Mathematics 260 Calculus for Social Sciences and Economics

Mathematics 260Calculus for Social Sciences and Economics

STUDY GUIDE(Revision 4)

Contents

Introduction i

Unit 1 Functions and Graphs 1Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Elementary Functions: Graphs and Transformations . . . . . . . . . . . . . . . . . . . . . . 8Linear and Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Polynomial and Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Finishing this Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Unit 2 Limits and the Derivative 34Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Infinite Limits and Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72Basic Differentiation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Basic Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80Marginal Analysis in Business and Economics . . . . . . . . . . . . . . . . . . . . . . . . 85Finishing this Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Unit 3 Additional Derivative Topics 89The Constant e and Continuous Compound Interest . . . . . . . . . . . . . . . . . . . . . . 89Derivatives of Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . 91Derivatives of Products and Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99Elasticity of Demand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Finishing This Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Unit 4 Graphing and Optimization 105First Derivative and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106Second Derivative and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Curve-sketching Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118Absolute Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123Finishing This Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Unit 5 Integration 128Antiderivatives and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131Differential Equations; Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . 134The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141Finishing This Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Unit 6 Additional Integration Topics 144Area between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144Applications in Business and Economics . . . . . . . . . . . . . . . . . . . . . . . . . . . 148Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148Other Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151Finishing This Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

Unit 7 Multivariable Calculus 158Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163Maxima and Minima Using Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . 164Finishing This Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166Practice Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

Index 169

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Introduction

Thinking skills and problem-solving activities are indispensable to every area of our lives. To some extent, weare all problem solvers. The problem solver’s work is mostly a tangle of guesswork, analogy, wishful thinking,observing patterns, and frustration. To become a master problem solver may be as inaccessible as acquiring theskills of a virtuoso, but everyone can become a better, more confident problem solver.1

In this course, you will learn to use a range of mathematical techniques to analyze data and solve problemsin the social sciences. The materials for this course consist of this Study Guide, the Course Orientation, thefollowing textbook:

Barnett, R. A., M. R. Ziegler, K. E. Byleen, and C. J. Stocker. Calculus for Business, Economics, LifeSciences, and Social Sciences: Brief Version. 14 Ed. Upper Saddle River, NJ: Pearson-Prentice Hall,2019.

and its accompanying student manual:

Barnett, R. A., M. R. Ziegler, K. E. Byleen, and C. J. Stocker. Student’s Solutions Manual: Brief Version.14 Ed. Upper Saddle River, NJ: Pearson-Prentice Hall, 2019.

Your primary learning resource for this course is the textbook. However, this Study Guide is designed todirect your study and provide complementary material, including additional examples and exercises. So donot disregard the Study Guide: success in this course will be much more difficult to attain if youconcentrate solely on the textbook. Take a few minutes now to familiarize yourself with the CourseOrientation and the Study Guide.

PrerequisitesTo succeed in this course, you must have a good grasp of high school algebra. You may find that the“Diagnostic Prerequisite Test,” on pages xvii–xviii of the textbook, can help you to identify areas ofweakness.

In some units of this guide, we give the prerequisites required to learn that particular unit. If you haveproblems with algebraic concepts, please see Appendix A: Basic Algebra Review in the textbook.

For more in-depth learning of algebra, we refer you to the seven books in the series The Language ofMathematics by Rahael K. Jalan, which you can purchase from the author (email [email protected]).Purchasing these materials, however, is optional.

In the prerequisites, we quote the book in Roman numerals followed by the section number in thatparticular book.

1 Blitzer, Robert. Introductory Algebra for College Students, 2nd ed., p. 123. Upper Saddle River, NJ: Prentice Hall, 1998.

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Recommendations for Learning MathematicsWe have some recommendations to help you to become a better, more confident problem solver.

• Be prepared to practise. To master mathematical concepts, all you need to do is practise, practise andpractise again. You must do as many as possible of the exercises assigned in this guide, and we alsoencourage you to try non-assigned exercises from your textbook. A steady pace, practice and patiencewill take you farther than any shortcuts. You may have to change your study habits, because those thatwere sufficient for success in high school are not good enough for university.

Answers to odd-numbered exercises are provided in the “Answers” section at the end of the textbook.Solutions to these exercises are provided in the Student’s Solutions Manual. Do not look at theseanswers or solutions before making at least one honest try to solve the exercise yourself. In otherwords, do not give up too easily. The process of learning from an exercise starts with asking the rightquestions: What do I want to know? What do I have to explain? What do I know about this? What isthis exercise saying? What is this exercise testing?

In mathematics, it pays to be persistent: people who succeed in mathematics are those who do not give upafter the first attempt at solving a problem, but keep asking themselves, “What else can I try?”

• Read your Study Guide and textbook. The textbook is not a collection of solved problems—you areexpected to understand the course material and put it to use to solve problems yourself. Huntingthrough a section to find a worked-out exercise that is similar to the assigned problem is totallyinappropriate. Solving a mathematical problem requires an understanding of the concepts discussed.

Never read mathematics without pencil and paper in hand. Read the assigned textbook section beforetrying the exercises. You should treat the examples as exercises and try to solve them without theauthors’ help. Be prepared to check all the details as you read, and above all, read critically. Takenothing for granted.

If you do not understand a statement, go back in the section, or to a previous section, to see if youmissed or misunderstood something. If you still fail to understand a concept after puzzling over it for awhile, mark the place, continue reading, and ask your tutor for help on that point.

Once again, do not limit yourself to the assigned exercises. Instead, try to do as many exercises as youhave time for, and do not avoid those that appear to be challenging—the fun in learning is thesatisfaction of being able to respond positively to a challenge. The time you spend in reading the StudyGuide and the textbook will save you time in the long run. You should note any difficulty you havewith the homework, and ask your tutor for help.

• Raise your expectations. In high school, students are accustomed to finding that the first few weeksof a new course are nothing more than a review of the last course, or even the last few courses. When anew idea is finally presented, the class spends several days working on it before moving ahead.Students quickly learn that there is no major crisis if they do not follow what is said the first time, or ifthey miss a class. The concept is repeated soon (and sometimes ad nauseam). In this course, however,the review of previous material is minimal; new concepts are introduced immediately. Moreover, everysubsequent concept assumes mastery of the previous ones, and involves the presentation of new ideas.This is why you must study steadily and constantly throughout each unit. Mathematics cannot belearned in a hurry.

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• Learn to evaluate your own work. Solutions for all odd-numbered exercises are provided in theStudent’s Solutions Manual. Solutions of even-numbered exercises are not given, because you areexpected to start relying more on your own knowledge and less on external verification to determinewhether an answer is correct.

