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Integrated Calculus/Pre-Calculus

Mark R. Woodard

November 30, 2000



Contents

0 Introduction and Notes to the Reader 110.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .130.2 Notes to the reader. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .150.3 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17

1 Functions and their Properties 191.1 Functions & Cartesian Coordinates. . . . . . . . . . . . . . . . . . . . . . . . .21

1.1.1 Functions and Functional Notation. . . . . . . . . . . . . . . . . . . . . 21


1.1.2 Cartesian Coordinates and the Graphical Representation of Functions. . 341.2 Circles, Distances, Completing the Square. . . . . . . . . . . . . . . . . . . . . 43

1.2.1 Distance Formula and Circles. . . . . . . . . . . . . . . . . . . . . . . 431.2.2 Completing the Square. . . . . . . . . . . . . . . . . . . . . . . . . . .44

1.3 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .481.3.1 General Equation and Slope-Intercept Form. . . . . . . . . . . . . . . . 481.3.2 More On Slope. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .491.3.3 Parallel and Perpendicular Lines. . . . . . . . . . . . . . . . . . . . . . 50

1.4 Catalogue of Familiar Functions. . . . . . . . . . . . . . . . . . . . . . . . . .561.4.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .561.4.2 Rational Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . .581.4.3 Algebraic Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . .591.4.4 The Absolute Value Function. . . . . . . . . . . . . . . . . . . . . . . . 591.4.5 Piecewise Defined Functions. . . . . . . . . . . . . . . . . . . . . . . . 621.4.6 The Greatest Integer Function. . . . . . . . . . . . . . . . . . . . . . . 62

1.5 New Functions From Old. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .661.5.1 New Functions Via Composition. . . . . . . . . . . . . . . . . . . . . . 661.5.2 New Functions via Algebra. . . . . . . . . . . . . . . . . . . . . . . . 701.5.3 Shifting and Scaling. . . . . . . . . . . . . . . . . . . . . . . . . . . .70


2 Limits 752.1 Secant Lines, Tangent Lines, and Velocities. . . . . . . . . . . . . . . . . . . . 77

2.1.1 The Tangent Line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .772.1.2 Instantaneous Velocity. . . . . . . . . . . . . . . . . . . . . . . . . . .79

2.2 Simplifying Difference Quotients. . . . . . . . . . . . . . . . . . . . . . . . . .842.3 Limits (Intuitive Approach). . . . . . . . . . . . . . . . . . . . . . . . . . . . .902.4 Intervals via Absolute Values and Inequalities. . . . . . . . . . . . . . . . . . .1022.5 If – Then Statements. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1072.6 Rigorous Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . .1112.7 Computing Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1172.8 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .126

2.8.1 Definitions Related to Continuity. . . . . . . . . . . . . . . . . . . . .1262.8.2 Basic Theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1292.8.3 The Intermediate Value Theorem. . . . . . . . . . . . . . . . . . . . . .130

3 The Derivative 1353.1 Definition of Derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .137

3.1.1 Slopes and Velocities Revisited. . . . . . . . . . . . . . . . . . . . . .1373.1.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1423.1.3 The Derivative as a Function. . . . . . . . . . . . . . . . . . . . . . . .144


3.2 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1473.3 Rules for Computing Derivatives. . . . . . . . . . . . . . . . . . . . . . . . . .153

3.3.1 Derivatives of Polynomials. . . . . . . . . . . . . . . . . . . . . . . . .1533.3.2 Product and Quotient Rules. . . . . . . . . . . . . . . . . . . . . . . .157

3.4 Algebraic Simplifications Involving Exponents. . . . . . . . . . . . . . . . . .1623.4.1 Rules of Exponents. . . . . . . . . . . . . . . . . . . . . . . . . . . . .1623.4.2 Simplifying the Result of the Product and Quotient Rules. . . . . . . . .166

3.5 Applications of Derivatives as Rates of Change. . . . . . . . . . . . . . . . . .1703.5.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1703.5.2 Some Specific Applications. . . . . . . . . . . . . . . . . . . . . . . .171

3.5.2.1 Rectilinear Motion. . . . . . . . . . . . . . . . . . . . . . . .1713.5.2.2 Biology Applications . . . . . . . . . . . . . . . . . . . . . .1733.5.2.3 Chemistry Applications. . . . . . . . . . . . . . . . . . . . .1743.5.2.4 Economics. . . . . . . . . . . . . . . . . . . . . . . . . . . .175

3.6 The Chain Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1773.7 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1813.8 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .187

3.8.1 The Second Derivative. . . . . . . . . . . . . . . . . . . . . . . . . . .1873.8.2 Third and Higher Derivatives. . . . . . . . . . . . . . . . . . . . . . . .188

3.9 Related Rates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .191


3.10 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1953.10.1 New Notation – Old Idea. . . . . . . . . . . . . . . . . . . . . . . . . .196

4 An Interlude – Trigonometric Functions 2034.1 Definition of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . . .205

4.1.1 Radian Measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2084.2 Trigonometric Functions at Special Values. . . . . . . . . . . . . . . . . . . . .2104.3 Properties of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . . .2134.4 Graphs of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . . . . .214

4.4.1 Graph of sin(x) and cos(x) . . . . . . . . . . . . . . . . . . . . . . . . .2144.4.2 Graph of tan(x) and cot(x) . . . . . . . . . . . . . . . . . . . . . . . . .2144.4.3 Graph of sec(x) and csc(x) . . . . . . . . . . . . . . . . . . . . . . . . .216

4.5 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2184.6 Derivatives of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . .220

4.6.1 Two Important Trigonometric Limits. . . . . . . . . . . . . . . . . . .2214.6.2 Derivatives of other Trigonometric Functions. . . . . . . . . . . . . . .223

5 Applications of the Derivative 2295.1 Definition of Maximum and Minimum Values. . . . . . . . . . . . . . . . . . .2315.2 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .242


5.3 Rolle’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2435.4 The Mean Value Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . .2475.5 An Application of the Mean Value Theorem. . . . . . . . . . . . . . . . . . . .2495.6 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2515.7 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2565.8 Vertical Asymptotes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2625.9 Limits at Infinity; Horizontal Asymptotes. . . . . . . . . . . . . . . . . . . . .264

5.9.1 A Simplification Technique. . . . . . . . . . . . . . . . . . . . . . . .2665.10 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2685.11 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2765.12 Newton’s Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2805.13 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2855.14 Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2885.15 Antiderivatives Mini-Quiz . . . . . . . . . . . . . . . . . . . . . . . . . . . . .290

6 Areas and Integrals 2936.1 Summation Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2956.2 Computing Some Sums. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2976.3 Areas of Planar Regions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .304

6.3.1 Approximating with Rectangles. . . . . . . . . . . . . . . . . . . . . .305


6.4 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3096.4.1 Definition and Notation . . . . . . . . . . . . . . . . . . . . . . . . . .3096.4.2 Properties ofŸ

b

af (x)dx. . . . . . . . . . . . . . . . . . . . . . . . . . .313

6.4.3 Relation of the Definite Integral to Area. . . . . . . . . . . . . . . . . .3146.4.4 Another Interpretation of the Definite Integral. . . . . . . . . . . . . . .315

6.5 The Fundamental Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . .3176.5.1 The Fundamental Theorem. . . . . . . . . . . . . . . . . . . . . . . . .3176.5.2 The Fundamental Theorem – Part II. . . . . . . . . . . . . . . . . . . .319

6.6 The Indefinite Integral and Substitution. . . . . . . . . . . . . . . . . . . . . .3226.6.1 The Indefinite Integral. . . . . . . . . . . . . . . . . . . . . . . . . . .3226.6.2 Substitution in Indefinite Integrals. . . . . . . . . . . . . . . . . . . . .3266.6.3 Substitution in Definite Integrals. . . . . . . . . . . . . . . . . . . . . .329

6.7 Areas Between Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3316.7.1 Vertically Oriented Areas. . . . . . . . . . . . . . . . . . . . . . . . . .3316.7.2 Horizontally Oriented Areas. . . . . . . . . . . . . . . . . . . . . . . .334

7 Appendix 3377.1 Ten – Minute Reviews. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .339

7.1.1 The Real Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . .3397.1.2 Sets, Set Notation, and Interval Notation. . . . . . . . . . . . . . . . . .341


Solutions to Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .344Solutions to Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .513


Chapter 0

Introduction and Notes to theReader

Contents

0.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13


0.2 Notes to the reader. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

0.3 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17


0.1. Preface

Biographical history, as taught in our public schools, is still largely a history of bone-heads: ridiculous kings and queens, paranoid political leaders, compulsive voyagers,ignorant generals – the flotsam and jetsam of historical currents. The men who radi-cally altered history, the great scientists and mathematicians, are seldom mentioned,if at all.–Martin Gardner

The above quotations suggests something—that the most important events which truly shapedwhat has become the world we live in today were those involving the work of great scientists andmathematicians. If this is true, then certainly the work related to the genesis of what we knowcall calculus in the 16th and 17th centuries is worth studying, as every modern branch of scienceuses this important tool. While it is certainly not true that everyone who takes a course such asthe one this book was written for will “need” or “use” calculus, it is certainly true that it is worthfinding out what the whole thing is all about, much as it is important to study other significanthistorical ideas, events, and happenings. It is hoped that those embarking on a study of calculuswill keep this in mind.

Every teacher of calculus has at one time or another remarked about a particular student orgroup of students “I think they understand the calculus concepts, but their algebra skills are so


weak, they can’t finish the problems.” There are many reasons that explain why this problem isso prevalent: some students never learned the precalculus material, some learned it but forgotmost of it, others are just a little rusty and need some extra practice. For these reasons, somecolleges and universities are starting so called “integrated” precalculus/calculus courses. Thistext is designed to be used in exactly those kinds of courses, although it could also be used in atraditional calculus course, with the precalculus material omitted by the instructor, but availablenonetheless to the motivated and interested student who needs or desires some extra amplifica-tions of, reminders about, or practice with the precalculus material. An attempt is made to havethe precalculus material appear “just in time”, that is, directly before it is needed for the firsttime.

In addition, this document is unusual in that the online version allows for reading in an inter-active style: hypertext links are available, solutions are offered to many examples and exercisesvia a hyperlink, and then the reader can then return to his or her place in the text via anotherjump. Miniature quizzes will also be available, which provide immediate feedback that willhopefully erase misunderstandings and misconceptions before they fester and grow. In order toget the reader acclimated to these features, there are some examples provided below.


0.2. Notes to the reader

The electronic version of this document has a number of interactive features, including hyper-linked jumps to solutions of many exercises, as well as interactive quizzes. Examples are below.Answers will be clear to anyone 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 bluehyperlink on the solutions page.)

EXERCISE0.2.1.How many children does the author of the document have? Hint: Readtheacknowledgements.

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

QuizWho is the author of the quotation contained in the preface?

(a) John Goodman (b) Martin Gardner(c) Martin Lawrence (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 andsee the answers, click on the button that says “Correct”.

Click to Initialize QuizAnswer 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, andmore effective for the student. Now it’s time to learn some mathematics!


0.3. Acknowledgments

The genesis of this project was funded in part by the Mellon Foundation through the MellonFurman-Wofford Program. Many of the illustrations were designed by Furman undergraduatestudent Brian Wagner using the PSTricks package. Thanks go to my wife Suzan, and my twodaughters, Hannah and Darby for patience while this project was being produced.



Chapter 1

Functions and their Properties

Contents

1.1 Functions & Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . 211.1.1 Functions and Functional Notation. . . . . . . . . . . . . . . . . . . 21

1.1.2 Cartesian Coordinates and the Graphical Representation of Functions34


1.2 Circles, Distances, Completing the Square. . . . . . . . . . . . . . . . . . 43

1.2.1 Distance Formula and Circles. . . . . . . . . . . . . . . . . . . . . 43

1.2.2 Completing the Square. . . . . . . . . . . . . . . . . . . . . . . . . 44

1.3 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.3.1 General Equation and Slope-Intercept Form. . . . . . . . . . . . . . 48

1.3.2 More On Slope. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

1.3.3 Parallel and Perpendicular Lines. . . . . . . . . . . . . . . . . . . . 50

1.4 Catalogue of Familiar Functions . . . . . . . . . . . . . . . . . . . . . . . 56

1.4.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

1.4.2 Rational Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.4.3 Algebraic Functions. . . . . . . . . . . . . . . . . . . . . . . . . . 59

1.4.4 The Absolute Value Function. . . . . . . . . . . . . . . . . . . . . . 59

1.4.5 Piecewise Defined Functions. . . . . . . . . . . . . . . . . . . . . . 62

1.4.6 The Greatest Integer Function. . . . . . . . . . . . . . . . . . . . . 62

1.5 New Functions From Old . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

1.5.1 New Functions Via Composition. . . . . . . . . . . . . . . . . . . . 66

1.5.2 New Functions via Algebra. . . . . . . . . . . . . . . . . . . . . . 70

1.5.3 Shifting and Scaling. . . . . . . . . . . . . . . . . . . . . . . . . . 70


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 importantto know what a function is. Many students erroneously think of a function as a formula – thisoversimplification and half-truth leads to problems later on in the course, so it is best to get thefollowing definition down pat at the outset.

Definition of FunctionDefinition 1.1.1 A function is a rule of correspondence between itemsin one set (called thedomain of the function) and items in another set(called therange of the function) so that every element in the domain ispaired with exactly one element in the range.

Note that the definition of function also includes the idea of the domain of a function andthe range of a function. Note also that the definition doesn’t say anything about formulas. Infact, a useful way to think about a function is as a machine. Domain elements are input to themachine, and then the function-machine turns them into (perhaps) something else, and these


outputs are the range elements. A simple example is illustrated in figure1.1. An animatedversion is availablehere

�

� ��

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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


�

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� �

��

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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 (oftendenoted¬), as will our range sets. (There is a review of the set of real numbers in section7.1.1.)Thus, we will think of functions as rules which transform certain numbers into other numbers.There are number of different ways to denote or describe a function. These include

1. A chart or arrow diagram.


2. A set of ordered pairs, specified by a list.

3. A verbal description.

4. A graph.

5. Functional notation.

An example of an arrow diagram is given in figure1.2. The “inputs” at the beginning of thearrows indicated a domain element, while the “outputs” at the end of the arrowhead indicate arange element. Thus, for example, in figure1.2we see that the domain element 1 is paired withthe range element-1. An example of a chart representing the same information can be seen intable1.1.1.

x y1 -14 -29 -3

Table 1.1: A function described by a table.


The chart or arrow diagram method of describing a function is effective when the domain issmall, but most of our domains will have an infinite number of elements, so this approach is notpractical.

Another method which is only effective when the domain is small is to write ordered pairs,with domain elements listed first and the corresponding range elements listed second. Thus, thefunction described by figure1.2could instead be denoted{(1,-1), (4,-2), (9,-3)}.

How do we handle the situation when the domain is infinite? Probably the best approach forthis situation is to usefunctional notation. In this notation, the rule of correspondence is givenby a formulaic expression, 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 whoserule is such that an input number is paired with the number which is 2 more than the square ofthe input. Thus, for example, the number 2 would be paired with 22

+ 2 = 6. We indicate this bywriting f (2) = 6, which you should think of as meaning that the number 2 goes into the functionmachinef , and 9 comes out. This notation has many advantages, but for now let’s highlight thedisadvantages. The range is not mentioned at all, the domain is easy to ignore, and it becomes thehabit of many to think of the function as a formula rather than a rule of correspondence between


elements in two sets. Resist this habit!.

Example 1.1.1If 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 thensubtract 4,f (x+ 1) should be be the result of doing those operations to the quantity(x+ 1), thus,the output should be 3(x+ 1)2 - 4, so

f (x+ 1) = 3(x+ 1)2 - 4

�

Example 1.1.2If f (x) = 2x2+ 5, what is the expanded form off (x+ h)?

Solution: f (x+ h) = 2(x+ h)2 + 5 = 2(x2+ 2hx+ h2

) + 5 = 2x2+ 4hx+ 2h2

+ 5. �

Note that in the previous example, no domain was specified at all! This will often be thecase, even though every function must have a domain. Thus, there must be a domainimpliedeven when it is not specified. Our rule of thumb for domains will be the following:

Rule of Thumb about DomainsUnless otherwise specified, the domainof a function written in functional notation will be taken to be the largestsubset of the real numbers for which the function is defined.


�

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Figure 1.3: An arrow diagram whichdoesrepresent a function.

Example 1.1.3What is the domain off (x) =x

x2 - 9?

Solution: Since there is no domain specified, therule of thumbapplies. Thus, the domain is theset of all real numbers wherex2

- 9 ê = 0. This the domain is the set of all real numbers exceptthose wherex = ±3. �

There is one aspect of thedefinition of functionwhich we haven’t discussed yet, and thatis the part which says that every domain element “is paired with exactly one element from therange.” Note that this doesn’t say that every range element is paired with exactly one domainelement, so an arrangement like that in figure1.3 is a function, while that in figure1.4 isn’t.

Example 1.1.4If the quantitiesu andw are related by the relationshipu2= w, is u a function of


�

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�

�

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Figure 1.4: An arrow diagram whichdoes notrepresent a function.

w? Isw a function ofu?

Solution: It is true thatw is a function ofu, because for every value ofu, there is only onepossible value foru2. On the other hand,u is not a function ofw, because for a given value ofw (like the value 9 for example) there could be two different values ofu (namely 3 and-3) forwhich the relationship is met. �

In the application of functional notation, we often use the variablex to represent quantitieswhich come from the domain set (this is more formally called theindependent variable) and weoften use the variabley to represent quantities which come from the range set (more formallycalled thedependent variable), but there is no reason why these letters necessarily must representthose quantities. For example, it is perfectly OK to write

x = y2+ 2y


and think ofx as a function ofy, thus makingy the independent variable andx the dependentvariable for this example.


We conclude with an example of a function described verbally. Consider the function whosedomain consists of any and all real numbersx, which has the property that the range valueassociated withx is the greatest integer less than or equal tox. This function is sometimesdenotedf (x) = dxt. Thus, for example, we haved3.1t = 3 andd-1.5t = -2. This functionis sometimes difficult to work with because there isn’t a “formula” which describes it, only theverbal definition.

Quiz: Functions and Functional Notation.

Click to Initialize QuizAnswer the following questions about functions and functional notation.When necessary, do your work on a scrap piece of paper, but record your answers on this docu-ment. Don’t forget to initialize the quiz by clicking in the appropriate place before starting, andthen click on the appropriate label 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

availablehere.-h

(z)(z+h)h

(z)(z+h)1h

3. For the functionh(w) = w2+9

w2-2w+1, what number or numbers aren’t in the domain?-1 1 2


4. What isd2.3+ 3.8t?5 6 d2.3t + d3.9t

Click to End Quiz


Exercises

EXERCISE1.1.1.If f (x) = 4x2- 3, what is f (2)? What isf (5)? What isf (x+ h)?

EXERCISE1.1.2.Suppose thatg(z) = -z2+ 4. What isg(z+ h) - g(z)?

EXERCISE1.1.3.If x andy are related by the expressionx = 3y- 5, isx a function ofy? Isy afunction ofx?

EXERCISE1.1.4.If x andy are so thatx = y2- 5, is x a function ofy? Isy a function ofx? If

your answer is no, find an example of two different outputs for a given input.

EXERCISE1.1.5.Would {(3,4), (4,3), (5,3), (6,2)} represent a function? Why or why not? If itis a function, what is the domain and what is the range?

EXERCISE1.1.6.Same questions as in the previous exercise, but using the set{(3,4), (4,3), (4,6)}.

EXERCISE1.1.7.Consider a point on the graph of the functiony = x2. Any such point wouldhave coordinates(x, x2

). Recalling the formula for the distance between two points in the plane,find the function which represents the distance from(x, x2

) to the point(1,3).

EXERCISE1.1.8.Consider the English sentence: For an input value ofx, the correspondingoutput value is obtained by adding 3 tox, then cubing this result, then subtracting 4 and dividing


this whole quantity byx. Write, using functional notation, the function which is described in thissentence.

EXERCISE1.1.9. This problem is like the previous one except in reverse. Write a standardEnglish sentence which describes the functionf (x) = x4

-52x+1.

EXERCISE1.1.10.What is the natural domain of the functiong(x) =0

4- x. Express youranswer using a standard English sentence.

EXERCISE1.1.11.What is the natural domain of the functionf (x) = (x+3)(x-3)(x-3) ?

EXERCISE1.1.12.Given the following data: Can you find a functionf (x) whose domain is the

x y4 56 97 118 1310 17

Table 1.2: Some(x, y) pairs which form part of a function.


set of all real numbers, so that this data meets the requirement thaty = f (x) for all the entries inthe table?

EXERCISE1.1.13.What isdp + 4t? If `xp stands for the ceiling function (i.e. the “least integergreater thanx function, what is p - 5p?

1.1.2. Cartesian Coordinates and the Graphical Representation of Func-tions

At some point in your mathematics education you have been exposed to the idea of the realline. This line associates to every real number, a point on the line and to every point on the linea real number. Thus numbers are given a geometric interpretation. An analogue of this idea(one dimension up) is theCartesian Coordinate System. This is also a system by which one canassociate numerical quantities with geometric ones, namely, a point in a plane is associated witha pair of real numbers. Our method of associating is as follows: we draw two perpendicularnumber lines, and call the point where they intersect(0,0) or, theorigin. If we wanted to seewhich point is associated with(2,4) for example, we would start at the origin and move 2 unitsto the right,and then 4 units up.

The first coordinate is related to horizontal movement and the second is related to vertical.The positive horizontal direction is to the right, and the positive vertical direction is up. Note that


the most common labels for the axes arex for the horizontal axis andy for the vertical, althoughthese 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, soa function can be thought of as a set of ordered pairs. By coloring the points which correspondto the ordered pairs which make up the function, we get what we refer to as thegraph of thefunction. A simple animation of this can be viewedhereNote that we can see both the domainand the range in a graphical representation of a function: for each point on the graph, if weproject straight down (or up) to the horizontal axis, the set of points on that copy of the realnumber line indicate the domain of the function. Likewise, if we project to the vertical axis, wecan see the range of the function.

Seeing the Domain and Range in a GraphTo see the domain of a function from its graph: Project onto thex axis.To see the range of a function from its graph: Project onto they axis.

Example 1.1.5Can you tell what the domain of the function shown in figure1.1.2is by project-ing onto thex axis?

Solution: The domain looks like it is the set[-3,3]. �

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


We can also graphically see whether or not a given curve really representsy as a functionof x via thevertical line test. This test says that if a vertical line can be drawn which intersectsthe curve in two distinct points (say(x1, y1) and(x1, y2) with y1 ê = y2), then the graph does notrepresent a function ofx, because there are two distincty values associated with the samex valuex1.

Vertical Line TestIf a vertical line can be drawn which intersects the curve in two distinctpoints (say(x1, y1) and(x1, y2) with y1 ê = y2), then the graph does notrepresent a function ofx.

Exercises

EXERCISE1.1.14.Plot the points(2,5) and (4,9) on a cartesian coordinate system. Draw aright triangle which has these two points as vertices, and which has two sides parallel to theaxes. What are the coordinates of the other vertex? Then, using the pythagorean theorem, findthe distance between the original two points.

EXERCISE1.1.15.Repeat the last problem, but use the arbitrary points(a, b) and(c, d). Whenyou are done, you will have the formula for thedistance between two points.


EXERCISE1.1.16.Graph the straight liney = 3x + 1, and find at least 3 different points onit. Then, taking the points two at a time, find the ratio of the difference in the y values to thedifference in thex values. You should get the same quantity every time. Why?

EXERCISE1.1.17.Graph the set of points(x, y) for which y = x3. Is y a function ofx? Isx afunction ofy? Can you tell what the range of this function is when consideringy to be a functionof x?

EXERCISE1.1.18.Draw the graph ofy =0

4- x + 5 by first finding the domain, and then bycarefully plotting points. Can you tell what the range of the function is by looking at your graph?(You should be able to.)

EXERCISE1.1.19.Draw a graph off (x) = x2+ 4. Note thatf (-1) = f (1), f (-2) = f (2),º .

In fact, f (-x) = f (x) for every value ofx, as can be seen algebraically by substituting-x for x.What does this tell us about thesymmetry of y = f (x)? Note: function with this property arecalledeven functions.

EXERCISE1.1.20.Think about the symmetry off (x) = x3. Note that f (-x) ê = f (x), for allvalues ofx exceptx = 0, but there is still something interesting you can say aboutf (-x) ascompared withf (x). What is it? Note: This kind of symmetry is calledsymmetry about theorigin , and functions with this property are calledodd functions.

EXERCISE1.1.21.Is f (x) = 5 an even function, an odd function, or neither?


EXERCISE1.1.22.Draw a sketch of the functionf (x) = 1x+1, making sure you have the proper

domain. Can you deduce the range of this function from your sketch?


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Figure 1.6: Graphing the functiony = x+ 1, or plotting set{(x, y)|y = x+ 1}.


Figure 1.7: Can you find the domain of function graphed here?


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Figure 1.8: The graph in the figure on the left passes the vertical line test and thusy is a functionof x, while the one on the right fails, soy is not a function ofx.


1.2. Circles, Distances, Completing the Square

1.2.1. Distance Formula and Circles

Given two points(x1, y1) and(x2, y2) in the plane, there will be occasions when you are interestedin knowing the distance between them. By plotting the two points along with the the point(x2, y1), you can see a right triangle whose hypotenuse has the length we are looking for. Thetwo sides of the triangle have lengthsx2 - x1 andy2 - y1. Thus, the pythagorean theorem givesthe required distance, and we have

The distance between the points(x1, y1) and(x2, y2) in the plane is givenby d =

0

(x2 - x1)2 + (y2 - y1)

2

A circle is the set of points a given distance, theradius, from a given point, thecenter.Using this fact and the above distance formula, we see that every point(x, y) on a circle of radiusr centered at(h, k) satisfies

0

(x- h)2 + (y- k)2 = r. Squaring both sides yields the standardequation for a circle, namely:

Equation of a Circle with Center at(x0, y0) and radiusr: (x- x0)2+ (y- y0)

2= r2


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

1.2.2. Completing the Square

A helpful algebraic technique when dealing with circles is the technique ofcompleting thesquare. This technique helps one transform an equation of a circle which is disguised so asto not look like one into the standard form, so one can tell the center and radius. In this tech-nique, an expression of the formx2

+ kx+º = c is changed by addingk22

to both sides, yielding

x2+ kx+ k

22+º = c+ k

22. This enables us to write(x+ k

2)2+º = c+ k

22, thus changing the

left hand side into a perfect square.

Example 1.2.1The equationx2+4x+y2

+6y = 10 is actually the equation of a circle. Completethe square for both thex expression and they expression in order to put this equation in standardform.

Solution: We need to add 4 to both sides, since half of 4 is 2, and 22= 4. (This will complete the

square in thex expression.) We also need to add 9 to both sides, since half of 6 is 3, and 32= 9.

(This will complete the square in they expression.) Thus we havex2+ 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

0

23. �

Exercises:

EXERCISE1.2.1.What is the center and radius of the circle whose equation isx2+2x+y2

-4y =25?

EXERCISE1.2.2.Let g(x) be the function which represents the distance between the point(0,2)and the point(x, f(x) on the graph off (x) = 3x2

- 4. Find and simplify a formula forg(x).

EXERCISE1.2.3.Graph the circle(x- 3)2 + (y+ 2)2 = 16 on a cartesian coordinate system. Ofcourse, this does not define a functional relationship betweeny andx, but if we were to restrictourselves to the upper semicircle, we would have such a relationship. Ify = f (x) is the uppersemicircle, find a formula forf (x).


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1.3. Lines

1.3.1. General Equation and Slope-Intercept Form

Recall that the general equation of a line is an equation of the formAx+By= C, whereA, BandC are real numbers. If we solve fory, (so that the expression will givey as a function ofx) wehavey = -A

Bx+ CB , which we usually write asy = mx+ b, where the expression-A

B = m can beseen to be theslopeof the line, and the expressionCB = b can be seen to be they - intercept. Inthe case whereA is zero, we have thatm is zero as well, and the line has the formy = b whichrepresents a horizontal line. IfB = 0 then we can’s solve fory as above, and we see that ourequation has the formx = C

A , which represents 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, wherem is the slope andb is they-intercept.


1.3.2. More On Slope

Theslopeof a line mentioned above plays a key role in the study of calculus. The slope of a lineis the numerical value which represents the “steepness” of the line. Given two points(x1, y1) and(x2, y2) on the line, 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 peopledescribe the slope is being the “rise over run”. From this definition of slope, it is easy to getpoint-slope form of the equation of a line. Given a point(x0, y0) on a line of slopem, we knowthat 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 bem. Multiplying both sides by theappropriate expression to clear denominators leads to the

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


Example 1.3.1A line has slope 5, and goes through the point(-2,-4). Find the equation of thisline, and indicate itsy - intercept.

Solution: 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 equationy+ 4 = 5(x+ 2), or y = 5x+ 10- 4,or y = 5x+ 6. Thus they - intercept is 6. �

Example 1.3.2Find the equation of the line that contains the points(-2,5) and(2,-3).

Solution: 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-interceptform: y = -2x- 1 �

1.3.3. Parallel and Perpendicular Lines

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


Lines which are not vertical are parallel if and only if they have the sameslope.Lines which are not vertical or horizontal are perpendicular if and onlyif their slopes multiply together to be-1.

Example 1.3.3Are the lines 3x+ y = 7 and-x- 3y = 9 parallel, perpendicular, or neither?

Solution: The line 3x+y = 7 can be writteny = -3x+7, so it has slope-3. The line-x-3y = 9can be written asy = -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. �

Exercises:

EXERCISE1.3.1.Find the equation of the line through the points(2,5) and(6,2).

EXERCISE1.3.2.There is a linear relationship between the Celsius and the Fahrenheit temper-ature scales. If water freezes at 32 degrees Fahrenheit and 0 degrees Celsius, and boils at 212Fahrenheit which is 100 Celsius, find the equation which relates the two scales.


EXERCISE1.3.3.On a recent edition of the ABC hit television showWho Wants to be a Mil-lionaire!, the one million dollar question was this: At what temperature reading is the Fahrenheittemperature the same as the Celsius temperature? Answer this question, using the result of thelast problem.

EXERCISE1.3.4.Find a line parallel to the line 3x + 6y = 4 but which goes through the point((1,2).

EXERCISE1.3.5.There are infinitely many lines which go through the point(1,4). Find a niceway to characterize them with an equation. There are also infinitely many lines with slope 3.Find an nice way to characterize these as well. Then find the one line which is in both of thesesets.

EXERCISE1.3.6.Find the equation of vertical line through(3,5). How about the horizontal linethrough the same point?


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1.4. Catalogue of Familiar Functions

1.4.1. Polynomials

We are interested in developing an informal “catalogue” of the functions we will be using, so wecan refer to them by name and understand some of the properties that they might posses. Ourfirst category will be the collection ofpolynomials.

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

anxn+ an-1xn-1

+º + a1x+ a0

where theai ’s are real numbers,an ê = 0, andn is a non-negative integer.

In the above definition, theai ’s are calledcoefficientsand the numbern is called thedegreeof the polynomial. Here are some familiar types of polynomials:

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


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

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

Quadratics have graphs which are polynomials which open up if the coefficient on the 2nddegree term is positive, and open down if it is negative. Thevertexof a parabola is the lowestpoint on the parabola if it opens down or is the highest point on the parabola if it opens up. Onecan find thex value of the vertex by completing the square. Iff (x) = ax2

+ bx+ c, we can writef (x) - c = a(x2

+ba), and then completing the square inside the parentheses yields

f (x) - c+b2

4a= a(x2

+ba+

b2

4a2),

so f (x) - c+ b2

4a = a(x+ b2a)

2. From this it is not too hard to see that the vertex will be atx = - b2a.

Example 1.4.1Find the vertex of the polynomialf (x) = 2x2- 8x+ 1.

Solution: We could just use the formula derived above, but instead (for the purposes of demon-stration), we re-derive the result. If we lety = 2x2

-8x+1, we can write this asy-1 = 2x2-8x,

or y- 1 = 2(x2- 4x). We will complete the square inside the parentheses on the right. We need

to add 4 to complete the square, but since this 4 is being multiplied by 2 outside the parentheses,we are actually adding 8 to the right-hand side, so we will need to add 8 to the left-hand side to


compensate. Thus we havey- 1+ 8 = 2(x2- 4x+ 4) = 2(x- 2)2. By subtracting 7 to both sides

of the result, we obtainy = 2(x- 2)-7. The vertex must occur wherex = 2, and at that point wehavey = -7. The vertex is thus(2,-7). �

Polynomials share many nice properties. For example, every polynomial has domain equalto R, the set of all real numbers.

1.4.2. Rational Functions

A rational functionis one which can be written as the ratio of two polynomials. For example,3x2+2x

5x-3 is a rational function. Alsof (x) = 1x +

3x2 is a rational function, because by simplifying it

can be written asf (x) = 3+xx2 . Rational functions have domain equal to the set of all real numbers

except for those numbers which cause the denominator to be zero.

The Domain of Polynomials and Rational FunctionsAll polynomials have domain equal to the setR of all real numbers.All rational function of the formg(x) = p(x)

q(x) where p(x) and q(x) arepolynomials have domain equal to the set of all real numbersexcept thosewhereq(x) = 0. That is, the domain ofg(x) is {x|g(x) ê = 0}.


1.4.3. Algebraic Functions

An algebraic functionis one which can be built out of the standard operations of addition, sub-traction, multiplication, division and roots. In particular, everything mentioned in this sectionso far is algebraic. Some examples of non-algebraic functions are the trigonometric functions( f (x) = sin(x), etc.), the exponential functions (f (x) = 2x, etc.), and the logarithmic functions( f (x) = logx). We will postpone our study of these types of functions until later.

1.4.4. The Absolute Value Function

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 thatx is from 0 on the number line. Ingeneral,|x- c| always gives the distance betweenx andc on the number line.

Example 1.4.2Describe, using absolute value, the set of all numbers whose distance from 4 isless than or equal to 1.


Solution: This is the set of numbersx so that|x- 4| < 1. �

The absolute value function plays

Figure 1.14: The graph of the absolute value function.

an important role when when simpli-fying certain expressions involving squareroots. This is because the simplest wayto write

0

x2 is to write it as|x|. Onecould write±x, but this choice of no-tation is not as explicit, because onemight not know when the+ applies andwhen the- applies, but when one writes|x|, there is no doubt, since the defini-tion of the absolute value contains spe-cific instructions about when to applythe- sign and when not to apply it.

Example 1.4.3Many students are taught the followingincorrect way to solvex2= 9. Taking

the square root of both sides, they obtainx = ±3. What is wrong with this “solution”?

Solution: First of all, the square root of 9 isnot ±3, since by “square root” we always meanwhat is sometimes called the principle square root. The square root of 9 is 3, period. The other


mistake is in thinking that0

x2 = x, which isn’t true. (Try plugging in-3, in light of the previoussentence.) What is true is that

0

x2 = |x|. Here is a correct method for solvingx2= 9: Take the

square root of both sides to obtain0

x2 = 3. This can be rewritten as|x| = 3. Interpreting thedefinition of |x|, we see that there are two solutions for|x| = 3, namely 3 and-3. �

Example 1.4.4Can the equation(y- 2)2 = 25- (x- 4)2 be solved fory? Do so, and find whatyvalue(s) correspond withx = 5?

Solution: In an attempt to solve fory, we take the square root of both sides. Doing so we obtain|y- 2| =

0

25- (x- 4)2. This means that

y- 2 =ÏÔÌÔÓ

0

25- (x- 4)2, if y- 2 ≥ 0;

-

0

25- (x- 4)2, if y- 2 < 0.(1.2)

Rewriting, we see that we have

y =ÏÔÌÔÓ

2+0

25- (x- 4)2, if y ≥ 2;

2-0

25- (x- 4)2, if y < 2.(1.3)

Letting x = 5, we obtain 2 values fory, namely 2+0

24 and 2-0

24. Note that we are reallylooking at a circle, and have identified the top semicircle as a function ofx and the bottomsemicircle as a function ofx. �


1.4.5. Piecewise Defined Functions

The absolute value function is an example of a piecewise defined function. By a piecewisedefined function we mean a rule of correspondence for which there are “different” sub-rulesgiven over different parts of the domain of the function. Here is another example, with anaccompanying graph.

