Integrated Calculus/Pre-Calculus - Furman Universitymath.furman.edu/~mwoodard/m141/docs/Math 11s/mybookpaper.pdf · The electronic version of this document has a number of interactive

Furman University

Integrated Calculus/Pre-Calculus

Mark R. Woodard

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Copyright c© 2000 [email protected] Revision Date: May 7, 2000

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Table of Contents

1 Functions and their Properties 91.1 Functions & Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.1 Functions and Functional Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2 Cartesian Coordinates and the Graphical Representation of Functions . . . . . . . . . . . . . 141.3 Circles, Distances, Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3.1 Distance Formula and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3.2 Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.1 General Equation and Slope-Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . 161.4.2 More On Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.3 Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Catalogue of Familiar Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.2 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.3 Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.5.4 The Absolute Value and Other Piecewise Defined Functions . . . . . . . . . . . . . . . 191.5.5 The Greatest Integer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.6 New Functions From Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.6.1 New Functions Via Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.7 New Functions via Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.8 Shifting and Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Limits 232.1 Secant Lines, Tangent Lines, and Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.1.1 The Tangent Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.1.2 Instantaneous Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2 Simplifying Difference Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 Limits (Intuitive Approach) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.4 Intervals via Absolute Values and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.5 If – Then Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.6 Rigorous Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.7 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.7.1 Definitions Related to Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.7.2 Basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.7.3 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Table of Contents (cont.) 3

3 The Derivative 393.1 Definition of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.1.1 Slopes and Velocities Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.1.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.1.3 The Derivative as a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2 Rules for Taking Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.3 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.4 Applications of Derivatives as Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.4.2 Some Specific Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47• Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47• Biology Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48• Chemistry Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48• Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.7 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.7.1 The Second Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.7.2 Third and Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.8 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.9 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.9.1 Mathematicizing the main idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4 An Interlude – Trigonometric Functions 574.1 Definition of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.1.1 Radian Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.2 Trigonometric Functions at Special Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.3 Properties of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.4 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.4.1 Graph of sin(x) and cos(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.4.2 Graph of tan(x) and cot(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.4.3 Graph of sec(x) and csc(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.5 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.6 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.6.1 Two Important Trigonometric Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.6.2 Derivatives of other Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . 64


5 Applications of the Derivative 675.1 Definition of Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.2 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.3 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.4 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.5 An Application of the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 735.6 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.7 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.8 Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.9 Limits at Infinity; Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5.9.1 A Simplification Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.10 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.11 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.12 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.13 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.14 Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.15 Antiderivatives Mini-Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6 Areas and Integrals 936.1 Summation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.2 Computing Some Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.3 Areas of Planar Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

6.3.1 Approximating with Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.4 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.4.1 Definition and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.4.2 Properties of

∫ b

af(x) dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.4.3 Relation of the Definite Integral to Area . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.4.4 Another Interpretation of the Definite Integral . . . . . . . . . . . . . . . . . . . . . . 103

6.5 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.5.1 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046.5.2 The Fundamental Theorem – Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6.6 The Indefinite Integral and Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.6.1 The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.6.2 Substitution in Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086.6.3 Substitution in Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6.7 Areas Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.7.1 Vertically Oriented Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.7.2 Horizontally Oriented Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113


7 Appendix 1157.1 3 – Minute Reviews . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.1.1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1167.1.2 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1167.1.3 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1167.1.4 Solving Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177.1.5 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.1.6 Quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.1.7 Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.1.8 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.1.9 Setting Up Word Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120Solutions to Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145


Preface

Every teacher of calculus has at one time or another remarked about a particular student or group of students“I think they understand the calculus concepts, but their algebra skills are so weak, they can’t finish theproblems.” There are many reasons that explain why this problem is so prevelant: some students neverlearned the precalculus material, some learned it but forgot most of it, others are just a little rusty and needsome extra practice. For these reasons, some colleges and universities are starting so called “integrated”precalculus/calculus courses. This text is designed to be used in exactly those kinds of courses, although itcould also be used in a traditional calculus course, with the precalculus material omitted by the instructor, butavailable nonetheless to the motivated and interested student who needs or desires some extra amplificationsof, reminders about, or practice with the precalculus material. An attempt is made to have the precalculusmaterial appear “just in time”, that is, directly before it is needed for the first time.

In addition, this document is unusual in that the online version allows for reading in an interactive style:hypertext links are available, solutions are offered to many examples and exercises via a hyperlink, and thenthe reader can then return to his or her place in the text via another jump. Miniature quizzes will also beavailable, which provide immediate feedback that will hopefully erase misunderstandings and misconceptionsbefore they fester and grow. In order to get the reader acclimated to these features, there are some examplesprovided below.

Notes to the reader

The electronic version of this document has a number of interactive features, including hyperlinked jumpsto solutions of many exercises, as well as interactive quizzes. Examples are below. Answers will be clear toanyone who has carefully read the dedication and acknowledgements.

An example of an exercise with a jump to the solution. (You can jump back by via the blue hyperlinkon the solutions page.)

Exercise 0.0.1. How many children does the author of the document have?

Here is an example of a short quiz. You will be taken to the “solution” after finding the right answer.You must first click on the colored label “Quiz” to initialize the quiz before it will work.

Quiz Who made most of the illustrations in this document?

(a) Bertolt Brecht (b) Brian Wagner(c) Babe Didrickson (d) Bebe Rebozo

Here is an example of a “graded” quiz. You must initialize the quiz by first clicking on “Begin Quiz,”and you must end the quiz by clicking on “End Quiz.” To correct your quiz and see the answers, click onthe button that says “Correct”.

Click to Initialize Quiz Answer the following questions about the author’s affiliations.


1. Which university employs the author of this document?Fordham Ferrum Ferris State Furman

2. In what state is the author’s university located?South Carolina North Carolina Georgia

Click to End Quiz

It is hoped that such interactivity will make the reading of the text both more enjoyable, and moreeffective for the student. Now it’s time to learn some mathematics!


Chapter 1

Functions and their Properties

Contents

1.1 Functions & Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.1 Functions and Functional Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Cartesian Coordinates and the Graphical Representation of Functions . . . . 14

1.3 Circles, Distances, Completing the Square . . . . . . . . . . . . . . . . . . . . . 15

1.3.1 Distance Formula and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3.2 Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4.1 General Equation and Slope-Intercept Form . . . . . . . . . . . . . . . . . . . . . . 16

1.4.2 More On Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4.3 Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Catalogue of Familiar Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5.2 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5.3 Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.5.4 The Absolute Value and Other Piecewise Defined Functions . . . . . . . . . . . . . 19

1.5.5 The Greatest Integer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.6 New Functions From Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.6.1 New Functions Via Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.7 New Functions via Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.8 Shifting and Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

9

Section 1.1: Functions & Cartesian Coordinates 10

1.1. Functions & Cartesian Coordinates

1.1.1. Functions and Functional Notation

The key object of study in a calculus course is a function. Since this is the case, it is important to knowwhat a function is. Many students erroneously think of a function as a formula – this oversimplification andhalf-truth leads to problems later on in the course, so it is best to get the following definition down pat atthe outset.

Definition of FunctionDefinition 1.1.1 A function is a rule of correspondence betweenitems in one set (called the domain of the function) and items inanother set (called the range of the function) so that every elementin the domain is paired with exactly one element in the range.

Note that the definition of function also includes the idea of the domain of a function and the range of afunction. Note also that the definition doesn’t say anything about formulas. In fact, a useful way to thinkabout a function is as a machine. Domain elements are input to the machine, and then the function-machineturns them into (perhaps) something else, and these outputs are the range elements. A simple example isillustrated in figure 1.1. An animated version is available at http://math.furman.edu/~mwoodard/math10/math/machine.html

�

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�� !#"$%!'&

Figure 1.1: An machine diagram for a function.

It is important to note that every function comes with three items:

http://math.furman.edu/~mwoodard/math10/math/machine.html

http://math.furman.edu/~mwoodard/math10/math/machine.html


�

�

�

� �

��

��

Figure 1.2: An arrow diagram for a function.

Every function consists of:

• A set called the domain.

• A set called the range.

• A rule of correspondence between elements in those sets

In this text, our domain sets will almost always be subsets of the set of real numbers (often denoted <),as will our range sets. (There is a review of the set of real numbers in section 7.1.1. Thus, we will thinkof functions as rules which transform certain numbers into other numbers. There are number of differentways to denote or describe a function; we will discuss a few, but will settle on a few to use for the restof this document. Since we are trying to indicate a relationship between quantities, we can use arrows toshow which domain element corresponds to which range element (see figure 1.2.) This is effective when thedomain is small, but most of our domains will have an infinite number of elements, so this approach is notpractical. Another method would be to just write ordered pairs, with domain elements listed first and thecorresponding range elements listed second. Thus, the function described by figure 1.2 could instead bedenoted {(1,−1), (4,−2), (9,−3)}.

Again, this approach is not practical when the domain is infinite. Probably the best approach for ourpurposes is to use functional notation. In this notation, the rule of correspondence is given by a formulaicexpression, and the domain is either specified directly or is implied. For example,

f(x) = x2 + 2, 0 ≤ x ≤ 5 (1.1)

represents the function whose domain consists of the real numbers between 0 and 5, and whose rule is suchthat an input number is paired with the number which is 5 more than the square of the input. Thus, forexample, the number 2 would be paired with 22 + 5 = 9. We indicate this by writing f(2) = 9, which youshould think of as meaning that the number 2 goes into the function machine f , and 9 comes out. Thisnotation has many advantages, but for now let’s highlight the disadvantages. The range is not mentionedat all, the domain is easy to ignore, and it becomes the habit of many to think of the function as a formularather than a rule of correspondence between elements in two sets. Resist this habit!.

Example 1.1.1. If f(x) = 3x2 − 4, what is f(x + 1)?


Solution: Since f is the rule which says that we should square the input, multiply by 3 and then subtract 4,f(x + 1) should be be the result of doing those operations to the quantity (x + 1), thus, the output shouldbe 3(x + 1)2 − 4, so

f(x + 1) = 3(x + 1)2 − 4

�

Note that in the previous example, no domain was specified at all! This will often be the case, eventhough every function must have a domain. Thus, there must be a domain implied even when it is notspecified. Our rule of thumb for domains will be the following:

Rule of Thumb about Domains Unless otherwise specified, thedomain of a function written in functional notation will be takento be the largest subset of the real numbers for which the functionis defined.

Example 1.1.2. What is the domain of f(x) =x

x2 − 9?

Solution: Since there is no domain specified, the rule of thumb applies. �

There is one aspect of the definition of function which we haven’t discussed yet, and that is the partwhich says that every domain element “is paired with exactly one element from the range.” Note that thisdoesn’t say that every range element is paired with exactly one domain element, so an arrangement like thatin figure 1.3 is a function, while that in figure 1.4 isn’t.

Example 1.1.3. If the quantities u and w are related by the relationship u2 = w, is u a function of w? Isw a function of u?

�

�

�

�

�

��

��

Figure 1.3: An arrow diagram which does represent a function.


�

�

�

�

�

��

Figure 1.4: An arrow diagram which does not represent a function.

Solution: It is true that w is a function of u, because for every value of u, there is only one possible value foru2. On the other hand, u is not a function of w, because for a given value of w (like the value 9 for example)there could be two different values of u (namely 3 and −3) for which the relationship is met. �

In the application of functional notation, we often use the variable x to represent quantities which comefrom the domain set (this is more formally called the independent variable) and we often use the variabley to represent quantities which come from the range set (more formally called the dependent variable), butthere is no reason why these letters necessarily must represent those quantities. For example, it is perfectlyOK to write

x = y2 + 2y

and think of x as a function of y, thus making y the independent variable and x the dependent variable forthis example.

Quiz: Functions and Functional Notation.

Click to Initialize Quiz Answer the following questions about functions and functional notation. Whennecessary, do you work on a scrap piece of paper, but record your answers on this document. Don’t forgetto initialize the quiz by clicking in the appropriate place before starting, and then click on the appropriatelabel to find you score.

1. If f(z) = 1z , what is f(z + h)?

1z + h h

z1

z+h

2. If f(y) = 1y , what is f(z + h)− f(z), in simplified form? If you need a hint, there is one available here.

−h(z)(z+h)

h(z)(z+h)

1h

3. What is the domain, written in interval notation, for h(w) = w2+9w2−2w+1?

(−∞, 1] ∩ (1,∞) (−∞, 1) ∪ (1,∞) (−∞,−1) ∪ (−1, 1) ∪ (1,∞)

Click to End Quiz

Section 1.2: Cartesian Coordinates and the Graphical Representation of Functions 14

1.2. Cartesian Coordinates and the Graphical Representation of

Functions

The Cartesian Coordinate System is a system by which one can associate numerical quantities with geometricones, namely, a point in a plane is associated with a pair of real numbers. Our method of associating isas follows: we draw two perpendicular number lines, and call the point where they intersect (0, 0) or, theorigin. If we wanted to see which point is associated with (2, 4) for example, we would start at the origin andmove 2 units to the right,and then 4 units up. The first coordinate is related to horizontal movement andthe second is related to vertical. The positive horizontal direction is to the right, and the positive verticaldirection is up. Note that the most common labels for the axes are x for the horizontal axis and y for thevertical, although these are somewhat arbitrary and on occasion we will have need for other labels.

As we have seen, a function can be thought of as an association of inputs with outputs, so a functioncan be thought of as a set of ordered pairs. By coloring the points which correspond to the ordered pairswhich make up the function, we get what we refer to as the graph of the function. Note that we can seeboth the domain and the range in this graphical representation: for each point on the graph, if we projectstraight down (or up) to the horizontal axis, the set of points on that copy of the real number line indicatethe domain of the function. Likewise, if we project to the vertical axis, we can see the range of the function.

To see the domain of a function from its graph:Project onto the x axis.

To see the range of a function from its graph:Project onto the y axis.

We can also graphically see whether or not a given curve really represents y as a function of x via thevertical line test. This test says that if a vertical line can be drawn which intersects the curve in two distinctpoints (say (x1, y1) and (x1, y2) with y1 6= y2), then the graph does not represent a function of x, becausethere are two distinct y values associated with the same x value x1.

Vertical Line TestIf a vertical line can be drawn which intersects the curve in twodistinct points (say (x1, y1) and (x1, y2) with y1 6= y2), then the

graph does not represent a function of x.

Section 1.3: Circles, Distances, Completing the Square 15

1.3. Circles, Distances, Completing the Square

1.3.1. Distance Formula and Circles

Given two points (x1, y1) and (x2, y2) in the plane, there will be occasions when you are interested in knowingthe distance between them. By plotting the two points along with the the point (x2, y1), you can see a righttriangle whose hypotenuse has the length we are looking for. The two sides of the triangle have lengthsx2 − x1 and y2 − y1. Thus, the pythagorean theorem gives the required distance, and we have

The distance between the points (x1, y1) and (x2, y2) in the planeis given by d =

√(x2 − x1)2 + (y2 − y1)2

A circle is the set of points a given distance, the radius, from a given point, the center. Using this fact andthe above distance formula, we see that every point (x, y) on a circle of radius r centered at (h, k) satisfies√

(x− h)2 + (y − k)2 = r. Squaring both sides yields the standard equation for a circle, namely:Equation of a Circle with Center at (x0, y0) and radius r: (x− x0)2 + (y − y0)2 = r2

Equation of a circle with center at (h, k) and radius r:(x− h)2 + (y − k)2 = r2

1.3.2. Completing the Square

A helpful algebraic technique when dealing with circles is the technique of completing the square. Thistechnique helps one transform an equation of a circle which is disguised so as to not look like one intothe standard form, so one can tell the center and radius. In this technique, an expression of the formx2 + kx + . . . = c is changed by adding k

2

2to both sides, yielding x2 + kx + k

2

2+ . . . = c + k

2

2. This enables

us to write (x + k2 )2 + . . . = c + k

2

2, thus changing the left hand side into a perfect square.

Exercise 1.3.1. The equation x2 + 4x + y2 + 6y = 10 is actually the equation of a circle. Complete thesquare for both the x expression and the y expression in order to put this equation in standard form.

Section 1.4: Lines 16

1.4. Lines

1.4.1. General Equation and Slope-Intercept Form

Recall that the general equation of a line is an equation of the form Ax + By = C, where A,B and C arereal numbers. If we solve for y, (so that the expression will give y as a function of x) we have y = −A

B x+ CA ,

which we usually write as y = mx + b, where the expression −AB = m can be seen to be the slope of the line,

and the expression CA = b can be seen to be the y - intercept. In the case where A is zero, we have that m

is zero as well, and the line has the form y = b which represents a horizontal line. If B = 0 then we can’ssolve for y as above, and we see that our equation has the form x = C

A , which reprsents a vertical line.

General Equation of a Line:Ax + By = C, where A,B and C are real numbers.

Slope-Intercept Form of a Line:y = mx + b, where m is the slope and b is the y-intercept.

1.4.2. More On Slope

The slope of a line mentioned above plays a key role in the study of calculus. The slope of a line is thenumerical value which represents the “steepness” of the line. Given two points (x1, y1) and (x2, y2) on theline, the slope is defined to be

m =y2 − y1

x2 − x1.

Note that the numerator of this fraction is the vertical “rise” as we move from the point (x1, y1) to(x2, y2), while the denominator is the horizontal “run”. Thus you will hear people describe the slope is beingthe “rise over run”. From this definition of slope, it is easy to get point-slope form of the equation of a line.Given a point (x0, y0) on a line of slope m, we know that any other point (x, y) on the line must satisfy

y − y0

x− x0= m,

since the “rise over run” from the point (x0, y0) to (x, y) must be m. Multiplying both sides by the appropriateexpression to clear denominators leads to the

Point-Slope Form of the Equation of a Line: y − y0 = m(x− x0)

Exercise 1.4.1. A line has slope 5, and goes through the point (−2,−4). Find the equation of this line,and indicate its y - intercept.

Section 1.4: Lines 17

Exercise 1.4.2. Find the equation of the line that contains the points (−2, 5) and (2,−3).

1.4.3. Parallel and Perpendicular Lines

There is a nice relationship between slopes of lines and whether or not they are perpendicular or parallel.

Lines which are not vertical are parallel if and only if they havethe same slope.Lines which are not vertical or horizontal are perpendicular if andonly if their slopes multiply together to be −1.

Exercise 1.4.3. Are the lines 3x + y = 7 and −x− 3y = 9 parallel, perpendicular, or neither?

Section 1.5: Catalogue of Familiar Functions 18

1.5. Catalogue of Familiar Functions

1.5.1. Polynomials

We are interested in devloping an informal “catalogue” of the functions we will be using, so we can refer tothem by name and understand some of the properties that they might posess. Our first category will be thecollection of polynomials.

Definition of PolynomialA polynomial is a function which can be written in the form

anxn + an−1xn−1 + . . . + a1x + a0

where the ai’s are real numbers, an 6= 0, and n is a non-negativeinteger.

In the above definition, the ai’s are called coefficients and the number n is called the degree of thepolynomial. Here are some familiar types of polynomials:

1. The constant function f(x) = k is a degree zero polynomial.

2. The line f(x) = mx + b is a first degree polynomial.

3. A quadratic function f(x) = ax2 + bx + c is a second degree polynomial.

Quadratics have graphs which are polynomals which open up if the coefficient on the 2nd degree term ispositive, and open down if it is negative. The vertex of a parabola is the lowest point on the parabola if itopens down or is the highest point on the parabola if it opens up. One can find the x value of the vertex bycompleting the square. If f(x) = ax2 + bx + c, we can write f(x)− c = a(x2 + b

a ), and then completing thesquare inside the parentheses yields

f(x)− c +b2

4a= a(x2 +

b

a+

b2

4a2),

so f(x)− c + b2

4a = a(x + b2x )2. From this it is not too hard to see that the vertex will be at x = − b

2a .Polynomials share many nice properties. For example, every polynomial has domain equal to R, the set

of all real numbers.

1.5.2. Rational Functions

A rational function is one which can be written as the ratio of two polynomials. For example, 3x2+2x5x−3 is a

rational function. Also f(x) = 1x + 3

x2 is a rational function, because by simplifying it can be written asf(x) = 3+x

x2 . Rational functions have domain equal to the set of all real numbers except for those numberswhich cause the denominator to be zero.

Section 1.5: Catalogue of Familiar Functions 19

1.5.3. Algebraic Functions

An algebraic function is one which can be built out of the standard operations of addition, subtraction,multiplication, division and roots. In particular, everything mentioned in this section so far is algebraic. Someexamples of non-algebraic functions are the trigonometric functions (f(x) = sin(x), etc.), the exponentialfunctions (f(x) = 2x, etc.), and the logarithmic functions (f(x) = log x). We will postpone our study ofthese types of functions until later.

1.5.4. The Absolute Value and Other Piecewise Defined Functions

The absolute value function, denoted |x| is defined by

|x| =

{x if x ≥ 0−x if x < 0

This function can be thought of as giving the distance that x is from 0 on the number line. In general,|x− c| always gives the distance between x and c on the number line.

The absolute value function is an example of a piecewise defined function. Here is another example

f(x) =

{16− x2 if x < 22x + 20 if x ≥ 2

Note that the domain of this function consists of all real numbers, but that the function is defineddifferently over the different “pieces” of the domain. To properly evaluate this function at x = 3 forexample, one must first ask which piece of the domain 3 falls into. A little thought reveals that 3 is in thecategory x ≥ 2, so we apply the 2x + 20 rule with input 3 and obtain that f(3) = 26.

1.5.5. The Greatest Integer Function

Another interesting and unusual function is the greatest integer function, denoted bxc. This function is notgiven by a formula, but rather a verbal rule of correspondance. The rule is that the output is always thegreatest integer that is less than or equal to the input. Thus b2.5c = 2,b−2.5c = −3, bπc = 3, and so on.This function is sometimes referred to as a step function, since its graph resembles stairs going up from leftto right.

Section 1.6: New Functions From Old 20

1.6. New Functions From Old

There are a number of ways to get new functions by combining old ones in various ways. These includecomposing two or more functions to get a new function, or combining familiar functions with addition,subtraction, multiplication or division. We will investigate all of these, and see how the graphs of familiarfunctions are changed by simple translations and scalings.

1.6.1. New Functions Via Composition

Recall that a function is a rule of correspondance which has inputs and outputs. If we put the output of aparticular function into a second function as an input, the resulting output (when thought of as a functionof the original input) is the composition of the two functions. Notationally, we write (f ◦ g)(x) = f(g(x)).This notations suggests that if we were to evaluate f ◦ g at a number like 2, we would compute f(g(2)),that is, compute g(2) first and then plug the resulting number into f as input. It is important to note thatf(g(x)) would not necessarily be equal to g(f(x), so we see that in general g ◦ f 6= f ◦ g.

The composition of f and g is defined by f ◦ g(x) = f(g(x)). Inthis definition we will call g the “inner” function and f the

“outer” function.This is not necessarily equal to the composition of g and f , which

is defined by g ◦ f(x) = g(f(x)). In this definition, f is the“inner” function and g is the “outer” function.

Exercise 1.6.1. For the functions f(x) = 3x + 2 and g(x) = x2 − 5, compute (f ◦ g)(x) and (g ◦ f)(x). Tosee that these are different, compute f ◦ g(2) and g ◦ f(2). Is this enough information to deduce that thetwo are different?

Exercise 1.6.2. Consider the function h(x) = (4x + 1)2 − 9. Can you find functions f and g so thath(x) = f(g(x))?

Click to Initialize Quiz Answer the following questions about compositions.

1. If f(x) = 1x , g(x) = x2, and h(x) = 2x + 1, what is f ◦ g ◦ h(x)?

12x+1

2 2x+1x2

x2

2x+1 2x + 1x2

2. If f(g(h(x))) =(2x + 1 + 1

2x+1

)2

, find possible representations for f , g, and h.

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

h(x) = 2x + 1g(x) = x2, f(x) = x + 1

x ,h(x) = 2x + 1

h(x) = x2, g(x) = x + 1x ,

h(x) = 2x + 1

Section 1.7: New Functions via Algebra 21

Click to End Quiz

1.7. New Functions via Algebra

One can also get new functions from old via the good old-fashioned operations of addition, subtraction,multiplication and division. For example, if f and g are functions, we can form

• (f + g)(x) = f(x) + g(x)

• (f − g)(x) = f(x)− g(x)

• (f · g)(x) = f(x) · g(x)

•(

fg

)(x) = f(x)

g(x) .

Note that if f and g in the above all have domain equal to the set of all real numbers, then each all of f + g,f − g and f · g do too, but that the domain of f

g could be only a subset of the set of all real numbers. Infact, any number a for which g(a) = 0 is definitely not in the domain of f

g .

1.8. Shifting and Scaling

Suppose you are given a function y = f(x) for which you are familiar with the graph, and a positive constantc. (You might want to think of f(x) = x2 and c = 3 as a good example during this discussion.) How are thegraphs of the following related to the graph of f(x)?

• f(x + c)

• f(x− c)

• f(x) + c

• f(x)− c

• cf(x)

• −cf(x)

• f(cx)

• f(−cx)

A little experimentation and thought should convince you of the following:

• f(x+ c) has the same shaped graph as f(x), but the whole thing is shifted c units to the left. (To helpyou remember this, note that plugging 0 into f(x) is the same as plugging −c into f(x + c).

Section 1.8: Shifting and Scaling 22

• f(x− c) has the same shaped graph as f(x), but the whole thing is shifted c units to the right.

• f(x) + c has a graph like f(x)’s but shifted c units up.

• f(x)− c has a graph like f(x)’s but shifted c units down.

• cf(x) is similar to f(x) but the y axis is scaled (if c > 1) to stretch the graph by a factor of c or (if0 < c < 1) to compress the graph.

• −cf(x) is similar to cf(x) but the whole thing is reflected about the x axis. Thus for example whenc = 1, the graphs of f(x) and −f(x) can be gotten from one another by reflecting about the x axis.

• f(cx) is similar to f(x) except that the x axis is scaled to stretch the graph horizontally (if c > 1) orcompress the graph horizontally (if c < 1.)