• Ask for help. When you cannot understand a concept, discuss the problem with your tutor. You mayalso have the opportunity of consulting other students or knowledgeable professionals, or you mayeven want to hire a private tutor. Discussing problems and course material with someone can help youto gain insight into difficult concepts. But take warning: others cannot understand the concept for you.It may be tempting to let others do your homework for you, but you will not learn if you do so. Youmust take the time to learn! It is unwise to pretend that you understand when you do not.

• Use the evaluation mechanisms. A test is a learning experience, not just a bureaucratic way toquantify your knowledge. Study your self-graded exercises and graded assignments carefully, andmake sure that you understand why you made mistakes, and how you can avoid making them again.The best time to understand missing concepts is right away, when the ideas are still fresh in your mind.

In a test or assignment, you must show that you have mastered the concepts and can solve problems.Rote memorization and regurgitation are not what we are looking for. Your grade reflects what youknow and how well you know it, as demonstrated by your personal performance, not on the basis of acomparison of your performance with that of others. Do not rely on the good will of the marker to getcredit for your work. The marker will grade what is on the paper, not what you might have intended todo.

• Prepare for the examinations. Practice examinations are provided at the end of each unit to allowyou to assess your mastery of the topics presented. These examinations consist of odd-numberedexercises from the textbook. You must try to complete these examinations within the time limitsuggested and without consulting your textbook. You will grade your work using the Student’sSolutions Manual.

Note that we do not want you to have false expectations because you did well in a practiceexamination. The best preparation for an examination is to try different exercises at random. At the endof each unit, we give you a list of exercises from the “Review” section of the textbook chapter. Use thislist to prepare for the examinations. Do the exercises in a random order.

Making summaries of each unit is another excellent way to prepare yourself for an examination. Andnever cram for a mathematics examination.

Note: All questions in assignments and examinations are worth several points, because several stepsmust be taken to solve them. So please show all your work and provide proper justification for all youranswers.

References in This Study GuideWithin the Study Guide, examples, theorem, definition, etc., are numbered; for example, “Theorem 3.23.”Note that the first number points to the unit where the item is found. Note as well that the numbering ofdefinitions and theorems is continuous in the Study Guide. So, for example, Theorem 3.1 is followed byDefinition 3.2. The same is true for the figures, examples, and exercises within a unit; each has its own

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counter. References are found either in this Study Guide or the textbook. Those in the textbook are soindicated.

Chapter ReviewYour textbook provides a “Review” section at the end of each chapter. We recommend that you study thesection titled “Important Symbols, Terms and Concepts” in each of these reviews.

When you come to the review, you should feel that you have learned enough to do the exercises presented.If you need assistance, consult your tutor.

TechnologyIn this course, we do not use technological aids in our work. You are not required to have a particularcalculator, or to use a computer algebra system or software program.

In examinations, you are allowed to use a simple calculator, but not a graphing calculator, and not aprogrammable calculator.

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UNIT 1

Functions and Graphs

Calculus is the area of mathematics that specializes in solving problems by establishing the function orfunctions that represent the quantities involved. In this calculus course, you will learn to apply differenttypes of functions to solve practical problems in economics, business, social sciences and life sciences. It isour hope that, in the process, you will come to appreciate how powerful the concept of functions is inproblem solving.

ObjectivesWhen you have completed this unit, you should be able to

1. apply the definition of a function to practical situations.

2. apply the vertical line test to determine whether a given curve corresponds to a function.

3. find the domain and range of a function.

4. sketch the graphs of elementary functions.

5. apply basic transformations (translation, expansion, contraction, reflection) to sketch graphs offunctions.

6. identify the solution set of algebraic inequalities.

7. graph linear equations.

8. identify the vertex and the axis of symmetry of the graph of a quadratic function.

9. calculate break-even points using linear and quadratic functions to model cost, revenue and profit.

10. identify polynomial and rational functions.

11. calculate compound interest.

12. apply exponential functions to solve problems of exponential growth.

13. apply exponential functions to solve problems of decay.

14. compute logarithmic regression models for given data sets.

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PrerequisitesTo learn this unit you must be able to

• apply the fractions properties◦ Section A.1 of Appendix A of the textbook

• do operations with polynomials◦ Section A.2 of Appendix A of the textbook◦ Book V sections 1.1, 1.1.1, 1.1.2, 1.2, 1.3 of Language of Mathematics

• apply the rules of exponents◦ Section A.5 of Appendix A of the textbook◦ Book I Section 3.6 of Language of Mathematics

• complete a perfect square◦ Section A.7 of Appendix A of the textbook◦ Book X Section Y.Z of Language of Mathematics

• apply the quadratic formula◦ Section A.7 of Appendix A of the textbook◦ Book VI section 2.7 of Language of Mathematics

FunctionsIndications1. Read Section 1.1 of the textbook, “Functions.” You may skip the paragraphs marked “Explore and

Discuss.”

2. Read the “Comments” section below.

3. Complete the exercises assigned at the end of the “Comments” section. If you have difficulty, consultyour tutor to discuss the problem.

CommentsTo solve problems using functions, we must identify the relation of dependency between the quantities(variables) involved. If the variable f depends on the variable x, then x is the independent variable and f isthe dependent variable, and we write f(x). From the list on page 4 of the textbook, we have

1. The annual income I depends on the person P ; thus, I(P ),or in words, to each person, there corresponds only one annual income.

2. The price P depends on the item i; thus, P (i),or in words, to each item, there corresponds only one price.

3. The grade point average g depends on the student s; thus, g(s),or in words, to each student, there corresponds only one grade.


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4. The maximum temperature T depends on the day d; thus, T (d),or in words, to each day, there corresponds only one maximum temperature.

5. The cost C depends on the manufacture of the item x; thus, C(x),or in words, to each item x, there corresponds only one cost.

6. The revenue R depends on the sale of x items; thus, R(x),or in words, to each x items, there corresponds only one revenue.

7. The area A depends on the square s; thus, A(s),or in words, to each square, there corresponds only one area.

8. The cube C depends on the number x; thus, C(x),or in words, to each number x, there corresponds only one cube C = x3.

From the definition of function on page 5, we see that to each independent variable, there corresponds oneand only one dependent variable. In terms of the graph of the function, the independent and dependentvariables are represented on the x-axis and y-axis, respectively. Hence, to each value on the x-axis, therecorresponds only one value on the y-axis. This fact explains why a vertical line must intersect the graph ofa function at only one point.

Example 1.1. If the pairs

(120500, 12000) and (120400, 12000)

are on the graph of the function I(P ) described in point 1, above, then to the person with SIN number120500, there corresponds an annual income of $12,000, and we write

I(120500) = 12, 000.

Similarly

I(120400) = 12, 000 = I(120400).