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 isdefined differently over the different “pieces” of the domain. To properly evaluate this functionat x = 3 for example, one must first ask which piece of the domain 3 falls into. A little thoughtreveals that 3 is in the categoryx ≥ 2, so we apply the 2x+ 20 rule with input 3 and obtain thatf (3) = 26.

1.4.6. The Greatest Integer Function

Another interesting and unusual function is the greatest integer function, denoteddxt. Thisfunction is not given by a formula, but rather a verbal rule of correspondence. The rule is thatthe output is always the greatest integer that is less than or equal to the input. Thusd2.5t =


2,d-2.5t = -3, dpt = 3, and so on. This function is sometimes referred to as astepfunction,since its graph resembles stairs going up from left to right.


Exercises:

EXERCISE1.4.1.Is y = x4+ px2

-

0

2 a polynomial, or a rational function? Explain.

EXERCISE1.4.2.Is 1x2 - 5x+1

x3 a polynomial? Is it a rational function?

EXERCISE1.4.3.What is the domain of3x2-4

16x-x2 ?

EXERCISE1.4.4.Use absolute values to write an inequality which would represent the collec-tion of x’s whose distance from 2 is less than or equal to.01. Then write this set in intervalnotation.

EXERCISE1.4.5.Use absolute values to write an inequality which would represent the collec-tion of x’s so that the distance fromf (x) to 5 is greater than.5, wheref (x) = 3x + 4. Can youwrite this set in interval notation?

EXERCISE1.4.6.Suppose a function is described verbally as follows: if the inputx is greaterthan or equal to 2, the output is 4x-3. If the input value is less than 2, the output isx2

+5. Writethis function in standard mathematical notation, and draw a sketch of its graph.

EXERCISE1.4.7.Suppose a function is defined by

y =ÏÔÌÔÓ

1x-1, if x £ 5;0

x- 5 if x > 5.


What is the domain of this function? What isf (6)? What isf (0)? Draw a sketch of the graph ofthis function.

EXERCISE1.4.8. Sketch a graph ofy = `xp, the ceiling function. Then sketch a graph ofy = d2xt.


1.5. New Functions From Old

There are a number of ways to get new functions by combining old ones in various ways. Theseinclude composing two or more functions to get a new function, or combining familiar functionswith addition, subtraction, multiplication or division. We will investigate all of these, and seehow the graphs of familiar functions are changed by simple translations and scalings.

1.5.1. New Functions Via Composition

Recall that a function is a rule of correspondence which has inputs and outputs. If we put theoutput of a particular function into a second function as an input, the resulting output (whenthought of as a function of the original input) is thecompositionof the two functions.


� ��

��

� � ��

Figure 1.17: The composition off (x) andg(x) to form ( f Î g)(x).

Notationally, we write( f Îg)(x) = f (g(x)). This notations suggests that if we were to evaluate( f Î g) at a number like 2, we would computef (g(2)), that is, computeg(2) first and then plugthe resulting number intof as input. It is important to note thatf (g(x)) would not necessarily beequal tog( f (x), so we see thatin general, (g Î f ) ê = ( f Î g).


Definition of CompositionThe compositionof f andg is defined by( f Î g)(x) = f (g(x)). In thisdefinition we will callg the “inner” function andf the “outer” function.This is not necessarily equal to the composition ofg and f , which isdefined by(gÎ f )(x) = g( f (x)). In this definition,f is the “inner” functionandg is the “outer” function.

Example 1.5.1For the functionsf (x) = 3x+2 andg(x) = x2-5, compute( f Îg)(x) and(gÎ f )(x).

To see that these are different, compute( f Î g)(2) and(g Î f )(2). Is this enough information todeduce that the two are different?

Solution: 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. Thusf (g(x)) = 3x2-13.

On the other hand,g( f (x)) = g(3x+ 2), and sinceg is the “square and subtract 5” rule, we haveg(3x + 2) = (3x + 2)2 - 5, and simplifying this gives(9x2

+ 12x + 4) - 5 = 9x2+ 12x - 1.

Thusg( f (x)) = 9x2+ 12x - 1. Now f (g(2)) = -1 andg( f (2)) = 59. Thisis enough to deduce

that f (g(x)) ê = g( f (x)), since for functions to be equal, the outputs must be equal forall theirinputs. �


Example 1.5.2Consider the functionh(x) = (4x + 1)2 - 9. Can you find functionsf andg sothath(x) = f (g(x))?

Solution: Note that the first things that are “done” byh to an inputx is thatx is multiplied by4 and then 1 is added. Later, this number is squared and 9 is subtracted. Thus one possibleanswer is to let the “inner” functiong(x) = 4x+ 1 and the “outer” functionf (x) = x2

- 9. Thenf (g(x)) = f (4x+ 1) = (4x+ 1)2 - 9. �

Click to Initialize QuizAnswer the following questions about compositions.

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

I

12x+1M

2 2x+1x2

x2

2x+1 2x+ 1x2

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

2, find possible representations forf , g, andh.

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 ,

f (x) = 2x+ 1

Click to End Quiz


1.5.2. 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, iff andg 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)

• J fg N (x) =

f (x)g(x) .

Note that if f andg in the above all have domain equal to the set of all real numbers, then eachall of f + g, f - g and f ◊ g do too, but that the domain offg could be only a subset of the set of

all real numbers. In fact, any numbera for whichg(a) = 0 is definitely not in the domain offg .

1.5.3. Shifting and Scaling

Suppose you are given a functiony = f (x) for which you are familiar with the graph, and apositive constantc. If we form the functiong(x) = f (x) + c, it turns out thatg(x) has almost


the same graph as that off (x), except that the whole graph has been shifted up byc units. (Youmight want to think of f (x) = x2 andc = 3 as a good example during this discussion.) Thequestion becomes: how are the graphs of the following related to the graph off (x)? f (x + c),f (x- c), f (x) + c, f (x) - c, c f(x), -c f(x), f (cx), f (-cx).

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

g(x) = Graph as compared withf (x), if c > 0.f (x+ c) shiftedc units to the leftf (x- c) shiftedc units to the rightf (x) + c shiftedc units upf (x) - c shiftedc units down

Table 1.3: Shiftingf (x).


g(x) = Graph as compared withf (x), if c > 1.c f(x) stretched vertically by a factor ofc1c f (x) compressed vertically by a factor ofc- f (x) reflected about thex axisf (cx) compressed horizontally by a factor ofcf I 1cxM stretched horizontally by a factor ofcf (-x) reflected about they axis

Table 1.4: Scaling and Reflectingf (x).

Note that if a function has the property that it issymmetric about the y axis, then f (x) = f (-x)for that function. Such functions are calledevenfunctions, because some examples of these aref (x) = x2, f (x) = x4, f (x) = x6, etc.. Functions for whichf (x) = - f (x) are said to besymmetricabout the origin, and are calledoddfunctions. Some examples of odd functions includef (x) = x,f (x) = x3, etc..

Example 1.5.3Is the sum of an odd function and another odd function always odd?

Solution: Yes. If f andg are both odd, then( f + g)(-x) = f (-x) + g(-x) = - f (x) + -g(x) =-( f (x) + g(x)) = -( f + g)(x). �


Exercises:

EXERCISE1.5.1.Draw a simple graph off (x) =0

x. Now graph all of the following forc = 3.f (x+ c), f (x- c), f (x) + c, f (x) - c, c f(x), -c f(x), f (cx), f (-cx).

EXERCISE1.5.2.If f (x) = 3x2+ 5 andg(x) =

0

x- 3+ 1, find ( f Î g)(x) and(g Î f )(x).

EXERCISE1.5.3.If H(x) = x2+1

2-x2 , find nontrivial functionsf (x) andg(x) so that( f Îg)(x) = H(x).

EXERCISE1.5.4.Draw a rough sketch off (x) = 1x , by plotting points. (Hint: the function is not

defined atx = 0.) Then sketchy = 1x+1 andy = 1

x + 1.

EXERCISE1.5.5.If f (x) =0

x, 4 £ x £ 9, draw a sketch ofg(x) = f (2x). (Hint: Be careful:what is the domain?) Then draw a sketch ofh(x) = 2 f (x). Lastly, draw a sketch ofj(x) = f I x2M.

EXERCISE1.5.6.Is f (x) = 3x+ 2 even, odd or neither? Explain.

EXERCISE1.5.7.Is g(x) = xx3-x even, odd or neither?



Chapter 2

Limits

Contents

2.1 Secant Lines, Tangent Lines, and Velocities. . . . . . . . . . . . . . . . . 77

2.1.1 The Tangent Line. . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

2.1.2 Instantaneous Velocity. . . . . . . . . . . . . . . . . . . . . . . . . 79


2.2 Simplifying Difference Quotients . . . . . . . . . . . . . . . . . . . . . . . 84

2.3 Limits (Intuitive Approach) . . . . . . . . . . . . . . . . . . . . . . . . . . 90

2.4 Intervals via Absolute Values and Inequalities . . . . . . . . . . . . . . . 102

2.5 If – Then Statements. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .107

2.6 Rigorous Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . .111

2.7 Computing Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .117

2.8 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .126

2.8.1 Definitions Related to Continuity. . . . . . . . . . . . . . . . . . . 126

2.8.2 Basic Theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . .129

2.8.3 The Intermediate Value Theorem. . . . . . . . . . . . . . . . . . . . 130


2.1. Secant Lines, Tangent Lines, and Velocities

One of the main problems that motivated the genesis of calculus was this: What does one meanby the slope of the tangent line to a curve at a given point, and how can we calculate it? Anotherimportant motivating question was: What does one mean by instantaneous velocity, and how canwe calculate it?

We will see in the following discussion that these two apparently unrelated questions are infact very closely 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 weare talking about, using the idea of the tangent line to a circle as a guide. We note that as soonas we know the slope 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) andm is the slope of the tangent line.It seems reasonable to approximate the 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 ash gets smaller. Thus wedefinethe tangentline to be the line whose slope is the result of “taking the limit ash Æ 0” in the above. Wewill discuss the technicalities of of this limiting procedure later, but for now will be content with“shrinkinghÆ 0” informally.

The Slope of the Tangent LineThe slope of the tangent line is obtained by first computing the slope ofa 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 ash Æ 0, so we will shrinkh to zero toobtain the slope of the tangent line.


Example 2.1.1Investigate the slope of the tangent line toy = x2 at x = 0. Then write theequation of the tangent line.

Solution: We must computef (0+h)- f (0)h . Of course, this is equivalent toh

2

h = h. And ash Æ 0,this is equal to 0. Thus, the tangent line is flat at this vertex of the parabola. The equation of thisline isy- 0 = 0(x- 0), or simplyy = 0 (thex-axis.) �

Example 2.1.2If you wanted to know the slope of the tangent line tof (x) = x3- 3x at x = 4,

how would you go about finding it?

Solution: We’d need to computef (4+h)- f (4)h , then simplify this expression, and then investigate

what happens ashÆ 0. �

2.1.2. Instantaneous Velocity

Suppose that you have a really accurate odometer and a very accurate stop watch but no speedome-ter, and you want to answer the question: how fast am I moving right now? (Assume you aretraveling in a straight line.) Everyone is familiar with the idea that average velocity (i.e. “rate”)is elapsed distance divided by elapsed time. (Think of that familiar formulad = r ◊ t.) Now itseems like a good approximation for instantaneous velocity would be average velocity over anextremely small time interval. And the approximation would get better as the time interval getssmaller. So iff (t) is our odometer reading at timet, and if we are interested in our instantaneous


velocity at timet = a, we could consider the average velocity between timest = a andt = a+ h.Our elapsed distance would bef (a + h) - f (a) and the elapsed time would beh, and thus theaverage velocity would be

f (a+ h) - f (a)h

.

And the approximation gets better ash Æ 0. This should look familiar! It is exactly the sameresult we obtained when investigating the slope of the tangent line.

Instantaneous VelocityThe instantaneous velocity at timet = a is obtained by first computingthe 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 ash Æ 0, so we will shrinkh to zero toobtain the instantaneous velocity.


Again, note that this idea of ‘taking the limit ash Æ 0” needs to be studied in more detail,which we will do in the next few sections.

Example 2.1.3Suppose that an object is moving in a straight line so that its position at timet isgiven bys(t) = t2

+ 2t. What is its average velocity between timet = 2 andt = 3? How aboutbetweent = 2 andt = 2.5? How about betweent = 2 andt = 2+ h?

Solution: The average velocity between times 2 and 3 is given byf (3)- f (2)3-2 =

15-81 = 7. The

average velocity between times 2 and 2.5 is given by f (2.5)- f (2)2.5-2 =

6.25-8.5 = 2(-1.75) = -3.5.

The minus sign indicates that the object is moving in the negative direction (on average) duringthis time interval. The average velocity between times 2 and 2+ h is given by f (2+h)- f (2)

2+h-2 =

(2+h)2+2(2+h)-8h . In the next section, we will discuss how to simplify such quantities. �

Example 2.1.4If a tennis ball is thrown straight up in the air from ground level (and there isno wind and if we neglect air resistance), it will travel in a straight line, up for a while and thenback down. In fact, its position at timet is given by an equation of the forms(t) = -16t2

+ v0t,wherev0 is theinitial velocity of the ball, that is, the speed it was going when released. If a ballis thrown up in the air with an initial velocity ofv0 = 96 feet per second, its position at timet iss(t) = -16t2

+96t. What is the average speed of the ball between timet = 1 and timet = 2? Howabout between timet = 1 and timet = 1+ h? How would one find the instantaneous velocity?

Solution: Between timet = 1 and timet = 2, the average velocity is given bys(2)-s(1)2-1 =


-64+192-(-16+96)1 = 48, so the average velocity is 48 feet per second. Between timet = 1 and time

t = 1+ h, the average velocity would be

s(1+ h) - s(1)1+ h- 1

=-16(1+ h)2 + 96(1+ h) - (-16+ 96)

h.

To find the instantaneous velocity, we would first simplify this expression, and then lethÆ 0.�


Exercises:

EXERCISE2.1.1.If f (x) = 3x+ 2, find f (2+h)- f (2)h . Can you work with this expression and try to

find the slope of the tangent line to this function atx = 2?

EXERCISE2.1.2.Draw a sketch of the functionf (x) = x3, and plot the point(1,1) and the point(2,8). Draw the line between these two points, and then find the slope of this line, labeling the“rise” and the “run” on the graph. Then, plot the point(1+ h,(1+ h)3) (for the purposes of yourpicture only, make the value ofh around.5.) Next, draw the line between the points(1,1) and(1+ h,(1+ h)3), and find the slope of this line.

EXERCISE2.1.3.A ball is thrown up in the air so that its position at timet is given bys(t) =-16t2

+ 32t. What is the average velocity between timet = .5 andt = 1? How about betweent = .5 andt = .5+ h?

EXERCISE2.1.4.The population of bacteria in a certain petri dish at timet is given byb(t) =t3+ t. What is the average rate of change of the population over the time interval[2,4]? What is

the average rate of change of the population over the time interval[2,2+ h]? If we lethÆ 0 inthis last quantity, what would the resulting number represent?


2.2. Simplifying Difference Quotients

By a difference quotientwe mean a quantity of the formf (x+h)- f (x)h or f (x)- f (c)

x-c . These kinds ofquantities arise when studying slopes of secant or tangent lines, and average and instantaneousvelocities. In this section 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 ordenominator and 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 ofthe same coin. “Multiplying out” and ”factoring” are inverse operations of each other. The ap-propriate one to do should be apparent from the context of the situation. The third item involvesusing theconjugateof an expression of the forms+ t or s- t, especially where either or both ofs andt involve square roots of expressions. The conjugate ofs+ t is s- t, and vice versa. For


example, the conjugate of0

x+ h- 5 is0

x+ h+ 5. The fourth item is especially useful wheneither the numerator or denominator of the difference quotient is a difference of quotients itself.

For example, an expression like1

x+h-1x

x would require the technique in item four.

Examples of each technique follow.

Example 2.2.1Find and simplify f (x+h)- f (x)h for f (x) = 3x2

+ 2x+ 3.

Solution:

We have thatf (x+ h) - f (x)

h=

3(x+ h)2 + 2(x+ h) + 3- (3x2+ 2x+ 3)

h. We note that tech-

nique one above seems to apply – that is, if we expand the expression in the numerator, thatwe might be able to simplify the expression. Using the fact that(x + h)2 = x2

+ 2xh + h2,

we arrive at the expression3x2+ 6xh+ 3h2

+ 2x+ 2h+ 3- 3x2- 2x- 3

h=

6xh+ 3h2+ 2h

h=

(h)(6x+ 3h+ 2)h

= 6x+ 3h+ 2.

�

Example 2.2.2Find and simplify f (x)- f (3)x-3 for f (x) = x3.

Solution:


We havef (x) - f (3)

x- 3=

x3-27

x-3 . To simplify this expression, we need to factor the numerator.

(We know that it has a factor ofx - 3 since the numerator has value 0 atx = 3.) Using the

difference of cubes formula, we have(x- 3)(x2

+ 3x+ 9)x- 3

= x2+ 3x+ 9. Recall that the general

formula for the difference of cubes isa3- b3

= (a- b)(a2+ ab+ b2

).�

Example 2.2.3Find and simplify f (x+h)- f (x)h for f (x) =

0

x- 2+ 3.

Solution:

We havef (x+ h) - f (x)

h=

0

x+ h- 2+ 3-0

x- 2+ 3h

=

0

x+ h- 2-0

x- 2h

. We need

to somehow get rid of the square roots in the numerator, so we try the technique of multiplyingthe numerator and the denominator by the conjugate of the numerator, namely

0

x+ h- 2 +0

x- 2. We have0

x+ h- 2-0

x- 2h

◊

0

x+ h- 2+0

x- 20

x+ h- 2+0

x- 2,

and if we multiply out the numerators (but not the denominators,) we arrive at

x+ h- 2- (x- 2)

h ◊ (0

x+ h- 2+0

x- 2),


and if we simplify the numerator and then cancel theh’s in the numerator and denominator, wearrive at

10

x+ h- 2+0

x- 2.

�

Example 2.2.4Find and simplify f (x)- f (4)x-4 for f (x) = 1

x .

Solution:

The quantityf (x) - f (4)

x- 4is equal to

1x -

14

x- 4.

The best approach seems to be simplifying the numerator by finding a common denominator

among1x

and14

. This would (of course) be 4x, and thus the numerator could be rewritten4- x4x

.

Thus, our original expression can be written

4-x4x

x- 4=

4- x(4x)(x- 4)

.


This last simplification comes from the fact that dividing a fraction by a quantity is the same as

multiplying by the reciprocal. Thus a quantity likeab

c=

ab◊

1c=

abc

. Now, getting back to our

simplification, we note that 4- x andx- 4 are opposites, so we can write 4- x = -(x- 4), andthen can cancel, leaving the minus sign. We have

-14x

as our final simplification.�

Exercises:

EXERCISE2.2.1.If f (x) = 13x, find and simplify f (x)- f (2)

x-2 .

EXERCISE2.2.2.If g(x) =0

x+ 3, find and simplifyg(x+h)-g(x)h .

EXERCISE2.2.3.For f (x) = x3+ x, find and simplify f (2+h)- f (2)

h .

EXERCISE2.2.4.Forh(x) = x2+ x+ 2, find and simplifyh(x)-h(2)

x-2 .


EXERCISE2.2.5.Suppose an object is moving in a straight line so that it’s position at timet is given by

f (t) = 3t+1. If you wanted to know the object’s instantaneous velocity at timet = 3, what

difference quotient would you need to simplify? Write it down, and then go ahead and simplifyit.

EXERCISE2.2.6.Find and simplify f (x)- f (4)x-4 for f (x) = 1

0

x.


2.3. Limits (Intuitive Approach)

We would like to understand this idea of “shrinkingh Æ 0” more formally. To this end, wewill define the important notion of alimit. We first give the notation, and then discuss the ideainformally and intuitively. A more formal approach will follow in a later section.

We will write

limxÆc

f (x) = L,

and will read it as “the limit asx approachesc of f (x) is L.”

The (informal) meaning of this is as follows:

Informal Definition of LimitWhen we write

limxÆc

f (x) = L,

we mean that whenx is very close to (but not equal to)c, the correspond-ing value of f (x) is very close to (and perhaps equal to)L.


��

��

�

Figure 2.2: Whenx is close to (but not equal to)c, f (x) is close toL.


Example 2.3.1Because it is true that whenx is close to 5,f (x) = 2x- 1 is close to 9, it wouldbe correct to write limxÆ5 2x- 1 = 9. In exercise2.1.1, we showed that the slope of the tangentline to y = 3x+ 2 atx = 2 is 3. In the process of doing that problem, we essentially showed thatif f (x) = 3x+ 2, that then limhÆ0

f (2+h)- f (2)h = 3.

Example 2.3.2Evaluate limxÆ9

0

x-3x-9 .

Solution: First note that it might pay to multiply both the numerator and denominator by theconjugate of the numerator. Thus we would have

limxÆ9

0

x- 3x- 9

◊

0

x+ 30

x+ 3= lim

xÆ9

x- 9

(x- 9)(0

x+ 3)= lim

xÆ9

10

x+ 3.

Now it isn’t hard to believe that whenx is really close to 9, 10

x+3is really close to1

6, so

limxÆ9

0

x-3x-9 =

16. �

Example 2.3.3Using figure2.3.3, identify limxÆ0 f (x) and limxÆ2 f (x).


� � � ��

��

��

��

�

�

�

�

�

��

��

��

��

Figure 2.3: What is limxÆ0 f (x) and limxÆ2 f (x)?


Solution: It is clear from the diagram that whenx is close to 0, f (x) is close to 0 too, solimxÆ0 f (x) = 0. However, whenx is close to but a little less than 2,f (x) is close to1

2, whilewhenx is close to but a little bigger than 2,f (x) is close to 2. Thusthere is not a single valuefor which it can be said that whenx is close to 2,f (x) is close to that value.Thus limxÆ2 f (x)does not exist.

�

Example 2.3.4Using figure2.3.4, identify limxÆ-1 f (x).


� � � ��

��

��

��

�

�

�

�

�

��

��

��

��

Figure 2.4: What is limxÆ-1 f (x)?


Solution:

Note that whenx is close to (but not equal to)-1, f (x) is close to 1. It doesn’t matter whatf (x) is defined to beat -1, or even whether or not it is defined at-1. It is still true that whenxis close to but not equal to-1, f (x) is close to 1. Thus limxÆ-1 f (x) = 1. �

In example2.3.3above, it seems that it might be useful to think about whenx is close to 2,but on specifically one side or the other of 2. Again, if one carefully analyzes the diagram, it canbe noted that when x is close to be a little to the left of 2, the function is close to1

2, but if x isclose to but a little to the right of 2, the function is close to 2. We will indicate this by sayingthat “the limit asx approaches 2 from the left is12 and the limit asx approaches 2 from the rightis 2.” More generally, we defineone-sided limits(and introduce the proper notation) as follows:


One-Sided Limits (Informal Definition)When we write

limxÆc+

f (x) = L,

we mean that whenx is very close to (but a little bigger than)c, thecorresponding value off (x) is very close to (and perhaps equal to)L.When we write

limxÆc-

f (x) = L

we mean that whenx is very close to (but a little smaller than)c, thecorresponding value off (x) is very close to (and perhaps equal to)L.

Thus in that example, even though limxÆ2 f (x) didn’t exist, we could correctly say thatlimxÆ2+ f (x) = 2 and limxÆ2- f (x) = 1

2

The following fact should be evident:


Relationship Between One-Sided and Two-Sided LimitsIn order for a limit to exist, the two corresponding one-sided limits mustboth 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 figure2.5, limxÆ0 f (x) fails to exist, because there is no particular number thatxis close to asx Æ 0. We will write limxÆ0 f (x) = •, but that notation will simply mean that thelimit doesn’t exist, because asxÆ 0, the function increases without bound.


� � � ��

��

��

��

�

�

�

�

�

��

��

��

��

Figure 2.5: A function without a limit atx = 0.


Example 2.3.5Does limxÆ3x+3x-3 exist?

Solution: No. In fact, both limxÆ3+x+3x-3 and limxÆ3-

x+3x-3 fail to exist, because limxÆ3+

x+3x-3 = •

and limxÆ3-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 whenx is close to but a little bigger than 3,the numerator is close to 6 and the denominator is close to 0 (but is a positive number.) Suchnumbers are very big, and get even bigger when choosing a value forx closer to 3. On theother hand, whenx is close to but a little less than 3, the numerator is still close to 6, and thedenominator is still close to 0, but the denominator is a negative number. Since the ratio of apositive number and a negative number is always a negative number, we know that the output is a“large” negative number. (“Large” in the sense that its absolute value is large,) and the numbersget even “larger” whenx is chosen closer to 3. Thus we see that both one-sided limits fail toexist.

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Exercises:

EXERCISE2.3.1.Translate the expression limxÆ5 4x - 1 = 19 into a sentence which containsonly English words. Then write a sentence using the words “close to” which describes thegeneral idea of what this statement means.

EXERCISE2.3.2.Translate the following into properly written symbols: “The limit ash ap-


proaches zero of the quotient of(x+ h) squared minusx2 andh is 2x.

EXERCISE2.3.3.A piecewise defined function is given by

f (x) =ÏÔÌÔÓ

3x- 1 if x > 2;

2- 3x if x £ 2.

Sketch this function and then answer the following questions: What is limxÆ2- f (x)? What islimxÆ2+ f (x)? What is limxÆ2 f (x)?

EXERCISE2.3.4.Draw the sketch of any functiong(x) for which the following is true: limxÆ-1- f (x) =3 and limxÆ-1+ f (x) = 4.

EXERCISE2.3.5.Consider the functionf (h) = |h|h . Draw a sketch of it and then investigate

limhÆ0 f (h). Hint: Think of f (h) as a piecewise defined function.

EXERCISE2.3.6.Supposep(x) is a polynomial with limxÆ3 p(x) = 4, and thatq(x) is a polyno-mial with limxÆ3 q(x) = 0. What (if anything) can you say about limxÆ3

p(x)q(x)? Can you generalize

your conclusion?

EXERCISE2.3.7.Find, in functional notation, a functionf (x)with the property that limxÆ4 f (x) =-•. Draw a sketch of your function. Hint: In order to be able to draw a nice sketch, you shouldtry to make the function you find as simple as possible.


2.4. Intervals via Absolute Values and Inequalities

Before we embark on an attempt to understand the rigorous definition of limits, we must firstmake sure we are comfortable with the relationship between certain inequalities involving ab-solute values and the corresponding intervals they describe. The key fact which is necessary tounderstand what follows is that

|a- b| represents the distance froma to b.

Because of this fact, we can think (for example) of|x- 2| as the distance fromx to 2, and theset of points with|x-2| < 1 as the set of pointsx so that the distance fromx to 2 is less than one.Thus this set represents the interval(1,3). In general, we have that the set of pointsx for which|x- a| < d is the set of pointsx so that the distance fromx to a is less thand, or in other words,the interval(a- d, a+ d).

If instead we were considering the set 0< |x- a| < d, we would be considering the union ofintervals(a-d, a)« (a, a+d), which doesn’t includea, since we are thinking about points whosedistance froma is strictly bigger than zero but less thand.

Example 2.4.1Describe the set of points 0< |x- 3| < .04.


Solution: The set of points 0< |x- 3| < .04 is the set of points(3- .04,3) « (3,3+ .04), or inother words,(2.96,3) « (3,3.04). �

Example 2.4.2Consider the set of points(2.4,2.6), and write it in the form|x- a| < d for somechoice ofa andd.

Solution: 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 allx so that|x- 2.5| < .1. �

Example 2.4.3It 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 andworking backwards to see that|x- 2| < .05?

Solution:

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.

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Example 2.4.4Building on the last problem, suppose you were told that whenever| f (x) - 3| < ethat there was then a corresponding numberd so that| f (x)-3| < ewhenever|x-2| < d. Mimic thesolution to the last problem in order to find the value ofd in terms ofe. Here againf (x) = 2x-1.

Solution:

Suppose that| f (x) - 3| < e. Then|2x- 1- 3| < e.Then|2x- 4| < e, so|x- 2| < e

2, sothe corresponding value ofd should bee2.

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Exercises:

EXERCISE2.4.1.Find (in interval notation) the set of pointsx for which it is true that|x-3| < .2.

EXERCISE2.4.2.find (in interal notation) the set of pointsx for which 0< |x- 4| < .3.

EXERCISE2.4.3.Find an inequality which describes the set of points(.75,1.25).


EXERCISE2.4.4. It turns out that forf (x) = 4x - 2, it is true that| f (x) - 6| < .2 whenever|x - 2| < .05. Can you show that this is true by starting with the inequality| f (x) - 6| < .2 andworking backwards to see that|x- 2| < .05?

EXERCISE2.4.5.Suppose that you were told that for any real positive numbere, it is possible tostart with the inequality|(5x- 1) - 14| < e and work backwards to get an inequality of the form|x- 3| < d for some real numberd which will depend one. Can you find the correct value ofd?

EXERCISE2.4.6.It happens to be true that ifx is in the set(2.9,3) « (3,3.1), then f (x) = 2x+ 1is within the set(6.8,7.2). On a set of coordinate axes draw the functionf (x) = 2x+ 1, and thenalong thex axis shade the set(2.9,3) « (3,3.1). Then along they axis, shade the set(6.8,7.2).Do you believe the statement that ifx is within the region you shaded along thex axis, that thenthe corresponding value off (x) is within the shaded region along they axis?

EXERCISE2.4.7. (The previous problem revisited.) Suppose you knew that you wantedf (x)to be within .2 of 7, that is, you wanted| f (x) - 7| < .2. The last problem says that this isaccomplished when|x-3| < .1. Show how to get this by starting with the inequality| f (x)-7| < .2and working towards|x- 3| < .1. Again, usef (x) = 2x+ 1.

EXERCISE2.4.8.(Revisiting the same problem yet again.) Suppose that instead of wantingf (x)to be within.2 of 7, you actually wanted it to be withine of 7, wheree represents an arbitrarypositive number. Thus, you want| f (x) - 7| < e. This is accomplished by having|x- 2| < d, for


some real numberd which depends one. Find a value ofd which accomplishes the task. Hint:start with| f (x) - 7| < e and work towards an inequality of the form|x- 2| < “something”. The“something” will be yourd.


2.5. If – Then Statements

Most of the statements of theorems in mathematics have the form “ifº thenº .” For example,everyone is familiar with the pythagorean theorem which states thatif a, b, andc are the sides of aright triangle withc opposite the right angle in the triangle,then a2

+b2= c2. The statement that

comes between the “if” and the “then” is sometimes called thehypothesis, while the statementthat comes after the “then” is called theconclusion. Thus, in the pythagorean theorem, thehypothesis is thata, b andc are the sides of a right triangle withc opposite the right angle andthe conclusion is thata2

+ b2= c2. Note that in an if – then statement the conclusion doesn’t

have to always hold in order for the statement to be true. It only needs to hold in the case thatthe hypotheses are met. For example, supposea = 2, b = 3 andc = 6. Note thata2

+ b2ê = c2.

Of course, this doesn’t mean that the pythagorean theorem isn’t valid, it only means that thesevalues ofa, b, andc aren’t the sides of a right triangle.

A statement like “ifp thenq”, is sometimes written asp Æ q. Thus,p ¨ q is actually thestatement “ifq thenp. We will sometimes be interested in statements of the formp W q whichrepresents the statements “ifp thenq and if q thenp.” This statement is often abbreviated as “pif and only if q” or even “p iff q.”

Truth tables are a convenient way to determine the truth value of a logical statement. In atruth table, we assign various values from the set{T, F} to the “variables”p andq, and deducethe value of the whole statement. For example, the truth table for “ifp thenq” would be:


p q pÆ qT T TT F FF T TF F T

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

Example 2.5.1Make a truth table for theq Æ p and compare it withp Æ q. Are they the sameor different?

Solution:

p q pÆ q qÆ pT T T TT F F TF T T FF F T T

We can see that they aren’t the same. Thus a statementpÆ q might be true but its converseqÆ p false. An example of this is the statement “ Iff (x) is differentiable on(a, b), then it is


continuous on(a, b). This statement is true but its converse “Iff (x) is continuous on(a, b) thenit is differentiable on(a, b) is false.

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Example 2.5.2Compare the statementpÆ q andqÆ p. Herep stands for the negation ofp. Ithas the opposite truth value of that ofp.

Solution:

p q pÆ q qÆ pT T T TT F F FF T T TF F T T

We see that these are the same. The statement ˜q Æ p is

called thecontrapositiveof pÆ q. �

Example 2.5.3How do the statements “p wheneverq” and “p only if q” compare with “if p thenq.” Also where does “p if and only if q” fit in?

Solution: “ p wheneverq” means the same thing as “q Æ p.”, and “p only if q” means the samething as “p Æ q.” The statement “p if and only if q” meansp Æ q AND q Æ p. This statementis often abbreviated “iff” and perhaps writtenpW q.

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Exercises:

EXERCISE2.5.1.Write the statement “f (x) is continuous atx = a whenever it is differentiableat x = a” in if – then form. Then write its converse and contrapositive in if – then format. Note:You do not need to know what the statement means in order to do this exercise.

EXERCISE2.5.2.If you were going to prove that|(2x- 1) - 3| < .2 whenever|x- 2| < .1, howwould you do it? First, identify the statement in if – then form, and then explain how you wouldgo about verifying such a statement.

EXERCISE2.5.3.Do problems 1 – 5 on page 128 of the companion to calculus.


2.6. Rigorous Definition of Limit

We want to make rigorous the idea of a limit. We want to say that limxÆ2 2x - 1 = 3 has aspecific meaning, more specific than “whenx is close to 2,f (x) is close to 3.” The main ideais the following: no matter how closef (x) is required to be to 3, we can assure that it is thatclose by requiringx to be within a certain distance of 2. Suppose we are required to makef (x)be within .01 of the value 3. We would be happy if there were a numberd so that whenx iswithin d units of 2 (i.e. |x - 2| < d), that it would then follow thatf (x) is within .01 unit of 3.Or more generally, for any real numbere > 0, if we are required to havef (x) be within e unitsof 3 (i.e. | f (x) = 3| < e), we would be satisfied if there was a real numberd (which would ofcourse depend one), so that as long as|x- 2| < d, it would then follow that| f (x) - 3| < e. Nowrecall that for limits, we don’t really requiref (x) to be near the limit whenx is equal to 2, weonly require it whenx is close to be not equal to 2. Thus we only really want to require that| f (x) - 3| < e whenever 0< |x- 2| < d. If in fact the inequality involvingf (x) is satisfied whenx = 2 then so be it, but it isn’t a requirement for the limit to exist.


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Figure 2.6: Whenx is within d units ofc, f (x) is within e units ofL.


In figure 2.6, note that limxÆc f (x) = L, and that for the given value ofe there is shown avalue ofd so that f (x) is within e of L wheneverx is within d of c. Or in other words, it is truethat if x is within the vertical shaded band, thenf (x) is within the horizontal shaded band.

To formalize all of this:

Rigorous Definition of LimitTo say that

limxÆc

f (x) = L,

means that for everye > 0 there existsd > 0 so that if0 < | x- c | < d, then| f (x) - L | < e.

There is a flash animation available which shows some of these ideas in a pictorial form. Itis available athere.

Example 2.6.1Prove rigorously that limxÆ2 2x- 1 = 3.