• f(−cx) is similar to f(cx), except that the whole thing is reflected about the y axis. Thus for example,f(−x) can be gotten from f(x) by reflecting about the y axis.

Note that if a function has the property that it is symmetric about the y axis, then f(x) = f(−x) for thatfunction. Such functions are called even functions, because some examples of these are f(x) = x2, f(x) = x4,f(x) = x6, etc.. Functions for which f(x) = −f(x) are said to be symmetric about the origin, and are calledodd functions. Some examples of odd functions include f(x) = x, f(x) = x3, etc..

Exercise 1.8.1. Is the sum of an odd function and another odd function always odd?

Exercise 1.8.2. Draw a simple graph of f(x) =√

x. Now graph all of the following for c = 3.

• f(x + c)

• f(x− c)

• f(x) + c

• f(x)− c

• cf(x)

• −cf(x)

• f(cx)

• f(−cx)

Chapter 2

Limits

Contents

2.1 Secant Lines, Tangent Lines, and Velocities . . . . . . . . . . . . . . . . . . . . . 24

2.1.1 The Tangent Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.1.2 Instantaneous Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2 Simplifying Difference Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.3 Limits (Intuitive Approach) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.4 Intervals via Absolute Values and Inequalities . . . . . . . . . . . . . . . . . . . 32

2.5 If – Then Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.6 Rigorous Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.7 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.7.1 Definitions Related to Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.7.2 Basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.7.3 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

23

Section 2.1: Secant Lines, Tangent Lines, and Velocities 24

2.1. Secant Lines, Tangent Lines, and Velocities

One of the main problems that motivated the genesis of calculus was this: What does one mean by the slopeof the tangent line to a curve at a given point, and how can we calculate it? Another important motivatingquestion was: What does one mean by instantaneous velocity, and how can we calculute it?

We will see in the following discussion that these two apparently unrelated questions are in fact veryclosely related.

2.1.1. The Tangent Line

We start by admitting that we haven’t defined a tangent line but have an intuitive idea of what we aretalking about, using the idea of the tangent line to a circle as a guide. We note that as soon as we know theslope of the tangent line, we can write down its equation. It is

y − b = m(x− a)

where the point of tangency is (a, b) and m is the slope of the tangent line. It seems reasonable to approximatethe slope of the tangent lines by using secant lines. More specifically, we consider the line through the point(a, f(a)) and (a + h, f(a + h)). The slope is clearly

f(a + h)− f(a)h

and we note that this approximation gets better as h gets smaller. Thus we define the tangent line to be theline whose slope is the result of “taking the limit as h → 0” in the above. We will discuss the technicalitiesof of this limiting procedure later, but for now will be content with “shrinking h → 0” informally.

The slope of the tangent line is obtained by first computing theslope of a secant line between the points (a, f(a)) and

(a + h, f(a + h)). This slope is given by

f(a + h)− f(a)h

.

This approximation gets better as h → 0, so we will shrink h tozero to obtain the slope of the tangent line.

Exercise 2.1.1.

If f(x) = x3 − 3x, what is the slope of the tangent line at x = 4? What is the equation of the tangentline at that point?


Exercise 2.1.2.

If f(x) =√

x + 3− 4, what is the slope of the tangent line at x = 6?

2.1.2. Instantaneous Velocity

Suppose that you have a really accurate odometer and a very accurate stop watch but no speedometer, andyou want to answer the question: how fast am I moving right now? (Assume you are traveling in a straightline.) Everyone is familiar with the idea that average velocity (i.e. “rate”) is elapsed distance divided byelapsed time. (Think of that familiar formula d = r · t.) Now it seems like a good approximation for instan-taneous velocity would be average velocity over an extremely small time interval. And the approximationwould get better as the time interval gets smaller. So if f(t) is our odometer reading at time t, and if we areinterested in our instantaneous velocity at time t = a, we could consider the average velocity between timest = a and t = a + h. Our elapsed distance would be f(a + h)− f(a) and the elapsed time would be h, andthus the average velocity would be

f(a + h)− f(a)h

.

And the approximation gets better as h → 0 so: HEY WAIT A MINUTE I’M GETTING DEJA VU!

The instantaneous velocity at time t = a is obtained by firstcomputing the average velocity between the points (a, f(a)) and

(a + h, f(a + h)). This average velocity is given by

f(a + h)− f(a)h

.

This approximation gets better as h → 0, so we will shrink h tozero to obtain the instantaneous velocity.

Note that this idea of ‘taking the limit as h → 0” needs to be studied in more detail, which we will do inthe next few sections.

Exercise 2.1.3.

Suppose an object is moving in a straight line so that it’s position at time t is given by f(t) = 3t+1 . What

is the object’s instantaneous velocity at time t = 3?


Exercise 2.1.4.

A ball dropped from the bell tower falls so that its position (in feet) above the ground is given bys(t) = −16t2 + 64. How fast is the ball moving after 1 second?

Section 2.2: Simplifying Difference Quotients 27

2.2. Simplifying Difference Quotients

By a difference quotient we mean a quantity of the form f(x+h)−f(x)h or f(x)−f(c)

x−c . These kinds of quantitiesarise when studying slopes of secant or tangent lines, and average and instantaneous velocities. In thissection we will focus on simplifying such quantities algebraically.

The four main techniques for simplifying difference quotients are:

1. The “multiply out then cancel and simplify” technique.

2. The “factor then cancel” technique.

3. The “multiply the numerator and denominator by the conjugate of either the numerator or denominatorand then simplify” technique.

4. The “find a common denominator, then combine and simplify” technique.

Some related comments: the first and second items can be thought of as two different sides of the same coin.“Multiplying out” and ”factoring” are inverse operations of each other. The appropriate one to do shouldbe apparent from the context of the situation. The third item involves using the conjugate of an expressionof the form s + t or s− t, especially where either or both of s and t involve square roots of expressions. Theconjugate of s+ t is s− t, and vice versa. For example, the conjugate of

√x + h−5 is

√x + h+5. The fourth

item is especially useful when either the numerator or denominator of the difference quotient is a differenceof quotients itself. For example, an expression like

1x+h−

1x

x would require the technique in item four.

Examples of each technique follow.

Exercise 2.2.1. Find and simplify f(x+h)−f(x)h for f(x) = 3x2 + 2x + 3.

Exercise 2.2.2. Find and simplify f(x)−f(3)x−3 for f(x) = x3 − x2 − 5x.

Exercise 2.2.3. Find and simplify f(x+h)−f(x)h for f(x) =

√x− 2 + 3.

Exercise 2.2.4. Find and simplify f(x)−f(4)x−4 for f(x) = 1√

x.

Section 2.3: Limits (Intuitive Approach) 28

2.3. Limits (Intuitive Approach)

We would like to understand this idea of “shrinking h → 0” more formally. To this end, we will define theimportant notion of a limit. We first give the notation, and then discuss the idea informally and intuitively.A more formal approach will follow in a later section.

We will writelimx→c

f(x) = L

and will read it as “the limit as x approaches c of f(x) is L.”The (informal) meaning of this is as follows:

When we write

limx→c

f(x) = L

we mean that when x is very close to (but not equal to) c, thecorresponding value of f(x) is very close to L.

Example: Since when x is close to 5, f(x) = 2x−1 is close to 9, it would be correct to write limx→5 2x−1 =9. In a previous problem, we showed that the slope of the tangent line to y =

√x + 3 − 4 at x = 6 is 1

6 .Thus it would be correct to write limh→0

f(6+h)−f(6)h = 1

6 .

Exercise 2.3.1. Evaluate limx→9

√x−3

x−9 .

Exercise 2.3.2. Using the diagram below, identify limx→0 f(x) and limx→2 f(x).


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Exercise 2.3.3. Using the following diagram, identify limx→−1 f(x).

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In the second exercise above, it seems that it might be useful to think about when x is close to 2, but onspecifically one side or the other of 2. To this end, we define the one-sided limits.

When we write

limx→c+

f(x) = L

we mean that when x is very close to (but a little bigger than) c,the corresponding value of f(x) is very close to L.

When we write

limx→c−

f(x) = L

we mean that when x is very close to (but a little smaller than) c,the corresponding value of f(x) is very close to L.

Thus in that exercise, even though limx→2 f(x) didn’t exist, we could correctly say that limx→2+ f(x) = 2and limx→2− f(x) = 1

2

The following fact should be evident:

In order for a limit to exist, the two corresponding one-sidedlimits must both exist and be equal. If either of the two one-sided

limits fail to exist, the total limit will also fail to exist.

Note that in the following diagram, limx→0 fails to exist, because there is no particular number that x

is getting close to as x → 0. We will write limx→0 f(x) = ∞, but that notation will simply mean that thelimit doesn’t exist, because as x → 0, the function increases without bound.


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Exercise 2.3.4. Does limx→3x+3x−3 exist?

Section 2.4: Intervals via Absolute Values and Inequalities 32

2.4. Intervals via Absolute Values and Inequalities

Before we embark on an attempt to understand the rigorous definition of limits, we must first make surewe are comfortable with the relationship between certain inequalities involving absolute values and thecorresponding intervals they describe. The key fact which is necessary to understand what follows is that

|a− b| represents the distance from a to b.

Because of this fact, we can think (for example) of |x − 2| as the distance from x to 2, and the set ofpoints with |x− 2| < 1 as the set of points x so that the distance from x to 2 is less than one. Thus this setrepresents the interval (1, 3). In general, we have |x− a| < δ is the set of points x so that the distance fromx to a is less than δ, or in other words, the interval (a− δ, a + δ).

If instead we were considering the set 0 < |x − a| < δ, we would be considering the union of intervals(a− δ, a) ∪ (a, a + δ), which doesn’t include a, since we are thinking about points whose distance from a isstrictly bigger than zero but less than δ.

Exercise 2.4.1. Describe the set of points 0 < |x− 3| < .04.

Exercise 2.4.2. Consider the set of points (2.4, 2.6), and write it in the form |x− a| < δ for some choice ofa and δ.

Exercise 2.4.3. It turns out that for f(x) = 2x− 1, it is true that |f(x)− 3| < .1 whenever |x− 2| < .05.Can you show that this is true by starting with the inequality |f(x)− 3| < .1 and working backwards to seethat |x− 2| < .05?

Exercise 2.4.4. Building on the last problem, suppose you were told that whenever |f(x) − 3| < ε thatthere was then a corresponding number δ so that |f(x)− 3| < ε whenever |x− 2| < δ. Mimic the solution tothe last problem in order to find the value of δ in terms of ε. Here again f(x) = 2x− 1.

Section 2.5: If – Then Statements 33

2.5. If – Then Statements

An example of an if..then statement is “If I do not understand this review section, then I will read it again.”An if..then statement can be phrased in a variety of ways without changing its meaning. A few examples

of how our first example can be rephrased are as follows:“My not understanding this review section implies that I will read it again.”“My not understanding this review section is a necessary condition for my reading it again.”“I do not understand this review section only if I will read it again.”“I will read this review section again if I don’t understand it.”An if..then statement consists of two parts, the “if” part called the hypothesis and the “then” part called

the conclusion. Each of these parts have a truth value. The truth values of the hypothesis and the conclusiondetermine the truth value of the if..then statement. The if..then statement is false only when the hypothesistrue but the conclusion is false.

Truth tables are a convenient way to determine the truth value of a logical statement according to

p q p → q

T T TT F FF T TF F T

When we proof a theorem in class, we are simply showing that the if..then statement is true. To dothis we will assume the “if” part is true and attempt to show the “then” part is also true. Since an if..thenstatement is false only when the hypothesis is true and the conclusion is false, we only need to show thatthe conclusion is always true when the hypothesis is true.

Exercise 2.5.1. Make a truth table for the q → p and compare it with p → q. Are they the same ordifferent?

Exercise 2.5.2. Compare the statement p → q and q̃ → p̃. Here p̃ stands for the negation of p. It has theopposite truth value of that of p.

Exercise 2.5.3. How do the statements “p whenever q” and “p only if q” compare with “if p then q.” Alsowhere does “p if and only if q” fit in?

Section 2.6: Rigorous Definition of Limit 34

2.6. Rigorous Definition of Limit

We want to make rigorous the idea of a limit. We want to say that limx→2 2x−1 = 3 has a specific meaning,more specific than “when x is close to 2, f(x) is close to 3.” The main idea is the following: no matter howclose f(x) is required to be to 3, we can assure that it is that close by requiring x to be within a certaindistance of 2. Suppose we are required to make f(x) be within .01 of the value 3. We would be happy ifthere were a number δ so that when x is within δ units of 2 (i.e. |x− 2| < δ), that it would then follow thatf(x) is within .01 unit of 3. Or more generally, for any real number ε > 0, if we are required to have f(x) bewithin ε units of 3 (i.e. |f(x) = 3| < ε), we would be satisfied if there was a real number δ (which would ofcourse depend on ε), so that as long as |x− 2| < δ, it would then follow that |f(x)− 3| < ε. Now recall thatfor limits, we don’t really require f(x) to be near the limit when x is equal to 2, we only require it when x isclose to be not equal to 2. Thus we only really want to require that |f(x)− 3| < ε whenever 0 < |x− 2| < δ.If in fact the inequality involving f(x) is satisfied when x = 2 so be it, but it isn’t a requirement for thelimit to exist.

To formalize all of this:

To say that

limx→c

f(x) = L

means that for every ε > 0 there exists δ > 0 so that if0 < |x− c | < δ, then | f(x)− L | < ε.

Exercise 2.6.1. Prove rigorously that limx→2 2x− 1 = 3.

Exercise 2.6.2. Prove rigorously that limx→1 4x + 2 = 6.

Section 2.7: Continuity 35

2.7. Continuity

2.7.1. Definitions Related to Continuity

Graphs of functions that have no holes, gaps or jumps in them seem to have especially nice properties. Wewould like to define a concept that would encapsulte these desirable properties. We say that such a functionis continuous at a point a if the value of f(a) is what we think it should be, based on what the function nearthe point is. Thus we say f(x) is continuous when the limit as we approach the point is equal to the valueof the function at the point.

We say that a function f(x) is continuous at a point x = a if thefollowing 3 things are true:

1. f(a) exists (i.e., the function is defined at a, or to put itanother way, a is in the domain of f(x).)

2. limx→a f(x) exists (i.e., the limit of f(x) exists at the pointin question.)

3. The two previously mentioned quantities are equal. That is

limx→a

f(x) = f(a).

Informally, we are calling a function continuous at a point if there are no holes, gaps, or jumps at thatpoint. If a function fails to be continuous at a point, it might be because the function isn’t defined at thatpoint, or because the the limit doesn’t exist at that point, or because these two things do exist but aren’tequal.

If a function is defined at a given point a, and if limx→a− f(x) = f(a), then we say the function iscontinuous from the left or left continuous. Likewise, if limx→a+ f(x) = f(a) then we say the function iscontinuous from the right or right continuous. Of course, in order to be continuous at a point a, the functionmust be both right continuous and left continuous at a that point.

If a function is continuous at every point in an interval (a, b) we say it is continuous on the interval (a, b).If it is also right continuous at a and left continuous at b, we say it is continuous on [a, b]. You should beable to guess what it means to be continuous on other intervals, like (−∞, 3) and [4,∞).

2.7.2. Basic Theorems

Using the limit theorems, it isn’t hard to show that if f and g are continuous at a point, then f + g, f − g,and fg are continuous at that point. The same is true for f

g as long as g(a) 6= 0. Also for compositions,


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Figure 2.1: This function is continuous from the left at x = 2, but is not continuous there.


If g(x) is continuous at a and f(x) is continuous at g(a), then f ◦ g is continuous at a. These facts followsimmediately from the various limit theorems. From these facts it is easy to build up a nice catalogue ofcontinuous functions. In particular, note the following:

Polynomials are continuous everywhere and rational functions arecontinuous wherever they are defined. (These follow directly fromthe plug-in theorem.) Also, the square root function is continuous

on its domain, as are other root functions. The absolute valuefunction is continuous everywhere, but the greatest integer

functions is discontinuous at each integer.

2.7.3. The Intermediate Value Theorem

One example of a consequence of continuity is the IVT. It says that if a function f(x) is continuous on aninterval [a, b], and if M is an intermediate value between f(a) and f(b), that there then exists a number c

between a and b so that f(c) = M . Thus if you are going 30 mph at one moment and 40 mph a little later,you must have been going 35 mph at some point in between. This wouldn’t follow necessarily if your speedweren’t a continuous function of time (which it is).

Exercise 2.7.1. Where is the function h(x) =

√(x+1x2−4

)2

+ 1 continuous?

Exercise 2.7.2. A removable discontinuity is a discontinuity where limx→c f(x) exists, but f(c) doesn’texist. It is called removable because one could define a new function f̃ which is equal to f everywhere exceptat c, and at this point has value equal to limx→c f(x). This new function is now continuous at c, we have“removed” the discontinuity. The function f(x) = x2−16

x−4 has a removable disconinuity at x = 4. Whatshould f̃(4) be defined to be in order to remove the discontinuity?


Chapter 3

The Derivative

Contents

3.1 Definition of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.1.1 Slopes and Velocities Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.1.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.1.3 The Derivative as a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2 Rules for Taking Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.3 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.4 Applications of Derivatives as Rates of Change . . . . . . . . . . . . . . . . . . 47

3.4.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4.2 Some Specific Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.7 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.7.1 The Second Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.7.2 Third and Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.8 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.9 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.9.1 Mathematicizing the main idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

39

Section 3.1: Definition of Derivative 40

3.1. Definition of Derivative

3.1.1. Slopes and Velocities Revisited

Recall that the expressionf(a + h)− f(a)

hrepresents the slope of a secant line between a and a + h, and that this slope approximates the slope of thetangent line at x = a, and this approximation gets better as h → 0. Thus it make sense to define the slopeof the tangent line as

limh→0

f(a + h)− f(a)h

.

By making the substition h = x− a, we can rewrite a + h as x. Also, as h → 0 we have x → a. Thus analternative but equivalent expression for the slope of the tangent line is

limx→a

f(x)− f(a)x− a

.

As discussed before, these quantities also represent the instantaneous rate of change of the function f

with respect to x.

Exercise 3.1.1. Find the equation of the tangent line at x = 9 for f(x) = x +√

x.

Exercise 3.1.2. Find the instantaneous rate of change of y = 3x2 − 5x + 3 at x = 3.

3.1.2. The Derivative

The quantity limh→0f(a+h)−f(a)

h is so important, it deserves its own name. We will call it the derivative off(x) at x = a. We will use the symbol f ′(a) to reprsent this quantity. Thus we have

The derivative of f(x) at x = a is denoted f ′(a) and is given by

f ′(a) = limh→0

f(a + h)− f(a)h

or

f ′(a) = limx→a

f(x)− f(a)x− a

.

Exercise 3.1.3. For f(x) =√

x + 2, find f ′(2).

Section 3.1: Definition of Derivative 41

3.1.3. The Derivative as a Function

There will be times when we want to think of the slope of the tangent line to the function f(x) at a generalpoint x. So it would be reasonable to define f ′(x) = limh→0

f(x+h)−f(x)h , or f ′(x) = limz→x

f(z)−f(x)z−x .

Exercise 3.1.4. For f(x) = 1x , find f ′(x).

Section 3.2: Rules for Taking Derivatives 42

3.2. Rules for Taking Derivatives

It turns out that there are a number of theorems which make the process of taking derivatives much easier.We will develop these now.

The first rule is that the derivative of a constant function f(x) = c is always zero. This is because forsuch a function, limh→0

f(x+h)−f(x)h = limh→0

c−c0 = limh→0

0h = 0.

The second rule involves the functions of the form f(x) = mx, whose graphs are straight lines throughthe origin. The derivative of such functions turns out to be the constant m, because limh→0

f(x+h)−f(x)h =

limh→0mx+mh−(mx)

h = limh→0mhh = limh→0 m = m.

The third rule shows us how to differentiate a power function of the form f(x) = xk. Before investigatingthis rule, we need to review the binomial theorem, which tells us how to expand an expression like (x + h)k.Actually, the only thing we need to know here is that

(x + h)k = c0xk + c1x

k−1h + c2xk−2h2 + c3x

k−3h3 + . . . + ck−1xhk−1 + ckhk

where the ci’s are constants. You may recall that there is a nice formula for the constants ci which involveseither Pascal’s triangle and factorials, but for our purposes, we only need to know the first two constants c0

and c1. It turns out that c0 = 1 and c1 = k. Thus we have

(x + h)k = xk + kxk−1h + c2xk−2h2 + c3x

k−3h3 + . . . + ck−1xhk−1 + ckhk.

Thus we can write f ′(x) = limh→0(x+h)k−xk

h as

f ′(x) = limh→0

xk + kxk−1h + some stuff with a factor of at least h2 − xk

h

= limh→0

kxk−1h + some stuff with a factor of at least h2

h

= limh→0

kxk−1 + some stuff with a factor of at least h

= kxk−1

We will refer to this theorem as the power rule.

Next we investigate the derivative of a funtion f(x) which has the form cg(x) for some constant c andsome function g(x). We note that limh→0

cg(x+h)−cg(x)h = c limh→0

g(x+h)−g(x)h , so that f ′(x) = cg′(x). We

will refer to this theorem as the constant multiple rule. Summarizing our work so far we have:

The derivative of a function f(x) = cxk is f ′(x) = ckxk−1. Thisalso holds in the case where k = 0 and k = 1

(where we interpret x0 as 1.)


How about a function f(x) which is the sum of two functions m(x) and n(x)? So the question is iff(x) = m(x) + n(x), what is f ′(x)? The answer (not surprisingly) turns out to be m′(x) + n′(x), sincelimh→0

m(x+h)+n(x+h)−(m(x)+n(x))h = limh→0

m(x+h)−m(x)h + n(x+h)−n(x)

h which of course can be written aslimh→0

m(x+h)−m(x)h +limh→0

n(x+h)−n(x)h which we recognize as m′(x)+n′(x). We will refer to this theorem

as the sum rule.

Note that using the rules so far, we can easily take the derivative of any polynomial. For example, todifferentiate f(x) = 4x5 + 2x2 − 4x + 2, we can (by the sum rule) think of the four terms 4x5, 2x2, −4x

and 2 separately. They have derivatives 20x4, 4x, −4 and 0 respectively, so (putting these back together)we have f ′(x) = 20x4 + 4x− 4.

Exercise 3.2.1. Find f ′(x) for f(x) = −3x6 + 2x3 − 12x− 123.

Once we understand the procedure in the last example, we should believe that

Every polynomial is differentiable everywhere.

The two rules which remain are a little less obviously true, and also a little harder to remember. Theserules involve functions which are the product or quotient of two other functions. To get us started, considerthe function f(x) = (2x+3)(3x− 1). By multiplying this out, we have that f(x) = 6x2 +7x− 3. Now usingthe above techniques, we see that f ′(x) = 12x + 7. Note that this result is not equal to what would beobtained by just multiplying the derivatives of 2x+3 and 3x−1 together. Thus whatever is going to becomeour product rule will have to be a little more complicated. It turns out that if f(x) = m(x)n(x), then the rightexpression for f ′(x) is m(x)n′(x)+n(x)m′(x), which probably can be remembered best as “the first functiontimes the derivative of the second plus the second times the derivative of the first” . In our example withm(x) = 2x+3 and n(x) = 3x−1, this expression would be (2x+3)(3)+(3x−1)(2) = 6x+9+6x−2 = 12x−7.The proof of the product rule (for the truly dedicated reader) can be found elsewhere.

Given that the product rule was a little messy and unexpected, you should now expect that the quotientrule will be a doozy. To help us toward the goal of understanding the quotient rule, consider the expressionf(x) = p(x)

q(x) = 2x+33x−1 . We could find f ′(x) by resorting to the definition.

Exercise 3.2.2. Use the definition of derivative to compute f ′(x) for f(x) = 2x+33x−1 .

Now note that the if we were to compute q(x)p′(x)−p(x)q′(x)(q(x))2 we would have (3x−1)(2)−(2x+3)(3)

(3x−1)2 = 6x−2−(6x+9)(3x−1)2 =

−11(3x−1)2 , which is f ′(x) as deduced in the previous exercise. This is (of course) not a proof, but hopefullygives you some evidence towards believing in the following quotient rule:


If f(x) = p(x)q(x) , then f ′(x) = q(x)p′(x)−p(x)q′(x)

(q(x))2 .

Note also that from this theorem follows the fact that all rational functions are differentiable at all pointsin their domain.

We now summarize all of the aforementioned derivative rules.

Derivative Rules.

1. If f(x) = c then f ′(x) = 0. (The constant rule.)

2. If f(x) = x, then f ′(x) = 1.

3. If f(x) = xk then f ′(x) = kxkn−1. (The power rule.)

4. If f(x) = cg(x), then f ′(x) = cg′(x). (The constant multi-plier rule.)

5. If f(x) = m(x)+n(x), then f ′(x) = m′(x)+n′(x). (The sumrule.)

6. If f(x) = m(x)n(x) then f ′(x) = m(x)n′(x) + mh(x)n′(x).(The product rule).

7. If f(x) = p(x)q(x) , then f ′(x) = q(x)p′(x)−p(x)q′(x)

(q(x))2 . (The quotientrule.)

Section 3.3: Differentiability 45

3.3. Differentiability

Recall that the expression

limh→0

f(x + h)− f(x)h

represents the derivative of the function f(x). Also recall that limits sometimes have the disturbing propertyof not existing. Thus there could be points in the domain of f(x) where f ′(x) doesn’t exist. At such a point,we say that f is not differentiable. Also, if a is in the domain of f(x), but f ′(a) doesn’t exist, we say that(a, f(a)) is a singular point of f(x). If f(x) is differentiable at every point in an interval (a, b), we say thatf(x) is differentiable on the interval (a, b).