Two different people may have the same annual income, but each person has one and only one annualincome. �

Example 1.2. If R(x) is the function described in point 6, above, then

R(35) = 4000

means that the sale of 35 items yields a revenue of $4,000.

Thus, the pair (35, 4000) is on the graph of this function. In this case, for an input of x = 35 we have anoutput of $4,000.

For this function, x cannot be negative. [Why?] That is, the revenue is not defined for x = −2, and we saythat R(−2) is undefined. �


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If no meaning is assigned to the variables f and x of a function f(x), then we consider f(x) to be theabstract representation of a function. It is useful to study function in an abstract form, because severalproblems may be represented by functions that share the same properties.

The functions we consider in this course are specified by equations or represented by algebraic expressionssuch as polynomial, rational, exponential and logarithmic expressions.

In “Applications” in Section 1.1 of the textbook, the cost C depends on the number x of items produced,and

C(x) = a+ bx

is a linear increasing function.

If x = 0, no item is produced, then the cost, C(0) = a, represents the fixed costs, such as overhead, set up,etc. The more items produced, the higher the absolute cost, although the cost per item declines.

The price per item p depends on the number of items x that can be sold at price p; thus,

p(x) = m− nx,

which is a decreasing linear function.

If x = 0, no item can be sold, then the price per item is m. That is, no item can be sold at a price m. Thegreater the number of items sold (the more demand increases), the lower the price. The number n is the rateat which the price is decreasing. For every item sold, the price decreases n dollars. Since the price ispositive,

p(x) = m− nx > 0 if m > nx and 0 ≤ x <m

n.

The revenue R depends on the number of items sold x at price p; thus,

R(x) = xp = x(m− nx),

which is a concave-down quadratic function. [We discuss quadratic functions in detail in a later section ofthis unit.]

Therefore, the revenue is at its maximum if x =m

2n; furthermore, the revenue increases for 0 < x <

m

2n,

and decreases form

2n< x <

m

n.

[If you have trouble seeing why this is so, try graphing a sample equation.]

The profit P depends on the number of item sold x, and is equal to the revenue minus the cost:

P (x) = R(x)− C(x) = x(m− nx)− (a+ bx) = −a+ (m− b)x− nx2,

which is a concave-down quadratic function.


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The profit is maximum if x =m− b

2n: the profit increases for 0 < x <

m− b2n

, and decreases form− b

2n< x.

If the revenue is equal to the cost, then the profit is zero, and

P (x) = −a+ (m− b)x− nx2 = 0

is solved using the quadratic formula.

If the revenue is greater than the cost, R > C, then the profit is positive, P > 0, and the operation is in theblack (makes a gain). If the cost is greater than the revenue, R < C, then the profit is negative, P < 0, andthe operation is in the red (makes a loss).

Example 1.3. In Example 7 in Section 1.1 of the textbook, and Matched Problem 7 in the same section,the cost function is

C(x) = 156 + 19.7x.

Note that the range of x is the interval [1, 15]. This statement says that this mathematical model is accuratefor 1 ≤ x ≤ 15, but for x > 15, the cost is not given by this function. [Remember that x refers to millionsof items.]

The fixed cost is 156. [Why?] The minimum cost is C(1) = 175.7 and the maximum cost isC(15) = 451.5, since the cost function is increasing. The price function is

p(x) = 94.8− 5x.

No item can be sold at a price of $94.80. [Why?] One million items can be sold at the price p(1) = 89.80.Fifteen million is the maximum number of items that can be sold, and the corresponding price isp(15) = 19.80.

Observe that the price p must be positive, and it is positive for any x ≤ 18; hence, our price-demandfunction for this problem is correct.

With the cost and price-demand functions, we obtain the revenue function

R(x) = xp(x) = x(94.8− 5x) = 94.8x− 5x2.

The graph is shown twice on page 12 of the textbook (see figures 10 and 11).

The maximum revenue is achieved when x =94.8

10= 9.48: the revenue increases for 1 ≤ x < 9.48 and

decreases for 9.48 < x ≤ 15.

The maximum revenue is R(9.48) = 449.35, or $449,350,000.

The profit function

P (x) = R(x)− C(x) = 94.8x− 5x2 − (156 + 19.7x) = −156 + 75.1x− 5x2


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is also a concave-down quadratic (parabola). The graph is shown twice on page 17 of the textbook (see thefigures labeled “D” and “E”).

The maximum profit is achieved when x =75.1

10= 7.51, and the maximum profit is P (7.51) = 126, or

$126,000,000.

As you should know by now, the profit increases for 1 ≤ x < 7.51 and decreases for 7.51 < x ≤ 15.

The revenue is equal to the cost for

−156 + 75.1x− 5x2 = 0,

which we solve as follows:

x =−75.1±

√75.12 − 4(−5)(−156)

−10=−75.1±

√50.2

−10

we have x ≈ 12.53 and x ≈ 2.49.

Hence, from the graph of the profit function, we see that revenue exceeds cost for 2.49 < x < 12.53; whileand 1 ≤ x < 2.49 and 12.53 < x ≤ 15, the cost exceeds the revenue.

In conclusion, the operation will be in the black as long as 3 ≤ x ≤ 12 and it should aim to maintainx = 7.5 to maximize its profit. If it does so, the price will be p(7.5) = 57.3, the cost will beC(7.5) = 303.75 and the revenue will be R(7.5) = 429.75. �

All functions have a domain and range. We must pay attention to these two sets.

Definition 1.1. The domain of a function f(x) is the set of independent variables and the range is theset of dependent variables.

The domain of a function is determined by either

• the meaning of the independent and dependent variables, or

• by the mathematical model of the function when the independent and dependent variables do not havean specific meaning.

If C(x) is the cost of a product for each x item produced. Then the domain is all values of x for whichC(x) makes sense. Thus, the independent variable x must be positive or zero. The fixed cost of producingcorresponds to x = 0. This is the cost before producing an item. There are no negative values forx—something must be produced. If the maximum production capacity is M , then x ≤M . Hence, thedomain of C(x) is 0 ≤ x ≤M , in interval notation is [0,M ]. In Example 1.3 on the preceding page, thedomain of the cost function is the interval [0,15].

See that, mathematically, we can evaluate the function C(x) = 156 + 19.7x at a negative number.

C(−2) = 156 + 19.7(−2) = 116.6.

However, C(−2) is incorrect because its interpretation is: the cost of producing −2 items, which does notmake sense. When evaluating a function we must make sure that what we are doing makes sense.


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If the function f(x) does not have a meaning, then we must pay attention to the independent variable x sothat f(x) is defined. If in the function

f(x) = 156 + 19.7x

the dependent variable f has no meaning whatsoever, then f(x) can be evaluated for any real number x. Itsdomain is all real numbers.