Solution:We must show that for anye > 0, there is a correspondingd > 0 so that|2x - 1 - 3| < e

whenever 0< |x- 2| < d. Working backwards,

http://math.furman.edu/~mwoodard/math10/demos/ed.html


Suppose that| f (x) - 3| < e. Then|2x- 1- 3| < e.Then|2x- 4| < e, so|x- 2| < e


Now to give a nice forwards formal presentation we could say:

Givene > 0, letd = e2.Then if 0< |x- 2| < e

2,then|2x- 4| < e, so|2x- 1- 3| < eso | f (x) - 3| < e.

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Example 2.6.2Prove rigorously that limxÆ1 4x+ 2 = 6.

Solution:We must show that for anye > 0, there is a correspondingd > 0 so that|4x + 2 - 6| < e

whenever 0< |x- 1| < d. Working backwards,

Suppose that| f (x) - 6| < e. Then|4x+ 2- 6| < e.


Then|4x- 4| < e, so|x- 1| < e


Now to give a nice forwards formal presentation we could say:

Givene > 0, letd = e4.Then if 0< |x- 1| < e

4,then|4x- 4| < e, so|4x+ 2- 6| < e,so | f (x) - 6| < e.

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Exercises:

EXERCISE2.6.1.Suppose you were going to prove rigorously that limxÆ4 5x- 1 = 19. What isthe corresponding if – then statement that you would need to prove?

EXERCISE2.6.2.Prove that if 0< |x- 3| < .1 then| f (x) - 10| < .3, wheref (x) = 3x+ 1. Youmay need to work backwards to see how to do it.

EXERCISE2.6.3.Draw the liney = 2x + 1, and plot the point(2,5). Suppose that you want itto be true that| f (x) - 5| < .4. For what value ofd would the statement “if 0< |x- 2| < d then


| f (x) - 5| < .4 be true? Use both your graph and some algebraic calculations to help you answerthe question.

EXERCISE2.6.4. Prove rigorously that limxÆ3 4x - 3 = 9. In the course of your proof youshould be able to find the value ofd for an arbitrarily given value ofe, so that the statement “if0 < |x- 3| < d then|4x- 3- 9| < e” is true.

EXERCISE2.6.5.Prove rigorously that limxÆ7 2x+ 1 = 15.

EXERCISE2.6.6.Prove rigorously that limxÆ-3 3x = -9.


2.7. Computing Limits

There are a number of facts (which we can proven using the rigorous definition of limit,) whichwe be useful in helping us compute limits. In the following, we will assume thatc andk are realnumbers, and that limxÆc f (x) and limxÆc g(x) both exist.

1. limxÆc

k = k. (The constant rule. )

2. limxÆc

x = c. (The identity rule.)

3. limxÆc

k f(x) = k limxÆc

f (x). (The constant multiplier rule.)

4. limxÆc

f (x) + g(x) = limxÆc

f (x) + limxÆc

g(x). (The sum rule.)

5. limxÆc

f (x) - g(x) = limxÆc

f (x) - limxÆc

g(x). (The difference rule.)

6. limxÆc

f (x)g(x) = limxÆc

f (x) ◊ limxÆc

g(x). (The product rule.)

7. limxÆc

f (x)g(x)

=limxÆc f (x)limxÆc g(x)

, as long as limxÆc g(x) ê = 0. (The quotient rule.)


8. limxÆc

f (x)n = (limxÆc

f (x))n, wheren is any number for which all expressions are defined. (The

power rule.)

Example 2.7.1Using only the above theorems, show that limxÆ3 x2+4x-4 = 32

+4(3)-4 = 17.

Solution:

limxÆ3

x2+ 4x- 4 = lim

xÆ3x2+ lim

xÆ34x- lim

xÆ34 Sum Rule

= limxÆ3

x2+ lim

xÆ34x- 4 Constant Rule

= limxÆ3

x2+ 4 lim

xÆ3x- 4 Constant Multiplier Rule

= KlimxÆ3

xO2

+ 4 limxÆ3

x- 4 Power Rule

= 32+ 4(3) - 4 The Identity Rule

= 17 Arithmetic!

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In fact, the result in the previous example can be extended to show that


The Plug-in Theorem for PolynomialsIf f (x) is a polynomial, then

limxÆc

f (x) = f (c).

Now using the limit theorem above known as the quotient rule, we can extend this result torational functions as follows:

The Plug-in Theorem for Rational FunctionsIf f (x) = p(x)

q(x) is a rational function and if limxÆc q(x) ê = 0, then

limxÆc f (x) = p(c)q(c) = f (c).

Example 2.7.2Compute limxÆ3x2+2x-15x+3 .

Solution: Sincex+ 3 is not zero forx = 3, the plug-in theorem applies. Thus, limxÆ3x2+2x-15x+3 =

32+2(3)-15

3+3 =06 = 0. �

Example 2.7.3Compute limxÆ3x2+2x-15

x2-5x+6


Solution: Note that the denominator is zero atx = 3, so the plug-in theorem does not apply.However, we can appeal to the fact that since these polynomials have a root of 3, they must havea factor ofx- 3. In fact, the numerator factors as(x- 3)(x+ 5) while the denominator factors as(x- 3)(x- 2). Thus the original problem is equivalent to

limxÆ3

(x- 3)(x+ 5)(x- 3)(x- 2)

= limxÆ3

x+ 5x- 2

=81= 8.

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Example 2.7.4Compute limxÆ16

0

x-416-x

Solution: Note that the plug-in theorem does not apply, since the denominator is zero atx = 16.However, we can rewrite the quotient in a much nicer way by multiplying the numerator and the


denominator by the conjugate of the numerator. We have

limxÆ16

0

x- 416- x

= limxÆ16

(0

x- 4)(0

x+ 4)

(16- x)(0

x+ 4)

= limxÆ16

x- 16

(16- x)(0

x+ 4)

= limxÆ16

-10

x+ 4

=-18

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Note that for the special functionsf (x) = |x| andg(x) = dxt, we have limxÆc |x| = |c| andlimxÆcdxt = dct when c is not an integer, but limxÆcdxt does not exist whenc is an integer,because the two one-sided limits don’t match up.

Example 2.7.5Compute limxÆ5|x-5|x-5


Solution: We must really use the fact that|x- 5| is defined differently on the different sides of 5.In fact

|x- 5| =ÏÔÌÔÓ

(x- 5) if x ≥ 5

(5- x) if x < 5.

Thus we must investigate the one-sided limits. First, consider the limit from the left. Note thatin this section of the domain,|x - 5| = 5 - x. Thus we have limxÆ5-

|x-5|x-5 = limxÆ5-

5-xx-5 =

limxÆ5- -1 = -1 On the other hand, in the portion of the domain to the right of 5, we have that|x- 5| = x- 5. Thus|x- 5| = 5- x. Thus we have limxÆ5+

|x-5|x-5 = limxÆ5+

x-5x-5 = limxÆ5- 1 = 1.

Since the two one-sided limits are different, the overall two-sided limit doesn’t exist. �

Example 2.7.6Let

f (x) = ;2x- 1 if x ≥ 23x2- 5 if x < 2

.

What is limxÆ2- f (x), limxÆ2+ f (x) and limxÆ2 f (x)?

Solution: In the portion of the domain which is to the left of 2, the function is defined to be 3x2-5.

Thus limxÆ2- f (x) = limxÆ2- 3x2- 5 = 7. On the other hand, in the portion of the domain which

is to the right of 2, the function is defined to be 2x- 1. Thus limxÆ2+ f (x) = limxÆ2+ 2x- 1 = 3.


Since these two one-sided limits are different, the two-sided limit limxÆ2 f (x) doesn’t exist.�

Exercises:

EXERCISE2.7.1.Compute limxÆ5 4x2+ 6x- 4. Show each step as in example2.7.1.

EXERCISE2.7.2.Using the rigorous definition of limit, prove that limxÆc x = c.

EXERCISE2.7.3.Compute limtÆ2t2-4

t2-2t+1.

EXERCISE2.7.4.Suppose you are considering limxÆ3p(x)q(x) , wherep(x) andq(x) are polynomials.

Suppose limxÆ3 p(x) = 0. What must be true about limxÆ3 q(x) so that you can say somethingdefinite about limxÆ3

p(x)q(x)?

EXERCISE2.7.5.Again, suppose you are interested in limxÆ3p(x)q(x) , but this time suppose that

you know that limxÆ3 q(x) = 0. What must be true about limxÆ3 p(x) in order for you to be ableto say something definite about limxÆ3

p(x)q(x) .

EXERCISE2.7.6. Suppose limxÆ4 f (x) = 3 and limxÆ4 g(x) = 8. What can you say about

limxÆ431

f (x)4 + f (x)g(x)?

EXERCISE2.7.7.Would it be possible for limxÆc f (x)-g(x) to exists without having limxÆc f (x)or limxÆc g(x) existing?


EXERCISE2.7.8.What is wrong with the following argument that all limits are equal to zero?Consider limxÆc f (x). It is equal to limxÆc x- c ◊ f (x)

x-c . By the limit theorem about the limit ofthe product equaling the product of the limits, we have limxÆc x- c ◊ limxÆc

f (x)x-c which is equal to

0 ◊ something= 0. Thus all limits are zero.

EXERCISE2.7.9.Compute limxÆ4

0

x-2x-4 .

EXERCISE2.7.10.Compute limhÆ0(3+h)2+3(3+h)-18

h .

EXERCISE2.7.11.Compute limxÆ2x3+x-10x-2 . Hint: Since plugging 2 into the numerator gives

zero, that tells you that the numerator should factor nicely, and one of the factors should bex-2.You can find the other factor by dividing.

EXERCISE2.7.12.Compute limyÆ1y2-1

y4+2y-3.

EXERCISE2.7.13.Suppose

f (x) =ÏÔÌÔÓ

3x2- 4 if x < 4;

11x if x ≥ 4.

Does limxÆ4 f (x) exist? If so, find it. If not, explain why not.


EXERCISE2.7.14.Suppose

f (x) =ÏÔÌÔÓ

3x2+ 4 if x < 4;

11x if x ≥ 4.

Does limxÆ4 f (x) exist? If so, find it. If not, explain why not.


2.8. Continuity

2.8.1. Definitions Related to Continuity

Graphs of functions that have no holes, gaps or jumps in them seem to have especially niceproperties. We would like to define a concept that would encapsulate these desirable properties.We say that such a function iscontinuousat a pointa if the value of f (a) is what we think itshould be, based on what the function near the point is. Thus we sayf (x) is continuous whenthe limit as we approach the point is equal to the value of the functionat the point.


Definition of Continuity at a PointWe say that a functionf (x) is continuousat a pointx = a if the following 3 things are true:

1. f (a) exists (i.e., the function is defined ata, or to put it another way,a is in thedomain of f (x).)

2. limxÆa f (x) exists (i.e., the limit off (x) exists at the point in question.)

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

limxÆa

f (x) = f (a).

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

If a function is defined at a given pointa, and if limxÆa- f (x) = f (a), then we say the functionis continuous from the leftor left continuous. Likewise, if limxÆa+ f (x) = f (a) then we say thefunction iscontinuous from the rightor right continuous. Of course, in order to be continuous ata pointa, the function must be both right continuous and left continuous at a that point.


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Figure 2.7: This function is continuous from the left atx = 2, but is not continuous there.


If a function is continuous at every point in an interval(a, b) we say it iscontinuous on theinterval (a, b). If it is also right continuous ata and left continuous atb, we say it is continuouson [a, b]. You should be able to guess what it means to be continuous on other intervals, like(-•,3) and[4,•).

2.8.2. Basic Theorems

Using the limit theorems, it isn’t hard to show that iff andg are continuous at a point, thenf +g,f - g, and f g are continuous at that point. The same is true forf

g as long asg(a) ê = 0. Also forcompositions, Ifg(x) is continuous ata and f (x) is continuous atg(a), then f Î g is continuous ata. These facts follows immediately from the various limit theorems. From these facts it is easyto build up a nice catalogue of continuous functions. In particular, note the following:


Basic Continuity TheoremsPolynomials are continuous everywhere and rational functions are con-tinuous wherever they are defined. (These follow directly from the plug-in theorem.) Also, the square root function is continuous on its domain,as are other root functions. The absolute value function is continuouseverywhere, but the greatest integer functions is discontinuous at eachinteger.

2.8.3. The Intermediate Value Theorem

One example of a consequence of continuity is the Intermediate Value Theorem, or IVT. It saysthat if a function f (x) is continuous on an interval[a, b], and if M is an intermediate valuebetweenf (a) and f (b), that there then exists a numberc betweena andb so that f (c) = M.Thus if you are going 30 mph at one moment and 40 mph a little later, you must have beengoing 35 mph at some point in between. This wouldn’t follow necessarily if your speed weren’ta continuous function of time (which it is).


The Intermediate Value TheoremIf f (x) is continuous on the interval[a, b], and if M is an intermediatevalue betweenf (a) and f (b), that there then exists a numberc betweenaandb so thatf (c) = M.

Example 2.8.1Where is the functionh(x) =1

I

x+1x2-4M

2+ 1 continuous?

Solution: Since the square root function is continous on its domain, and the expression under-neath the square root is positive, this doesn’t present any problems. Also,x+1

x2-4 is a rationalfunction, which is continuous wherever it is defined, but note that it isn’t defined atx = 2 orx = -2. Thus, at these pointsh(x) is definitely not continuous, since it is not defined there. Atevery other pointh(x) is the composition of continuous functions, and is thus continuous. So theonly discontinuities are atx = 2,-2. �

Example 2.8.2A removable discontinuityis a discontinuity where limxÆc f (x) exists, butf (c)doesn’t exist. It is called removable because one could define a new functionf which is equalto f everywhere except atc, and at this point has value equal to limxÆc f (x). This new functionis now continuous atc, we have “removed” the discontinuity. The functionf (x) = x2

-16x-4 has

a removable discontinuity atx = 4. What shouldf (4) be defined to be in order to remove thediscontinuity? Note: A discontinuity which is not removable is calledessential.


Solution: We need to havef (4) = limxÆ4 f (x), and this can be shown to be 8. (Do it!) Thus if wedefine f (4) = 8, then f and f are the same everywhere except at 4, andf is continuous atx = 4.

�

Exercises:

EXERCISE2.8.1.Where is the functionf (x) = x2-9

x-3 discontinuous? Is it a removable or anessential discontinuity?

EXERCISE2.8.2.Where is the functionf (x) = |0

x- 3- 18| continuous?

EXERCISE2.8.3.The functionf (x) = x3-8

x-2 has a removable discontinuity atx = 2. How shouldthe function be redefined atx = 2 in order to “remove” the discontinuity?

EXERCISE2.8.4.Consider the function

f (x) =

ÏÔÔÔÌÔÔÔÓ

2x+ 3 if x £ 0;

3- x if 0 < x < 5;

x+ 5 if x ≥ 5.

Where isf (x) not continuous? Justify your answer.


EXERCISE2.8.5.One can use the Intermediate Value Theorem to show that the functionf (x) =x3- x- 1 has a root on[1,2]. Show how to do this, by identifying the value ofa, b, f (a), f (b)

andM in the statement of the IVT.

EXERCISE2.8.6.How could you show that the equationx = x4- 3x + 1 has at at least 2 real

solutions?

EXERCISE2.8.7.Explain why the functiong(h) = |h|h is discontinuous ath = 0.

EXERCISE2.8.8.Where is the functionf (x) = d2xt discontinuous? Hint: A sketch might help.

EXERCISE2.8.9.Draw the sketch of any function which is continuous on(-•,3), is continuouson [3,6] and is continuous on(6,•), but is not continuous atx = 3 or atx = 6.



Chapter 3

The Derivative

Contents

3.1 Definition of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . .137

3.1.1 Slopes and Velocities Revisited. . . . . . . . . . . . . . . . . . . . 137

3.1.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . .142


3.1.3 The Derivative as a Function. . . . . . . . . . . . . . . . . . . . . .144

3.2 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1473.3 Rules for Computing Derivatives . . . . . . . . . . . . . . . . . . . . . . .153

3.3.1 Derivatives of Polynomials. . . . . . . . . . . . . . . . . . . . . . .153

3.3.2 Product and Quotient Rules. . . . . . . . . . . . . . . . . . . . . .157

3.4 Algebraic Simplifications Involving Exponents . . . . . . . . . . . . . . . 1623.4.1 Rules of Exponents. . . . . . . . . . . . . . . . . . . . . . . . . . .162

3.4.2 Simplifying the Result of the Product and Quotient Rules. . . . . . . 166

3.5 Applications of Derivatives as Rates of Change. . . . . . . . . . . . . . . 1703.5.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .170

3.5.2 Some Specific Applications. . . . . . . . . . . . . . . . . . . . . .171

3.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1773.7 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . .1813.8 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .187

3.8.1 The Second Derivative. . . . . . . . . . . . . . . . . . . . . . . . .187

3.8.2 Third and Higher Derivatives. . . . . . . . . . . . . . . . . . . . . .188

3.9 Related Rates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1913.10 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .195

3.10.1 New Notation – Old Idea. . . . . . . . . . . . . . . . . . . . . . . .196


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 betweena anda + h, and that this slope approximates theslope of the tangent line atx = a, and this approximation gets better ash Æ 0. Thus it makesense to define the slope of the tangent line as

limhÆ0

f (a+ h) - f (a)h

.

By making the substitutionh = x - a, we can rewritea + h asx. Also, ash Æ 0 we havexÆ a. Thus an alternative but equivalent expression for the slope of the tangent line is

limxÆa

f (x) - f (a)x- a

.

As discussed before, these quantities also represent the instantaneous rate of change of thefunction f with respect tox.


Example 3.1.1Find the instantaneous rate of change off (x) with respect tox for f (x) = x3, atx = 3.

Solution: We must compute limxÆ3f (x)- f (3)

x-3 or limhÆ0f (3+h)- f (3)

h . Let’s do the former. We have

limxÆ3

f (x) - f (3)x- 3

= limxÆ3

x3- 27

x- 3

= limxÆ3

(x- 3)(x2+ 3x+ 9)

x- 3Difference of cubes

= limxÆ3

x2+ 3x+ 9

= 27 By the Plug-In Theorem

�

Example 3.1.2Find the instantaneous rate of change ofy = 3x2- 5x+ 3 atx = 3.

Solution:

We must compute limhÆ0

f (3+ h) - f (3)h

. This is equal to limhÆ0

3(3+ h)2 - 5(3+ h) + 3- 15h

.

Thus we have


limhÆ0

3(3+ h)2 - 5(3+ h) + 3- 15h

= limhÆ0

3(9+ 6h+ h2) - 15- 5h+ -12

h

= limhÆ0

(27+ 18h+ 3h2) - 5h- 27

h

= limhÆ0

12h+ 3h2

h= lim

hÆ012+ 3h

= 12

Thus the instantaneous rate of change ofy with respect tox is 12. �

Example 3.1.3Find the equation of the tangent line atx = 9 for f (x) = x+0

x.

Solution:

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),


wherem is the slope of the tangent line. Thus we must compute limxÆ9f (x)- f (9)

x-9 .

We have


limxÆ9

f (x) - f (9)x- 9

= limxÆ9

x+0

x- 12x- 9

= limxÆ9

0

x+ x- 12x- 9

= limxÆ9

0

x+ x- 12x- 9

◊

0

x- (x- 12)0

x- (x- 12)

= limxÆ9

x- (x- 12)2

(x- 9)(0

x- (x- 12))

= limxÆ9

x- (x2- 24x+ 144)

(x- 9)(0

x- (x- 12))

= limxÆ9

-(x2+ 25x- 144)

(x- 9)(0

x- (x- 12))

= limxÆ9

-(x2- 25x+ 144)

(x- 9)(0

x- (x- 12))

= limxÆ9

-(x- 9)(x- 16)

(x- 9)(0

x- (x- 12))

= limxÆ9

-(x- 16)

(0

x- (x- 12))

=76


Thus the equation of the tangent line is given byy- 12= 76(x- 9). �

3.1.2. The Derivative

The quantity limhÆ0f (a+h)- f (a)

h is so important, it deserves its own name. We will call it thederivative of f(x) at x= a. We will use the symbolf ¢(a) to reprsent this quantity. Thus we have

Definition of DerivativeThe derivative off (x) atx = a is denotedf ¢(a) and is given by

f ¢(a) = limhÆ0

f (a+ h) - f (a)h

or

f ¢(a) = limxÆa

f (x) - f (a)x- a

.

Example 3.1.4For f (x) =0

x+ 2, find f ¢(2).

Solution:



f (2+ h) - f (2)h


0

4+ h- 2h

. Thus we have

limhÆ0

0

4+ h- 2h

= limhÆ0

0

4+ h- 2h

◊

0

4+ h+ 20

4+ h+ 2

= limhÆ0

4+ h- 4

h(0

4+ h+ 2)

= limhÆ0

h

h(0

4+ h+ 2)

= limhÆ0

10

4+ h+ 2

=14

Thus f ¢(2) = 14. �


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 functionf (x) at a general pointx. So it would be reasonable to definef ¢(x) = limhÆ0

f (x+h)- f (x)h , or

f ¢(x) = limzÆxf (z)- f (x)

z-x .

Example 3.1.5For f (x) = 1x , find f ¢(x).

Solution:


f (x+ h) - f (x)h


1x+h -

1x

h. We have


limhÆ0

1x+h -

1x

h= lim

hÆ0

x-(x+h)(x)(x+h)

h

= limhÆ0

x- (x+ h)hx(x+ h)

= limhÆ0

-hhx(x+ h)

= limhÆ0

(-1)(x)(x+ h)

=-1x2

Thus f ¢(x) = -1x2 . �

Exercises:

EXERCISE3.1.1. Compute, using the definition of derivative, the value off ¢(2) for f (x) =x2+ x+ 1.

EXERCISE3.1.2.Compare each of the following:


1. limxÆzf (x)- f (z)

x-z

2. limxÆzf (z)- f (x)

z-x

3. limzÆxf (x)- f (z)

x-z

4. limzÆxf (z)- f (x)

z-x

In particular, each of the above is the derivative off (x) at eitherx or atz. Which is which?

EXERCISE3.1.3.Find f ¢(2) for f (x) =0

x+ 2.

EXERCISE3.1.4.limhÆ0(3+h)2+2(3+h)-15

h representsf ¢(a) for some functionf and some numbera. What is a possible choice forf (x) and fora?

EXERCISE3.1.5.Supposef (x) = ax2, wherea is a constant. Findf ¢(x).

EXERCISE3.1.6.Supposey = (2x+ 1)2. What wouldy¢ be?


3.2. Differentiability

Recall that the expression

limhÆ0

f (x+ h) - f (x)h

represents the derivative of the functionf (x). Also recall that limits sometimes have the disturb-ing property of not existing. Thus there could be points in the domain off (x)wheref ¢(x) doesn’texist. At such a point, we say thatf is not differentiable. Also, if a is in the domain off (x), butf ¢(a) doesn’t exist, we say that(a, f(a)) is a singular pointof f (x). If f (x) is differentiable atevery 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 functionf (x) = |x| at a = 0. To see thatthis is a singular point, we will see that limhÆ0

|0+h|-|0|h doesn’t exist. Of course, this expression

is the same as limhÆ0|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 limhÆ0-|h|h and limhÆ0+

|h|h . Now

recall that|h| = -h if h < 0, so limhÆ0-|h|h = limhÆ0-

-hh = limhÆ0- -1 = -1 (since the limit of

a constant is that constant.) On the other hand (sorry, that was a bad pun) we know that whenh > 0, |h| = h, so we have limhÆ0+

|h|h = limhÆ0+

hh = limhÆ0+ 1 = 1. Thus the two one-sided limits

are different, so the overall limit doesn’t exist. Thus(0,0) is a singular value forf (x) = |x|.

Example 3.2.1Show that(0,0) is a singular point forf (x) = 30

x2.


Solution:

Our job is to show that limhÆ030(0+h)2- 300

h

Figure 3.1: The absolute value function is continuousatx = 0 but is not differentiable atx = 0.

doesn’t exist. We can write30

h2 ash23 ,

and using the rules of exponents, we

can write h23

h ash23-1= h

-13 =

1

h13=

130h

. Now limhÆ0130h

doesn’t exist, so(0,0) is a singular point for this func-tion.

�

There is an important theorem whichrelates the differentiability off (x) atx = a to the continuity off (x) atx = a.The theorem is as follows:


Differentiability at x = a implies continuity at x = a.If f (x) is differentiable atx = a then it is continuous atx = a. However,it is possible for a function to be continuous atx = a without beingdifferentiable there.

This theorem states that if you know thatf (x) is differentiable atx = a, then you know thatit is continuous there as well. Or, (using the contrapositive statement,) Iff (x) is not continuousatx = a, then f (x) is not differentiable atx = a.

Here is the proof of the theorem: Suppose you know thatf (x) is differentiable atx = a sof ¢(a) = limxÆa

f (x)- f (a)x-a exists. In particular you know thatf (a) exists, or else this expression

wouldn’t make sense. Now to show thatf (x) is continuous atx = a, we only need to show thatlimxÆa f (x) = f (a), or (equivalently) show that limxÆa f (x) - f (a) = 0. But we can performthe following somewhat devious trick of multiplying by one in a clever way: we can rewritef (x) - f (a) as( f (x) - f (a)) ◊ x-a

x-a Now taking the limit asxÆ a we get limxÆa( f (x) - f (a)) ◊ x-ax-a =

limxÆaf (x)- f (a)

x-a ◊ limxÆa(x- a) = f ¢(a) ◊ 0 = 0. Of course, in this last string of equalities we usedthe fact that the limit of a product is the product of the limitsas long as both limits exist.Thefact that limxÆa

f (x)- f (a)x-a exists is exactly our hypothesis thatf (x) is differentiable atx = a.

To finish up the proof, we note that since limxÆa f (x) - f (a) = 0, then we also have thatlimxÆa f (x) = f (a), so f (x) must be continuous atx = a.


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

0

x2 are continuous at(0,0), but are not differentiable there.

Continuity does not imply differentiability.It is possible for a function to be continuous at a point without beingdifferentiable at that point. One example is the functionf (x) = |x| atx = 0, and another example isf (x) = 3

0

x2

atx = 0.

Exercises:

EXERCISE3.2.1.Does the functionf (x) = |x - 2| have a singular point atx = 2? That is, isx = 2 a point wheref (x) is not differentiable? Hint: write down the limit that would have toexist in order for the function to be differentiable atx = 2, and then look at this limit ash Æ 0-

and ashÆ 0+.

EXERCISE3.2.2.We will see in the next section that all polynomials are differentiable every-where. However, rational functions will sometimes have points where they aren’t differentiable


in the same way that they sometimes have points where they aren’t continuous. Formulate astatement about how to describe the points at which a rational function isn’t differentiable.


f (x) =ÏÔÌÔÓ

xx-1 if x ê = 1;

3 if x = 1.

This function is not differentiable atx = 1, because it is not continuous there. Show that thisfunction is not differentiable atx = 1 directly, that is, show that the limit of the differencequotient does not exist.


f (x) =ÏÔÌÔÓ

2x- 1 if x < 0;

-2x- 1 if x ≥ 0.

This function has a singular point atx = 0. Show that this is true. Is this function continuous atx = 0?

EXERCISE3.2.5.Show that the functionf (x) = xx-1 is differentiable atx = 0 by computing

f ¢(0).



f (x) =ÏÔÌÔÓ

x2 if x < 0;

4x2 if x ≥ 0.

Is this function differentiable atx = 0? Is it continuous there?


3.3. Rules for Computing Derivatives

3.3.1. Derivatives of Polynomials

It turns out that there are a number of theorems which make the process of computing derivativesmuch easier. We will develop these now, focusing first on some results which will make thecomputation of the derivative of any polynomial almost trivial.

The first rule is that the derivative of a constant functionf (x) = c is always zero. This isbecause for such a function, limhÆ0

f (x+h)- f (x)h = limhÆ0

c-ch = limhÆ0

0h = 0.

The second rule involves the functions of the formf (x) = mx, whose graphs are straightlines through the origin. The derivative of such functions turns out to be the constantm, becauselimhÆ0

f (x+h)- f (x)h = limhÆ0

mx+mh-(mx)h = limhÆ0

mhh = limhÆ0 m= m.

The third rule shows us how to differentiate a power function of the formf (x) = xk, wherek is a positive integer. Before investigating this 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 toknow here is that

(x+ h)k = c0xk+ c1xk-1h+ c2xk-2h2

+ c3xk-3h3+º + ck-1xhk-1

+ ckhk

where theci ’s are constants. You may recall that there is a nice formula for the constantsci


which involves either Pascal’s triangle or factorials, but for our purposes, we only need to knowthe first two constantsc0 andc1. It turns out thatc0 = 1 andc1 = k. Thus we have

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

+ c3xk-3h3+º + ck-1xhk-1

+ ckhk.

Thus we can writef ¢(x) = limhÆ0(x+h)k-xk

h as

f ¢(x) = limhÆ0

xk+ kxk-1h+ some stuff with a factor of at leasth2

- xk

h

= limhÆ0

kxk-1h+ some stuff with a factor of at leasth2

h

= limhÆ0

kxk-1+ some stuff with a factor of at leasth

= kxk-1

We will refer to this theorem as thepower rule.We will see in a later section that this rule isvalid for fractional values ofk as well, but that will require another proof. The rule is also validfor negative values ofk. Summarizing, we have:


The Power Rule for DerivativesThe derivative of a functionf (x) = xk is f ¢(x) = kxk-1. This also holds inthe case wherek = 0 andk = 1 (where we interpretx0 as 1.)

Next we investigate the derivative of a functionf (x) which has the formcg(x) for someconstantc and some functiong(x). We note that limhÆ0

cg(x+h)-cg(x)h = c limhÆ0

g(x+h)-g(x)h , so that

f ¢(x) = cg¢(x). We will refer to this theorem as theconstant multiplier rule.

The Constant Multiplier Rule for DerivativesThe derivative of a functionf (x) = cg(x) is f ¢(x) = cg¢(x).

How about a functionf (x) which is the sum of two functionsm(x) andn(x)? So the questionis if f (x) = m(x) +n(x), what is f ¢(x)? The answer (not surprisingly) turns out to bem¢(x) +n¢(x),since limhÆ0

m(x+h)+n(x+h)-(m(x)+n(x))h = limhÆ0

m(x+h)-m(x)h +

n(x+h)-n(x)h which of course can be written

as limhÆ0m(x+h)-m(x)

h + limhÆ0n(x+h)-n(x)

h which we recognize asm¢(x) +n¢(x). We will refer to thistheorem as thesum rule.


Note that using the rules so far, we can easily take the derivative of any polynomial. Forexample, to differentiatef (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 havef ¢(x) = 20x4

+ 4x- 4.

Example 3.3.1Find f ¢(x) for f (x) = -3x6+ 2x3

- 12x- 123.

Solution: For f (x) = -3x6+ 2x3

- 12x- 123, we havef ¢(x) = -18x5+ 6x2

- 12. �

This last example demonstrate that one should be able to differentiate any polynomial withlittle or no difficulty. In fact, we have

Differentiation of PolynomialsEvery polynomial is differentiable everywhere. The derivative of

f (x) = anxn+ an-1xn-1

+º + a1x+ a0

isf ¢(x) = nanxn-1

+ (n- 1)an-1xn-2+º + 2a2x+ a1.


3.3.2. Product and Quotient Rules

The next two rules 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 usstarted, consider the functionf (x) = (2x + 3)(3x - 1). By multiplying this out, we have thatf (x) = 6x2

+ 7x - 3. Now using the above techniques, we see thatf ¢(x) = 12x + 7. Note thatthis result isnot equal to what would be obtained by just multiplying the derivatives of 2x + 3and 3x- 1 together. Thus whatever is going to become ourproduct rulewill have to be a littlemore complicated. It turns out that iff (x) = m(x)n(x), then the right expression forf ¢(x) ism(x)n¢(x) + n(x)m¢(x), which probably can be remembered best as“the first function times thederivative of the second plus the second times the derivative of the first”. In our example withm(x) = 2x+3 andn(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.Summarizing, we have:

The Product Rule for DerivativesIf f (x) = m(x)n(x), then f ¢(x) = m(x)n¢(x) + n(x)m¢(x). Also, if m(x)is differentiable everywhere andn(x) is differentiable everywhere thenf (x) = m(x)n(x) is differentiable everywhere as well.


Given that the product rule was a little messy and unexpected, you should now expect thatthe quotient rule 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 findf ¢(x) by resorting to the definition.

Example 3.3.2Use the definition of derivative to computef ¢(x) for f (x) = 2x+33x-1.

Solution:


limhÆ0

f (x+ h) - f (x)h

= limhÆ0

2x+2h+33x+3h-1 -

2x+33x-1

h

= limhÆ0

2x+2h+33x+3h-1 ◊

3x-13x-1 -

2x+33x-1 ◊

3x+3x-13x+3h-1

h

= limhÆ0

(2x+2h+3)(3x-1)-(3x+3x-1)(2x+3)(3x-1)(3x+3h-1)

h

= limhÆ0

(6x2+ 6hx+ 9x- 2x- 2h- 3) - (6x2

+ 6hx- 2x+ 9x9h- 3)(h)(3x- 1)(3x+ 3h- 1)

= limhÆ0

-11h(h)(3x- 1)(3x+ 3h- 1)

= limhÆ0

-11(3x- 1)(3x+ 3h- 1)

=-11(3x- 1)2

�

Now note that the if we were to compute the expressionq(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 example.

This is (of course) not a proof, but hopefully gives you some evidence towards believing in thefollowing quotient rule:

The Quotient Rule for DerivativesIf f (x) = p(x)

q(x) , then f ¢(x) = q(x)p¢(x)-p(x)q¢(x)(q(x))2 . Also, if p(x) andq(x) are

differentiable everywhere, thenf (x) = p(x)q(x) is differentiable everywhere

except where q(x) = 0.

Note also that from this theorem follows the fact that all rational functions are differentiableat all points in 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 multiplier rule.)

5. If f (x) = m(x) + n(x), then f ¢(x) = m¢(x) + n¢(x). (The sum rule.)

6. If f (x) = m(x) ◊ n(x) then f ¢(x) = m(x) ◊ n¢(x) + m(x) ◊ n¢(x). (Theproduct rule).

7. If f (x) = p(x)q(x) , then f ¢(x) = q(x)p¢(x)-p(x)q¢(x)

(q(x))2 . (The quotient rule.)

Exercises:

EXERCISE3.3.1.In Stewart section 3.3, do problems 1 – 39 odd.


3.4. Algebraic Simplifications Involving Exponents

It is time to review some algebraic facts which will help to make the process of simplifying theoutput of the differentiation process easier. We start with a review of the rules of exponents.

3.4.1. Rules of Exponents

Many students have trouble working with expressions involving exponents because they forgetwhat the definition ofan is. Let us start with a reminder of this definition.


Definition of an.If a is a non-negative real number andn is a positive integer, thenan

=

n factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúûa ◊ a ◊ aµa . On the other hand,a

1n =

n0

a, anda-n=

1an . Also, if n is

not an integer, but is the rational numberpq wherep andq are integers,

thenan= a

pq =

q0

ap. Finally, we mention that ifa is negative, the aboveare also true as long as they represent real numbers. For example, ifa isnegative thena

12 is not defined.

Example 3.4.1Compute 823 , 16-

32 , and 3

32

23 .

Solution:

1. 823 =

30

82, and since the cube root of 8 is 2, this expression is equal to 22, which is 4.

2. 16-32 =

10

163 =

143 =

164.

3. 332

23=

0

33(

23, and

0

33=

0

30

30

3 = 30

3. Thus,0

33(

23 = 3

0

323=

31

30

3 ◊ 30

3 =30

27= 3. Of course, we could also get this more easily from the rule of exponents, which


we now review.

�

In order to remember the rules of exponentiation, like the fact thatan◊ am= an+m, it is useful

to recall that whenn is an integer,an means

n factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúûa ◊ a ◊ aµa. Thus

an◊ am=

n factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúûa ◊ a ◊ aµa ◊

m factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúûa ◊ a ◊ aµa =

n+m factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúûa ◊ a ◊ aµa = an+m

.Of course, this rule works whether or notn andm are integers. The companion rule to this

one is thatan

am = an-m.