Here is an example of a singular point: consider the function f(x) = |x| at a = 0. To see that thisis a singular point, we will see that limh→0

|0+h|−|0|h doesn’t exist. Of course, this expression is the same

as limh→0|h|h , and since the absolute value function is defined differently on the two sides of the number

zero, it would behoove us to investigate limh→0−|h|h and limh→0+

|h|h . Now recall that |h| = −h if h < 0, so

limh→0−|h|h = limh→0−

−hh = limh→0− −1 = −1 (since the limit of a constant is that constant.) On the other

hand (sorry, that was a bad pun) we know that when h > 0, |h| = h, so we have limh→0+|h|h = limh→0+

hh =

limh→0+ 1 = 1. Thus the two one-sided limits are different, so the overall limit doesn’t exist. Thus (0, 0) isa singular value for f(x) = |x|.

Exercise 3.3.1. Show that (0, 0) is a singular value for f(x) = 3√

x2.

There is an important theorem which relates the differentiability of f(x) at x = a to the continuity off(x) at x = a. The theorem states that if you know that f(x) is differentiable at x = a, then you know thatit is continuous there as well. The reason is as follows: suppose you know that f(x) is differentiable at x = a

so f ′(a) = limx→af(x)−f(a)

x−a exists. In particular you know that f(a) exists, or else this expression wouldn’tmake sense. Now to show that f(x) is continuous at x = a, we only need to show that limx→a f(x) = f(a),or (equivalently) show that limx→a f(x) − f(a) = 0. But we can perform the following somewhat devioustrick of multiplying by one in a clever way: we can rewrite f(x) − f(a) as (f(x) − f(a)) · x−a

x−a Now takingthe limit as x → a we get limx→a(f(x) − f(a)) · x−a

x−a = limx→af(x)−f(a)

x−a · limx→a(x − a) = f ′(a) · 0 = 0.Thus putting these ideas together, we have limx→a f(x)− f(a) = 0, so limx→a f(x) = f(a), so f(x) must becontinuous at x = a.

Note that for the if – then statment above, the converse doesn’t hold. This is because it isn’t necessarilytrue that if a function is continuous at a point that it must also be differentiable there. In fact, we’ve alreadyseen two such examples. Both the absolute value function and the function f(x) = 3

√x2 are continuous at

(0, 0), but are not differentiable there.To summarize:

Section 3.3: Differentiability 46

If f(x) is differentiable at x = a then it is continuous at x = a.However, it is possible for a function to be continuous at x = a

without being differentiable there.

Section 3.4: Applications of Derivatives as Rates of Change 47

3.4. Applications of Derivatives as Rates of Change

3.4.1. Overview

Recall that one of the motivations toward the definition of the derivative was in the investigating of theconcept of instantaneous velocity as the limit of average velocity as the time interval shrank to zero. Ingeneral, no matter what quantities are represented by y = f(x), we can think of the derivative as a “rate”in the sense that as the quantity x changes, so does the quantity y, and if we write h = (x + h)− x) as ∆x

and write ∆y = f(x + h) − f(x), we can think of ∆y∆x = f(x+h)−f(x)

h as an average rate of change of y withrespect to x, since we have the ratio of a change in y to a change in x over a small interval. The limit of thisquantity as ∆x → 0 (which of course is the derivative) can be written lim∆x→0

∆y∆x , can be thought of as the

limit of the average rate of change y with respect to x as the x interval shrinks, so we can think of this asan instantaneous rate of change of y with respect to x. To make our derivative look more like the limit of achange in y over a change in x, we will sometimes write f ′(x) as dy

dx .Thus

f ′(x) = lim∆x→0

∆y

∆x=

dy

dx

is the derivative of f(x) with respect to x, and can be thought ofas the instantaneous rate of change of y = f(x) with respect to x.

3.4.2. Some Specific Applications

• Rectilinear Motion

Suppose an object is moving in a straight line (either vertically or horizontally) and its position at time t

is given by s(t). he theory developed so far indicates that the derivative s′(t) should give us the objectsvelocity at time t, since the instantaneous rate of change of position with respect to time is velocity. Thus,we will write v(t) = s′(t). Also, since the rate of change of velocity with respect to time is commonlycalled acceleration, we can write v′(t) = a(t). Now since F = ma where F is force and m is the massof an object, we can think of the acceleration at time t as determining the force acting on our object. Ifvelocity were zero at a given time, we would know that the object was stopped at that time. Thus, nonzerovelocity indicates movement; we will consider positive velocity to indicate movement to the right (if our lineis oriented horizontally) or upward movement (if our line is oriented vertically.) Likewise, positive or negativeacceleration will be thought of as acting in those directions. When an object’s velocity and acceleration bothhave the same sign, then the object will speed up. Note that this might occur when they are both positive orboth negative (as in the case of a ball thrown downward from a high tower, both its velocity and acceleration


are negative, so the ball speeds up.) When velocity and acceleration have opposite signs, the object will beslowing down.

If an object’s position at time t is given by s(t), thenv(t) = s′(t) = ds

dt represents the objects velocity at time t. Also,v′(t) = a(t) represents the object’s acceleration at time t.

Exercise 3.4.1. An object is moving along a straight line so that its position at time t is given by s(t) =2t3 − 24t + 2. Analyze the movement of this object.

• Biology Applications

Many biologists are interested in the growth of populations, of humans, animals, cells, etc.. If p(t) is the sizeof a given population at time t, then p′(t) is the population growth rate at time t.

Exercise 3.4.2. A population of bacterial cells in a petri dish is growing so that at time t, there are√t(t2 + 3t) cells present.

1. What is the change in the size of population between time t = 4 and t = 9?

2. What is the instantaneous growth rate of the population at time t = 4?

• Chemistry Applications

Chemists are often interested in rates of chemical reactions, as well as the relationship between variousphysical quantities of various types of matter. For example, one form of Boyle’s law states that the absolutetemperature T , pressure P and volume V of an ideal gas of fixed mass m are related by PV = mRT whereR is constant.

Exercise 3.4.3. In the above equation for Boyle’s law, suppose you were able to hold temperature T

constant. What would happen to pressure P as volume V increased? What would be the instantaneous rateof change of pressure with respect to volume?

• Economics

Suppose C(x) represents the cost of producing x units of a certain good. The quantity C(n + 1) − C(n)would represent the cost of producing the (n + 1)st unit after already producing n units. This item issometimes referred to as the marginal cost of producing the (n + 1)st item. It can be approximated by the


derivative, since C(n+1)−C(n)1 ≈ limh→0

C(n+h)−C(n)h . Thus economists often refer to C ′(x) as the marginal

cost function.

Exercise 3.4.4. Suppose that the cost of producing x units of a certain good is 2x+.004x2+.008x3

2000 . Find theactual cost of producing the 101st unit, and compare with C ′(100).

Section 3.5: The Chain Rule 50

3.5. The Chain Rule

We need one more shortcut rule for taking derivatives. The question is: If the function f(x) is really theresult of composing functions m(x) and n(x), what is the derivative of f in terms of information about m

and n? For example, how can we take the derivative of (2x+3)2 without expanding the expression first. Thiswould be useful to know, because we wouldn’t want to have to expand something like (2x+3)100 in order todifferentiate it! Going back to the example of f(x) = (2x+3)2, if we do expand it we get f(x) = 4x2+12x+9,so f ′(x) = 8x + 12. Note that this is not the same thing as 2(2x + 3)1, so that can’t be the right thing. Butusing the fact that a derivative is almost a quotient, how about if we write ∆m

∆x = ∆m∆n · ∆n

∆x . (Here we arepretending that ∆n 6= 0, so that this makes sense.) Now by shrinking appropriate quantities to zero, onemight hope that we would have

f ′(x) =dm

dx=

dm

dn· dn

dx.

Let’s see if this works for the case f(x) = (2x + 3)2. We can write f(x) = m(n(x)) if m(x) = x2 andn(x) = 2x + 3. Then dm

dn would be the derivative of the squaring function at (2x + 3), so we would get2(2x + 3)1 for this portion of the result. Now dn

dx would be n′(x) which would be the constant 2. Thus thetotal derivative f ′(x) would be 2(2x+3)1 ·2 which would be 8x+12, which is the result desired as we noticedabove.

To summarize:

If f(x) = m(n(x)), then dfdx = dm

dn · dndx , or in other notation,

f ′(x) = m′(n(x)) · n′(x).

Exercise 3.5.1. Find f ′(x) for f(x) =√

12x5 − 3x2 + 4x + 3.

Exercise 3.5.2. Find the derivative of√

x +√

x.

Exercise 3.5.3. By using the chain rule twice in sucession, find a formula for the derivative of a functionof the form f(x) = r(s(t(x))). (This could be called the triple chain rule.)

Section 3.6: Implicit Differentiation 51

3.6. Implicit Differentiation

Consider the function f(x)2. The chain rule tells us that the derivative of this (with respect to x) is2f(x)1f ′(x). Suppose we were to call the function f(x) by the alternative name y, and write y2 for theexpression f(x)2. Then the derivative would have to be 2yy′ or 2y dy

dx , again by the chain rule. If you werethinking that the derivative of y2 should be simply 2y, you were probably thinking of the derivative of theexpression y2 with respect to y, which would indeed be simply 2y. Thus we see that when differentiating, wemust always be aware of which quantity represents the variable we are differentiating with respect to, andwhich quantity or quantities are actually functions of that variable.

This comes into play when considering an equation like the circle x2 + y2 = 1. Of course, we could solvethis for y in terms of x (and consider the top half and the bottom half of the circle separately), but it maybe more convenient to deal with this equation in its standard form. Now even though the circle doesn’trepresent y as a function of x (since it fails the vertical line test), it is still the case that there is a tangentline at every point on the circle and all but two of those points have non-vertical tangent lines. Thus, weshould be able to compute the slope of the tangent line dy

dx at all points except the two with vertical tangents.When dealing with a function which is not in the usual y = f(x) form with y written explicitely in termsof the variable x, we say we have an implicit function. The process of finding the derivative of a function(without resorting to solving for y explicitely in terms of x) is called implicit differentiation. This is carriedout by simply differentiating both sides of the equation with respect to x, and recalling (as above) that weare to think of y as a function of x. Thus, for example, in the case of the implicit equation x2 + y2 = 1,if we differentiate both sides of the equation with respect to x, we would obtain 2x + 2y dy

dx = 0, since thederivative of x2 is 2x, the derivative of y2 with respect to x is 2y dy

dx (by the chain rule) and the derivativeof the constant 1 is 0. This equation can now be solved for dy

dx to obtain dydx = −x

y . Thus the slope of the

tangent line at a point like (0, 1) is 0, while at a point like (√

22 ,

√2

2 ), the slope is −1. We can even see thatthe points on the circle where the tangent line has undefined slope are the points where y = 0, namely (1, 0)and (−1, 0).

Exercise 3.6.1. Consider the curve y = 3x , but for practice with implicit differentiation, write it as xy = 3.

Find dydx by differentiating both sides with respect to x. Be sure to use the product rule on the left hand side

of the equation. When finished, check your answer by finding dydx for the explicit function y = 3

x .

When differentiating a reasonably complex implicit function, the result often involves several terms, somewhich have factors of dy

dx and some of which don’t, and these factors could be on either side of the equalsign. To solve such an equation for dy

dx , first identify which terms have factors of dydx , and which ones don’t.

Gather all the terms which do have a dydx factor on one side of the equal sign, and put all other terms on the

other side. Then factor out the dydx factor, and finish solving by dividing both sides of the equation by the

appropriate factor. For example, consider the function x3y2 + 4x + 5y2 − 2 = xy. Differentiating both sideswith respect to x yields

x3 · 2ydy

dx+ y2 · 3x2 + 4 + 10y

dy

dx− 0 = x

dy

dx+ y(1).

Section 3.6: Implicit Differentiation 52

Analyzing this equation, we see that there are 4 non-zero terms on the left hand side, and 2 terms on theright hand side. Of these terms, there are two terms on the left hand side which have a dy

dx factor, and oneterm on the right hand side which has one. Gathering all the terms with a dy

dx factor on the left (and theothers on the right) we obtain

2x3ydy

dx+ 10y

dy

dx− x

dy

dx= y − 3x2y2 − 4.

Factoring out the dydx factor yields

dy

dx

(2x3y + 10y − x

)= y − 3x2y2 − 4,

sody

dx=

y − 3x2y2 − 42x3y + 10y − x

.

Exercise 3.6.2. Suppose√

xy = x + y + 1. Find dydx .

Section 3.7: Higher Derivatives 53

3.7. Higher Derivatives

3.7.1. The Second Derivative

Since the derivative of a function is another function, it makes sense to think about differentiating thederivative to get another function, which we will call the second derivative of the original function. We willwrite f ′′(x) or d2y

dx2 for the second derivative of y = f(x) with respect to x. Because of our previous dealingwith the derivative, we should expect that there is both a geometric significance to the second derivative(related to the graph of the function) and a physical intepretation (related to the rate of change of thefunction.) We now explore the latter, leaving the discussion of the significance of the second derivative inrelation to geometry for a later chapter.

Suppose y = f(t) represents the position of an object at time t. We have already seen that f ′(t)represents the instantaneous rate of change of f (position) with respect to t (time.) Thus f ′(t) representsvelocity. Likewise, f ′′(t) should represent the instantaneous rate of change of f ′ (velocity) with respect to t

(time.) The rate of change of velcocity with respect to time is what we generally call acceleration. Thus wecan think of f ′′(t) = a(t) as the acceleration of the object at time t.

3.7.2. Third and Higher Derivatives

Of course, we can differentiate f ′′(x) to obtain f ′′′(x) and so on. Since it is impractical to write f ′′′′′′′′′′′′′′′′(x)for the sixteenth derivative of f with respect to x, we instead write f (16)(x). Of course, this could lead toconfusion with the expression (f(x))16, but hopefully we can use the context to “differentiate” between thetwo. (Sorry, bad pun.) In Leibniz notation, we write dny

dxn to stand for the nth derivative of y with respectto x.

Exercise 3.7.1. Explain why the nth derivative of a degree k polynomial is always the constant zero, ifn > k.

Exercise 3.7.2. Find the first few derivatives of f(x) = 1x2 . See if you can find a pattern, so that you can

write down a general formula for dkfdxk . Hint: the number (1) · (2) · (3) · . . . · (n) is sometimes written as n!.

This is read “n factorial.” For example, 8! = 8 · 7 · 6 · . . . · 1 = 40320.

Section 3.8: Related Rates 54

3.8. Related Rates

One application of implicit differentiation and the chain rule is in the solution to problems like the following:Suppose two cars leave an intersection at the same time, with one car headed north and the other car headedeast, with the northbound car travelling at 50 mph and the eastbound car travelling at 30 mph. How far arethey separating after 30 minutes? We can solve such a problem by trying to find a relationship between therate at which the two cars are separating and the rates they are going. To this end, we could let x representthe distance the northbound car has gone after t hours, and y the distance the eastbound car has gone aftert hours. Note that these are both functions of time t. Also, we can related the distance between the cars bythe pythagorean theorem: x2 + y2 = z2, where z is the distance between the cars at time t. Differentiatingimplicitely with respect to t yields 2xdx

dt + 2y dydt = 2z dz

dt , and solving for dzdt yields dz

dt = x dxdt +y dy

dt

z . Nowdxdt = 50 and dy

dt = 30 and when t = .5, x = 25 and y = 15 and thus (using the pythagorean theorem)z =

√252 + 152 =

√850. Thus at the moment in question, dz

dt = 25·50+15·30√850

.

Exercise 3.8.1. A right circular cylinder with radius 2 feet and length 10 feet is being filled with water ata rate of 3 cubic feet per minute. At what rate is the height of the water in the cylinder increasing?

Exercise 3.8.2. A swimming pool which is 3 feet deep at the shallow end and 8 feet deep at the deep endis being filled with water at the rate of 10 cubic feet per minute. The pool is 25 feet long and 10 feet wide,and the shallow end descends along a straight line to the deep end. How fast is the water level rising whenthe water level (measure in the deep end) is 3 feet?

Here are some suggestions which you might find useful when trying to solve related rates problems.

Section 3.8: Related Rates 55

Steps for Solving Related Rates Problems

1. Draw a picture if appropriate.

2. Label things in the picture which are functions of time witha letter, and label things that are constants with the appro-priate number.

3. Write down the rates which you know and the rate you wantto know. This helps keep you focused on the task at hand.

4. Write down an equation which relates the variables which isvalid for all time t.

5. Differentiate both sides of the equation with respect to t,using the chain rule.

6. After differentiating, substitute the values of the quantitiesthat are known at the desired time.

7. Solve for the desired rate.

Section 3.9: Linearization 56

3.9. Linearization

Near the point of tangency, a given curve and its tangent line look quite similar. This can be seen using the“zoom” feature on a graphing calculator or computer on which a curve and a tangent line to the curve havebeen graphed together. If we zoom in on the point of tangency, the curve and the tangent line are hard todistinguish from each other. Thus we could say that the tangent line is a good approximation to the curvenear the point of tangency. In fact, it can be proven that the tangent line is the best linear approximationto the curve which goes through the given point.

3.9.1. Mathematicizing the main idea

Recall that the tangent line to y = f(x) at the point x = a is given by

y − f(a) = f ′(a)(x− a).

Now instead of calling this line “y”, let’s call it L(x). Thus we have

L(x) = f(a) + f ′(a)(x− a).

And the above main idea says that when x is close to a,

f(x) ≈ f(a) + f ′(a)(x− a).

The quantity (x − a) in the above discussion is sometimes called ∆x or dx, since it represents a smallchange in x when x is near a. The quantity (f(x) − f(a)) in the above discussion is sometimes called ∆y.It represents the corresponding change in y for the function f(x). The quantity f ′(a)(x − a) = f ′(a)dx issometimes called dy. This represents the corresponding change in y when looking at the tangent line ratherthan the curve. Thus for a given point a near x, dx is determined, and this produces ∆y (which is a differencebetween y values on the curve,) and dy (which is a difference between y values on the tangent line.) Oneway of re-expressing the main idea can then be rewritten as

∆y ≈ dy

because this is equivalent tof(x)− f(a) ≈ f ′(a)dx.

Exercise 3.9.1. Show how to use the linear approximation of the square root function near the point (36, 6)to approximate

√36.05 without the use of a calculator.

Exercise 3.9.2. Draw a sketch of y = x2, and draw the tangent line at x = 1. Using a value of dx = .05,find dx, dy and ∆y, and draw them at appropriate places on the sketch.

Chapter 4

An Interlude – Trigonometric

Functions

Contents

4.1 Definition of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . 58

4.1.1 Radian Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.2 Trigonometric Functions at Special Values . . . . . . . . . . . . . . . . . . . . . 59

4.3 Properties of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . 60

4.4 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.4.1 Graph of sin(x) and cos(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.4.2 Graph of tan(x) and cot(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.4.3 Graph of sec(x) and csc(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.5 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.6 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . 63

4.6.1 Two Important Trigonometric Limits . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.6.2 Derivatives of other Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 64

57

Section 4.1: Definition of Trigonometric Functions 58

4.1. Definition of Trigonometric Functions

You may at some point in your mathematical career have defined the trigonometric functions using ratios ofsides of a right triangle. Under this definition, you will recall that the sine of an angle θ in a right triangle isthe ratio of the side opposite θ to the hypotenuse. The other trigonometric ratios are defined in box below.

If a is the length of the side opposite θ in a right triangle, and b

is the length of the side adjacent to θ, and c is the length of thehypotenuse, then

sin(θ) = ac

cos(θ) = bc

tan(θ) = ab

cot(θ) = ba

sec(θ) = cb

csc(θ) = ca

Although these definitions are accurate, there is a sense in which they are lacking, because the angleθ in a right triangle can only have a measure between 0◦ and 90◦. We need a definition which will allowthe domain of the sine function to be the set of all real numbers. Our definition will make use of the unitcircle, x2 + y2 = 1. We first associate every real number t with a point on the unit circle. This is done by“wrapping” the real line around the circle so that the number zero on the real line gets associated with thepoint (0, 1) on the circle. The positive portion of the number line is wrapped counterclockwise around thecircle, while the negative portion is wrapped counterclockwise. Another way of describing this association isto say that for a given t, if t > 0 we simply start at the point (0, 1) and move our pencil counterclockwisearound the circle until the tip has moved t units. The point we stop at is the point associated with thenumber t. If t < 0, we do the same thing except we move clockwise. If t = 0, we simply put our pencil on(0, 1) and don’t move. Using this association, we can now define cos(t) and sin(t).

Using the above association of t with a point (x(t), y(t)) on theunit circle, we define cos(t) to be the function x(t), and sin(t) to

be the function y(t), that is, we define cos(t) to be the x

coordinate of the point on the unit circle obtained in the aboveassociation, and define sin(t) to be the y coordinate of the point

on the unit circle obtained in the above assocation.

Section 4.2: Trigonometric Functions at Special Values 59

Using the above definition of sin(t) and cos(t), we can define

tan(t) = sin(t)cos(t)

cot(t) = cos(t)sin(t)

sec(t) = 1cos(t)

csc(t) = 1sin(t)

Exercise 4.1.1. What point on the unit circle corresponds with t = π? What therefore is cos(π) and sin(π).

Exercise 4.1.2. What point on the unit circle correspond with t = 3π2 ? What therefore is cos

(3π2

)and

sin(

3π2

)?

4.1.1. Radian Measure

In calculus, we will use radian measure rather than degree measure. There are numerous reasons for this,most of them based on the fact that radians are more “natural” (even if it may not seem that way to you atfirst.) A radian is defined to be the measure of an angle cut of in the circle of radius one by an arc of lengthone. Thus, a 90◦ angle corresponds to an angle of radian measure π

2 , since the distance one fourth of theway around the unit circle is π

2 . It is also useful to note that an angle of measure 1◦ corresponds with anangle of radian measure π

180 , since 90 of these would correspond to a right angle. Also, an angle of radianmeasure 1 would correspond to an angle of measure

(180π

)◦, since π2 of these would correspond to a right

angle. These facts are enough to help you convert from degrees to radians and back, when necessary.

Exercise 4.1.3. What is the degree measure of the angle θ = π6 ?

Exercise 4.1.4. What is the radian measure of the angle 225◦?

4.2. Trigonometric Functions at Special Values

There are a few special angles for which you should know the values of the trigonometric functions, withouthaving to resort to a table or a calculator. These are summarized in the following table.

θ cos(θ) sin(θ)

0 1 0π6

√3

212

π4

√2

2

√2

2π3

12

√3

2π2 0 1

You should also be able to use reference angles along with these values to compute the values of thetrigonometric functions at related angles in the second, third and fourth quadrants. For example, the point

Section 4.3: Properties of Trigonometric Functions 60

associated with t = 5π6 is directly across the unit circle from the point associated with π

6 . (In this case wesay that we are using π

6 as a reference angle.) Thus the coordinates of the point associated with 5π6 has the

same y value and the opposite x value of the point associated with π6 . Thus cos( 5π

6 ) = − cos(π6 ) = −

√3

2 ,and sin( 5π

6 ) = sin(π6 ) = 1

2 .

Exercise 4.2.1. Evaluate sin( 7π6 ).

Exercise 4.2.2. Evaluate tan(−3π4 ).

Exercise 4.2.3. Solve sin(x) + cos(x) = 0 for x.

Exercise 4.2.4. Solve 2 cos(2x) + 1 = 0 for x.

4.3. Properties of Trigonometric Functions

1. Pythagorean Identity: sin2(x) + cos2(x) = 1.

2. Pythagorean Identity Rephrased: 1+tan2(x) = sec2(x), 1+cot2(x) = csc2(x).

3. Range of sin(x) and cos(x): −1 ≤ sin(x) ≤ 1, −1 ≤ cos(x) ≤ 1.

4. cos(x) is Even: cos(−x) = cos(x).

5. sin(x) is Odd: sin(−x) = − sin(x).

6. Periodicity: sin(x + 2π) = sin(x), cos(x + 2π) = cos(x).

4.4. Graphs of Trigonometric Functions

4.4.1. Graph of sin(x) and cos(x)

Note that from the definition of sine and cosine, it is clear that the domain of each of these is the set of allreal numbers. Also, from the properties above, we know that the range of both of these is the set of numbersbetween −1 and 1, and that the functions are periodic. This information, together with a few points plottedas a guide, are enough to graph the two functions. Note that if we shift the graph of the sine function by π

2

units to the left, we get the graph of the cosine function. This is related to the fact that sin(x− π2 ) = cos(x).

Section 4.5: Trigonometric Identities 61

4.4.2. Graph of tan(x) and cot(x)

Note that the domain of tan(x) is the set of all real numbers except those at which cos(x) = 0. Thus, thepoints π

2 , 3π2 , and so on aren’t in the domain of tan(x). An easy way to characterize these points is to say

that these are all the points which have the form π2 + kπ, where k is any integer. Thus the domain of the

tangent function is {x : x ∈ R, x 6= π2 + kπ, k ∈ Z}.

Exercise 4.4.1. What is the domain of cot(x)?

We can get a good grasp on the graph of tan(x) by plotting a few points and doing a careful analysis ofthe limiting behavior when x is near π

2 and the other points that aren’t in the domain. Note that when x

is a little less than π2 , sin(x) is close to 1, while cos(x) is close to zero (but is positive.) Thus, the ratio of

these two things should be big. Thus limx→π2− tan(x) = +∞. A similar analysis of points to the right of −π

2

helps us believe that limx→−π

2+ tan(x) = −∞. This information together with a few strategically plotted

points (like tan(−π4 = −1 and tan(0) = 0 and tan(π

4 ) = 1) should help us to graph the tangent function onthe interval (−π

2 , π2 ). Now using the fact that the tangent function is periodic with period π, we can graph

the whole tangent function.

Exercise 4.4.2. Do a similar analysis to graph the cotangent function.

4.4.3. Graph of sec(x) and csc(x)

Like the tangent function, the domain of the secant function is the set of all real numbers except thosewhich make cos(x) equal to zero. Thus the domain of the secant function is the same as the domain of thetangent function. Also, the fact that the cosine function always has values between −1 and 1 tells us thatsec(x) = 1

cos(x) always has values less than −1 or greater than 1. An analysis of the limiting behavior ofsec(x) near x = π

2 and −π2 and a few strategically plotted points leads to the graph of y = sec(x).

