continuous absolute value function (Example 7).
Because the sine function is everywhere-continuous (Exercise 70), the numerator term x sin x is the product of continuous functions, and the denominator term x? + 2 is an everywhere-positive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.43). E
Theorem 9 is actually a consequence of a more general result, which we now state
and prove.
THEOREM 10—Limits of Continuous Functions If g is continuous at the point b and lim, f(x) = b, then
lim, g(f(x)) = g(b) = g(lim,—, f(x).
Proof Let € > 0 be given. Since g is continuous at b, there exists a number 6, > 0 such that
le) — g(b)| <e whenever 0 < |y — b| < ài.
Since lim, ,. f(x) = b, there exists a 6 > O such that
f(x) — b| < ô whenever 0 < |x - c| < 6.
If welet y — f(x), we then have that
ly - b| < ô whenever 0 < |x - c| < 6,
which implies from the first statement that | gly) — g(b)| = | e(f(x)) B g(b)| < e whenever 0 « |x — c| < 6. From the definition of limit, this proves that lim, ,.g(f(x) = g(b). E
EXAMPLE 9 As an application of Theorem 10, we have the following calculation:
lim cos (z + sin (# + 3) = cos( lim 2x + lim sin (2 + 3) xml2 2 xm/2 x—ml2 2
= cos (m + sin27) = cos = - 1. E
3- $— — — ——e 2- 1- i 1 i >x 0 1 2 3 4
FIGURE 2.44 The function - [2x2 Isya fo = [ 2<x<4 does not take on all values between fC) = 0 and f(4) = 3; it misses all the values between 2 and 3.
2.5 Continuity 81
Intermediate Value Theorem for Continuous Functions
Functions that are continuous on intervals have properties that make them particularly use- ful in mathematics and its applications. One of these is the Intermediate Value Property. A function is said to have the Intermediate Value Property if whenever it takes on two values, it also takes on all the values in between.
THEOREM 11—The Intermediate Value Theorem for Continuous Functions If f is a continuous function on a closed interval [ a, b ], and if yọ is any value between f(a) and f(b), then yọ = f(c) for some c in [ a, b ].
y ^
y —fQ)
f(b)
>x
Theorem 11 says that continuous functions over finite closed intervals have the Inter- mediate Value Property. Geometrically, the Intermediate Value Theorem says that any horizontal line y = yo crossing the y-axis between the numbers f(a) and f(b) will cross the curve y = f(x) at least once over the interval |a, b].
The proof of the Intermediate Value Theorem depends on the completeness property of the real number system (Appendix 7) and can be found in more advanced texts.
The continuity of f on the interval is essential to Theorem 11. If f is discontinuous at even one point of the interval, the theorem's conclusion may fail, as it does for the func- tion graphed in Figure 2.44 (choose y, as any number between 2 and 3).
A Consequence for Graphing: Connectedness Theorem 11 implies that the graph of a function continuous on an interval cannot have any breaks over the interval. It will be connected—a single, unbroken curve. It will not have jumps like the graph of the greatest integer function (Figure 2.39), or separate branches like the graph of 1/x (Figure 2.41).
A Consequence for Root Finding We call a solution of the equation f(x) = 0 a root of the equation or zero of the function f. The Intermediate Value Theorem tells us that if f is continuous, then any interval on which f changes sign contains a zero of the function.
In practical terms, when we see the graph of a continuous function cross the horizon- tal axis on a computer screen, we know it is not stepping across. There really is a point where the function's value is zero.
EXAMPLE 10 Show that there is a root of the equation x? — x — 1 = 0 between 1 and 2.
Solution Let f(x) = x? — x — 1. Since f(1)= 1 — 1 — 1 =-—1 < 0 and f(2) = 23 — 2 — 1 = 5 > 0, we see that yọ = 0 is a value between f(1) and f(2). Since f is continuous, the Intermediate Value Theorem says there is a zero of f between 1 and 2. Figure 2.45 shows the result of zooming in to locate the root near x = 1.32. E
82 Chapter 2: Limits and Continuity
(a) (b)
1.320 1.3248
—0.02 (c) (d)
FIGURE 2.45 Zooming in on a zero of the function f(x) = x? — x — 1. The zero is near x = 1.3247 (Example 10).
EXAMPLE 11 Use the Intermediate Value Theorem to prove that the equation
V2x+5=4- x
><
has a solution (Figure 2.46).
y=4-x?
Solution We rewrite the equation as
V2x + 54+ x? A, and set f(x) = V2x + 5 + x°. Now g(x) = V2x + 5 is continuous on the interval
l
l
l y=V2x+5 | i [—5/2, oo) since it is the composite of the square root function with the nonnegative linear | function y = 2x + 5. Then f is the sum of the function g and the quadratic function y = x’, 0 c 3 7^ and the quadratic function is continuous for all values of x. It follows that f(x) = V 2x + 5 + x? is continuous on the interval [ —5/2, oo). By trial and error, we find the function values FIGURE 2.46 The curves fO) = V5 & 2.24 and f(2) = V9 + 4 = 7, and note that f is also continuous on the y= Vix + 5andy=4- x finite closed interval [0,2] C [—5/2, oo). Since the value yọ = 4 is between the numbers have the same value at x — c where 2.24 and 7, by the Intermediate Value Theorem there is a number ce [0, 2] such that V2x + 5 = 4 — x? (Example 11). f(c) = 4. That is, the number c solves the original equation. a
Continuous Extension to a Point
Sometimes the formula that describes a function f does not make sense at a point x = c. It might nevertheless be possible to extend the domain of f, to include x = c, creating a new function that is continuous at x = c. For example, the function y = f(x) = (sin x)/x is continuous at every point except x = 0, since the origin is not in its domain. Since y = (sinx)/x has a finite limit as x — 0 (Theorem 7), we can extend the function's domain to include the point x = 0 in such a way that the extended function is continuous at x — 0. We define the new function
Fx =4 7^
(b)
FIGURE 2.48
(a) The graph
of f(x) and (b) the graph of its continuous extension F(x)
(Example 12).
2.5 Continuity 83
The function F(x) is continuous at x = 0 because .. Sinx liga = FO,
so it meets the requirements for continuity (Figure 2.47).
> ><
(a) (b)
FIGURE 2.47 The graph (a) of f(x) = (sinx)/x for ^7/2 = x = w/2 does not include the point (0, 1) because the function is not defined at x = 0. (b) We can remove the discon- tinuity from the graph by defining the new function F(x) with F(0) = 1 and F(x) = f(x) everywhere else. Note that F(0) = lim, o f(x).
More generally, a function (such as a rational function) may have a limit at a point where it is not defined. If f(c) is not defined, but lim, ,. f(x) = L exists, we can define a new function F(x) by the rule
feo, if x is in the domain of f F(x) = L, if s= g.
The function F is continuous at x = c. It is called the continuous extension of f to x = c. For rational functions f, continuous extensions are often found by canceling com- mon factors in the numerator and denominator.
EXAMPLE 12 Show that
xXx+x-6
2.4 5972 2
f(x) = has a continuous extension to x = 2, and find that extension.
Solution Although f(2) is not defined, if x # 2 we have
ca Rad - He MT LR Ko-"Uu-4 GIEF] x42
'The new function
x-3 KD
F(x) =
is equal to f(x) for x # 2, but is continuous at x = 2, having there the value of 5/4. Thus F is the continuous extension of f to x = 2, and
. x-x-6 ; 5
rag IO ay
The graph of f is shown in Figure 2.48. The continuous extension F has the same graph except with no hole at (2, 5/4). Effectively, F is the function f with its point of disconti-
nuity at x = 2 removed. E
84 Chapter 2: Limits and Continuity
Exercises 25
Continuity from Graphs In Exercises 1—4, say whether the function graphed is continuous on [—1, 3]. If not, where does it fail to be continuous and why?
1. 2.
i-i 3e ed
y =f
y = g@)
>x 3. 4 ; x y= k(x) y= ka) 2} 2r 1 1 | [— Sy | | yx -1 0 1 p) 3 =] 0 1 2 3 Exercises 5—10 refer to the function x)-1, -1zx«0 2x. 0<x<1 f(x) = 1, y= —2x + 4, 1«xc«2 0, 2<x<3
graphed in the accompanying figure.
-X
y= x?-1 The graph for Exercises 5-10.
5. a. Does f(- 1) exist?
b. Does lim,_,_)+ f(x) exist?
c. Does lim, ,- f(x) = f(-1)?
d. Is f continuous at x = —1? 6. a. Does f(1) exist?
b. Does lim,.,, f(x) exist?
c. Does lim,., f(x) = fC)?
d. Is f continuous at x — 1?
7. a. Is f defined at x = 2? (Look at the definition of f.) b. Is f continuous at x — 2? 8. At what values of x is f continuous?
9. What value should be assigned to f(2) to make the extended function continuous at x — 2?
10. To what new value should f(1) be changed to remove the discon- tinuity?
Applying the Continuity Test
At which points do the functions in Exercises 11 and 12 fail to be con- tinuous? At which points, if any, are the discontinuities removable? Not removable? Give reasons for your answers.
11. Exercise 1, Section 2.4 12. Exercise 2, Section 2.4
At what points are the functions in Exercises 13-30 continuous?
1 1 13. y= -3 14. y = ———— +4 ee A ED & x-3 15. y = ————— 16. y= PU edd TOM ur . 1 x 17. y= x ^ I| + sinx 18. y= = »-|x-i| "Wei 2 cos x x 2 19. y = —— 20. y = too 21. y = csc 2x 22. y = tan tan x xtt+ 1 23. y - € 24. y = 22 y wet) y ] + sin? x 25. y= Vx +3 26. y= W3x-1 27. y = (2x — 1)" 28. y = (2 — wn! o Raf T 29. g(x) = x— 3 5, x=3 6- VOS ausus 30. fay = 4% ^ Par 3. =) 4, =-2
Limits Involving Trigonometric Functions Find the limits in Exercises 31—38. Are the functions continuous at the
point being approached? T
31. lim sin(x — sin x) 32. lim sn(3 cos (tan D) pum
Xm
33. lim sec (y sec? y — tan? y — 1) e
34. lim (Z cos (sin 8)
x
35.
36.
37.
38.
. T lim cos (—=—) 120 V 19 — 3 sec 2t lim V esc? x + 5V3 tan x
x—>7/6
|. Jcos? x — cos x lim sin ,|-—— -——— x0
. (ree 2x — sin 2) lim sec | ——— —— x0 3x
Continuous Extensions
39.
40.
41.
42.
43.
44.
45.
46.
47.
Define g(3) in a way that extends g(x) = (x? — 9)/(x — 3) to be continuous at x — 3.
Define h(2) in a way that extends h(t) = (t + 3t to be continuous at f — 2.
Define f(1) in a way that extends f(s) = ($? — 1)/(s? — 1) to be continuous at s = 1.
10/( — 2)
Define g(4) in a way that extends a(x) = (x7 — 16)/ (2? —
to be continuous at x — 4.
3x — 4)
For what value of a is fo) d -1, x«3 Xx) = 2ax, x23
continuous at every x?
For what value of b is
(x) u E: x) = d bx?, zx —2 continuous at every x? For what values of a is fo) dx-—2a, x 22 x) = 12; x«2 continuous at every x? For what value of b is x—b , x«0 g(x) = b+ 1 r+b, x>0 continuous at every x? For what values of a and b is —2, xs-l f(xy) = ax—b, -1«x«1 3, xe 1
continuous at every x?
48.
2.5 Continuity 85
For what values of a and b is
ax + 2b, x= 0 ex) = 4x2 + 3a-—b, O<x=2 ay c. x2
continuous at every x?
In Exercises 49—52, graph the function f to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at x = 0. If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function’s value(s) should be?
49.
51.
z TM
EE so. fo) = E - 1
f(x) = T 52. f(x) = (1 + 23)!/* X
Theory and Examples
53.
54. 55.
56.
57.
58.
59.
60.
A continuous function y — f(x) is known to be negative at x = 0 and positive at x = 1. Why does the equation f(x) = 0 have at least one solution between x — 0 and x — 1? Illustrate with a sketch.
Explain why the equation cosx — x has at least one solution.
Roots of a cubic Show that the equation x? — 15x + 1 — 0 has three solutions in the interval [ —4, 4].
A function value Show that the function F(x) = (x — ay (x — b)? + x takes on the value (a + b)/2 for some value of x.
Solving an equation If f(x) = x? — 8x + 10, show that there are values c for which f(c) equals (a) m; (b) — V3: (c) 5,000,000.
Explain why the following five statements ask for the same infor- mation.
a. Find the roots of f(x) = x? — 3x — 1.
b. Find the x-coordinates of the points where the curve y — x? crosses the line y = 3x + 1.
c. Find all the values of x for which x? — 3x = 1.
d. Find the x-coordinates of the points where the cubic curve y = x? — 3x crosses the line y = 1.
e. Solve the equation x? — 3x — 1 — 0.
Removable discontinuity Give an example of a function f(x) that is continuous for all values of x except x — 2, where it has a removable discontinuity. Explain how you know that f is dis- continuous at x — 2, and how you know the discontinuity is removable.
Nonremovable discontinuity Give an example of a function g(x) that is continuous for all values of x except x = —1, where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.
86
61.
62.
63.
64.
65.
66.
67.
Chapter 2: Limits and Continuity
A function discontinuous at every point
a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function
f@) = T if x is rational ^ 0, if x is irrational
is discontinuous at every point. b. Is f right-continuous or left-continuous at any point?
If functions f(x) and g(x) are continuous for 0 = x = 1, could f(x)/g(x) possibly be discontinuous at a point of [0,1]? Give reasons for your answer.
If the product function h(x) = f(x): g(x) is continuous at x = 0, must f(x) and g(x) be continuous at x = 0? Give reasons for your answer.
Discontinuous composite of continuous functions Give an example of functions f and g, both continuous at x = 0, for which the composite f ° g is discontinuous at x = 0. Does this contradict Theorem 9? Give reasons for your answer.
Never-zero continuous functions Is it true that a continuous function that is never zero on an interval never changes sign on that interval? Give reasons for your answer.
Stretching a rubber band Is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for your answer.
A fixed point theorem Suppose that a function f is continuous on the closed interval [0,1] and that 0 = f(x) = 1 for every x in [0, 1]. Show that there must exist a number c in [0,1] such that f(c) = c (c is called a fixed point of f).
68.
69.
70.
The sign-preserving property of continuous functions Let f be defined on an interval (a, b) and suppose that f(c) # 0 at some c where f is continuous. Show that there is an interval (c — 6,c + 6) about c where f has the same sign as f(c).
Prove that f is continuous at c if and only if lim f(c + h) = fto. mm
Use Exercise 69 together with the identities
sin(h + c) =sinhcosc + cos h sinc,
cos(h + c) = cos h cosc — sinhsinc
to prove that both f(x) = sin x and g(x) = cos x are continuous at every point x = c.
Solving Equations Graphically
Use the Intermediate Value Theorem in Exercises 71—76 to prove that each equation has a solution. Then use a graphing calculator or com- puter grapher to solve the equations.
71. 72. 73. 74. 75. 76.
x-—-3x—120 2x3 — 2x7 2x +1=0
x(x — 1) = 1 (one root) Vrt+Vitx=4 cosx = x (one root). Make sure you are using radian mode.
2sinx =x (three roots). Make sure you are using radian mode.
2.6 Limits Involving Infinity; Asymptotes of Graphs
FIGURE 2.49 The graph of y = 1/x approaches 0 as x — œ or x — —oo.
Finite Limits as x — +00
In this section we investigate the behavior of a function when the magnitude of the inde- pendent variable x becomes increasingly large, or x — +00. We further extend the con- cept of limit to infinite limits, which are not limits as before, but rather a new use of the term limit. Infinite limits provide useful symbols and language for describing the behavior of functions whose values become arbitrarily large in magnitude. We use these limit ideas to analyze the graphs of functions having horizontal or vertical asymptotes.
The symbol for infinity (CO) does not represent a real number. We use co to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. - For example, the function f(x) = 1/x is defined for all x # 0 (Figure 2.49). When x is positive and becomes increasingly large, 1/x becomes increasingly small. When x is negative and its magnitude becomes increasingly large, 1/x again becomes small. We
summarize these observations by saying that f(x) = 1/x has limit 0 as x — œ or
cise definitions.
Xx — —œ, or that 0 is a limit of f(x) = 1/x at infinity and negative infinity. Here are pre-
No matter what positive number e is, the graph enters
1 this band at x = 2
x and stays.
><
No matter what positive number e is, the graph enters
this band at x = —z and stays.
FIGURE 2.50 The geometry behind the argument in Example 1.
2.6 Limits Involving Infinity; Asymptotes of Graphs 87
DEFINITIONS 1. We say that f(x) has the limit L as x approaches infinity and write
lim. feo-L if, for every number e > 0, there exists a corresponding number M such that for all x x>M => lf) — L| < e. 2. We say that f(x) has the limit L as x approaches minus infinity and write lim. feo-L
if, for every number e > 0, there exists a corresponding number N such that for all x
x<N => fœ- L| < e.
Intuitively, lim,—oo f(x) = L if, as x moves increasingly far from the origin in the positive direction, f(x) gets arbitrarily close to L. Similarly, lim,., f(x) = L if, as x moves increasingly far from the origin in the negative direction, f(x) gets arbitrarily close to L. The strategy for calculating limits of functions as x — +00 is similar to the one for finite limits in Section 2.2. There we first found the limits of the constant and identity functions y — k and y — x. We then extended these results to other functions by applying Theorem 1 on limits of algebraic combinations. Here we do the same thing, except that the starting functions are y = k and y = 1/x instead of y = k and y = x. The basic facts to be verified by applying the formal definition are lim k=k and lim Rc. 0. (1)
x too x +00 X
We prove the second result in Example 1, and leave the first to Exercises 87 and 88.
EXAMPLE 1 Show that
(a) lim 2 —0 (b) lim z = 0.
Solution
(a) Let e > 0 be given. We must find a number M such that for all x x>M => : — 0] = I « e.
The implication will hold if M — 1/e or any larger positive number (Figure 2.50). This proves lim, ,4; (1/x) = 0. (b) Let e > 0 be given. We must find a number N such that for all x
« e.
XN => l
1 t-0| =
The implication will hold if N = —1/e or any number less than —1/e (Figure 2.50). This proves lim,—. oo (1/x) = 0. [as]
Limits at infinity have properties similar to those of finite limits.
THEOREM 12 All the Limit Laws in Theorem 1 are true when we replace lim,—{, by lim, 5 or lim, o. That is, the variable x may approach a finite number c or £00.
88 Chapter 2: Limits and Continuity
—2 NOT TO SCALE
FIGURE 2.51 The graph of the func- tion in Example 3a. The graph approaches the line y = 5/3 as |x| increases.
FIGURE 2.52 The graph of the function in Example 3b. The graph
approaches the x-axis as |x| increases.
EXAMPLE 2 The properties in Theorem 12 are used to calculate limits in the same way as when x approaches a finite number c.
(a) lim (s + 1) = lim 5+ lim 1 Sum Rule
XxX—00 x—00 x—0o Ñ =5+0=5 Known limits W tim 73 = pm íV3.l.l xc-—00 X X-—»—00
. : 1 . 1 = lim rm V3 * lim x' lim x Product Rule x——00 Xx——00 x—-—00
m V3 :0:0-20 Known limits [|
Limits at Infinity of Rational Functions
To determine the limit of a rational function as x — +00, we first divide the numerator and denominator by the highest power of x in the denominator. The result then depends on the degrees of the polynomials involved.
EXAMPLE 3 These examples illustrate what happens when the degree of the numera- tor is less than or equal to the degree of the denominator.
5x + 8x — 3 = ļi S (8/x) = 3/x?) Divide numerator and
(a) lm 3x2 +2 pare 34+ (2 r x2) denominator by x?. gs i ^S LER à See Fig. 2.51. lix -2. a (11/x?) + Q/x?) Divide numerator and (b) lm. 2x? 1 " lim 2 a / x3) denominator by x?. = : 2 a = 0 See Fig. 2.52. oO
Cases for which the degree of the numerator is greater than the degree of the denomi- nator are illustrated in Examples 9 and 13.
Horizontal Asymptotes
If the distance between the graph of a function and some fixed line approaches zero as a point on the graph moves increasingly far from the origin, we say that the graph approaches the line asymptotically and that the line is an asymptote of the graph.
Looking at f(x) = 1/x (see Figure 2.49), we observe that the x-axis is an asymptote of the curve on the right because
lim y = 0 and on the left because lim l =0
We say that the x-axis is a horizontal asymptote of the graph of f(x) = 1/x.
DEFINITION A line y — b is a horizontal asymptote of the graph of a func- tion y — f(x) if either
lim f@)=b or — lim fQ)- b.
l 1 7 i l x y=-l1
| 3-2
^» f= 41
FIGURE 2.53 The graph of the function in Example 4 has two
horizontal asymptotes.
FIGURE 2.54 Theliney = lisa horizontal asymptote of the function graphed here (Example 5b).