Another rule is thatanm= anm. This can be remembered by thinking ofanm as

m factorsõúúúúúúúúúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúúúúúúúúúû

an◊ an◊ anµan

which is the same thing as having (n factors ofa) m times, which of course is the same thing ashavingnmfactors ofa.

Also, we mention that(ab)n = anbn, since

(ab)n =

n factorsõúúúúúúúúúúúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúúúúúúúúúúúû

ab ◊ ab ◊ abµab=

n factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúûa ◊ a ◊ aµa ◊

n factorsõúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúû

b ◊ b ◊ bµb = anbn.


The companion rule to this one is thatI abMn=

an

bn . We summarize these as follows:

In all cases, assume thata, b andc are such that the expressions involvingthem represent real numbers. Then

• anam= an+m

• an

am = an-m

• (an)m= anm

• (ab)n = anbn

• I abMn=

an

bn

Note, however, that the functionf (x) = Jx12N

2is not the same function asg(x) = x, since

they don’t have the same domain. (Although they do agree on the domain off (x).) Also notethat the functionh(x) = (x2

)12 is different from eitherf (x) or g(x), since it has a different domain

from f (x) and is different fromg(x) at the negative numbers. For example,g(-1) = -1, whereash(-1) =

0

(-1)2 =0

1 = 1. The best way to writeh(x) is as|x|, since it agrees with the functiong(x) = x on the positive numbers, but is the opposite ofg(x) on the negative numbers.


In general whenm andn are even, we have that(xm)

1n = |x|

mn .

Example 3.4.2Simplify the following expressions without affecting the domain.f (x) = ((2x)12 ◊

x-14 )

8 andg(x) = ((2x)12 + x

-14 )

2.

Example 3.4.3Suppose you wanted to compute the derivative off (x) = ( 30

x2 +50

x3)2. Rewritethe functionf (x) first, and then differentiate it.

3.4.2. Simplifying the Result of the Product and Quotient Rules

After computing a derivative via either the product or quotient rules, we often need to simplifythe result. Our goal in such simplifications is to put the function in a form in which is is easyto see where it is equal to zero. Thus, for a polynomial or similar expression, we need to factorit completely, whereas for a rational expression we need to express it as the ratio of factoredexpressions.

For example, suppose you computed a derivative via the product rule and obtainedf ¢(x) =(x2- 4)(6)(2x + 1)5(2) + (2x + 1)6(2x). Our goal would be to factor this completely, but many

well-intentioned students have difficulty with doing this because they make the fatal mistake oftrying to multiply the result out first. Instead, note that the expression is already half-factored.In fact, we can complete the factorization by noting that the expression has two terms, both of


which have numerous factors of 2x+ 1. In fact, to the left of the+ sign there are 5 factors, whileto the right there are 6. Thus we can factor out 5 factors of(2x+ 1) and also a common factor of2. This yields 2(2x+1)5 A6(x2

- 4) + (2x+ 1)(x)E. This simplifies to 2(2x+1)5(8x2+ x-24). We

can then see that this expression is zero whenx = -12 or whenx = -1±

0

1-(4)(-8)(24)16 .

Example 3.4.4Find the derivative of(6x2- 1)(4x3

) via the product rule, and then simplify theresult.

After computing a derivative via the quotient rule, we often need to either factor the numer-ator, or perhaps multiply by one in a way so as to simplify the numerator.

Example 3.4.5If g(x) =0

xx2+8x, computeg¢(x) and simplify the result.

Solution: g¢(x) =(x2+8x) 12 x

-12 -0

x(2x+8)(x2+8x)2 . Note that the numerator has two terms, and thatboth would

be “nicer” if we were to multiply by 2x12 . Also, the denominator wouldn’t be much worse, as it


would still be factored. Thus we multiplyg¢(x) by 2x12

2x12. We obtain

g¢(x) =(x2+ 8x) 12x

-12 -0

x(2x+ 8)

(x2 + 8x)2◊

2x12

2x12

=(x2+ 8x) - 2x(2x+ 8)

(x2 + 8x)2 ◊ 20

x

=-3x2

- 8x

(x2 + 8x)2 ◊ 20

x

=(x)(-3x- 8)

(x2 + 8x)2 ◊ 20

x.

Note that we can easily see thatg¢(x) is zero atx = - 83. Note that the expression is not zero at

x = 0, but rather is undefined there. �

Example 3.4.6Find the derivative off (x) = 3 30

x+20

x

x45

, and then simplify.

Exercises:

EXERCISE3.4.1.Compute the derivative off (x) = 4x2(3x3-x) using the product rule, and then

simplify the result.


EXERCISE3.4.2.Suppose that the derivative ofh(x) is h¢(x) = (3x2+ 5)2(4)(x3

- 11)3(3x2) +

(x3- 11)42(3x2

+ 5)(6x). Simplify h¢(x).

EXERCISE3.4.3.Find and simplifyf ¢(x) for f (x) =0

x4x2-1.

EXERCISE3.4.4.Find and simplifyg¢(x) for g(x) = x1+x2 .

EXERCISE3.4.5.Find f ¢(x) for f (x) = 30

x ◊ (3x+0

x), by simplifying before differentiating.


3.5. Applications of Derivatives as Rates of Change

3.5.1. Overview

Recall that one of the motivations toward the definition of the derivative was in the investigatingof the concept of instantaneous velocity as the limit of average velocity as the time intervalshrank to zero. In general, no matter what quantities are represented byy = f (x), we can think ofthe derivative as a “rate” in the sense that as the quantityx changes, so does the quantityy, andif we write h = (x+ h) - x) asDx and writeDy = f (x+ h) - f (x), we can think ofDy

Dx =f (x+h)- f (x)

has an average rate of change ofy with respect tox, since we have the ratio of a change iny toa change inx over a small interval. The limit of this quantity asDx Æ 0 (which of course isthe derivative) can be written limDxÆ0

DyDx, can be thought of as the limit of the average rate of

changey with respect tox as thex interval shrinks, so we can think of this as an instantaneousrate of change ofy with respect tox. To make our derivative look more like the limit of a changein y over a change inx, we will sometimes writef ¢(x) as dy

dx. This is sometimes calledLeibnizNotationfor the derivative.

Thus


Leibniz Notation for the Derivative

f ¢(x) = limDxÆ0

DyDx=

dydx

is the derivative off (x) with respect tox, and can be thought of as theinstantaneous rate of change ofy = f (x) with respect tox.

3.5.2. Some Specific Applications

3.5.2.1. Rectilinear Motion

Suppose an object is moving in a straight line (either vertically or horizontally) and its positionat time t is given bys(t). he theory developed so far indicates that the derivatives¢(t) shouldgive us the objects velocity at timet, since the instantaneous rate of change of position withrespect to time is velocity. Thus, we will writev(t) = s¢(t). Also, since the rate of changeof velocity with respect to time is commonly called acceleration, we can writev¢(t) = a(t).Now sinceF = ma whereF is force andm is the mass of an object, we can think of theacceleration at timet as determining the force acting on our object. If velocity were zero at a


given time, we would know that the object was stopped at that time. Thus, nonzero velocityindicates movement; we will consider positive velocity to indicate movement to the right (if ourline is oriented horizontally) or upward movement (if our line is oriented vertically.) Likewise,positive or negative acceleration will be thought of as acting in those directions. When an object’svelocity and acceleration both havethe same sign, then the object will speed up. Note thatthis might occur when they are both positive or both negative (as in the case of a ball throwndownward from a high tower, both its velocity and acceleration are negative, so the ball speedsup.) When velocity and acceleration have opposite signs, the object will be slowing down.

Relationship Between Position, Velocity and AccelerationIf an object’s position at timet is given bys(t), thenv(t) = s¢(t) = ds

dtrepresents the object’s velocity at timet. Also,v¢(t) = a(t) represents theobject’s acceleration at timet.

Example 3.5.1An object is moving along a straight line so that its position at timet is given bys(t) = 2t3

- 24t + 2. Analyze the movement of this object.

Solution: s¢(t) = v(t) = 6t2- 24. Now, this equation is zero whent = 2 or t = -2, so those times

represent when the object is stopped. An analysis of the sign ofv(t) shows thatv(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 timet = 0. A quick check shows that the accelerationis positive fort > 0 and negative fort < 0. Thus during the time interval(-•,-2) the object ismoving to the right but slowing down. Att = -2, the object is stopped. Fromt = -2 to t = 0 theobject is moving to the left and speeding up. At timet = 0 the acceleration becomes positive, sothe object continues to move to the left, but starts to slow down. Att = 2, the object is stopped,and thereafter moves to the right and speeds up. �

3.5.2.2. Biology Applications

Many biologists are interested in the growth of populations, of humans, animals, cells, etc.. Ifp(t) is the size of a given population at timet, thenp¢(t) is the population growth rate at timet.

Example 3.5.2A population of bacterial cells in a petri dish is growing so that at timet, thereare0

t(t2+ 3t) cells present.

1. What is the change in the size of population between timet = 4 andt = 9?

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


Solution:

1. At time t = 4, the population has sizep(4) =0

4(42+ 3(4)) = 2(16+ 12) = 56. At time

t = 9, the population has sizep(9) =0

9(92+3(9) = 3(81+27) = 324. Thus over that time

interval of length 5, the change inp is 324- 56= 268. Thus the average rate of change ofthe population is268

5 .

2. The instantaneous rate of change of the population would be given byp¢(4). We havep¢(t) =

0

t(2t + 3) + (t2+ 3t) 12t

-12 . Thusp¢(4) =

0

4 ◊ 11+ (28) ◊ 12(0

4)= 22+ 28

4 = 29.

�

3.5.2.3. Chemistry Applications

Chemists are often interested in rates of chemical reactions, as well as the relationship betweenvarious physical quantities of various types of matter. For example, one form of Boyle’s lawstates that the absolute temperatureT, pressureP and volumeV of an ideal gas of fixed massmare related byPV = mRTwhereR is constant.

Example 3.5.3In the above equation for Boyle’s law, suppose you were able to hold temperatureT constant. What would happen to pressureP as volumeV increased? What would be theinstantaneous rate of change of pressure with respect to volume?


Solution: We can rewrite the equation asP = mRTV . Recall that all ofm, R, andT are being

treated as constants in this discussion. Thus we can finddPdV = -(mRT)V-2.

�

3.5.2.4. Economics

SupposeC(x) represents the cost of producingx units of a certain good. The quantityC(n+ 1) -C(n) would represent the cost of producing the(n + 1)st unit after already producingn units.This item is sometimes referred to as themarginal costof producing the(n + 1)st item. It canbe approximated by the derivative, sinceC(n+1)-C(n)

1 ª limhÆ0C(n+h)-C(n)

h . Thus economists oftenrefer toC¢(x) as themarginal cost function.

Example 3.5.4Suppose that the cost of producingx units of a certain good is2x+.004x2+.008x3

2000 .Find the actual cost of producing the 101st unit, and compare withC¢(100).

Solution: C(101) = 4.242606 (Use your calculator.)C(100) = 4.12 Thus the difference is.122606. NowC¢(x) = 1

2000(2+ .008x+ .024x2). SoC¢(100) = 1

2000(2+ .8+240) = 242.8200 = .1214.

�

Exercises:

EXERCISE3.5.1.Do 1, 3, 5, 11, 13, 17, 27, and 29 from Stewart, section 3.4. Also do 47 and


49 (parts a and c) from section 3.8 on page 198 of Stewart.


3.6. The Chain Rule

We need one more shortcut rule for taking derivatives. The question is: If the functionf (x) isreally the result of composing functionsm(x) andn(x), what is the derivative off in terms ofinformation aboutm andn? For example, how can we take the derivative of(2x + 3)2 withoutexpanding the expression first. This would be useful to know, because we wouldn’t want to haveto expand something like(2x + 3)100 in order to differentiate it! Going back to the example off (x) = (2x+ 3)2, if we do expand it we getf (x) = 4x2

+ 12x+ 9, so f ¢(x) = 8x+ 12. Note thatthis isnot the same thing as 2(2x+ 3)1, so that can’t be the right answer. But using the fact that aderivative isalmosta quotient, how about if we writeDm

Dx =DmDn ◊

DnDx . (Here we are pretending that

Dn ê = 0, so that this makes sense.) Now by shrinking appropriate quantities to zero, one mighthope that we would have

f ¢(x) =dmdx=

dmdn◊

dndx

.

Let’s see if this works for the casef (x) = (2x+3)2. We can writef (x) = m(n(x)) if m(x) = x2

andn(x) = 2x+ 3. Thendmdn would be the derivative of the squaring functionat (2x+ 3), so we

would get 2(2x + 3)1 for this portion of the result. Nowdndx would ben¢(x) which would be the

constant 2. Thus the total derivativef ¢(x) would be 2(2x+ 3)1 ◊ 2 which would be 8x+ 12, whichis the result desired as we noticed above.

To summarize:


The Chain RuleIf f (x) = m(n(x)), then d f

dx =dmdn ◊

dndx, or in other notation,

f ¢(x) = m¢(n(x)) ◊ n¢(x).

In the above, we refer ton(x) as theinsidefunction andm(x) as theoutsidefunction. Thus thechain rule can be stated verbally as: “The derivative of the composition of the outside functionwith the inside function is the derivative of the outside function evaluated at the inside function,times the derivative of the inside function.

Example 3.6.1Find f ¢(x) for f (x) =0

12x5 - 3x2 + 4x+ 3.

Solution: For f (x) =0

12x5 - 3x2 + 4x+ 3, the inside functionn(x) is 12x5- 3x2

+ 4x + 3,while the outside function is

0

x. Thus the expressionm¢(n(x) would be 120

12x5-3x2+4x+3, while

n¢(x) = 60x4- 6x+ 4. Thus f ¢(x) = 60x4

-6x+420

12x5-3x2+4x+3. �

Example 3.6.2Find the derivative of1

x+0

x.

Solution: Let f (x) = (x+ x12 )

12 . The inside function is given byx+ x

12 while the outer function is

given byx12 . Thus the derivative off is given by1

2 Jx+ x12N

-12◊ J1+ 1

20

xN. �


Example 3.6.3By using the chain rule twice in sucession, find a formula for the derivative of afunction of the formf (x) = r(s(t(x))). (This could be called the triple chain rule.)

Solution: Let s(t(x)) be calledn(x) for the moment. Thenf (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).

�

Exercises:

EXERCISE3.6.1.Suppose thatf (x) = (3x+ 1)3. Identify m(x) andn(x) so that f (x) = m(n(x)).Then computem¢(x), m¢(n(x)) andn¢(x). Then computef ¢(x) = m¢(n(x)) ◊ n¢(x). Note: You willusually not do this much work to computef ¢(x), but it is good to do this while first learning thechain rule.

EXERCISE3.6.2.Do the same thing as above for the functionf (x) =0

x2 + 1.

EXERCISE3.6.3.Use the result of the previous problem to compute the derivative ofg(x) =0

x2+14x-2 .


EXERCISE3.6.4.Without knowing anything aboutm(x) other than the fact that it is differen-tiable everywhere, what is the best representation ofg¢(x) if g(x) = f (m(x))?

EXERCISE3.6.5.Use the product rule to find the derivative off (x) = (2x+ 1)3 ◊ (4x2+ 4x)5. Be

sure to simplify your result.

EXERCISE3.6.6.Suppose thatf (2) = 1, f ¢(2) = 3, g(3) = 2, g¢(3) = 4, andg¢(2) = 8. Computeh¢(3) if h(x) = f (g(x)). Hint: You may not need all of the given information.

EXERCISE3.6.7.Recall that the functiony = |x| is not differentiable atx = 0, but has derivative1 for x > 0 and has derivative-1 for x < 0. This can be written succinctly as follows:

y¢ =|x|x

,

since that expression is exactly what we just described. Use this fact and the chain rule tocomputef ¢(x) for f (x) = |6x- 3|. Can you generalize to a formula for the derivative of|n(x)|?

EXERCISE3.6.8.Do problems 1, 7, 11, 17, 19, 23, 25, 43, 53, and 55 from Stewart, page 183.


3.7. Implicit Differentiation

Consider the functionf (x)2. The chain rule tells us that the derivative of this (with respect tox) is2 f (x)1 f ¢(x). Suppose we were to call the functionf (x) by the alternative namey, and writey2 forthe expressionf (x)2. Then the derivative would have to be 2yy¢ or 2ydy

dx, again by the chain rule.If you were thinking that the derivative ofy2 should be simply 2y, you were probably thinking ofthe derivative of the expressiony2 with respect to y, which would indeed be simply 2y. Thus wesee that when differentiating, we must always be aware of which quantity represents the variablewe are differentiating with respect to, and which quantity or quantities are actually functions ofthat variable.

This comes into play when considering an equation like the circlex2+ y2

= 1. Of course,we could solve this fory in terms ofx (and consider the top half and the bottom half of thecircle separately), but it may be more convenient to deal with this equation in its standard form.Now even though the circle doesn’t representy as a function ofx (since it fails the vertical linetest), it is still the case that there is a tangent line at every point on the circle and all but two ofthose points have non-vertical tangent lines. Thus, we should be able to compute the slope of thetangent linedy

dx at all points except the two with vertical tangents. When dealing with a functionwhich is not in the usualy = f (x) form with y written explicitly in terms of the variablex, wesay we have animplicit function. The process of finding the derivative of a function (withoutresorting to solving fory explicitly in terms ofx) is calledimplicit differentiation.


Implicit Functions and Implicit DifferentiationWe say thaty is animplicit functionof x if there is an equation relatingyandx which defines a functional relationship between the quantities, butwhich does not have the formy = f (x) with the independent variableyby itself on one side of the equation. We performimplicit differentiationon such a function by taking the derivative of both sides of the equationwhich defines the implicit function, keeping close track of the fact thatwe are consideringy to be a function ofx.

For example, in the case of the implicit equationx2+ y2= 1, if we differentiate both sides of

the equation with respect tox, we would obtain 2x + 2ydydx = 0, since the derivative ofx2 is 2x,

the derivative ofy2 with respect tox is 2ydydx (by the chain rule) and the derivative of the constant

1 is 0. This equation can now be solved fordydx to obtaindy

dx =-xy . Thus the slope of the tangent

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

22 ,0

22 ), the slope is-1. We can even see that

the points on the circle where the tangent line has undefined slope are the points wherey = 0,namely(1,0) and(-1,0).

Example 3.7.1Consider the curvey = 3x , but for practice with implicit differentiation, write it

asxy= 3. Find dydx by differentiating both sides with respect tox. Be sure to use the product rule


on the left hand side of the equation. When finished, check your answer by findingdydx for the

explicit functiony = 3x .

Solution:If xy= 3, then differentiating the left hand side with respect tox (recalling thaty is a function

of x) we getxdydx + y(1) from the product rule. The derivative of the right hand side is zero, since

we have a constant on that side of the equation. Thus we have

xdydx+ y = 0,

sodydx=-yx

.

As suggested, since this function is easy to write explicitly asy = 3x , we can check our answer

by differentiating this function. We getdydx =

-3x2 . But if y = -3

x , then the expressions-yx and -3

x2

are the same. �

When differentiating a reasonably complex implicit function, the result often involves severalterms, some which have factors ofdy

dx and some of which don’t, and these factors could be on

either side of the equal sign. To solve such an equation fordydx, first identify which terms have


factors ofdydx, and which ones don’t. Gather all the terms which do have ady

dx factor on one side of

the equal sign, and put all other terms on the other side. Then factor out thedydx factor, and finish

solving by dividing both sides of the equation by the appropriate factor. For example, considerthe functionx3y2

+ 4x+ 5y2- 2 = xy. Differentiating both sides with respect tox yields

x3◊ 2y

dydx+ y2◊ 3x2

+ 4+ 10ydydx- 0 = x

dydx+ y(1).

Analyzing this equation, we see that there are 4 non-zero terms on the left hand side, and 2 termson the right hand side. Of these terms, there are two terms on the left hand side which have adydx factor, and one term on the right hand side which has one. Gathering all the terms with ady

dxfactor on the left (and the others on the right) we obtain

2x3ydydx+ 10y

dydx- x

dydx= y- 3x2y2

- 4.

Factoring out thedydx factor yields

dydxI2x3y+ 10y- xM = y- 3x2y2

- 4,

sodydx=

y- 3x2y2- 4

2x3y+ 10y- x.


Example 3.7.2Suppose0

xy= x+ y+ 1. Find dydx.

Solution:Differentiating

0

xy= x+ y+ 1 with respect tox yields

12(xy)

-12Kx

dydx+ yO = 1+

dydx

.

There are two terms on the left hand side, one of which has a factor ofdydx. That is also true for

the right hand side. Gathering terms with adydx factor on the left yields

12(xy)

-12 ◊ x

dydx-

dydx= 1-

12(xy)

-12 y.

Factoring out thedydx factor and solving yields

dydx=

1- 12(xy)

-12 y

12(xy)

-12 x- 1

.

�


Exercises:

EXERCISE3.7.1.Consider the equationx23 +y

23 = 4, which defines our beloved Furman symbol.

Find the slope of the tangent line to this curve atx = 1.

EXERCISE3.7.2.Find dydx for the implicit functionx2y+ y2

+ 3 = xy.

EXERCISE3.7.3.Findy¢ (the derivative ofy with respect tox) for the function0

x+ y = x2y3.

EXERCISE3.7.4.Suppose that you had an implicit function involving the variableszandv, andwere asked to finddz

dv. If your equation had az2 term, would its derivative be 2zor 2zdzdv? Explain.

EXERCISE3.7.5.Repeat the previous question, except suppose that your were asked to finddvdz.

EXERCISE3.7.6.Suppose thatz2+ (zv)3 -

0

zv= 5. Find dzdv.

EXERCISE3.7.7.Suppose thatz2+ (zv)3 -

0

zv= 5. Find dvdz.

EXERCISE3.7.8.Do problems 7, 11, 13, 23, 25, 27 from Stewart, section 3.7.


3.8. Higher Derivatives

3.8.1. The Second Derivative

Since the derivative of a function is another function, it makes sense to think about differentiat-ing the derivative to get another function, which we will call thesecond derivativeof the originalfunction. We will write f ¢¢(x) or d2y

dx2 for the second derivative ofy = f (x) with respect tox. Be-cause of our previous dealing with the derivative, we should expect that there is both a geometricsignificance to the second derivative (related to the graph of the function) and a physical inter-pretation (related to the rate of change of the function.) We now explore the latter, leaving thediscussion of the significance of the second derivative in relation to geometry for a later chapter.

Supposey = f (t) represents the position of an object at timet. We have already seen thatf ¢(t) represents the instantaneous rate of change off (position) with respect tot (time.) Thusf ¢(t) represents velocity. Likewise,f ¢¢(t) should represent the instantaneous rate of change off ¢

(velocity) with respect tot (time.) The rate of change of velcocity with respect to time is what wegenerally callacceleration. Thus we can think off ¢¢(t) = a(t) as the acceleration of the object attime t.


3.8.2. Third and Higher Derivatives

Of course, we can differentiatef ¢¢(x) to obtain f ¢¢¢(x) and so on. Since it is impractical to writef ¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢(x) for the sixteenth derivative off with respect tox, we instead writef (16)

(x). Ofcourse, this could lead to confusion with the expression( f (x))16, but hopefully we can use thecontext to “differentiate” between the two. (Sorry, bad pun.) In Leibniz notation, we writedny

dxn

to stand for thenth derivative ofy with respect tox.

Example 3.8.1Explain why thenth derivative of a degreek polynomial is always the constantzero, ifn > k.

Solution: Suppose you have a degreek polynomial like f (x) = akxk+ º + a1x + a0. Each

derivative of this will result in a polynomial of lower degree, in fact it will have degree one lessthan what came before. So the first derivative has degreek- 1, the second has degreek- 2 andso on. Thekth derivative would thus have degreek - k = 0. But a zeroth degree polynomial isa constant function, and the derivative of a constant function is zero. Thus thek+ 1st derivative(and thus all those after that) result in the constant zero. �

Example 3.8.2Find the first few derivatives off (x) = 1x2 . See if you can find a pattern, so that

you can write down a general formula fordk fdxk . Hint: the number(1) ◊ (2) ◊ (3) ◊º◊(n) is sometimes

written asn!. This is read “n factorial.” For example, 8!= 8 ◊ 7 ◊ 6 ◊ º ◊ 1 = 40320.


Solution: 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 looksto be positive whenk is even and negative whenk is odd, so a factor of-1k would take care ofthis alternating sign. The denominator is alwaysx to a power, and the power looks to bek + 2for thekthderivative. Thus it looks like we have the following formula for thekth derivative:

dkydxk=(-1)k(k+ 1)!

xk+2.

�

Exercises:

EXERCISE3.8.1.Compute the second derivative with respect tox of y = 2x2

x2-1. You will probablywant to simplify after taking the first derivative and before taking the second derivative.

EXERCISE3.8.2.Computey¢¢ for y =0

x+0

x- 2.

EXERCISE3.8.3.Consider the functionf (x) = 1x . Take the first few derivatives off (x) and see

if you can find a nice way to represent thenth derivative.

EXERCISE3.8.4.When a ball is thrown up in the air from the surface of the earth, its positionat timet is given by the equations(t) = -16t2

+v0t, wherev0 represents the ball’s initial velocity


when released. Find a formula for the object’s velocity at timet and its acceleration at timet.Can you explain the significance of the fact that the acceleration is a constant?

EXERCISE3.8.5.Suppose that a ball is thrown upwards from the surface of the earth with aninitial velocity of 96 feet per second. When is the ball speeding up? When is it slowing down?How high does it go?

EXERCISE3.8.6.Find a formula for thenth derivative off (x) = 12(2x+1)3 .

EXERCISE3.8.7.Consider the implicit functiony2+ xy = 4. How could you findd2y

dx2 for this

function? Hint: finddydx by implicit differentiation first, and then think about how to findd2y

dx2 .

EXERCISE3.8.8.The 15th derivative of a certain polynomial is a constant. What can you sayabout the polynomial?

EXERCISE3.8.9.Do problems 9, 11, 25, 29, and 25 from section 3.8 of Stewart


3.9. Related Rates

One application of implicit differentiation and the chain rule is in the solution to problems likethe following: Suppose two cars leave an intersection at the same time, with one car headed northand the other car headed east, with the northbound car traveling at 50 mph and the eastbound cartraveling at 30 mph. How far are they separating after 30 minutes? We can solve such a problemby trying to find a relationship between the rate at which the two cars are separating and the ratesthey are going. To this end, we could letx represent the distance the northbound car has goneafter t hours, andy the distance the eastbound car has gone aftert hours. Note that these areboth functions of timet. Also, we can related the distance between the cars by the pythagoreantheorem:x2

+y2= z2, wherez is the distance between the cars at timet. Differentiating implicitly

with respect tot yields 2xdxdt + 2ydy

dt = 2zdzdt , and solving fordz

dt yields dzdt =

xdxdt +ydy

dtz . Now dx

dt = 50

and dydt = 30 and whent = .5, x = 25 andy = 15 and thus (using the pythagorean theorem)

z=0

252 + 152 =0

850. Thus at the moment in question,dzdt =

25◊50+15◊300

850.

Example 3.9.1A right circular cylinder with radius 2 feet and length 10 feet is being filled withwater at a rate of 3 cubic feet per minute. At what rate is the height of the water in the cylinderincreasing?

Solution: Draw a picture of such a cylinder, and label the heighth. The volume of the water in


the cylinder is the area of the base times the height, so it isV = 4ph. ThusdVdt = 4p dh

dt , and sincedVdt = 3, we havedh

dt = 3 ◊ 4p = 12p. This would be measured in feet per minute. �

Example 3.9.2A swimming pool which is 3 feet deep at the shallow end and 8 feet deep at thedeep end is being filled with water at the rate of 10 cubic feet per minute. The pool is 25 feetlong and 10 feet wide, and the shallow end descends along a straight line to the deep end. Howfast is the water level rising when the water level (measured in the deep end) is 3 feet?

Solution: Note that when the water ish feet deep, the volume of water is the area of a triangularcross section times the width (which is 10 feet). Letx be the length of water in the pool. Then thearea of the triangle is12xh, so the volume of water is given byV = 1

2 ◊ 10xh = 5xh. What do wedo about the fact that we don’t know anything aboutdx

dt ? Using similar triangles, we can writexh =

255 , sox = 5h. Thus we can rewrite the volume asV = 5(5h)h = 25h2. ThusdV

dt = 50hdhdt , so

3 = 50hdhdt , so dh

dt =3

50h. Whenh = 3, we havedhdt =

150. �

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


Steps for Solving Related Rates Problems

1. Draw a picture if appropriate.

2. Label things in the picture which are functions of time with a letter,and label things that are constants with the appropriate number.

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

4. Write down an equation which relates the variables which is validfor all time t.

5. Differentiate both sides of the equation with respect tot, using thechain rule.

6. After differentiating, substitute the values of the quantities that areknown at the desired time.

7. Solve for the desired rate.


Exercises:

EXERCISE3.9.1.Do problems 1 – 15 odd from section 3.9 of Stewart.


3.10. Linearization

Near the point of tangency, a given curve and its tangent line look quite similar. This can be seenusing the “zoom” feature on a graphing calculator or computer on which a curve and a tangentline to the curve have been graphed together. If we zoom in on the point of tangency, the curveand the tangent line are hard to distinguish from each other. Thus we could say that the tangentline is a good approximation to the curve near the point of tangency. In fact, it can be proventhat the tangent line is thebest linear approximation to the curve which goes through the givenpoint.


Linearization – Main IdeaNear the point of tangencyx = a, the functionf (x) and the tangent lineL(x) resemble each other. Thus we will write

L(x) ª f (x)

for values ofx neara. It can be shown that the tangent line is in factthebestlinear approximation tof (x). That is, of all the lines which gothrough the pointx = a, the one which most resembles the graph off (x)is L(x).

3.10.1. New Notation – Old Idea

Recall that the tangent line toy = f (x) at the pointx = a is given by

y- f (a) = f ¢(a)(x- a),

ory = 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 whenx is close toa,

f (x) ª f (a) + f ¢(a)(x- a).

Now suppose thata is the fixed point of tangency, and the valuex is allowed to vary slightlyfrom a. We could let the quantity(x- a) be given a name: it is sometimes calledDx or dx, sinceit represents a small difference inx values. Note that we can then think of the variablex as beingrepresented bya + dx. The quantityf (x) - f (a) = f (a + dx) - f (a) in the above discussionis sometimes calledDy. It represents the corresponding change iny for the function f (x). Thequantity f ¢(a)(x-a) = f ¢(a)dx is sometimes calleddy. This represents the corresponding changein y when looking at the tangent line rather than the curve. Thus for a given pointa nearx, dx isdetermined, and this producesDy (which is a difference betweeny values on the curve,) anddy(which is a difference betweeny values on the tangent line.) One way of re-expressing the mainidea can then be rewritten as

Dy ª dy

because this is equivalent tof (x) - f (a) ª f ¢(a)dx.


Definition of dyand Dy.Given a functionf (x), a point of tangencyx = a and a value ofdx, thequantitydy is defined to bedy= f ¢(a)dx, and the quantityDy is definedto be f (a+dx)- f (a). Dy represents the difference between twoy values,both of which are on the curve, whiledy represents the difference of twoy values, both of which are on the tangent line. The fact that

Dy ª dy

is due to the fact that the tangent line is a close approximation to thecurve, near the point of tangency.

Example 3.10.1Show how to use the linear approximation of the square root function near thepoint (36,6) to approximate

0

36.05 without the use of a calculator.

Solution: We will computeL(36.05) instead off (36.05). Since these are approximately equal, wewill have an approximation. Using the value ofa = 36, we have thatL(x) = f (36)+ f ¢(36)(x-36),so

L(x) = 6+112(x- 36).


(Here we used the fact thatf ¢(x) = 12x

-12 , so f ¢(36) = 1

20

36=

112.) To computeL(36.05), we

simply plug 36.05 into L(x), obtainingL(36.05) = 6 + 112(36.05 - 36) = 6 + 1

12 ◊5

100 Thus

L(36.05) = 6+ 51200 ª

0

35.05. �

Example 3.10.2Draw a sketch ofy = x2, and draw the tangent line atx = 1. Using a value ofdx= .05, finddx, dyandDy, and draw them at appropriate places on the sketch.


Exercises:

EXERCISE3.10.1.Let f (x) = 30

2x- 1. Find the linearizationL(x) to f (x) at x = 14. Then usethis to approximatef (14.01).

EXERCISE3.10.2.Explain how to use the ideas of this section to approximate0

399 withoutusing a calculator.

EXERCISE3.10.3.Suppose that you were interested in doing a calculation that had to do with

the function f (x) =0

x-12x2+5 nearx = 2, but you only needed to do an approximate calculation, so

you decided to instead use the functionL(x) which is the linearization off (x) atx = 2. FindL(x).

EXERCISE3.10.4.Sketch the graph off (x) = 1x , andL(x), the linearization off (x) at x = 1.

Suppose thatdx = 1.1 - 1 = .1 is given. Find the corresponding value ofdy, and the value ofDy, and draw all of these on a graph.

EXERCISE3.10.5.Repeat the last exercise, using the functionf (x) = 3- x2, with the lineariza-tion being taken atx = 1, anddx= 1.2- 1 = .2.

EXERCISE3.10.6.Explain why it is correct to say that0

1+ x ª 1+ x2, at least for values ofx

near 0.

EXERCISE3.10.7.Compute an approximation, using a linearization, to(1.01)40. Note that in


order to do this, you must decide on the appropriate functionf (x), and the point of tangencyx = a.



Chapter 4

An Interlude – TrigonometricFunctions

Contents

4.1 Definition of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . 205


4.1.1 Radian Measure. . . . . . . . . . . . . . . . . . . . . . . . . . . .208

4.2 Trigonometric Functions at Special Values . . . . . . . . . . . . . . . . . 210

4.3 Properties of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . 213

4.4 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 214

4.4.1 Graph of sin(x) and cos(x) . . . . . . . . . . . . . . . . . . . . . . .214

4.4.2 Graph of tan(x) and cot(x) . . . . . . . . . . . . . . . . . . . . . . .214

4.4.3 Graph of sec(x) and csc(x) . . . . . . . . . . . . . . . . . . . . . . .216

4.5 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . .218

4.6 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . 220

4.6.1 Two Important Trigonometric Limits. . . . . . . . . . . . . . . . . 221

4.6.2 Derivatives of other Trigonometric Functions. . . . . . . . . . . . . 223


4.1. Definition of Trigonometric Functions

You may at some point in your mathematical career have defined the trigonometric functionsusing ratios of sides of a right triangle. Under this definition, you will recall that the sine ofan angleq in a right triangle is the ratio of the side oppositeq to the hypotenuse. The othertrigonometric ratios are defined in box below.

If a is the length of the side oppositeq in a right triangle, andb is thelength of the side adjacent toq, andc is the length of the hypotenuse,then

sin(q) = ac

cos(q) = bc

tan(q) = ab

cot(q) = ba

sec(q) = cb

csc(q) = ca

Although these definitions are accurate, there is a sense in which they are lacking, because


the angleq in a right triangle can only have a measure between 0Î and 90Î. We need a definitionwhich will allow the domain of the sine function to be the set of all real numbers. Our definitionwill make use of the unit circle,x2

+ y2= 1. We first associate every real numbert with a point

on the unit circle. This is done by “wrapping” the real line around the circle so that the numberzero on the real line gets associated with the point(0,1) on the circle. The positive portion ofthe number line is wrapped counterclockwise around the circle, while the negative portion iswrapped counterclockwise. Another way of describing this association is to say that for a givent, if t > 0 we simply start at the point(0,1) and move our pencil counterclockwise around thecircle until the tip has movedt units. The point we stop at is the point associated with the numbert. If t < 0, we do the same thing except we move clockwise. Ift = 0, we simply put our pencilon (0,1) and don’t move. Using this association, we can now define cos(t) and sin(t).