Exercise 4.4.3. Finish this analysis and draw a graph of y = sec(x). Then do a similar analysis and graphy = csc(x).

4.5. Trigonometric Identities

1. Addition Identity for sin(x): sin(x + y) = sin(x) cos(y) + sin(y) cos(x).

2. Addition Identity for cos(x): cos(x + y) = cos(x) cos(y)− sin(x) sin(y).

From these, we can get other identities like the double angle formulas. For example,

sin(2x) = sin(x + x) = sin(x) cos(x) + sin(x) cos(x) = 2 sin(x).

Section 4.5: Trigonometric Identities 62

Exercise 4.5.1. Show that cos(2x) = cos2(x)− sin2(x) in a manner similar to the above.

We can also show that sin2(x) = 1−cos(2x)2 as follows: since cos(2x) = cos2(x) − sin2(x), we have

1−cos(2x)2 = 1−(cos2(x)−sin2(x))

2 = 1−cos2(x)+sin2(x)2 . Now using the fact that 1 − cos2(x) = sin2(x), we have

that the above expression is equal to sin2(x)+sin2(x)2 = sin2(x).

Exercise 4.5.2. Show that cos2(x) = 1+cos(2x)2 , using methods similar to the above.

Section 4.6: Derivatives of Trigonometric Functions 63

4.6. Derivatives of Trigonometric Functions

If f(x) = sin(x), what is f ′(x)? We have no alternative but to use the definition of derivative. Thus we havethat

f ′(x) = limh→0

sin(x + h)− sin(x)h

= limh→0

sin(x) cos(h)− sin(h) cos(x)− sin(x)h

= limh→0

sin(x) cos(h)− sin(x)− sin(h) cos(x)x

= limh→0

sin(x) (cos(h)− 1)− cos(x) sin(h)h

= limh→0

sin(x)(

(cos(h)− 1)h

)− cos(x)

(sin(h)

h

)= sin(x)

(limh→0

cos(h)− 1h

)− cos(x)

(limh→0

sin(h)h

).

If we denote limh→0cos(h)−1

h by a and limh→0sin(h)

h by b, then we have deduced that the derivative ofsin(x) is a sin(x) + b cos(x). It remains to compute these two limits.

4.6.1. Two Important Trigonometric Limits

We will need to use a little geometry to help us compute limh→0sin(h)

h . Consider the picture below:

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Recall that the area of a sector of a circle of radius r and angle h is A = 12hr2. Now consider the sector

OSQ. It has area 12h cos2(h). On the other hand, triangle OPQ has area 1

2 cos(h) sin(h), and sector OPR

has area 12h. From the picture we can tell that even when h is small,

area OSQ ≤ area OPQ ≤ area OPR

so12h cos2(h) ≤ 1

2cos(h) sin(h) ≤ 1

2h.

If we multiply through by 2 and divide by cos(h) · h, we see that

cos(h) ≤ sin(h)h

≤ 1cos(h)

.

Now as h → 0, cos(h) → 1, so this inequality together with the squeeze theorem tell us that limh→0sin(h)

h =1.

Exercise 4.6.1. Use the now established fact that limh→0sin(h)

h = 1, to show that limh→0cos(h)−1

h = 0.Hint: Try multiplying the numerator and the denominator by the conjugate of the numerator.

4.6.2. Derivatives of other Trigonometric Functions

Knowing that ddx sin(x) = cos(x) and d

dx cos(x) = − sin(x), and knowing the quotient rule is enough to enableone to compute the derivatives of the other trigonometric functions, since they are all quotients involvingthese two trigonometric functions. For example

ddx tan(x) = d

dxsin(x)cos(x)

= cos(x) cos(x)−sin(x)(− sin(x))cos2(x)

= 1cos2(x)

= sec2(x)

Exercise 4.6.2. Compute ddx sec(x), d

dx cot(x), and ddx csc(x) in a manner similar to the previous example.

Of course, it would be a good idea to practice these new trigonometric derivatives in conjuction with outfamilar rules such as the chain rule, quotient rule, product rule, etc..

Exercise 4.6.3. Compute the derivative with respect to x of f(x) = sin(

x2+12x−3

).

Exercise 4.6.4. Compute g′(x) for g(x) = (√

x2 − 4) · tan(√

x2 − 4).


Click to Initialize Quiz Answer the following questions about trigonometric derivatives.

1. What is the derivative of f(x) = tan(x)cos(x) ?

sec2(x)cos(x)

cos(x)sin(2x)

sin2(x)+1cos3(x)

2. What is the derivative of f(x) = tan(cos(x))?sec2(cos(x) tan(− sin(x)) sec2(cos(x)) · − sin(x)

3. If f(x) = (sin(x2 + cos(x2)))2, what is f ′(x)?2 sin(x2 + cos(x2)) 2 sin(x2 + cos(x2)) · (cos(x2 +

− sin(x2))2 sin(x2 + cos(x2)) · (cos(x2 +cos(x2)) · (2x +− sin(x2) · 2x)

4. Find ddx

√cot(2x + 1).√

− csc2(2x + 1) 12 (cot(2x+1)

−12 ·− csc2(2x+1)·2 1

2 (cot(2x + 1)−12 · 1

Click to End Quiz


Chapter 5

Applications of the Derivative

Contents

5.1 Definition of Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . 68

5.2 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.3 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.4 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.5 An Application of the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . 73

5.6 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.7 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

5.8 Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.9 Limits at Infinity; Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . 80

5.9.1 A Simplification Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.10 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5.11 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.12 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

5.13 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.14 Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.15 Antiderivatives Mini-Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

67

Section 5.1: Definition of Maximum and Minimum Values 68

5.1. Definition of Maximum and Minimum Values

An important application of calculus is that it can be used to maximize and minimize functions. One mightbe trying to minimize the cost of producing some item, maximize the area or volume of a geometric region,or minimize the time it takes for a certain biological reaction to occur. In all of these, the theoreticalmathematical basis is the same.

We begin with the definition of what we mean by a maximum and minimum value.

A function f has a maximum value of M at x = c if

f(c) = M, and f(x) ≤ f(c) for allx in the domain of f(x).

Likewise, we say that f has a minimum value of m at x = c if

f(c) = m, and f(x) ≥ f(c) for allx in the domain of f(x).

Exercise 5.1.1. Using what you know about the graph of y = x2, find the absolute minimum of y =(x− 3)2 + 2 and the absolute maximum of y = −3− (x + 2)2.

Sometimes the fact that the maximum is begin considered over the entire domain is emphasized byputting the adjectives global or absolute in front of the words maximum or minimum. Thus one might say“there is a global maximum of 3 at x = 4” or “there is an absolute minimum of 2 at x = 5.” In contrast,you might sometimes want to discuss the fact that a certain function value is larger than those near it, evenif it isn’t the largest in the whole domain. We call such points relative or local maximums or minimums.

We say that there is a relative or local maximum value M at x = c

if there exists an interval I containing c so that

f(c) = M, and f(x) ≤ f(c) for allx in I.

Likewise we say that here is a relative or local minimum value m

at x = c if there exists an interval I containing c so that

f(c) = m, and f(x) ≤ f(c) for allx in I.

Exercise 5.1.2. Consider the function pictured. Where does it have relative extrema? (Note: the wordextrema means “maximum or minimum.”)


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Note that not all functions have maximums or minimums (local or global.) For example, the straight liney = 2x+1 has neither kind of maximum or minimum over its entire domain. Even if you restrict the domainto be a finite interval like (2, 5), there is still no maximum or minimum value. If we restrict the domain tobe a closed interval like [2, 5], this example does have a maximum of 11 at x = 5 and a minimum of 5 atx = 2, but a function like

g(x) =

x + 2 if 2 ≤ x ≤ 3

−20x + 80 if 3 < x < 4

x + 2 if 4 ≤ x ≤ 5

has no maximum or minimum on [2, 5].

Exercise 5.1.3. Explain why f(x) = 2x + 1 has no maximum or minimum on (2, 5). Then explain why thefunction g(x) defined in the previous paragraph has no maximum or minimum on [2, 5].

Of course, the fact that g(x) above has no maximum or minimum on [2, 5] is related to the fact that itis not continuous. What turns out to be true (but is not obvious and is actually quite difficult to prove) isthat if we restrict ourselves to continuous functions defined on closed intervals, we can be sure we have amaximum and a minimum. This is called the extreme value theorem.

The Extreme Value TheoremA continuous function defined on a closed interval always has amaximum and a minimum value on that interval.

The examples above show the necessity of the hypotheses: if the interval is open, the maximum orminimum might not exist. If the function is not continuous, the maximum or minimum might not exist.


Of course, the question still remains: how can one find the value of x which gives a maximum or aminimum? Here are some examples that you are already familiar with: the function f(x) = x2 has aminimum at x = 0. Of course, the interesting thing about this function is that the tangent line at x = 0 is ahorizontal line. In other words, the slope of the tangent line at this point is zero. Thus we see that extremacan occur at points where the derivative is zero. We will call such points stationary points. Another exampleis h(x) = |x|, which also has a minimum at x = 0. However, this function does not have a horizontal tangentline at x = 0, in fact, there is no tangent line there at all! Recall that h(x) = |x| is not differentiable atx = 0, so we see that extrema can occur at points where the derivative doesn’t exist. We call such pointssingular points. A third example is to reconsider the function f(x) = 2x + 1 on the interval [2, 5]. We havealready seen that the maximum value for this function occurs at the right endpoint, while the minimumoccurs at the left endpoint. Thus we see that extrema can occur at endpoints. What is not obvious (but istrue nonetheless,) is that this is the complete list of types of points where extrema can occur. This fact iscalled “Fermat’s Theorem.”

Fermat’s Theorem:If f(x) has a maximum or a minimum at a point (c, f(c)) wherec is not an endpoint of the domain, then either f ′(c) = 0 or f ′(c)doesn’t exist.

We can apply this theorem as follows: given a continuous function f(x) and a closed interval [a, b], we canfind all singular and stationary points by studying f ′(x) and finding where it is equal to zero or where it isundefined. Since the maximum and minimum must occur at either a stationary point, a singular point, or anendpoint, by comparing the function values at these points, we can determine the maximum and minimumfor the function.

Exercise 5.1.4. Find the absolute maximum and absolute minimum of f(x) = 4x5−5x4 +3 on the interval[−3, 2].


Click to Initialize Quiz Answer the following questions about maximum and minimum values.

1. What are the values of the relative extrema of f(x) = 2x3 + 3x2 − 120x + 30?−5, 4 455,−274 4, 0

2. What is the maximum value of f(x) = |x− 3| on [−1, 5]?There is no maximumvalue.

0 4 2

3. Does the function f(x) = 3x23 + 2 have a minimum on [0, 4]? If so, what is it?

Yes, it is 2. No, there isn’t one. Yes, it is 5.

4. Does f(x) = −x2 + 4x have a maximum on (1, 5)? How about a minimum?There is a maximum but nominimum.

There is neither a maximumnor a minimum.

There is a minimum but nomaximum.

Click to End Quiz

Section 5.2: Overview 72

5.2. Overview

The Mean Value Theorem is an enigma, in that it seems to be a somewhat innocuous statement of fact, butturns out to be a key element in the proof of many important theorems. It is important because it relatesthe derivative (which is not actually a quotient but is rather the limit of a certain quotient) to a particularquotient. As a warm-up, we first investigate a related theorem due to the French mathematician MichelRolle.

5.3. Rolle’s Theorem

Try to draw a function f(x) with the following characteristics: It is defined on the interval [2, 5], it iscontinuous and differentiable on its domain, and it satisfies f(2) = 0 = f(5). No matter how you drawsuch a function, you will find that you there must have been someplace on the domain where there was ahorizontal tangent line. In other words, there must have been a point c between 2 and 5 where f ′(c) = 0.This fact is called Rolle’s Theorem:

Rolle’s TheoremIf f(x) is continous on [a, b] and differentiable on (a, b), and iff(a) = 0 = f(b), then there is a point c with a < c < b andf ′(c) = 0.

The reason Rolle’s Theorem is true is quite simple. Since f(x) is a continuous function defined ona closed interval, the Extreme Value Theorem says that it has a maximum and a minimum on [a, b]. Ifeither the maximum or minimum or the minimum occur at some point c other than the endpoints, thenFermat’s Theorem says that either f ′(c) = 0 or f ′(c) doesn’t exist. But one of our hypotheses is that f(x)is differentiable on [a, b], so f ′(c) must exist, so f ′(c) = 0. If neither the maximum nor the minimum occurat a point other than an endpoint, then f(a) = 0 = f(b) is both the maximum and the minimum. But thismeans that our function is the constant function f(x) = 0, and this function has lots of values where thederivative is zero! (In fact, the derivative is zero at every point on (a, b).)

Exercise 5.3.1. Consider the function f(x) = x2−2x−3x+8 on the interval [−1, 3]. Does Rolle’s theorem apply?

If not, explain why not. If so, find one or more of the promised values of c where f ′(c) = 0.

Exercise 5.3.2. The next two exercises will show you that the hypotheses for Rolle’s theorem are necessary.Consider the function defined by g(x) = x

23 on the interval [−1, 1]. Show that there is no value of c so that

g′(c) = 0 on [−1, 1]. Why does this not violate Rolle’s Theorem?

Section 5.4: The Mean Value Theorem 73

Exercise 5.3.3. Consider the function defined by

g(x) =

0 if x = 0

−2x + 2 if 0 < x ≤ 1.

Does Rolle’s Theorem apply to this function on [0, 1]?

5.4. The Mean Value Theorem

One way to think of Rolle’s Theorem is this: if a function f(x) meets the hypotheses of the theorem, thenthere must be a point c between a and b where the tangent line is parallel to the x axis. The Mean ValueTheorem is a generalization of Rolle’s Theorem. It doesn’t require that f(a) = 0 = f(b), but instead saysthat as long as f(x) is continuous on [a, b] and differentiable on (a, b), there will be a point c between a andb where the tangent line is parallel to the line containing (a, f(a)) and (b, f(b)). Stated algebraically we have

The Mean Value TheoremIf f(x) is continuous on [a, b] and differentiable on (a, b), then thereis a point c so that a < c < b with

f ′(c) =f(b)− (a)

b− a.

Note that you can think of the right hand side of the above equation as representing the slope of the linebetween (a, f(a)) and (b, f(b)), so the above says that there is a point c where the slope of the tangent lineis equal to the slope of the secant line, so the two lines are parallel.

The Mean Value Theorem can be seen to be true using Rolle’s Theorem and a clever trick. Given f(x)which meets the hypotheses of the theorem, consider the function g(x) = f(x) − f(a) − f(b)−f(a)

b−a · (x − a).One can check that this function is continuous on [a, b] and differentiable on (a, b), and g(a) = 0 = g(b).(You should check this yourself!) Thus, g(x) meets the hypotheses of Rolle’s theorem. So there exists c withg′(c) = 0. But g′(c) = f ′(c) − f(b)−f(a)

b−a , so if g′(c) = 0, we get the conclusion of the Mean Value Theorem,that f ′(c) = f(b)−f(a)

b−a . This function g(x) seems to be quite mysterious, it can be derived geometrically,however, by taking the picture for the Mean Value Theorem and trying to make it look like the picture forRolle’s Theorem.

5.5. An Application of the Mean Value Theorem

We claimed above that the Mean Value Theorem is often used to proof other interesting and importanttheorems. We will now see an example of one such theorem. By way of introduction, note that there are

Section 5.5: An Application of the Mean Value Theorem 74

many functions which have derivative equal to 2x. Some of them are y = x2, y = x2 + 4, y = x2 − 123, andso on. The question is: do all functions whose derivative is equal to 2x have the form y = x2 + C where C

is a constant? The answer turns out to be yes, and we will prove it using the Mean Value Theorem.

The “+C” TheoremIf f(x) and h(x) are differentiable functions, and if f ′(x) = h′(x),then f(x) = h(x) + C.

We can see that the “+C” theorem holds by considering g(x) = f(x)− h(x). Note that g′(x) = f ′(x)−h′(x) = 0. We will show that g must be a constant function. Suppose x1 is any point in the domain of g, andg(x1) = k. Let x2 be any other point in the domain of g with x1 6= x2. Consider the quantity g(x2)−g(x1)

x2−x1.

By the Mean Value Theorem, there must be a number c between x1 and x2 with g′(c) = g(x2)−g(x1)x2−x1

. But

g′(x) = 0 for all x, so in particular g′(c) = 0. Thus we have g(x2)−g(x1)x2−x1

= 0, but then g(x2)− g(x1) = 0, sog(x2) = g(x1) = k. We have shown that for any point x2 in the domain of g, the output is the same. Thusg(x) = f(x)− h(x) is a constant function. So f(x) = h(x) + C for some constant C.

Section 5.6: Monotonicity 75

5.6. Monotonicity

We all have an intuitive idea about what it means for a function to “go up” or increase (or to do the opposite:“go down” or decrease.) Let’s formalize this idea with a definition:

We say that a function is increasing on an interval I if for allx1, x2 ∈ I with x1 < x2 we have f(x1) < f(x2). We say that f(x)is decreasing on an interval I if for all x1, x2 ∈ I with x1 < x2 wehave f(x1) > f(x2)

For example, f(x) = x2 is increasing on the interval [0,∞) and decreasing on the interval (−∞, 0]. Afunction is said to be monotonic on an interval I if it is strictly increasing or strictly decreasing on I. Inattempting to understand the geometry of the graph of a particular function f(x), it is desirable to findthe intervals on which the function is monotonic. Of course, it is most interesting to find the largest suchintervals. There is a clear and almost obvious relationship between the derivative of a function and themonotonicity. This can be seen by drawing a number of strictly increasing functions, and drawing tangentlines on any of them at any point, and thinking about what is always true about the slopes of the tangentlines. You should be able to informally convince yourself pretty quickly that they all have positive slopes.Thus there seems to be a relationship between an increasing function and a positive derivative. Likewise,there seems to be a relationship between a decreasing function and a negative derivative. The followingseems plausible:

Monotonicity TheoremIf f(x) is so that f ′(x) > 0 for all x in the interval I, then f(x) isincreasing on I. If f ′(x) < 0 for all x in I, then f(x) is decreasingon I.

The proof of the Monotonicity Theorem uses the Mean Value Theorem. The idea is that if f ′(x) > 0on I, and if x1 < x2 on I, then there is a point c between x1 and x2 with f(x2)−f(x1)

x2−x1= f ′(c). But if

f ′(c) > 0, we must have f(x2)− f(x1) > 0, since the denominator x2− x1 > 0. So f(x1) < f(x2), so f(x) isincreasing. A similar proof works for f(x) decreasing when f ′(x) < 0 on I. This result can be used to findthe intervals of monotonicity for a given function: by finding the largest intervals on which the derivative off(x) is positive, we are also finding the largest intervals on which f(x) is increasing. A similar statementcan be made replacing the word “increasing” by “decreasing” and the word “positive” by “negative.”

Example 5.6.1. Find the largest intervals on which f(x) = xx2+2 is monotonic.

Section 5.6: Monotonicity 76

Solution: Find the largest intervals of monotonicity for f(x) = xx2+2 .

First we need to compute and simplify f ′(x). We have

f ′(x) =(x2 + 2)(1)− (x)(2x)

(x2 + 2)2Quotient Rule

=2− x2

(x2 + 2)2Simplifying

Note that the numerator of f ′(x) is zero when x = ±√

2, and the denominator is never zero. Thus, f ′(x)is either strictly positive or strictly negative on the intervals (−∞,−

√2), (−

√2,√

2), and (√

2,∞). Usingtest values, we see that f ′(x) is positive on (−

√2,√

2) and negative on the other two intervals. Thus f(x)is increasing on (−

√2,√

2) and decreasing on (−∞,−√

2) and on (√

2,∞). �

One upshot of the Monotonicity Theorem is that we can now see how to use the first derivative to help uslocate relative maximum and relative minimums. This is because if f(x) is continuous on the interval (a, b),and is increasing on an interval (a, c) and decreasing on (c, b) where c is between a and b, then there mustclearly be at least a relative maximum at x = c. A similar statement can be made about relative minimumsif f(x) changes from decreasing to increasing in the middle of some interval on which f is continuous. Theseideas are collectively know as

The First Derivative TestSuppose that f(x) is continuous on (a, b) and differentiable on (a, b)except perhaps at a point c with a < c < b. Suppose also that eitherf ′(c) = 0 or f ′(c) doesn’t exist. Then

1. If f ′(x) < 0 on (a, c) and f ′(x) > 0 on (c, b), then there is arelative minimum value at x = c.

2. If f ′(x) > 0 on (a, c) and f ′(x) < 0 on (c, b), then there is arelative maximum value at x = c.

3. If the sign of f ′(x) is the same on both (a, c) and (c, b), thenthere is no relative extremum at x = c.

Exercise 5.6.1. Use the analysis done previously to find the relative extrema for f(x) = xx2+2 .

Section 5.7: Concavity 77

5.7. Concavity

We’ve had such good luck relating the sign of the derivative of a function to information about the graphof the function that one can’t help but wonder whether or not the sign of the second derivative would yieldany nice information about the graph of the function. Consider the following graph of y = sin(x).

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Note that between −π and 0, the slopes of the tangent lines are increasing as we go from left to right.(The lines start out having negative slopes, and then the slopes increase to zero, and then the lines continueto get steeper.) Since the derivative of y = sin(x) tells us the slope of the tangent line at a given point, wecan see that the derivative of the derivative of y must be positive here. Thus the second derivative is positive.In this situation, we say that y is concave up on the interval (−π, 0). On the other hand, between 0 and π

we see that the slopes of the tangent lines are decreasing as we move from left to right, so the derivative ofthe derivative must be negative here, so y′′ < 0 on (0, π). Summarizing:

We say that f(x) is concave up on an interval I iff f ′′(x) > 0 on I.We say that f(x) is concave down on I iff f ′′(x) < 0 on I. If c is apoint in the domain of f(x), and if f(x) has one type of concavityon (a, c) and the other type on (c, b), then we say that the point(c, f(c)) is a point of inflection of f(x).

Exercise 5.7.1. Continue studying f(x) = xx2+2 by analyzing its concavity.

It turns out that for stationary points, we can use the second derivative to help analyze whether a givenpoint is a relative maximum or a relative minimum. This is because if f(x) has a relative minimum and ahorizontal tangent line at x = c, then it must be concave up there. (Try drawing an example and see foryourself!) On the other hand if it has a relative maximum at x = c and a horizontal tangent line, it must beconcave down there. Thus we have:

Section 5.7: Concavity 78

The Second Derivative TestSuppose that f(x) is so that f ′′(x) exists in some interval contain-ing c, and f ′(c) = 0.

1. If f ′′(c) < 0 then there is a relative maximum at (c, f(c)).

2. If f ′′(c) > 0 then there is a relative minimum at (c, f(c)).

3. If f ′′(c) = 0, we make no conclusion.

Exercise 5.7.2. Use the analysis done previously to find the relative extrema for f(x) = xx2+2 using the

second derivative test.

Section 5.8: Vertical Asymptotes 79

5.8. Vertical Asymptotes

We have already noted in this course that sometimes the limit of a function as x → c doesn’t exist becausewhen x is close to but not equal to c, the function gets very large. In this case we write limx→c f(x) = ∞.Geometrically, this means that there is a vertical asymptote at x = c. For example, the function f(x) = 1

x2

has a vertical asymptote at x = 0, since limx→01x2 = ∞. For the functions we have studied, vertical

asymptotes tend to occur when we have a rational expression of the form f(x) = n(x)d(x) , and there is a value

c for which d(c) = 0, but n(c) 6= 0. Then as x → c, the ratio f(x) would increase without bound. Of course,it could be the case that both n(c) and d(c) are zero, but the ratio could increase without bound, as theexample below shows.

Exercise 5.8.1. Does f(x) = sin(x)x2 have a vertical asymptote at x = 0 or not?

Section 5.9: Limits at Infinity; Horizontal Asymptotes 80

Click to Initialize QuizFor each of the following, find all of the vertical asymptotes.

1. f(x) = 5xx−3

x = 3 x = 0 x = 0, x = 3

2. f(x) = x3

x2−1

x = 1 x = 0, x = 1 x = ±1 x = ±1, x = 0

3. f(x) = x cot(x), x ∈ [−4, 4].

There aren’t any. x = ±π x = ±π, x = 0

Click to End Quiz

5.9. Limits at Infinity; Horizontal Asymptotes

We now study a different kind of limit involving infinity, which will be related to horizontal asymptotes. Ahorizontal asymptote occurs when a curve levels off at a certain y value as x gets very large in either thepostive or negative direction. Thus, it seems natural to define limx→∞ f(x). We need a technical definitionakin to the epsilon/delta definition of ordinary limits. We want to say that when x is really big, the functionf(x) should be close to the constant function g(x) = L. Thus, we want to say that for any ε > 0, if x is largeenough, we should have |f(x)− L| < ε. Formally we have:

Technical Definition of limx→∞ f(x) = L.

We say that limx→∞ f(x) = L if it is true that for any ε > 0 thereexists M so that if x > M , then |f(x)− L| < ε.

Exercise 5.9.1. Use the above definition to show that limx→∞1x = 0. What does this tell us about horizontal

asymptotes?

The definition above doesn’t really address the question of “how does one compute limits as x → ∞?”Fortunately, our limit theorems are still applicable, and this with the result of the previous exercise allowus to draw some nice deductions. Also, we will introduce a simplification technique that will proof to beadvantageous.

Exercise 5.9.2. Use the limit theorems and the above result to show that limx→∞1xr = 0, for r a rational

number, r ≥ 0.

Section 5.9: Limits at Infinity; Horizontal Asymptotes 81

5.9.1. A Simplification Technique

When trying to compute limx→∞ f(x) where f(x) is a rational expression like n(x)d(x) , it is often useful to

simplify the rational expression by dividing both the numerator and denominator by the same quantity.This is especially useful if you can find a quantity q(x) so that you know the limit of either n(x)

q(x) or d(x)q(x) .