2.6 Limits Involving Infinity; Asymptotes of Graphs 89
The graph of the function
sketched in Figure 2.51 (Example 3a) has the line y = 5/3 as a horizontal asymptote on both the right and the left because
lim f(x) -3 and — lim f(x) = 5,
EXAMPLE 4 Find the horizontal asymptotes of the graph of
xX- 2
fœ) = II
Solution We calculate the limits as x — -- oo.
3 3. 1 — (2/x3) Forx = 0: lim 2? = jim *— 2 = lim E - x9 |x| +1 x93 +1 ^ x91 + (0/x) S 3. 1 — (2/33) eee: dus 2 ia lim / -1.
xx 1 >=% (=x) +1 ia-9-1-(0/53)
The horizontal asymptotes are y = —1 and y = 1. The graph is displayed in Figure 2.53. Notice that the graph crosses the horizontal asymptote y — —1 for a positive value of x. El
EXAMPLE 5 Find (a) lim. sin(1/x) and (b) lim x sin(1/x).
Solution (a) We introduce the new variable t = 1/x. From Example 1, we know that t— 0* as x — oo (see Figure 2.49). Therefore,
. . 1 P lim siny = lim sint = 0. P. t0*
x00 (b) We calculate the limits as x — oo and x — —oo:
lim xsint = lim 94 =1 and lim xsinl- lim 9B = 1, x—00 x t-0* t x—-—00 x t0 is
The graph is shown in Figure 2.54, and we see that the line y = 1 is a horizontal
asymptote. E
The Sandwich Theorem also holds for limits as x — X oo. You must be sure, though, that the function whose limit you are trying to find stays between the bounding functions at very large values of x in magnitude consistent with whether x — œ or x > —o9.
EXAMPLE 6 Using the Sandwich Theorem, find the horizontal asymptote of the curve
sin x y-2?--—.
90 Chapter 2: Limits and Continuity
—3m —-2m —m 0 a 27 3m
FIGURE 2.55 A curve may cross one of its asymptotes infinitely often (Example 6).
y A y=t/" Pd ^ ^ 3r P rud ra " d 2} s—o Pd P t 1 y=tr- 1- #— ` r d 4 l l eo l l t > =2 =i uL 3 US rd 4 ELA ^ 4 ^ 4 e——o —2r- ^ 4 ^ 4
FIGURE 2.56 The graph of the greatest integer function y — [z] is sandwiched between y = t — l andy = t.
Solution We are interested in the behavior as x— +00. Since
sin x
1 0= 1 X
and lim, ,.5, |1/x| = 0, we have lim,.+.0.(sinx)/x = 0 by the Sandwich Theorem. Hence,
lim (2+ SB) =2+0=2,
x too
and the line y — 2 is a horizontal asymptote of the curve on both left and right (Figure 2.55). This example illustrates that a curve may cross one of its horizontal asymptotes many times. E
We can investigate the behavior of y = f(1/x) as x — 0 by investigating y = f(f) as t— +00, where t = 1/x.
EXAMPLE 7 Find imat]. Solution We let t = 1/x so that
lim "HI — lim HH
x—*
From the graph in Figure 2.56, we see that t — 1 = |e] = f, which gives
j= 1 < Ll <1 Multiply inequalities by + > 0. It follows from the Sandwich Theorem that lim H t| 5t so 1 is the value of the limit we seek. |
EXAMPLE 8 Find lim (x — Vx? + 16).
Solution Both of the terms x and Vx? + 16 approach infinity as x — 00, so what hap- pens to the difference in the limit is unclear (we cannot subtract Co from oo because the symbol does not represent a real number). In this situation we can multiply the numerator and the denominator by the conjugate radical expression to obtain an equivalent algebraic result:
Jim (x- Vx? + 16) = lim (x - Va? + 6): wus
de = AIG). —16 = lim = lim ; 4*0. + Ve +16 ar V +16 As x — œ, the denominator in this last expression becomes arbitrarily large, so we see that the limit is 0. We can also obtain this result by a direct calculation using the Limit Laws:
The vertical distance between curve and
E line goes to zero as x > oo 5r- 4 Y Oblique 3 asymptote 2
>x
FIGURE 2.57 The graph of the function
in Example 9 has an oblique asymptote.
You can get as high as you want by taking x close enough to0. No matter how high B is, the graph goes higher.
You can get as low as| @ you want by taking x close enough to 0.
>x No matter how low —B is, the graph goes lower. —B
FIGURE 2.58 One-sided infinite limits:
lim L oo and
x0*
2.6 Limits Involving Infinity; Asymptotes of Graphs 91
Oblique Asymptotes
If the degree of the numerator of a rational function is 1 greater than the degree of the denominator, the graph has an oblique or slant line asymptote. We find an equation for the asymptote by dividing numerator by denominator to express f as a linear function plus a remainder that goes to zero as x — +00,
EXAMPLE 9 Find the oblique asymptote of the graph of | :2—3 f (x) s 2x —4 in Figure 2.57.
Solution We are interested in the behavior as x — «oo. We divide (2x — 4) into (x2 — 3):
2*1 2x — 42 — 3 x? — 2x 2x — 3 2x — 4 1 This tells us that 2 = fa) = 74 (3+ E (x4). —— — linear g(x) remainder
As x— +00, the remainder, whose magnitude gives the vertical distance between the graphs of f and g, goes to zero, making the slanted line
49-24
an asymptote of the graph of f (Figure 2.57). The line y = g(x) is an asymptote both to the right and to the left. The next subsection will confirm that the function f(x) grows arbitrarily large in absolute value as x — 2 (where the denominator is zero), as shown in the graph. Bl
Notice in Example 9 that if the degree of the numerator in a rational function is greater than the degree of the denominator, then the limit as |x| becomes large is +00 or —oo, depending on the signs assumed by the numerator and denominator.
Infinite Limits
Let us look again at the function f(x) = 1/x. As x — 0*, the values of f grow without bound, eventually reaching and surpassing every positive real number. That is, given any positive real number B, however large, the values of f become larger still (Figure 2.58). Thus, f has no limit as x — 0°. It is nevertheless convenient to describe the behavior of f by saying that f(x) approaches oo as x — 0*. We write
. xd lim f(x) = lim 4 = oo. x0* PC ) xoor* In writing this equation, we are not saying that the limit exists. Nor are we saying that there
is a real number co, for there is no such number. Rather, we are saying that lim,_.9+ (1/x) does not exist because 1/x becomes arbitrarily large and positive as x > 0*.
92 Chapter 2: Limits and Continuity
><
FIGURE 2.59 Near x = 1, the func- tion y = 1/(« — 1) behaves the way the function y = 1/x behaves near x = 0. Its graph is the graph of y = 1/x shifted
1 unit to the right (Example 10).
No matter how B high B is, the graph goes higher.
FIGURE 2.60 The graph of f(x) in Example 11 approaches infinity as x — 0.
As x — 0^, the values of f(x) = 1/x become arbitrarily large and negative. Given any negative real number —B, the values of f eventually lie below —B. (See Figure 2.58.) We write
, xcd lim f(x) = lim z = ~. x—0- A ) x0-* Again, we are not saying that the limit exists and equals the number —oo. There is no real
number —oo. We are describing the behavior of a function whose limit as x — 0^ does not exist because its values become arbitrarily large and negative.
EXAMPLE 10 Find lim —!— and lim xrx-1 xl xX
Geometric Solution The graph of y = 1/(x — 1) is the graph of y = 1/x shifted 1
unit to the right (Figure 2.59). Therefore, y — 1/(x — 1) behaves near 1 exactly the way
y = 1/x behaves near 0:
lim —_ = oo and lim = —OO0, xox — | =i x= I
Analytic Solution Think about the number x — 1 and its reciprocal. As x > 1*, we have (x — 1) ^0* and l/(x — 1) 0o. As x— LE, we have (x — 1) —0^ and 1/(x — 1) ^ —oo. mg
EXAMPLE 11 Discuss the behavior of
fix) = A as x 0. x
Solution As x approaches zero from either side, the values of 1/x? are positive and become arbitrarily large (Figure 2.60). This means that
; s | 1 = lim 4 = oo. Prae, feo 230 x? The function y = 1/x shows no consistent behavior as x — 0. We have 1/x — co if x — 0*, but 1/x — —oo if x > 0. All we can say about lim,—ọo (1/x) is that it does not exist. The function y = 1/x? is different. Its values approach infinity as x approaches zero from either side, so we can say that lim, (1/x?) = oo. E
EXAMPLE 12 These examples illustrate that rational functions can behave in various ways near zeros of the denominator.
ji (x -—2y i (x 2y lim 22 = 0 WM gd. POG Dat® 25062 Pi : Xx i 1 1 ®) ime 4 M5G-2G*2 LÀxt2 4 zx dé» Tox x—3 E The values are negative (c) lim, xl—4 lim, (x — 2) + 2) = fory > 2,.x near 2. x — 3 The values are positive
(d) lim 5—5 = lim G= Dar a = oo for x < 2, x near 2.
o E————————
FIGURE 2.61 Forc- ó«x«c +ð, the graph of f(x) lies above the line y = B.
><
FIGURE 2.62 Forc-ó«x«c- 6, the graph of f(x) lies below the line y-7-B.
2.6 Limits Involving Infinity; Asymptotes of Graphs 93
(e) ji = lim 2 x2 x —4 x2 (x S 2)(x + 2)
does not exist. See parts (c) and (d).
2-x _, -«-2) , -] m im 1m i—2»(x 295 x92(x—2) x2(x — 27
In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f), where cancellation still leaves a zero factor in the denominator. E
^ RN 4 EXAMPLE 13 Find lim © +1. xo-o0 3x +x — 7
Solution We are asked to find the limit of a rational function as x — —0©0, so we divide the numerator and denominator by x?, the highest power of x in the denominator:
20 —6x + 1 . 2x3 — 6x7 + x? lim 5 = lim =4 = x>-00 3x* +x —7 xo-o 3+ x^ — "Ix
. 2x2(x4-3)94x? — ]im = = x>-œ 3 +x 1— TX 2
= —00, x" 0, x — 3-00
because the numerator tends to —oo while the denominator approaches 3 as x —^ —oo. BH
Precise Definitions of Infinite Limits
Instead of requiring f(x) to lie arbitrarily close to a finite number L for all x sufficiently close to c, the definitions of infinite limits require f(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seen before. Figures 2.61 and 2.62 accompany these definitions.
DEFINITIONS 1. We say that f(x) approaches infinity as x approaches c, and write
lim f(x) = oo, if for every positive real number B there exists a corresponding 6 > 0 such that for all x 0< |x-c| <6 => JG > B. 2. We say that f(x) approaches minus infinity as x approaches c, and write lim f(x) = —oo, if for every negative real number —B there exists a corresponding 6 > 0
such that for all x
0< |x-cl «6 => fa<=B,
The precise definitions of one-sided infinite limits at c are similar and are stated in the exercises.
94 Chapter 2: Limits and Continuity
Vertical asymptote
Horizontal 1 asymptote
Horizontal asymptote, y-0
Vertical asymptote, x=0
FIGURE 2.63 The coordinate axes are asymptotes of both branches of the hyper- bola y = 1/x.
EXAMPLE 14 Prove that lim l 209 x0 x”
Solution Given B > 0, we want to find 6 > 0 such that
0 < |x — 0| <6 implies Isp.
x
Now,
Ee o AES S
m y B or, equivalently,
lx| < ET
VB Thus, choosing 8 — 1/ VB (or any smaller positive number), we see that 1 2 ae B. 82
|x| « 8. implies >
Therefore, by definition,
ree | lim 757 09. [| x0 X
Vertical Asymptotes
Notice that the distance between a point on the graph of f(x) = 1/x and the y-axis approaches zero as the point moves vertically along the graph and away from the origin (Figure 2.63). The function f(x) = 1/x is unbounded as x approaches 0 because
lim l- —00,
. 1 lim 4 = oo and x x0
x0
We say that the line x = 0 (the y-axis) is a vertical asymptote of the graph of f(x) = 1/x. Observe that the denominator is zero at x = 0 and the function is undefined there.
DEFINITION A line x = a is a vertical asymptote of the graph of a function y = f(x) if either
lim, f(x) = 00 or lim f(x) = too.
x-*ü
EXAMPLE 15 Find the horizontal and vertical asymptotes of the curve _x+t3 Tx»
Solution We are interested in the behavior as x — +00 and the behavior as x > —2, where the denominator is zero.
The asymptotes are quickly revealed if we recast the rational function as a polynomial with a remainder, by dividing (x + 2) into (x + 3):
1 x+2)x +3 x+2 1
Vertical asymptote, xc———2 Horizontal asymptote, 2 yal 1 i ii gy 12 3 FIGURE 2.64 The lines y = 1 and X — —2 are asymptotes of the curve in
Example 15.
y=- S x?—4 Vertical Vertical asymptote, x — 2 asymptote, Horizontal x=-2
asymptote, y = 0
FIGURE 2.65 Graph of the function in Example 16. Notice that the curve approaches the x-axis from only one side. Asymptotes do not have to be two-sided.
2.6 Limits Involving Infinity; Asymptotes of Graphs 95
This result enables us to rewrite y as:
E 1 TE x+2° As x — +00, the curve approaches the horizontal asymptote y = 1; as x — —2, the curve approaches the vertical asymptote x = —2. We see that the curve in question is the graph of f(x) = 1/x shifted 1 unit up and 2 units left (Figure 2.64). The asymptotes, instead of
being the coordinate axes, are now the lines y = 1 and x = —2. Oo
EXAMPLE 16 Find the horizontal and vertical asymptotes of the graph of 8
vr - 4"
f@) =-
Solution We are interested in the behavior as x — +00 and as x — £2, where the denominator is zero. Notice that f is an even function of x, so its graph is symmetric with respect to the y-axis.
(a) The behavior as x — X oo. Since lim, 4; f(x) = 0, the line y = 0 is a horizontal asymptote of the graph to the right. By symmetry it is an asymptote to the left as well (Figure 2.65). Notice that the curve approaches the x-axis from only the negative side (or from below). Also, f(0) — 2.
(b) The behavior as x > £2. Since
lim. f(x) = —oo and lim f(x) = oo, the line x = 2 is a vertical asymptote both from the right and from the left. By sym- metry, the line x = —2 is also a vertical asymptote. There are no other asymptotes because f has a finite limit at all other points. E
EXAMPLE 17 The curves
sin x COS X
y = secx = gos x and y = tanx =
both have vertical asymptotes at odd-integer multiples of 7 /2, where cos x = 0 (Figure 2.66).
y = " zn A y= sec x y= tanx
>x
T rts} 3 | Ij Ww EI
SpA
FIGURE 2.66 The graphs of sec x and tan x have infinitely many vertical asymptotes (Example 17). [|
Dominant Terms
In Example 9 we saw that by long division we could rewrite the function
2 10) = X74
96 Chapter 2: Limits and Continuity
500,000
300,000
100,000
—20 -10 0 10 20 — 100,000 +
(b)
FIGURE 2.67 The graphs of f and g are (a) distinct for |x| small, and (b) nearly identical for |x| large (Example 18).
as a linear function plus a remainder term:
[x 1 fe) = G t 1) + (<4)
This tells us immediately that
fæ = > + 1 For |x| large, is near 0.
1 2x— 4
1
f@ = 2x — 4 For x near 2, this term is very large in absolute value.
If we want to know how f behaves, this is the way to find out. It behaves like y = (x/2) + 1 when |x| is large and the contribution of 1/(2x — 4) to the total value of f is insignificant. It behaves like 1/(2x — 4) when x is so close to 2 that 1/(2x — 4) makes the dominant contribution.
We say that (x/2) + 1 dominates when x is numerically large, and we say that 1/(2x — 4) dominates when x is near 2. Dominant terms like these help us predict a function's behavior.
EXAMPLE 18 Let f(x) = 3x^ — 2x8 + 3x? — 5x + 6 and g(x) = 3x*. Show that although f and g are quite different for numerically small values of x, they are virtually identical for |x| very large, in the sense that their ratios approach 1 as x — oo or x > —oo.
Solution The graphs of f and g behave quite differently near the origin (Figure 2.67a), but appear as virtually identical on a larger scale (Figure 2.67b).
We can test that the term 3x* in f, represented graphically by g, dominates the poly- nomial f for numerically large values of x by examining the ratio of the two functions as x — +00, We find that
f(x) . 3x — 2x3 + 3x2 — 5x + 6 lim
E g(x) ~ x too 3x4 BT 2 1 5 2 B EXC à x^ Bx? 2) zm] which means that f and g appear nearly identical when |x| is large. E
Summary
In this chapter we presented several important calculus ideas that are made meaningful and precise by the concept of the limit. These include the three ideas of the exact rate of change of a function, the slope of the graph of a function at a point, and the continuity of a function. The primary methods used for calculating limits of many functions are captured in the algebraic Limit Laws of Theorem 1 and in the Sandwich Theorem, all of which are proved from the pre- cise definition of the limit. We saw that these computational rules also apply to one-sided limits and to limits at infinity. However, to calculate complicated limits such as
. LY . X — sinx . lim | 1), lim ———., and lim x!/*, x0 x0 x x0
techniques other than simple algebra are required. The derivative is one of the tools we need to calculate limits such as these, and this notion is the central subject of our next chapter.
Exercises 2.6 |
Finding Limits 1. For the function f whose graph is given, determine the following limits. a. lim fe») b. lim, fe») c. lim. fe») d. lim. fe) e. lim fœ f. lim Foo g. lim feo h. lim f(x) i lim f(x)
-6-5-4-3 -2
2. For the function f whose graph is given, determine the following
limits.
a. lim f(x) b. dim, f(x) c. Jim f(x) d. lim feo e. lim, foo f. lim foo g. m f (x) h. Jim, feo i. im. FO) j lim f@) k. lim fœ) L lim fœ)
In Exercises 3-8, find the limit of each function (a) as x— oo and (b) as x —^ —oo. (You may wish to visualize your answer with a graphing calculator or computer.)
3. fa) 72 2-3 4. fo-2--2 5. g(x) = — 6. g(x) = a 2 + (1/x) 8 — (5/x?) —5 + (7/x) 3 — Q/x 7. h(x) = —————— 8. Ax) = —————— O= 3- 0/3 "ir (V2/x2) Find the limits in Exercises 9-12. ig 10. lim SE
X00 0 -—oo 30
2.6 Limits Involving Infinity; Asymptotes of Graphs 97
. 2-—rt- sint ; r+ sinr 11. a t + cost 12. ims; t7-—5sinr Limits of Rational Functions In Exercises 13-22, find the limit of each rational function (a) as x — oo and (b) as x > —oo.
13. fo) = 273 M. fe) = 51 — 15. f(x) = Eur 16. f(x) = 5 + :
Meho) = E Brig 2x* Qu +6 19. g(x) = Bon 20. 8%) = EE
21. f(x) = Di 22. h(x) — Bait
Limits as x — o» or x —%
The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23-36.
| 2 _ 1/3 23. dig. | 9 di dtc x N 2x2 + x x—>—00 8x2 — 3
— 3 \5 3 c 25. lim (4 3 26. lim || ——À—
4 27. tim VA ta 28. lim 2+ V* x0 3x-7 x92 — Vx 3 Lc -1 4 29. lim Vx — Vx 306. dum 5 — 5 EFE Dex - x3 5/3 _ M/3 3. 31. lim 2x" x" +7 32. lim Wig ede ded x20 x8/5 + 3x + Vx 02x + x? — 4 : +1 ] 241 33, m x+1 mh un x+1 = _ 433 Bs; tn a 36. dui 9t a> V/Ay? + 25 ao V/x® + 9 Infinite Limits Find the limits in Exercises 37-48. T "ES: s ium. 3x m = 2x m PT —2 s m EV . 2X n 3x Sb rdg 42. dim d. 10 . 4 : =1 43. lim ———— 44. lim x1 (x — 7X x0 x(x + 1) . : 2 45. a. Jim, ENT b. lim 3xl/5
98 Chapter 2: Limits and Continuity
: 2 How mos olm s 47 lim ET 48
Find the limits in Exercises 49—52.
49. lim tanx 50. x(m[2) 5]. um (1 + csc 0) 52.
Find the limits in Exercises 53—58.
lm secx x(-m/2)*
lim (2 — cot 0) 0—0
53. lim " as "um a. x— 2t b. x>2X7 € x—-—2* d. x—> -27 54. lim — as x. a. x— 1t b. x—17 € x -1* d. x— -17 2 55. im( — 3 as a. x—0* b. x07 c. x V2 d. x—> —1 ye 56. lim5- 743s a. x—-27 b. x—-2- €. x—]1* d. x—0* o n 57. iu Se x3 — 2x? a. x—0* b. x—2* G. XA d. x2
e. What, if anything, can be said about the limit as x — 0?
P 58. m ee as
xX — 4x a. x— 2t b. €. X —0- d.
e. What, if anything, can be said about the limit as x — 0?
Find the limits in Exercises 59-62.