Using the above association oft with a point(x(t), y(t)) on the unitcircle, we define cos(t) to be the functionx(t), and sin(t) to be the

functiony(t), that is, we define cos(t) to be thex coordinate of the pointon the unit circle obtained in the above association, and define sin(t) tobe they coordinate of the point on the unit circle obtained in the above

assocation.


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)

Example 4.1.1What point on the unit circle corresponds witht = p? What therefore is cos(p)and sin(p).

Solution:

Since it is 2p units around the unit circle, half-way around would correspond witht = p.Thus the point on the unit circle which corresponds witht = p would be(-1,0). Therefore,cos(p) = -1 and sin(p) = 0. �

Example 4.1.2What point on the unit circle correspond witht = 3p2 ? What therefore is cosI 3p2 M

and sinI 3p2 M?

Solution:

The point on the circle corresponding witht = 3p2 would be the point that is34 of the way


around the circle, since3p2 is three fourths of 2p. Thus the point in question is(0,-1), and thussinI 3p2 M = -1 and cosI 3p2 M = 0 �

4.1.1. Radian Measure

In calculus, we will use radian measure rather than degree measure. There are numerous reasonsfor this, most of them based on the fact that radians are more “natural” (even if it may not seemthat way to you at first.) A radian is defined to be the measure of an angle cut of in the circle ofradius one by an arc of length one. Thus, a 90Î angle corresponds to an angle of radian measurep

2 , since the distance one fourth of the way around the unit circle isp

2 . It is also useful to note thatan angle of measure 1Î corresponds with an angle of radian measurep180, since 90 of these wouldcorrespond to a right angle. Also, an angle of radian measure 1 would correspond to an angleof measureI 180

pM

Î, sincep2 of these would correspond to a right angle. These facts are enough to

help you convert from degrees to radians and back, when necessary.

Example 4.1.3What is the degree measure of the angleq = p6?

Solution:

Since one radian corresponds with180p

degrees,p6 radians would correspond withp6 ◊180p= 30

degrees. �


Example 4.1.4What is the radian measure of the angle 225Î?

Solution:

Since one degree corresponds withp180 radians, 225 degrees would correspond with 225◊p

180 =5p4 radians. �


4.2. Trigonometric Functions at Special Values

There are a few special angles for which you should know the values of the trigonometric func-tions, without having to resort to a table or a calculator. These are summarized in the followingtable.

q cos(q) sin(q)

0 1 0p

6

0

32

12

p

4

0

22

0

22

p

312

0

32

p

2 0 1

You should also be able to use reference angles along with these values to compute thevalues of the trigonometric functions at related angles in the second, third and fourth quadrants.For example, the point associated witht = 5p

6 is directly across the unit circle from the pointassociated withp6 . (In this case we say that we are usingp6 as a reference angle.) Thus thecoordinates of the point associated with5p

6 has the samey value and the oppositex value of the


point associated withp6 . Thus cos( 5p6 ) = - cos( p6) = -0

32 , and sin( 5p6 ) = sin( p6) =

12.

Example 4.2.1Evaluate sin( 7p6 ).

Solution:

The point on the unit circle corresponding witht = 7p6 is directly opposite the point on the

unit circle corresponding witht = p6 . This is because7p6 = p +p

6 . The point associated withp6 is

(

0

32 , 1

2), so the point directly opposite the circle from this point is(-0

32 ,- 1

2). Thus sinI p6M =-12 .�

Example 4.2.2Evaluate tan(-3p4 ).

Solution:

The point on the unit circle associated with-3p4 is directly opposite the point associated with

t = p

4 . This is because-3p4 + p =

p

4 . The point associated witht = p

4 is (0

22 ,0

22 ). Thus the

point associated with-3p4 is (-

0

22 ,-

0

22 ), and tanI-3p

4 M = 1, since it is the ratio of these twocoordinates. �

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


Solution:

If sin(x)+cos(x) = 0, then sin(x) = - cos(x). Of course, we also know that sin2(x)+cos2(x) =

1, so by 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)| =0

22 . Thus we are looking at two points on the unit circle:(-

0

22 ,0

22 )

and(0

22 ,-

0

22 ). These points are directl opposite the circle from each other, and the first one is

the reflection about they axis of the point(0

22 ,0

22 ), which corresponds withx = p4 . Thus we are

looking atx = 3p4 , and the sum of this withp,-p,2p,-2p,º. Thus we can say that the solutions

all have the formx = 3p4 + kp, k Œ Ÿ. �

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

Solution: 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 with2p3 and 4p

3 . Thus, since we arelooking for where cos(2x) is- 1

2, we must have 2x = 2p3 and 2x = 4p

3 . Thus we get thatx = p3 andx = 2p

3 . Also, if z is a solution, thenz+ p is a solution, because cos(2(z+ p)) = cos(2z+ 2p) =cos(2z). Thus the solutions have the formp3 + kp, k Œ Ÿ and 2p

3 + kp, k Œ Ÿ. �


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+ 2p) = sin(x), cos(x+ 2p) = 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 theset of all real numbers. Also, from the properties above, we know that the range of both of theseis the set of numbers between-1 and 1, and that the functions are periodic. This information,together with a few points plotted as a guide, are enough to graph the two functions. Note thatif we shift the graph of the sine function byp2 units to the left, we get the graph of the cosinefunction. This is related to the fact that sin(x- p2) = cos(x).

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, the pointsp2 , 3p

2 , and so on aren’t in the domain of tan(x). An easy way to characterizethese points is to say that these are all the points which have the formp

2 + kp, wherek is anyinteger. Thus the domain of the tangent function is{x : x Œ —, x ê = p2 + kp, k Œ Ÿ}.

Example 4.4.1What is the domain of cot(x)?

Solution:


The cotangent function is defined as cot(x) = cos(x)sin(x) . This will be undefined when sin(x) = 0,

which occurs at 0,±p,±2p,º. Every other real number is in the domain of both sin(x) andcos(x), and so is in the domain of the quotient of these two. Thus the domain is{x : x ê = kp, k ŒŸ}. �

We can get a good grasp on the graph of tan(x) by plotting a few points and doing a carefulanalysis of the limiting behavior whenx is nearp2 and the other points that aren’t in the domain.Note that whenx is a little less thanp2 , sin(x) is close to 1, while cos(x) is close to zero (but ispositive.) Thus, the ratio of these two things should be big. Thus limxÆ p2

- tan(x) = +•. A similaranalysis of points to the right of-p2 helps us believe that limxÆ -p2 + tan(x) = -•. This informationtogether with a few strategically plotted points (like tan(-p4 = -1 and tan(0) = 0 and tan( p4) = 1)should help us to graph the tangent function on the interval(-

p

2 , p2). Now using the fact that thetangent function is periodic with periodp, we can graph the whole tangent function.

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

Solution:

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 ofp. If x is a little greater than 0, then cos(x) is nearone, while sin(x) is very small (but is positive.) Thus the ratio is a large number. We can see


from this analysis that limxÆ0+ cot(x) = +•. On the other hand, ifx is a litle less than 0, thensin(x) is negative, while cos(x) is still close to 1, so is positive. Thus the ratio of the two, whilelarge in absolute value, is negative. Thus limxÆ0- cot(x) = -•. Using this information alongwith a few key plotted points (like cot( p2) = 0 and cot( p4) = 1) leads to the graph. It is also usefulto note that cot(x) is periodic with periodp.

�

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 exceptthose which make cos(x) equal to zero. Thus the domain of the secant function is the same asthe domain of the tangent function. Also, the fact that the cosine function always has valuesbetween-1 and 1 tells us that sec(x) = 1

cos(x) always has valuesless than-1 orgreater than 1.An analysis of the limiting behavior of sec(x) nearx = p2 and -p2 and a few strategically plottedpoints leads to the graph ofy = sec(x).

Example 4.4.3Finish this analysis and draw a graph ofy = sec(x). Then do a similar analysisand graphy = csc(x).

Solution:


The sketches are coming soon. �


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).

Example 4.5.1Show 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 tosin2(x)+sin2

(x)2 = sin2

(x).

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

Solution:


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. �


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 have that

f ¢(x) = limhÆ0

sin(x+ h) - sin(x)h

= limhÆ0

sin(x) cos(h) - sin(h) cos(x) - sin(x)h

= limhÆ0

sin(x) cos(h) - sin(x) - sin(h) cos(x)x

= limhÆ0

sin(x) (cos(h) - 1) - cos(x) sin(h)h

= limhÆ0

sin(x) K(cos(h) - 1)

hO - cos(x) K

sin(h)hO

= sin(x) KlimhÆ0

cos(h) - 1h

O - cos(x) KlimhÆ0

sin(h)hO .


If we denote limhÆ0cos(h)-1

h by a and limhÆ0sin(h)

h by b, then we have deduced that the deriva-tive of sin(x) is asin(x) + bcos(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 limhÆ0sin(h)

h . Consider the picture below:

Figure 4.1: This is the wrong picture. I need to refind the correct one.


Recall that the area of a sector of a circle of radiusr and angleh is A = 12hr2. Now consider

the sectorOSQ. It has area12hcos2(h). On the other hand, triangleOPQhas area12 cos(h) sin(h),

and sectorOPRhas area12h. From the picture we can tell that even whenh is small,

areaOSQ£ areaOPQ£ areaOPR

so12

hcos2(h) £12

cos(h) sin(h) £12

h.

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

cos(h) £sin(h)

h£

1cos(h)

.

Now ash Æ 0, cos(h) Æ 1, so this inequality together with the squeeze theorem tell us thatlimhÆ0

sin(h)h = 1.

Example 4.6.1Use the now established fact that limhÆ0sin(h)

h = 1, to show that limhÆ0cos(h)-1

h =

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


Solution:

limhÆ0

cos(h) - 1h

= limhÆ0

cos(h) - 1h

◊cos(h) + 1cos(h) + 1

= limhÆ0

cos2(h) - 1h ◊ (cos(h) + 1)

= limhÆ0

- sin2(h)

h ◊ (cos(h) + 1)

= limhÆ0

sin(h)h◊ lim

hÆ0

- sin(h)cos(h) + 1

= 1 ◊-0

1+ 1= 0

�

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 isenough to enable one to compute the derivatives of the other trigonometric functions, since they


are all quotients involving these two trigonometric functions. For example

ddx tan(x) = d

dxsin(x)cos(x)

=cos(x) cos(x)-sin(x)(- sin(x))

cos2(x)

=1

cos2(x)= sec2(x)

Example 4.6.2Compute ddx sec(x), d

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

example.

Solution:

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)

=-1

sin2(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)

�

Of course, it would be a good idea to practice these new trigonometric derivatives in conjuc-tion with out familar rules such as the chain rule, quotient rule, product rule, etc..

Example 4.6.3Compute the derivative with respect tox of f (x) = sinJ x2+1

2x-3N.


Solution:

If f (x) = sinJ x2+1

2x-3N, then f ¢(x) = cosJ x2+1

2x-3N ◊(2x-3)(2x)-(x2

+1)(2)(2x-3)2 .

�

Example 4.6.4Computeg¢(x) for g(x) = (0

x2 - 4) ◊ tan(0

x2 - 4).

Solution:

If g(x) = (0

x2 - 4) ◊ tan(0

x2 - 4), theng¢(x) = (0

x2 - 4) ◊ sec2(0

x2 - 4) ◊ 12(x

2- 4)

-12 ◊ 2x+

tan(0

x2 - 4) ◊ 12(x

2- 4)

-12 ◊ 2x.

�


Click to Initialize QuizAnswer the following questions about trigonometric derivatives.

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

sec2(x)cos(x)

cos(x)sin(2x)

sin2(x)+1

cos3(x)

2. What is the derivative off (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

0

cot(2x+ 1).0

- csc2(2x+ 1) 12(cot(2x+ 1)

-12 ◊

- csc2(2x+ 1) ◊ 2

12(cot(2x+ 1)

-12 ◊ 1


Click to End Quiz


Chapter 5

Applications of the Derivative

Contents

5.1 Definition of Maximum and Minimum Values . . . . . . . . . . . . . . . . 2315.2 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2425.3 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .243


5.4 The Mean Value Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . .247

5.5 An Application of the Mean Value Theorem. . . . . . . . . . . . . . . . . 249

5.6 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .251

5.7 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .256

5.8 Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .262

5.9 Limits at Infinity; Horizontal Asymptotes . . . . . . . . . . . . . . . . . . 264

5.9.1 A Simplification Technique. . . . . . . . . . . . . . . . . . . . . .266

5.10 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .268

5.11 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .276

5.12 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .280

5.13 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .285

5.14 Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .288

5.15 Antiderivatives Mini-Quiz . . . . . . . . . . . . . . . . . . . . . . . . . .290


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 might be trying to minimize the cost of producing some item, maximize the area or volumeof a geometric region, or minimize the time it takes for a certain biological reaction to occur. Inall of these, the theoretical mathematical basis is the same.

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

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

f (c) = M, and f (x) £ f (c) for all x in the domain off (x).

Likewise, we say thatf has aminimum value of m at x= c if

f (c) = m, and f (x) ≥ f (c) for all x in the domain off (x).

Example 5.1.1Using what you know about the graph ofy = x2, find the absolute minimum ofy = (x- 3)2 + 2 and the absolute maximum ofy = -3- (x+ 2)2.


Solution: The graph ofy = (x - 3)2 + 2 has the same shape as the graph ofy = x2, except thatit is shifted 3 units to the right and 2 units up. Thus, the vertex is now at(3,2), and the graphopens upward. Thus there is an absolute minimum of 2 atx = 3. On the other hand, the graphof y = -3- (x+ 2)2 is the same as the graph ofy = x2 except that it is flipped about thex axis,then shifted 2 units to the left and then 3 units down. Thus the vertex is now at(-2,-3), and theparabola opens downward. Thus the absolute maximum value is-3 atx = -2. �

Sometimes the fact that the maximum is begin considered over the entire domain is empha-sized by putting the adjectivesglobal or absolutein front of the words maximum or minimum.Thus one might say “there is a global maximum of 3 atx = 4” or “there is an absolute minimumof 2 atx = 5.” In contrast, you might sometimes want to discuss the fact that a certain functionvalue is larger than those near it, even if it isn’t the largest in the whole domain. We call suchpointsrelativeor local maximums or minimums.


We say that there is arelativeor local maximum valueM atx = c if thereexists an intervalI containingc so that

f (c) = M, and f (x) £ f (c) for all x in I.

Likewise we say that here is arelativeor local minimum valuematx = cif there exists an intervalI containingc so that

f (c) = m, and f (x) £ f (c) for all x in I.

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


� � � � ��

��

��

��

��

�

�

�

�

�

��

��

��

��

Solution:


� � � � ��

��

��

��

��

�

�

�

�

�

��

��

��

��

It looks like the pictured function has local minimums at aboutx = -3.1, x = .8 andx = 4.7.(It is hard to be too precise on such a graph.) It looks like there are local maximums atx = 0 andx = 1.6. �

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 domain to be a finite interval like(2,5), there is still no maximum orminimum value. If we restrict the domain to be a closed interval like[2,5], this example doeshave a maximum of 11 atx = 5 and a minimum of 5 atx = 2, but a function like

g(x) =

ÏÔÔÔÌÔÔÔÓ

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].

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

Solution: Note that the function never reaches a value of 11, although it gets very close. Sinceit doesn’t actually reach the value of 11, this cannot be the maximum value. It is tempting tosay something like: “the maximum value is whatever real number is the closest real numberto 11 which is less than 11”, but a little thought about the nature of real numbers leads one torealize that that statement is silly. There is no such real number, so this function does not havea maximum. Likewise, the function is always greater than 5, but asx Æ 2+, we see that we canget as close to a value of 5 as we desire. Thus there is no minimum value of this function.

The functiong(x) as similar problems. The function almost gets as high as 20, and almost getsas low as 0, but never quite makes it to these values. Thus, there is no maximum or minimum.


�

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

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

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

Of course, the question still remains: how can one find the value ofx which gives a maximumor a minimum? Here are some examples that you are already familiar with: the functionf (x) =x2 has a minimum atx = 0. Of course, the interesting thing about this function is that the tangentline at x = 0 is a horizontal line. In other words, the slope of the tangent line at this point iszero. Thus we see that extrema can occur at points where the derivative is zero. We will callsuch pointsstationary points. Another example ish(x) = |x|, which also has a minimum atx = 0.


However, this function does not have a horizontal tangent line atx = 0, in fact, there is no tangentline there at all! Recall thath(x) = |x| is not differentiable atx = 0, so we see that extrema canoccur at points where the derivative doesn’t exist. We call such pointssingular points. A thirdexample is to reconsider the functionf (x) = 2x+ 1 on the interval[2,5]. We have already seenthat the maximum value for this function occurs at the right endpoint, while the minimum occursat the left endpoint. Thus we see that extrema can occur at endpoints. What is not obvious (butis true nonetheless,) is that this is the complete list of types of points where extrema can occur.This fact is called “Fermat’s Theorem.”

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

We can apply this theorem as follows: given a continuous functionf (x) and a closed interval[a, b], we can find all singular and stationary points by studyingf ¢(x) and finding where it isequal to zero or where it is undefined. Since the maximum and minimum must occur at eithera stationary point, a singular point, or an endpoint, by comparing the function values at thesepoints, we can determine the maximum and minimum for the function.

Example 5.1.4Find the absolute maximum and absolute minimum off (x) = 4x5- 5x4

+ 3 on


the interval[-3,2].

Solution: 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.The derivative isf ¢(x) = 20x4

- 20x3= 20x3

(x- 1). Setting f ¢(x) = 0, we get stationary pointsatx = 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 atx = 2 and a minimumvalue of-1374 atx = -3. �


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

1. What are the values of the relative extrema off (x) = 2x3+ 3x2

- 120x+ 30?-5,4 455,-274 4,0

2. What is the maximum value off (x) = |x- 3| on [-1,5]?There is nomaximum value.

0 4 2

3. Does the functionf (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 butno minimum.

There is neither amaximum nor a minimum.

There is a minimum but nomaximum.

Click to End Quiz



5.2. Overview

The Mean Value Theorem is an enigma, in that it seems to be a somewhat innocuous statementof fact, but turns out to be a key element in the proof of many important theorems. It is importantbecause it relates the derivative (which is not actually a quotient but is rather the limit of a certainquotient) to a particular quotient. As a warm-up, we first investigate a related theorem due to theFrench mathematician Michel Rolle.


5.3. Rolle’s Theorem

Try to draw a functionf (x) with the following characteristics: It is defined on the interval[2,5],it is continuous and differentiable on its domain, and it satisfiesf (2) = 0 = f (5). No matterhow you draw such a function, you will find that you there must have been someplace on thedomain where there was a horizontal tangent line. In other words, there must have been a pointc between 2 and 5 wheref ¢(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 if f (a) =0 = f (b), then there is a pointc with a < c < b and f ¢(c) = 0.

The reason Rolle’s Theorem is true is quite simple. Sincef (x) is a continuous functiondefined on a closed interval, the Extreme Value Theorem says that it has a maximum and aminimum on[a, b]. If either the maximum or minimum or the minimum occur at some pointc other than the endpoints, then Fermat’s Theorem says that eitherf ¢(c) = 0 or f ¢(c) doesn’texist. But one of our hypotheses is thatf (x) is differentiable on[a, b], so f ¢(c) must exist, sof ¢(c) = 0. If neither the maximum nor the minimum occur at a point other than an endpoint, thenf (a) = 0 = f (b) is both the maximum and the minimum. But this means that our function is the


constant functionf (x) = 0, and this function has lots of values where the derivative is zero! (Infact, the derivative is zero at every point on(a, b).)

Example 5.3.1Consider the functionf (x) = x2-2x-3x+8 on the interval[-1,3]. Does Rolle’s theo-

rem apply? If not, explain why not. If so, find one or more of the promised values ofc wheref ¢(c) = 0.

Solution:

Yes, Rolle’s Theorem does apply because since this is a rational function, it is continuous anddifferentiable on its domain, which consists of all real numbers exceptx = -8. Since-8 isn’tpart 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)2

Quotient Rule

=(2x2+ 14x- 16) - (x2

- 2x- 3)(x+ 8)2

Expanding in the numerator

=x2+ 16x- 13(x+ 8)2

Combining like terms


This derivative is zero when its numerator is zero, which is whenx2+ 16x- 13 = 0, which

(by the quadratic formula is whenx = -8±0

77. Now according to the theorem, at least one ofthese must be in the interval[-1,3], and since-8 +

0

77 ª .775, we see that-8 +0

77 is thepromised value ofc. Note that-8-

0

77 isn’t in the interval[-1,3], so we disregard it. �

Example 5.3.2The next two examples will show you that the hypotheses for Rolle’s theorem arenecessary. Consider the function defined byg(x) = x

23 on the interval[-1,1]. Show that there is

no value ofc so thatg¢(c) = 0 on[-1,1]. Why does this not violate Rolle’s Theorem?

Solution:

Note thatg¢(x) = 23x

-13 =

23 30

x. This is never zero since its numerator is never zero. This does

not violate Rolle’s Theorem, becauseg(x) is not differentiable atx = 0, so it is not differentiableon (-1,1). Rolle’s Theorem only applies to functions which are continuous on the given closedinterval, and differentiable on the corresponding open interval. �

Example 5.3.3Consider the function defined by

g(x) =ÏÔÌÔÓ

0 if x = 0

-2x+ 2 if 0 < x £ 1.

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


Solution:

No, it does not apply. The function is differentiable on(0,1), but it is not continuous at 0, sois not continuous on[0,1]. Note thatg¢(x) = -2, which is never zero. �


5.4. The Mean Value Theorem

One way to think of Rolle’s Theorem is this: if a functionf (x) meets the hypotheses of thetheorem, then there must be a pointc betweena andb where the tangent line is parallel to thex axis. The Mean Value Theorem is a generalization of Rolle’s Theorem. It doesn’t require thatf (a) = 0 = f (b), but instead says that as long asf (x) is continuous on[a, b] and differentiableon (a, b), there will be a pointc betweena andb where the tangent line is parallel to the linecontaining(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 there is apointc so thata < 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 slopeof the line between(a, f(a)) and(b, f(b)), so the above says that there is a pointc where the slopeof the tangent line is 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 functiong(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), andg(a) = 0 = g(b). (You should check this yourself!) Thus,g(x)meets the hypotheses ofRolle’s theorem. So there existsc with g¢(c) = 0. Butg¢(c) = f ¢(c) - f (b)- f (a)

b-a , so if g¢(c) = 0, we

get the conclusion of the Mean Value Theorem, thatf ¢(c) = f (b)- f (a)b-a . This functiong(x) seems to

be quite mysterious, it can be derived geometrically, however, by taking the picture for the MeanValue Theorem and trying to make it look like the picture for Rolle’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 andimportant theorems. We will now see an example of one such theorem. By way of introduction,note that there are many functions which have derivative equal to 2x. Some of them arey = x2,y = x2

+ 4, y = x2- 123, and so on. The question is: do all functions whose derivative is equal

to 2x have the formy = x2+C whereC 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) andh(x) are differentiable functions, and iff ¢(x) = h¢(x), thenf (x) = h(x) +C.

We can see that the “+C” theorem holds by consideringg(x) = f (x) - h(x). Note thatg¢(x) =f ¢(x) - h¢(x) = 0. We will show thatg must be a constant function. Supposex1 is any point inthe domain ofg, andg(x1) = k. Let x2 be any other point in the domain ofg with x1 ê = x2.Consider the quantityg(x2)-g(x1)

x2-x1. By the Mean Value Theorem, there must be a numberc between

x1 andx2 with g¢(c) = g(x2)-g(x1)

x2-x1. But g¢(x) = 0 for all x, so in particularg¢(c) = 0. Thus we have

g(x2)-g(x1)

x2-x1= 0, but theng(x2) - g(x1) = 0, sog(x2) = g(x1) = k. We have shown that for any point


x2 in the domain ofg, the output is the same. Thusg(x) = f (x) - h(x) is a constant function. Sof (x) = h(x) +C for some constantC.


5.6. Monotonicity

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

We say that a function isincreasing on an interval Iif for all x1, x2 Œ Iwith x1 < x2 we havef (x1) < f (x2). We say thatf (x) is decreasing onan interval I if for all x1, x2 Œ I with x1 < x2 we havef (x1) > f (x2)

For example,f (x) = x2 is increasing on the interval[0,•) and decreasing on the interval(-•,0]. A function is said to bemonotonic on an interval Iif it is strictly increasing or strictlydecreasing onI . In attempting to understand the geometry of the graph of a particular functionf (x), it is desirable to find the intervals on which the function is monotonic. Of course, it is mostinteresting to find the largest such intervals. There is a clear and almost obvious relationshipbetween the derivative of a function and the monotonicity. This can be seen by drawing a numberof strictly increasing functions, and drawing tangent lines on any of them at any point, andthinking about what is always true about the slopes of the tangent lines. You should be ableto informally convince yourself pretty quickly that they all have positive slopes. Thus thereseems 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. Thefollowing seems plausible:

Monotonicity TheoremIf f (x) is so thatf ¢(x) > 0 for all x in the intervalI , then f (x) is increasingon I . If f ¢(x) < 0 for all x in I , then f (x) is decreasing onI .

The proof of the Monotonicity Theorem uses the Mean Value Theorem. The idea is that iff ¢(x) > 0 onI , and ifx1 < x2 on I , then there is a pointc betweenx1 andx2 with f (x2)- f (x1)

x2-x1= f ¢(c).

But if f ¢(c) > 0, we must havef (x2) - f (x1) > 0, since the denominatorx2 - x1 > 0. Sof (x1) < f (x2), so f (x) is increasing. A similar proof works forf (x) decreasing whenf ¢(x) < 0on I . This result can be used to find the intervals of monotonicity for a given function: byfinding the largest intervals on which the derivative off (x) is positive, we are also finding thelargest intervals on whichf (x) is increasing. A similar statement can be made replacing the word“increasing” by “decreasing” and the word “positive” by “negative.”

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

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


First we need to compute and simplifyf ¢(x). We have

f ¢(x) =(x2+ 2)(1) - (x)(2x)(x2 + 2)2

Quotient Rule

=2- x2

(x2 + 2)2Simplifying

Note that the numerator off ¢(x) is zero whenx = ±0

2, and the denominator is never zero.Thus, f ¢(x) is either strictly positive or strictly negative on the intervals(-•,-

0

2), (-0

2,0

2),and(

0

2,•). Using test values, we see thatf ¢(x) is positive on(-0

2,0

2) and negative on theother two intervals. Thusf (x) is increasing on(-

0

2,0

2) and decreasing on(-•,-0

2) and on(

0

2,•). �

One upshot of the Monotonicity Theorem is that we can now see how to use the first deriva-tive to help us locate relative maximum and relative minimums. This is because iff (x) is contin-uous on the interval(a, b), and is increasing on an interval(a, c) and decreasing on(c, b) wherec is betweena andb, then there must clearly be at least a relative maximum atx = c. A similarstatement can be made about relative minimums iff (x) changes from decreasing to increasingin the middle of some interval on whichf is continuous. These ideas are collectively know as


The First Derivative TestSuppose thatf (x) is continuous on(a, b) and differentiable on(a, b) ex-cept perhaps at a pointc 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 a relativeminimum value atx = c.

2. If f ¢(x) > 0 on(a, c) and f ¢(x) < 0 on(c, b), then there is a relativemaximum value atx = c.

3. If the sign of f ¢(x) is the same on both(a, c) and(c, b), then thereis no relative extremum atx = c.

Example 5.6.2Use the analysis done previously to find the relative extrema forf (x) = xx2+2.

Solution:

Use the analysis done previously to find the relative extrema forf (x) = xx2+2.


We already found that there were stationary points atx = ±0

2, and thatf (x) is increasingon (-

0

2,0

2) and decreasing on(-•,-0

2) and on(0

2,•). Using the first derivative test, weknow see that there is a local maximum atx =

0

2 and a local minimum atx = -0

2. The localmaximum value is

0

24 while the local minimum value is-

0

24 . �


5.7. Concavity

We’ve had such good luck relating the sign of the derivative of a function to information aboutthe graph of the function that one can’t help but wonder whether or not the sign of the sec-ond derivative would yield any nice information about the graph of the function. Consider thefollowing graph ofy = sin(x).

��

�

��

�

� �

�

��

Note that between-p and 0, the slopes of the tangent lines are increasing as we go from leftto right. (The lines start out having negative slopes, and then the slopes increase to zero, andthen the lines continue to get steeper.) Since the derivative ofy = sin(x) tells us the slope of the


tangent line at a given point, we can see that the derivative of the derivative ofy must be positivehere. Thus the second derivative is positive. In this situation, we say thaty is concave upon theinterval (-p,0). On the other hand, between 0 andp we see that the slopes of the tangent linesare decreasing as we move from left to right, so the derivative of the derivative must be negativehere, soy¢¢ < 0 on(0,p). Summarizing:

We say thatf (x) is concave upon an intervalI iff f ¢¢(x) > 0 on I . Wesay thatf (x) is concave downon I iff f ¢¢(x) < 0 on I . If c is a point inthe domain off (x), and if f (x) has one type of concavity on(a, c) andthe other type on(c, b), then we say that the point(c, f(c)) is apoint ofinflectionof f (x).

Example 5.7.1Continue studyingf (x) = xx2+2 by analyzing its concavity.

Solution:

Continue studyingf (x) = xx2+2 by analyzing its concavity.

Recall thatf ¢(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)3Combining Like Terms

=(2x)(x2

- 6)(x2 + 2)3

Factoring the Numerator

Now an analysis of this second derivative for places where the sign might change shows thatx = 0 andx = ±

0

6 are possibilites, since that is where the numerator (and thusf ¢¢(x)) is zero.Also note that there are no places where the denominator is zero. The potential intervals of


concavity are thus(-•,-0

6), (-0

6,0), (0,0

6), and(0

6,•). Using test values we see thatthe second derivative is negative on(-•,-

0

6) and on(0,0

6), and is positive on(-0

6,0) andon (0

6,•). Thus f (x) is concave down on(-•,-0

6) and on(0,0

6), and is concave up on

(-

0

6,0) and on(0

6,•). There are points of inflection at(0,0), (-0

6, -0

68 ), and(

0

6,0

68 ).

�

It turns out that for stationary points, we can use the second derivative to help analyzewhether a given point is a relative maximum or a relative minimum. This is because iff (x)has a relative minimum and a horizontal tangent line atx = c, then it must be concave up there.(Try drawing an example and see for yourself!) On the other hand if it has a relative maximumatx = c and a horizontal tangent line, it must be concave down there. Thus we have:


The Second Derivative TestSuppose thatf (x) is so that f ¢¢(x) exists in some interval containingc,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.

Example 5.7.2Use the analysis done previously to find the relative extrema forf (x) = xx2+2

using the second derivative test.

Solution:

Use the analysis done previously to find the relative extrema forf (x) = xx2+2.

We already found that there were stationary points atx = ±0

2, and we foundf ¢¢(x) =(2x)(x2

-6)(x2+2)3 . Since f ¢¢(-

0

2) = -20

2(-4)43 > 0 there must be a relative minimum atx = -

0

2. Since


f ¢¢(0

2) = 20

2(-4)43 < 0, there must be a relative maximum atx =

0

2. Note that this coincideswith what we obtained using the first derivative test. �


5.8. Vertical Asymptotes

We have already noted in this course that sometimes the limit of a function asx Æ c doesn’texist because whenx is close to but not equal toc, the function gets very large. In this case wewrite limxÆc f (x) = •. Geometrically, this means that there is a vertical asymptote atx = c. Forexample, the functionf (x) = 1

x2 has a vertical asymptote atx = 0, since limxÆ01x2 = •. For the

functions we have studied, vertical asymptotes tend to occur when we have a rational expressionof the form f (x) = n(x)

d(x) , and there is a valuec for which d(c) = 0, butn(c) ê = 0. Then asx Æ c,the ratio f (x) would increase without bound. Of course, it could be the case that bothn(c) andd(c) are zero, but the ratio could increase without bound, as the example below shows.

Example 5.8.1Does f (x) = sin(x)x2 have a vertical asymptote atx = 0 or not?

Solution:

We must compute limxÆ0sin(x)

x2 . Let’s first focus on the limit asxÆ 0+. Recall that limxÆ0sin(x)

x =

1, so it would be useful to writesin(x)x2 as

sin(x)xx . Then asxÆ 0+, the numerator is approaching 1,

while the denominator is approaching zero (but is positive), thus the ratio is increasing without

bound. Thus limxÆ0+sin(x)

xx = •. This is enough to show that there is a vertical asymptote atx = 0.

A careful analysis of the limit asxÆ 0- shows that that limit is-•. �


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) = xcot(x), x Œ [-4,4].

There aren’t any. x = ±p x = ±p, 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 horizontalasymptotes. A horizontal asymptote occurs when a curve levels off at a certainy value asx getsvery large in either the postive or negative direction. Thus, it seems natural to define limxÆ• f (x).We need a technical definition akin to the epsilon/delta definition of ordinary limits. We want tosay that whenx is really big, the functionf (x) should be close to the constant functiong(x) = L.Thus, we want to say that for anye > 0, if x is large enough, we should have| f (x) - L| < e.Formally we have:

Technical Definition of limxÆ• f (x) = L.We say that limxÆ• f (x) = L if it is true that for anye > 0 there existsMso that ifx > M, then| f (x) - L| < e.

Example 5.9.1Use the above definition to show that limxÆ•1x = 0. What does this tell us about

horizontal asymptotes?

Solution:


We are trying to show that limxÆ• 1x = 0. We must show that for anye > 0, there is a value

of M so that ifx > M, | f (x) - 0| < e. Working backwards from this last expression, we see thatwe need1

x < e, or in other words, we needx > 1e. So let’s takeM = 1

e. Then (starting over) if

x > M, we havex > 1e

so 1x < e. Also sincex is some positive number, we can write

ƒƒƒƒ

1xƒƒƒƒ

< e orƒƒƒƒ

1x - 0

ƒƒƒƒ

< e. Thus we have shown what we set out to show.

Since limxÆ•1x = 0, the liney = 0 must be a horizontal asymptote.

�

The definition above doesn’t really address the question of “how does one compute limits asx Æ •?” Fortunately, our limit theorems are still applicable, and this with the result of the pre-vious example allow us to draw some nice deductions. Also, we will introduce a simplificationtechnique that will proof to be advantageous.

Example 5.9.2Use the limit theorems and the above result to show that limxÆ•1xr = 0, for r a

rational number,r ≥ 0.

Solution: Note that 1xr = I

1x M

r. Thus limxÆ•

1xr = IlimxÆ•

1x M

r= 0r

= 0. �


5.9.1. A Simplification Technique

When trying to compute limxÆ• f (x) where f (x) is a rational expression liken(x)d(x) , it is oftenuseful to simplify the rational expression by dividing both the numerator and denominator bythe same quantity. This is especially useful if you can find a quantityq(x) so that you know thelimit of either n(x)

q(x) or d(x)q(x) . For example, consider limxÆ• x2

3x2+2x+1. As the reader will shortly see,

we know the limits of bothx2

x2 (of course) and3x2+2x+1x2 , so it would be prudent to divide both the


numerator and denominator byx2. Thus we have

limxÆ•

x2

3x2 + 2x+ 1= lim

xÆ•

x2

3x2 + 2x+ 1◊

1x2

1x2

divide by one in a clever way

= limxÆ•

x2

x2

3x2+2x+1x2

Rewriting

= limxÆ•

13x2

x2 +2xx2 +

1x2

Simplifying

= limxÆ•

1

3+ 2x +

1x2

Simplifying

=1

3+ 0+ 0Taking the Limit

=13

Arithmetic!


5.10. Graphing

We have now devised techniques which, when put together, enable us to graph almost any func-tion in our catalogue. To do this, we carefully analyze our function for domain, asymptotes,monotonicity, concavity and extrema. We then plot any “important” points such as inflectionpoints, relative maximum and minimum points, and indicate any asymptotes on our graph, andthen draw a sketch, connecting the points we have plotted and being sure to respect the informa-tion 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 care-fully noted. Any isolated points which are not in the domain should appear on anysign chart created while analyzing this function.