There are two cases we must keep in mind when working with functions.

• Rational Functions. The domain of a rational function R(x) =p(x)

q(x)is all x so that q(x) 6= 0.

• Root Functions. The domain of a root function R(x) = n√g(x) where n is even, is all x so that g(x)

is defined and g(x) ≥ 0. For n odd the domain is all x so that g(x) is well defined.

Example 1.4. We want to determine the domain of the functions

a. f(x) =3x+ 1√

3x

b. g(x) =

√3x2 + 1

2x2 + 1

c. h(x) =3x+ 13√x2 − 4

For the function f the domain is all x so that√

3x 6= 0 and 3x > 0. Hence, it is all x so that x > 0.

For the function g the domain is all x so that 3x2 + 1 ≥ 0 and 2x2 + 1 6= 0. We see first that 2x2 + 1 6= 0for any x. For the other inequality

3x2 + 1 > 0⇒ 3x2 > −1⇒ x2 >1

3⇒ |x| > 1√

3

Hence, x < − 1√3

and x >1√3

. Inequalities are discussed in more detail in Section 1.3 of the textbook.

For the function h the domain is all x so that x2 − 4 6= 0. These are all x except 2 and −2. �

Exercises1. Do parts (A)–(C) of the exercise titled “Explore and Discuss 1” in Section 1.1 of the textbook.

2. Do the exercise titled “Explore and Discuss 2” in the same section.

3. Do odd-numbered exercises 1–7, 15–35, and 39–69 in Section 1.1 of the textbook.

4. Do exercises 81–84 in Section 1.1 of the textbook.

5. Solve odd-numbered problems 85–91 in the same section.


(Rev. C4)

Elementary Functions: Graphs and TransformationsIndications1. Read Section 1.2 of the textbook, “Elementary Functions: Graphs and Transformations.” You may skip

the paragraphs marked “Explore and Discuss.”



CommentsIn this section, we will use the notation of the basic elementary functions listed on page 17 of the textbook.You must know the graphs of the six basic functions shown on page 18.

The shifting of a function’s graph is caused by the addition or substraction of a constant.

Let C be a positive constant and u(x) be a function.

• If the constant C is added to the function u, then the graph of

U(x) = u(x) + C

is a vertical shift C units up of the graph of u.

• If the constant C is subtracted from the function u, then the graph of

U(x) = u(x)− C

is a vertical shift C units down of the graph of u.

• If the constant C is added to the independent variable x, then the graph of

U(x) = u(x+ C)

is a horizontal shift C units to the left of the graph of u.

• If the constant C is subtracted from the independent variable x, then the graph of

U(x) = u(x− C)

is a vertical shift C units to the right of the graph of u.

Example 1.5. If the graph of the function u(x) is as shown on Figure 1.1 on the next page,

then U(x) = u(x) + 2 is the shift 2 units up of u(x), as shown in Figure 1.2 on the following page;

and U(x) = u(x− 2) is the shift 2 units to the right of u(x), as shown in Figure 1.3 on the next page.


(Rev. C4)

Figure 1.1 Basic graph of the function u

Figure 1.2 Graph of the function U(x) = u(x) + 2

Figure 1.3 Graph of the function U(x) = u(x− 2)

To identify the shifts in the function U(x) = u(x− 2) + 2, we look at the transformations applied to obtainU .

• Subtract 2 from x and apply u to obtain u(x− 2)—a 2-unit shift to the right.

• Add 2 to u(x− 2)—a shift of the graph of u(x− 2) by 2 units up.

Summarizing: U consists of a 2 unit shift to the right followed by a shift of 2 units up. The graph is shownin Figure 1.4 on the following page.

The vertical stretching and contraction of a function’s graph results from multiplication by a constant.

Let C > 1 be a positive constant and u(x) be a function.


(Rev. C4)

Figure 1.4 Graph of the function U(x) = u(x− 2) + 2 �

• If the constant C multiplies the function u, then the graph of

U(x) = Cu(x)

is a vertical stretch of the graph of u by C units.

• If the constant1

Cmultiplies the function u, then the graph of

U(x) =1

Cu(x)

is a vertical compression of the graph of u by1

Cunits.

• If the constant C multiplies the independent variable x, then the graph of

U(x) = u(Cx)

is a horizontal contraction of the graph of u by1

Cunits.

• If the constant1

Cmultiplies the independent variable x, then the graph of

U(x) = u( xC

)is a horizontal stretch of the graph of u by C units.

Example 1.6. If the graph of the function u(x) is as shown on Figure 1.1 on the previous page, thenU(x) = 2u(x) is a vertical stretch of the graph of u by 2 units, as shown in Figure 1.5 on the followingpage.

The function U(x) = u(x

2

)is a horizontal stretch of the graph of u by 2 units, as shown in Figure 1.6 on

the next page.

To identify the transformations in the function U(x) = 2u(x

2

), we again look at the transformations

applied to obtain U .


(Rev. C4)

Figure 1.5 Graph of the function U(x) = 2u(x)

Figure 1.6 Graph of the function U(x) = u(x

2

)

• Multiply x by1

2, and apply u to obtain u

(x2

)—a horizontal stretch of the graph of u by 2 units.

• Multiply 2u(x

2

)by 2—a vertical stretch of the graph of u

(x2

)by 2 units.

Summarizing: U consists of a horizontal stretch of the graph of u by 2 units, followed by a vertical stretchby 2 units. The graph is shown in Figure 1.7.

Figure 1.7 Graph of the function U(x) = 2u(x

2

)�

The reflection of a function’s graph results from multiplication by −1.

Let u(x) be a function.


(Rev. C4)

• If the function u is multiplied by −1, then the graph of

U(x) = −u(x)

is a reflection of the graph of u with respect to the x-axis.

• If the independent variable x is multiplied by −1, then the graph of

U(x) = u(−x)

is a reflection of the graph of u with respect to the y-axis.

Example 1.7. If the graph of the function u(x) is as shown in Figure 1.1 on page 9, then U(x) = −u(x)is the reflection of the graph of u(x) with respect to the x-axis, as shown in Figure 1.8.

Figure 1.8 Graph of the function U(x) = −u(x)

The graph of the function U(x) = u(−x) is the reflection of the graph of u with respect to the y-axis asshown in Figure 1.9.

Figure 1.9 Graph of the function U(x) = u(−x)

To identify the transformations in the function U(x) = −u(−x), we again look at the transformationsapplied to obtain U .

• Multiply x by −1 and apply u to obtain u(−x)—a reflection of the graph of u with respect to thex-axis.