For example, consider limx→∞x2

3x2+2x+1 . As the reader will shortly see, we know the limits of both x2

x2 (of

course) and 3x2+2x+1x2 , so it would be prudent to divide both the numerator and denominator by x2. Thus

we have

limx→∞

x2

3x2 + 2x + 1= lim

x→∞

x2

3x2 + 2x + 1·

1x2

1x2

divide by one in a clever way

= limx→∞

x2

x2

3x2+2x+1x2

Rewriting

= limx→∞

13x2

x2 + 2xx2 + 1

x2

Simplifying

= limx→∞

13 + 2

x + 1x2

Simplifying

=1

3 + 0 + 0Taking the Limit

=13

Arithmetic!

Section 5.10: Graphing 82

5.10. Graphing

We have now devised techniques which, when put together, enable us to graph almost any function in ourcatalogue. To do this, we carefully analyze our function for domain, asymptotes, monotonicity, concavity andextrema. We then plot any “important” points such as inflection points, relative maximum and minimumpoints, and indicate any asymptotes on our graph, and then draw a sketch, connecting the points we haveplotted and being sure to respect the information obtained so far.

A checklist to make sure we don’t miss anything in this somewhat long process follows:


Steps for Graphing f(x).

1. Note the domain of f(x). Any points or intervals not in the domain should be carefullynoted. Any isolated points which are not in the domain should appear on any sign chartcreated while analyzing this function.

2. Look for vertical and horizontal asymptotes. Recall that a vertical asymptote occurs atx = c if limx→c± f(x) = ±∞. Such things often (but not always) occur in rational expressionswhere the denominator is zero. Horizontal asymptotes occur when limx→±∞ f(x) = L, inwhich case there is a horizontal asymptote at y = L.

3. Investigate the Monotonicity of f(x). This will involve a study of the sign of f ′(x).Be sure to carefully make a sign chart over the domain which includes any values at whichf ′(x) = 0 or at which f ′(x) doesn’t exist.

4. Locate Relative Extrema of f(x). Since you have already made a sign chart for f ′(x), youcan easily apply the first derivative test to locate the relative extrema of f(x).

5. Investigate the Concavity of f(x). This will involve a study of the sign of f ′′(x). Be sureto make a sign chart. From the sign chart you should be able to find points of inflection. Itwould also be prudent to double check your conclusions about relative extrema by applyingthe second derivative test to any points where f ′(x) = 0.

6. Preparation for Graphing.

(a) Plot a few important points. Plot all extrema and inflection points. (Be sure to usef(x) and not f ′(x) or f ′′(x) to determine the y values of points you are plotting.)

(b) Intercepts. Plot the point (0, f(0)), since this is usually easy to compute, and representsthe y-intercept. If easy to do so, set f(x) = 0 and solve, and then plot any of these x-intercepts that you find.

(c) Asymptotes. Using a dashed line, indicate the asymptotes on the graph, and for verticalasymptotes, think about whether the function is approaching +∞ or −∞ as the x valueis approached from either the left or the right.

(d) Symmetry. Think for a moment about symmetry. If f(−x) = f(x), then the functionis symmetric about the x axis. If f(−x) = −f(x), then the function is symmetric aboutthe origin. In most cases, of course, the function will have neither of these two kinds ofsymmetry.

7. Sketch the Graph. Using all the information gathered so far, draw a sketch of the function.Be sure to respect the domain, all the intervals of monotonicty and concavity, the extrema andso forth. If it does not seem possible to draw a sketch which respects all the information youhave, there is likely a mistake in some calculation. Check over your work to try and find themistake before proceeding.


Exercise 5.10.1. Sketch a graph of f(x) = x2−2x3 , following the above guidelines.

Section 5.11: Optimization 85

5.11. Optimization

Suppose that you wanted to make a rectangular aquarium out of 4 pieces of glass, and you wanted it to holdexactly 32 cubic feet of water. Also, the aquarium is going to have two square sides (not the bottom), andthe aquarium doesn’t need a top. What dimensions should you make it, in order to minimize the amountof glass needed? This kind of question is a constrained optimization problem: a problem in which your goalis to maximize or minimize some quantity subject to one or more given constraints. In this problem, we areconstrained by the fact that our aquarium must have a certain volume. Our objective is to minimize theamount of glass used. In such problems, we need to give variable names to the quantities involved, and thencome up with an equation which represents each constraint, and one which represents the objective to bemaximized or minimized. Then typically, we use the constraint equation to reduce the number of variablesin the objective equation, and then use the techniques of this chapter to maximize or minimize the objective.For example, if we let the base of our aquarium have dimensions x×y, then we should let the third dimensionalso be x, so that there is an x× x square on two of the sides. Note that x2y = 32, since the volume is x2y.The surface area represents the amount of glass we require, and the bottom has dimensions x× y, and twoof the sides have dimensions x × x and two have dimensions y × x. A rough sketch of the aquarium willhelp in making this determination. Thus, our objective is to minimize S = xy + 2x2 + 2xy = 2x2 + 3xy.This function has two variables in it, but our constraint tells us that y = 32

x2 , so we can rewrite S asS = 2x2 + 3x( 32

x2 ) = 2x2 + 96x . This function is continuous for x > 0, and we shall shortly see that it only

has one critical number for x > 0, and that it is strictly decreasing from 0 to this critical number, andstrictly increasing from this critical number on. Thus, this only critical number will yield a minimum, asdesired. Calculating f ′(x), we have that f ′(x) = 4x − 96

x2 , which is zero only when x3 = 24, that is, whenx = 2 3

√3. Note that f ′(x) < 0 on (0, 2 3

√3) and f ′(x) > 0 on (2 3

√3,∞). Thus, there is a absolute minimum

at x = 2 3√

3. At this value of x, we have that y = 32x2 = 32

2 3√3. Thus, the dimensions of the aquarium should

be 2 3√

3× 2 3√

3× 163√3

.

Section 5.11: Optimization 86

Steps for Solving Optimization Problems

1. Read the problem carefully. Draw a diagram if appropiate. As youread, think about the constraints and the objective. You may need to readthe problem thorugh more than once.

2. Write down a constraint equation. In doing so, be sure that you havecarefully identified whatever variables you are using. There might be morethan one constraint equation for a given problem.

3. Write down an objective function. This is the function to be maximizedor minimized.

4. Simplify the objective function, using the constraint equation. Dothis especially if your objective function has more than one variable.

5. Maximize or Minimize the objective function. This involves findingcritical numbers of the objective function. Be sure to pay attention to thenatural domain of the objective function.

6. Be sure that your solution is a maximum or a minimum, as desired.This will likely involve the first derivative test, or similar analysis of wherethe function is monotonic.

7. Be sure to answer the question that is asked. Sometimes the criticalnumbers are desired, other times the value of the objective function at thesevalues will be required.

Exercise 5.11.1. Farmer Brown wants to build a rectangular pen, and then subdivide it into 4 identicalpens by building two parallel walls as shown. He has 200 feet of fence at his disposal. What dimensionsshould he use in order to maximize the area of the entire enclosed rectangle?

Section 5.12: Newton’s Method 87

5.12. Newton’s Method

Newton’s Method is a method for finding roots of an equation. We study it now simply because it is anotherinteresting and useful application of the derivative. The necessary information to begin the method consistsof the function you are trying to find the root of, and an initial “guess”, or approximation to the root. Theend result of the method (or algorithm) is a sequence of numbers which in many cases will be converging toa root of the equation.

The outline of the steps in the process are as follows:

1. Call the initial estimate x0.

2. Consider the tangent line to y = f(x) at the point (x0, f(x0)). As long as this tangent line isn’t parallelto the x−axis, it should intersect the x−axis at some point. Call this point of intersection (x1, 0).

3. Consider the tangent line to y = f(x) at the point (x1, f(x1)). As long as this tangent line isn’t parallelto the x−axis, it should intersect the x−axis at some point. Call this point (x2, 0).

4. Repeat this process. If the sequence x0, x1, x2, . . . converges, it likely is converging to a root r.

An animation demonstrating this idea and the following derivation of the iterative formula is at thefollowing link:

http://math.furman.edu/ mwoodard/math11/demos/newton.html

To generate a formula for this process, the key fact is that the slope of the tangent line at x0 is given byf ′(x0), while the slope of the line between (x1, 0) and (x0, f(x0)) is given by 0−f(x0)

x1−x0. Thus we need to set

these equal to each other and solve for x1, to find out what it is in terms of information about x0. We have

0− f(x0)x1 − x0

= f ′(x0),

and multiplying both side by the denominator on the right yields

−f(x0) = f ′(x0)(x1 − x0).

We then have

−f(x0) = f ′(x0)(x1 − x0)

−f(x0) = f ′(x0)(x1)− f ′(x0)(x0) Distributing

f ′(x0)(x0)− f(x0) = f ′(x0)(x1) Adding f ′(x0)(x0) to both sides

f ′(x0)(x0)− f(x0)f ′(x0)

= (x1) Dividing

In general, our recursive formula is given by

Section 5.12: Newton’s Method 88

f ′(xn)(xn)− f(xn)f ′(xn)

= (xn+1)

For example, suppose you wanted to find a root of f(x) = x2 − 4, and decided to start with an initialestimate of x = 3. (Of course, we already know that the roots of this are ±2, but this example is just toillustrate the method.) Since f ′(x) = 2x, we have that f ′(xn) = 2xn. Thus we have

xn+1 =f ′(xn)(xn)− f(xn)

f ′(xn)=

(2xn)(xn)− (x2n − 4)

2xn.

If we simplify this, we see that xn+1 = x2n+42xn

. Starting with x0 = 3, we get the following table:

i xi f(xi)

0 3 51 2.16667 0.6944442 2.00641 0.02568213 2.00001 0.0000409602

From this is it pretty clear that the xi’s are converging to 2, as expected.

Exercise 5.12.1. Use Newton’s Method to find a root for f(x) = x3 + x− 4.

Section 5.13: Antiderivatives 89

5.13. Antiderivatives

We have spent much effort learning how to carry out the following problem: given y = f(x), find y′ = f ′(x),and interpret its significance. In this section, we consider the inverse problem: given y = f(x), find afunction F (x) so that F ′(x) = f(x). Such a function is called an antiderivative of f(x). For example, anantiderivative of y = 2x is F (x) = x2, since the derivative of F (x) = x2 is F ′(x) = 2x. An antiderivativeof y = cos2(x) sin(x) is F (x) = −1

3 cos3(x), since then F ′(x) = −13 · 3 cos2(x) · − sin(x) = cos2(x) sin(x). Of

course, this last example may leave you scratching your head . . . , you can see that it is correct, but howdoes one figure it out based on the function you were given. That will be discussed at length in the sectionsto come. For now, we will focus on some basic facts about antiderivatives, and will think a little about whatthey might be good for.

It is worth noting at the outset that if one knows a single antiderivative of a function, he in fact knowsall of them, by the “+C” theorem. Thus, since F (x) = x2 represents a single antiderivative of y = 2x,the collection x2 + C represents all possible antiderivatives of y = 2x. This is often called the generalantiderivative of y = 2x.

Exercise 5.13.1. Find the general antiderivative of y = 2x + 8.

In the preceeding exercise we used the idea that the antiderivative of a sum should be the sum of theantiderivatives. This can be stated more formally as follows:

The Sum Rule and Constant Multiplier Rule forAntiderivatives

If F (x) is an antiderivative of f(x), and G(x) is an antiderivative ofg(x), then F (x) + G(x) + C is the general antiderivative of f(x) +g(x). Also, the if k is a constant, the general antiderivative of kf(x)is kF (x) + C.

Exercise 5.13.2. What is the general antiderivative of y = 5x3 + 8x + 10?

The previous exercise suggests a rule.

Power Rule for AntiderivativesIf f(x) = xn where n is an a rational number other than −1, thenthe general antiderivative of f(x) is F (x) = xn+1

n+1 .

Section 5.14: Rectilinear Motion 90

The above is easily checked by differentiating. Note that the exclusion of −1 is necessary, as otherwisethe formula is undefined.

5.14. Rectilinear Motion

Suppose that a particle is moving in a straight line, and the function a(t) which describes the object’sacceleration at time t is given. This is a reasonable scenario, since the physics formula F = ma tells us thatthe acceleration is essentially given by the forces acting on the particle. Thus, if one were to understand allof the forces acting on the particle, he would know a(t). The question, is: given a(t), can we recover v(t)and s(t), the objects velocity at time t and position at time t?

The answer turns out to be yes, as long as we have a few more pieces of information. For example,suppose that a(t) = 2t+1. Since acceleration is the derivative of velocity, velocity must be the antiderivativeof acceleration. Thus v(t) = t2 + t + C for some constant C. If we know the velocity at time t = 0 (let’ssay that it was 5 feet per second) then we could compute C. This is because v(0) = 5 = 02 + 0 + C, soC = 5. The fact that v(0) = 5 is called an initial condition. If we also had an initial condition for position,we could compute it as well. For example, if s(0) = 2, then since s(t) is an antiderivative of v(t), we wouldhave s(t) = t3

3 + t2

2 + 5t + C1, where C1 is some constant which is potentially different from C. The initialcondition tells us that s(0) = 2 = 03

3 + 02

2 + 5(0) + C1, so C1 = 2.

Exercise 5.14.1. The acceleration due to gravity near the surface of the earth is approximately −32 feetper second per second. (We sometimes say “‘feet per second squared.”) If an object is thrown upward froma height of 3 feet with an initial velocity of 12 feet per second, find a formula for s(t), the position of theobject at time t. How high does the ball go?

Exercise 5.14.2. Find s(t) if a(t) = cos t + sin t, s(0) = 4, v(0) = 1.

5.15. Antiderivatives Mini-Quiz

Click to Initialize Quiz Answers the following questions about antiderivatives.

1. For the initial value problem “Find s(t), given a(t), where v(0) = m and s(0) = n,” the following isalways true:

The constant of integration arem and n.

We can use m and n to findthe constants of integration.

m and n have nothing to dowith the constants ofintegration.

2. The “+C” Theorem implies which of the following:

Every function has anantiderivative.

The antiderivative of the zerofunction is a constant.

The derivative of a polynomialhas lower degree than theoriginal function.

Section 5.15: Antiderivatives Mini-Quiz 91

3. If m(x) = k1g(x) − k2f(x), and if G(x) is an antiderivative of g(x), and F (x) is an antiderivative off(x), then the general antiderivative of m(x) is:

k1G(x)− k2F (x) + C K1G(x)−K2F (x) + C G(x)− F (x) + C

4. If m(x) = f(x)g(x), and F ′(x) = f(x) and G′(x) = g(x), then an antiderivative of m(x) is:

M(x) = F (x)G(x) M(x) = F (x)g(x) + G(x)f(x) None of the above.

Click to End Quiz

Section 5.15: Antiderivatives Mini-Quiz 92

Chapter 6

Areas and Integrals

Contents

6.1 Summation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6.2 Computing Some Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6.3 Areas of Planar Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

6.3.1 Approximating with Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

6.4 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.4.1 Definition and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.4.2 Properties of∫ b

af(x) dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.4.3 Relation of the Definite Integral to Area . . . . . . . . . . . . . . . . . . . . . . . . 102

6.4.4 Another Interpretation of the Definite Integral . . . . . . . . . . . . . . . . . . . . 103

6.5 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

6.5.1 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

6.5.2 The Fundamental Theorem – Part II . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6.6 The Indefinite Integral and Substitution . . . . . . . . . . . . . . . . . . . . . . 107

6.6.1 The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6.6.2 Substitution in Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

6.6.3 Substitution in Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6.7 Areas Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.7.1 Vertically Oriented Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.7.2 Horizontally Oriented Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

93

Section 6.1: Summation Notation 94

6.1. Summation Notation

We all know how to add. In this section, we will simply talk about adding, but will focus on the case wherewe have many terms in our sum, each of which can be described by a formula of some sort. For example,consider the sum

1 + 2 + 3 + 4 + . . . + 100

which has as its ith term the number i. We will discuss a convenient way to write this sum, and also a wayto compute this sum.

We will use the greek capital letter Σ to stand for “sum” (they both have something to do with the letter“s”,) and will use subscripts and superscripts to indicate where the sum begins and ends. Thus, if we write

25∑i=1

i2

we are talking about the sum whose ith term is i2, and which starts with 1 = 12 and ends with 625 = 252.Thus

25∑i=1

i2 = 1 + 4 + 9 + . . . + 625.

In general, we havem∑

i=n

ai = an + an+1 + an+2 + . . . + am.

Exercise 6.1.1. How would you write the sum 4 + 9 + 16 + 25 + . . . + 100 in summation notation?

Exercise 6.1.2. Write the following sum out in the form a1 + a2 + . . . + an:∑8

i=3 i2 + i + 1.

6.2. Computing Some Sums

There is an interesting story about the mathematician Gauss(1777-1855) which goes something like this:Gauss was such a child prodigy that he sometimes was difficult to control during class, so one day histeacher decided to keep him busy by having him add up the first 100 integers. Gauss went to work and wasdone almost immediately! His teacher was perplexed and asked Gauss how he could have done it so quickly,and Gauss answered that he noticed that if he wrote the numbers this way:

1 2 3 . . . 50100 99 98 . . . 51

that the columns added to 101. Since there are 50 columns, the answers must be (101)(50) = 5050. Thisamazingly insightful (for a child) concept can be generalized to compute the sum of the first n integers asfollows. Assume that n is odd. Then write the first n− 1 terms as follows:

1 2 3 . . . n−12

n− 1 n− 2 n− 3 . . . n+12

Section 6.2: Computing Some Sums 95

and note that the columns add to be n, and there are n−12 of them. Thus the sum of the first n− 1 terms is

(n−12 )(n). Now if we add the last term n to this sum, we obtain(

n− 12

)(n) + n = (n)

(n− 1

2+ 1)

= (n)(

n + 12

).

Exercise 6.2.1. Explain why in the previous example, the 2nd row ends with n+12 . Then show that the

formula∑n

i=1 i = n(n+1)2 holds for all n by showing it is true in the case where n is even. (This case is more

like Gauss’ calculation than the previous example is.)

It turns out that there are few more nice formulas like Gauss’ for∑n

i=1 i2 and∑n

i=1 i3, although theseare not nearly as easy to prove. We will need them anyway, so we will list them here, together with theprevoius result about

∑ni=1 i.

n∑i=1

i =n(n + 1)

2(6.1)

n∑i=1

i2 =(n)(n + 1)(2n + 1)

6(6.2)

n∑i=1

i3 =(

n(n + 1)2

)2

(6.3)

Suppose that you wanted to compute something like∑50

i=1 2i2 + 4i + 8. We can use the above formulastogether with some basic facts about sums to compute this. The facts we need are simply that sums arecommutative, that we can factor constants out of sums, and that the sum of n terms of a constant k is justthe product of n with k. For completeness, we write this ideas using summation notation.

n∑i=1

ai + bi =n∑

i=1

ai +n∑

i=1

bi (6.4)

n∑i=1

kai = kn∑

i=1

ai (6.5)

n∑i=1

k =

n terms︷︸︸︷k + k + k + . . . + k = nk (6.6)

Section 6.2: Computing Some Sums 96

Thus we can compute∑50

i=1 2i2 + 4i + 8 as follows:50∑

i=1

2i2 + 4i + 8 =50∑

i=1

2i2 +50∑

i=1

4i +50∑

i=1

8 By 6.4

= 250∑

i=1

i2 + 450∑

i=1

i +50∑

i=1

8 By 6.5

= 250∑

i=1

i2 + 450∑

i=1

i + 400 By 6.6

= 2(

(50)(51)(101)6

)+ 4

((50)(51)

2

)+ 400 By 6.1 and 6.2

= 91350 By arithmetic.

Exercise 6.2.2. Compute∑n

i=1 4i3 + 6i− 1. Your answer will be a function of n.

Example 6.2.1. Suppose that f(x) = x2+1 and xi = 2+2in is a sequence. Compute both: a)

n∑i=1

f(xi) ·2n

,and

b) The limit as n →∞ of the answer to a).

Solution: Since f(x) = x2 + 1 and xi = 2 + 2in , we have that f(xi) =

(2 + 2i

n

)2+ 1. If we expand this

expression we get that f(xi) = 4 + 4in + 4i

n + 4i2

n2 + 1 which is the same expression as 5 + 8in + 4i2

n2 . We are

supposed to sum the product of this expression with 2n , so we must compute

n∑i=1

10n

+16i

n2+

8i2

n3. We see

thatn∑

i=1

10n

+16i

n2+

8i2

n3=

10n

n∑i=1

1 +16n2

n∑i=1

i +8n3

n∑i=1

i2

=10n· n +

16n2

n(n + 1)2

+8n3

n(n + 1)(2n + 1)6

= 10 +8(n + 1)

n+

(4)(n + 1)(2n + 1)3n2

This is in a form which makes it reasonably easy to compute the limit as n →∞. We have

limn→∞

(10 +

8(n + 1)n

+(4)(n + 1)(2n + 1)

3n2

)=

10 + 8 limn→∞

n + 1n

+ 4 limn→∞

2n2 + 3n + 13n2

= 10 + 8(1) + 4(

23

)=

623

�

Section 6.3: Areas of Planar Regions 97

6.3. Areas of Planar Regions

In the course of your human experience, you have encountered and learned some basic facts about areas ofplanar regions. You know, for example, that the area of a rectangle is the length times the width, and ifyou think for a moment about this fact, it makes perfect sense. For example, in a 2× 3 rectangle, you canactually see the 6 square units that make up the region. You also know that the area of triangle is 1

2 of thebase times the height, which again makes sense if you draw a rectangle with the same base and the sameheight as the triangle, and work to convince yourself that there is just as much area outside the triangle butinside the rectangle is there is inside the triangle, and thus the area of the triangle is one half the area ofthe corresponding rectangle. You also “know” that the area of a circle is πr2 where r is the radius, but itis healthy to admit that this is a different kind of knowledge. In fact, the only reason you probably “know”this fact is that someone who you trusted convinced you to believe it on blind faith! If true confessions arein order, you probably will have to admit that you don’t really know anything about the area of anythingwith a curved side. In this section, we will investigate such area, and use calculus to see how to computethem.

We will begin by considering area of regions which are curved on only one “side.” Suppose f(x) is afunction with f(x) > 0 on the interval (a, b). Let R be the region under y = f(x), above the x axis, andbetween the lines x = a and x = b. This region has 3 straight “sides” and one curved “side”. We willattempt to find the area of such a region.

6.3.1. Approximating with Rectangles

In order to have a specifice region R to work with, let’s let f(x) = x2 + 1 and let a = 2 and b = 4. The firstgood idea we will introduce toward finding the area of R is to approximate the area with something we knowthe area of. We will use rectangles, since (as mentioned before) we understand the areas of those. Let’s use4 rectangles to start. If we divide the interval from a to b up into 4 subintervals, the endpoints would be at2, 2.5, 3, 3.5 and 4. If we were to draw four rectangles whose bases are on these four subintervals, and whoseheight is determined by the function value using the endpoint on the right-hand side of the subinterval, wewould get a decent approximation to the area of R. The good news is that we can compute the sum of theareas of these four rectangles, and it would be approximately the area of R. The bad news is that it wouldn’tbe exactly right. To compute the area we would need to compute the widths of the rectangles, which is 1

2

for each, and the height of each which are f(2.5), f(3), f(3.5), and f(4) respectively. These values can becomputed to be 29

4 , 10, 534 , and 17 respectively. If we multiply each of these by the width 1

2 and add wearrive at 95

4 . This is a very rough approximation, but at least it’s a start. Now, what if we were to increasethe number of rectangles to eight? A quick sketch reveals what looks like a better approximation. There isa nice demo of this at http://math.furman.edu/ dcs/java/quad.html.

One can’t help but extrapolate along these lines: if eight rectangles yields a better approximation thanfour, and 16 yields a better approximation than eight, why not use infinitely many rectangles? Of course,this statement doesn’t really make sense as stated, but we can make it precise by requiring n rectangles, andthen taking the limit as n →∞.

http://math.furman.edu/~dcs/java/quad.html


Toward this end, we want to divide the the interval from a to b up into n distinct subintervals. It will beuseful to make them all the same width, and to call this width ∆x. A little thought shows that ∆x = b−a

n .It will also be useful to let a = x0 and b = xn, and the other endpoints of the subintervals be called xi. Inthis case we would have xi = a + i∆x. This is summarized as

∆x = b−an is the width of each subinterval. The collection xi =

a + i∆x are the endpoints of the subintervals.

Now each rectangle has width ∆x, and height f(xi) for some endpoint xi. To encompass all of the n

rectangles and using the right-hand side of each subinterval, we would have the approximate area given by

f(x1)∆x + f(x2)∆x + f(x3)∆x + . . . + f(xn)∆x

which is better written asn∑

i=1

f(xi)∆x =n∑

i=1

f

(a + i

b− a

n

)(b− a

n

).

Note that in this sum the f(x − i) factor of each term represents the height of the ith rectangle, while ∆x

represents the width of the ith rectangle. In our current example, we would haven∑

i=1

f

(2 + i

2n

)(2n

)=

n∑i=1

((2 + i

2n

)2

+ 1

)(2n

).

This sum is exactly the one we computed in the previous section. If we let An be this approximation usingn rectangles, you will recall that we computed An = 10 + 8(n+1)

n + 4(n+1)(2n+1)3n2 .

Finally, to compute the area, we need to find limn→∞ An, which we previously saw was 623 . Thus, the

area under y = x2 + 1, above the x axis and between a = 2 and b = 4 is 623 .

To summarize, we note the steps in this somewhat arduous process.


Computing Areas of Simple RegionsThese are the steps to compute the area under y = f(x), above thex axis and between x = a and x = b when f(x) > 0 on (a, b). Thisis sometimes called the Right-Hand Rule.

1. Use rectangles of width ∆x = b−an .

2. Let xi = a + i∆x be the endpoints of the subintervals.

3. Let An =∑n

i=1 f(xi)∆x be the approximation using n rect-angles. Simplify and compute An using the techniques of theprevious section.