59. im(2 — 3) as
a. t—0* b.
a. t—0* b 2 61 tim( + e s) a a. x—0* b €. x—]1* d ; 1 1 62. tim( I. = s) as a. x 0* b
x-—2*
x—1*
t0
. t0
Me
k
« x—0-
.x—l.
Graphing Simple Rational Functions Graph the rational functions in Exercises 63—68. Include the graphs and equations of the asymptotes and dominant terms.
1 gc 64. P... _ 1 2 29 oh: e A OS =a xb 2x 67. y= F3 68. y= 11
Inventing Graphs and Functions In Exercises 69—72, sketch the graph of a function y — f(x) that satis- fies the given conditions. No formulas are required—just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.) 69. f(0) = 0, f(1) = 2, f- 1) = -2, lim fœ) = —], and
lim fœ@ = 1 `
X00 70. f(0) = 0, lim f(x) = 0, lim f(x) = 2, and
x too x0*
lim f(x) = -2 71. f(0) = 0, dim fœ) = 0, lim. fe) = Jim, fo») = oo,
lim fo) = —oo, and lim _ f(x) = =% 72. fQ) = 1, f-1) = 0, lim f(x) = 0, lim, fa) = o». lim f(x) = —00, and lim, f(x) = 1
In Exercises 73-76, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.)
73. lim f(x) = 0, lim f(x) = œ, and lim f(x) = co x— too x2: x2* . B _ : a : A 74 lim. g(x) = 0, lim g(x) i and lim, g(x) 75. lim h(x) = —1, lim A(x) = 1, lim A(x) = —1, and x——00 x— 00 x0- lim h(x) = 1 x0* 76. lim k(x) = 1, lim k(x) = oo, and lim k(x) = —oo x too x1 x1* 77. Suppose that f(x) and g(x) are polynomials in x and that
lim, s (fG)/2g(x)) = 2. Can you conclude anything about lim, oc (f(x)/g(x))? Give reasons for your answer.
78. Suppose that f(x) and g(x) are polynomials in x. Can the graph of f(x)/g(x) have an asymptote if g(x) is never zero? Give reasons for your answer.
79. How many horizontal asymptotes can the graph of a given ratio- nal function have? Give reasons for your answer.
Finding Limits of Differences When x— +00 Find the limits in Exercises 80—86.
80. lim (Vx + 9 - Vx +4) 81. Jim (Ve +25- Væ- 1) 82. lim (Vax? +3 + x)
83. lim (2x + V4x2 + 3x — 2)
x —00
84. lim (V 9x? — x — 3x)
x00
85. lim (Và + 3x — Væ — 2x) 86. lim (V3? +x- Vg -x)
Using the Formal Definitions Use the formal definitions of limits as x — +00 to establish the limits in Exercises 87 and 88.
87. If f has the constant value f(x) = k, then lim. f(x) = k. 88. If f has the constant value f(x) = k, then lim f(x) = k.
Use formal definitions to prove the limit statements in Exercises 89-92.
89. lim 1. = —oo 90. lim + = oo x0 x0 |x| rudi MEE 91. lim — 4 = —oo 92. lim ——— — oo x3 (x — 3) x--5(x-^.35)
93. Here is the definition of infinite right-hand limit.
We say that f(x) approaches infinity as x approaches c from the right, and write
lim f(x) — oo,
xc* if, for every positive real number B, there exists a correspond- ing number 6 > 0 such that for all x
c<x<ct+6 > f(x) > B.
Modify the definition to cover the following cases.
a. lim f(x) = oo
b. lim f(x) = —oo c. lim f(x) = —oo
Use the formal definitions from Exercise 93 to prove the limit state- ments in Exercises 94—98.
94. lim + = oo 95. lim 4 = —oo x0" x0
Chapter 2 Questions to Guide Your Review 99
Wm t--* 97 dp icem 98. lim 3799 xrl x
Oblique Asymptotes Graph the rational functions in Exercises 99-104. Include the graphs and equations of the asymptotes.
s X xl 99. y= — 100. y= =O | x-—4 el 10. y = 102. y = 2—34 X. 3 103. y = I 104, y = * 4 x
Additional Graphing Exercises Graph the curves in Exercises 105-108. Explain the relationship between the curve’s formula and what you see.
105. y = ——~ 106. y = ——— y =
1 = 2/3 107. y = x” +B
108. v— si T 08. y i55 )
Graph the functions in Exercises 109 and 110. Then answer the follow- ing questions.
a. How does the graph behave as x — 0*? b. How does the graph behave as x — +00? c. How does the graph behave near x = 1 and x = —1?
Give reasons for your answers.
2/3 2/3 109. y = i 2) 110. y= X E z)
Chapter PX Questions to Guide Your Review
1. What is the average rate of change of the function g(t) over the interval from t = a to t = b? How is it related to a secant line?
2. What limit must be calculated to find the rate of change of a func- tion g(t) at t = 19?
3. Give an informal or intuitive definition of the limit lim f(x) = L. XH
Why is the definition "informal"? Give examples.
4. Does the existence and value of the limit of a function f(x) as x approaches c ever depend on what happens at x — c? Explain and give examples.
5. What function behaviors might occur for which the limit may fail to exist? Give examples.
6. What theorems are available for calculating limits? Give exam- ples of how the theorems are used.
7. How are one-sided limits related to limits? How can this relation- ship sometimes be used to calculate a limit or prove it does not exist? Give examples.
8. What is the value of limp—o ((sin 0)/0)? Does it matter whether 0 is measured in degrees or radians? Explain.
100 Chapter 2: Limits and Continuity
9. What exactly does lim,_,. f(x) = L mean? Give an example in which you find a 6 > 0 for a given f, L, c, and e > 0 in the pre- cise definition of limit.
10. Give precise definitions of the following statements.
a. lim, f(x) = 5 b. lim,» f(x) = 5 €. lim,.. f(x) = oo d. lim,— f(x) = —oo
11. What conditions must be satisfied by a function if it is to be con- tinuous at an interior point of its domain? At an endpoint?
12. How can looking at the graph of a function help you tell where the function is continuous?
13. What does it mean for a function to be right-continuous at a point? Left-continuous? How are continuity and one-sided conti- nuity related?
14. What does it mean for a function to be continuous on an interval? Give examples to illustrate the fact that a function that is not con-
tinuous on its entire domain may still be continuous on selected intervals within the domain.
Chapter PM Practice Exercises
Limits and Continuity 1. Graph the function
1, xs-l a —l <x = 0
fœ = 1; x0 Sy, «x«t
1, xml.
Then discuss, in detail, limits, one-sided limits, continuity, and one-sided continuity of f at x — —1, 0, and 1. Are any of the discontinuities removable? Explain.
2. Repeat the instructions of Exercise 1 for
0, xc] üAx . desse d f(x) = D wl b XI.
3. Suppose that f(t) and f(r) are defined for all t and that lim,..,, f(t) = —7 and lim,.,, g(r) = 0. Find the limit as t— tọ of the following functions.
a. 3f(d) b. FOF
FO c. f()-8(0 d. 85-3 e. cos (g(f)) f. IOl
g. fO + g(t) h. 1/f()
15.
16.
17.
18.
19.
20.
21.
What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example.
What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graph- ing and solving the equation f(x) — 0?
Under what circumstances can you extend a function f(x) to be continuous at a point x — c? Give an example.
What exactly do lim,—o.o f(x) = L and lim, f(x) = L mean? Give examples.
What are lim,— +00 k (k a constant) and lim,— +00 (1/x)? How do you extend these results to other functions? Give examples.
How do you find the limit of a rational function as x — +00? Give examples.
What are horizontal and vertical asymptotes? Give examples.
. Suppose the functions f(x) and g(x) are defined for all x and that
lim, f(x) = 1/2 and lim, g(x) = V2. Find the limits as x — 0 of the following functions.
a. —g(x) b. g(x): f(x) c. f(x) + ga) d. 1/fQo) e. x + f(x) f. fe: et
In Exercises 5 and 6, find the value that lim, 9 g(x) must have if the given limit statements hold.
, (4 — 8&9 . ; 5. lim| —,—_]} = 1 6. lim { x lim g(x) } = 2 x0 x-4 x0 7. On what intervals are the following functions continuous? a. f(x) = x1? b. g(x) = x^ €. A(x) = x25? d. k(x) = x 1/6 8. On what intervals are the following functions continuous? a. f(x) = tanx b. g(x) = csc x cos x sinx c. h(x) = En d. k(x) = I Finding Limits In Exercises 9—28, find the limit or explain why it does not exist. 2: 9. ia t4 x + 5x7 — 14x a. as x —^ 0 b. as x 52
. x)? tx Wnt oe +e a. as x —^ 0 b. as x — —1 E à i 1. lim L— VE 12. lim *,—2 x1 =x xa x = a’ | (thy - x Gtr- 13. lim — — — — 14. lim — — — — h0 h x0 h lp. t , 2+xy—-8 15. dms 4 inq = x0 x0 1/3 _ 2/3 _ i fa t 18. lim% is x>l Vx — 1 x—>64 x—8 . tan(2x) : 19. lin — — 20. lim cscx x10 tan (arx) XT 21. lim sin G + sin x) 22. lim cos? (x — tan x) cH 2 xT 33. lim — 24; Jim eee
x03 sinx — x x0 sinx
In Exercises 25-28, find the limit of g(x) as x approaches the indi- cated value.
25. lim (4g())'? = 2
1
26. Ns x + ew)? 2 AS 27. lim 3x | = oo x1 g(x) d 23; duo eg
A92 AS / g(x)
Roots
29. Let f(x) = x? - x — 1.
a. Use the Intermediate Value Theorem to show that f has a Zero between —1 and 2.
b. Solve the equation f(x) = 0 graphically with an error of magnitude at most 107°.
€. It can be shown that the exact value of the solution in part (b) is
GP)"
2° 18 2 18
Evaluate this exact answer and compare it with the value you found in part (b).
30. Let f(0) = 6? — 20 + 2.
a. Use the Intermediate Value Theorem to show that f has a Zero between —2 and 0.
b. Solve the equation f(0) = 0 graphically with an error of magnitude at most 1074.
€. It can be shown that the exact value of the solution in part (b) is
19 1/3 19 ys fol seda
Evaluate this exact answer and compare it with the value you found in part (b).
Chapter 2 Practice Exercises 101
Continuous Extension
31. Can f(x) = x(x? — 1)/|x? — 1| be extended to be continuous at x = 1 or —1? Give reasons for your answers. (Graph the func- tion—you will find the graph interesting.)
32. Explain why the function f(x) = sin(1/x) has no continuous extension to x = 0.
In Exercises 33-36, graph the function to see whether it appears to have
a continuous extension to the given point a. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at a. If the function does not appear to have a continuous extension, can it be extended to be continuous from the right or left? If so, what do you think the extended function’s value should be?
xm
33. fœ) = nani x— Nx
34. (0) = AC. a —m/2
35. A(t) = (1 + |), a-0
36. k(x) = ' SL
Limits at Infinity Find the limits in Exercises 37-46.
lc. Di ess 37. dt 38. lim Z x00 DX T 7 X—-—00 5x2 +7 o a 39. lim t8 40. lim ——?—— x—-00 3x xoox?— Tx + 1 x2 — 7x 42. lim x* ox)
41. lim CEN x x>% 12x? + 128
sin x (If you have a grapher, try graphing the function for-5 =x =5.) 4 (if you have a grapher, try graphing 44. lim — — f(x) = x(cos (1/x) — 1) near the origin to "see" the limit at infinity.) . x+ sinx + 2Vx a ae x + sinx
43. lim
32/3 4 cl 46. lim M x06 x?/5 + cos?x
Horizontal and Vertical Asymptotes 47. Use limits to determine the equations for all vertical asymptotes.
| x4 a y=- Se e b. Ma M= oJ i x)-x-6 x)-2x—- 8 48. Use limits to determine the equations for all horizontal asymptotes. 1-2 Vx + a y= b. ye fœ = = rt+4 249 € 3(07- x
aci
102 Chapter 2: Limits and Continuity
Chapter PAM Additional and Advanced Exercises
[T] 1.
Assigning a value to 0° The rules of exponents tell us that a? = 1 if a is any number different from zero. They also tell us that 0" = 0 if n is any positive number.
If we tried to extend these rules to include the case 0°, we would get conflicting results. The first rule would say 0° = 1, whereas the second would say 0° = 0.
We are not dealing with a question of right or wrong here. Neither rule applies as it stands, so there is no contradiction. We could, in fact, define 0° to have any value we wanted as long as we could persuade others to agree.
What value would you like 0° to have? Here is an example that might help you to decide. (See Exercise 2 below for another example.)
a. Calculate x* for x = 0.1, 0.01, 0.001, and so on as far as your calculator can go. Record the values you get. What pattern do you see?
b. Graph the function y = x“ for 0 < x = 1. Even though the function is not defined for x = 0, the graph will approach the y-axis from the right. Toward what y-value does it seem to be headed? Zoom in to further support your idea.
. A reason you might want 0° to be something other than 0 or 1
As the number x increases through positive values, the numbers 1/x and 1 / (In x) both approach zero. What happens to the number
i 1/(Inx) f(x) = B
as x increases? Here are two ways to find out.
a. Evaluate f for x — 10, 100, 1000, and so on as far as your calculator can reasonably go. What pattern do you see?
b. Graph f in a variety of graphing windows, including win- dows that contain the origin. What do you see? Trace the y-values along the graph. What do you find?
. Lorentz contraction In relativity theory, the length of an
object, say a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket's length as Lo at rest, then at speed v the length will appear to be
This equation is the Lorentz contraction formula. Here, c is the speed of light in a vacuum, about 3 X 10? m/sec. What happens to L as v increases? Find lim, ,,- L. Why was the left-hand limit needed?
4. Controlling the flow from a draining tank Torricelli's law
says that if you drain a tank like the one in the figure shown, the rate y at which water runs out is a constant times the square root of the water's depth x. The constant depends on the size and shape of the exit valve.
Exit rate y ft?/min
Suppose that y — Vx/ 2 for a certain tank. You are trying to maintain a fairly constant exit rate by adding water to the tank with a hose from time to time. How deep must you keep the water if you want to maintain the exit rate
ll
a. within 0.2 ft/min of the rate yọ = 1 ft/min? b. within 0.1 ft/min of the rate yọ = 1 f /min?
. Thermal expansion in precise equipment As you may know,
most metals expand when heated and contract when cooled. The dimensions of a piece of laboratory equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laboratory where the equipment is to be used. A typical aluminum bar that is 10 cm wide at 70?F will be
y = 10 + (t — 70) x 107+
centimeters wide at a nearby temperature t. Suppose that you are using a bar like this in a gravity wave detector, where its width must stay within 0.0005 cm of the ideal 10 cm. How close to fy = 70°F must you maintain the temperature to ensure that this tolerance is not exceeded?
. Stripes on a measuring cup The interior of a typical 1-L mea-
suring cup is a right circular cylinder of radius 6 cm (see accom- panying figure). The volume of water we put in the cup is there- fore a function of the level h to which the cup is filled, the formula being
V = q6h = 36mh.
How closely must we measure h to measure out 1 L of water (1000 cm?) with an error of no more than 196 (10 cm?)?
Stripes about
] mm wide
(a)
r=6cm
Liquid volume V = 367th
A 1-L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6cm
Precise Definition of Limit In Exercises 7-10, use the formal definition of limit to prove that the function is continuous at c.
7. 9. 11.
12.
13.
14.
fox -—g é-1 8. g(x) = 1/Qx), c= 1/4 h(x) = V2x -3, c=2 10. Fx)- V9—-x c=5 Uniqueness of limits Show that a function cannot have two dif-
ferent limits at the same point. That is, if lim, ,, f(x) = Lı and lim, f(x) = L», then Lı = Ly.
Prove the limit Constant Multiple Rule:
lim kf(x) = k lim f(x) for any constant k.
One-sided limits If lim, - f(x) = A and lim, .$- f(x) = B, find
a. limy f(? — x) b. limy fæ — x)
d. limyo- f(x? — x?)
Limits and continuity Which of the following statements are true, and which are false? If true, say why; if false, give a counter- example (that is, an example confirming the falsehood).
c. limyso- fG? — x?)
a. If lim, ,, f(x) exists but lim, ,, g(x) does not exist, then lim, , (f(x) + g(x)) does not exist.
b. If neither lim,_,. f(x) nor lim,_,.. g(x) exists, then lim,- (f(x) + g(x)) does not exist.
c. If f is continuous at x, then so is |f].
d. If |f| is continuous at c, then so is f.
In Exercises 15 and 16, use the formal definition of limit to prove that the function has a continuous extension to the given value of x.
15.
x-—1
fy = Sh, —2x—3
2x — 6
2 x=-l1 16. g(x) =~ , x=3
17.
18.
19.
20.
21.
22.
23.
103
Chapter 2 Additional and Advanced Exercises
A function continuous at only one point Let
x, ifxisrational
fe = {
O, if x is irrational.
a. Show that f is continuous at x = 0.
b. Use the fact that every nonempty open interval of real num- bers contains both rational and irrational numbers to show that f is not continuous at any nonzero value of x.
The Dirichlet ruler function If x is a rational number, then x can be written in a unique way as a quotient of integers m/n where n > 0 and m and n have no common factors greater than 1. (We say that such a fraction is in lowest terms. For example, 6/4 written in lowest terms is 3/2.) Let f(x) be defined for all x in the interval [ 0, 1] by
1 f(x) -f 2n
0, if x is irrational.
if x = m/n isa rational number in lowest terms
For instance, f(0) = f(1) = 1, f(1/2) = 1/2, f(1/3) = f(2/3) = 1/3, f(1/4) = f(3/4) = 1/4, and so on.
a. Show that f is discontinuous at every rational number in [ 0, 1].
b. Show that f is continuous at every irrational number in [ 0, 1 ]. (Hint: If € is a given positive number, show that there are only finitely many rational numbers r in [ 0, 1 ] such that f(r) = e.)
€. Sketch the graph of f. Why do you think f is called the "ruler function"?
Antipodal points Is there any reason to believe that there is always a pair of antipodal (diametrically opposite) points on Earth's equator where the temperatures are the same? Explain.
If lim,—, FŒ) + g(@)) = 3 and lim, (f(x) — g(x) = — 1, find
lim,, f(x)g(x) .
Roots of a quadratic equation that is almostlinear The equa- tion ax? + 2x — 1 = 0, where a is a constant, has two roots if a — —1 anda # 0, one positive and one negative:
-l+ Vita = S
L= yiká
r(a) = r(a) = ——>
a. What happens to r,(a) as a — 0? Asa — —1^? b. What happens to r (a) as a — 0? Asa —1*?
c. Support your conclusions by graphing r,(a) and r (a) as functions of a. Describe what you see.
d. For added support, graph f(x) = ax? + 2x — 1 simultane- ously for a — 1, 0.5, 0.2, 0.1, and 0.05.
Root of an equation Show that the equation x + 2 cosx = 0 has at least one solution.
Bounded functions A real-valued function f is bounded from above on a set D if there exists a number N such that f(x) = N for all x in D. We call N, when it exists, an upper bound for f on D and say that f is bounded from above by N. In a similar man- ner, we say that f is bounded from below on D if there exists a number M such that f(x) = M for all x in D. We call M, when it exists, a lower bound for f on D and say that f is bounded from below by M. We say that f is bounded on D if it is bounded from both above and below.
a. Show that f is bounded on D if and only if there exists a number B such that | f(x)| = B for all x in D.
104 Chapter 2: Limits and Continuity
b. Suppose that f is bounded from above by N. Show that if . sin(x? — x — 2) . sin(x? — x — 2) lim,., f(x) = L, then L < N. C n xel Tr M cx- c. Suppose that f is bounded from below by M. Show that if M e E wc @ +t DO-2 _ lim, f(x) = L, then L = M. lim es a 1- lim FI 3 x-1 x x-i x 24. Max {a,b} and min (a,b aes tat] . sin(1— Vx) : sin(1—- Vx)1- VA a. Show that the expression d. lim — lim = | b| x1 xc xl 1—- Vx x s a+b, |@— max {a, b} 2 7523 "TN E um 1-x d equals a if a = b and equals b if b = a. In other words, XcE qx p(1 T Vx) er he (1 + Vx) 2 max {a,b} gives the larger of the two numbers a and b. i EN i . Hp . i Find the limits in Exercises 25-30. b. Find a similar expression for min {a, b}, the smaller of a ] . db . sin(l — cos x) ; sin x and b. 25. lim — —, ——— 26. lim — —- . x0 x20 sin V/x sin Generalized Limits Involving x xw om sin (sin x) 28. li sin (x? + x) . lim — — im — — —— The formula limg.,9(sin0)/0 = 1 can be generalized. If lim, x9 c zo) r f(x) = 0 and f(x) is never zero in an open interval containing the . sin(3? — 4) . sin( Vx = 3) point x = c, except possibly c itself, then 29. lim— 1-2 30. im- 38 sinf(x) _ rue d Oblique Asymptotes Here are several examples. Find all possible oblique asymptotes in Exercises 31—34. inae 2x2 + Be 3 E a. lim "> = j 31. y — Gai 32. y=xtxsing b lim SX. — lim SI E uu E = 1-0=0 33. y= Vx? 1 34. y= Vx? + 2x ` x0 * xo0 x? x0 x
Chapter PM Technology Application Projects
Mathematica/Maple Modules:
Take It to the Limit
Part I
Part II (Zero Raised to the Power Zero: What Does It Mean?)