• Multiply u(−x) by −1—a reflection of the graph of u(−x) with respect to the y-axis.


(Rev. C4)

Summarizing: U consists of a reflection of the graph of u with respect to the x-axis, followed by areflection of the graph of u(−x) with respect to the y-axis, as shown in Figure 1.10. �

Figure 1.10 Graph of the function U(x) = −u(−x)

The difficulty in sketching the graph of a function U from a basic function u is in identifying eachtransformation and the order in which it is applied. Proceed as if you were to use a calculator to evaluate U .

Example 1.8. Let u be the function whose graph is as shown on Figure 1.1 on page 9. To identify thetransformations in the function

U(x) = −2u(2x)− 3,

we again look, step-by-step, at the transformations applied to obtain U .

Let us begin with the original graph of u(x):

Figure 1.11 Graph of the function U(x) = −2u(2x)− 3

• u(2x): Multiply the independent variable x by 2 and apply u—the graph of u is horizontally

compressed by1

2units.

• 2u(2x): Multiply u(2x) by 2—the graph of u(2x) is vertically stretched by 2 units.

• −2u(2x): Multiply 2u(2x) by −1—the graph of 2u(2x) is reflected with respect to the x-axis.

• −2u(2x)− 3: Subtract 3 from −2u(2x)—the graph of −2u(2x) is shifted 3 units down.

Figure 1.15 on page 15 shows all four transformations used to obtain the graph of U from the graph of u. �


(Rev. C4)

Figure 1.12 Graph of the function u(2x)

Figure 1.13 Graph of the function 2u(2x)

Figure 1.14 Graph of the function U(x) = −2u(2x)

Go over the tutorial “Transformations.”

A function F (x) is “piecewise defined” when the mathematical model of the dependent variable F variesaccording to the different ranges of the independent variable x.

The evaluation of a piecewise function depends on the location of the independent variable. Its domain isthe union of all the ranges of the independent variable.


(Rev. C4)

Figure 1.15 Graph of the function U(x) = −2u(2x)− 3

Example 1.9. Let C(x) be the cost function of Example 7 in Section 1.2 of the textbook. To evaluateC(35.5), we locate 35.5 in the range 5 < 35.5 ≤ 40 and we have

C(35.5) = 3.933 + 0.4601(35.5− 5) = 17.96605

The domain of this function is zero and all positive numbers, because the first range of x gives the interval[0, 5], the second (5, 40] and the third (40,∞). The union of these 3 intervals is all x ≥ 0. �

Example 1.10. The domain of the piecewise function

g(x) =

3

4x− 1if x <

1

4

4x− 1 if1

4≤ x < 6

2x2 + 35 if 6 < x < 24

is the union of the intervals(−∞, 1

4

),[

1

4, 6

)and (6, 24). Hence, is all real numbers less than 24

except 6. To evaluate this function at −3 we choose the first range of x.

f(−3) =3

4(−3)− 1= − 3

13.

For x = 1 we choose the second

f(1) = 4(1)− 1 = 3.

For x = 6.01 we choose the third

f(6.01) = 2(6.01)2 + 35 = 107.2402. �

Exercises6. Do the exercises titled “Explore and Discuss 2,” in Section 1.2 of the textbook.

7. Do odd-numbered exercises 1–25, 35–41, 45, 47, 49, 51, in Section 1.2 of the textbook.

8. Solve odd-numbered problems 67, 71, 73, 75, in the same section.


(Rev. C4)

Linear and Quadratic FunctionsIndications1. Read the subsection “Linear Functions, Equations, and Inequalities” in Section 1.3 of the textbook.



Comments

Remember that a fractiona

bis equivalent to the fraction

c

dif and only if ad = bc; that is,

a

b=c

dif and only if ad = bc.

Hence, if

5

x− 1=

6

3− 2x

we can obtain the linear equation

(3− 2x)5 = 6(x− 1).

Solving, we have

5(3− 2x) = 6(x− 1)

15− 10x = 6x− 6

15 + 6 = 6x+ 10x

21 = 16x

x =21

16

Observe that a linear equation has only one solution.

To verify a solution we should replace the value of the solution on each side of the equation and compare toensure that both are equal. We have

5

x− 1=

6

3− 2x

and the solution we obtain is x =21

16.


(Rev. C4)

So, on the left side we get

521

16− 1

=55

16

=5× 16

5= 16,

and on the right side we get

6

3− 2

(21

16

) =6

3− 21

8

=63

8

=6× 8

3= 16.

These results shows that

521

16− 1

=6

3− 2

(21

16

) ,and our solution is correct.

Example 1.11. Solve and check

4x− 3

3− 6x=

4

7.

You should verify each of the following steps.

4x− 3

3− 6x=

4

7

7(4x− 3) = 4(3− 6x)

28x− 21 = 12− 24x

28x+ 24x = 12 + 21

52x = 33

x =33

52

Check

4

(33

52

)− 3

3− 6

(33

52

) =− 6

13

−21

26

=6× 26

13× 21=

4

7�

It is very useful to be able to visualize concepts in mathematics. For inequalities, consider the real line withpositive numbers to the right and negative numbers to the left of 0.

Remember that the farther to the right a positive number is, the larger it is; and the farther to the left anegative number is, the smaller it is. Conversely, the closer to zero a positive number is, the smaller it is,and the closer to zero a negative number is, the larger it is.


(Rev. C4)

Also observe that a < b if and only if either b− a > 0 or a− b < 0.

Example 1.12. It is true that5

13<

7

8because

5

13− 7

8=

5(8)− 7(13)

8(13)= − 51

104< 0. �

Inequality Properties

1. If c > 0 and a < b, then ac < bc.

2. If c < 0 and a < b, then ac > bc.

Property 1 is saying that when we multiply an inequality by a positive number, the sense or direction of <remains the same. Property 2 says that when we multiply an inequality by a negative number the sense ordirection of < changes to >.

Remember that division by the number d is the same as multiplying by the number1

d. Hence, properties 1

and 2 also apply for division. As a consequence, we have

1. For either b > 0 and d > 0, or b < 0 and d < 0

a

b<c

dif and only if ad < bc.

2. For either b > 0 and d < 0, or b < 0 and d > 0

a

b<c

dif and only if ad > bc.

From the properties of the inequalities we can see why this is so.

Ifa

b<c

dand we apply property 1 with the numbers b, d > 0, then the direction of the inequality does not

change, and

a

b<c

d⇒ abd

b<cbd

d⇒ ad < cb.