2. Look for vertical and horizontal asymptotes. Recall that a vertical asymptoteoccurs atx = c if lim xÆc± f (x) = ±•. Such things often (but not always) occurin rational expressions where the denominator is zero. Horizontal asymptotes occurwhen limxÆ±• f (x) = L, in which case there is a horizontal asymptote aty = L.

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

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


Steps for Graphing f (x), continued.

5. Investigate the Concavity of f (x). This will involve a study of the sign off ¢¢(x). Besure to make a sign chart. From the sign chart you should be able to find points ofinflection. It would also be prudent to double check your conclusions about relativeextrema by applying the second derivative test to any points wheref ¢(x) = 0.

6. Preparation for Graphing.

(a) Plot a few important points. Plot all extrema and inflection points. (Be sureto use f (x) and not f ¢(x) or f ¢¢(x) to determine they values of points you areplotting.)

(b) Intercepts. Plot the point(0, f (0)), since this is usually easy to compute, andrepresents they-intercept. If easy to do so, setf (x) = 0 and solve, and then plotany of thesex-intercepts that you find.


Steps for Graphing f (x), continued.

6. Preparation for Graphing. (Continued.)

(c) Asymptotes. Using a dashed line, indicate the asymptotes on the graph, andfor vertical asymptotes, think about whether the function is approaching+• or-• as thex value is approached from either the left or the right.

(d) Symmetry. Think for a moment about symmetry. Iff (-x) = f (x), then thefunction is symmetric about thex axis. If f (-x) = - f (x), then the function issymmetric about the origin. In most cases, of course, the function will haveneither of these two kinds of symmetry.

7. Sketch the Graph. Using all the information gathered so far, draw a sketch of thefunction. Be sure to respect the domain, all the intervals of monotonicty and con-cavity, the extrema and so forth. If it does not seem possible to draw a sketch whichrespects all the information you have, there is likely a mistake in some calculation.Check over your work to try and find the mistake before proceeding.

Example 5.10.1Sketch a graph off (x) = x2-2

x3 , following the above guidelines.


Solution:

Sketch a graph off (x) = x2-2

x3 , 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 realnumbers except those which make the denominator 0. Thus any value for whichx3

= 0 isnot in the domain, but only 0 meets this requirement. Thus the domain consists of all realnumbers except 0. We will be sure to include 0 on any sign chart we make.

2. Look for vertical and horizontal asymptotes. We need to consider the limits as we ap-proach 0 from each side. Consider limxÆ0+

x2-2

x3 . Since the numerator is approaching anumber other than 0, and the denominator is approaching zero, the ratio is getting “big”,but is a negative number since the numerator is near-2 and the denominator is postive.Thus limxÆ0+

x2-2

x3 = -•. On the other hand, limxÆ0-x2-2

x3 = •. We have a vertical asymp-

tote atx = 0. To look for horizontal asymptotes, consider limxÆ•x2-2

x3 . A quick checkshows that this limit is 0. (Divide the numerator and denominator byx3 to see this.) Thusy = 0 (thex-axis) is a horizontal asymptote.

3. Investigate the Monotonicity of f (x). A little work (do it!) shows thatf ¢(x) = 6-x2

x4 . This


is zero atx = ±0

6. By making the sign chart, we see thatf (x) is decreasing on(-•,-0

6)and on(

0

6,•). It is increasing on(0

6,0) and on(0,0

6).

4. Locate Relative Extrema of f (x). Using the sign charts, we see that there is a localmaximum atx =

0

6 and a local minimum atx = -0

6.

5. Investigate the Concavity of f (x). A little work (do it!) shows thatf ¢¢(x) = 2x2-24

x5 . This

is zero atx = ±0

12. By making the sign chart forf ¢¢(x), we see thatf (x) is concave downon (-•,-

0

12) and on(0,0

12). It is concave up on(-0

12,0) and on(0

12,•). Thereare inflection points atx = ±

0

12. Also, applying the second derivative test tox = ±0

6confirms our conclusions that there a local minimum atx = -

0

6 and a local maximum atx =0

6.

6. Preparation for Graphing.

(a) Plot a few important points. We will want to plot all the points(-0

12,- 10120

12),

(

0

12, 10120

12), (-0

6,- 460

6), (0

6, 460

6).

(b) Intercepts. There is noy intercept sincex = 0 is not in the domain off (x). A quickcheck forx intercepts shows thatx = ±

0

2 are intercepts.


(c) Asymptotes.Since the asmptotes are the axes, these are already drawn. Recall thatf (x) is approaching+• asxÆ 0- and f (x) is approaching-• asxÆ 0+.

(d) Symmetry. Sincef (-x) = (-x)2-2(-x)3 = -

x2-2

x3 = - f (x), this function is symmetric aboutthe origin. 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


�


5.11. Optimization

Suppose that you wanted to make a rectangular aquarium out of 4 pieces of glass, and you wantedit to hold exactly 32 cubic feet of water. Also, the aquarium is going to have two square sides (notthe bottom), and the aquarium doesn’t need a top. What dimensions should you make it, in orderto minimize the amount of glass needed? This kind of question is aconstrained optimizationproblem:a problem in which your goal is to maximize or minimize some quantity subject to oneor more given constraints. In this problem, we are constrained by the fact that our aquarium musthave a certain volume. Ourobjectiveis to minimize the amount of glass used. In such problems,we need to give variable names to the quantities involved, and then come up with an equationwhich represents each constraint, and one which represents the objective to be maximized orminimized. Then typically, we use the constraint equation to reduce the number of variables inthe objective equation, and then use the techniques of this chapter to maximize or minimize theobjective. For example, if we let the base of our aquarium have dimensionsx¥y, then we shouldlet the third dimension also bex, so that there is anx ¥ x square on two of the sides. Note thatx2y = 32, since the volume isx2y. The surface area represents the amount of glass we require,and the bottom has dimensionsx¥ y, and two of the sides have dimensionsx¥ x and two havedimensionsy¥ x. A rough sketch of the aquarium will help in making this determination. Thus,our objective is to minimizeS= xy+ 2x2

+ 2xy= 2x2+ 3xy. This function has two variables in

it, but our constraint tells us thaty = 32x2 , so we can rewriteSasS= 2x2

+3x( 32x2 ) = 2x2

+96x . This


function is continuous forx > 0, and we shall shortly see that it only has one critical numberfor x > 0, and that it is strictly decreasing from 0 to this critical number, and strictly increasingfrom this critical number on. Thus, this only critical number will yield a minimum, as desired.Calculatingf ¢(x), we have thatf ¢(x) = 4x- 96

x2 , which is zero only whenx3= 24, that is, when

x = 2 30

3. Note thatf ¢(x) < 0 on (0,2 30

3) and f ¢(x) > 0 on (2 30

3,•). Thus, there is a absoluteminimum atx = 2 3

0

3. At this value ofx, we have thaty = 32x2 =

322 303

. Thus, the dimensions of

the aquarium should be 230

3¥ 2 30

3¥ 16303

.


Steps for Solving Optimization Problems

1. Read the problem carefully. Draw a diagram if appropiate. As you read, thinkabout the constraints and the objective. You may need to read the problem thorughmore than once.

2. Write down a constraint equation. In doing so, be sure that you have carefullyidentified whatever variables you are using. There might be more than one constraintequation for a given problem.

3. Write down an objective function. This is the function to be maximized or mini-mized.

4. Simplify the objective function, using the constraint equation.Do this especiallyif your objective function has more than one variable.

5. Maximize or Minimize the objective function. This involves finding critical num-bers of the objective function. Be sure to pay attention to the natural domain of theobjective function.

6. Be sure that your solution is a maximum or a minimum, as desired.This willlikely involve the first derivative test, or similar analysis of where the function ismonotonic.

7. Be sure to answer the question that is asked.Sometimes the critical numbersare desired, other times the value of the objective function at these values will berequired.


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

Solution:

Let the outside of the entire pen structure have dimensionsx ¥ y. Then the objective is tomaximize the areaA = xy, subject to the constraint that there is only 200 feet of fence available.A diagram shows that 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 = 200 is the constraint equation. Solving this foryyields y = 100- 2x, and substituting this into the area equation gives an objective functionA = x(100- 2x) = 100x - 2x2. It is quite easy to maximize this downward opening parabola— the vertex is at thex value whereA¢(x) = 0, so we needA¢(x) = 100- 4x to be zero, whichoccurs whenx = 25. Note that while the natural domain for this parabola is the set of all realnumbers, it is only true thatx ≥ 0 andy ≥ 0 if x is between 0 and 50. One can easily checkthat this is a maximum, in fact, a global maximum, since the area functionA(x) is increasing on[0,25] and decreasing on[25,50]. Of course whenx = 25, the constraint equation tells us thaty = 100- 2(25) = 50. Thus the outer dimensions of the pen should be 25¥ 50. �


5.12. Newton’s Method

Newton’s Method is a method for finding roots of an equation. We study it now simply because itis another interesting and useful application of the derivative. The necessary information to beginthe method consists of the function you are trying to find the root of, and an initial “guess”, orapproximation to the root. The end result of the method (oralgorithm) is a sequence of numberswhich in many cases will be converging to a root of the equation.

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

1. Call the initial estimatex0.

2. Consider the tangent line toy = f (x) at the point(x0, f (x0)). As long as this tangent lineisn’t parallel to thex-axis, it should intersect thex-axis at some point. Call this point ofintersection(x1,0).

3. Consider the tangent line toy = f (x) at the point(x1, f (x1)). As long as this tangent lineisn’t parallel to thex-axis, it should intersect thex-axis at some point. Call this point(x2,0).

4. Repeat this process. If the sequencex0, x1, x2,º converges, it likely is converging to aroot r.


An animation demonstrating this idea and the following derivation of the iterative formula isat the following link:

Click Here to Visit The Newton Animation

To generate a formula for this process, the key fact is that the slope of the tangent line atx0

is given by f ¢(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 forx1, to find out what it is in terms ofinformation aboutx0. 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).

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


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

f ¢(xn)(xn) - f (xn)

f ¢(xn)= (xn+1)

For example, suppose you wanted to find a root off (x) = x2- 4, and decided to start with

an initial estimate ofx = 3. (Of course, we already know that the roots of this are±2, but thisexample is just to illustrate the method.) Sincef ¢(x) = 2x, we have thatf ¢(xn) = 2xn. Thus wehave

xn+1 =f ¢(xn)(xn) - f (xn)

f ¢(xn)=(2xn)(xn) - (x2

n - 4)2xn

.


If we simplify this, we see thatxn+1 =x2

n+42xn

. Starting withx0 = 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 thexi ’s are converging to 2, as expected.

Example 5.12.1Use Newton’s Method to find a root forf (x) = x3+ x- 4.

Solution: 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 ofx0 = 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. �


5.13. Antiderivatives

We have spent much effort learning how to carry out the following problem: giveny = f (x), findy¢ = f ¢(x), and interpret its significance. In this section, we consider the inverse problem: giveny = f (x), find a functionF(x) so thatF ¢(x) = f (x). Such a function is called anantiderivativeof f(x). For example, an antiderivative ofy = 2x is F(x) = x2, since the derivative ofF(x) = x2

is F ¢(x) = 2x. An antiderivative ofy = cos2(x) sin(x) is F(x) = -13 cos3(x), since thenF ¢(x) =

-13 ◊ 3 cos2(x) ◊ - sin(x) = cos2(x) sin(x). Of course, this last example may leave you scratching

your headº , you can see that it is correct, but how does one figure it out based on the functionyou were given. That will be discussed at length in the sections to come. For now, we will focuson some basic facts about antiderivatives, and will think a little about what they might be goodfor.

It is worth noting at the outset that if one knows a single antiderivative of a function, he in factknows all of them, by the “+C” theorem. Thus, sinceF(x) = x2 represents a single antiderivativeof y = 2x, the collectionx2

+C represents all possible antiderivatives ofy = 2x. This is oftencalledthe general antiderivativeof y = 2x.

Example 5.13.1Find the general antiderivative ofy = 2x+ 8.

Solution:


A general antiderivative ofy = 2x + 8 should involve the sum of the antiderivatives of 2xand 8. We have already seen that an antiderivative of 2x is x2, and a little thought shows that anantiderivative of a constant like “8” is the linear function 8x. Thus one specific antiderivative ofy = 2x+ 8 would beF(x) = x2

+ 8x. By the “+C” theorem, the general antiderivative would bex2+ 8x+C. �

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

The Sum Rule and Constant Multiplier Rule forAntiderivatives

If F(x) is an antiderivative off (x), andG(x) is an antiderivative ofg(x),thenF(x) +G(x) +C is the general antiderivative off (x) +g(x). Also, theif k is a constant, the general antiderivative ofk f(x) is kF(x) +C.

Example 5.13.2What is the general antiderivative ofy = 5x3+ 8x+ 10?

Solution:


By the sum and constant multiplier rules, we can simply try to find antiderivatives forx3

andx, and then string these together with the proper constant factors and plus signs. A littlethought reveals that an antiderivative ofx3 should be a 4th degree monomial, and some trial anderror yields the correct result ofx

4

4 . Similarly, an antiderivative ofx is x2

2 . Thus the general

antiderivative ofy = 5x3+ 8x+ 10 is 5◊ x4

4 + 8 ◊ x2

2 + 10x+C = 54x4+ 4x2

+ 10x+C. �

The previous example suggests a rule.

Power Rule for AntiderivativesIf f (x) = xn wheren is an a rational number other than-1, then thegeneral antiderivative off (x) is F(x) = xn+1

n+1 .

The above is easily checked by differentiating. Note that the exclusion of-1 is necessary, asotherwise the formula is undefined.


5.14. Rectilinear Motion

Suppose that a particle is moving in a straight line, and the functiona(t) which describes theobject’s acceleration at timet is given. This is a reasonable scenario, since the physics formulaF = ma tells us that the acceleration is essentially given by the forces acting on the particle.Thus, if one were to understand all of the forces acting on the particle, he would knowa(t). Thequestion, is: givena(t), can we recoverv(t) ands(t), the objects velocity at timet and position attime t?

The answer turns out to be yes, as long as we have a few more pieces of information. Forexample, suppose thata(t) = 2t+1. Since acceleration is the derivative of velocity, velocity mustbe the antiderivative of acceleration. Thusv(t) = t2

+ t +C for some constantC. If we know thevelocity at timet = 0 (let’s say that it was 5 feet per second) then we could computeC. This isbecausev(0) = 5 = 02

+ 0 +C, soC = 5. The fact thatv(0) = 5 is called aninitial condition.If we also had an initial condition for position, we could compute it as well. For example, ifs(0) = 2, then sinces(t) is an antiderivative ofv(t), we would haves(t) = t3

3 +t2

2 + 5t + C1,whereC1 is some constant which is potentially different fromC. The initial condition tells usthats(0) = 2 = 03

3 +02

2 + 5(0) +C1, soC1 = 2.

Example 5.14.1The acceleration due to gravity near the surface of the earth is approximately-32 feet per second per second. (We sometimes say “‘feet per second squared.”) If an object


is thrown upward from a height of 3 feet with an initial velocity of 12 feet per second, find aformula fors(t), the position of the object at timet. How high does the ball go?

Solution:

We are givena(t) = -32, sov(t) = -32t +C1. since the initial velocity is 12, we know thatv(0) = 12. Thusv(0) = 12= -32(0)+C1, soC1 = 12. We then have thats(t) = -32t2

2 +12t+C2 =

-16t2+12t +C2. Also,s(0) = 3 = -16(0)2+12(0) +C2, soC2 = 3. Thuss(t) = -16t2

+12t +3.The ball is at its maximum height when its velocity is zero. Thus, the maximum occurs when-32t + 12= 0, or whent = 12

32 =38. The maximum is-16 9

64 + 1238 + 3 = 21

4 . �

Example 5.14.2Finds(t) if a(t) = cost + sint, s(0) = 4, v(0) = 1.

Solution:

We havev(t) = sint + - cost + C1. Also, v(0) = 1 = sin(0) + cos(0) + C1 = 1 + C1, soC1 = 0. Thusv(t) = sint + - cost. Therefore,s(t) = - cost + - sint +C2. Also, s(0) = 4 =- cos(0) + - sin(0) +C2 = -1+C2, ThusC2 = 5. We therefore haves(t) = - cost + - sint + 5

�


5.15. Antiderivatives Mini-Quiz

Click to Initialize QuizAnswers the following questions about antiderivatives.

1. For the initial value problem “Finds(t), given a(t), wherev(0) = m and s(0) = n,” thefollowing is always true:

The constant of integrationarem andn.

We can usemandn to findthe constants ofintegration.

mandn have nothing to dowith the constants ofintegration.

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

Every function has anantiderivative.

The antiderivative of thezero function is a constant.

The derivative of apolynomial has lowerdegree than the originalfunction.

3. If m(x) = k1g(x)-k2 f (x), and ifG(x) is an antiderivative ofg(x), andF(x) is an antiderivativeof f (x), then thegeneral antiderivativeof 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), andF ¢(x) = f (x) andG¢(x) = g(x), then an antiderivative ofm(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


Chapter 6

Areas and Integrals

Contents

6.1 Summation Notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2956.2 Computing Some Sums. . . . . . . . . . . . . . . . . . . . . . . . . . . .2976.3 Areas of Planar Regions. . . . . . . . . . . . . . . . . . . . . . . . . . . .304


6.3.1 Approximating with Rectangles. . . . . . . . . . . . . . . . . . . . 305

6.4 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .309

6.4.1 Definition and Notation. . . . . . . . . . . . . . . . . . . . . . . . .309

6.4.2 Properties ofŸb

af (x)dx. . . . . . . . . . . . . . . . . . . . . . . . .313

6.4.3 Relation of the Definite Integral to Area. . . . . . . . . . . . . . . . 314

6.4.4 Another Interpretation of the Definite Integral. . . . . . . . . . . . . 315

6.5 The Fundamental Theorem . . . . . . . . . . . . . . . . . . . . . . . . .317

6.5.1 The Fundamental Theorem. . . . . . . . . . . . . . . . . . . . . . .317

6.5.2 The Fundamental Theorem – Part II. . . . . . . . . . . . . . . . . . 319

6.6 The Indefinite Integral and Substitution . . . . . . . . . . . . . . . . . . . 322

6.6.1 The Indefinite Integral. . . . . . . . . . . . . . . . . . . . . . . . .322

6.6.2 Substitution in Indefinite Integrals. . . . . . . . . . . . . . . . . . . 326

6.6.3 Substitution in Definite Integrals. . . . . . . . . . . . . . . . . . . . 329

6.7 Areas Between Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . .331

6.7.1 Vertically Oriented Areas. . . . . . . . . . . . . . . . . . . . . . . .331

6.7.2 Horizontally Oriented Areas. . . . . . . . . . . . . . . . . . . . . .334


6.1. Summation Notation

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

1+ 2+ 3+ 4+º + 100

which has as itsith term the numberi. We will discuss a convenient way to write this sum, andalso a way to compute this sum.

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

25

‚

i=1

i2

we are talking about the sum whoseith term isi2, and which starts with 1= 12 and ends with625= 252. Thus

25

‚

i=1

i2 = 1+ 4+ 9+º + 625.


In general, we havem

‚

i=n

ai = an + an+1 + an+2 +º + am.

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

Solution:

We recognize the numbers 4, 9, 16 and 25 as the squares of the integers 2, 3, 4, and 5repsectively. Thus, we suspect that the terms involve the square ofi. We must start withi = 2and end withi = 10 to get the correct sum. We have

4+ 9+ 16+º + 100=10

‚

i=2

i2.

�

Example 6.1.2Write the following sum out in the forma1 + a2 +º + an: ⁄8i=3 i2 + i + 1.

Solution:

⁄

8i=3 i2+ i+1 = (32

+3+1)+(42+4+1)+º+(82

+8+1) = 13+21+31+43+57+73. �


6.2. Computing Some Sums

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

1 2 3 º 50100 99 98 º 51

that the columns added to 101. Since there are 50 columns, the answers must be(101)(50) =5050. This amazingly insightful (for a child) concept can be generalized to compute the sum ofthe firstn integers as follows. Assume thatn is odd. Then write the firstn- 1 terms as follows:

1 2 3 ºn-1

2n- 1 n- 2 n- 3 º

n+12


and note that the columns add to ben, and there aren-12 of them. Thus the sum of the firstn- 1

terms is( n-12 )(n). Now if we add the last termn to this sum, we obtain

K

n- 12O (n) + n = (n) K

n- 12+ 1O = (n) K

n+ 12O .

Example 6.2.1Explain why in the previous example, the 2nd row ends withn+12 . Then show

that the formula⁄ni=1 i = n(n+1)

2 holds for alln by showing it is true in the case wheren is even.(This case is more like Gauss’ calculation than the previous example is.)

Solution:

First of all, the reason the 2nd row ends withn+12 is that that is the number which is one more

than what the first row ends with. Specifically,

n- 12+ 1 =

n- 12+

22=

n+ 12

.

To see that the formula⁄ni=1 i = n(n+1)

2 holds whenn is even, we write


1 2 3 ºn2

n n- 1 n- 2 ºn+2

2

and note that the columns add ton + 1, and that there aren2 such columns. Thus, the sum isn(n+1)

2 . �

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

ni=1 i2 and⁄n

i=1 i3, althoughthese are not nearly as easy to prove. We will need them anyway, so we will list them here,together with the prevoius result about⁄n

i=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 = Kn(n+ 1)

2O

2

(6.3)


Suppose that you wanted to compute something like⁄50i=1 2i2 + 4i + 8. We can use the above

formulas together with some basic facts about sums to compute this. The facts we need aresimply that sums are commutative, that we can factor constants out of sums, and that the sumof n terms of a constantk is just the product ofn with k. For completeness, we write this ideasusing 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 termsõúúúúúúúúúúúúúúúúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúúúúúúúúúúúúúúúúû

k+ k+ k+º + k = nk (6.6)


Thus we can compute⁄50i=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 By6.6

= 2K(50)(51)(101)

6O + 4K

(50)(51)2

O + 400 By6.1and6.2

= 91350 By arithmetic.

Example 6.2.2Compute⁄ni=1 4i3 + 6i - 1. Your answer will be a function ofn.

Solution:


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

= 4K(n)(n+ 1)(2n+ 1)

6O + 6K

(n)(n+ 1)2

O - n By 6.1and6.2

�

Example 6.2.3Suppose thatf (x) = x2+ 1 andxi = 2 + 2i

n is a sequence. Compute both: a)n

‚

i=1

f (xi) ◊2n

,and b) The limit asnÆ • of the answer to a).

Solution: Since f (x) = x2+ 1 andxi = 2 + 2i

n , we have thatf (xi) = I2+ 2in M

2+ 1. If we

expand this expression we get thatf (xi) = 4+ 4in +

4in +

4i2

n2 + 1 which is the same expression as


5+ 8in +

4i2

n2 . We are supposed to sum the product of this expression with2n, so we must compute

n

‚

i=1

10n+

16in2+

8i2

n3. We see that

n

‚

i=1

10n+

16in2+

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 asnÆ •. We have

limnÆ•K10+

8(n+ 1)n

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

3n2 O =

10+ 8 limnÆ•

n+ 1n+ 4 lim

nÆ•

2n2+ 3n+ 13n2

= 10+ 8(1) + 4K23O

=623

�


6.3. Areas of Planar Regions

In the course of your human experience, you have encountered and learned some basic factsabout areas of planar regions. You know, for example, that the area of a rectangle is the lengthtimes the width, and if you think for a moment about this fact, it makes perfect sense. Forexample, in a 2¥3 rectangle, you can actually see the 6 square units that make up the region. Youalso know that the area of triangle is1

2 of the base times the height, which again makes sense ifyou draw a rectangle with the same base and the same height as the triangle, and work to convinceyourself that there is just as much area outside the triangle but inside the rectangle is there isinside the triangle, and thus the area of the triangle is one half the area of the correspondingrectangle. You also “know” that the area of a circle ispr2 wherer is the radius, but it is healthyto 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 trueconfessions are in order, you probably will have to admit that you don’t really know anythingabout the area of anything with a curved side. In this section, we will investigate such area, anduse calculus to see how to compute them.

We will begin by considering area of regions which are curved on only one “side.” Supposef (x) is a function withf (x) > 0 on the interval(a, b). Let R be the region undery = f (x), abovethe x axis, and between the linesx = a andx = b. This region has 3 straight “sides” and onecurved “side”. We will attempt to find the area of such a region.


6.3.1. Approximating with Rectangles

In order to have a specifice regionR to work with, let’s let f (x) = x2+ 1 and leta = 2 and

b = 4. The first good idea we will introduce toward finding the area ofR is to approximate thearea with something we know the area of. We will use rectangles, since (as mentioned before)we understand the areas of those. Let’s use 4 rectangles to start. If we divide the interval froma to b up into 4 subintervals, the endpoints would be at 2,2.5,3,3.5 and 4. If we were to drawfour rectangles whose bases are on these four subintervals, and whose height is determined bythe function value using the endpoint on the right-hand side of the subinterval, we would get adecent approximation to the area ofR. The good news is that we can compute the sum of theareas of these four rectangles, and it would be approximately the area ofR. The bad news is thatit wouldn’t be exactly right. To compute the area we would need to compute the widths of therectangles, which is12 for each, and the height of each which aref (2.5), f (3), f (3.5), and f (4)respectively. These values can be computed to be29

4 ,10, 534 , and 17 respectively. If we multiply

each of these by the width12 and add we arrive at954 . This is a very rough approximation, but

at least it’s a start. Now, what if we were to increase the number of rectangles to eight? Aquick sketch reveals what looks like a better approximation. There is a nice demo of this athttp://math.furman.edu/ dcs/java/quad.html.

One can’t help but extrapolate along these lines: if eight rectangles yields a better approxi-mation than four, and 16 yields a better approximation than eight, why not use infinitely many

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


rectangles? Of course, this statement doesn’t really make sense as stated, but we can make itprecise by requiringn rectangles, and then taking the limit asnÆ •.

Toward this end, we want to divide the the interval froma to b up inton distinct subintervals.It will be useful to make them all the same width, and to call this widthDx. A little thoughtshows thatDx = b-a

n . It will also be useful to leta = x0 andb = xn, and the other endpoints ofthe subintervals be calledxi . In this case we would havexi = a+ iDx. This is summarized as

Dx = b-an is the width of each subinterval. The collectionxi = a + iDx

are the endpoints of the subintervals.

Now each rectangle has widthDx, and heightf (xi) for some endpointxi . To encompassall of the n rectangles and using the right-hand side of each subinterval, we would have theapproximate area given by

f (x1)Dx+ f (x2)Dx+ f (x3)Dx+º + f (xn)Dx

which is better written asn

‚

i=1

f (xi)Dx =n

‚

i=1

f Ka+ ib- a

nO K

b- anO .


Note that in this sum thef (x - i) factor of each term represents the height of theith rectangle,while Dx represents the width of theith rectangle. In our current example, we would have

n

‚

i=1

f K2+ i2nO K

2nO =

n

‚

i=1

KK2+ i2nO

2

+ 1O K2nO .

This sum is exactly the one we computed in the previous section. If we letAn be this approxi-mation usingn rectangles, you will recall that we computedAn = 10+ 8(n+1)

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

3n2 .

Finally, to compute the area, we need to find limnÆ• An, which we previously saw was623 .

Thus, the area undery = x2+ 1, above thex axis and betweena = 2 andb = 4 is 62

3 .

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


Computing Areas of Simple RegionsThese are the steps to compute the area undery = f (x), above thex axisand betweenx = a andx = b when f (x) > 0 on(a, b). This is sometimescalled theRight-Hand Rule.

1. Use rectangles of widthDx = b-an .

2. Let xi = a+ iDx be the endpoints of the subintervals.

3. Let An = ⁄ni=1 f (xi)Dx be the approximation usingn rectangles.

Simplify and computeAn using the techniques of the previous sec-tion.

4. Compute limnÆ• An, yielding the area.

Example 6.3.1Compute the area underf (x) = 4x2+ 2x+ 1 betweena = 3 andb = 6.

Example 6.3.2Compute the area underf (x) = 3x + 5 betweena = 0 andb = 4 two differentways: by using the techniques of this section and by simply drawing the region and using thefact that it is made up of straight sides. Note that answer is the same either way.


6.4. The Definite Integral

6.4.1. Definition and Notation

In the previous section we have seen that for a given functionf (x) and interval[a, b], then

quantity limnÆ•

n

‚

i=1

xiDx is an important quantity, since it gives the area undery = f (x) and above

the x axis betweenx = a andx = b if f (x) > 0 on [a, b]. This “limit of a sum of areas ofrectangles” is important enough that it deserves its own special name and notation. We willcall this quantity thedefinite integral of f(x) with respect to x from a to b.Our definition willactually be slightly more general. We don’t actually have to have all subintervals the same with,and we don’t have to use the right-hand side of each subinterval. As long as we have pointsa = xo, x1, x2, ldots, xn = b, we can letDxi = xi+1 - xi be the width of theith subinterval, and

we can letx*i be any point on theith subinterval. Then we can consider limnÆ•

n

‚

i=1

f (x*i )Dxi as the

limit of a sum of areas of rectangles with various widths and (perhaps) using various points todetermine the heights of the rectangles. Of course, the important theorem (whose proof we willomit) is that we get the same result whether thexi ’s are left-hand endpoints, right-hand endpoints,midpoints, or whatever, as long as the widthsDxi shrink to zero asnÆ •. Thus we have:


Definition of Definite IntegralLet the interval[a, b] be divided up into subintervals by a set of pointsxi

so thata = xo < x1 < x2 < µ < xn = b.

Let Dxi = xi+1 - xi be the width of theith subinterval. Letx*i be anynumber so thatxi £ x*i £ xi+1. Then the definite integral off (x) withrespect tox from a to b is given by

limnÆ•

n

‚

i=1

f (x*i ) Dxi .

This quantity is a number which doesn’t depend on the choice of end-pointsxi or the choice ofx*i . The definite integral is defined whether ornot f (x) > 0 on[a, b], but only represents the area underf (x) and abovethex axis betweena andb when f (x) ≥ 0 on(a, b).

Example 6.4.1Compute the definite integral from 2 to 5 off (x) = 5x3+ 3x. Does this represent

an area?


Since it is somewhat painful to write out the quantity represented by the definite integral, itwill be useful to introduce some notation to represent it. The notation will necessarily have toinvolve the functionf (x) and the endpointsa andb. We choose a mathematical symbol whichwill remind us of an “S” for “sum.” The symbol is actually two symbols which are always

used in conjuction: the integral sign “Ÿ ” and the differential “dx.” Thus we writeŸb

af (x)dx or

‡

b

af (x)dx. Thus we have:


Notation for the Definite IntegralWe write the definite integral off (x)with respect tox from x = a to x = bas

‡

b

af (x)dx.

The symbolŸ is called theintegral sign, and thea andb are called thelower limit of integrationand theupper limit of integration, respectively.The symboldx is called thedifferential, and is always used in conjuc-tion with the integral sign. Thex in the symboldx represents that thedefinite integral is to be considered with respect tox. The functionf (x),which is placed between the integral sign and the differential, is calledthe integrand.

Thus, for example, the previous example could have been worded: findŸ

5

25x3+ 3x dx.

Example 6.4.2ComputeŸ3

-15x+ 2dx.


6.4.2. Properties of Ÿb

af (x)dx.

Because the definite integral is defined to be a limit of a sum of areas of rectangles, and becauselimits and sums have nice properties, we might expect for the definite integral to have some niceproperties, especially as related to constant multipliers and sums. We have the following:

Properties of the Definite Integral

1. Ÿb

ak f(x)dx= k Ÿ

b

af (x)dx, wherek is a constant.

2. Ÿb

af (x) ± g(x)dx= Ÿ

b

af (x)dx± Ÿ

b

ag(x)dx.

3. Ÿb

af (x)dx+ Ÿ

c

bf (x)dx= Ÿ

c

af (x)dx

4. Ÿb

af (x)dx= - Ÿ

a

bf (x)dx

Thus, for example, to computeŸ5

13x2+ 3x + 1dx we could instead compute 3Ÿ

5

1x2 dx+

3Ÿ5

1x dx+ Ÿ

5

11dx.


6.4.3. Relation of the Definite Integral to Area

We have seen that whenf (x) ≥ 0 on[a, b] thatŸb

af (x)dx is the area underf (x), above thex-axis

and betweenx = a andx = b. What if f (x) is not greater than or equal to zero on[a, b]? Iff (x) £ 0 on [a, b], In this case, think about the functiong(x) = - f (x). This function would be

greater than or equal to 0 on[a, b], soŸb

ag(x)dx would be the area underg(x). But the graph of

g(x) = - f (x) is obtained from the graph off (x) by flipping around thex-axis. Thus the area underg(x) betweena andb is the same as the areaabove f(x), below thex-axis and betweenx = a andx = b. If a functionh(x) is above thex-axis between 0 and 1 and is below thex-axis between 1and 2, thenŸ

2

0f (x)dx, which by the above properties is equal toŸ

1

0f (x)dx+ Ÿ

2

1f (x)dx, which

yields the area belowf (x) between 0 and 1 minus the area abovef (x) between 1 and 2. Thisidea can be generalized to see that in general, the definite integral gives the area above thex-axis minus the area below thex-axis. To get the total area bounded byf (x), one would need to

computeŸb

a| f (x)|dx. Summarizing:


If Aup is the area belowf (x) and above thex-axis, andAdown is the areaabovef (x) and below thex-axis, then

‡

b

af (x)dx= Aup - Adown.

The total area bounded byf (x) is given byŸb

a| f (x)|dx.


-1x3dx. Does this represent an area? Explain.

Example 6.4.4In a previous example we computedŸ3

-15x+ 2dx. Thinking only in terms of the

area(s) represented here, recompute this using a sketch as your guide. (Hint: This works becauseall sides of the region 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 areother applications and interpretations of this entity. Just like the two main interpretations of


the derivative (slope of tangent line and instantaneous rate of change) there is an interpretation

of Ÿb

af (x)dx in terms of “physics” in addition to the geometric interpretation we have already

discussed. Suppose that an object is moving in a straight line, but with a velocity that varieswith time t. Let v(t) represent the velocity at timet. Our intuition tells us that if we know thevelocity atany time t, we should be able to recover the distance the object travels between timea andb. For the sake of simplicity, let’s assume the object is always moving to the right, so itsvelocity is always positive on[a, b]. If v(t) were constant, we know that distance= rate◊ time,but here the rate (velocity) isn’t constant. However, on a very small time interval[ti , ti+1] thevelocity is almost constant, and we could approximate it byv(t*i ) for somet*i on [ti , ti+1]. Thusfor that small time interval, the distance is given approximately byv(t*i ) ◊Dti . (This simply comesfrom distance= rate◊ time.) Adding up all of the distances over the various small time intervalsgives⁄n

i=1 v(t*i ) ◊ Dti , and this approximation will get better as the time intervals get shorter, so

the exact distance is given by limnÆ• sumni=1v(t*i ) ◊ Dti , but this is exactlyŸ

b

av(t)dt. Thus in this

context, the definite integral of velocity gives distance travelled. Ifv(t) is sometimes positiveand sometimes negative on[a, b], the definite integral gives the difference between the distancetravelled to the right and the distance travelled to the left, sometimes calleddisplacement. To get

the total distance travelled, one would have to computeŸb

a|v(t)|dt.


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 ofCalculus. This theorem will connect the theory of the definite integral with differentiation in afundamental way, by linking the definite integral of a function with information related to itsantiderivative. The key idea is the following: consider a small subinterval of the closed interval[a, b], say the interval fromxi to xi+1. The portion of the definite integral that has something todo with this interval is the term of the formf (x*i )Dxi , whereDxi = xi+1 - xi . Now, suppose thatF(x) is any antiderivative off (x), and apply the Mean Value Theorem toF(x) on the interval[xi , xi+1]. The MVT says that there is a pointci betweenxi andxi+1 with

F(xi+1) - F(xi) = F ¢(ci) ◊ Dxi = f (ci)Dxi .