Part III (One-Sided Limits)
Visualize and interpret the limit concept through graphical and numerical explorations. Part IV (What a Difference a Power Makes)
See how sensitive limits can be with various powers of x.
Going to Infinity Part I (Exploring Function Behavior as x > © or x > — %0) This module provides four examples to explore the behavior of a function as x — 0o or x — —oo.
Part II (Rates of Growth) Observe graphs that appear to be continuous, yet the function is not continuous. Several issues of continuity are explored to obtain results that you
may find surprising.
Derivatives
OVERVIEW In the beginning of Chapter 2, we discussed how to determine the slope of a curve at a point and how to measure the rate at which a function changes. Now that we have studied limits, we can define these ideas precisely and see that both are interpretations of the derivative of a function at a point. We then extend this concept from a single point to the derivative function, and we develop rules for finding this derivative function easily, without having to calculate any limits directly. These rules are used to find derivatives of many of the common functions reviewed in Chapter 1, as well as various combinations of them.
The derivative is one of the key ideas in calculus, and is used to study a wide range of problems in mathematics, science, economics, and medicine. These problems include finding points where a continuous function is zero, calculating the velocity and accelera- tion of a moving object, determining how the rate of flow of a liquid into a container changes the level of the liquid within it, describing the path followed by a light ray going from a point in air to a point in water, finding the number of items a manufacturing com- pany should produce in order to maximize its profits, studying the spread of an infectious disease within a given population, or calculating the amount of blood the heart pumps in a minute based on how well the lungs are functioning.
3. I Tangents and the Derivative at a Point
y =f Q(xo + h, f(x + h))
i f(Xq + h) — f(xg)
>X
FIGURE 3.1 The slope of the tangent fG + h) — fao) AS ESL
line at P is lim h0
In this section we define the slope and tangent to a curve at a point, and the derivative of a function at a point. The derivative gives a way to find both the slope of a graph and the instantaneous rate of change of a function.
Finding a Tangent to the Graph of a Function
To find a tangent to an arbitrary curve y = f(x) at a point P(xo, f(x9)), we use the procedure introduced in Section 2.1. We calculate the slope of the secant through P and a nearby point Q(x) + h, f(% + h)). We then investigate the limit of the slope as h — 0 (Figure 3.1). If the limit exists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope.
DEFINITIONS The slope of the curve y = f(x) at the point P(x, f(xo)) is the number
o fo + h) — foo) m lim
lim 5 (provided the limit exists).
The tangent line to the curve at P is the line through P with this slope.
105
106 Chapter 3: Derivatives
In Section 2.1, Example 3, we applied these definitions to find the slope of the parab- ola f(x) — x? at the point P(2, 4) and the tangent line to the parabola at P. Let's look at another example.
EXAMPLE 1 (a) Find the slope of the curve y = 1/x at any point x = a # 0. What is the slope at the point x = —1?
(b) Where does the slope equal —1/4? (c) What happens to the tangent to the curve at the point (a, 1/a) as a changes?
i
slope is — 1
aema Solution
(a) Here f(x) = 1/x. The slope at (a, 1/a) is
FIGURE 3.2 The tangent slopes, steep 1 et near the origin, become more gradual as _ f(a + h) — fa) _ ath 4 . Lla-—(ath) ; lim = lim ————— = lim the point of tangency moves away h0 h0 h h>oh ala + h) Example 1). i = lim ——"— = jim——1__--1 n0 hala + h) p70 a(a + h) a?
Notice how we had to keep writing “lim,—.9” before each fraction until the stage at which we could evaluate the limit by substituting h = 0. The number a may be positive or negative, but not 0. When a — —1, the slope is —1/(-1)? = —
slope is -i (Figure 3.2). (b) The slope of y = 1/x at the point where x = a is —1/a?. It will be — 1/4 provided 1 that slope is E^
——
dU OX This equation is equivalent to à? = 4, so a = 2 or a = —2. The curve has slope
— 1/4 at the two points (2, 1/2) and (-2, —1/2) (Figure 3.3).
FIGURE 3.3 The two tangent lines to (c) The slope —1/a? is always negative if a # 0. As a — 0”, the slope approaches —oo
and the tangent becomes increasingly steep (Figure 3.2). We see this situation again as a — O~. As a moves away from the origin in either direction, the slope approaches 0 and the tangent levels off becoming more and more horizontal. E
y = 1/x having slope —1/4 (Example 1).
Rates of Change: Derivative at a Point The expression
fo + hA) — fo) h ,
hz 0
is called the difference quotient of f at xy with increment h. If the difference quotient has a limit as ^ approaches zero, that limit is given a special name and notation.
DEFINITION The derivative of a function f at a point x9, denoted f'(xo), is fe + h) — fe) h
The notation f'(xo) is read “f prime of xj." f'x)- lim h—0
provided this limit exists.
3.1 Tangents and the Derivative at a Point 107
If we interpret the difference quotient as the slope of a secant line, then the derivative gives the slope of the curve y = f(x) at the point P(%, f(x9)). Exercise 33 shows that the derivative of the linear function f(x) = mx + b at any point x, is simply the slope of the line, so
f'(x) = m,
which is consistent with our definition of slope.
If we interpret the difference quotient as an average rate of change (Section 2.1), the derivative gives the function’s instantaneous rate of change with respect to x at the point X = xg. We study this interpretation in Section 3.4.
EXAMPLE 2 In Examples 1 and 2 in Section 2.1, we studied the speed of a rock fall- ing freely from rest near the surface of the earth. We knew that the rock fell y = 167? feet during the first t sec, and we used a sequence of average rates over increasingly short inter- vals to estimate the rock’s speed at the instant t = 1. What was the rock's exact speed at this time?
Solution We let f(t) = 16r. The average speed of the rock over the interval between t = l and t = 1 + h seconds, for h > 0, was found to be
fü +h) - fd) = 16(1 + hy — 16(1)? = 16(4? + 2h)
h n à = 16(h + 2). The rock’s speed at the instant t = 1 is then f'G) = lim 16(h + 2) = 16(0 + 2) = 32 ft/sec. Our original estimate of 32 ft/sec in Section 2.1 was right. E
Summary
We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, and the derivative of a function at a point. All of these ideas refer to the same limit.
The following are all interpretations for the limit of the difference quotient,
2o fog + h) — fo) lim . h—0 h
. The slope of the graph of y = f(x) at x = xy . The slope of the tangent to the curve y = f(x) at x = xo
. The rate of change of f(x) with respect to x at x = xp
A U Nme
. The derivative f'(xọ) at a point
In the next sections, we allow the point x, to vary across the domain of the function f.
108 Chapter 3: Derivatives
Exercises
Slopes and Tangent Lines In Exercises 1—4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points P, and P».
1. 2.
><
x 3. 4 y y A A "n 3 2 1 B, i P; } 10 i 3]^" 0 FT
In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together.
5. y=4- x, (1,3) 6 y-G-1?-*1 (LD
7. y 2Vx, (1,2) 8. y= x (71, 1) EN" i 1 9. y=x, (-2,—8) 10. y — zs EA R 8
In Exercises 11-18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
11. fo) - x2 +1, (2,5) 12. fü =x- 2x7, (1,-1)
X 8 oC). Jig.
15. A(t) = P, (2,8) 17. fc) = Vx, (4,2)
13. g(x)
(2, 2)
16. hA =P 3t (1,4) 18. f(xy) = Vx +1, (83)
In Exercises 19—22, find the slope of the curve at the point indicated. 19. y = 5x - 32, x=1 20. y 22 -2x +7, x 2-2 1 xc
queque ASP 2 y= FT
21. y=
x=0
Interpreting Derivative Values
23. Growth of yeast cells In a controlled laboratory experiment, yeast cells are grown in an automated cell culture system that counts the number P of cells present at hourly intervals. The num- ber after t hours is shown in the accompanying figure.
250 200 150 100
50
0 123456 7
>t
a. Explain what is meant by the derivative P'(5). What are its units?
b. Which is larger, P'(2) or P'(3)? Give a reason for your answer.
€. The quadratic curve capturing the trend of the data points (see Section 1.4) is given by P(t) = 6.1027 — 9.28t + 16.43. Find the instantaneous rate of growth when t = 5 hours.
24. Effectiveness of a drug On a scale from 0 to 1, the effective- ness E of a pain-killing drug ¢ hours after entering the blood- stream is displayed in the accompanying figure.
E
>t
a. At what times does the effectiveness appear to be increasing? What is true about the derivative at those times?
b. At what time would you estimate that the drug reaches its maximum effectiveness? What is true about the derivative at that time? What is true about the derivative as time increases in the 1 hour before your estimated time?
At what points do the graphs of the functions in Exercises 25 and 26 have horizontal tangents?
25. f(x) =x? + 4x-1 26. gx) = xX — 3x
27. Find equations of all lines having slope —1 that are tangent to the curve y = 1/(x — 1).
28. Find an equation of the straight line having slope 1/4 that is tan- gent to the curve y — Vx.
Rates of Change
29. Object dropped from a tower An object is dropped from the top of a 100-m-high tower. Its height above ground after t sec is 100 — 4.9? m. How fast is it falling 2 sec after it is dropped?
30.
31.
32.
33.
34.
Speed of a rocket At f sec after liftoff, the height of a rocket is 3? ft. How fast is the rocket climbing 10 sec after liftoff?
Circle's changing area What is the rate of change of the area of a circle (A = zr?) with respect to the radius when the radius is r = 3?
Ball’s changing volume What is the rate of change of the vol- ume of a ball (V = (4/3)mr?) with respect to the radius when the radius is r — 2?
Show that the line y = mx + b is its own tangent line at any point (xo, mxg + D).
Find the slope of the tangent to the curve y — 1/ Vx at the point where x — 4.
Testing for Tangents
35.
36.
Does the graph of xisin(l/x, x70 have a tangent at the origin? Give reasons for your answer. Does the graph of . [xsin(1/x) x70 m= b x=0
have a tangent at the origin? Give reasons for your answer.
Vertical Tangents
We say that a continuous curve y = f(x) has a vertical tangent at the point where x = x if the limit of the difference quotient is CO or —oo. For example, y = x! has a vertical tangent at x = 0 (see accompa-
nying figure): . f0+h-fO ,. m5-0 lim = lim h—0 h n0 h -= liim | = i Im 5 i
VERTICAL TANGENT AT ORIGIN
However, y = x?? has no vertical tangent at x = 0 (see next figure):
&0c7m-—380 .. p^-0 im = lim h—0 h
h>0 h
= in ie
109
3.1 Tangents and the Derivative at a Point
does not exist, because the limit is co from the right and —co from the
left.
37.
38.
NO VERTICAL TANGENT AT ORIGIN Does the graph of =], x«0 fe) = 0 x=0 l x>0
have a vertical tangent at the origin? Give reasons for your answer.
ue) = f x«0 ^ 1; x-—40
Does the graph of
have a vertical tangent at the point (0, 1)? Give reasons for your answer.
Graph the curves in Exercises 39—48.
a. Where do the graphs appear to have vertical tangents?
b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38.
Ly =x 40. y = x^^ .y — xl 42. y = x3/5 y = 4x7/5 — 2x 44, y = 9/8 — sy? y= LP — (y — p^ 46. y — xl/5 + (x — 1) -Vix. «<0 —= y= {yr sg TENES
COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exer- cises 49-52:
49.
51. 52.
a. Plot y = f(x) over the interval (xj — 1/2) € x = (x + 3). b. Holding x, fixed, the difference quotient
+h) - ay) = fe ) fo)
at xy becomes a function of the step size h. Enter this function into your CAS workspace.
c. Find the limit of q as h — 0.
d. Define the secant lines y = f(x9) + q* (x — Xo) for h = 3,2, and 1. Graph them together with f and the tangent line over the interval in part (a).
5
fW} *2x x 2-0 50 fi)-xt*X, x«-71
f(x) = x + sin(2x), f(x) = cos x + 4 sin(2x),
Xo = 7/2
Xo = 7
110
Chapter 3: Derivatives
3.2 The Derivative as a Function
HISTORICAL ESSAY The Derivative
y =f)
fe = feo
ZX
Derivative of fat x is
fj lid f(x +h) — fe) h—0 h = lim f@ - fe) Lo Xe
Lx
FIGURE 3.4 Two forms for the differ- ence quotient.
Derivative of the Reciprocal Function
díl 1 TE
Secant slope is
In the last section we defined the derivative of y = f(x) at the point x = x to be the limit
fG + h) — fx)
f (xo) Lr lim h
We now investigate the derivative as a function derived from f by considering the limit at each point x in the domain of f.
DEFINITION The derivative of the function f(x) with respect to the variable x is the function f' whose value at x is
fæ + h) — feo 5 ,
f'G) = lim
provided the limit exists.
We use the notation f(x) in the definition to emphasize the independent variable x with respect to which the derivative function f'(x) is being defined. The domain of f' is the set of points in the domain of f for which the limit exists, which means that the domain may be the same as or smaller than the domain of f. If f' exists at a particular x, we say that f is differentiable (has a derivative) at x. If f' exists at every point in the domain of f. we call f differentiable.
If we write z = x + h, then h = z — x and h approaches 0 if and only if z approaches x. Therefore, an equivalent definition of the derivative is as follows (see Figure 3.4). This formula is sometimes more convenient to use when finding a derivative function, and focuses on the point z that approaches x.
Alternative Formula for the Derivative
f@ - feo
f'G) = lim £o
Calculating Derivatives from the Definition
The process of calculating a derivative is called differentiation. To emphasize the idea that differentiation is an operation performed on a function y — f(x), we use the notation
PRI)
as another way to denote the derivative f'(x). Example 1 of Section 3.1 illustrated the dif- ferentiation process for the function y — 1/x when x — a. For x representing any point in the domain, we get the formula
dafi) dl
dx AX x»
Here are two more examples in which we allow x to be any point in the domain of f.
Derivative of the Square Root Function
d E Yt =
—
2Vx
><
0 4
FIGURE 3.5 The curve y = Vx and its tangent at (4, 2). The tangent's slope is found by evaluating the derivative at x — 4 (Example 2).
3.2 The Derivative as a Function 111
EXAMPLE 1 Differentiate f(x) = —*—.
x—1
Solution We use the definition of derivative, which requires us to calculate f(x + h) and then subtract f(x) to obtain the numerator in the difference quotient. We have
= —*— and dde _@th fa) =z ad fa Gap ai FQ) = har h Definition x+h x xc h^h-—1 x— 1 = lim h0 h sgh TD 1) - xa +h — 1) a e ad=gb hooh (x +h- DG = 1) b d bd = nmn (x*h—Dx-10 Simplify. E =i Cancel h # 0. Hi
prk- ia- G-
EXAMPLE 2 (a) Find the derivative of f(x) = Vx for x > 0.
(b) Find the tangent line to the curve y = Vx at x = 4.
Solution (a) We use the alternative formula to calculate f": ; . fQ- fe» (Osim 7, |n Vg — VX A EDI l vz- vz = lim e»(vz- vx)(vz- vx) — lim l = : zox Vz + Vx AVX
(b) The slope of the curve at x = 4 is
2v4 ^4 The tangent is the line through the point (4, 2) with slope 1/4 (Figure 3.5):
ra= =!
= Ls y-2-t4( 4)
yid a
Notations
There are many ways to denote the derivative of a function y = f(x), where the indepen- dent variable is x and the dependent variable is y. Some common alternative notations for the derivative are
d d rey oy =F = T= 6 = pto) = Df).
112 Chapter 3: Derivatives
>%
— [UU A T se ll in € wa
A D' Y 10 15 = 1 = t B' -2r Vertical coordinate — 1
>x
(b)
FIGURE 3.6 We made the graph of
y = f'(x) in (b) by plotting slopes from the graph of y = f(x) in (a). The vertical coordinate of B’ is the slope at B and so on. The slope at E is approximately
8/4 — 2. In (b) we see that the rate of change of f is negative for x between A' and D'; the rate of change is positive for x to the right of D'.
Slope = lim f(b + h) — fib) h—07 h
Slope = fla + h) — fla)
FIGURE 3.7 Derivatives at endpoints of a closed interval are one-sided limits.
The symbols d/dx and D indicate the operation of differentiation. We read dy/dx as “the derivative of y with respect to x," and df /dx and (d/dx) f(x) as “the derivative of f with respect to x." The "prime" notations y' and f' come from notations that Newton used for derivatives. The d/dx notations are similar to those used by Leibniz. The sym- bol dy/dx should not be regarded as a ratio (until we introduce the idea of “differen- tials" in Section 3.9).
To indicate the value of a derivative at a specified number x — a, we use the notation
vy D| d _ d f m dx x=a dx x=a 4x1 x=a For instance, in Example 2 d 1 1 1 f 4 aa Nae: == — — x f ( ms x-4 2Vx x=4 2VA 4
Graphing the Derivative
We can often make a reasonable plot of the derivative of y — f(x) by estimating the slopes on the graph of f. That is, we plot the points (x, f'(x)) in the xy-plane and connect them with a smooth curve, which represents y = f'(x).
EXAMPLE 3 Graph the derivative of the function y — f(x) in Figure 3.6a.
Solution We sketch the tangents to the graph of f at frequent intervals and use their slopes to estimate the values of f'(x) at these points. We plot the corresponding (x, f'(x)) pairs and connect them with a smooth curve as sketched in Figure 3.6b. ig
What can we learn from the graph of y — f'(x)? Ata glance we can see
1. where the rate of change of f is positive, negative, or zero; the rough size of the growth rate at any x and its size in relation to the size of f(x);
3. where the rate of change itself is increasing or decreasing.
Differentiable on an Interval; One-Sided Derivatives
A function y — f(x) is differentiable on an open interval (finite or infinite) if it has a derivative at each point of the interval. It is differentiable on a closed interval [ a, b | if it is differentiable on the interior (a, b) and if the limits
+ h) — m Ha Top qu) Right-hand derivative at a
h—0* h b + h) — f(b lim 105 IW Left-hand derivative at b
exist at the endpoints (Figure 3.7).
Right-hand and left-hand derivatives may be defined at any point of a function’s domain. Because of Theorem 6, Section 2.4, a function has a derivative at a point if and only if it has left-hand and right-hand derivatives there, and these one-sided derivatives are equal.
EXAMPLE 4 Show that the function y = |x| is differentiable on (—0°, 0) and (0, 00) but has no derivative at x = 0.
Solution From Section 3.1, the derivative of y = mx + b is the slope m. Thus, to the right of the origin,
d d d d (ll) -4 07402271 qt b) = ml =x
y' not defined at x = 0: right-hand derivative # left-hand derivative
FIGURE 3.8 The function y — |x| is not differentiable at the origin where the graph has a "corner" (Example 4).
3.2 The Derivative as a Function 113
To the left,
Lix) = ev= cips- hs
(Figure 3.8). There is no derivative at the origin because the one-sided derivatives differ there:
lo+al-lol_ dal
Right-hand derivative of |x| at zero = lim
h—0* h n0: h
oar
= lim > |a| = hwhenh > 0 n—orh
= lm1l =1 h—0*
"- lox Al —jol a Left-hand derivative of |x] at zero = =
h—0- h n0 h
m Mu
= lim ;- |h| = —h when h < 0 h0- h
= lim-1 = -1. L| h0
EXAMPLE 5 In Example 2 we found that for x > 0,
Since the (right-hand) limit is not finite, there is no derivative at x — 0. Since the slopes of the secant lines joining the origin to the points (h, Vh) on a graph of y = Vx approach c6, the graph has a vertical tangent at the origin. (See Figure 1.17 on page 9.) lii
When Does a Function Not Have a Derivative at a Point?
A function has a derivative at a point xy if the slopes of the secant lines through P(xo, f(x)) and a nearby point Q on the graph approach a finite limit as Q approaches P. Whenever the secants fail to take up a limiting position or become vertical as Q approaches P, the deriva- tive does not exist. Thus differentiability is a “smoothness” condition on the graph of f. A function can fail to have a derivative at a point for many reasons, including the existence of points where the graph has
0 a aa
1. a corner, where the one-sided 2. a cusp, where the slope of PQ approaches derivatives differ. OO from one side and —C9 from the other.
114
Chapter 3: Derivatives
3. a vertical tangent, 4. a discontinuity (two examples shown). where the slope of PQ approaches oo from both sides or approaches —oo from both sides (here, —oo).
Another case in which the derivative may fail to exist occurs when the function's slope is oscillating rapidly near P, as with f(x) — sin (1/x) near the origin, where it is discontinu- ous (see Figure 2.31).