Conversely, if ab < cd and we apply property 1 with the numbers1

b,

1

d> 0, then the direction of the

inequality does not change, and

ab < cd⇒ ab

b<cd

b⇒ a <

cd

b⇒ a

d<cd

db=c

b

Ifa

b<c

dand we apply property 1 with the numbers b, d < 0, then the direction of the inequality changes

twice, and

ab

b>cb

d⇒ a >

cb

d⇒ ad <

cbd

d= cb.


(Rev. C4)

We leave the converse of this statement and the second statement above to you as an exercise.

Example 1.13. It is true that

5

13<

7

8,

because 5(8) < 7(13); and it is not true that

− 4

15< − 3

11,

because −4(11) > −3(15). �

For fractions we also have

a

b> 0 if either a > 0, b > 0 or a < 0, b < 0

and

a

b< 0 if either a < 0, b > 0 or a > 0, b < 0.

Example 1.14. To solve the inequality

1

x+ 3<

4

5,

we could multiply both sides by x+ 3, but we would then have to consider two cases: x+ 3 positive andx+ 3 negative. [Why?] Instead, we consider the difference

1

x+ 3− 4

5=

5− 4(x+ 3)

5(x+ 3)=−7− 4x

5x+ 15< 0.

To see for which x this fraction is negative, we consider two cases:

Case 1−7− 4x < 0 and 5x+ 15 > 0.

Solving, we have x > −7

4and x > −3.

Since −7

4> −3, both inequalities are satisfied if x > −7

4.

Case 2−7− 4x > 0 and 5x+ 15 < 0.

Solving, we have x < −7

4and x < −3.

Since −7

4> −3, both inequalities are satisfied if x < −3.


(Rev. C4)

Hence, the solution set of this inequality is the union of these two intervals; that is, (−∞,−3)∪ (−7/4,∞)(see Figure 1.16). Neither endpoint of these intervals (i.e., −3 and −7/4) satisfies the inequality. [Why?] �

−3

74

– 0

Figure 1.16 Solution set

It is not necessary to use a graphing utility to sketch a linear equation. It is enough to plot two points, sinceone and only one line passes through two distinct points.

If, in the linear equation Ax+By = C, the constant C is nonzero, we plot the

x-intercept(C

A, 0

)and the

y-intercept(

0,C

B

).

These two points determine the line completely.

If the constant C is zero, then Ax+By = 0 and the origin (0, 0) is one point on the line. [Why?]

We then choose any other point by setting one of the variables to equal any nonzero value, and solving forthe other.

If, for example, we set x = 1, then A+ yB = 0 and y = −AB

.

The point(

1,−AB

), together with (0, 0), determines the line completely.

Example 1.15. To sketch the line given by the equation

3x− y =2

3

we have the x-intercepts(

2

9, 0

)and the y-intercept

(0,−2

3

).

Plot these two points, and draw the line passing through them. �

Example 1.16. To sketch the line given by the equation

y = 3x+ 5,

we have the y-intercept (0, 5) and the x-intercept(−5

3, 0

).


(Rev. C4)

Plot these two points and draw the line passing through them. �

Table 2 in Section 1.3 of the textbook lists the different forms of an equation of a line. From any one ofthese forms it is possible to obtain the others.

Example 1.17. We obtain the slope-intercept form of a line with standard equation 2x− 5y = 7 bysolving for the variable y

2x− 5y = 7

2x− 7 = 5y

y =2x

5− 7

5

The point-slope form is obtained from the slope-intercept form by factoring the slope m

y =2x

5− 7

5

y =2

5

(x− 7

2

)y − 0 =

2

5

(x− 7

2

)�

Example 1.18. We obtain the standard form of a line with the slope-intercept form

y =3

2x− 3

by solving for −3. Thus,

y − 3

2x = −3.

We then multiply by 2 and obtain the equivalent form

2y − 3x = −6. �

Exercises9. Do odd-numbered exercises 1–21 in Section 1.3 of the textbook.

Indications1. Read subsections “Quadratic Functions, Equations, and Inequalities” and “Properties of Quadratic

Functions and Their Graphs,” in Section 1.3 of the textbook. You may skip the paragraph marked“Explore and Discuss 1.”

2. Read the “Comments” section on the following page.


(Rev. C4)


CommentsTechniques for sketching a quadratic function, and the properties of a quadratic function are presented inthe “Summary” in Section 1.3 of the textbook. It is important that you know how to sketch these functionsproperly and without the aid of a calculator.

It is useful to know that a quadratic function

f(x) = ax2 + bx+ c

has a vertex at

x = − b

2a

and that the maximum or minimum value is

f

(− b

2a

).

The interpretation of the vertex is also important. The revenue and profit functions are quadratic if the costand price-demand functions are linear. The revenue function is concave down, and the vertex is themaximum revenue, the profit function is also concave down, and the vertex is the maximum profit.

Example 1.19. Let

f(x) = −0.0117x2 + 0.32x+ 17.9

be the market share function described in Exercise 80 in Section 1.3 of the textbook.

This function is concave down; therefore, it has a maximum at the vertex which corresponds to

x = − 0.32

2(−0.0117)≈ 13.67.

Since x = 13.37 corresponds to the year 1980 + 13.37 ≈ 1993, we see that early in 1993, the market sharewas maximum at approximately f(13) ≈ 20.08%.

Observe that f(13) = 20.0827 < 20.2, and 20.2% corresponds to the year 2000. This discrepancy occursbecause the mathematical model f is no more than a close approximation to the data. �

Example 1.20. Let

R(x) = 2000x− 60x2

be the revenue function described in Exercise 84 in Section 1.3 of the textbook.


(Rev. C4)

This function is concave down, and has a maximum at

x = − 2000

2(−60)=

100

6≈ 16.66.

That is, if 16.66 thousand notebook computers are sold, the revenue is at a maximum, and the maximumrevenue is

R

(100

6

)≈ $16, 666.66.

On the other hand, 16.66 thousand notebooks can be sold if the price is

p

(100

6

)= $1000.00 �

Every parabola is the transformation of the basic quadratic function f(x) = x2. See that if a > 0,

q(x) = a(x− h)2 + k

is the shift to the right of x2 by h units, the vertical stretch (if a > 1) or a vertical compression (if a < 1) bya units, followed by a shit up of k units.

Example 1.21. In the quadratic function

q(x) =1

4(x− 6)2 + 3

the vertex is (6, 3) and we obtain its graph by shifting the graph of x2 six units to the right, compress it by aquarter and shift it up 3 units.

12

3

6

Figure 1.17 The graph of1

4(x− 6)2 + 3 �


(Rev. C4)

It is then useful to obtain the vertex form of a quadratic. This is done by completing the perfect square. SeeExample 4 in Section 1.3 of the textbook and the tutorial “Quadratic Functions.”