In other words, there is a pointci so that the area of the rectangle with baseDxi and heightdetermined by the functionf evaluated atci is exactly the difference between the antiderivativeF evalauated at the right-side endpoint andF evaluated at the left side endpoint. If we do this on


each subinterval, then by choosing appropriate values ofci , it must follow that

n

‚

i=1

f (ci)Dxi =

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) +º + F(xn) - F(xn-1),

which simplifies toF(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], thenŸ

b

af (x)dx= F(b) -F(a), whereF(x)

is anyantiderivative off (x).

Note that the fundamental theorem draws a connection between the definite integral offand its antiderivative,F . It also provides amucheasier way to calculate definite integrals, sinceantiderivatives are often easier to find than it is to go through the ardous process of finding thelimit of a sum of areas of rectangles. Here is an example:

Example 6.5.1Use the fundamental theorem to calculateŸ3

1x2+ 4x dx.


Solution: Since an antiderivative off (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 expressionF(b) -F(a) is so important in the practice of using

the fundamental theorem to computeŸb

af (x)dx, it will be useful to introduce notation for it. We

will use the notationF(x)|ba to representeF(b) - F(a).

Example 6.5.2Use the above notation to computeŸp

0sin(x)dx.

Solution: Ÿp

0sin(x)dx= - cos(x)|p0 = - cos(p) - - 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 definiteintegrals and antiderivatives. The second part of the fundamental theorem will reiterate thisrelationship, but in a slightly different way. We will create a kind of “definite integral function”,and see that it is actually the antiderivative of the integrand. Specifically, consider the functionA(x) = Ÿ

x

0f (t)dt. Note that in this definition, the independent variable for the functionA is x.


We choose the letterA because the function is sort of an area function – iff (t) is above thet axis,thenA(x) gives the area underf (t) betweent = 0 andt = x. Of course, ifF(t) is an antiderivative(with respect tot) of f (t), then another name forA(x) is F(x) - F(0). Then the derivative withrespect tox of A(x) is F ¢(x) - 0 (sinceF(0) is a constant), but sinceF is an antiderivative off ,we have thatF ¢(x) = f (x). Thus,A¢(x) = f (x). This is the Fundamental Theorem – Part II.

Fundamental Theorem of Calculus – Part IISuppose thatA(x) is the function defined byA(x) = Ÿ

x

af (t)dt. Then

A¢(x) = f (x). That is, the derivative of a function defined so that theindependent variable is the upper limit of integration of a definite integralhas as its derivative the integrand of the definite integral as a function ofthat upper limit of integration.

Example 6.5.3What is the derivative with respect tox of Ÿx

2

0

sin(t)dt?

Solution: The FTOC – Part II applies. The derivative is simply0

sin(x). �

Example 6.5.4If A(x) = Ÿ4

x1t dt with x > 0, what isA¢(x)?


Solution: We can’t apply the FTOC – Part II toA(x) as written, but we could apply it to-A(x) =

- Ÿ4

x1t dt = Ÿ

x

41t dt. Thus the derivative of-A(x) is 1

x , so the derivative ofA(x) = - 1x . �

Example 6.5.5What is ddx KŸ

x2

1sec(t)dtO?

Solution: The chain rule applies, together with the FTOC – Part II. Note that the function is theresult of composing the functionA(x) = Ÿ

x

1sec(t)dt with the functiong(x) = x2. In fact, the

original function is thenA(g(x)). Thus, by the chain rule, the derivative must beA¢(g(x))g¢(x).By the FTOC – Part II,A¢(x) = sec(x), soA¢(g(x)) = sec(x2

). Also, g¢(x) = 2x, so the derivativeof the original function must be sec(x2

)2x. �


6.6. The Indefinite Integral and Substitution

One thing that arises from the fundamental theorem is the idea that antiderivatives are quiteimportant, since they help us to evaluate definite integrals. Thus, we probably should study themin more detail. In this section, we will introduce a new notation for the general antiderivative,based on the fact that it is closely related to the definite integral. We also learn a technique forfinding antiderivatives, and thus definite integrals which is essentially the inverse of the chainrule.

6.6.1. The Indefinite Integral

The FTOC indicates that antiderivatives and integrals are closely related. Because of this, andbecause of our need for better notation for the antiderivative of a function, we introduce thefollowing new notation for the general antiderivative.


The Indefinite IntegralWe will call the general antiderivative of a functionf (x) the indefiniteintegralof f (x), and will denote it byŸ f (x)dx. Thus

‡f (x)dx= F(x) +C,

whereF ¢(x) = f (x). We choose this notation (which is similar to thenotation for the definite integral) because the fundamental theorem ofcalculus tells us that the general antiderivative is related to the definiteintegral.

Note the difference betweenŸb

af (x)dx and Ÿ 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 butrather their symbols are similar because they are related. To reiterate: they are related because

one can computeŸb

af (x)dx by using any of the functions from the collectionŸ f (x)dx, and

evaluating atb anda and subtracting.

Example 6.6.1ComputeŸ sec2(x) + x dx.


Solution: Ÿ sec2(x) + x dx= tan(x) + x2

2 +C. �

Example 6.6.2ComputeŸ (x2+ 3x+ 1)(2x+ 8)dx.

Solution:

‡(x2+ 3x+ 1)(2x+ 8)dx=

‡2x3+ 6x2

+ 2x+ 16x2+ 24x+ 8dx Multiplying Out the Integrand

=‡

2x3+ 22x2

+ 26x+ 8dx 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) Ÿ f (x)dx

sinx - cosx+C

cosx sinx+C

sec2 x tanx+C

csc2 x - cotx+C

secx tanx secx+C

cscxcotx - cscx+C

k, ka constant kx+C

xn, n ê = -1 xn+1

n+1 +C

kg(x) k Ÿ g(x)dx

g(x) + h(x) Ÿ g(x)dx+ Ÿ h(x)dx


6.6.2. Substitution in Indefinite Integrals

We now encounter a technique of integration which will help us to integrate more complexfunctions, helping us to expand the above list of functions we know how to integrate. Thetechnique is calledsubstitution, because in the technique we will replace (or substitute) somequantities for other quantities. The goal is to perform a substitution which will simplify theintegrand, hopefully to one on the above list. The motivation for the technique comes from thechain rule. Recall that the chain rules says that

ddx

f (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 thatŸ 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 equivalentŸ 200(3x+1)(3x2+2x)99 dx?

The idea behind the technique of substitution is the following: suppose that we gave the quantity


3x2+ 2x a simpler name. What if we called it simplyu? Then sincedu

dx = 6x+ 2, it might be agood idea to callduby the name “(6x+ 2) ◊ dx.” Then we could write the integral

‡100(

uõúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúû

(3x2+ 2x))99

duõúúúúúúúúúúúúúúúúúùúúúúúúúúúúúúúúúúúû

(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,

we should report our answer in terms ofx, since the original problem was in terms ofx. Thus wecould write(3x2

+ 2x)100+C as our answer.

Substitution Technique – Indefinite IntegralsGiven an integral of the formŸ f ¢(g(x))g¢(x)dx, we let u = g(x) anddu= g¢(x)dx. Then we write the integral asŸ f ¢(u)du, which integratesto be f (u)+C. This should be reported asf (g(x))+C, which is the correctantiderivative off ¢(g(x))g¢(x), as can be checked by the chain rule.


Example 6.6.3ComputeŸ 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 wesubstitute, we obtain the integralŸ sin(u) ◊ -du, which can be written as- Ÿ sin(u)du. This isequal to-(- cos(u)) +C, so the desired antiderivative is cos(sin(x)) +C. �

Example 6.6.4ComputeŸ x0

5x2 + 8dx.

Solution: If we let u = 5x2+ 8, thendu= 10x dx. It might be helpful to write this asdu

10 = x dx.Thus our integral can be rewritten asŸ 1

10

0

u du = 110 Ÿ0

u du. Upon integrating, we obtain110

23u

32 +C = 1

15(5x2+ 8)

32 +C. �

How do I know what to let u equal?This is a common question asked by many beginning calculus students.The best answer is: letu be something so that by substitutingu anddu, you are able to completely transform the integral into an integral interms ofu which you know how to compute. A more practical answermay be this: try to letu be something whose derivative is a factor ofthe integrand. This is a necessary (but not sufficient) ingredient of asuccessful substitution!


6.6.3. Substitution in Definite Integrals

Can the technique of substitution be used to evaluate definite integrals? Of course the answer isyes – since it helps us to find antiderivatives and that is what is needed to use the FTOC. Caremust be taken, however, to be sure write the substitution correctly in the presence of the limitsof integration.

Substitution Technique – Definite IntegralsGiven an integral of the formŸ

b

af ¢(g(x))g¢(x)dx, we let u = g(x) and

du = g¢(x)dx. Then we write the integral asŸg(b)

g(a)f ¢(u)du, which inte-

grates to bef (u)ƒƒƒ

g(b)g(a) = f (g(b)) - f (g(a)). This is the numerical value of

the original definite integral, so no further modifications are necessary.


0xsin(x2

+ 4)dx.

Solution: It seems reasonable to letu = x2+4, since the derivative of this expression is 2x which

is a factor of the integrand (if we think ofx as 2xx .) Sincedu = 2x dx, we can writex dx = du

2 .

Substituting we obtain the integralŸ20

412 sin(u)du.The 1

2 factor comes from thedu2 substitution,


and the limits of integration come from the fact thatu has value 4 whenx = 0 and has value 20whenx = 4. This new integral has exactly the same numerical value as the original, so I needonly compute it. But we have

‡

20

4

12

sin(u)du=12- cos(u)|20

4

= -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 the functions.


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 kindsof areas and related definite integrals. In fact, we have seen that in general, the definite integralof f (x) from x = a to x = b gives us thesignedarea trapped between thex axis and the curve,and between the linesx = a andx = b. We now try to extend this theory to enable us to computearea of regions like the following.

Let R be the region bounded by the curvesg(x) = x2- 5 and f (x) = 1- x2. Note that these

curves cross whenx2- 5 = 1 - x2, which is when 2x2

= 6, or whenx = ±0

3. Now imag-ine taking the interval from[-

0

3,0

3], and dividing it up into subintervals and then computinglimnÆ•⁄

ni=1( f (x

*

i ) - g(x*i )) Dxi . This should be thought of as a limit of a sum of areas of rect-angles, which are approximating the areabetweenthe curvesf (x) andg(x). It is important tonote that over the interval[-

0

3,0

3], f (x) is greater thang(x), so thatf (xi) - g(xi) is a positivenumber, 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 asŸ0

3

-

0

3( f (x) - g(x))dx. This can be computed as

‡

0

3

-

0

3( f (x) - g(x))dx=

‡

0

3

-

0

31- x2

- (x2- 5)dx

=‡

0

3

-

0

36- 2x2 dx Simplifying the Integrand

= 6x- 2x3

3

ƒƒƒƒƒƒƒƒ

0

3

-

0

3

By the FTOC

= 60

3- 230

33- (6(-

0

3- 2-30

33) Simplifying

= 120

3- 40

3 More Simplifying

= 80

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 byx = a, x =

b, f(x),andg(x) is Ÿb

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 planar regionbounded above byf (x) and below byg(x). We call such an areaver-tically orientedbecause the approximating rectangles are upright.

Example 6.7.1Let a be the smallest positive number where cos(x) = sin(x), and letb > a be thenext smallest positive number where cos(x) = sin(x). Compute the area bounded by the curvesf (x) = sin(x) andg(x) = cos(x) betweena andb.

Solution: A quick analysis of the unit circle leads us to believe thata is p4 andb = 5p4 . Between

a andb, it is clear that sinx > cosx. Thus we need to computeŸ5p4p

4sinx- cosx dx, which by the

FTOC is equal to

- cosx- sinx|5p4p

4= -(-

0

22-

0

22) - (-

0

22-

0

22) = 2

0

2.

�


6.7.2. Horizontally Oriented Areas

For certain planar regions, it is easier to think of the approximating rectangles as laying on theirsides – that is, we will use horizontal rectangles to yield our definite integral. An example of sucha planar region could be the region bounded byx = 1- y2 andx+ y = -1. This region is mosteasily thought of as boiunded on the left by the linex+y = -1 and on the right by the “sideways”parabolax = 1- y2. If instead you look for 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” ratherthan “up and down” for this region. Because we are thinking horizontally, we imagine dividingthey axis up into subintervals rather than thex axis. This will lead to an integral with respect tox rather than one with respect toy. Since the two functions cross wheny = -1 and wheny = 2,we want to divide the interval from[-1,2] along they axis into subintervals. The horizontalapproximating rectangles we want to use to find the area are bounded on the right by 1- y2 andon the left by-1- y. Thus our approximating sum has the form⁄n

i=1(1- y2i ) - (-1- yi)Dyi , and

after taking the limit asnÆ •, 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 byx = g(y) and on the right byx = h(y) betweeny = c andy = d is given by

‡

d

ch(y) - g(y)dy.

Example 6.7.2Find the area bounded by the curvesx = 3- y2 andy = x- 1.

Solution: These curves cross wheny+ 1 = 3- y2, which is wheny2+ y- 2 = 0. Thus they cross

aty = -2 andy = 1. Over this region, the left-hand boundary of the region is formed by the linex = y+ 1, and the right-hand boundary is formed by the parabolax = 3- y2. Thus the area is of


the region is given byŸ1

-23- y2

- (y+ 1)dy. Computing we get

‡

1

-23- y2

- (y+ 1)dy=‡

1

-23- y2

- y- 1dy

=‡

1

-22- y- y2 dy

= 2y-y2

2-

y3

3

ƒƒƒƒƒƒƒƒ

1

-2

= 2-12-

13- K-4- 2-

-83O

= 4.5

�

Assignment:Stewart: Section 6.1, 1 - 29 odd.


Chapter 7

Appendix

Contents

7.1 Ten – Minute Reviews. . . . . . . . . . . . . . . . . . . . . . . . . . . . .3397.1.1 The Real Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . .339

7.1.2 Sets, Set Notation, and Interval Notation. . . . . . . . . . . . . . . . 341


Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .344

Solutions to Quizzes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .513


7.1. Ten – Minute Reviews

7.1.1. The Real Numbers

It is convenient to think of the real numbers as consisting of all of the points on the number line.For every point on the number line (we will henceforth call it thereal line), there is a real numberand for every real number there is a corresponding point on the number line. We denote the set ofreal numbersR, although on the blackboard your instructor may write—. Of course, the familiarset of whole numbers{0,1,2,º} is a subset of the set of reals. If we take the set of wholenumbers and throw in their opposites, we have the set of integersZ = {º,-2,-1,0,1,2,º}.Another important subset of the setR is the setQ of rational numbers. It is useful to think ofQ as consisting of all quotients you can form by taking ratios of integers, with the denominatornot being zero. Thus,Q = { p

q | p is an integer andq is an integer andq ê = 0}. Note that everyinteger is also a rational numbers (since we can writep = p

1 ), and every decimal which eitherterminates or repeats is rational, as the next few examples demonstrate.

Example 7.1.1Show that the real number 5.341 can be written as the ratio of two integers.

Solution: 5.431= 5+ 410 +

3100+

11000. If we look for a common denominator, we see that a good

one is 1000. Thus we can write 5.431= 50001000+

4001000+

301000+

11000 =

54311000. �

Example 7.1.2Consider the number 3.212121º which we sometimes write as 3.21. Show that


this is a rational number.

Solution: Note that since there are two digits in the “repeat”, if we multiply this number by 100,the stuff after the decimal stays the same. We will exploit this idea in the following manner: letx = 3.2121º. Then 100x = 321.2121º. If we subtract these expressions, we get 99x = 318.Thusx = 318

99 . �

Of course, there are numbers which can’t be written as the ratio of two integers. An exampleis0

2. Such numbers are calledirrational .Note that the sum of any two rationals is a rational, sincea

b +cd =

ad+bcbd , and if the quantities

a, b, c, dare all integers, thenad+ bcandbd are too.An important property that the set of real numbers possesses is what we will call thebe-

tweenness property. This property states that between any two real numbers is a third realnumber. A consequence of this property is that there is no such thing as “the smallest positivereal number”, because as soon as you have a candidatea, you can find a smaller candidatea2(among others). In fact, for any two real numbersa andb, there are infinitely many real numbersbetween them.

Exercises

EXERCISE7.1.1.When you hit thep button on your calculator, it says something like 3.14159265359.Given thatp is irrational, is there any way that this representation ofp is exactly correct?


EXERCISE7.1.2.After thinking about the previous question a little, try answering this one: Ifyou know that a certain problem has answerx =

0

2, what would be wrong with instead givingyour calculator’s value ofx = 1.41432 instead?

EXERCISE7.1.3.Does it make sense to talk about “the largest real number less than 5”? Whyor why not?

EXERCISE7.1.4.Is x = 5.321321º a rational number? If so, find the integersp andq so thatx = p

q .

7.1.2. Sets, Set Notation, and Interval Notation

A set is a collection of objects specified by a list or a rule. When a set is small, it is usually spec-ified by a list. For example, the set of the first five positive integers could be written{1,2,3,4,5}.Note the use of the set braces,{ and}.

Of course, if a set is infinite in size, it is not practical to list all of the elements in it. Instead,we might specify its contents via a rule. For example, we might write the set of all real numbersbetween 2 and 5 and including those endpoints as{x |2 £ x £ 5}. If we were to write{x | x <0 or x ≥ 2}, then we are thinking about all real numbers which are either negative or greater thanor equal to two.


If one were to take the real number line and remove finitely many numbers from it, whatremains is a collection of sets calledintervals. An interval might consist of all the real numbersbetween-3 and 5, or it could consist of all real numbers greater than 6, etc. The former exampleis an example of an interval with finite length, while the latter has infinite length. We can expressthese sets via a rule, but it is also convenient to introduce another notation for such sets calledinterval notation. In interval notation, we don’t use the curly braces{ and }, but instead useparentheses( and) and/or square brackets,[ and]. If we write (-3,5), we mean the same thingas if we write{x | - 3 < x < 5}. If instead we wanted to indicate the set{x | - 3 £ x £ 5}, wewould write [-3,5]. Thus the “open” parentheses indicate that the endpoint is not included inthe set, while the “closed” brackets indicate that the endpoint is included.

We must still discuss how to indicate an infinite length interval. If, for example, we wantedto specify the set of all real numbers greater than or equal to 6, we could start our interval bywriting [6,º but we would need a symbol to indicate that this interval doesn’t end, but keepsgoing forever. We choose the “infinity” symbol• to indicate this. Thus we write[6,•) toindicate the set of all real numbers greater than or equal to 6. If we wanted to indicate the entirereal liner, we could write(-•,•). Note that wealwaysuse open parentheses rather than closedbrackets with the• symbol.

When we want to throw the elements in two or more sets together into a larger set, the correctoperation is called theunion of the sets, and the corresponding symbol is«. Thus, ifA is a setandB is a set thenA« B is the set of elements which belong to eitherA or B (or both).


Exercises

EXERCISE7.1.5.Describe the set of positive real numbers using set notation. Then write it withinterval notation.

EXERCISE7.1.6.Find the solution to the inequalityx2-9 < 0, and write it in set notation. Then

write the solution to the inequalityx2- 9 > 0 in set notation. For both of these, also write your

answer in interval notation.

EXERCISE7.1.7. For what values ofx is 2x0

9-xdefined? That is, what is the domain of this

function? Write your answer in either set notation or interval notation.

EXERCISE7.1.8. Write the set of all real numbers which are between-5 and 9, and whichincludes-5 but doesn’t include 9 in both set notation and interval notation.

EXERCISE7.1.9.Translate{x | x < -3 orx > 6} into interval notation.

EXERCISE7.1.10.Explain why the following is true: the union of the intervals(2,4) and(6,9)consists of all the points between 2 and 4and all the points between 6 and 9. However, whenwe write this set in set notation we use the wordor, like so: {x |2 < x < 4 or 6< x < 9}.

EXERCISE7.1.11.What is the domain of the functionf (z) =0

z2 - 4z? Write your answer inset notation and in interval notation.


Solutions to Exercises

Exercise 0.2.1.The author of the document is blessed to have two wonderful daughters, Hannahand Darby. Exercise 0.2.1


Exercise 1.1.1. f (2) = 13. f (5) = 97. f (x + h) = 4(x + h)2 - 3 = 4(x2+ 2hx+ h2

) - 3 =4x2+ 8hx+ 4h2

- 3. Exercise 1.1.1


Exercise 1.1.2.g(z+ h) - g(z) = -(z+ h)2 + 4 - I-z2+ 4M = - Iz2

+ 2zh+ h2M + 4 + z2

- 4 =-z2- 2zh- h2

+ z2= -2zh- h2.

Exercise 1.1.2


Exercise 1.1.3.Yes, these are both functions of each other, because for each it is true that everyinput value determines a unique output value. Exercise 1.1.3


Exercise 1.1.4.x is a function ofy, but y is not a function ofx. For example, ifx = 4, theny2= 9, but theny could be either 3 or-3, so there would be two outputs for the one input.

Exercise 1.1.4


Exercise 1.1.5.Yes, it is a function because the ordered pairs are defining a rule of correspon-dence between elements in the domain setD = {3,4,5,6} and the range setR= {2,3,4}.

Exercise 1.1.5


Exercise 1.1.6.No, there are two different outputs (3 and 6) associated with the one input valueof 4, so this is not a function. Exercise 1.1.6


Exercise 1.1.7.The distance between(x, x2) and(1,3) is given byd(x) =

0

(x- 1)2 + (x2 - 3)2.Exercise 1.1.7


Exercise 1.1.8.f (x) = (x+3)2-4x . Exercise 1.1.8


Exercise 1.1.9. For an input value ofx, the quantity is raised to the fourth power, and 5 issubtracted from the result. This result is then divided by the result of adding one to twice theinput value. Exercise 1.1.9


Exercise 1.1.10.The set of all real numbers less than or equal to 4. Exercise 1.1.10


Exercise 1.1.11.The set of all real numbers except for the real number 3. Exercise 1.1.11


Exercise 1.1.12.f (x) = 2x- 3. Exercise 1.1.12


Exercise 1.1.13.dp + 4t = 7, while`p - 5p = -1. Exercise 1.1.13


Exercise 1.1.14.The other vertex could have coordinates(4,5) or (2,9) depending on how youdraw the triangle. The distance between the original two points is then

0

(4- 2)2 + (9- 5)2 =0

20. Exercise 1.1.14


Exercise 1.1.15.The other vertex should have coordinates either(c, b) or (a, d). In either case,the distance is

0

(c- a)2 + (d - b)2. Exercise 1.1.15


Exercise 1.1.16.Some points on it include(0,1), (1,4), and(2,7). The ratio of the difference inx’s to the distance iny’s is always 3, for example:

4- 11- 0

=7- 42- 1

=7- 12- 0

This is because this quantity represents the slope, which is always a constant. A proof: If(x1,3x1 + 1) is one point and(x2,3x2 + 1) is another, then

3x2 + 1- (3x1 + 1)x2 - x1

=3(x2 - x1)

x2 - x1= 3.

Exercise 1.1.16


Exercise 1.1.17.Yes,y is a function ofx andx = 30

y is a function ofy. The domain and rangeof both of these functions is the setR, the set of all real numbers. Exercise 1.1.17


Exercise 1.1.18.The domain of this function is the set of all real numbers less than or equal to4. The range is the set of all real numbers greater than or equal to 5. Exercise 1.1.18


Exercise 1.1.19.This function is symmetric about they axis. That is, if you were to “fold” thepaper along they axis, the two sides would match up. Any function which has the property thatf (-x) = f (x) for all x has this property. Exercise 1.1.19


Exercise 1.1.20.What is true aboutf (x) = x3 is that f (-x) = - f (x) for all values ofx. Func-tions with this kind of symmetry are said to be symmetric about the origin, and are called oddfunctions. Exercise 1.1.20


Exercise 1.1.21.This function is even, sincef (-x) = 5 = f (x) for all x. Exercise 1.1.21


Exercise 1.1.22.The range is the set of all real numbers except for 0. The domain is the set ofall real numbers except for the real number 1. Exercise 1.1.22


Exercise 1.2.1.To complete the squares, we add 1 and 4 to both sides, obtainingx2+ 2x+ 1+

y2- 4y+ 4 = 25+ 1+ 4. This can be rewritten as(x+ 1)2 + (y- 2)2 = 30, and thus our circle

has center(-1,2) and radius0

30. Exercise 1.2.1


Exercise 1.2.2.The distance between(0,2) and(x,3x2-4)would be given byg(x) =

0

(x- 0)2 + (3x2 - 4- 2)2.This can be simplified tog(x) =

0

9x4 - 35x2 + 36. Exercise 1.2.2


Exercise 1.2.3.Suppose we were to solve fory. We would have(y+2)=16- (x-3)2, and takingsquare roots of both sides we have

|y+ 2| =1

16- (x- 3)2.

Now according to the definition of absolute value,|y+ 2| will be y+ 2 wheny ≥ 2, but it will bethe opposite of this wheny < 2. Thus we have the following cases:If y ≥ 2, we gety = -2+

0

16- (x- 3)2,If y < 2, we gety = -2-

0

16- (x- 3)2.A close examination of the graph leads us to the conclusion that the first of these is the uppersemicircle, while the second is the lower. Thus, the upper semicircle has equation given by

y = -2+1

16- (x- 3)2.

Exercise 1.2.3


Exercise 1.3.1.The line (in slope-intercept form) is given byy = -34 x+ 13

2 . Exercise 1.3.1


Exercise 1.3.2.F = 95C+ 32. Exercise 1.3.2


Exercise 1.3.3.The temperature is-40. Note that95 ◊ -40= -72, and-72+ 32= -40.Exercise 1.3.3


Exercise 1.3.4.The line should have slope-12 , so in point-slope form it should bey - 2 =

-12 (x- 1). Exercise 1.3.4


Exercise 1.3.5.Note that all lines which go through the point(1,4) have the formy-4 = m(x-1)for some slopem. This characterizes all lines through(1,4). The lines with slope 3 all have theform y = 3x+ b, for some value ofb. The only line in both of these sets isy- 4 = 3(x- 1), ory = 3x+ 1. Exercise 1.3.5


Exercise 1.3.6.The vertical line through that point is the linex = 3. The horizontal line throughthat point is the liney = 5. Exercise 1.3.6


Exercise 1.4.1.It is a polynomial. It has the forma4x4+ a3x3

+ a2x2+ a1x+ a0, wherea4 = 1,

a3 = 0, a2 = p, a1 = 0, anda0 = -0

2. Note that these are all real number coefficients.Exercise 1.4.1


Exercise 1.4.2. It is not a polynomial, because it does not have the form of one, nor can itbe manipulated to have the form of one. It is a rational function, because it can be written asthe ratio of two polynomials by finding a common denominator. That is, we can write it asxx3 -

5x+5x3 =

-4x-5x3 , and this last expression is the ratio of two polynomials. Exercise 1.4.2


Exercise 1.4.3. It is the set of all real numbers except for 0 and 16. We could write this as{x|x ê = 0, x ê = 16} or as(-•,0) « (0,16) « (16,•). Exercise 1.4.3


Exercise 1.4.4.This would be thosex’s for which |x - 2| £ .01. In interval notation this is theset[1.99,2.01]. Exercise 1.4.4


Exercise 1.4.5. We need| f (x) - 5| > .5, so we need|3x + 4 - 5| > .5, or in other words,|3x - 1| > .5. Of course|3x - 1| is .5 when 3x - 1 = .5 or when 3x - 1 = -.5, so that is whenx = .5 or whenx = 1

6. Now we want to know when|3x - 1| > .5, but this is the same thing aswanting to know when|3x-1|- .5 > 0. If we make a sign chart (using16 and 1

2 as landmarks) wesee that this expression is positive forx’s less than1

6 and greater than12. So the set in question is(-•, 1

6) « (12,•). Exercise 1.4.5


Exercise 1.4.6.

f (x) =ÏÔÌÔÓ

4x- 3 if x ≥ 2;

x2+ 5 if x < 2.

Exercise 1.4.6


Exercise 1.4.7.The domain is the set of all real numbers except forx = 1. In interval notation itis (-•,1) « (1,•). f (6) = 1 and f (0) = -1. Exercise 1.4.7


Exercise 1.4.8.Hint: The ceiling function looks somewhat (but not exactly) like the floor func-tion. The graph ofy = d2xt has value 0 betweenx = 0 andx = 1

2, but has value 1 betweenx = 1

2 andx = 1. Try drawing your sketch using interval of length12 in the domain to guide

you. Exercise 1.4.8


Exercise 1.5.1.Hint: make sure you are translating or scaling correctly. Use table1.5.3on page71and table1.5.3on page72 to guide you. Exercise 1.5.1


Exercise 1.5.2. ( f Î g)(x) = f (0

x- 3 + 1) = 3(0

x- 3 + 1)2 + 5. This can be simplified to60

x- 3+3x-1. (gÎ f )(x) = g(3x2+5) =

0

3x2 + 5- 3+1 =0

3x2 + 2+1. Exercise 1.5.2


Exercise 1.5.3.One possible answer is the following: Letg(x) = x2 and let f (x) = x+12-x. Then

( f Î g)(x) = H(x). You should perform this calculation to assure yourself that it is correct.Exercise 1.5.3


Exercise 1.5.4.Hint: The graph ofy = 1x+1 should be the same as that off (x) except shifted one

unit to the left (so that the function is now undefined atx = -1 rather than atx = 0.) On theother handy = 1

x + 1 is the same asf (x) except that the whole thing is shifted up one unit.Exercise 1.5.4


Exercise 1.5.5.The domain ofg(x) is the set ofx’s between 2 and92. Thus, the graph ofg(x) iscompressed horizontally as compared withf (x). The graph ofh(x) should be stretched verticallyas compared withf (x) (note: same domain, different range,) and the graph ofj(x) is stretchedhorizontally as compared withf (x). In particular, note that the domain ofj(x) is the set[8,18].

Exercise 1.5.5


Exercise 1.5.6.Note that f (-x) = -3x + 2, which is neither equal tof (x) nor- f (x), so this isneither even nor odd. You could also graph it and note its lack of symmetry.Exercise 1.5.6


Exercise 1.5.7. Note thatg(-x) = -x(-x)3-(-x) =

-x-(x3-x) =

xx3-x = g(x). Thus,g(x) is an even

function. Exercise 1.5.7


Exercise 2.1.1.The expressionf (2+h)- f (2)h would equal3(2+h)+2-8

h =6+3h+2-8

h =3hh = 3. Thus the

slope of any secant line (regardless of the value ofh) is 3, and so the slope of the tangent line is3 also. Exercise 2.1.1


Exercise 2.1.2.The slope of the line between(1,1) and(2,8) is 8-12-1 = 7. The slope between

(1,1) and(1+h,(1+h)3) is (1+h)3-11+h-1 =

(1+h)3-1h . If you want to simplify the above expression, you

must multiply out the numerator. Since(1+ h)3 = 1+ 3h+ 3h2+ h3, we have

1+ 3h+ 3h2+ h3- 1

h=(h)(3+ 3h+ h2

h= 3+ 3h+ 3h2.

Exercise 2.1.2


Exercise 2.1.3.Betweent = .5 andt = 1, the average velocity is given bys(1)-s(.5)1-.5 =

-16+32-(-4+16).5 =

4.5 = 8. Betweent = .5 andt = .5 + h, the average velocity would be given bys(.5+h)-s(.5)

.5+h-.5 =

-16(.5+h)2+32(.5+h)-(-4+16)h . In order to simplify this expression, one would need to multiply out the

numerator. The result would be

-16(.25+ h+ h2) + 16+ 32h+ 4- 16

h=-4- 16h- 16h2

+ 16+ 32h+ 4- 16h

,

and this can be simplified to be equal to

16h- 16h2

h= 16- 16h.

Note that whenh = .5, the result is 8, which is what we got in the first portion of the problem.Exercise 2.1.3


Exercise 2.1.4.The average rate of change of the population over the time interval[2,4] wouldbe given byb(4)-b(2)

4-2 =64+4-(8+2)

4-2 =582 = 29. Over the time interval[2,2+ h], the average rate of

change would be given byb(2+h)-b(2)h . This would be

(2+ h)3 + (2+ h) - (8+ 2)2+ h- 2

=(2+ h)3 + 2+ h- 10

h.

If we were to lethÆ 0, this quantity would represent the instantaneous rate of change or “growthrate” of the population at the timet = 2. In order to compute this, we would need to simplify theexpression. Note that(2+ h)3 = 8+ 12h+ 6h2

+ h3. Thus our expression would be

8+ 12h+ 6h2+ h3+ 2+ h- 10

h=

13h+ 6h2+ h3

h= 13+ 6h+ h2.

If hÆ 0, we would get a growth rate of 13. Exercise 2.1.4


Exercise 2.2.1.We would havef (x)- f (2)x-2 =

13x-

16

x-2 . A common denominator (of the terms in the

numerator) would be 6x. Thus we would have26x-

x6x

x-2 =2-x

(6x)(x-2) . Now sincex - 2 and 2- x are

opposites, we can cancel them, leaving a minus sign. Thus we are left with-16x .

Exercise 2.2.1


Exercise 2.2.2.The difference quotientg(x+h)-g(x)h would be

0

x+ h+ 3-0

x+ 3h

.

To simplify this, we need to multiply the numerator and denominator by the conjugate of thenumerator. This conjugate would be

0

x+ h+ 3+0

x+ 3. Thus we have

0

x+ h+ 3-0

x+ 3h

◊

0

x+ h+ 3+0

x+ 30

x+ h+ 3+0

x+ 3

which would equalx+ h+ 3- (x+ 3)

(h)(0

x+ h+ 3+0

x+ 3).

This simplifies to be 10

x+h+3+0

x+3. Exercise 2.2.2


Exercise 2.2.3.We would havef (2+h)- f (2)h =

(2+h)3+2+h-10h . If we expand(2 + h)3 we have 8+

12h+ 6h2+ h3. Thus our difference quotient is equal to

8+ 12h+ 6h2+ h3+ 2+ h- 10

h=

13h+ 6h2+ h3

h.

If we factor anh out of the numerator and cancel with theh factor in the denominator, we have13+ 6h+ h2. Exercise 2.2.3


Exercise 2.2.4.h(x)-h(2)x-2 =

x2+x+2-(8)

x-2 =x2+x-6x-2 . Now the numerator can be factored as(x-2)(x+3),

so our difference quotient becomes(x+3)(x-2)x-2 = x+ 3. Exercise 2.2.4


Exercise 2.2.5.

We need to computef (3+h)- f (3)h . We havef (3+ h) = 3

3+h+1 =3

4+h. Thus we have

f (3+ h) - f (3)h

=

34+h -

34

h.

To simplify this last expression, we should find a common denominator between 4+ h and 4,which should be 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=

-3h4(4+ h)h

.

Cancelling theh’s leaves -34(4+h) . ShrinkinghÆ 0 yields a value of-3

16 .

Exercise 2.2.5


Exercise 2.2.6.f (x)- f (4)x-4 =

10

x-

12

x-4 . To simplify, we need to find a common denominator among thetwo rational expressions in the numerator. The common denominator would be 2

0

x. Thus wehave

220

x-

0

x20

x

x- 4,

which can be written as2-0

x

20

x(x- 4).

To simplify this expression, we need to multiply the numerator and the denominator by theconjugate of the numerator. Thus we have

2-0

x

20

x(x- 4)◊

2+0

x

2+0

x=

4- x

20

x(2+0

x)(x- 4),

and the(4 - x) factor in the numerator cancels the(x - 4) factor in the denominator, leaving afactor of-1. Thus we end up with

-1

20

x(2+0

x).