Differentiable Functions Are Continuous
A function is continuous at every point where it has a derivative.
THEOREM 1—Differentiability Implies Continuity If f has a derivative at x = c, then f is continuous at x = c.
Proof Given that f'(c) exists, we must show that lim, ,. f(x) = f(c), or equivalently, that lim, f(c + h) = f(c). If h # 0, then
f(c + h) = f(c) + (f(c + h) — fe) -qo + EF PTOL, Now take limits as h — 0. By Theorem 1 of Section 2.2, f(c + h) — fo. hi
ae os PM = fle) vfu) = f(c) + 0 = f(c). H
Similar arguments with one-sided limits show that if f has a derivative from one side (right or left) at x = c, then f is continuous from that side at x = c.
Theorem 1 says that if a function has a discontinuity at a point (for instance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function y = | x | fails to be differentiable at every integer x — n (Example 4, Section 2.5).
Caution The converse of Theorem 1 is false. A function need not have a derivative at a point where it is continuous, as we saw with the absolute value function in Example 4.
Exercises
Finding Derivative Functions and Values Using the definition, calculate the derivatives of the functions in Exercises 1—6. Then find the values of the derivatives as specified.
1. f@=4-= x; fC-3.f 0. F'O) 2. Fœ) = œ- 1? +1; FC), F(0), F'Q)
3. 9) = z g'C D, g'Q), e'( V3)
l1-z 2z
5. p(0) = V30; p'(1), p'G), p'Q/3) 6. r(s) = V2s + 1; r'(0) r'(1), r'(1/2)
4. kZ) = ; KED, kO), k’'(-V2)
In Exercises 7—12, find the indicated derivatives.
dy dr 3
i = 2x3 i = 24 7. di if y= 2x 8. dE if r=s 2s 3 ds . _ ot dv . Pre 9. a if s 2r4 1 10. di if v-t 7 dp 7 dz P — Jn . _ 11. da if p=q 12. Tw if z CERE
Slopes and Tangent Lines In Exercises 13-16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable.
13. f) x42, x--3 14. ka) =;
15. 5- P —nD, t=-1 16. y =:
] ex'
In Exercises 17-18, differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function.
8 Vx—2 18. w = gz) = 1 + V4-z (zw) = (3,2)
17. y 2 fΠ= (x, y) = (6, 4)
In Exercises 19-22, find the values of the derivatives.
ds ) 1 19. — if s-1—3P? 20. — if y21-x dti V3 * dr 2 dw 21. — f = 22. -4 df =zt Blige age EN
Using the Alternative Formula for Derivatives Use the formula
fü — feo
fœ = lim TER
to find the derivative of the functions in Exercises 23-26.
23. fe) = — 24. f(x) =x —3x+4 25. g(x) = a 26. gx) = 1+ Vx
3.2 The Derivative as a Function 115
Graphs Match the functions graphed in Exercises 27—30 with the derivatives graphed in the accompanying figures (a)-(d).
y y 0 X 0 -X (a) (b) y y! (c) (d) 27. 28. y y ^ y = fi) y —f300 0 >x 0 >x 29 30. y = fae) -X
31. a. The graph in the accompanying figure is made of line seg- ments joined end to end. At which points of the interval [—4, 6] is f’ not defined? Give reasons for your answer. y A (0, 2) (6, 2) y —fQ)
la, —2) (4, —2)
b. Graph the derivative of f. The graph should show a step function.
116 Chapter 3: Derivatives
32. Recovering a function from its derivative
33.
34.
a. Use the following information to graph the function f over the closed interval [—2, 5].
i) The graph of f is made of closed line segments joined end to end.
ii) The graph starts at the point (—2, 3).
iii) The derivative of f is the step function in the figure
shown here. i ^ y =f'@)
l1- oo
D] RE NS NE RN CNN x =2 0 1 3 3
= o———e
e—— -2
b. Repeat part (a), assuming that the graph starts at (—2, 0) instead of (—2, 3).
Growth in the economy The graph in the accompanying figure shows the average annual percentage change y = f(t) in the US. gross national product (GNP) for the years 2005-2011. Graph dy/ dt (where defined).
ER] m
6 5 4 3 2 1 0
2005 2006 2007 2008 2009 2010 2011
Fruit flies (Continuation of Example 4, Section 2.1.) Popula- tions starting out in closed environments grow slowly at first, when there are relatively few members, then more rapidly as the number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment.
a. Use the graphical technique of Example 3 to graph the deriv-
ative of the fruit fly population. The graph of the population is reproduced here.
Number of flies
>t
0 10 20 30 40 50 Time (days)
b. During what days does the population seem to be increasing fastest? Slowest?
35. Temperature The given graph shows the temperature T in °F at Davis, CA, on April 18, 2008, between 6 A.M. and 6 P.M.
T
80 70 60
50
Temperature (°F)
40
>t
0 3 6 9 12 6 AM. 9AM. 12NOON 3PM. 6 PM.
Time (hr) a. Estimate the rate of temperature change at the times i) 7 A.M. ii) 9 A.M. iii) 2 P.M. iv) 4 P.M. b. At what time does the temperature increase most rapidly? Decrease most rapidly? What is the rate for each of those times? c. Use the graphical technique of Example 3 to graph the deriv- ative of temperature T versus time t.
36. Weight loss Jared Fogle, also known as the “Subway Sandwich Guy,” weighed 425 Ib in 1997 before losing more than 240 Ib in 12 months (http://en.wikipedia.org/wiki/Jared_Fogle). A chart showing his possible dramatic weight loss is given in the accom- panying figure.
W
^ 500 425 300
200
Weight (Ib)
100
0 123 45 67 8 9 1011 12
Time (months)
a. Estimate Jared’s rate of weight loss when i t=1 ii) +=4 iii) : = 11
b. When does Jared lose weight most rapidly and what is this rate of weight loss?
c. Use the graphical technique of Example 3 to graph the deriv- ative of weight W.
One-Sided Derivatives Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37—40 are not differentiable at the point P.
P(0, 0) : 0 1 2
39. 40.
In Exercises 41 and 42, determine if the piecewise-defined function is
differentiable at the origin.
4 yc x20 ex xc-2x 7, x«0
LP, xm 42. g(x) = ee x«0
Differentiability and Continuity on an Interval Each figure in Exercises 43-48 shows the graph of a function over a closed interval D. At what domain points does the function appear to be
a. differentiable? b. continuous but not differentiable?
c. neither continuous nor differentiable?
Give reasons for your answers. 43.
y=fo) y D:-—3z54-2
3.2 The Derivative as a Function 117
Theory and Examples In Exercises 49—52, a. Find the derivative f'(x) of the given function y = f(x). b. Graph y = f(x) and y = f'(x) side by side using separate sets of coordinate axes, and answer the following questions. c. For what values of x, if any, is f' positive? Zero? Negative?
d. Over what intervals of x-values, if any, does the function y — f(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.) 49. y = —? 50. y = —1/x 51. y = 9/3 52. y = x4/4 53. Tangent to a parabola Does the parabola y = 2x7 — 13x + 5 have a tangent whose slope is —1? If so, find an equation for the line and the point of tangency. If not, why not?
54. Tangent to y — Vx Does any tangent to the curve y — Vx cross the x-axis at x — —1? If so, find an equation for the line and the point of tangency. If not, why not?
55. Derivative of — f Does knowing that a function f(x) is differ- entiable at x = xy tell you anything about the differentiability of the function —f at x = xg? Give reasons for your answer.
56. Derivative of multiples Does knowing that a function g(f) is differentiable at £ = 7 tell you anything about the differentiabil- ity of the function 3g at t = 7? Give reasons for your answer.
57. Limit of a quotient Suppose that functions g(t) and A(t) are defined for all values of ¢ and g(0) = A(O) = 0. Can lim, (g()/(A()) exist? If it does exist, must it equal zero? Give reasons for your answers.
58. a. Let f(x) be a function satisfying | F(x)| <=xfor-l<=x<=1.
Show that f is differentiable at x = 0 and find f’(0).
b. Show that
fe) = 0, x=0 is differentiable at x = 0 and find f'(0).
59. Graph y = 1/(2Vx) in a window that has 0 = x = 2. Then, on the same screen, graph
coOM ARE - Vx m h
for h = 1,0.5, 0.1. Then try h = —1, —0.5, —0.1. Explain what is going on.
60. Graph y = 3x? in a window that has -2z x 2,0 y x 3. Then, on the same screen, graph
Ay - aa h
for h = 2,1,0.2. Then try h = —2, —1, —0.2. Explain what is going on.
61. Derivative of y = |x| Graph the derivative of f(x) = |x|. Then graph y = (|x| — 0)/(x — 0) = |x|/x. What can you conclude?
118
Chapter 3: Derivatives
62. Weierstrass’s nowhere differentiable continuous function
The sum of the first eight terms of the Weierstrass function f(x) = X59 (2/3)" cos (93) is g(x) = cos(mx) + (2/3)! cos (99x) + (2/3)? cos (9777x) + (2/3 cos(9mx) + +++ + (2/3) cos (977x). Graph this sum. Zoom in several times. How wiggly and bumpy
is this graph? Specify a viewing window in which the displayed portion of the graph is smooth.
COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exer- cises 63-68.
a. Plot y — f(x) to see that function's global behavior.
b. Define the difference quotient q at a general point x, with general step size h.
c. Take the limit as ^ — 0. What formula does this give?
d. Substitute the value x = x, and plot the function y = f(x) together with its tangent line at that point.
3 i 3 Differentiation Rules
63. 64.
65.
66.
67. 68.
e. Substitute various values for x larger and smaller than x, into the formula obtained in part (c). Do the numbers make sense with your picture?
f. Graph the formula obtained in part (c). What does it mean when its values are negative? Zero? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer.
f) =x +x - x, f(x) = x^ 4 325, x21 4x
Xy = 1
f= a7? Xp = 2 x1 = x=] f= zyr ^ f(x) = sin2x, xo m/2 f(x) = x cosx, xo = 7/4
This section introduces several rules that allow us to differentiate constant functions, power functions, polynomials, rational functions, and certain combinations of them, sim- ply and directly, without having to take limits each time.
Powers, Multiples, Sums, and Differences
A simple rule of differentiation is that the derivative of every constant function is zero.
y t Derivative of a Constant Function c (x, c) (x + h, c) 36 If f has the constant value f(x) = c, then l l i I df d l | — = — = | | dx dx 7 9 P. i - 0 x xth Proof We apply the definition of the derivative to f(x) = c, the function whose outputs
FIGURE 3.9 The rule (d/dx)(c) = 0
is another way to say that the values of
constant functions never change and that the slope of a horizontal line is zero at
every point.
have the constant value c (Figure 3.9). At every value of x, we find that
Fœ = lim
fa + h- fo h
"HER ^ = lim =
in limO = 0. [|
h—0
From Section 3.1, we know that
j=- x?
d yt) = y2 or aU ) as
From Example 2 of the last section we also know that
1 E 12) = Lyn,
= , or (x 2Vx dx
These two examples illustrate a general rule for differentiating a power x". We first prove the rule when n is a positive integer.
HISTORICAL BIOGRAPHY Richard Courant
(1888-1972)
Applying the Power Rule Subtract 1 from the exponent and multiply
the result by the original exponent.
3.3 Differentiation Rules 119
Derivative of a Positive Integer Power
If n is a positive integer, then
Proof of the Positive Integer Power Rule The formula zn — x = (z PS xc! + z"? x der dele oe zz? + xh
can be verified by multiplying out the right-hand side. Then from the alternative formula for the definition of the derivative,
f'(x) = lim 2 ur mta
x4 x = lim(z" ! + z^?7y Fore + gax? + xh) n terms Ix = nx" |, a
The Power Rule is actually valid for all real numbers n. We have seen examples for a negative integer and fractional power, but n could be an irrational number as well. To apply the Power Rule, we subtract 1 from the original exponent n and multiply the result by n. Here we state the general version of the rule, but postpone its proof until Chapter 7.
Power Rule (General Version) If n is any real number, then
d = nx!
dx :
for all x where the powers x" and x"^! are defined.
EXAMPLE 1 Differentiate the following powers of x.
() 3 (x^ (94V? @) 5, (9x5 — (9 Vier
Solution
(a) £ (x) = 3x37! = 3x?
d 2 " 2. 2/3) = £.Q/3-1 = 1/3 (b) d (x2/3) 3* 3*
O E (a2) = vaa
dx (d) 2(1) m 4 (x) 4x ^12 -4x5- E (e) 4 (x48) = - ix um _ - $77 e a gem) = (artem) = E + zem i lo icy "
The next rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.
120
Chapter 3: Derivatives
we
Slope = 3(2x) = 6x 3- = 6(1) = 2 yur 2- Slope = 2x = 2(1) =2 (1, 1) (1) l »-X 0 1 2
FIGURE 3.10 The graphs of y = x? and y = 3x7. Tripling the y-coordinate triples the slope (Example 2).
Denoting Functions by u and v
The functions we are working with when we need a differentiation formula are likely to be denoted by letters like f and g. We do not want to use these same letters when stating general differentiation rules, so instead we use letters like u and v that are not likely to be already in use.
Derivative Constant Multiple Rule
If u is a differentiable function of x, and c is a constant, then
d NEU! 4; 9 TE
In particular, if n is any real number, then
d
qc «x = nx" |,
Proof d cu — lim cu(x + h) — cu(x) Derivative definition dx h0 h with f(x) = cu(x) u(x + h) — u(x) = clim P Constant Multiple Limit Property 1—3 = ott u is differentiable. E dx EXAMPLE 2
(a) The derivative formula
d (3,2) = 3.2% = di (3x^) = 3+2x = 6x says that if we rescale the graph of y = x? by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3 (Figure 3.10). (b) Negative of a function
The derivative of the negative of a differentiable function u is the negative of the func-
tion's derivative. The Constant Multiple Rule with c — —1 gives d ER ET .d _ du dx fri dx pig l dx w dx' "
The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.
Derivative Sum Rule
If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable. At such points,
For example, if y = x^ + 12x, then y is the sum of u(x) = x^ and v(x) = 12x. We then have dy d
d = Z (y4 e cu rhe h (x^) + ES (12x) = 4x? + 12.
(0. 2)
1 l >x
FIGURE 3.11 The curve in Example 4 and its horizontal tangents.
3.3 Differentiation Rules 121
Proof We apply the definition of the derivative to f(x) = u(x) + v(x):
(x + h) + v(x + hj — [uw + vw) 2 (uc) + v(a)) = lim E v(x - [u(x v(x) | . [u(x + h) —u(x) v(x t+ h) — v(x) — lim + h—0 h h tie u(x + h) — u(x) "T v(x + h) — v(x) _ du dv = h0 2 h—0 h dx dx
Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives:
d |. d _ du dv du dv d e — Y= glut CDe] = CO = ae ae The Sum Rule also extends to finite sums of more than two functions. If u;, wu», . . . , Up, are differentiable at x, then so is u, + u t: + u,, and d dd us d: " | du, " du; $ á du, dx Un F t un) = dx dx dx
For instance, to see that the rule holds for three functions we compute
E d d du, du du du; Ji s + ws) = SE (Qn + wy) + wy) = Fy + e) + qn ı gp. HM.
A proof by mathematical induction for any finite number of terms is given in Appendix 2.
EXAMPLE 3 Find the derivative of the polynomial y = x? + Fx? — 5x + 1.
d Solution H = ae + £L (42) s (5x) + A (1) Sum and Difference Rules
=32 + F-2r-5 40-324 Sr -5 m
We can differentiate any polynomial term by term, the way we differentiated the poly- nomial in Example 3. All polynomials are differentiable at all values of x.
EXAMPLE 4 Does the curve y = x* — 2x? + 2 have any horizontal tangents? If so, where?
Solution The horizontal tangents, if any, occur where the slope dy/dx is zero. We have
dy d, 2 = A qct — 2x^ + 2) = 4x — 4x. . dy Now solve the equation f E 0 for x: 48 — 4x = 0 A4x(x? — 1) = 0 x =0,1,-1.
The curve y = xf — 2x? + 2 has horizontal tangents at x = 0, 1, and —1. The corre- sponding points on the curve are (0, 2), (1, 1), and (—1, 1). See Figure 3.11. We will see in Chapter 4 that finding the values of x where the derivative of a function is equal to zero is an important and useful procedure. E
122
Chapter 3: Derivatives
Equation (3) is equivalent to saying that
(9 '-f'g-fg.
This form of the Product Rule is useful and applies to dot products and cross products of vector-valued functions, studied in Chapter 13.
Picturing the Product Rule Suppose u(x) and v(x) are positive and
increase when x increases, and h > 0.
v(x + h) M Av u(x + h) Av yt TEA Poa edt v(x) u(x)v(x) (x) Au 0 / Au \ u(x) u(x + h)
Then the change in the product uv is the difference in areas of the larger and smaller “squares,” which is the sum of the upper and right-hand reddish-shaded rectangles. That is,
A(uv) = u(x + h)v(x + h) — u(x)v(x) = u(x + h)Av + v(x)Au. Division by h gives A(uv) _ «ga he Au h u(x + h) A v(x) ho
The limit as h — 0* gives the Product Rule.
Products and Quotients
While the derivative of the sum of two functions is the sum of their derivatives, the deriva- tive of the product of two functions is not the product of their derivatives. For instance,
d, vy _ don _ f d .d uut dx (x * x) di (x^) = 2x, while d (x) ET (x) = 1:1 1.
The derivative of a product of two functions is the sum of two products, as we now explain.
Derivative Product Rule
If u and v are differentiable at x, then so is their product uv, and
d dv du dx Um = Wk + vx
The derivative of the product uv is u times the derivative of v plus v times the deriva- tive of u. In prime notation, (uv)' = uv' + vu'. In function notation,
[f@g@)] = fg a) + gCOF Go. (3) EXAMPLE 5 Find the derivative of y = (32 + 1)G8 + 3). Solution (a) From the Product Rule with u = x? + 1 and v = x? + 3, we find f[G + G8 +3)] = 62+ G2) * G -3)99 Sony sult v
= 3x4 + 3x? + 2x4 + 6x = 5x4 + 3x? + 6x. (b) This particular product can be differentiated as well (perhaps better) by multiplying
out the original expression for y and differentiating the resulting polynomial:
y = (2? + 13 + 3) = 28 + PF + 3x 4+ 3
d a 5x* + 3x? + 6x. dx This is in agreement with our first calculation. E
Proof of the Derivative Product Rule
d . u(x + h(x + h) — u(x)v(x) EN (uv) = lim h
h->0
To change this fraction into an equivalent one that contains difference quotients for the derivatives of u and v, we subtract and add u(x + h)v(x) in the numerator:
d . ux + hy(x + h) — u(x + h)v(x) + u(x + h)v(x) — u(x)v(x) (uv) — lim dx h—0 h + h) — +h) - Sil ep g E
v(x + h) — v(x)
u(x + h) — u(x) n Ie
h
li + h)-li + v(x). li S MEUM T va)
123
3.3 Differentiation Rules
As h approaches zero, u(x + h) approaches u(x) because u, being differentiable at x, is con- tinuous at x. The two fractions approach the values of dv /dx at x and du/dx at x. Therefore, L|
dv ,
du v
d » dx pan cu dx
The derivative of the quotient of two functions is given by the Quotient Rule.
dx
# 0, then the quotient u/v is dif-
|
Derivative Quotient Rule If u and v are differentiable at x and if v(x) ferentiable at x, and du dv dfu) dx dx dx \ V V In function notation, d |f e . g@f'@) — feds’) dx | g(x) gx) i 2—1 p+
Find the derivative of y =
EXAMPLE 6
Solution We apply the Quotient Rule with u = f
N
— landv = PD 1: v(du/dt) — u(dv/dt)
dy (+1) 2t- (fF — 1) 3P afa dt (P ES 1) dt\v ui _ 2r* + 2t — 344 + 3p (8 + 1) _ —ft + 3P 2t (B-1 oC 3 Proof of the Derivative Quotient Rule ux h) uw) dí(uY. v(x +h) v(x) dx\v} — i0 h
v(x)u(x + h) — u(x)v(x + h)
= lim
hv(x + h)v(x)
h0
To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and v, we subtract and add v(x)u(x) in the numerator. We then get v(x)u(x + h) — v(x)u(x) + voux) — u(x)v(x + h)
u v
d dx
( ) = lim h0
u(x
hv(x + h)v(x) v(x + h) — v(x) u(x)
+ h) — u(x) h
v(x)
= lim
h0
Taking the limits in the numerator and denominator now gives the Quotient Rule. Exercise
74 outlines another proof.
h v(x + h)v(x) |
The choice of which rules to use in solving a differentiation problem can make a dif-
ference in how much work you have to do. Here is an example.