The linear equation y = mx+ b is a line and the solution set of the inequalities mx+ b > 0 or mx+ b < 0correspond to the values of x to the left or right of this line. To determine which one, it is enough to test theinequality at one value of x.

Example 1.22. Sketch the line y = 3x− 7. The solution set given by the inequality 3x− 7 ≥ 0 is on theright or left of the line, and correspond to all points x so that y = 3x− 7 ≥ 0. To determine the region wechoose x = 0 (you may choose another). Hence, for x = 0

y = 3(0)− 7 = −7 6≥ 0.

The solution set does not include x = 0, which is on the left of the line.

7/3

-7

solu�on set of y = − 7 ≥ 0

Figure 1.18 The solution set of 3x− 7 ≥ 0

The solution set is all x ≥ 7

3on the right side of the line. From Figure 1.18, we see that y ≥ 0 for these

values of x. �

Similarly, the solution set of a quadratic ax2 + bx+ c > 0 or ax2 + bx+ c < 0 corresponds to the x insideor outside the parabola. To determine it, we again choose a value of x.

Example 1.23. Sketch the parabola y = −2x2 + 12x− 16. The solution set given by the inequality−2x2 + 12x− 16 ≤ 0 are the values of x inside or outside the parabola. To determine which one, wechoose x = 0. Hence,

−2(0) + 12(0)− 16 = −16 ≤ 0.


(Rev. C4)

2

2 3 4

Figure 1.19 The solution set of −2x2 + 3x+ 5 ≤ 0

The value of 0 is in the solution set, and the solution set is outside the parabola. From Figure 1.19, thesolution set is the union of the intervals (−∞, 2] ∪ [4,∞). Check that the values of x = 2, 4 satisfy theinequality. �

Exercises10. Do odd-numbered exercises 23–53 on in Section 1.3 of the textbook.

Indications1. Read the subsections, “Applications” and “Linear and Quadratic Regression,” pages 39–44 of the

textbook.



CommentsIn this section, the mathematical model used to solve the problems presented is a linear model (as in theprevious section). To obtain this linear model we use “regression analysis,” a process through with whichwe find a function that best fits the data provided. Intuitively, the graph of the mathematical model is theline that is closest to or includes the data plotted on the coordinate plane. In all the examples we consider,the regression model is provided and obtained with the help of the linear regression function of a calculator.If you have a suitable calculator, you may want to obtain the equation yourself, but it is not necessary thatyou do so. What is important in all these examples is to identify and establish

• the variables (which variable depends on what), and

• the correct interpretation of the slope.

From the data provided, we must decide which quantity depends on the other.


(Rev. C4)

From Table 4 on page 43 of the textbook, we can see that the price p of the diamonds depends on theirweight in carats c; hence, p(c). From the slope of linear equation

y = 6137.4x− 478.9,

on page 43, we can see that the price of the diamonds increases at a rate of $6,137.00 dollars per carat.

Example 1.24. Let w be the ideal body weight of man and h be the man’s height, in inches, over 5 feet.The weight depends on the height; thus, w(h). The linear model is

w(h) = 1.9h+ 52.

The ideal body weight of a man who is 5 feet tall is 52 kg.

The slope indicates that for each inch in height over 5 feet, the ideal weight increases by 1.9 kg.

The ideal body weight of a 5′8′′ tall man is w(8) = 1.9(8) + 52 = 67.2 kg.

To determine the height of a man with an ideal body weight of 70 kg, we would use

70 = 1.9h+ 52

and would solve for h

h = 9.47.

Thus, his height should be approximately 5′9.5′′. �

Problem solving skills are developed with practice. The recommended steps on page 29 of the textbookmay help you to acquire these skills. Note that you must aim to develop your own strategies for solvingproblems.

Example 1.25. A jazz concert brought in $165,000 on the sale of 8,000 tickets. If the tickets sold for $15and $25 each, how many of each type of ticket were sold?

We read the problem and express it using our own words.

There are two types of tickets: those of $15.00 and those of $25.00. The total of tickets sold was 8,000 andthe monetary value of all the tickets sold is $165,000, we want to know how many tickets of each type weresold.

Thus, we are looking for two unknown quantities: x, the number of tickets of $15.00, and y, the number oftickets of $25.00.

The information given indicates that

x+ y = 8000 and 15x+ 25y = 165000.

These two equations are the mathematical model of the problem. The answer is the solution of these twoequations.


(Rev. C4)

The equation 15x+ 25y = 165000 is equivalent to 3x+ 5y = 33000; and from the first equation,x = 8000− y. Hence,

3(8000− y) + 5y = 33000

24000 + 2y = 33000

12000 + y = 16500

y = 4500

We then have x = 8000− 4500 = 3500.

The answer is: 3,500 tickets of $15.00 and 4,500 tickets of $25.00. �

Example 1.26. The publisher of a new book called Muscle-Powered Sports figures fixed costs at $92,000and variable costs at $2.10 for each book produced. If the book is sold to distributors for $15 each, howmany must be sold for the publisher to break even?

Express this problem using your own words.

From the question, we are looking for the number of books the publisher would need to sell, say x, in orderto break even. We need to understand what we mean by “break even”: at the break-even point, the cost ofproducing the books is equal to the total amount received from the sale of the books.

The cost is

C(x) = 92000 + 2.10x

and the total amount received from the sale of the books is

T (x) = 15x.

The break even point is

92000 + 2.10x = 15x.

Solving this equation,

92000 + 2.10x = 15x

92000 = 15x− 2.10x

92000 = 12.90x

x = 7131.78,

we find that the publisher needs to sell 7,132 books to almost break even, without losing any money. If thepublisher sells 7,131 books, they almost break even but lose a small amount of money. �

Exercises11. Do odd-numbered exercises 65–81 on pages 46–48 of the textbook.


(Rev. C4)

Polynomial and Rational FunctionsIndications1. Read Section 1.4 of the textbook, “Polynomial and Rational Functions.”



CommentsA polynomial of degree zero is a constant function: f(x) = a0.

A polynomial of degree one is a linear function: f(x) = ax+ b.

A polynomial of degree two is a quadratic function: f(x) = ax2 + bx = c.

Observe that the graphs of polynomial functions are continuous; that is, their graphs do not break.Therefore, polynomials do not have vertical asymptotes.

The domain of a rational function

f(x) =n(x)

d(x),

where n(x) and d(x) are polynomials, is given by the set of all x such that d(x) 6= 0.