4. Compute limn→∞ An, yielding the area.

Exercise 6.3.1. Compute the area under f(x) = 4x2 + 2x + 1 between a = 3 and b = 6.

Exercise 6.3.2. Compute the area under f(x) = 3x + 5 between a = 0 and b = 4 two different ways: byusing the techniques of this section and by simply drawing the region and using the fact that it is made upof straight sides. Note that answer is the same either way.

Section 6.4: The Definite Integral 100

6.4. The Definite Integral

6.4.1. Definition and Notation

In the previous section we have seen that for a given function f(x) and interval [a, b], then quantity

limn→∞

n∑i=1

xi∆x is an important quantity, since it gives the area under y = f(x) and above the x axis between

x = a and x = b if f(x) > 0 on [a, b]. This “limit of a sum of areas of rectangles” is important enough thatit deserves its own special name and notation. We will call this quantity the definite integral of f(x) withrespect to x from a to b. Our definition will actually be slightly more general. We don’t actually have to haveall subintervals the same with, and we don’t have to use the right-hand side of each subinterval. As long aswe have points a = xo, x1, x2, ldots, xn = b, we can let ∆xi = xi+1 − xi be the width of the ith subinterval,

and we can let x∗i be any point on the ith subinterval. Then we can consider limn→∞

n∑i=1

f(x∗i )∆xi as the limit

of a sum of areas of rectangles with various widths and (perhaps) using various points to determine theheights of the rectangles. Of course, the important theorem (whose proof we will omit) is that we get thesame result whether the xi’s are left-hand endpoints, right-hand endpoints, midpoints, or whatever, as longas the widths ∆xi shrink to zero as n →∞. Thus we have:

Definition of Definite IntegralLet the interval [a, b] be divided up into subintervals by a set ofpoints xi so that

a = xo < x1 < x2 < · · · < xn = b.

Let ∆xi = xi+1−xi be the width of the ith subinterval. Let x∗i beany number so that xi ≤ x∗i ≤ xi+1. Then the definite integral off(x) with respect to x from a to b is given by

limn→∞

n∑i=1

f(x∗i ) ∆xi.

This quantity is a number which doesn’t depend on the choice ofendpoints xi or the choice of x∗i . The definite integral is definedwhether or not f(x) > 0 on [a, b], but only represents the areaunder f(x) and above the x axis between a and b when f(x) ≥ 0on (a, b).

Exercise 6.4.1. Compute the definite integral from 2 to 5 of f(x) = 5x3 +3x. Does this represent an area?


Since it is somewhat painful to write out the quantity represented by the definite integral, it will be usefulto introduce some notation to represent it. The notation will necessarily have to involve the function f(x)and the endpoints a and b. We choose a mathematical symbol which will remind us of an “S” for “sum.”The symbol is actually two symbols which are always used in conjuction: the integral sign “

∫” and the

differential “dx.” Thus we write∫ b

af(x) dx or

∫ b

a

f(x) dx. Thus we have:

Notation for the Definite IntegralWe write the definite integral of f(x) with respect to x from x = a

to x = b as ∫ b

a

f(x) dx.

The symbol∫

is called the integral sign, and the a and b are calledthe lower limit of integration and the upper limit of integration,respectively. The symbol dx is called the differential, and is alwaysused in conjuction with the integral sign. The x in the symbol dx

represents that the definite integral is to be considered with respectto x. The function f(x), which is placed between the integral signand the differential, is called the integrand.

Thus, for example, the previous exercise could have been worded: find∫ 5

25x3 + 3x dx.

Exercise 6.4.2. Compute∫ 3

−15x + 2 dx.

6.4.2. Properties of∫ b

af(x) dx.

Because the definite integral is defined to be a limit of a sum of areas of rectangles, and because limits andsums have nice properties, we might expect for the definite integral to have some nice properties, especiallyas related to constant multipliers and sums. We have the following:


Properties of the Definite Integral

1.∫ b

akf(x) dx = k

∫ b

af(x) dx, where k is a constant.

2.∫ b

af(x)± g(x) dx =

∫ b

af(x) dx±

∫ b

ag(x) dx.

3.∫ b

af(x) dx +

∫ c

bf(x) dx =

∫ c

af(x) dx

4.∫ b

af(x) dx = −

∫ a

bf(x) dx

Thus, for example, to compute∫ 5

13x2+3x+1 dx we could instead compute 3

∫ 5

1x2 dx+3

∫ 5

1x dx+

∫ 5

11 dx.

6.4.3. Relation of the Definite Integral to Area

We have seen that when f(x) ≥ 0 on [a, b] that∫ b

af(x) dx is the area under f(x), above the x-axis and

between x = a and x = b. What if f(x) is not greater than or equal to zero on [a, b]? If f(x) ≤ 0 on [a, b],In this case, think about the function g(x) = −f(x). This function would be greater than or equal to 0 on[a, b], so

∫ b

ag(x) dx would be the area under g(x). But the graph of g(x) = −f(x) is obtained from the graph

of f(x) by flipping around the x-axis. Thus the area under g(x) between a and b is the same as the areaabove f(x), below the x-axis and between x = a and x = b. If a function h(x) is above the x-axis between0 and 1 and is below the x-axis between 1 and 2, then

∫ 2

0f(x) dx, which by the above properties is equal

to∫ 1

0f(x) dx +

∫ 2

1f(x) dx, which yields the area below f(x) between 0 and 1 minus the area above f(x)

between 1 and 2. This idea can be generalized to see that in general, the definite integral gives the areaabove the x-axis minus the area below the x-axis. To get the total area bounded by f(x), one would needto compute

∫ b

a|f(x)| dx. Summarizing:

If Aup is the area below f(x) and above the x-axis, and Adown isthe area above f(x) and below the x-axis, then∫ b

a

f(x) dx = Aup −Adown.

The total area bounded by f(x) is given by∫ b

a|f(x)| dx.

Exercise 6.4.3. Compute∫ 1

−1x3dx. Does this represent an area? Explain.

Exercise 6.4.4. In a previous exercise we computed∫ 3

−15x + 2 dx. Thinking only in terms of the area(s)

represented here, recompute this using a sketch as your guide. (Hint: This works because all sides of theregion are straight lines.)


6.4.4. Another Interpretation of the Definite Integral

We used area to motivate the definition of the definite integral, but it turns out that there are otherapplications and interpretations of this entity. Just like the two main interpretations of the derivative (slopeof tangent line and instantaneous rate of change) there is an interpretation of

∫ b

af(x) dx in terms of “physics”

in addition to the geometric interpretation we have already discussed. Suppose that an object is moving ina straight line, but with a velocity that varies with time t. Let v(t) represent the velocity at time t. Ourintuition tells us that if we know the velocity at any time t, we should be able to recover the distance theobject travels between time a and b. For the sake of simplicity, let’s assume the object is always moving to theright, so its velocity is always positive on [a, b]. If v(t) were constant, we know that distance = rate ·time, buthere the rate (velocity) isn’t constant. However, on a very small time interval [ti, ti+1] the velocity is almostconstant, and we could approximate it by v(t∗i ) for some t∗i on [ti, ti+1]. Thus for that small time interval,the distance is given approximately by v(t∗i ) ·∆ti. (This simply comes from distance = rate · time.) Addingup all of the distances over the various small time intervals gives

∑ni=1 v(t∗i ) ·∆ti, and this approximation

will get better as the time intervals get shorter, so the exact distance is given by limn→∞ sumni=1v(t∗i ) ·∆ti,

but this is exactly∫ b

av(t) dt. Thus in this context, the definite integral of velocity gives distance travelled. If

v(t) is sometimes positive and sometimes negative on [a, b], the definite integral gives the difference betweenthe distance travelled to the right and the distance travelled to the left, sometimes called displacement. Toget the total distance travelled, one would have to compute

∫ b

a|v(t)| dt.

Section 6.5: The Fundamental Theorem 104

6.5. The Fundamental Theorem

6.5.1. The Fundamental Theorem

We turn now to a theorem with an ominous sounding name – The Fundamental Theorem of Calculus. Thistheorem will connect the theory of the definite integral with differentiation in a fundamental way, by linkingthe definite integral of a function with information related to its antiderivative. The key idea is the following:consider a small subinterval of the closed interval [a, b], say the interval from xi to xi+1. The portion ofthe definite integral that has something to do with this interval is the term of the form f(x∗i )∆xi, where∆xi = xi+1− xi. Now, suppose that F (x) is any antiderivative of f(x), and apply the Mean Value Theoremto F (x) on the interval [xi, xi+1]. The MVT says that there is a point ci between xi and xi+1 with

F (xi+1)− F (xi) = F ′(ci) ·∆xi = f(ci)∆xi.

In other words, there is a point ci so that the area of the rectangle with base ∆xi and height determinedby the function f evaluated at ci is exactly the difference between the antiderivative F evalauated at theright-side endpoint and F evaluated at the left side endpoint. If we do this on each subinterval, then bychoosing appropriate values of ci, it must follow that

n∑i=1

f(ci)∆xi =n∑

i=1

F (xi)− F (xi−1),

but this sum on the right can be expanded to be

F (x1)− F (x0) + F (x2)− F (x1) + . . . + F (xn)− F (xn−1),

which simplifies to F (xn)− F (x0) = F (b)− F (a). Thus we have

The Fundamental Theorem of Calculus (Part I)If f(x) is continuous on [a, b], then

∫ b

af(x) dx = F (b)−F (a), where

F (x) is any antiderivative of f(x).

Note that the fundamental theorem draws a connection between the definite integral of f and its an-tiderivative, F . It also provides a much easier way to calculate definite integrals, since antiderivatives areoften easier to find than it is to go through the ardous process of finding the limit of a sum of areas ofrectangles. Here is an example:

Example 6.5.1. Use the fundamental theorem to calculate∫ 3

1x2 + 4x dx.

Solution: Since an antiderivative of f(x) = x2 + 4x is F (x) = x3

3 + 2x2, we see that

F (3)− F (1) = 9 + 18− (13

+ 2) = 225− 13

=743


. �

A word on notation: since the expression F (b) − F (a) is so important in the practice of using thefundamental theorem to compute

∫ b

af(x) dx, it will be useful to introduce notation for it. We will use the

notation F (x)|ba to represente F (b)− F (a).

Example 6.5.2. Use the above notation to compute∫ π

0sin(x) dx.

Solution:∫ π

0sin(x) dx = − cos(x)|π0 = − cos(π)−− cos(0) = −(−1) + 1 = 2. �

6.5.2. The Fundamental Theorem – Part II

The first part of the fundamental theorem says that there is a close relationship between definite integralsand antiderivatives. The second part of the fundamental theorem will reiterate this relationship, but ina slightly different way. We will create a kind of “definite integral function”, and see that it is actuallythe antiderivative of the integrand. Specifically, consider the function A(x) =

∫ x

0f(t) dt. Note that in this

definition, the independent variable for the function A is x. We choose the letter A because the functionis sort of an area function – if f(t) is above the t axis, then A(x) gives the area under f(t) between t = 0and t = x. Of course, if F (t) is an antiderivative (with respect to t) of f(t), then another name for A(x)is F (x) − F (0). Then the derivative with respect to x of A(x) is F ′(x) − 0 (since F (0) is a constant), butsince F is an antiderivative of f , we have that F ′(x) = f(x). Thus, A′(x) = f(x). This is the FundamentalTheorem – Part II.

Fundamental Theorem of Calculus – Part IISuppose that A(x) is the function defined by A(x) =

∫ x

af(t) dt.

Then A′(x) = f(x). That is, the derivative of a function defined sothat the independent variable is the upper limit of integration of adefinite integral has as its derivative the integrand of the definiteintegral as a function of that upper limit of integration.

Example 6.5.3. What is the derivative with respect to x of∫ x

2

√sin(t) dt?

Solution: The FTOC – Part II applies. The derivative is simply√

sin(x). �

Example 6.5.4. If A(x) =∫ 4

x1t dt with x > 0, what is A′(x)?

Solution: We can’t apply the FTOC – Part II to A(x) as written, but we could apply it to −A(x) =−∫ 4

x1t dt =

∫ x

41t dt. Thus the derivative of −A(x) is 1

x , so the derivative of A(x) = − 1x . �

Example 6.5.5. What is ddx

(∫ x2

1sec(t) dt

)?


Solution: The chain rule applies, together with the FTOC – Part II. Note that the function is the resultof composing the function A(x) =

∫ x

1sec(t) dt with the function g(x) = x2. In fact, the original function

is then A(g(x)). Thus, by the chain rule, the derivative must be A′(g(x))g′(x). By the FTOC – Part II,A′(x) = sec(x), so A′(g(x)) = sec(x2). Also, g′(x) = 2x, so the derivative of the original function must besec(x2)2x. �

Section 6.6: The Indefinite Integral and Substitution 107

6.6. The Indefinite Integral and Substitution

One thing that arises from the fundamental theorem is the idea that antiderivatives are quite important,since they help us to evaluate definite integrals. Thus, we probably should study them in more detail. Inthis section, we will introduce a new notation for the general antiderivative, based on the fact that it isclosely related to the definite integral. We also learn a technique for finding antiderivatives, and thus definiteintegrals which is essentially the inverse of the chain rule.

6.6.1. The Indefinite Integral

The FTOC indicates that antiderivatives and integrals are closely related. Because of this, and because ofour need for better notation for the antiderivative of a function, we introduce the following new notation forthe general antiderivative.

The Indefinite IntegralWe will call the general antiderivative of a function f(x) the indef-inite integral of f(x), and will denote it by

∫f(x) dx. Thus∫

f(x) dx = F (x) + C,

where F ′(x) = f(x). We choose this notation (which is similarto the notation for the definite integral) because the fundamentaltheorem of calculus tells us that the general antiderivative is relatedto the definite integral.

Note the difference between∫ b

af(x) dx and

∫f(x) dx. The former is a number, while the latter is a

collection of functions. They are related not because their symbols are similar but rather their symbols aresimilar because they are related. To reiterate: they are related because one can compute

∫ b

af(x) dx by using

any of the functions from the collection∫

f(x) dx, and evaluating at b and a and subtracting.

Example 6.6.1. Compute∫

sec2(x) + x dx.

Solution:∫

sec2(x) + x dx = tan(x) + x2

2 + C. �


(x2 + 3x + 1)(2x + 8) dx.


Solution:∫(x2 + 3x + 1)(2x + 8) dx =

∫2x3 + 6x2 + 2x + 16x2 + 24x + 8 dx Multiplying Out the Integrand

=∫

2x3 + 22x2 + 26x + 8 dx Combining Like Terms

=2x4

4+

22x3

3+

26x2

2+ 8x + C Integrating

=x4

2+

22x3

3+ 13x2 + 8x + C Simplifying

�

It might be useful to list everything we know about antiderivatives to this point. We have

f(x)∫

f(x)dx

sinx − cos x + C

cos x sinx + C

sec2 x tanx + C

csc2 x − cot x + C

sec x tanx sec x + C

csc x cot x − csc x + C

k, k a constant kx + C

xn, n 6= −1 xn+1

n+1 + C

kg(x) k∫

g(x) dx

g(x) + h(x)∫

g(x) dx +∫

h(x) dx

6.6.2. Substitution in Indefinite Integrals

We now encounter a technique of integration which will help us to integrate more complex functions, helpingus to expand the above list of functions we know how to integrate. The technique is called substitution,because in the technique we will replace (or substitute) some quantities for other quantities. The goal is toperform a substitution which will simplify the integrand, hopefully to one on the above list. The motivationfor the technique comes from the chain rule. Recall that the chain rules says that

d

dxf(g(x)) = f ′(g(x))g′(x).

Thus, it must be true that ∫f ′(g(x))g′(x) = f(g(x)) + C


Since these statements are equivalent. For example, the derivative of (3x2+2x)100 is 100(3x2+2x)99(6x+2). Thus it must be true that

∫100(3x2 + 2x)99(6x + 2) dx = (3x3 + 2x)100 + C, but how does one recognize

this? Or worse, what if we were given the equivalent∫

200(3x + 1)(3x2 + 2x)99 dx? The idea behind thetechnique of substitution is the following: suppose that we gave the quantity 3x2 + 2x a simpler name.What if we called it simply u? Then since du

dx = 6x + 2, it might be a good idea to call du by the name“(6x + 2) · dx.” Then we could write the integral∫

100(

u︷︸︸︷(3x2 + 2x))99

du︷︸︸︷(6x + 2) dx

as ∫100u99 du,

and this last integral is one we know how to do. It is equal to 100u100

100 + C = u100 + C. Of course, weshould report our answer in terms of x, since the original problem was in terms of x. Thus we could write(3x2 + 2x)100 + C as our answer.

Substitution Technique – Indefinite IntegralsGiven an integral of the form

∫f ′(g(x))g′(x) dx, we let u = g(x)

and du = g′(x) dx. Then we write the integral as∫

f ′(u) du, whichintegrates to be f(u) + C. This should be reported as f(g(x)) +C, which is the correct antiderivative of f ′(g(x))g′(x), as can bechecked by the chain rule.


sin(cos(x)) · sin(x) dx.

Solution: Let’s try letting u = cos(x). Then du = − sin(x) dx, so sin(x) dx = −du. If we substitute, weobtain the integral

∫sin(u) · −du, which can be written as −

∫sin(u) du. This is equal to −(− cos(u)) + C,

so the desired antiderivative is cos(sin(x)) + C. �


x√

5x2 + 8 dx.

Solution: If we let u = 5x2 + 8, then du = 10x dx. It might be helpful to write this as du10 = x dx. Thus

our integral can be rewritten as∫

110

√u du = 1

10

∫ √u du. Upon integrating, we obtain 1

1023u

32 + C =

115 (5x2 + 8)

32 + C. �


How do I know what to let u equal?This is a common question asked by many beginning calculus stu-dents. The best answer is: let u be something so that by substi-tuting u and du, you are able to completely transform the integralinto an integral in terms of u which you know how to compute.A more practical answer may be this: try to let u be somethingwhose derivative is a factor of the integrand. This is a necessary(but not sufficient) ingredient of a successful substitution!

6.6.3. Substitution in Definite Integrals

Can the technique of substitution be used to evaluate definite integrals? Of course the answer is yes – sinceit helps us to find antiderivatives and that is what is needed to use the FTOC. Care must be taken, however,to be sure write the substitution correctly in the presence of the limits of integration.

Substitution Technique – Definite IntegralsGiven an integral of the form

∫ b

af ′(g(x))g′(x) dx, we let u = g(x)

and du = g′(x) dx. Then we write the integral as∫ g(b)

g(a)f ′(u) du,

which integrates to be f(u)|g(b)g(a) = f(g(b)) − f(g(a)). This is the

numerical value of the original definite integral, so no further mod-ifications are necessary.

Example 6.6.5. Compute∫ 4

0x sin(x2 + 4) dx.

Solution: It seems reasonable to let u = x2 + 4, since the derivative of this expression is 2x which is a factorof the integrand (if we think of x as 2x

x .) Since du = 2x dx, we can write dx = du2 . Substituting we obtain

the integral∫ 20

412 sin(u) du. The 1

2 factor comes from the du2 substitution, and the limits of integration come

from the fact that u has value 4 when x = 0 and has value 20 when x = 4. This new integral has exactlythe same numerical value as the original, so I need only compute it. But we have∫ 20

4

12

sin(u) du =12− cos(u)|204

= −12

[cos(20)− cos(4)] ,

so this is the value of the original integral. �


Assignment:

1. Stewart, section 5.4, 1 - 41 odd.

2. Stewart, section 5.5, 1 - 51 every other odd. You do not need to graph thefunctions.

Section 6.7: Areas Between Curves 112

6.7. Areas Between Curves

6.7.1. Vertically Oriented Areas

The theory developed in this chapter so far indicates a close relationship between certain kinds of areas andrelated definite integrals. In fact, we have seen that in general, the definite integral of f(x) from x = a tox = b gives us the signed area trapped between the x axis and the curve, and between the lines x = a andx = b. We now try to extend this theory to enable us to compute area of regions like the following.

Let R be the region bounded by the curves g(x) = x2−5 and f(x) = 1−x2. Note that these curves crosswhen x2 − 5 = 1 − x2, which is when 2x2 = 6, or when x = ±

√3. Now imagine taking the interval from

[−√

3,√

3], and dividing it up into subintervals and then computing limn→∞∑n

i=1(f(x∗i )− g(x∗i ))∆xi. Thisshould be thought of as a limit of a sum of areas of rectangles, which are approximating the area betweenthe curves f(x) and g(x). It is important to note that over the interval [−

√3,√

3], f(x) is greater than g(x),so that f(xi) − g(xi) is a positive number, and we really are getting the area (and not the signed area).

Of course, this limit of a sum of areas of rectangles can be written as∫√3

−√

3(f(x) − g(x)) dx. This can be

computed as∫ √3

−√

3

(f(x)− g(x)) dx =∫ √

3

−√

3

1− x2 − (x2 − 5) dx

=∫ √

3

−√

3

6− 2x2 dx Simplifying the Integrand

= 6x− 2x3

3

∣∣∣∣√

3

−√

3

By the FTOC

= 6√

3− 23√

33

− (6(−√

3− 2−3√

33

) Simplifying

= 12√

3− 4√

3 More Simplifying

= 8√

3

The general principal illustrated in the previous example can be summarized as follows.


Vertically Oriented Areas Between CurvesIf f(x) ≥ g(x) over [a, b], then the area bounded by x = a, x =b, f(x), andg(x) is

∫ b

af(x) − g(x) dx. If in addition, f(a) = g(a)

and f(b) = g(b), then this represents the area of a single planarregion bounded above by f(x) and below by g(x). We call such anarea vertically oriented because the approximating rectangles areupright.

Example 6.7.1. Let a be the smallest positive number where cos(x) = sin(x), and let b > a be the nextsmallest positive number where cos(x) = sin(x). Compute the area bounded by the curves f(x) = sin(x)and g(x) = cos(x) between a and b.

Solution: A quick analysis of the unit circle leads us to believe that a is π4 and b = 5π

4 . Between a and b, it

is clear that sinx > cos x. Thus we need to compute∫ 5π

4π4

sinx− cos x dx, which by the FTOC is equal to

− cos x− sinx|5π4

π4

= −(−√

22−√

22

)− (−√

22−√

22

) = 2√

2.

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6.7.2. Horizontally Oriented Areas

For certain planar regions, it is easier to think of the approximating rectangles as laying on their sides – thatis, we will use horizontal rectangles to yield our definite integral. An example of such a planar region couldbe the region bounded by x = 1− y2 and x + y = −1. This region is most easily thought of as boiunded onthe left by the line x + y = −1 and on the right by the “sideways” parabola x = 1− y2. If instead you lookfor the function on top and the function on the bottom, you find that it switches part of the way through.Thus, it is best to think “left and right” rather than “up and down” for this region. Because we are thinkinghorizontally, we imagine dividing the y axis up into subintervals rather than the x axis. This will lead to anintegral with respect to x rather than one with respect to y. Since the two functions cross when y = −1 andwhen y = 2, we want to divide the interval from [−1, 2] along the y axis into subintervals. The horizontalapproximating rectangles we want to use to find the area are bounded on the right by 1− y2 and on the leftby −1 − y. Thus our approximating sum has the form

∑ni=1(1 − y2

i ) − (−1 − yi)∆yi, and after taking thelimit as n →∞, we end up with the integral∫ 2

−1

(1− y2)− (−1− y) dy.

In general, we have


Area of Regions usings Horizontal ApproximationsThe area of a region bounded on the left by x = g(y) and on theright by x = h(y) between y = c and y = d is given by∫ d

c

h(y)− g(y) dy.

Example 6.7.2. Find the area bounded by the curves x = 3− y2 and y = x− 1.

Solution: These curves cross when y + 1 = 3− y2, which is when y2 + y− 2 = 0. Thus they cross at y = −2and y = 1. Over this region, the left-hand boundary of the region is formed by the line x = y + 1, andthe right-hand boundary is formed by the parabola x = 3 − y2. Thus the area is of the region is given by∫ 1

−23− y2 − (y + 1) dy. Computing we get∫ 1

−2

3− y2 − (y + 1) dy =∫ 1

−2

3− y2 − y − 1 dy

=∫ 1

−2

2− y − y2 dy

= 2y − y2

2− y3

3

∣∣∣∣1−2

= 2− 12− 1

3−(−4− 2− −8

3

)= 4.5

�

Assignment:Stewart: Section 6.1, 1 - 29 odd.

Chapter 7

Appendix

Contents

7.1 3 – Minute Reviews . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.1.1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.1.2 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.1.3 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.1.4 Solving Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

7.1.5 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

7.1.6 Quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

7.1.7 Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

7.1.8 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

7.1.9 Setting Up Word Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Solutions to Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

115

Section 7.1: 3 – Minute Reviews 116

7.1. 3 – Minute Reviews

7.1.1. The Real Numbers

It is convenient to think of the real numbers as being points on a line. Points on the number line representreal numbers.

The real numbers can be further divided into two distinct sets of numbers–rational numbers and irrationalnumbers. Rational numbers are numbers that can be written in the form a

b where a and b are integers andb 6= 0. They include terminating decimals and repeating decimals. Irrational numbers are those which arenot rational. Some examples of irrational numbers include

√3 and π. A point to keep in mind is that it

is impossible to write an irrational number as a finite decimal. Irrationals can only be approximated by afinite decimal. When it becomes necessary to approximate an irrational number be certain to use correctnotation. For example, never write

√2 = 1.414. Instead, write

√2 ≈ 1.414.

The operations of addition, subtraction, and multiplication can be performed on pairs of real numberswith the result also being a real number. Division can be performed on pairs of real numbers with theresulting quotient being real as long as the divisor is not 0. Division by zero is undefined for the realnumbers. The square roots of negative numbers are also not real numbers. More generally, the even indexedroots of negative numbers are not real numbers.

Some examples that are not real numbers are√−5 and 4

x when x = 0.