124 Chapter 3: Derivatives
How to Read the Symbols for Derivatives
y “y prime"
y" "y double prime" d?y d “d squared y dx squared" x y" “y triple prime" y? “Y super n” PY. li de d to the n of y by dx to the n
D" “D to the n”
EXAMPLE 7 Find the derivative of
(x — DG? — 2x) y= ;
x4
Solution Using the Quotient Rule here will result in a complicated expression with many terms. Instead, use some algebra to simplify the expression. First expand the numer- ator and divide by x*:
(x — DG? - 2x) _ a= Ge oe = yl
x4 x4
— 3x ? + 2x3,
Then use the Sum and Power Rules:
dy E " E E rx T 3(—2)x > + 2(-3)x 1 6 6 3) t 3 f E
Second- and Higher-Order Derivatives
If y = f(x) is a differentiable function, then its derivative f'(x) is also a function. If f' is also differentiable, then we can differentiate f' to get a new function of x denoted by f". So f" = (f')'. The function f" is called the second derivative of f because it is the deriv- ative of the first derivative. It is written in several ways:
i d dy' fl) = 5 2(2) = c» = Dy = DSO.
The symbol D? means the operation of differentiation is performed twice. If y = x$, then y' = 63? and we have o_O _d imo umo - (6x5) = 30x4.
Thus D?(x°) = 30x*. If y" is differentiable, its derivative, y" = dy"/dx = d*y/dx°, is the third derivative of y with respect to x. The names continue as you imagine, with
d a-p ZY (n) — G= — == n dx? dx" di:
y
denoting the nth derivative of y with respect to x for any positive integer n.
We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y — f(x) at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.
EXAMPLE 8 The first four derivatives of y = x? — 3x? + 2 are
First derivative: y = 3x? — 6x Second derivative: y” = 6x — 6 Third derivative: y" =6 Fourth derivative: — y? = 0.
All polynomial functions have derivatives of all orders. In this example, the fifth and later derivatives are all zero. El
Exercises
Derivative Calculations In Exercises 1—12, find the first and second derivatives.
1 y=-x? +3 2By=xr+x4+8
3. s = 5P — 3p 4. w = 3g! — 723 + 2122 | 4x3 ui ee
5. y= 3 x 6 y= 3 2*4
T w = 3¢2-4 8 s=-2F1 +5
9. y = 62 — 10x — 5x? 10. y = 4 — 2x - x? 15 1% 41
Igel: 12. r 6 PE g4
In Exercises 13-16, find y' (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate.
13. y = (3 - x)(2
15. y = (x? 4 DE + 54 J 16. y = (1 + x2)(x3/4 — x3)
x+ 1) 14. y = (2x + 3)(5x* — 4x)
Find the derivatives of the functions in Exercises 17—28.
uda m : : DUE e
19. go) - EE 20. fy = =o
21. v= (1- 9(1* 2)? 22. w= Qx - Tx + 5) 23. f(s) = “ 2 24. u = EN
3j poi Ge 26. roo ve) 27. y= Ta DES p y= Dem
Find the derivatives of all orders of the functions in Exercises 29-32.
EF 23 29. y — 2 5* x 30. y= 120 31. y= (x - D(x-- 2)(x -3) 32. y = (4? + 3X2 — xxx
Find the first and second derivatives of the functions in Exercises 33-40.
3 2 = 33, y = 17 34,9 = 6-1(07+0+1 2+ xQ?—x4+ 1 35. pe X ) 36. pot Or —x + 1) 0? x
37. w= ( JE =f SEG eps ded ei
2+ 3\/qi-1 2 + 3 39. »-(t yt - ) 40. p = - - 12q q (q — D? + (q+ 1»
3.3 Differentiation Rules 125
41. Suppose u and v are functions of x that are differentiable at
x = 0 and that u(0) = 5, u'(0 = —3, w0)—--1, v'(0) = 2.
Find the values of the following derivatives at x = 0.
d dfu dí(v d a. qx m b. a(s) c. a(x) d. g UP — 2u)
42. Suppose u and v are differentiable functions of x and that u(1) —2, u'(1)-0, v1) =5, v'(1) = -I.
Find the values of the following derivatives at x — 1.
dfu dí(v d b. a(s) c. a(x) d. aoe — 2u)
Slopes and Tangents 43. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x? — 4x + 1 at the point (2, 1).
d a. d (uv)
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope?
c. Tangents having specified slope Find equations for the tan- gents to the curve at the points where the slope of the curve is 8.
44. a. Horizontal tangents Find equations for the horizontal tan- gents to the curve y — x? — 3x — 2. Also find equations for the lines that are perpendicular to these tangents at the points of tangency.
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve's tangent at this point.
45. Find the tangents to Newton's serpentine (graphed here) at the origin and the point (1, 2).
46. Find the tangent to the Witch of Agnesi (graphed here) at the point (2, 1).
126
47.
48.
49. 50. 51.
52.
53.
54.
Chapter 3: Derivatives
Quadratic tangent to identity function The curve y= ax? + bx + c passes through the point (1, 2) and is tangent to the line y = x at the origin. Find a, b, and c.
Quadratics having a common tangent The curves y — x? + ax + b and y = cx — x? have a common tangent line at the point (1, 0). Find a, b, and c.
Find all points (x, y) on the graph of f(x) = 3x? — 4x with tan- gent lines parallel to the line y = 8x + 5.
Find all points (x, y) on the graph of g(x) — Ip = ix + 1 with tangent lines parallel to the line 8x — 2y = 1.
Find all points (x, y) on the graph of y = x/(x — 2) with tangent lines perpendicular to the line y = 2x + 3.
Find all points (x, y) on the graph of f(x) = x? with tangent lines passing through the point (3, 8).
1 f(x) = x? 10- 7^
a. Find an equation for the line that is tangent to the curve y = x3 — x at the point (—1, 0).
b. Graph the curve and tangent line together. The tangent inter- sects the curve at another point. Use Zoom and Trace to esti- mate the point's coordinates.
c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
a. Find an equation for the line that is tangent to the curve y = xX — 6x? + 5x at the origin.
b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordinates.
€. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
Theory and Examples For Exercises 55 and 56 evaluate each limit by first converting each to a derivative at a particular x-value.
55.
57.
yu e Ww ur im a a rar rae Find the value of a that makes the following function differentia-
ble for all x-values.
w= C ifx <0 pe xX — 3x, ifx=0
58.
59.
60.
6l.
62.
63.
64.
Find the values of a and b that make the following function dif- ferentiable for all x-values.
The general polynomial of degree n has the form
P(x) = a,x" + a, ax" | oss + aax 4
ax + ag where a, # 0. Find P'(x).
The body's reaction to medicine The reaction of the body to a dose of medicine can sometimes be represented by an equation of the form
where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pres- sure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on.
Find dR/dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see how to find the amount of medicine to which the body is most sensitive.
Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?
The Reciprocal Rule
a. The Reciprocal Rule says that at any point where the function v(x) is differentiable and different from zero,
d(1|. ldv dx NU v? dx`
Show that the Reciprocal Rule is a special case of the Deriva- tive Quotient Rule.
b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.
Generalizing the Product Rule The Derivative Product Rule gives the formula
dv du
d. (uy) = dx “Y Max Yd
for the derivative of the product uv of two differentiable func- tions of x.
a. Whatis the analogous formula for the derivative of the prod- uct uvw of three differentiable functions of x?
b. What is the formula for the derivative of the product uu; usu, of four differentiable functions of x?
c. What is the formula for the derivative of a product uj uus * * * Un of a finite number n of differentiable functions of x?
Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d
di (x) — —mx i!
where m is a positive integer.
3.4 The Derivative as a Rate of Change 127
65. Cylinder pressure If gas in a cylinder is maintained at a con- 66. The best quantity to order One of the formulas for inventory stant temperature 7, the pressure P is related to the volume V by a management says that the average weekly cost of ordering, pay- formula of the form ing for, and holding merchandise is
_ nRT an? _ km hq
Pe eu y? AQ =g * em * y, in which a, b, n, and R are constants. Find dP/dV. (See accompa- where q is the quantity you order when things run low (shoes, nying figure.) TVs, brooms, or whatever the item might be); k is the cost of
placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dA /dq and d?A / dq.
3.4 The Derivative as a Rate of Change
In Section 2.1 we introduced average and instantaneous rates of change. In this section we study further applications in which derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.
Instantaneous Rates of Change
If we interpret the difference quotient (f(x + h) — f(x))/h as the average rate of change in f over the interval from x to x + h, we can interpret its limit as h — 0 as the rate at which f is changing at the point x.
DEFINITION The instantaneous rate of change of f with respect to x at xy is the derivative
fo + h) — fo)
! = li f (Xo) Pa, h ,
provided the limit exists.
Thus, instantaneous rates are limits of average rates.
It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instanta- neous rate of change.
EXAMPLE 1 The area A of a circle is related to its diameter by the equation
How fast does the area change with respect to the diameter when the diameter is 10 m?
128 Chapter 3: Derivatives
Position at time t ... and at time t + At | As
bd —X e >s s —f(t) s t As = f(t + Af) FIGURE 3.12 The positions of a body moving along a coordinate line at time t and shortly later at time ¢ + At. Here the coordinate line is horizontal.
>
0 >t (a) s increasing: positive slope so moving upward s ^ 0 >t
(b) s decreasing: negative slope so moving downward
FIGURE 3.13 For motion s = f(t) along a straight line (the vertical axis), v — ds/dt is (a) positive when s increases and (b) negative when s decreases.
Solution The rate of change of the area with respect to the diameter is
dA T 7D dp 444
When D = 10 m, the area is changing with respect to the diameter at the rate of (7/2)10 = 5z m?/m - 15.71 m?/m. m
Motion Along a Line: Displacement, Velocity, Speed, Acceleration, and Jerk
Suppose that an object (or body, considered as a whole mass) is moving along a coordinate line (an s-axis), usually horizontal or vertical, so that we know its position s on that line as a function of time f:
s = fQ. The displacement of the object over the time interval from t to t + At (Figure 3.12) is As = f(t + At) — fi, and the average velocity of the object over that time interval is
displacement Ay f(t+ Ant — fO travel time — Ar — At i
Vav —
To find the body's velocity at the exact instant t, we take the limit of the average velocity over the interval from f to t + At as At shrinks to zero. This limit is the deriva- tive of f with respect to f.
DEFINITION Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body's position at time t is s = f(t), then the body's velocity at time t is
_ f@ + AD — fO
m f
ds w= g um At
Besides telling how fast an object is moving along the horizontal line in Figure 3.12, its velocity tells the direction of motion. When the object is moving forward (s increasing), the velocity is positive; when the object is moving backward (s decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity. The blue curves in Figure 3.13 represent position along the line over time; they do not portray the path of motion, which lies along the vertical s-axis.
If we drive to a friend's house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show —30 on the way back, even though our distance from home is decreasing. The speedometer always shows speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.
DEFINITION Speed is the absolute value of velocity. ds
Speed = |w(n| = di
EXAMPLE 2 Figure 3.14 shows the graph of the velocity v = f'(r) of a particle moving along a horizontal line (as opposed to showing a position function s = f(f) such as in Figure 3.13). In the graph of the velocity function, it's not the slope of the curve that tells us if the par- ticle is moving forward or backward along the line (which is not shown in the figure), but rather
HISTORICAL BIOGRAPHY Bernard Bolzano
(1781-1848)
3.4 The Derivative as a Rate of Change 129
I | MOVES FORWARD | FORWARD |
(v — const), |
(v = 0) AGAIN Velocity v = f'(f) (UR l I i i I l L Speeds SIE Steady RS Slows Y We Speeds | up l | down up |
t (sec)
MOVES BACKWARD (v <0)
FIGURE 3.14 The velocity graph of a particle moving along a horizontal line, discussed in Example 2.
the sign of the velocity. Looking at Figure 3.14, we see that the particle moves forward for the first 3 sec (when the velocity is positive), moves backward for the next 2 sec (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at t = 3 sec (when the velocity is zero) and reverses direc- tion as the velocity starts to become negative. The particle is now moving backward and gain- ing in speed until t = 4 sec, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time t = 4, the particle starts to slow down again until it finally stops at time ? = 5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at t = 6 sec, speeding up during the final second of the forward motion indicated in the velocity graph. a
The rate at which a body’s velocity changes is the body’s acceleration. The acceleration measures how quickly the body picks up or loses speed. In Chapter 13 we will study motion in the plane and in space, where acceleration of an object may also lead to a change in direction.
A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in accel- eration are abrupt.
DEFINITIONS Acceleration is the derivative of velocity with respect to time. If a body's position at time tis s = f(t), then the body's acceleration at time t is
— dv ds
a(t) — du dg
Jerk is the derivative of acceleration with respect to time:
: da _ d?s I) = ade de
130 Chapter 3: Derivatives
t (seconds) 5 (meters) t=0 0 -0 t=1 0C? F5
— 10 a5 t=2 © +20 25 = 30 - 35 r- 40 t=3 0C? +45
v
FIGURE 3.15 A ball bearing falling from rest (Example 3).
Near the surface of Earth all bodies fall with the same constant acceleration. Galileo's experiments with free fall (see Section 2.1) lead to the equation
s= bgp,
where s is the distance fallen and g is the acceleration due to Earth's gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are significant.
The value of g in the equation s = (1/2)gf? depends on the units used to measure t and s. With ¢ in seconds (the usual unit), the value of g determined by measurement at sea level is approximately 32 ft/sec? (feet per second squared) in English units, and g — 9.8 m/sec? (meters per second squared) in metric units. (These gravitational con- stants depend on the distance from Earth's center of mass, and are slightly lower on top of Mt. Everest, for example.)
The jerk associated with the constant acceleration of gravity (g — 32 ft/sec?) is zero:
. d Jm An object does not exhibit jerkiness during free fall.
EXAMPLE 3 Figure 3.15 shows the free fall of a heavy ball bearing released from rest at time ¢ = 0 sec.
(a) How many meters does the ball fall in the first 3 sec?
(b) What is its velocity, speed, and acceleration when t = 3?
Solution (a) The metric free-fall equation is s = 4.9£?. During the first 3 sec, the ball falls
s(3) = 4.933)? = 44.1 m.
(b) At any time f, velocity is the derivative of position: wo) = s() = 5 (4.97) = oar.
At t = 3, the velocity is v(3) = 29.4 m/sec in the downward (increasing s) direction. The speed at t = 3 is speed = |v(3)| = 29.4 m/sec. The acceleration at any time t is a(t) = v'(t) = s"(t) = 9.8 m/sec?.
At t = 3, the acceleration is 9.8 m/sec?. E
EXAMPLE 4 A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec (about 109 mph) (Figure 3.162). It reaches a height of s = 160r — 162? ft after t sec.
(a) How high does the rock go?
(b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
(c) What is the acceleration of the rock at any time f during its flight (after the blast)? (d) When does the rock hit the ground again?
Height (ft)
s = 160r — 16?
160
= ds _ —160 v= i 160
FIGURE 3.16 (a) The rock in Example 4. (b) The graphs of s and v as functions of time; s is largest when v — ds/dt — 0. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock's velocity, graphed here as a straight line.
3.4 The Derivative as a Rate of Change 131
Solution
(a)
(b
—
(c)
(d)
In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is O. To find the maximum height, all we need to do is to find when v = 0 and evaluate s at this time.
At any time f during the rock's motion, its velocity is
ds d s ger = z (1602 161^) = 160 — 32t ft/sec.
The velocity is zero when 160 — 32t = 0 or t = 5sec. The rock’s height at t = 5 sec is Smax = 8(5) = 160(5) — 16(5)* = 800 — 400 = 400 ft. See Figure 3.16b.
To find the rock’s velocity at 256 ft on the way up and again on the way down, we first find the two values of t for which
s(t) = 160t — 16? = 256. To solve this equation, we write 16? — 160r + 256 = 0 16(? — 10t + 16) = 0 (t — 2) — 8) = 0 t = 2sec, t = 8sec. The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock’s velocities at these times are v(2) = 160 — 32(2) = 160 — 64 = 96 ft/sec. v(8) = 160 — 32(8) = 160 — 256 = —96 ft/sec.
At both instants, the rock's speed is 96 ft/sec. Since v(2) > 0, the rock is moving upward (s is increasing) at t = 2sec; it is moving downward (s is decreasing) at t = 8 because v(8) « 0.
At any time during its flight following the explosion, the rock’s acceleration is a constant
— dv _ dt
a P (160 — 324) = —32 ft/sec?. The acceleration is always downward and is the effect of gravity on the rock. As the rock rises, it slows down; as it falls, it speeds up.
The rock hits the ground at the positive time ¢ for which s = 0. The equation 160r — 16:7? = 0 factors to give 16110 — f) = 0, so it has solutions t = 0 and t = 10. At t = O, the blast occurred and the rock was thrown upward. It returned to the ground 10 sec later. E
Derivatives in Economics
Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals.
In a manufacturing operation, the cost of production c(x) is a function of x, the num-
ber of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc/ dx.
132 Chapter 3: Derivatives
Cost y (dollars) ^
Slope = marginal cost
y = c(x)
l
l
l
|
l
|
l
I
l
l 0 x eh Production (tons/week)
FIGURE 3.17 Weekly steel production: c(x) is the cost of producing x tons per week. The cost of producing an additional htons is c(x + h) — c(x).
y ^
l l l I l l l I l l l l i x
-X
0
FIGURE 3.18 The marginal cost dc/dx is approximately the extra cost Ac of producing Ax — 1 more unit.
Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton:
c(x + h) — c(x) average cost of each of the additional h h tons of steel produced.
The limit of this ratio as h — 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.17): dc . c(x + h) — c(x)
gem lim h = marginal cost of production.
Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit:
Ac _ c(x + 1) — c(x) Ax 1 à
which is approximated by the value of dc/dx at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc/dx, which is the rise in the tangent line if Ax — 1 (Figure 3.18). The approximation works best for large values of x.
Economists often represent a total cost function by a cubic polynomial
c(x) = ax? + Bx? + yx + 6
where 6 represents fixed costs, such as rent, heat, equipment capitalization, and manage- ment costs. The other terms represent variable costs, such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.
EXAMPLE 5 Suppose that it costs c(x) = xX — 6x? + 15x
dollars to produce x radiators when 8 to 30 radiators are produced and that r(x) = x3 — 3x? + 12x
gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling 11 radiators a day?
Solution The cost of producing one more radiator a day when 10 are produced is about c’(10):
quy — 4643 = 2 c'(x) = Am — 6x? + 15x) = 332 — 12x + 15 c'(10) = 3(100) — 12(10) + 15 = 195. The additional cost will be about $195. The marginal revenue is r'(x) = 1 — 3x? + 12x) = 3x? — 6x + 12.
The marginal revenue function estimates the increase in revenue that will result from sell- ing one additional unit. If you currently sell 10 radiators a day, you can expect your reve- nue to increase by about
r'(10) = 3(100) — 6(10) + 12 = $252
if you increase sales to 11 radiators a day. [|
3.4 The Derivative as a Rate of Change 133
EXAMPLE 6 To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 2896 and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 2896 of your entire income in taxes. It just means that at your current income level /, the rate of increase of taxes T with respect to income is dT/dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase. i
Sensitivity to Change
When a small change in x produces a large change in the value of a function f(x), we say that the function is relatively sensitive to changes in x. The derivative f'(x) is a measure of
this sensitivity.
dy/dp EXAMPLE 7 Genetic Data and Sensitivity to Change
2 The Austrian monk Gregor Johann Mendel (1822-1884), working with garden peas and other plants, provided the first scientific explanation of hybridization.
His careful records showed that if p (a number between 0 and 1) is the frequency of the
E =2- 2p gene for smooth skin in peas (dominant) and (1 — p) is the frequency of the gene for wrin- kled skin in peas, then the proportion of smooth-skinned peas in the next generation will be y = 2pd — p) + p? = 2p — p.
0 1 >P The graph of y versus p in Figure 3.19a suggests that the value of y is more sensitive to a b) change in p when p is small than when p is large. Indeed, this fact is borne out by the derivative graph in Figure 3.19b, which shows that dy/dp is close to 2 when p is near 0
FIGURE 3.19 (a) The graph of and close to 0 when p is near 1. y = 2p — p’, describing the proportion of The implication for genetics is that introducing a few more smooth skin genes into a smooth-skinned peas in the next genera- population where the frequency of wrinkled skin peas is large will have a more dramatic tion. (b) The graph of dy/dp effect on later generations than will a similar increase when the population has a large pro- (Example 7). portion of smooth skin peas. [|
Exercises
Motion Along a Coordinate Line Exercises 1—6 give the positions s = f(t) of a body moving on a coor- dinate line, with s in meters and f in seconds.
a. Find the body's displacement and average velocity for the given time interval.
b. Find the body's speed and acceleration at the endpoints of the interval.
c. When, if ever, during the interval does the body change direction?
1.s-2-3t-2, 0xt-2
28s=6-f, 0xtx6
3.5——P -32—3t, 0zxtx3
4 s-(n/4-P-P, 0ztz3 25 3
mode n ]- 4-9
6. 5 = 2 -—4ztz0
7. Particle motion At time f, the position of a body moving along the s-axis is s = £ — 6f + 9t m.
a. Find the body’s acceleration each time the velocity is zero. b. Find the body’s speed each time the acceleration is zero. c. Find the total distance traveled by the body from t = 0 to t = 2.