Exercise 2.2.6


Exercise 2.3.1.The sentence would be “The limit asx approaches 5 of 4x- 1 is 19.” The idea isthat whenx is close to 5, 4x- 1 is close to 19. Exercise 2.3.1


Exercise 2.3.2.This would be written

limhÆ0

(x+ h)2 - x2

h= 2.

Exercise 2.3.2


Exercise 2.3.3.For this function, limxÆ2- f (x) = -4, and limxÆ2+ f (x) = 5. Exercise 2.3.3


Exercise 2.3.4.A good example of what you could sketch would be the function

f (x) =ÏÔÌÔÓ

3 if x < -1;

4 if x > -1.

Exercise 2.3.4


Exercise 2.3.5.Note thatf (h) can be thought of as the function

f (h) =ÏÔÌÔÓ

-1 if x < 0,

1 if x > 0.

This function has a limit ash approaches 0 from the left of-1 and a limit ash approaches 0from the right of 1, and so it is correct to say that the limit ash approaches 0 off (h) does notexist. Exercise 2.3.5


Exercise 2.3.6. Since the numerator is very close to 4 and the denominator is a very smallnumber (i.e., close to 0), the whole quotient is very large. (At least its absolute value is large –it may actually be a negative number.) Thus it is correct to say that the limit doesn’t exist. Thecorrect generalization is this: if limxÆa p(x) = k wherek is any real number other than zero,and if limxÆa q(x) = 0, then limxÆa

p(x)q(x) doesn’t exist. In fact, limxÆa+

p(x)q(x) = • or limxÆa+

p(x)q(x) =

-•, and the same can be said about the limits from the left.Exercise 2.3.6


Exercise 2.3.7.One possible answer isf (x) = -1(x-4)2 .

Exercise 2.3.7


Exercise 2.4.1.This would be the set(2.8,3.2). Exercise 2.4.1


Exercise 2.4.2.This would be the set(3.7,4) « (4,4.3). Exercise 2.4.2


Exercise 2.4.3.This would be the set of allx so that|x- 1| < .25. Exercise 2.4.3


Exercise 2.4.4.

We have| f (x) - 6| < .2, so|4x- 2- 6| < .2, so|4x- 8| < .2, and thus|x- 2| < .2

4 = .05.

Exercise 2.4.4


Exercise 2.4.5.

We are to start with|5x- 1- 14| < e, so then|5x- 15| < e, so also|x- 3| < e

5.

Thus, we should taked to be e5. Exercise 2.4.5


Exercise 2.4.6.Hint: After making the sketch and shading the appropriate regions, try drawingthe vertical linesx = 2.9 andx = 3.1, as well as the horizontal linesy = 6.8 andy = 7.2.If you sketchedf (x) carefully, it should be true that whenx is between the vertical lines, thecorrespondingy value is between the horizontal lines. Exercise 2.4.6


Exercise 2.4.7.Starting with| f (x) - 7| < .2, we have

|2x+ 1- 7| < .2, so|2x- 6| < .2, so|x- 3| < .1,

and all of these steps are reversible. Thus it is correct to say that if|x-3| < .1, then| f (x)-7| < .2.Exercise 2.4.7


Exercise 2.4.8.Our solution will mirror that of the last problem. In fact, starting with| f (x)-7| <e, we have

|2x+ 1- 7| < e, so|2x- 6| < e, so|x- 3| < e

2,

and all of these steps are reversible. Thus it is correct to say that if|x-3| < e

2, then| f (x)-7| < e.Thus we need to taked = e2. Exercise 2.4.8


Exercise 2.5.1.The original statement written in if – then form is: Iff (x) is differentiable atx = a the it is continuous atx = a. The converse would be: Iff (x) is continuous atx = a then itis differentiable atx = a. The contrapositive would be: Iff (x) is not continuous atx = a, thenit is not differentiable atx = a. It turns out to be true for this statement that the original (andtherefore its contrapositive) is true, but the converse is false. Exercise 2.5.1


Exercise 2.5.2.In if – then form, the statement is: if|x - 2| < .1 then|(2x - 1) - 3| < .2. Toprove this statement, you would have to assume that|x - 2| < .1 and then show that it followsthat |(2x- 1) - 3| < .2. Exercise 2.5.2


Exercise 2.5.3.


Exercise 2.6.1.The if – then statement is: if 0< |x- 4| < d then|5x- 1- 19| < e.Exercise 2.6.1


Exercise 2.6.2.Here is the “forwards” work:

Suppose 0< |x- 3| < .1. Then it is also true that|x- 3| < .1. Then|3x- 9| < .3. So it then follows that|3x+ 1- 10| < .3, which is the statement| f (x) - 10| < .3.

Exercise 2.6.2


Exercise 2.6.3.Here are the algebraic calculations which we can get by working backwardsfrom the statement| f (x) - L| < e. We have

| f (x) - L| < e, so therefore|2x+ 1- 5| < e. Therefore|2x- 4| < e and thus|x- 2| < e

2. Thus it looks like we would needd =

e

2.

Exercise 2.6.3


Exercise 2.6.4.SCRATCH WORK: We need|4x- 3- 9| < e. Thus we need|4x- 12| < e, or inother words we need|x- 3| < e

4. Thus it looks like we needd = e4.FORMAL PROOF: Lete > 0 be given. Letd = e4.

Then suppose that 0< |x- 3| < d. Then|x- 3| < e

4, so|4x- 12| < e, so then|4x- 3- 9| < e, and this is the statement| f (x) - L| < e, so we are done.

Exercise 2.6.4


Exercise 2.6.5.This time we present only the rigorous “forwards” proof.Let e > 0 be given. Letd = e2. Then if we suppose that

0 < |x- 7| < d, then it follows that|x- 7| < e

2, so then|2x- 14| < e, so also|2x+ 1- 15| < e, which is the statement| f (x) - L| < e, so we are done.

Exercise 2.6.5


Exercise 2.6.6.This time we present only the rigorous “forwards” proof.Let e > 0 be given. Letd = e3. Then if we suppose that

0 < |x- -3| < d, then it follows that|x+ 3| < e

3, so then|3x+ 9| < e, so also|3x- -9| < e, which is the statement| f (x) - L| < e, so we are done.

Exercise 2.6.6


Exercise 2.7.1.

limxÆ5

4x2+ 6x- 4 = lim

xÆ54x2+ lim

xÆ56x- lim

xÆ54 Sum Rule

= limxÆ5

4x2+ lim

xÆ56x- 4 Constant Rule

= 4 limxÆ5

x2+ 6 lim

xÆ5x- 4 Constant Multiplier Rule

= 4KlimxÆ5

xO2

+ 6 limxÆ5

x- 4 Power Rule

= 4 ◊ 52+ 6(5) - 4 The Identity Rule

= 126 Arithmetic!

Exercise 2.7.1


Exercise 2.7.2.Let e > 0 be given, and letd = e. Now suppose that 0< |x - c| < d. Thensincee = d, we also have|x- c| < e. But this actually is the statement that| f (x) - L| < e, wheref (x) = x andL = c, so we are done!. Exercise 2.7.2


Exercise 2.7.3.Since the limit of the numerator is 0 and the limit of the denominator is not zero,we actually can use the plug-in theorem. Thus we have limtÆ0

t2-4

t2-2t+1 =01 = 0.

Exercise 2.7.3


Exercise 2.7.4.It would have to be true that limxÆ3 q(x) ê = 0. If this were true, you could deducethat limxÆ3

p(x)q(x) = 0. If limxÆ3 q(x) = 0, then there is no conclusion without further information.

Exercise 2.7.4


Exercise 2.7.5. If you knew that limxÆ3 p(x) ê = 0, then you could conclude that limxÆ3p(x)q(x)

does not exist. If however it were true that limxÆ3 p(x) = 0, then you could make no conclusionwithout further information. Exercise 2.7.5


Exercise 2.7.6.Using the limit theorems you could deduce that limxÆ431

f (x)4 + f (x)g(x) =

31

34 +38

Exercise 2.7.6


Exercise 2.7.7.Yes. For example, supposef (x) = 1x andg(x) = 1+ 1

x , neither of which have alimit asxÆ 0. But f (x) -g(x) = 1

x - (1+1x ) = -1. So limxÆ0 f (x) -g(x) = -1. So the difference

of these two has a limit even though neither of them have a limit individually.Exercise 2.7.7


Exercise 2.7.8.The key thing that is wrong with this argument is that the limit of a product isonly equal the the product of the limits whenboth limits exist. It is likely that limxÆc

f (x)x-c doesn’t

exist, so this step is bogus. In fact, note that if limxÆc f (x) ê = 0, then it is definitely true thatlimxÆc

f (x)x-c does not exist!

It is probably worth mentioning also that the statement 0◊ something= 0 is somewhatmisleading – it is only valid when the “something” is a real number. In particular, whatever0 ◊ • means, it definitely isn’t necessarily zero.

Exercise 2.7.8


Exercise 2.7.9.We have

limxÆ4

0

x- 2x- 4

= limxÆ4

0

x- 2x- 4

◊

0

x+ 20

x+ 2

= limxÆ4

x- 4

(x- 4)(0

x+ 2)

= limxÆ4

10

x+ 2

=14

Exercise 2.7.9


Exercise 2.7.10.If we expand the numerator, we arrive at

limhÆ0

9+ 6h+ h2+ 9+ 3h- 18h

= limhÆ0

9h+ h2

h= lim

hÆ09+ h

= 9

Exercise 2.7.10


Exercise 2.7.11.The numerator factors as(x - 2)(x2+ 2x + 5). This can be deduced by long

division. (Try dividingx- 2 intox3+ x- 10 and see what you get.) Thus we have

limxÆ2

x3+ x- 10x- 2

= limxÆ2

(x- 2)(x2+ 2x+ 5)

x- 2= lim

xÆ2x2+ 2x+ 5

= 13

Exercise 2.7.11


Exercise 2.7.12.The numerator factors into(y + 1)(y - 1) and the denominator factors into(y- 1)(y3

+ y2+ y+ 3). Again, this last factor can be obtained by long division. Thus we have

limyÆ1

y2- 1

y4 + 2y- 3= lim

yÆ1

(y+ 1)(y- 1)(y- 1)(y3 + y2 + y+ 3)

= limyÆ1

y+ 1y3 + y2 + y+ 3

=26

=13

Exercise 2.7.12


Exercise 2.7.13.Yes, limxÆ4 f (x) does exist, and has value 44, because limxÆ4- f (x) = limxÆ4+ f (x) =44. More specifically, limxÆ4- f (x) = limxÆ4- 3x2

-4 = 44. And also limxÆ4+ f (x) = limxÆ4+ 11x =44. Since the two one-sided limits both exist and are equal, the corresponding two-sided limitalso exists. Exercise 2.7.13


Exercise 2.7.14.No, in this case limxÆ4- f (x) = limxÆ4- 3x2+ 4 = 52, and this does not equal

limxÆ4+ f (x) (which has value 44 as in the previous problem.) Exercise 2.7.14


Exercise 2.8.1.This function is continuous on its domain, which is the set of all real numbersexcluding 3. In set notation, it is{x : x ê = 3}. Since limxÆ3 f (x) = 6, it is in fact a removablediscontinuity. Exercise 2.8.1


Exercise 2.8.2.It is continuous on its domain, which is the set of all real numbers greater thanor equal to 3. In interval notation, it is[3,•). Exercise 2.8.2


Exercise 2.8.3.It must be redefined to be equal to limxÆ2x3-8

x-2 . This limit is equal to limxÆ2 x2+

2x+ 4 = 12. Exercise 2.8.3


Exercise 2.8.4.This function is not continuous atx = 5, and that is the only point. It is clearlycontinuous everywhere except perhaps atx = 0 andx = 5, and a quick check atx = 0 shows thatthe limit does exist (since the two one-sided limits exist and both equal 3), and the limit is equalto f (0) = 3. However, atx = 5 the limit doesn’t exist, since the limit asx Æ 5- is -2, while thelimit as xÆ 5+ is 10. Exercise 2.8.4


Exercise 2.8.5.Let a = 1, b = 2 andM = 0. Note thatM is betweenf (a) = f (1) = -1 andf (b) = f (2) = 5. Thus, there must be a root off (x) (that is, a value ofc where f (c) = 0) between1 and 2. Exercise 2.8.5


Exercise 2.8.6.First letg(x) = x4-4x+1. Then roots ofg(x) are solutions to the given equation.

Now notice thatg(0) = 1 which is greater than zero, whileg(1) = -2, which is less than zero.Thus there is at least one root between 0 and 1. Now also note thatg(2) = 9, which is greaterthan zero. Thus there is also a root between 1 and 2. Thus there are at least two real roots.

Exercise 2.8.6


Exercise 2.8.7.This function is discontinuous ath = 0 because limhÆ0|h|h does not exist. In

particular limhÆ0-|h|h = -1 while limhÆ0+

|h|h = 1. Thus the function is discontinuous ath = 0.

Exercise 2.8.7


Exercise 2.8.8.It is discontinuous at all the integers and all the “half integers”, that is, at allpoints of the formk

2 wherek is an integer. For example, it is discontinuous atx = 12, since

limxÆ 12-d2xt = 0 while limxÆ 1

2+d2xt = 1. Exercise 2.8.8


Exercise 2.8.9.Hint: The graph you sketch should be right continuous atx = 3 and left contin-uous atx = 6, but should not be continuous at those points. One of the infinitely many possiblesketches would be the sketch of the function

f (x) =

ÏÔÔÔÌÔÔÔÓ

1 if x < 3;

2 if 3 £ x £ 6;

3 if x > 6.

Exercise 2.8.9


Exercise 3.1.1.Since f (x) = x2+ x+ 1, we have

f ¢(2) = limhÆ0

f (2+ h) - f (2)h

= limhÆ0

(2+ h)2 + 2+ h+ 1- 7h

= limhÆ0

4+ 4h+ h2+ 1+ h+ 1- 7h

= limhÆ0

5h+ h2

h= lim

hÆ05+ h

= 5.

So f ¢(2) = 5. Exercise 3.1.1


Exercise 3.1.2.Both the first and second items above are equal to each other and are, in fact,f ¢(z). The third and fourth items are both equal tof ¢(x). Exercise 3.1.2



f ¢(2) = limhÆ0

f (2+ h) - f (2)h

= limhÆ0

0

4+ h- 2h

= limhÆ0

0

4+ h- 2h

◊

0

4+ h+ 20

4+ h+ 2

= limhÆ0

4+ h- 4

h(0

4+ h+ 2)

= limhÆ0

10

4+ h+ 2

=14

Exercise 3.1.3


Exercise 3.1.4.If f (x) = x2+ 2x anda = 3 then f ¢(a) = f ¢(3) = limhÆ0

(3+h)2+2(3+h)-(32+2◊3)

h asdesired. Exercise 3.1.4



f ¢(x) = limzÆx

az2- ax2

z- x

= limzÆx

a ◊(z+ x)(z- x)

z- x= a lim

zÆxz+ x

= a(2x)

= 2ax.

Exercise 3.1.5


Exercise 3.1.6.We would have

y¢ = limhÆ0

(2(x+ h) + 1)2 - (2x+ 1)2

h

= limhÆ0

(2x+ 2h+ 1+ 2x+ 1)(2x+ 2h+ 1- (2x+ 1))h

Difference of Squares!

= limhÆ0

(4x+ 2h+ 2)(2h)h

= limhÆ0(4x+ 2h+ 2)(2)

= (4x+ 2)(2)

= 8x+ 4

Exercise 3.1.6


Exercise 3.2.1.Yes, the functionf (x) = |x - 2| has a singular point atx = 2. This is becauselimhÆ0

|2+h-2|-|2-2|h does not exist. We know that this limit does not exist because it is the same

thing as limhÆ0|h|h , which was discussed in the beginning of this section. Exercise 3.2.1


Exercise 3.2.2.Rational functions are not differentiable at the points where they are not defined(because they are not continuous there,) but they are differentiable at all points in their domains.In fact, in the next section we will encounter the “quotient rule” which will tell us the derivativeof a rational function in terms of information about the numerator, the denominator, and thederivatives of the numerator and the denominator. Exercise 3.2.2


Exercise 3.2.3.We need to show that limxÆ1f (x)- f (1)

x-1 does not exist. We have

limxÆ1

f (x) - f (1)x- 1

= limxÆ1

xx-1 - 3

x- 1

= limxÆ1

x- (3(x- 1))(x- 1)2

= limxÆ1

3- 2x(x- 1)2

= +•

So f ¢(1) does not exist. Exercise 3.2.3


Exercise 3.2.4. In order to show that there is a singular point at(0,-1), we must show thatlimxÆ0

f (x)--1x-0 does not exist. Since the function is defined differently to the right and left of

zero, we should probably investigate the two one-sided limits. Focusing first on the limit asxÆ 0+ we have

limxÆ0+

f (x) - -1x- 0

= limxÆ0+

-2x- 1- -1x

= limxÆ0+

-2xx

= -2.

However, if we investigate the limit from the left we have

limxÆ0-

f (x) - -1x- 0

= limxÆ0-

2x- 1- -1x

= limxÆ0+

2xx

= 2.

Since the two one-sided limits off (x)- f (0)x-0 are different, the total limit does not exist. Thus the

function is not differentiable atx = 0. Exercise 3.2.4


Exercise 3.2.5.We would need to compute limxÆ0

xx-1-0x-0 , which is the same thing as limxÆ0

1x-1.

But this can be computed by “plugging in”, to give a result of-1. Thus, f ¢(0) = -1, andthereforef (x) is differentiable atx = 0. Exercise 3.2.5


Exercise 3.2.6. First note that the function is continuous atx = 0, because limxÆ0+ f (x) =limxÆ0- f (x) = 0 = f (0). Of course, that doesn’t tell us anything about its differentiability. Tosee if the function is differentiable atx = 0, we must see whether or not the limxÆ0

f (x)- f (0)x-0 exists.

We have

limxÆ0+

f (x) - f (0)x- 0

= limxÆ0+

4x2- 0

x- 0= lim

xÆ0+4x

= 0

and we also have

limxÆ0-

f (x) - f (0)x- 0

= limxÆ0-

x2- 0

x- 0= lim

xÆ0+x

= 0.

Since both one-sided limits off (x)- f (0)x-0 exist and are equal, the functionf (x) is differentiable at

x = 0. Exercise 3.2.6


Exercise 3.3.1.


Exercise 3.4.1.The derivative is 4x2(9x2- 1) + (3x3

- x)(8x), and there is a factor of 4x2 oneither side of the plus sign. Thus we can write this as 4x2

(9x2- 1+ 6x2

- 2) = 4x2(15x2

- 3) =12x2(5x2- 1). Exercise 3.4.1


Exercise 3.4.2. Note that on either side of the main plus sign, there are common factors of(x3- 11) (actually there are at least 3 on each side), as well as a common factor of(3x2

+ 5) andalso 12x. Thus we can factor these out, leaving

12x(x3- 11)3(3x2

+ 5)[(3x2+ 5)x+ (x3

- 11)]

which simplifies to12x(x3

- 11)3(3x2+ 5)[4x3

+ 5x+ -11]

Exercise 3.4.2


Exercise 3.4.3.To diffeentiate, we use the quotient rule. We have

f ¢(x) =(4x2- 1) ◊ 1

2x-12 -0

x ◊ 8x

(4x2 - 1)2.

In order to simplify this, we note that both terms in the numerator will be much nicer if multipliedby x

12 , and the denominator will only get a little worse. In fact, we can get rid of all fractions in

the numerator by multiplying by 20

x. We therefore compute

=(4x2- 1) ◊ 1

2x-12 -0

x ◊ 8x

(4x2 - 1)2◊

2x12

2x12

=(4x2- 1) - (16x2

)

(4x2 - 1)2(20

x)

=-12x2

- 1

(4x2 - 1)220

x

Exercise 3.4.3


Exercise 3.4.4.Using the quotient rule, the derivative would bef ¢(x) = (1+x2)(1)-(x)(2x)(1+x2)2

and this is

the same thing as1-x2

(1+x2)2, which doesn’t simplify any further. Exercise 3.4.4


Exercise 3.4.5.We can writef (x) as f (x) = x13 (3x+ x

12 ), and my multiplying this out we arrive

at f (x) = 3x43 + x

56 . Thus the derivative isf ¢(x) = 4x

13 +

56x

-16 . Exercise 3.4.5


Exercise 3.5.1.


Exercise 3.6.1.We could takem(x) = x3 andn(x) = 3x+ 1). Thenm¢(x) = 3x2 andn¢(x) = 3, som¢(n(x)) ◊ n¢(x) = 3(3x+ 1)2 ◊ 3.

Exercise 3.6.1


Exercise 3.6.2.We could takem(x) =0

x andn(x) = x2+ 1. Thenm¢(x) = 1

2x-12 andn¢(x) = 2x.

Thus,m¢(n(x)) ◊ n¢(x) = 12(x

2+ 1)

-12 ◊ (2x). Exercise 3.6.2


Exercise 3.6.3.We must use the quotient rule. We have

g¢(x) =(4x- 2)( 12(x

2+ 1)

-12 ◊ (2x)) -

0

x2 + 1 ◊ 4

(x2 + 1)2.

If you wanted to simplify this, a good first step would be to multiply by(x2+1)

12

(x2+1)12.

Exercise 3.6.3


Exercise 3.6.4.g¢(x) = f ¢(m(x)) ◊m¢(x), by the chain rule. Exercise 3.6.4


Exercise 3.6.5.We havef ¢(x) = (2x + 1)35(4x2+ 4x)4(8x + 4) + (4x2

+ 4x)53(2x + 1)2(2). Weshould try to factor this by looking for common factors on either side of the main+ sign. We seethat each term has at least 2 factors of(2x+ 1) and at least 4 factors of 4x2

+ 4x. Also, there is afactor of 2. Thus we can writef ¢(x) as

2(2x+ 1)2(4x2+ 4x)4[(2x+ 1)(5)(8x+ 4) + (4x2

+ 4x)(3)].

Upon further investigation, we note that there is still a factor of 4 which can be factored out, sowe arrive at

8(2x+ 1)2(4x2+ 4x)4[5(2x+ 1)(2x+ 1) + (3x2

+ 3x)].

Now if we simplify what is in the brackets, we finally arrive at

8(2x+ 1)2(4x2+ 4x)4[23x2

+ 23x+ 5].

Exercise 3.6.5


Exercise 3.6.6.h¢(3) = f ¢(g(3)) ◊ g¢(3), which is equal tof ¢(2) ◊ g¢(3) = 3 ◊ 4 = 12.Exercise 3.6.6


Exercise 3.6.7.The derivative of|6x-3| would be |6x-3|6x-3 ◊6. In general, the derivative ofy = |n(x)|

would bey¢ = |n(x)|n(x) ◊ n¢(x).

Exercise 3.6.7


Exercise 3.6.8.


Exercise 3.7.1.Differentiating both sides with respect tox yields

23

x-13 +

23

y-13dydx= 0.

If we multiply both sides by32 and the solve fordydx, we arrive at

dydx= - K

yxO

13

.

Now whenx = 1, we can see (using the original equation) thaty = 30

3. Thus at the point

(1,30

3), we have thatdydx = -

31

30

3. Exercise 3.7.1


Exercise 3.7.2.We differentiate both sides, noting that we must use the product rule on bothsides of the equation. We have

x2y+ y2+ 3 = xy

x2 dydx+ 2xy+ 2y

dydx= x

dydx+ y

x2 dydx+ 2y

dydx- x

dydx= y- 2xy

dydx(x2+ 2y- x) = y- 2xy

dydx=

y- 2xyx2 + 2y- x

.

Exercise 3.7.2



12(x+ y)

-12 (1+ y¢) = x23y2y¢ + y32x

12(x+ y)

-12 - 2xy3

= 3x2y2y¢ -12(x+ y)

-12 y¢

12(x+ y)

-12 - 2xy3

= (3x2y2-

12(x+ y)

-12 )y¢

12(x+ y)

-12 - 2xy3

(3x2y2 -12(x+ y)

-12 )

= y¢

Exercise 3.7.3


Exercise 3.7.4.If one is asked to computedzdv, then we are to think ofz as being a function of

v. Thusz is the name of the function, whilev is the independent variable. Thus, to differentiatez2 we would need the chain rule (since we really have the composition of the squaring functionwith the function called “z”), so the derivative would be 2zdz

dv. Exercise 3.7.4


Exercise 3.7.5.In this case, we would be thinking ofv as the function andz as the independentvariable, so the derivative with respect toz of z2 would be simply 2z. Exercise 3.7.5


Exercise 3.7.6. We note that we are thinking ofz as the name of the function andv as theindependent variable. We have

2zdzdv+ 3(zv)2(z+ v

dzdv) - (0

z+ v12

z-12

dzdv) = 0

2zdzdv+ 3z2v3 dz

dv-

v

20

z

dzdv=0

z- 3z3v2

dzdv(2z+ 3z2v3

-v

20

z) =0

z- 3z3v2

dzdv=

0

z- 3z3v2

(2z+ 3z2v3 -v

20

z)

Exercise 3.7.6


Exercise 3.7.7.This time we are thinking ofv as the name of the function and are thinking ofzas the independent variable. We have

2z+ 3(zv)2(zdvdz+ v) - (

0

zdvdz+ v

1

20

z) = 0

3z3v2 dvdz-0

zdvdz=

v

20

z- 2z- 3z2v3

dvdz(3z3v2

-0

z) =v

20

z- 2z- 3z2v3

dvdz=

v20

z- 2z- 3z2v3

(3z3v2 -0

z)

Exercise 3.7.7


Exercise 3.7.8.


Exercise 3.8.1.We must compute the first derivative first. We havey¢ = (x2-1)(4x)-(2x2

)(2x)(x2-1)2 . It

would pay to simplify this before attempting to differentiate again. We can factor out a factor ofx from the numerator, yieldingy¢ = (x)(4x2

-4-(4x2))

(x2-1)2 , which simplifies toy¢ = -4x(x2-1)2 . Now we can

differentiate this to gety¢¢. We havey¢¢ = (x2-1)2◊(-4)-(-4x)◊2(x2

-1)(2)(x2-1)4 . This could be simplified by

factoring out a common factor of(-4)(x2-1), yieldingy¢¢ = (-4)(x2

-1)((x2-1)-(4x))

(x2-1)4 which simplifies

to y¢¢ = (-4)(x2-4x-1)

(x2-1)3 . Exercise 3.8.1


Exercise 3.8.2.Again, we must first computey¢. Let us first writey asy = x12 + (x- 2)

12 . Then

y¢ = 12x-

12 +

12(x- 2)-

12 . The second derivative is theny¢¢ = - 1

4x-32 + -

14(x- 2)-

32 .Exercise 3.8.2


Exercise 3.8.3.It turns out the thenth derivative is

f (n)(x) = (-1)nn!

xn+1.

Exercise 3.8.3


Exercise 3.8.4. Since velocity is the derivative of position,v(t) = s¢(t) = -32t + v0. Sinceacceleration is the derivative of velocity, we havea(t) = v¢(t) = -32. The acceleration is constantsince the only force acting on the object is the pull of gravity, which is a constant near the earth’ssurface. Thus, sinceF = ma and since the object’s mass is constant, it must be true thata isconstant also. Exercise 3.8.4


Exercise 3.8.5.According to the previous problem, the object’s position at timet is given bys(t) = -16t2

+ 96t. Thus the velocity at timet is given byv(t) = s¢(t) = -32t + 96. Note thatsince the graph ofs(t) is a parabola which opens downward, the maximum will be at the vertex,which is where the tangent line tos has slope zero. Thus we need to look for the point wherev(t) = 0. (Note: this also makes sense because when the object is at its maximum height, itsvelocity is zero.) Nowv(t) = -32t + 96 is zero whent = 3, so that is when the maximum heightoccurs. The actual height is thens(3) = -16(9) + 96(3) = 144 feet. Exercise 3.8.5


Exercise 3.8.6.Let’s compute a few derivatives and look for a pattern.

y¢ =12◊ (-3)(2x+ 1)-4

◊ 2,

and since the12 and the factor of 2 cancel, we get

y¢ = (-3)(2x+ 1)-4.

Nowy¢¢ = (3)(4)(2x+ 1)-5

◊ 2,

y¢¢¢ = -(3)(4)(5)(2x+ 1)-6◊ 2 ◊ 2.

It looks like each derivative has(2x + 1) raised to a negative exponent, and has a factorial likething in the front, and has some factors of 2 as well. A closer inspection reveals that for thenthderivative, the exponent on(2x+ 1) would be-(n+ 3), the factorial thing out front looks like itstarts withn+ 2 (although note that it doesn’t proceed all the way down to 2, but ends at 3), andthere aren- 1 factors of 2. Now also note that

(n+ 2)(n+ 1)(n)µ(4)(3) =(n+ 2)(n+ 1)(n)µ(3)(2)

2=(n+ 2)!

2.


Thus it looks like the coefficient in front for thenth derivative should ben+22 . Finally, we need to

make the signs alternate, so we have

f (n)(x) =(-1)n ◊ (n+ 2)! ◊ 2(n-1)

2 ◊ (2x+ 1)n+3.

Exercise 3.8.6


Exercise 3.8.7.We will first find dydx by implicit differentiation. We have

2yy¢ + xy¢ + y = 0

2yy¢ + xy¢ = -y

y¢(2y+ x) = -y

y¢ =-y

2y+ x

Now, to findy¢¢, we must differentiate both sides of this last equation, remembering thaty is afunction of x.We get

y¢¢ =(2y+ x)(-y¢) - (-y)(2y¢ + 1)

(2y+ x)2

y¢¢ =-2yy¢ - xy¢ + 2yy¢ + y

(2y+ x)2

y¢¢ =-xy¢ + y(2y+ x)2

.

Now there is something less than desirable about this answer – it is in terms ofx, y, andy¢ ratherthan being simply in terms ofx andy. Of course, we do know whaty¢ is in terms ofx andy, so


substituting, we have

y¢¢ =-x J -y

2y+xN + y

(2y+ x)2

which can be simplified to be

y¢¢ =2xy+ 2y2

(2y+ x)3.

Exercise 3.8.7


Exercise 3.8.8.The polynomial must have had degree 15 to start with, since the degree of apolynomial goes down by one with each derivative. Exercise 3.8.8


Exercise 3.8.9.


Exercise 3.9.1.


Exercise 3.10.1.The linearization is given byL(x) = f (14)+ f ¢(14)(x-14), and sincef (14) = 3,we haveL(x) = 3+ f ¢(14)(x-14). It remains to findf ¢(14). We have thatf ¢(x) = 1

3(2x-1)-23 ◊2,

so f ¢(14) = 2

3◊(2723 )

=227. ThusL(x) = 3+ 2

27(x- 14). Exercise 3.10.1


Exercise 3.10.2.One reasonable approach would be to letf (x) =0

x, and let the pointa = 400.Then f (399) ª L(399), whereL(x) is the linearization neara = 400. We haveL(x) =

0

400+f ¢(400)(x - 400), soL(x) = 20+ 1

40(x - 400). Thus,L(399) = 20+ - 140 =

79940 . A quick check

with a calculator shows that the square root of 40 is about 19.97498435544, while the value of79940 is 19.975, so these are pretty close. Exercise 3.10.2


Exercise 3.10.3.The linearization off (x) is given byL(x) = f (2)+ f ¢(2)(x-2) = 113+ f ¢(2)(x-2).

It remains to computef ¢(2). We have f ¢(x) =(2x2+5)◊ 12 (x-1)

-12 -0

x-1◊4x(2x2+5)2 . Now we only need to

compute f ¢(2), so rather than simplify we simply evaluate this expression atx = 2, obtaining-3338. This,L(x) = 1

13 +-3338(x- 2). Exercise 3.10.3


Exercise 3.10.4. We have thatDy = f (1.1) - f (1) = 11.1 - 1 = - 1

11. On the other hand,dy= f ¢(1)(.1) = -1

12 ◊ .1 = -.1. Note thatdyª Dy. Exercise 3.10.4


Exercise 3.10.5.Dy = f (1.2) - f (1) = 3- (1.2)2 - (3- 12) = 1- (1.2)2 = -.44. On the other

hand,dy= f ¢(1)(.2) = -2(1)(.2) = -.4. Again, note thatdyª Dy. Exercise 3.10.5


Exercise 3.10.6.The reason is that 1+ x2 is the linearization off (x) =

0

1+ x at x = 0, as youcan check. Exercise 3.10.6


Exercise 3.10.7.Let’s let y = x40, and find the linearization atx = 1. We would haveL(x) =1+ f ¢(1)(x- 1), which is equal to 1+ 40(x- 1). Now L(1.01) would be 1+ 40(1.01- 1) = 1.4.The calculator value of(1.01)40 is about 1.4888. The reason these aren’t super close is that thefunctiony = x40 is so steep that the tangent line isn’t quite as good an approximation as it is forother curves. Exercise 3.10.7


Exercise 7.1.1.No, it can’t possibly be exactly correct, since the number 3.14159265359 is arational number (since it is a decimal which ends), whilep is irrational. The calculator is justgiving you the best rational approximation that it knows. Exercise 7.1.1


Exercise 7.1.2.Because 1.41432ê =0

2. This can be seen because 1.41432 is rational, while0

2 is irrational. Exercise 7.1.2


Exercise 7.1.3.No, there is no such thing. In fact, if you nominate a candidate (let’s call itc), then I can form5+c

2 which is larger thanc and still less than 5. Here we are exploiting thebetweeness property of the real numbers. Exercise 7.1.3


Exercise 7.1.4.Yes, it is rational. In fact, it is equal to5316999 =

1772333 . Exercise 7.1.4


Exercise 7.1.5.The set of positive real numbers is the set of numbers which are strictly greaterthan zero, so the set could be described by{x : x > 0}. In interval notation, we would have(0,•).

Exercise 7.1.5


Exercise 7.1.6.Note thatx2- 9 = 0 whenx = 3 andx = -3, so in between these numbers the

expression is either greater than 0 or less than 0. A quick check reveals that between the numbersis whenx2

-9 is negative. Thus the solution set for the first given inequality is{x : -3 < x < 3},while a solution set for the second inequality is{x : x < -3 or x > 3}. In interval notation, thefirst inequality could be written(-3,3), while the second could be written(-•,-3) « (3,•).

Exercise 7.1.6


Exercise 7.1.7.Note that the square root function isn’t defined when its input is less than zero,so we are interested in knowing when 9-x < 0 (these pointswon’t be in the domain.) Of course,these points are those for whichx > 9, so these points are eliminated. Also, we know thatfractions aren’t defined when their denominators are zero, so we must ask whether or not

0

9- xis ever equal to zero. Setting them equal and solving yieldsx = 9, so this point is eliminated.Every other real number would be ok, so the domain is{x : x < 9, or (in interval notation)(-•,9). Exercise 7.1.7


Exercise 7.1.8.In set notation:{x| - 5 £ x £ 9}. In interval notation:[-5,9). Exercise 7.1.8


Exercise 7.1.9.In interval notation, the set is(-•,-3) « (6,•). Exercise 7.1.9


Exercise 7.1.10.We want to include the numbers between 2 and 4and the numbers between 6and 9, so when we set up the condition that needs to be met, we say that we wantx to have eitherone property or the other. The key thing is that in our set notation, the stuff that comes after thevertical line is a condition which has to be met in order to be included in the set. Thus, to includeboth thing one and thing two in our set, we say that the condition to be met is that what is beingincluded is either thing one or thing two. Exercise 7.1.10


Exercise 7.1.11.Sincez2- 4z≥ 0 whenz is less than or equal to 0 and whenz is greater than or

equal to 4, the domain is{z|z£ 0 or z≥ 4}. In interval notation:(-•,0] « [4,•).Exercise 7.1.11


Solutions to Quizzes

Solution to Quiz: The quotation is from Martin Gardner. End Quiz

Documents

Integrated Calculus/Pre-Calculus - Furman Universitymath.furman.edu/~mwoodard/m140/docs/currentweb.pdfcolleges and universities are starting so called “integrated” precalculus/calculus