Exercise 7.1.1. Which of the following are not real numbers? π, 5√−243, 4

√−16, −5√

2

Exercise 7.1.2. Let a be a rational number and let β and γ be irrational. What can be said abouta + β? aβ? β + γ?

7.1.2. Factoring

Common Factors: ab + ac = a(b + c)Difference of Squares: a2 − b2 = (a + b)(a− b)Difference of Cubes: a3 − b3 = (a− b)(a2 + ab + b2)Sum of Cubes: a3 + b3 = (a + b)(a2 − ab + b2)Trinomial Squares: a2 ± 2ab + b2 = (a± b)2

Factor Theorem: Let P (x) be a function. If r is a real number such that P (r) = 0, then (x − r) is afactor of P (x).

Exercise 7.1.3. Factor 2x3 + 3x2 − 8x− 12 completely.

Exercise 7.1.4. Factor x4 − x2 − 6x− 9 completely.

7.1.3. Solving Equations

Exercise 7.1.5. Solve for x:3x + 5 =

x− 72


Exercise 7.1.6. Solve for x:

b2y

xy − ax= C

7.1.4. Solving Systems of Equations

For example, solve the system of equations:

x + 2y = 10 (7.1)

3x− 5y = 8 (7.2)

Solve equation (1) for x:x = 10− 2y

Now substitute 10− 2y in for x in equation (2):3(10− 2y)− 5y = 8. Now solve for y:30− 6y − 5y = 8−11y = −22So y = 2Now substitute 2 in for y in the equation: x = 10− 2y.x = 10− 2(2)So x = 6Hence the solution to the system is (6, 2).

For example, solve the system of equations:

3x− 2y = 5 (7.3)

4x− 3y = 9 (7.4)

Multiply equation (3) by −4 and equation (4) by 3.

− 12x + 8y = −20 (7.5)

12x− 9y = 27 (7.6)

Add equations (5) and (6) and solve for y:−y = 7So y = −7.Now substitute y = −7 into equation (1) and solve for x:3x− 2(−7) = 53x = −9So x = −3. Hence the solution to the system of equations is (−3,−7)

Exercise 7.1.7. Solve the systems of equations:

7x + 4y = 8 (7.7)

6x + 3y = 4 (7.8)


7.1.5. Absolute Value

|x| =

{x if x ≥ 0

−x if x < 0Properties of Absolute Value: Let x and y be real numbers.

Property 1: |x||y| = |xy|Property 2: | − x| = |x|Property 3: |x|+ |y| ≥ |x + y|

The greatest integer function which is represented JxK is the function that sends each real number x tothe largest integer that is less than x. For example, J5.71K = 5 and J−23.9K = −24.

Exercise 7.1.8. Evaluate:a) J| − 4.5|Kb) |J−4.5K|c) JπK

Exercise 7.1.9. For each of the following statements, either show the statement is true and provide a ex-ample when it is false.Let x and y be real numbers.a) JxKJyK ≤ JxyKb) |JxK|JyK ≤ JxyK

Exercise 7.1.10. Show or provide a counterexample:J|x|KJ|y|K ≤ J|xy|K

7.1.6. Quadratics

Quadratic Formula: Let ax2 + bx + c = 0 where a,b, and c are real numbers and a 6= 0. Then

x =−b±

√b2 − 4ac

2a.

Equation of Parabola with Vertex (h, k): y − k = a(x− h)2

If a > 0 then the parabola opens up. If a < 0 then the parabola opens down.

7.1.7. Exponents

Let a,b,c,d be real numbers where a > 0 and b > 0.Property 1: a−c = 1

ac

Property 2: acad = ac+d

Property 3: (ac)d = acd

Property 4: (ab)c = acbc


Note that in general (a + b)c 6= ac + bc.Property 5: ac

ad = ac−d

Property 6: (ab )c = ac

bc

7.1.8. Inequalities

7.1.9. Setting Up Word Problems

Hints

1. You should have the difference between two fractional expressions. To combine them, you would needto find a common denominator (which would be the product of the two denominators) and combinethem. Return to the problem.

Solutions to Exercises 120

Solutions to Exercises

Exercise 0.0.1. The author of the document is blessed to have two wonderful daughters, Hannah andDarby. Exercise 0.0.1

Exercise 1.3.1. We need to add 4 to both sides, since half of 4 is 2, and 22 = 4. (This will complete thesquare in the x expression.) We also need to add 9 to both sides, since half of 6 is 3, and 32 = 9. (This willcomplete the square in the y expression.) Thus we have x2 +4x+4+y2 +6y+9 = 10+4+9, which becomes(x+2)2 +(y +3)2 = 23, which we recognize as a circle centered at (−2,−3) of radius

√23. Exercise 1.3.1

Exercise 1.4.1. According to the point-slope form of the equation of a line, our line must have the formy−(−4) = 5(x−(−2). Simplifying this yields the equation y+4 = 5(x+2), or y = 5x+10−4, or y = 5x+6.Thus the y - intercept is 6. Exercise 1.4.1

Exercise 1.4.2. First find the slope. m = 5−(−3)−2−2 = −2

Now take the slope and one of the points (2,−3) and put them into the point-slope formula. y− (−3) =−2(x− 2) By simplifying and solving for y, we get the equation in slope-intercept form: y = −2x− 1

Exercise 1.4.2

Exercise 1.4.3. The line 3x + y = 7 can be written y = −3x + 7, so it has slope −3. The line −x− 3y = 9can be written as y = −1

3 x− 3, so it has slope −13 . These two slopes are not the same, nor do they multiply

together to be −1, so these lines are neither parallel nor perpendicular. Exercise 1.4.3

Exercise 1.6.1. We have f(g(x)) = f(x2 − 5). Now since f is the “multiply by 3 and add 2” rule,f(x2−5) = 3(x2−5)+2. Simplifying this gives 3x2−15+2 = 3x2−13. Thus f(g(x)) = 3x2−13. On the other

hand, g(f(x)) = g(3x+2), and since g is the “square and subtract 5” rule, we have g(3x+2) = (3x+2)2−5,and simplifying this gives (9x2 + 12x + 4) − 5 = 9x2 + 12x − 1. Thus g(f(x)) = 9x2 + 12x − 1. Nowf(g(2)) = −1 and g(f(2)) = 59. This is enough to deduce that f(g(x)) 6= g(f(x)), since for functions to beequal, the outputs must be equal for all their inputs. Exercise 1.6.1

Exercise 1.6.2. Note that the first things that are “done” by h to an input x is that x is multiplied by 4and then 1 is added. Later, this number is squared and 9 is subtracted. Thus one possible answer is to letthe “inner” function g(x) = 4x + 1 and the “outer” function f(x) = x2 − 9. Then f(g(x)) = f(4x + 1) =(4x + 1)2 − 9. Exercise 1.6.2

Exercise 1.8.1. Yes. If f and g are both odd, then (f + g)(−x) = f(−x) + g(−x) = −f(x) + −g(x) =−(f(x) + g(x)) = −(f + g)(x). Exercise 1.8.1

Exercise 1.8.2. Exercise 2.1.1.

First compute f(4+h)−f(4)h . We have f(4+h) = (4+h)3−3(4+h), which can be expanded to 64+48h+

12h2 + h3 − 12− 3h = h3 + 12h2 + 45h + 52. Thusf(4 + h)− f(4)

h=

h3 + 12h2 + 45h + 52− 52h

=(h)(h2 + 12h + 45)

h= h2 + 12h + 45.


Now if we shrink h → 0, we are left with 45. Thus the slope of the tangent line at x = 4 is 45. The equationof the tangent line would thus be y − 52 = 45(x− 4).

Exercise 2.1.1

Exercise 2.1.2.

The slope of the secant line is given by f(6+h)−f(6)h =

√6+h+3−4−(−1)

h . Simplifying this yields√

9+h−3h To

simplify this expression further, we need the technique of multiplying by the conjugate. The conjugate of theexpression

√9 + h− 3 is the expression

√9 + h + 3. So we multiply both the numerator and the denomator

by this conjugate quantity, yielding√

9 + h− 3h

·√

9 + h + 3√9 + h + 3

=(9 + h)− 9

h · (√

9 + h + 3)=

h

h · (√

9 + h + 3)=

1√9 + h + 3

.

Then as h → 0, we obtain a result of 16 . Thus the slope of the tangent line at x = 6 is 1

6 .

Exercise 2.1.2

Exercise 2.1.3.

We need to compute f(3+h)−f(3)h . We have f(3 + h) = 3

3+h+1 = 34+h . Thus we have

f(3 + h)− f(3)h

=3

4+h −34

h.

To simplify this last expression, we should find a common denominator between 4 + h and 4, which shouldbe 4(4 + h). Thus, the above expression is equal to

124(4+h) −

(4+h)(3)4(4+h)

h=

12−12−3h4(4+h)

h=

−3h

4(4 + h)h.

Cancelling the h’s leaves −34(4+h) . Shrinking h → 0 yields a value of −3

16 .

Exercise 2.1.3

Exercise 2.1.4.

If we compute f(1+h)−f(1)h we get −16(1+h)2+64−48

h which simpifies to −16−32h−16h2+64−48h = −32h−16h2

h =(h)(−32−16h)

h = −32− 16h. Shrinking h → 0 yields a value of −32.


Exercise 2.1.4

Exercise 2.2.1.

Exercise 2.2.1

Exercise 2.2.2.

Exercise 2.2.2

Exercise 2.2.3.

Exercise 2.2.3

Exercise 2.2.4.

Exercise 2.2.4

Exercise 2.3.1.

First note that it might pay to multiply both the numerator and denominator by the conjugate of thenumerator. Thus we would have

limx→9

√x− 3

x− 9·√

x + 3√x + 3

= limx→9

x− 9(x− 9)(

√x + 3)

= limx→9

1√x + 3

.

Now it isn’t hard to believe that when x is really close to 9, 1√x+3

is really close to 16 , so limx→9

√x−3

x−9 = 16 .

Exercise 2.3.1

Exercise 2.3.2.


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It is clear from the diagram that when x is close to 0, f(x) is close to 0 too, so limx→0 f(x) = 0. However,when x is close to but a little less than 2, f(x) is 1

2 , while when x is close to but a little bigger than 2, f(x)is close to 2. Thus there is not a single value for which it can be said that when x is close to 2,f(x) is close to that value. Thus limx→2 f(x) does not exist.

Exercise 2.3.2

Exercise 2.3.3.


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Note that when x is close to (but not equal to) −1, f(x) is close to 1. It doesn’t matter what f(x) isdefined to be at −1, or even whether or not it is defined at −1. It is still true that when x is close to butnot equal to −1, f(x) is close to 1. Thus limx→−1 f(x) = −1.

Exercise 2.3.3

Exercise 2.3.4.

No. In fact, both limx→3+x+3x−3 and limx→3−

x+3x−3 fail to exist, because limx→3+

x+3x−3 = ∞ and limx→3−

x+3x−3 =

−∞. We can see this either from the graph of this function, or by analyzing the quotient algebraically nearx = 3. We see that when x is close to but a little bighger than 3, the numerator is close to 6 and thedenominator is close to 0 (but is a positive number.) Such numbers are very big, and get even bigger whenchoosing a value for x closer to 6. On the other hand, when x is close to but a little less than 3, the numeratoris still close to 6, and the denominator is still close to 0, but the denominator is a negative number. Sincethe ratio of a positive number and a negative number is always a negative number, we know that the outputis a “large” negative number. (“Large” in the sense that its absolute value is large,) and the numbers geteven “larger” when x is chosen closer to 3. Thus we see that both one-sided limits fail to exist.

Exercise 2.3.4

Exercise 2.4.1.The set of points 0 < |x − 3| < .04 is the set of points (3 − .04, 3) ∪ (3, 3 + .04), or in other words,

(2.96, 3) ∪ (3, 3.04). Exercise 2.4.1


Exercise 2.4.2.Note that the midpoint of the interval (2.4, 2.6) is 2.5, and the interval has “radius” .1. Thus this set can

be described as the set of all x so that |x− 2.5| < .1. Exercise 2.4.2

Exercise 2.4.3.

Suppose that |f(x)− 3| < .1. Then |2x− 1− 3| < .1.Then |2x− 4| < .1, so|x− 2| < .1

2 , so|x− 2| < .05, and these steps can all be reversed.

Exercise 2.4.3

Exercise 2.4.4.

Suppose that |f(x)− 3| < ε. Then |2x− 1− 3| < ε.Then |2x− 4| < ε, so|x− 2| < ε

2 , sothe corresponding value of δ should be ε

2 .

Exercise 2.4.4

Exercise 2.5.1.

p q p → q q → p

T T T TT F F TF T T FF F T T

We can see that they aren’t the same. Thus a statement p → q

might be true but its converse q → p false. An example of this is the statement “ If f(x) is differentiable on(a, b), then it is continuous on (a, b). This statment is true but its converse “If f(x) is continuous on (a, b)then it is differentiable on (a, b) is false.

Exercise 2.5.1

Exercise 2.5.2.

p q p → q q̃ → p̃

T T T TT F F FF T T TF F T T

We see that these are the same. The statement q̃ → p̃ is called

the contrapositive of p → q. Exercise 2.5.2

Exercise 2.5.3. “p whenever q” means the same thing as “q → p.”, and “p only if q” means the samething as “p → q.” The statement “p if and only if q” means p → q AND q → p. This statement is oftenabbreviated “iff” and perhaps written p ↔ q.

Exercise 2.5.3


Exercise 2.6.1.We must show that for any ε > 0, there is a corresponding δ > 0 so that |2x − 1 − 3| < ε whenever

0 < |x− 2| < δ. Working backwards,

Suppose that |f(x)− 3| < ε. Then |2x− 1− 3| < ε.Then |2x− 4| < ε, so|x− 2| < ε


2 .

Now to give a nice forwards formal presentation we could say:

Given ε > 0, let δ = ε2 .

Then if 0 < |x− 2| < ε2 ,

Then |2x− 4| < ε, so|2x− 1− 3| < ε

so |f(x)− 3| < ε.

Exercise 2.6.1

Exercise 2.6.2.We must show that for any ε > 0, there is a corresponding δ > 0 so that |4x + 2 − 6| < ε whenever

0 < |x− 1| < δ. Working backwards,

Suppose that |f(x)− 6| < ε. Then |4x + 2− 6| < ε.Then |4x− 4| < ε, so|x− 1| < ε


4 .

Now to give a nice forwards formal presentation we could say:

Given ε > 0, let δ = ε4 .

Then if 0 < |x− 1| < ε4 ,

Then |4x− 4| < ε, so|4x + 2− 6| < ε

so |f(x)− 6| < ε.

Exercise 2.6.2

Exercise 2.7.1. Since the square root function is continous on its domain, and the expression underneaththe square root is positive, this doesn’t present any problems. Also, x+1

x2−4 is a rational function, which iscontinuous wherever it is defined, but note that it isn’t defined at x = 2 or x = −2. Thus, at these pointsh(x) is definitely not continuous, since it is not defined there. At every other point h(x) is the composition ofcontinuous functions, and is thus continuous. So the only discontinuities are at x = 2,−2. Exercise 2.7.1


Exercise 2.7.2. We need to have f̃(4) = limx→4 f(x), and this can be shown to be 8. (Do it!) Thus if wedefine f̃(4) = 8, then f and f̃ are the same everywhere except at 4, and f̃ is continuous at x = 4.

Exercise 2.7.2

Exercise 3.1.1.

First note that the point (9, 12) is on the tangent line, so the equation of the tangent line is

y − 12 = m(x− 9),

where m is the slope of the tangent line. Thus we must compute limx→9f(x)−f(9)

x−9 .

We have

limx→9

f(x)− f(9)x− 9

= limx→9

x +√

x− 12x− 9

= limx→9

√x + x− 12

x− 9

= limx→9

√x + x− 12

x− 9·√

x− (x− 12)√x− (x− 12)

= limx→9

x− (x− 12)2

(x− 9)(√

x− (x− 12))

= limx→9

x− (x2 − 24x + 144)(x− 9)(

√x− (x− 12))

= limx→9

−(x2 + 25x− 144)(x− 9)(

√x− (x− 12))

= limx→9

−(x2 − 25x + 144)(x− 9)(

√x− (x− 12))

= limx→9

−(x− 9)(x− 16)(x− 9)(

√x− (x− 12))

= limx→9

−(x− 16)(√

x− (x− 12))=

76

Thus the equation of the tangent line is given by y − 12 = 76 (x− 9). Exercise 3.1.1

Exercise 3.1.2.


We must compute limh→0

f(3 + h)− f(3)h

. This is equal to limh→0

3(3 + h)2 − 5(3 + h) + 3− 15h

limh→0

3(3 + h)2 − 5(3 + h) + 3− 15h

= limh→0

3(9 + 6h + h2)− 15− 5h +−12h

= limh→0

(27 + 18h + 3h2)− 5h− 27h

= limh→0

12h + 3h2

h

= limh→0

12 + 3h = 12

Thus the instantaneous rate of change of y with respect to x is 12. Exercise 3.1.2

Exercise 3.1.3.


f(2 + h)− f(2)h


√4 + h− 2

h. We have

limh→0

√4 + h− 2

h= lim

h→0

√4 + h− 2

h·√

4 + h + 2√4 + h + 2

= limh→0

4 + h− 4h(√

4 + h + 2)

= limh→0

h

h(√

4 + h + 2)

= limh→0

1√4 + h + 2

=14

Thus f ′(2) = 14 . Exercise 3.1.3

Exercise 3.1.4.


f(x + h)− f(x)h


1x+h −

1x

h. We have


limh→0

1x+h −

1x

h= lim

h→0

x−(x+h)(x)(x+h)

h

= limh→0

x− (x + h)hx(x + h)

= limh→0

−h

hx(x + h)

= limh→0

(−1)(x)(x + h)

=−1x2

Thus f ′(x) = −1x2 . Exercise 3.1.4

Exercise 3.2.1. For f(x) = −3x6 + 2x3 − 12x− 123, we have f ′(x) = −18x5 + 6x− 12. Exercise 3.2.1

Exercise 3.2.2.

limh→0

2x+2h+33x+3h−1 −

2x+33x−1

h

= limh→0

2x+2h+33x+3h−1 ·

3x−13x−1 −

2x+33x−1 ·

3x+3x−13x+3h−1

h

= limh→0

(2x+2h+3)(3x−1)−(3x+3x−1)(2x+3)(3x−1)(3x+3h−1)

h

= limh→0

(6x2 + 6hx + 9x− 2x− 2h− 3)− (6x2 + 6hx− 2x + 9x9h− 3)(h)(3x− 1)(3x + 3h− 1)

= limh→0

−11h

(h)(3x− 1)(3x + 3h− 1)

= limh→0

−11(3x− 1)(3x + 3h− 1)

=−11

(3x− 1)2

Exercise 3.2.2

Exercise 3.3.1.

Our job is to show that limh→0

3√

(0+h)2− 3√0

h doesn’t exist. We can write 3√

h2 as h23 , and using the rules

of exponents, we can write h23

h as h23−1 = h

−13 = 1

h13

= 13√

h. Now limh→0

13√

hdoesn’t exist, so (0, 0) is a

singular point for this function. Exercise 3.3.1


Exercise 3.4.1. s′(t) = v(t) = 6t2 − 24. Now, this equation is zero when t = 2 or t = −2, so those timesrepresent when the object is stopped. An analysis of the sign of v(t) shows that v(t) is positive on the interval(−∞,−2) and on (2,∞). Thus the object is moving to the right during those intervals. On the interval(−2, 2), the object is moving to the left, since velocity is negative on that interval. Now a(t) = v′(t) = 12t,which is zero at time t = 0. A quick check shows that the acceleration is positive for t > 0 and negativefor t < 0. Thus during the time interval (−∞,−2) the object is moving to the right but slowing down. Att = −2, the object is stopped. From t = −2 to t = 0 the object is moving to the left and speeding up. Attime t = 0 the acceleration becomes positive, so the object continues to move to the left, but starts to slowdown. At t = 2, the object is stopped, and thereafter moves to the right and speeds up. Exercise 3.4.1

Exercise 3.4.2.

1. At time t = 4, the population has size p(4) =√

4(42 + 3(4)) = 2(16 + 12) = 56. At time t = 9, thepopulation has size p(9) =

√9(92 + 3(9) = 3(81 + 27) = 324. Thus over that time interval of length 5,

the change in p is 324− 56 = 268. Thus the average rate of change of the population is 2685 .

2. The instantaneous rate of change of the population would be given by p′(4). We have p′(t) =√

t(2t +3) + (t2 + 3t) 1

2 t−12 . Thus p′(4) =

√4 · 11 + (28) · 1

2(√

4)= 22 + 28

4 = 29.

Exercise 3.4.2

Exercise 3.4.3. We can rewrite the equation as P = mRTV . Recall that all of m, R, and T are being treated

as constants in this discussion. Thus we can find dPdV = −(mRT )V −2.

Exercise 3.4.3

Exercise 3.4.4. C(101) = 4.242606 (Use your calculator.) C(100) = 4.12 Thus the difference is .122606.Now C ′(x) = 1

2000 (2 + .008x + .024x2). So C ′(100) = 12000 (2 + .8 + 240) = 242.8

200 = .1214. Exercise 3.4.4

Exercise 3.5.1. For f(x) =√

12x5 − 3x2 + 4x + 3, the inside function n(x) is 12x5−3x2 +4x+3, while theoutside function is

√x. Thus the expression m′(n(x) would be 1

2√

12x5−3x2+4x+3, while n′(x) = 60x4−6x+4.

Thus f ′(x) = 60x4−6x+42√

12x5−3x2+4x+3. Exercise 3.5.1

Exercise 3.5.2. Let f(x) = (x + x12 )

12 . The inside function is given by x + x

12 while the outer function is

given by x12 . Thus the derivative of f is given by 1

2

(x + x

12

)−12 ·(1 + 1

2√

x

). Exercise 3.5.2

Exercise 3.5.3. Let s(t(x)) be called n(x) for the moment. Then f(x) = r(n(x)), so f ′(x) = r′(n(x)) ·n′(x).Now n′(x) = s′(t(x)) · t′(x), by the chain rule. Thus (substituting) we get

f ′(x) = r′(s(t(x)) · s′(t(x)) · t′(x).

Exercise 3.5.3

Exercise 3.6.1.


If xy = 3, then differentiating the left hand side with respect to x (recalling that y is a function of x) weget x dy

dx + y(1) from the product rule. The derivative of the right hand side is zero, since we have a constanton that side of the equation. Thus we have

xdy

dx+ y = 0,

sody

dx=−y

x.

As suggested, since this function is easy to write explicitly as y = 3x , we can check our answer by differentiaing

this expression. We get dydx = −3

x2 . But if y = −3x , then the expressions −y

x and −3x2 are the same.

Exercise 3.6.1

Exercise 3.6.2.Differentiating

√xy = x + y + 1 with respect to x yields

12(xy)

−12

(x

dy

dx+ y

)= 1 +

dy

dx.

There are two terms on the left hand side, one of which has a factor of dydx . That is also true for the right

hand side. Gathering terms with a dydx factor on the left yields

12(xy)

−12 · xdy

dx− dy

dx= 1− 1

2(xy)

−12 y.

Factoring out the dydx factor and solving yields

dy

dx=

1− 12 (xy)

−12 y

12 (xy)

−12 x− 1

.

Exercise 3.6.2

Exercise 3.7.1. Suppose you have a degree k polynomial like f(x) = akxk + . . .+a1x+a0. Each derivativeof this will result in a polynomial of lower degree, in fact it will have degree one less than what came before.So the first derivative has degree k − 1, the second has degree k − 2 and so on. The kth derivative wouldthus have degree k − k = 0. But a zeroth degree polynomial is a constant function, and the derivative of aconstant function is zero. Thus the k + 1st derivative (and thus all those after that) result in the constantzero. Exercise 3.7.1

Exercise 3.7.2. Note that f ′(x) = −2x3 , f ′′(x) = 2·3

x4 , f ′′′(x) = −2·3·4x5 . It looks like the numerator for the

kth derivative has a (k + 1)! in it, as well as either a plus or minus 1 factor. In fact, the sign looks to bepositive when k is even and negative when k is odd, so a factor of −1k would take care of this alternatingsign. The denominator is always x to a power, and the power looks to be k + 2 for the kth derivative. Thusit looks like we have the following formula for the kth derivative:

dky

dxk=

(−1)k(k + 1)!xk+2

.

Exercise 3.7.2


Exercise 3.8.1. Draw a picture of such a cylinder, and label the height h. The volume of the water in thecylinder is the area of the base times the height, so it is V = 4πh. Thus dV

dt = 4π dhdt , and since dV

dt = 3, wehave dh

dt = 3 · 4π = 12π. This would be measured in feet per minute. Exercise 3.8.1

Exercise 3.8.2. Note that when the water is h feet deep, the volume of water is the area of a triangularcross section times the width (which is 10 feet). Let x be the length of water in the pool. Then the area ofthe triangle is 1

2xh, so the volume of water is given by V = 12 · 10xh = 5xh. What do we do about the fact

that we don’t know anything about dxdt ? Using similar triangles, we can write x

h = 255 , so x = 5h. Thus we

can rewrite the volume as V = 5(5h)h = 25h2. Thus dVdt = 50hdh

dt , so 3 = 50hdhdt , so dh

dt = 350h . When h = 3,

we have dhdt = 1

50 . Exercise 3.8.2

Exercise 3.9.1. Exercise 3.9.2. Exercise 4.1.1.

Since it is 2π units around the unit circle, half-way around would correspond with t = π. Thus the pointon the unit circle which corresponds with t = π would be (−1, 0). Therefore, cos(π) = −1 and sin(π) = 0.

Exercise 4.1.1

Exercise 4.1.2.

The point on the circle corresponding with t = 3π2 would be the point that is 3

4 of the way around thecircle, since 3π

2 is three fourths of 2π. Thus the point in question is (0,−1), and thus sin(

3π2

)= −1 and

cos(

3π2

)= 0 Exercise 4.1.2

Exercise 4.1.3.