8. Particle motion At time ¢ = 0, the velocity of a body moving along the horizontal s-axis is v = 1? — 4t + 3.
a. Find the body’s acceleration each time the velocity is zero. b. When is the body moving forward? Backward?
c. When is the body’s velocity increasing? Decreasing?
Free-Fall Applications 9. Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.8627 on Mars and s = 11.4417 on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m/sec (about 100 km/h) on each planet?
134
10.
11.
12.
13.
14.
Chapter 3: Derivatives
Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m/sec (about 86 km/h) reaches a height of s = 24t — 0.8? m in t sec.
a. Find the rock’s velocity and acceleration at time t. (The accel- eration in this case is the acceleration of gravity on the moon.)
. How long does it take the rock to reach its highest point? How high does the rock go?
. How long does it take the rock to reach half its maximum height?
? £05 cz
How long is the rock aloft?
Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m/ sec. Because the accel- eration of gravity at the planet’s surface was g, m/sec?, the explorers expected the ball bearing to reach a height of s = 15t — (1/2)g,? m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of g,?
Speeding bullet A 45-caliber bullet shot straight up from the surface of the moon would reach a height of s = 832r — 2.66? ft after t sec. On Earth, in the absence of air, its height would be s = 832t — 16r ft after t sec. How long will the bullet be aloft in each case? How high will the bullet go?
Free fall from the Tower of Pisa Had Galileo dropped a can-
nonball from the Tower of Pisa, 179 ft above the ground, the
ball's height above the ground f sec into the fall would have been
s = 179 — 160.
a. What would have been the ball’s velocity, speed, and acceler- ation at time /?
b. About how long would it have taken the ball to hit the ground? c. What would have been the ball’s velocity at the moment of impact? Galileo's free-fall formula Galileo developed a formula for a body's velocity during free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball's behavior when the plank was vertical and the ball fell freely; see part (a) of the accom- panying figure. He found that, for any given angle of the plank, the balls velocity t sec into motion was a constant mul- tiple of t. That is, the velocity was given by a formula of the form v = kt. The value of the constant k depended on the inclination of the plank.
In modern notation—part (b) of the figure—with distance in meters and time in seconds, what Galileo determined by experi- ment was that, for any given angle 0, the ball’s velocity ¢ sec into the roll was
v = 9.8(sin0)t m/sec.
Free-fall position
3 10
. 9 V Q SI À
“a
(a) (b)
a. What is the equation for the ball’s velocity during free fall?
b. Building on your work in part (a), what constant acceleration does a freely falling body experience near the surface of Earth?
Understanding Motion from Graphs
15.
16.
17.
The accompanying figure shows the velocity v = ds/dt = f(t) (m/sec) of a body moving along a coordinate line.
v (m/sec) A uL, 2570 L I/I > t (sec) 0 4 6/8 10 —3r
a. When does the body reverse direction?
b. When (approximately) is the body moving at a constant speed?
c. Graph the body's speed for 0 = t = 10.
d. Graph the acceleration, where defined.
A particle P moves on the number line shown in part (a) of the accompanying figure. Part (b) shows the position of P as a func- tion of time f.
P 7 < —.e— > > s (cm) (a) s (cm) A s = f(t) ^ é > t (sec) -2 (6, —4) —4 -
(b)
a. When is P moving to the left? Moving to the right? Standing still?
b. Graph the particle's velocity and speed (where defined).
Launching a rocket When a model rocket is launched, the pro- pellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands.
The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.
a. How fast was the rocket climbing when the engine stopped?
b. For how many seconds did the engine burn?
18.
19.
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8 LOOFREREEE So HN $ o EHN XE Cec D IIT Cert O TTT TTT Tt ttt 8 UMELECCEEECEEELLEI EEEEBEEL EHE ic EEE REE EEE EEEE RENEE EEEEE EEE = Oee eN TTT I TTT TTT TT LLLI TT TTT TTT ty jx pese] | ER E EEG EE ER E EIER EI INT TTT TT I | N A S M A N E E E S ee
50 ENEE pupussp up NE EEE EEE EE EEE EEE EEE EEE
1005 2 4 6 8 10 12
Time after launch (sec)
When did the rocket reach its highest point? What was its velocity then?
When did the parachute pop out? How fast was the rocket falling then?
How long did the rocket fall before the parachute opened? When was the rocket's acceleration greatest?
When was the acceleration constant? What was its value then (to the nearest integer)?
The accompanying figure shows the velocity v = f(t) of a particle moving on a horizontal coordinate line.
c. d.
>e
> f (sec)
When does the particle move forward? Move backward? Speed up? Slow down?
When is the particle’s acceleration positive? Negative? Zero?
When does the particle move at its greatest speed?
When does the particle stand still for more than an instant?
Two falling balls The multiflash photograph in the accompa- nying figure shows two balls falling from rest. The vertical rulers are marked in centimeters. Use the equation s = 490r (the free-fall equation for s in centimeters and t in seconds) to answer the following questions. (Source: PSSC Physics, 2nd ed., Reprinted by permission of Education Development Cen- ter, Inc.)
3.4 The Derivative as a Rate of Change 135
E
?,» 9» 8 9? 9?^*9339U0«€
Oe eee
a. How long did it take the balls to fall the first 160 cm? What was their average velocity for the period?
b. How fast were the balls falling when they reached the 160-cm mark? What was their acceleration then?
c. About how fast was the light flashing (flashes per second)?
20. A traveling truck The accompanying graph shows the position
s of a truck traveling on a highway. The truck starts at t = 0 and
returns 15 h later at £ = 15.
a. Use the technique described in Section 3.2, Example 3, to graph the truck's velocity v = ds/dtfor0 = t = 15. Then repeat the process, with the velocity curve, to graph the truck's acceleration dv/dt.
b. Suppose that s = 152? — P. Graph ds/dt and d?s/ d? and compare your graphs with those in part (a).
Position, s (km)
0 5 10 15 Elapsed time, t (hr)
136
21.
22.
Chapter 3: Derivatives
The graphs in the accompanying figure show the position s, velocity v = ds/dt, and acceleration a = d?s/di? of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers.
y ^
® ® ©
>t
The graphs in the accompanying figure show the position s, the velocity v = ds/dt, and the acceleration a = d?s/df? of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers.
y A
®
>t
©
Economics
23.
24.
Marginal cost Suppose that the dollar cost of producing x washing machines is c(x) = 2000 + 100x — 0.1x?.
a. Find the average cost per machine of producing the first 100 washing machines.
b. Find the marginal cost when 100 washing machines are produced.
€. Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by cal- culating the latter cost directly.
Marginal revenue Suppose that the revenue from selling x washing machines is
r(x) — 20,000( 1 = J
dollars.
a. Find the marginal revenue when 100 machines are produced.
b. Use the function r'(x) to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week.
c. Find the limit of r'(x) as x — oo. How would you interpret this number?
Additional Applications
25.
26.
27.
28.
29.
30.
Bacterium population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped grow- ing and began to decline. The size of the population at time f (hours) was b = 10° + 10*r — 10?2.. Find the growth rates at
a. t = 0 hours. b. t Cc. ¢ = 10 hours.
5 hours.
Body surface area A typical male’s body surface area S in square meters is often modeled by the formula $ = di V/wh, where / is the height in cm, and w the weight in kg, of the person. Find the rate of change of body surface area with respect to weight for males of constant height h = 180 cm (roughly 5'9"). Does S increase more rapidly with respect to weight at lower or higher body weights? Explain.
Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula
a2 y=6(1- 4) m.
a. Find the rate dy/dt (m/h) at which the tank is draining at time f.
b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy/dt at these times?
c. Graph y and dy/dt together and discuss the behavior of y in relation to the signs and values of dy/dt.
Draining a tank The number of gallons of water in a tank t minutes after the tank has started to drain is O(t) = 200(30 — £y. How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min?
Vehicular stopping distance Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is
s = Ll1v + 0.054v?,
where s is measured in ft and v in mph. The linear term 1.1v models the distance the car travels during the time the driver per- ceives a need to stop until the brakes are applied, and the qua- dratic term 0.054v? models the additional braking distance once they are applied. Find ds/dv at v = 35 and v = 70 mph, and interpret the meaning of the derivative.
Inflating a balloon The volume V = (4/3)zr? of a spherical
balloon changes with the radius.
a. At what rate (ft?/ft) does the volume change with respect to the radius when r = 2 ft?
b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 ft?
3.5 Derivatives of Trigonometric Functions 137
31. Airplane takeoff Suppose that the distance an aircraft travels velocity function v(t) = ds/dt = f'(t) and the acceleration function along a runway before takeoff is given by D = (10/9)1?, where D is a(t) = d’s/dt? = f"(t). Comment on the object’s behavior in relation measured in meters from the starting point and t is measured in sec- to the signs and values of v and a. Include in your commentary such onds from the time the brakes are released. The aircraft will become topics as the following:
airborne when its speed reaches 200 km/h. How long will it take to become airborne, and what distance will it travel in that time?
32. Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains £ along the wall of the crater, activity was later confined to a single d. When does it speed up and slow down? vent in the crater’s floor, which at one point shot lava 1900 ft e straight into the air (a Hawaiian record). What was the lava's exit f. When is it farthest from the axis origin? velocity in feet per second? In miles per hour? (Hint: If vp is the exit velocity of a particle of lava, its height t sec later will be 33. s = 200: — 16P, 0 = 1 = 12.5 (a heavy object fired straight s = uot — 16f? ft. Begin by finding the time at which ds/dt = 0. up from Earth's surface at 200 ft/sec) Neglect air resistance.) 34.5—-2—3: * 2, OSt<5
3 _ 62 Analyzing Motion Using Graphs 35. s=1 rh Qeures Exercises 33-36 give the position function s = f(t) of an object mov- 36. s=4-7t+6°-P, OSts4 ing along the s-axis as a function of time t. Graph f together with the
a. When is the object momentarily at rest? . When does it move to the left (down) or to the right (up)?
. When does it change direction?
. When is it moving fastest (highest speed)? Slowest?
ll
ll
ll
3.5 Derivatives of Trigonometric Functions
Many phenomena of nature are approximately periodic (electromagnetic fields, heart rhythms, tides, weather). The derivatives of sines and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.
Derivative of the Sine Function
To calculate the derivative of f(x) — sin x, for x measured in radians, we combine the limits in Example 5a and Theorem 7 in Section 2.4 with the angle sum identity for the sine function:
sin(x + h) = sinxcos h + cos x sin A. If f(x) — sin x, then 2o fx th) — fx) . sin(x + h) — sinx m = lim
P . " FN x) = li Derivative definition Po h—0 h h—0 h . (sinxcos h + cos xsin A) — sin x = lim h—0 h li sin x(cos h — 1) + cos x sin h = ]i h>0 h ET , cosh — 1 ; sin A = lim | sin x: ——,—— | + lim| cosx. h—0 h h0 h , . cosh — 1 . sinh . = sin x* lim ~~~ — + cos x* lim) = sinx:0 + cosx:1 = cos x. : MR Example 5a and limit 0 limit 1 Theorem 7, Section 2.4
The derivative of the sine function is the cosine function:
d _ Ay (sin x) = cos x.
138 Chapter 3: Derivatives
EXAMPLE 1 We find derivatives of the sine function involving differences, products, and quotients.
dy (a y x — sinx: ES = 2x — £ (sin x) Difference Rule = 2x — cosx dy d (b) y = x? sin x: "iE x z Gino) + 2x sin x Product Rule = x? cosx + 2x sin x. d. : . x. (sin x) — sinx: 1 (c) _ Sinx, dy _ pni ) actuated y x M dx 32 uotien ule
| xcosx — sinx x2 Derivative of the Cosine Function With the help of the angle sum formula for the cosine function, cos(x + h) = cosxcosh — sinx sinh,
we can compute the limit of the difference quotient:
d . cos(x + h) — cos x — (cos x) = lim Derivative definition dx h0 h (cos x cos h — sin x sinh) — cos x Cosine angle sum = lim identity h—0 h . cosx(cosh — 1) — sin xsin A — lim h—0 h : cosh — 1 -— sin A = lim cos x: ———— = lim sin x: —— h—0 h h—0 h . cosh— 1 . . sinh = cos x* lim — — ——- — sin x: lim h—0 h h—0 » . Example 5a and n m = cos x*0 — sinx’ 1 Theorem 7, Section 2.4 1 aoe š = —sin x. | | d =A | 0 T | -—1- | | | | | , | y . . . H H . . . | NET | y= -sinx The derivative of the cosine function is the negative of the sine function: | | | |
d LM di (cos x) = —sin x.
FIGURE 3.20 The curve y' = —sinx as the graph of the slopes of the tangents to Figure 3.20 shows a way to visualize this result in the same way we did for graphing
the curve y — cos x. derivatives in Section 3.2, Figure 3.6. EXAMPLE 2 We find derivatives of the cosine function in combinations with other functions.
(a) y = 5x + cos x: dy d dx dx
= 5 — sinx.
(5x) + 4. (cos x) Sum Rule
Rest position
| POTITUS
Position at t=0
RY
FIGURE 3.21 A weight hanging from a vertical spring and then displaced oscil- lates above and below its rest position (Example 3).
=)
FIGURE 3.22 The graphs of the position and velocity of the weight in Example 3.
139
3.5 Derivatives of Trigonometric Functions
(b) y = sin xcos x:
dy = sin ge (cos x) + cos me (sin x) Product Rule dx dx dx = sinx(—sin x) + cos x(cos x) = cos? x — sin? x COS X c = n (© 1 — sinx
F d d : dy (1 — sin x) a (cos x) — cos x "x (1 — sin x) = Quotient Rule
dx (1 — sin xy (1 — sin x)(—sin x) — cos x(0 — cos x) i (1 — sin x 1 — sinx m" F = Q — sinx? sin^x + cos*x = 1 1 ~ T= sinx z
Simple Harmonic Motion
The motion of an object or weight bobbing freely up and down with no resistance on the end of a spring is an example of simple harmonic motion. The motion is periodic and repeats indefinitely, so we represent it using trigonometric functions. The next example describes a case in which there are no opposing forces such as friction to slow the motion.
EXAMPLE 3 A weight hanging from a spring (Figure 3.21) is stretched down 5 units beyond its rest position and released at time f = 0 to bob up and down. Its position at any later time f is
s = 5cost.
What are its velocity and acceleration at time t?
Solution We have
Position: s = 5cost for _ds_d = 5g Velocity: dix mcs (5cos f) 5sinf ; dv d ; u Acceleration: a N di (—5 sin f) 5cos t.
Notice how much we can learn from these equations:
1. As time passes, the weight moves down and up between s = —5 and s = 5 on the s-axis. The amplitude of the motion is 5. The period of the motion is 27, the period of the cosine function.
2. The velocity v = —5 sin t attains its greatest magnitude, 5, when cos t = 0, as the graphs show in Figure 3.22. Hence, the speed of the weight, v| =5 | sin f|, is greatest when cos ¢ = 0, that is, when s = O (the rest position). The speed of the weight is zero when sin t = 0. This occurs when s = 5cost = t5, at the endpoints of the interval of motion.
3. The weight is acted on by the spring and by gravity. When the weight is below the rest position, the combined forces pull it up, and when it is above the rest position, they pull it down. The weight's acceleration is always proportional to the negative of its displacement. This property of springs is called Hooke’s Law, and is studied further in Section 6.5.
140
Chapter 3: Derivatives
4. The acceleration, a = —5cos t, is zero only at the rest position, where cos t = 0 and the force of gravity and the force from the spring balance each other. When the weight is anywhere else, the two forces are unequal and acceleration is nonzero. The accel- eration is greatest in magnitude at the points farthest from the rest position, where cost = +1. a
EXAMPLE 4 The jerk associated with the simple harmonic motion in Example 3 is
. da d, ELS. j= d g 5 cos f) = 5 sin t.
It has its greatest magnitude when sint = +1, not at the extremes of the displacement but at the rest position, where the acceleration changes direction and sign. Oo Derivatives of the Other Basic Trigonometric Functions
Because sin x and cos x are differentiable functions of x, the related functions
COS X
Lt cot x = sec x = -— and csc x = COS x? . sin x’ COS X^ sin x
tan x —
are differentiable at every value of x at which they are defined. Their derivatives, calcu- lated from the Quotient Rule, are given by the following formulas. Notice the negative signs in the derivative formulas for the cofunctions.
The derivatives of the other trigonometric functions:
d — d DS 4, an X) — sec^x qx Oot) csc* x
d d dx SEE x) = sec x tan x dx 5€ X) = —csc x cot x
To show a typical calculation, we find the derivative of the tangent function. The other derivations are left to Exercise 60.
EXAMPLE 5 Find d(tan x)/dx.
Solution We use the Derivative Quotient Rule to calculate the derivative:
d,. . d COS x—— (sin x) — sin x—- (cos x)
sin x dx dx Quotient Rule
2: (tan x) — E ( dx dx \ COS x mem COS X COS x — sin x(—sin x)
cos?x
. cos?x + sin?x cos?x
1 x = sec?x. El
EXAMPLE 6 Find y" if y = sec x.
Solution Finding the second derivative involves a combination of trigonometric derivatives.
y = secx
= sec x tan x Derivative rule for secant function
< |
Exercises
Derivatives In Exercises 1-18, find dy/dx.
1. y = —10x + 3cosx 3. y =x’ cosx
5. y = cscx - AVx +7 7. f(x) = sinxtanx
9. y=xsecx++
cot x 1. y= 1 + cotx
4 1 13. y = cosx * janx
141
3.5 Derivatives of Trigonometric Functions
n
-4 =k (sec x tan x)
d d = secx di (tan x) + tan "n (sec x) Derivative Product Rule
= sec x(sec? x) + tanx(secxtan x) Derivative rules
sec? x + sec x tan? x a
The differentiability of the trigonometric functions throughout their domains gives another proof of their continuity at every point in their domains (Theorem 1, Section 3.2). So we can calculate limits of algebraic combinations and composites of trigonometric functions by direct substitution.
EXAMPLE 7 We can use direct substitution in computing limits provided there is no division by zero, which is algebraically undefined.
V2 + secx V2 + sec 0 V24+1
4b cos(z — tanx) cos(z — tan0) cos(r — 0) -
Ven E
2. y=245sinx 4. y= Vx sec x + 3
1 6. y = x cotx — 5 ©
10. y = (sinx + cos x)sec x
cos x 12. y = ——— — y 1 + sinx cos x x 14. y= ^y^ + cosx
15. y = (sec x + tan x)(sec x — tan x)
16. y = x? cos x — 2xsinx — 2cosx
17. f(x) = x? sin x cos x In Exercises 19-22, find ds/dt. 19. s = tant — t
1 + csct
= 5 T= escri
In Exercises 23-26, find dr/d0.
23. r = 4 — 0? sin0 25. r = sec 0 csc 0
In Exercises 27-32, find dp/dq.
1
27. ni + otg
18. g(x) = (2 — x) tan?x
20. s = ? — sect + 1
sin t 22. s = ———— 1 — cost
24. r = 0sin 0 + cos 0 26. r = (1 + sec 0)sin 0
28. p = (1 + cse q)cosq
d gcc. 9d | 1 30. p= Clu vag 31. p= a 32. p= d 33. Find y" if
a. y = csc x. b. y = sec x. 34. Find y? = d* y/dx* if
a. y — —2sin x. b. y = 9cos x.
Tangent Lines
In Exercises 35-38, graph the curves over the given intervals, together with their tangents at the given values of x. Label each curve and tan- gent with its equation.
35. y= sinx, —37/2 x x € 2v x = —7, 0, 37/2 36. y= tanx, —7/2 «€ x € 7/2 x = —7/3,0, 7/3 37. y= secx, —m/2« x € 7/2 x = —m/3,m/4 38. y= 1 + cosx, —37/2 Sx € 2m x = —7/3, 37/2
Do the graphs of the functions in Exercises 39-42 have any horizontal tangents in the interval 0 = x = 27? If so, where? If not, why not?
Visualize your findings by graphing the functions with a grapher. 39. y = x + sinx 40. y = 2x + sinx
41. y =x — cotx 42. y =x + 2cosx
142
43
44.
In
Chapter 3: Derivatives
. Find all points on the curve y = tan x, —7/2 < x < q/2, where the tangent line is parallel to the line y — 2x. Sketch the curve and tangent(s) together, labeling each with its equation.
Find all points on the curve y = cot x, 0 € x < m, where the tangent line is parallel to the line y = —x. Sketch the curve and tangent(s) together, labeling each with its equation.
Exercises 45 and 46, find an equation for (a) the tangent to the
curve at P and (b) the horizontal tangent to the curve at Q.
45.
46.
Se <
2 = Er T Mel 0 Q y =4 + cotx — 2cscx LI I Lx 0 71 2 3 4 y=1 + V2 csc x + cot x Trigonometric Limits
Find the limits in Exercises 47—54.
47.
48.
49.
51.
52.
53.
Th
peg
XM
tan T 1 a 4 sec x
. . T + tanx lim sin -n = x0 tanx — 2secx
. sin f lim tan ( ]— — 10 t
eory and Examples
lim sed cos xd
54. lim cos (25) 00 sin 0
The equations in Exercises 55 and 56 give the position s = f(t) of a body moving on a coordinate line (s in meters, ¢ in seconds). Find the body's velocity, speed, acceleration, and jerk at time t = 7/4sec.
55. 57.
s=2-—2sint 56. s = sint + cost Is there a value of c that will make Q2 sin E x0 f(x) = * C, x=0
continuous at x = 0? Give reasons for your answer.
T
58.
59.
60.
61.
62.
63.
Is there a value of b that will make
x t b, G(x) = {
COS X,
x«0
x20
continuous at x = 0? Differentiable at x = 0? Give reasons for your answers.
By computing the first few derivatives and looking for a pattern, find d°” / dx??? (cos x).
Derive the formula for the derivative with respect to x of
€. cot x.
a. sec x. b. csc x.
A weight is attached to a spring and reaches its equilibrium posi- tion (x = 0). It is then set in motion resulting in a displacement of
x = 10cos t,
where x is measured in centimeters and ¢ is measured in seconds. See the accompanying figure.
r- —10
Equilibrium position atx =0
--2--ro
r 10
a. Find the spring’s displacement when t = 0,7 = 7/3, and t = 3/4.
b. Find the spring’s velocity when t = 0, t = 7/3, and t = 30/4.
Assume that a particle’s position on the x-axis is given by x = 3cost+ 4sint,
where x is measured in feet and ¢ is measured in seconds.
a. Find the particle's position when t = 0, t = 7/2, and p= T:
b. Find the particle's velocity when t = 0, t = 7/2, and t= T.
Graph y = cos x for —7 = x = 27. On the same screen, graph
sin(x + A) — sin x im h
for h = 1,0.5,0.3, and 0.1. Then, in a new window, try h = —1,—0.5, and —0.3. What happens as h > 0*? As h 07? What phenomenon is being illustrated here?
64.
65.
Graph y = —sin x for —7 = x = 27. On the same screen, graph
cos(x + h) — cosx i h
for h — 1,0.5,0.3, and 0.1. Then, in a new window, try h = —1, —0.5, and —0.3. What happens as h > 0+? As h — 0? What phenomenon is being illustrated here?
Centered difference quotients The centered difference quotient
fœ + h) - fœ- h) 2h
is used to approximate f'(x) in numerical work because (1) its limit as h — 0 equals f'(x) when f'(x) exists, and (2) it usually gives a better approximation of f'(x) for a given value of h than the difference quotient
fG + h) - f% mr o
See the accompanying figure.
y ^ Slope = f'(x) "EE ILLE C P h B A | dos TOD e d) P 2h y =f) | | | | h h - 0 x—-h x xth
a. To see how rapidly the centered difference quotient for f(x) = sin x converges to f'(x) = cos x, graph y = cos x together with
_ sin(x + A) — sin(x — h) 2h
over the interval [ —7, 27] for h = 1, 0.5, and 0.3. Com- pare the results with those obtained in Exercise 63 for the same values of h.
b. To see how rapidly the centered difference quotient for f(x) = cos x converges to f'(x) = —sin x, graph y = —sin x together with | cos(x + h) — cos(x — h) 2h
over the interval [ —7, 27] for h = 1, 0.5, and 0.3. Compare the results with those obtained in Exercise 64 for the same values of h.
66.
67.
68.
69.
70.
143
3.5 Derivatives of Trigonometric Functions
A caution about centered difference quotients | (Continuation
of Exercise 65.) The quotient
fe +h) - fœ- h) 2h
may have a limit as h — 0 when f has no derivative at x. As a case in point, take f(x) = |x| and calculate
. |O+ Al — |0 — Al D 2h i
As you will see, the limit exists even though f(x) = |x| has no derivative at x — 0. Moral: Before using a centered difference quotient, be sure the derivative exists.
Slopes on the graph of the tangent function Graph y = tan x and its derivative together on (~m /2, 77/2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.
Slopes on the graph of the cotangent function Graph y = cot x and its derivative together for 0 < x < m. Does the graph of the cotangent function appear to have a smallest slope? A largest slope? Is the slope ever positive? Give reasons for your answers.
Exploring (sin kx) /x Graph y = (sin x)/x, y = (sin 2x)/x, and y = (sin 4x)/x together over the interval —2 < x < 2. Where does each graph appear to cross the y-axis? Do the graphs really intersect the axis? What would you expect the graphs of y = (sin 5x)/x and y = (sin(—3x))/x to do as x —^0? Why? What about the graph of y = (sin kx)/x for other values of k? Give reasons for your answers.
Radians versus degrees: degree mode derivatives What hap- pens to the derivatives of sin x and cos x if x is measured in degrees instead of radians? To find out, take the following steps.
a. With your graphing calculator or computer grapher in degree mode, graph
sin h
fy) = 95
and estimate lim; o f(/1). Compare your estimate with 7/180. Is there any reason to believe the limit should be 7/180?
b. With your grapher still in degree mode, estimate
lim $95 h—1 h—0 h c. Now go back to the derivation of the formula for the deriva- tive of sin x in the text and carry out the steps of the deriva- tion using degree-mode limits. What formula do you obtain for the derivative?
d. Work through the derivation of the formula for the derivative of cos x using degree-mode limits. What formula do you obtain for the derivative?
e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of sin x and cos x?
144 Chapter 3: Derivatives
3.6 The Chain Rule
: i C: y turns B: u turns A: x turns
FIGURE 3.23 When gear A makes x turns, gear B makes u turns and gear C makes y turns. By comparing cir- cumferences or counting teeth, we see that y = u/2 (C turns one-half turn for each B turn) and u = 3x (B turns
three times for A's one), so y = 3x/2.
Thus, dy/dx = 3/2 = (1/2)(3) = (dy /du)(du/dx).
How do we differentiate F(x) = sin (x? — 4)? This function is the composite f ° g of two functions y = f(u) = sin u and u = g(x) = x? — 4 that we know how to differentiate. The answer, given by the Chain Rule, says that the derivative is the product of the deriva- tives of f and g. We develop the rule in this section.
Derivative of a Composite Function
The function y = ls 5x) is the composite of the functions y = m and u — 3x.
E" 2 We have dy 3 dy 1| d HL, dx 2’ du 2 us dx : ; 3.- : ; Since 257 3, we see in this case that
dy | dy du dx du dx’
If we think of the derivative as a rate of change, our intuition allows us to see that this rela- tionship is reasonable. If y = f(u) changes half as fast as u and u = g(x) changes three times as fast as x, then we expect y to change 3/2 times as fast as x. This effect is much like that of a multiple gear train (Figure 3.23). Let's look at another example.
EXAMPLE 1
The function y = (32 + 1)? is the composite of y = f(u) = i? and u = g(x) = 3x? + 1. Calculating derivatives, we see that Ott — ay 6x = 2(3x* + 1)-6x = 36x? + 12x.
Substitute for u
Calculating the derivative from the expanded formula (3x3? + 1)? = 9x* + 6x? + 1 gives the same result:
dy d dx dx = 36) + 12x. |
(9x^ + 6x? + 1)
The derivative of the composite function f(g(x)) at x is the derivative of f at g(x) times the derivative of g at x. This is known as the Chain Rule (Figure 3.24).
Composite f° g
Rate of change at xis f'(gQ)) - g'G).
f
Rate of change at xis g'(x). x u = g(x)
Rate of change at g(x) is f'( go).
— = —— y =f) = f(g)
FIGURE 3.24 Rates of change multiply: The derivative of f ° g at x is the derivative of f at g(x) times the derivative of g at x.
3.6 The Chain Rule 145
THEOREM 2—The Chain Rule If f(u) is differentiable at the point u = g(x) and g(x) is differentiable at x, then the composite function (f ° g) (x) = f(g(x)) is differentiable at x, and (f © g)’(x) = f'(aG0)* g'CO. In Leibniz’s notation, if y = f(u) and u = g(x), then dy _ dy du
dx du dx’
where dy/du is evaluated at u = g(x).
A Proof of One Case of the Chain Rule: Let Au be the change in u when x changes by Ax, so that Au = g(x + Ax) — g(x). Then the corresponding change in y is Ay = f(u + Au) — f(u). If Au ¥ 0, we can write the fraction Ay/Ax as the product
Ay Ay Au Ax Au’ Ax d) and take the limit as Ax — 0: dou dx Ao Ax = tim OYA Ax>0 Au Ax Ay . Au
= lim : lim Ax0 Au Ax—0 Ax
A Au (Note that Au — 0 as Ax > 0
: y oq > Jim, Au . Aum Ax since g is continuous.)
_ dy. du ~ du dx’
The problem with this argument is that if the function g(x) oscillates rapidly near x, then Au can be zero even when Ax # 0, so the cancelation of Au in Equation (1) would be invalid. A complete proof requires a different approach that avoids this problem, and we give one such proof in Section 3.9. a
EXAMPLE 2 An object moves along the x-axis so that its position at any time t = 0 is given by x(t) = cos(t? + 1). Find the velocity of the object as a function of t.
Solution We know that the velocity is dx/dt. In this instance, x is a composite function: x = cos(u) and u = £ + 1. We have
dx du du dt
= —sin(u) x = cos(u)
146 Chapter 3: Derivatives
Ways to Write the Chain Rule
(fog) = f'(eGO)* g'GO
dy _ dy du dx du dx dy
dio TEO gw
do „du OOF
HISTORICAL BIOGRAPHY Johann Bernoulli
(1667-1748)
By the Chain Rule, dx _ dx du dt du dt = —sin(u)* 2t a evaluated at u = —sin(t? + 1)*2t = —Otsin(? + 1). a
“Outside-Inside” Rule
A difficulty with the Leibniz notation is that it doesn’t state specifically where the deriva- tives in the Chain Rule are supposed to be evaluated. So it sometimes helps to think about the Chain Rule using functional notation. If y = f(g(x)), then
y cogi i
dx ^ 4 (07 g0).
In words, differentiate the “outside” function f and evaluate it at the “inside” function g(x) left alone; then multiply by the derivative of the "inside function."
EXAMPLE 3 Differentiate sin (x? + x) with respect to x.
Solution We apply the Chain Rule directly and find
5 into des] easet d a) (ip T). dx
ES inside inside derivative of leftalone the inside i
Repeated Use of the Chain Rule
We sometimes have to use the Chain Rule two or more times to find a derivative.
EXAMPLE 4 Find the derivative of g(f) = tan(5 — sin 2%).
Solution Notice here that the tangent is a function of 5 — sin 2t, whereas the sine is a function of 2t, which is itself a function of t. Therefore, by the Chain Rule, na ad , g(t)- 4; tan — sin 21))
i d . Derivative of tan u with = sec?(5 — sin 21) Go ~ sin 2n aed clu
d Derivative of 5 — sinu
= sec?(5 — sin27)- (o — cos 2t* do) E Y
= sec?(5 — sin 21) * (-cos 24) + 2 = —2(cos 2t)sec?(5 — sin 2f). E
The Chain Rule with Powers of a Function
If f is a differentiable function of u and if u is a differentiable function of x, then substitut- ing y = f(u) into the Chain Rule formula
dy _ dy du
dx du dx
leads to the formula
d , du sf) = fra).
3.6 The Chain Rule 147
If n is any real number and f is a power function, f(u) = u”, the Power Rule tells us that f'(u) = nu" !. If u is a differentiable function of x, then we can use the Chain Rule to extend this to the Power Chain Rule:
n-1du dx'
m
A (u") = nu fw" = nu EXAMPLE 5 The Power Chain Rule simplifies computing the derivative of a power of an expression.
Power Chain Rule with
(a) £ (5x3 xt)! = 7(5x3 x52 (so x) u = 5x8 — x n=7
= 7(5x — x4)6(5 -3x2 — 4x3) = 7(5x — x*)e(15x? — 4x?)
d( 1 d " (b) race = z) -a 08-27
E d Power Chain Rule with = Ax 2) dx (3x 2) u-c3x—-2,n--1 = -1Gx — 2)?Q)
Ze WEE (3x — 2)/°
In part (b) we could also find the derivative with the Derivative Quotient Rule.
a . 5 ) = 5şin .d . Power Chain Rule with u = sin x, n = 5, (c) dx BID co SI dx sim X because sin" x means (sin x)", n # —1.
— 5sin^xcosx |
EXAMPLE 6 In Section 3.2, we saw that the absolute value function y = |x| is not differentiable at x = 0. However, the function is differentiable at all other real numbers, - of the as we now show. Since |x| — Vx, we can derive the following formula: Absolute Value Function aD = ave Power Chain Rule with d 5) u=x,n=1/2,x #0
" x0 "WE -1
, EL =l. 2 = |x| 2|x| =~ o x20 E x
EXAMPLE 7 Show that the slope of every line tangent to the curve y = 1/(1 — 2x)? is positive.
Solution We find the derivative:
dy d E R (1 = 2x) = —3(1 2x) 4 $ £ (1 2x) Power Chain Rule with u = (1 — 2x), n = —3 = —3(1 - 2xy*-(-2) 6
(1 — 23)*
148
Exercises
Chapter 3: Derivatives
Derivative Calculations
At any point (x, y) on the curve, the coordinate x is not 1/2 and the slope of the tangent line is
CHE M dx (1— 2x)" which is the quotient of two positive numbers. a
EXAMPLE 8 The formulas for the derivatives of both sin x and cos x were obtained under the assumption that x is measured in radians, not degrees. The Chain Rule gives us new insight into the difference between the two. Since 180° = 7 radians, x° = ax/180 radians where x° is the size of the angle measured in degrees.
By the Chain Rule,
ZE out °) = a sin| 255. | = Z cos| Z ]) = -Z os (x?) dP? dx? 80) 7 180 1 1807 — 180 ^ > See Figure 3.25. Similarly, the derivative of cos (x°) is — (7/180) sin (X^).
The factor 7/180 would compound with repeated differentiation, showing an advan- tage for the use of radian measure in computations. El
3t
TX
180
A CTO LAA IRE P UT
ps 180 y =sinx
y = sin(x^) = sin
FIGURE 3.25 The function sin (x°) oscillates only 77/180 times as often as sin x oscillates. Its maximum slope is 77/180 at x = 0 (Example 8).
Find the derivatives of the functions in Exercises 19-40.
In Exercises 1-8, given y = f(u) and u = g(x), find dy/dx = 19. p= V3-t 20. q— War — r f'(gG))g' C9. 4 4 Bart 3i 1. y=6u-9, u-(1/2$1 2. y - 286, u- & - 1 21. s = 3 tsin3t + =~ cos 5r 22, s = sin( 224) - cos( 224) 3. y= sinu, u= 3x +1 4. y=cosu, u=—x/3 23. r = (csc 0 + cot 6)! 24. r = 6(sec 0 — tan 0)? 5. y= Vu, u = sinx 6. y ^ sinu, u =x=— cosx 1 " i 25. y = x° sin'x + xcos?x — 26. y= zsin°x — 3 cos x 7. y = tanu, u= qx? 8. y 2 —secu, u =x 7x ži 27. y = Lox - 25 4 (4 3 In Exercises 9-18, write the function in the form y — f(u) and 18 2x u = g(x). Then find dy/dx as a function of x. 1/2 4 5 6 28. y = (5 — 2x)? 4 - 1 9. y= (2x + 1) 10. y = (4 — 3x) g\x E (1 7 2u i = e 7 1)" 29. y = (Ax + 3'x + 1? | 30. y = Qx — 5) M(x? — 5x)® T ` 2 "o nu 31. h(x) = x tan (2Vx) +7 32. k(x) = x? sec (P)
B.y=(F+x-4) 14. y = V3x? — 4x + 6
1 33. f(x) = V7 + xsecx 34. g(x) = EE 15. y = sec(tan x) 16. y — cot (= = i) x )
sin 0 2 1 + sin3r\!
17. y = tan?x 18. y = 5cos ^x 35. f(0) — 1 cos) 36. g(t) = “3 =
37. r — sin(0?)cos (20)
i f 39. q4 = E ) 2 Vt--1
In Exercises 41—58, find dy/dt.
4l. y = sin? (at — 2) 43. y = (1 + cos2r) * 45. y = (ttan r)'?
Boy 47. y — BA
49. y = sin(cos(2t — 5))
£a 51. y — (1 + av (1)) 53. y = V1 + cos(t*) 55. y = tan" (sin? t)
57. y = 3at(2 — 5)*
Second Derivatives Find y" in Exercises 59-64.
1\3 1 61. y = g cot (3x — 1)
63. y = xx + 1)4
Finding Derivative Values
38.
64.
= sec V6 tan (5)
i sin f cot =
sec? mt (1 + cot(t/2))? (3/4 sin t4?
(221) cos(ssin(5))
1
ei + cos? (71) P
4sin( VI + vi)
cost (sec?31)
ETEVE
(1- vx)!
9 tan (3
x2(x3 - 1)
In Exercises 65-70, find the value of (f ° g)' at the given value of x.
65. flu) — i? +1,
u = g(x) = Vx,
x= 1
66. fu) = 1
1 w 4780-7
fh. —X«
x—-—l
67. f(u) — cot To u = g(x) =5Vx, x-1
68. f(u) =u 4 L 1
69. dec
u = g(x) = mx,
x= 1/4
w+.
2 u—l
u = g(x) = 10x? 4
x+1, x=0
70. f(u) = (: F L) , u= g(x) 7 1, x=-l
71. Assume that f'(3) =
What is y' at x = 2?
1, g'(2) = 5, g(2) = 3, and y = f(g(x)).
72. If r = sin(f())), f(0) = 7/3, and f'(0) = 4, then what is dr/dt
att = 0?
73. Suppose that functions f and g and their derivatives with respect
to x have the following values at x = 2 and x = 3.
x f(x) 2 8
g(x)
2
-4
f'(x) g' (x) 1/3 E 2T 5
3.6 The Chain Rule 149
Find the derivatives with respect to x of the following combina- tions at the given value of x.
a. 2f(x), x=2 b. f(x) + g(x), x-3 c. fœ) g(x), x =3 d. f(x)/g(), x =2 e. f(g(x)), x = 2 f Vf@, x=2
g 1/¢(x), x =3 h. Vf^(x) + g(x) x = 2 74. Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.
x f(x) g(x) f' (x) g'(x) 0 1 1 5 1/3 1 3 -4 -1/3 -8/3
Find the derivatives with respect to x of the following combina- tions at the given value of x.
a. 5f() — gx) x=1 b. fG)g(), x-0 f(x) u _
Cc. a(x) + 1 x= 1 d. f(g), x=0
e. g(f(x)), x =0 f. xl! - f@)?, x= 1
g fat ga) x=0
75. Find ds/dt when 0 = 37/2 if s = cos@ and d0/dt = 5. 76. Find dy/dt when x = 1 if y = x? + 7x — 5 and dx/dt = 1/3.
Theory and Examples
What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 77 and 78.
77. Find dy/dx if y = x by using the Chain Rule with y as a compos- ite of
a. y= (u/5) - 7. and u= 5x — 35 b. y=1+(1/u) and u= l/(x — 1). 78. Find dy/dx if y = x?? by using the Chain Rule with y as a com-
posite of a. y=w and u= Vx b. y= Vu and u=x.
79. Find the tangent to y = ((x — D)/(x + 1)? at x = 0. 80. Find the tangent to y = Vx? — x + 7atx = 2. 81. a.
b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval —2 « x « 2? Give reasons for your answer.
Find the tangent to the curve y = 2 tan(zx/4) at x = 1.
82. Slopes on sine curves
a. Find equations for the tangents to the curves y — sin 2x and y = —sin (x/2) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer.
b. Can anything be said about the tangents to the curves y = sin mx and y = —sin(x/m) at the origin (m a constant # 0)? Give reasons for your answer.
c. Fora given m, what are the largest values the slopes of the curves y = sin mx and y = —sin(x/m) can ever have? Give reasons for your answer.
150
83.
84.
85.
86.
87.
88.
89.
Chapter 3: Derivatives
d. The function y — sin x completes one period on the interval [ 0, 27 ], the function y = sin 2x completes two periods, the function y = sin(x/2) completes half a period, and so on. Is there any relation between the number of periods y — sin mx completes on [ 0, 277 ] and the slope of the curve y = sin mx at the origin? Give reasons for your answer.
Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time f sec is
s = Acos(27bt),
with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What effect does doubling the fre- quency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)
Temperatures in Fairbanks, Alaska The graph in the accom- panying figure shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365-day year. The equation that approximates the temperature on day x is
= zm ; y 37sin| 2c D + 25
and is graphed in the accompanying figure. a. On what day is the temperature increasing the fastest? b. About how many degrees per day is the temperature increas-
ing when it is increasing at its fastest?
y D
D c
B ©
Temperature (°F) 8
e
— = N TX
20L SS & we JS Pk S SR cem S YS we FI we
Particle motion The position of a particle moving along a coordinate