Since one radian corresponds with 180π degrees, π

6 radians would correspond with π6 ·

180π = 30 degrees.

Exercise 4.1.3

Exercise 4.1.4.

Since one degree corresponds with π180 radians, 225 degrees would correspond with 225 · π

180 = 5π4 radians.

Exercise 4.1.4

Exercise 4.2.1.

The point on the unit circle corresponding with t = 7π6 is directly opposite the point on the unit circle

corresponding with t = π6 . This is because 7π

6 = π + π6 . The point associated with π

6 is (√

32 , 1

2 ), so the pointdirectly opposite the circle from this point is (−

√3

2 ,− 12 ). Thus sin

(π6

)= −1

2 . Exercise 4.2.1

Exercise 4.2.2.

The point on the unit circle associated with −3π4 is directly opposite the point associated with t = π

4 .This is because −3π

4 + π = π4 . The point associated with t = π

4 is (√

22 ,

√2

2 ). Thus the point associated with−3π

4 is (−√

22 ,−

√2

2 ), and tan(−3π

4

)= 1, since it is the ratio of these two coordinates. Exercise 4.2.2


Exercise 4.2.3.

If sin(x) + cos(x) = 0, then sin(x) = − cos(x). Of course, we also know that sin2(x) + cos2(x) = 1, soby substituting) we have that (− cos(x))2 + cos2(x) = 1, so 2 cos2(x) = 1, from which it is easy to see that| cos(x)| =

√2

2 . Thus we are looking at two points on the unit circle: (−√

22 ,

√2

2 ) and (√

22 ,−

√2

2 ). Thesepoints are directl opposite the circle from each other, and the first one is the reflection about the y axis ofthe point (

√2

2 ,√

22 ), which corresponds with x = π

4 . Thus we are looking at x = 3π4 , and the sum of this with

π,−π, 2π,−2π, . . .. Thus we can say that the solutions all have the form x = 3π4 + kπ, k ∈ Z.

Exercise 4.2.3

Exercise 4.2.4. Clearly if 2 cos(2x) + 1 = 0, then cos(2x) = − 12 . But an analysis of the unit circle shows

that the points where the cosine function is − 12 are associated with 2π

3 and 4π3 . Thus, since we are looking

for where cos(2x) is − 12 , we must have 2x = 2π

3 and 2x = 4π3 . Thus we get that x = π

3 and x = 2π3 . Also, if

z is a solution, then z + π is a solution, because cos(2(z + π)) = cos(2z + 2π) = cos(2z). Thus the solutionshave the form π

3 + kπ, k ∈ Z and 2π3 + kπ, k ∈ Z. Exercise 4.2.4

Exercise 4.4.1.

The cotangent function is defined as cot(x) = cos(x)sin(x) . This will be undefined when sin(x) = 0, which

occurs at 0,±π,±2π, . . .. Every other real number is in the domain of both sin(x) and cos(x), and so is inthe domain of the quotient of these two. Thus the domain is {x : x 6= kπ, k ∈ Z}. Exercise 4.4.1

Exercise 4.4.2.

The cotangent function is defined by cot(x) = cos(x)sin(x) . This function is undefined where sin(x) = 0, which

is at all multiples of π. If x is a little greater than 0, then cos(x) is near one, while sin(x) is very small (butis positive.) Thus the ratio is a large number. We can see from this analysis that limx→0+ cot(x) = +∞. Onthe other hand, if x is a litle less than 0, then sin(x) is negative, while cos(x) is still close to 1, so is positive.Thus the ratio of the two, while large in absolute value, is negative. Thus limx→0− cot(x) = −∞. Using thisinformation along with a few key plotted points (like cot(π

2 ) = 0 and cot(π4 ) = 1) leads to the graph. It is

also useful to note that cot(x) is periodic with period π.Exercise 4.4.2

Exercise 4.4.3.

The sketches are coming soon. Exercise 4.4.3

Exercise 4.5.1. Exercise 4.5.2.

First note that we can rewrite cos(2x) as follows:


cos(2x) = cos(x + x)= cos(x) cos(x)− sin(x) sin(x)= cos2(x)− (1− cos2(x))= 2 cos2(x)− 1

Thus we have that cos(2x) = 2 cos2(x)−1, and solving this expression for cos2(x) yields the desired identity.Exercise 4.5.2

Exercise 4.6.1.

limh→0

cos(h)− 1h

= limh→0

cos(h)− 1h

· cos(h) + 1cos(h) + 1

= limh→0

cos2(h)− 1h · (cos(h) + 1)

= limh→0

− sin2(h)h · (cos(h) + 1)

= limh→0

sin(h)h

· limh→0

− sin(h)cos(h) + 1

= 1 · −01 + 1

= 0

Exercise 4.6.1

Exercise 4.6.2.

ddx sec(x) = d

dx1

cos(x)

= cos(x)·0−1·(− sin(x))cos2(x)

= sin(x)cos2(x)

= sec(x) tan(x)

ddx cot(x) = d

dxcos(x)sin(x)

= sin(x)(− sin(x)−cos(x) cos(x))sin2(x)

= −1sin2(x)

= − csc2(x)

ddx csc(x) = d

dx1

sin(x)

= sin(x)·0−1·cos(x)sin2(x)

= − cos(x)sin2(x)

= − cot(x) csc(x)


Exercise 4.6.2

Exercise 4.6.3.

If f(x) = sin(

x2+12x−3

), then f ′(x) = cos

(x2+12x−3

)· (2x−3)(2x)−(x2+1)(2)

(2x−3)2 .

Exercise 4.6.3

Exercise 4.6.4.

If g(x) = (√

x2 − 4)·tan(√

x2 − 4), then g′(x) = (√

x2 − 4)·sec2(√

x2 − 4)· 12 (x2−4)−12 ·2x+tan(

√x2 − 4)·

12 (x2 − 4)

−12 · 2x.

Exercise 4.6.4

Exercise 5.1.1. The graph of y = (x− 3)2 + 2 has the same shape as the graph of y = x2, except that it isshifted 3 units to the right and 2 units up. Thus, the vertex is now at (3, 2), and the graph opens upward.Thus there is an absolute minimum of 2 at x = 3. On the other hand, the graph of y = −3 − (x + 2)2 isthe same as the graph of y = x2 except that it is flipped about the x axis, then shifted 2 units to the leftand then 3 units down. Thus the vertex is now at (−2,−3), and the parabola opens downward. Thus theabsolute maximum value is −3 at x = −2. Exercise 5.1.1

Exercise 5.1.2.

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It looks like the pictured function has local minimums at about x = −3.1, x = .8 and x = 4.7. (It is hardto be too precise on such a graph.) It looks like there are local maximums at x = 0 and x = 1.6.

Exercise 5.1.2

Exercise 5.1.3. Note that the function never reaches a value of 11, although it gets very close. Since itdoesn’t actually reach the value of 11, this cannot be the maximum value. It is tempting to say something


like: “the maximum value is whatever real number is the closest real number to 11 which is less than 11”,but a little thought about the nature of real numbers leads one to realize that that statement is silly. Thereis no such real number, so this function does not have a maximum. Likewise, the function is always greaterthan 5, but as x → 2+, we see that we can get as close to a value of 5 as we desire. Thus there is no minimumvalue of this function.

The function g(x) as similar problems. The function almost gets as high as 20, and almost gets as lowas 0, but never quite makes it to these values. Thus, there is no maximum or minimum.

Exercise 5.1.3

Exercise 5.1.4. First note that the function is continuous (since it is a polynomial), and we are given aclosed interval. Thus we know that this function has a maximum and a minimum on this interval. Thederivative is f ′(x) = 20x4 − 20x3 = 20x3(x − 1). Setting f ′(x) = 0, we get stationary points at x = 0 andx = 1. Comparing the values of the function at these points we have:

x f(x)−3 −13740 31 22 51

From this chart we can see that there is a maximum value of 51 at x = 2 and a minimum value of −1374at x = −3. Exercise 5.1.4

Exercise 5.3.1.

Yes, Rolle’s Theorem does apply because since this is a rational function, it is continuous and differentiableon its domain, which consists of all real numbers except x = −8. Since −8 isn’t part of the interval −1, 3],this function is continuous and differentiable on the given interval. Differentiating we have

f ′(x) =(x + 8)(2x− 2)− (x2 − 2x− 3)(1)

(x + 8)2Quotient Rule

=(2x2 + 14x− 16)− (x2 − 2x− 3)

(x + 8)2Expanding in the numerator

=x2 + 16x− 13

(x + 8)2Combining like terms

This derivative is zero when its numerator is zero, which is when x2 + 16x − 13 = 0, which (by thequadratic formula is when x = −8 ±

√77. Now according to the theorem, at least one of these must be in

the interval [−1, 3], and since −8 +√

77 ≈ .775, we see that −8 +√

77 is the promised value of c. Note that−8−

√77 isn’t in the interval [−1, 3], so we disregard it. Exercise 5.3.1

Exercise 5.3.2.


Note that g′(x) = 23x

−13 = 2

3 3√x. This is never zero since its numerator is never zero. This does not violate

Rolle’s Theorem, because g(x) is not differentiable at x = 0, so it is not differentiable on (−1, 1). Rolle’sTheorem only applies to functions which are continuous on the given closed interval, and differentiable onthe corresponding open interval. Exercise 5.3.2

Exercise 5.3.3.

No, it does not apply. The function is differentiable on (0, 1), but it is not continuous at 0, so is notcontinuous on [0, 1]. Note that g′(x) = −2, which is never zero. Exercise 5.3.3

Exercise 5.6.1.

Use the analysis done previously to find the relative extrema for f(x) = xx2+2 .

We already found that there were stationary points at x = ±√

2, and that f(x) is increasing on (−√

2,√

2)and decreasing on (−∞,−

√2) and on (

√2,∞). Using the first derivative test, we know see that there is a

local maximum at x =√

2 and a local minimum at x = −√

2. The local maximum value is√

24 while the

local minimum value is −√

24 . Exercise 5.6.1

Exercise 5.7.1.

Continue studying f(x) = xx2+2 by analyzing its concavity.

Recall that f ′(x) = 2−x2

(x2+2)2 . So

f ′′(x) =(x2 + 2)2(−2x)− (2− x2)(2)(x2 + 2)(2x)

(x2 + 2)4Quotient Rule

=(x2 + 2)((x2 + 2)(−2x)− (4x)(2− x2))

(x2 + 2)4Factoring the Numerator

=(x2 + 2)((−2x3 +−4x)− (8x− 4x3))

(x2 + 2)4Distributing

=(−2x3 +−4x)− (8x− 4x3)

(x2 + 2)3Cancelling Factors

=(2x3 − 12x)(x2 + 2)3

Combining Like Terms

=(2x)(x2 − 6)

(x2 + 2)3Factoring the Numerator

Now an analysis of this second derivative for places where the sign might change shows that x = 0 andx = ±

√6 are possibilites, since that is where the numerator (and thus f ′′(x)) is zero. Also note that

there are no places where the denominator is zero. The potential intervals of concavity are thus (−∞,−√

6),(−√

6, 0), (0,√

6), and (√

6,∞). Using test values we see that the second derivative is negative on (−∞,−√

6)


and on (0,√

6), and is positive on (−√

6, 0) and on (√

6,∞). Thus f(x) is concave down on (−∞,−√

6)and on (0,

√6), and is concave up on (−

√6, 0) and on (

√6,∞). There are points of inflection at (0, 0),

(−√

6, −√

68 ), and (

√6,

√6

8 ). Exercise 5.7.1

Exercise 5.7.2.

Use the analysis done previously to find the relative extrema for f(x) = xx2+2 .

We already found that there were stationary points at x = ±√

2, and we found f ′′(x) = (2x)(x2−6)(x2+2)3 . Since

f ′′(−√

2) = −2√

2(−4)43 > 0 there must be a relative minimum at x = −

√2. Since f ′′(

√2) = 2

√2(−4)43 < 0,

there must be a relative maximum at x =√

2. Note that this coincides with what we obtained using thefirst derivative test. Exercise 5.7.2

Exercise 5.8.1.

We must compute limx→0sin(x)

x2 . Let’s first focus on the limit as x → 0+. Recall that limx→0sin(x)

x = 1,

so it would be useful to write sin(x)x2 as

sin(x)x

x . Then as x → 0+, the numerator is approaching 1, whilethe denominator is approaching zero (but is positive), thus the ratio is increasing without bound. Thus

limx→0+

sin(x)x

x = ∞. This is enough to show that there is a vertical asymptote at x = 0. A careful analysisof the limit as x → 0− shows that that limit is −∞. Exercise 5.8.1

Exercise 5.9.1.

We are trying to show that limx→∞1x = 0. We must show that for any ε > 0, there is a value of M so

that if x > M , |f(x)−0| < ε. Working backwards from this last expression, we see that we need 1x < ε, or in

other words, we need x > 1ε . So let’s take M = 1

ε . Then (starting over) if x > M , we have x > 1ε so 1

x < ε.Also since x is some positive number, we can write

∣∣ 1x

∣∣ < ε or∣∣ 1x − 0

∣∣ < ε. Thus we have shown what we setout to show.

Since limx→∞1x = 0, the line y = 0 must be a horizontal asymptote.

Exercise 5.9.1

Exercise 5.9.2. Note that 1xr =

(1x

)r. Thus limx→∞1xr =

(limx→∞

1x

)r = 0r = 0. Exercise 5.9.2

Exercise 5.10.1.

Sketch a graph of f(x) = x2−2x3 , following the above guidelines.

1. Note the domain of f(x). Since f(x) is a rational function, its domain is the set of all real numbersexcept those which make the denominator 0. Thus any value for which x3 = 0 is not in the domain,but only 0 meets this requirement. Thus the domain consists of all real numbers except 0. We will besure to include 0 on any sign chart we make.


2. Look for vertical and horizontal asymptotes. We need to consider the limits as we approach 0from each side. Consider limx→0+

x2−2x3 . Since the numerator is approaching a number other than 0,

and the denominator is approaching zero, the ratio is getting “big”, but is a negative number since thenumerator is near −2 and the denominator is postive. Thus limx→0+

x2−2x3 = −∞. On the other hand,

limx→0−x2−2

x3 = ∞. We have a vertical asymptote at x = 0. To look for horizontal asymptotes, considerlimx→∞

x2−2x3 . A quick check shows that this limit is 0. (Divide the numerator and denominator by x3

to see this.) Thus y = 0 (the x-axis) is a horizontal asymptote.

3. Investigate the Monotonicity of f(x). A little work (do it!) shows that f ′(x) = 6−x2

x4 . This is zeroat x = ±

√6. By making the sign chart, we see that f(x) is decreasing on (−∞,−

√6) and on (

√6,∞).

It is increasing on (√

6, 0) and on (0,√

6).

4. Locate Relative Extrema of f(x). Using the sign charts, we see that there is a local maximum atx =

√6 and a local minimum at x = −

√6.

5. Investigate the Concavity of f(x). A little work (do it!) shows that f ′′(x) = 2x2−24x5 . This is zero

at x = ±√

12. By making the sign chart for f ′′(x), we see that f(x) is concave down on (−∞,−√

12)and on (0,

√12). It is concave up on (−

√12, 0) and on (

√12,∞). There are inflection points at

x = ±√

12. Also, applying the second derivative test to x = ±√

6 confirms our conclusions that therea local minimum at x = −

√6 and a local maximum at x =

√6.

6. Preparation for Graphing.

(a) Plot a few important points. We will want to plot all the points (−√

12,− 1012√

12), (√

12, 1012√

12),

(−√

6,− 46√

6), (

√6, 4

6√

6).

(b) Intercepts. There is no y intercept since x = 0 is not in the domain of f(x). A quick check forx intercepts shows that x = ±

√2 are intercepts.

(c) Asymptotes. Since the asmptotes are the axes, these are already drawn. Recall that f(x) isapproaching +∞ as x → 0− and f(x) is approaching −∞ as x → 0+.

(d) Symmetry. Since f(−x) = (−x)2−2(−x)3 = −x2−2

x3 = −f(x), this function is symmetric about theorigin. We will think about this when sketching.

7. Sketch the Graph. It should look like this:


-4 -2 2 4

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

Exercise 5.10.1

Exercise 5.11.1.

Let the outside of the entire pen structure have dimensions x × y. Then the objective is to maximizethe area A = xy, subject to the constraint that there is only 200 feet of fence available. A diagram showsthat in building this pen, the farmer will use 4x feet to build the “vertical” walls and 2y feet to build the“horizontal” walls. Thus the total number of feet used is 4x+2y, which must equal 200. Thus 4x+2y = 200is the constraint equation. Solving this for y yields y = 100−2x, and substituting this into the area equationgives an objective function A = x(100 − 2x) = 100x − 2x2. It is quite easy to maximize this downwardopening parabola — the vertex is at the x value where A′(x) = 0, so we need A′(x) = 100 − 4x to bezero, which occurs when x = 25. Note that while the natural domain for this parabola is the set of all realnumbers, it is only true that x ≥ 0 and y ≥ 0 if x is between 0 and 50. One can easily check that this is amaximum, in fact, a global maximum, since the area function A(x) is increasing on [0, 25] and decreasing on[25, 50]. Of course when x = 25, the constraint equation tells us that y = 100− 2(25) = 50. Thus the outerdimensions of the pen should be 25× 50. Exercise 5.11.1

Exercise 5.12.1. The formula f ′(xn)(xn)−f(xn)f ′(xn) = xn+1 becomes

xn+1 =(3x2

n + 1)(xn)− (x3n + xn − 4)

3x2n + 1

which simplifies to

xn+1 =2x3

n + 43x2

n + 1.

If we start with an initial guess of x0 = 1, we have the following table:


i xi f(xi)

0 1 -21 1.5 0.8752 1.38709677 0.05592292973 1.37883895 0.0002832023464 1.3787967 0.00000000738259587

So the root seems to be pretty close to 1.3787. Exercise 5.12.1

Exercise 5.13.1.

A general antiderivative of y = 2x+8 should involve the sum of the antiderivatives of 2x and 8. We havealready seen that an antiderivative of 2x is x2, and a little thought shows that an antiderivative of a constantlike “8” is the linear function 8x. Thus one specific antiderivative of y = 2x + 8 would be F (x) = x2 + 8x.By the “+C” theorem, the general antiderivative would be x2 + 8x + C. Exercise 5.13.1

Exercise 5.13.2.

By the sum and constant multiplier rules, we can simply try to find antiderivatives for x3 and x, andthen string these together with the proper constant factors and plus signs. A little thought reveals thatan antiderivative of x3 should be a 4th degree monomial, and some trial and error yields the correct resultof x4

4 . Similarly, an antiderivative of x is x2

2 . Thus the general antiderivative of y = 5x3 + 8x + 10 is5 · x4

4 + 8 · x2

2 + 10x + C = 54x4 + 4x2 + 10x + C. Exercise 5.13.2

Exercise 5.14.1.

We are given a(t) = −32, so v(t) = −32t + C1. since the initial velocity is 12, we know that v(0) = 12.Thus v(0) = 12 = −32(0)+C1, so C1 = 12. We then have that s(t) = −32 t2

2 +12t+C2 = −16t2 +12t+C2.Also, s(0) = 3 = −16(0)2 + 12(0) + C2, so C2 = 3. Thus s(t) = −16t2 + 12t + 3. The ball is at its maximumheight when its velocity is zero. Thus, the maximum occurs when −32t + 12 = 0, or when t = 12

32 = 38 . The

maximum is −16 964 + 12 3

8 + 3 = 214 . Exercise 5.14.1

Exercise 5.14.2.

We have v(t) = sin t + − cos t + C1. Also, v(0) = 1 = sin(0) + cos(0) + C1 = 1 + C1, so C1 = 0. Thusv(t) = sin t + − cos t. Therefore, s(t) = − cos t + − sin t + C2. Also, s(0) = 4 = − cos(0) + − sin(0) + C2 =−1 + C2, Thus C2 = 5. We therefore have s(t) = − cos t +− sin t + 5 Exercise 5.14.2

Exercise 6.1.1.

We recognize the numbers 4, 9, 16 and 25 as the squares of the integers 2, 3, 4, and 5 repsectively. Thus,we suspect that the terms involve the square of i. We must start with i = 2 and end with i = 10 to get the


correct sum. We have

4 + 9 + 16 + . . . + 100 =10∑

i=2

i2.

Exercise 6.1.1

Exercise 6.1.2.∑8i=3 i2+i+1 = (32+3+1)+(42+4+1)+. . .+(82+8+1) = 13+21+31+43+57+73. Exercise 6.1.2

Exercise 6.2.1.

First of all, the reason the 2nd row ends with n+12 is that that is the number which is one more than

what the first row ends with. Specifically,n− 1

2+ 1 =

n− 12

+22

=n + 1

2.

To see that the formula∑n

i=1 i = n(n+1)2 holds when n is even, we write

1 2 3 . . . n2

n n− 1 n− 2 . . . n+22

and note that the columns add to n + 1, and that there are n2 such columns. Thus, the sum is n(n+1)

2 .Exercise 6.2.1

Exercise 6.2.2.

n∑i=1

4i3 + 6i− 1 =n∑

i=1

4i3 +n∑

i=1

6i−n∑

i=1

1 By 6.4

= 4n∑

i=1

i3 + 6n∑

i=1

i−n∑

i=1

1 By 6.5

= 4n∑

i=1

i3 + 6n∑

i=1

i− n By 6.6

= 4(

(n)(n + 1)(2n + 1)6

)+ 6

((n)(n + 1)

2

)− n By 6.1 and 6.2

Exercise 6.2.2

Exercise 6.3.1. Exercise 6.3.2. Exercise 6.4.1. Exercise 6.4.2. Exercise 6.4.3. Exercise 6.4.4.Exercise 7.1.1. Only 4

√−16 is not a real number. 5

√−243 = −3 which is an integer. Both π and −5√

2are

irrational numbers. Exercise 7.1.1

Exercise 7.1.2. a + β is an irrational number. aβ can be either rational or irrational depending on thevalues of a and β. For example, if a = 1 and β =

√2, then aβ =

√2 which is irrational. If a = 0 and


β =√

3, then aβ = 0 which is rational. Similarly, β + γ can be either rational or irrational. For example, ifβ = π and γ = −π, then β + γ = 0 which is rational. However, if β =

√2 and γ =

√2, then β + γ = 2

√2

which is irrational. Exercise 7.1.2

Exercise 7.1.3. Group the first two terms and the last two terms together. (2x3 + 3x2) + (−8x − 12)Notice that an x2 can be factored out of the first group and a −4 can be factored out of the second group.x2(2x + 3)− 4(2x + 3) Now notice that each group has the common factor (2x + 3). Factor out the (2x + 3).(2x + 3)(x2 − 4) The factor (x2 − 4) is a difference of squares which can be factored into (x + 2)(x − 2).Finally we end with (2x + 3)(x + 2)(x− 2) as the factored form. Exercise 7.1.3

Exercise 7.1.4. Group the second, third, and final terms together. x4 + (−x2 − 6x − 9) Factor out a −1from the group. x4 − (x2 + 6x + 9) Notice that x2 + 6x + 9 is a trinomial square which can be factored into(x + 3)2. So now we have x4 − (x + 3)2. Since x4 = (x2)2, x4 − (x + 3)2 is a difference of squares. So it maybe factored into (x2 + x + 3)(x2 − x− 3). Exercise 7.1.4

Exercise 7.1.5. Multiply both sides of the equation by 2. 2(3x + 5) = x− 76x + 10 = x− 7Now subtract x from both sides:5x + 10 = −7Now subtract 10 from both sides:5x = −17x = − 17

5 Exercise 7.1.5

Exercise 7.1.6. Multiply both sides of the equation by xy − ax:b2y = C(xy − ax)Factor out the x on the left side of the equation:b2y = Cx(y − a)Now divide both sides of the equation by C(y − a):

x =b2y

C(y − a)Exercise 7.1.6

Exercise 7.1.7. Multiply equation (7) by −3 and equation (8) by 4:

− 21x− 12y = −24 (7.9)

24x + 12y = 16 (7.10)

Add equations (9) and (10) and solve for x:3x = −8So x = − 8

3 .Now substitute x = − 8

3 into equation (8) and solve for y:


6(− 83 ) + 3y = 4

3y = 20So y = 20

3 .Hence the solution to the system is (− 8

3 , 203 ).

Exercise 7.1.7

Exercise 7.1.8.a) J| − 4.5|K = J4.5K = 4b) |J−4.5K| = | − 5| = 5c) JπK = 3 Exercise 7.1.8

Exercise 7.1.9.

a) False, let x = − 12 and y = − 1

2 .So J− 1

2KJ− 12K = (−1)(−1) = 1

But J(− 12 )(− 1

2 )K = J 14K = 0 < 1

b) False, let x = 2.5 and y = −2.5.So |J2.5K|J−2.5K = |2|(−3) = −6But J(2.5)(−2.5)K = J−6.25K = −7 < −6

Exercise 7.1.9

Exercise 7.1.10.

True.Proof: By the definition of absolute value, |x| > 0 and y > 0.

Then |x| ≥ J|x|K and |y| ≥ J|y|K.Since J|x|K ≥ 0, take the inequality |y| ≥ J|y|K and multiplythe leftside by |x| and the right side by J|x|K.This yields the inequality: |x||y| ≥ J|x|KJ|y|KBy a property of absolute value, |x||y|=|xy|Now apply the greatest integer function to both sides ofthe inequality. The resulting inequality isJ|xy|K ≥ J(J|x|KJ|y|K)K.Since J|x|KJ|y|K is an integer,J|xy|K ≥ J(J|x|KJ|y|K)K = J|x|KJ|y|KHence J|xy|K ≥ J|x|KJ|y|K.

Exercise 7.1.10

Solutions to Quizzes 145

Solutions to Quizzes

Solution to Quiz: Most of the illustrations were done by Brian Wagner, an undergraduate at FurmanUniversity. See the acknowledgments for the credits. End Quiz

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