Example 1.27. The domain of the function

h(x) =3x− 6

2x2 − 2x− 12

is the set of all x such that 2x2 − 2x− 12 6= 0; that is,

2(x2 − x− 6) = 2(x− 3)(x+ 2) 6= 0

The domain is all x except 3 and −2. �

In Example 2 in Section 1.4 of the textbook, the authors of your textbook suggest steps (A)–(E) to sketchthe graph of rational functions. These are only for rational functions. The same is true for the procedure onpage 56, to find vertical and horizontal asymptotes for rational functions. This procedure does not apply toother type of functions.

Exercises12. Do odd-numbered exercises 1–9, 23–27, 33–39, in Section 1.4 of the textbook.

13. Do parts (A)–(C) of exercises 47, 49 and 51 in Section 1.4 of the textbook.


(Rev. C4)

14. Solve problems 57 and 59, in the same section of the textbook.

Exponential FunctionsIndications1. Read Section 1.5 of the textbook, “Exponential Functions.” You may skip the paragraphs marked

“Explore and Discuss.”



CommentsFor each positive real number b, we define an exponential function y = bx. This exponential function is ofbase b.

In Unit 4, you will see why the number e is most suitable for solving problems whose mathematical modelrequires an exponential function.

The mathematical model of both exponential population growth and exponential radioactive decay is theexponential function

f(x) = Pekt

where P is the initial population or initial amount of radioactive substance, and k is a constant indicatingthe rate of growth or decay.

Hence for k > 0 the model corresponds to a exponential growth, and for k < 0 the model is of exponentialdecay. Observe that for k > 0 the function is increasing, for k < 0, the function is decreasing as in Figure 2on page 63 of the textbook.

In the same way that a linear regression is the linear function that best fits a data set, so the exponentialregression is the function of the form y = abx that best fits a data set.

Observe that for this model, y(0) = a and y(1) = ab.

On page 69, the authors of your textbook present a summary of compound interest. Observe the differencebetween compound and continuous compound interest. In compound interest, the money is compounded mtimes a year, and in continuous compound interest the money is compounded continuously throughout theyear.

Exercises15. Do the exercises titled “Explore and Discuss 1” and “Explore and Discuss 2” in Section 1.5 of the

textbook.


(Rev. C4)

16. Do odd-numbered exercises 1–19 and 39–45 in the same section.

17. Solve odd-numbered problems 51–59, 63–67 in the same section.

Logarithmic FunctionsIndications1. Read Section 1.6 of the textbook, “Logarithmic Functions.” You may skip the paragraphs marked

“Explore and Discuss.”



The concept of one-to-one functions and inverse functions is introduced in this section, because the inversefunction of an exponential function of base b is the logarithmic function of base b.

Visually, a function f is one-to-one if any horizontal line intersects the graph of f at one and only one point.

In Figure 1.20, we show a function that is not one-to-one. This function is not invertible. Note that thehorizontal line intersects the graph of the function at more than one point.

Figure 1.20 Graph of a function that is not one-to-one

When a function f is invertible (one-to-one), its inverse is a function g whose graph is a reflection of thegraph of f with respect to the line y = x. In Figure 1.21 on the following page, we have the graphs of aninvertible function f and its inverse g.

Observe that if (a, b) is on the graph of f—that is, if f(a) = b—then the pair (b, a) is on the graph ofg—that is, g(b) = a. Therefore, the composition of these two functions gives

f(g(b)) = f(a) = b and g(f(a)) = g(b) = a.

The logarithmic function of base b is the inverse of the exponential function of base b. Thus, the followingcancellation laws hold

blogb x = x and logb(bx) = x


(Rev. C4)

Graph of fGraph of g

Figure 1.21 Graph of an invertible function and its inverse

For the particular exponential function ex, the inverse is the natural logarithm lnx. Hence,

elnx = x and ln(ex) = x.

The cancellation law says that

logb x is the power to which we must raise b in order to obtain x.

Thus, log2(64) = 6 because 26 = 64.

Moreover, the properties of the exponential functions listed on page 64 of the textbook, and the propertiesof the logarithmic functions listed on page 76 of the textbook correspond to one another. For example,

b0 = 1 corresponds to logb 1 = 0

and

bxby = bx+y corresponds to logb x+ logb y = logb(xy).

We leave discovering the others as an exercise.

We use the cancellation laws and the properties of the exponential and logarithmic functions to solveequations.

Example 1.28. You must identify the properties used. See how the cancellation law is applied.

ex2−3 = 6

ln(ex2−3) = ln(6)

x2 − 3 = ln 6

x = ±√

ln 6 + 3 �

Example 1.29. To solve

log8(x2 + 3) = log2(1 + x),


(Rev. C4)

we apply the “Remark” on page 79 of the textbook, and use the fact that

ln 8 = ln(23) = 3 ln 2.

log8(x3 + 3) = log2(1 + x)

ln(x3 + 3)

ln 8=

ln(1 + x)

ln 2

ln(x3 + 3)

3 ln 2=

ln(1 + x)

ln 2

ln(x3 + 3) = 3 ln(1 + x)

ln(x3 + 3) = ln[(1 + x)3]

x3 + 3 = x3 + 3x2 + 3x+ 1

3x2 + 3x− 2 = 0

x =−3±

√33

6�

Exercises18. Do odd-numbered exercises 1–53 and 65–75, in Section 1.6 of the textbook.

19. Solve odd-numbered problems 83–87, in the same section of the textbook.

Finishing this Unit1. Review the objectives of this unit and make sure you are able to meet each of them.

2. Study the section of the “Chapter 1: Summary and Review” titled “Important Terms, Symbols andConcepts” at the end of Chapter 1 of the textbook.

3. If there is concept, definition, example or exercise that it is not yet clear to you, go back and re-read it.Contact your tutor if you need help.

4. You may want to do problems 1–3, 5, 7, 11, 22, 24, 27, 29, 30, 33, 37, 41, 44, 49, 68, 69, 77, 79, 82,84, 87, and 91 from the “Review Exercises” section at the end of Chapter 1.

5. Complete the “Practice Examination” provided for this unit. Evaluate yourself, first checking youranswers against those provided in the “Answers” section at the end of the textbook, and thencomparing your solutions with those provided on pages 1–37, 1–50 of the Student’s Solutions Manual.The number of points in a question may indicate the number of steps in the solution. Give yourself fullcredit if your answer is correct and you give a complete solution, even if your solution differs from thatshown in the Student’s Solutions Manual.


(Rev. C4)

Practice ExaminationTime: 1.5 hoursTotal Points: 46Passing grade: 55%

First, answer the question below.

1. Explain “exponential regression.” Marks: 2 pts

Then, do the following exercises from the “Review Exercises” at the end of Chapter 1 of the textbook.

To obtain full credit you must justify all your answers and show your work.

2. Exercise 25 Marks: 2/2/2/2 for a total of 8 pts

3. Exercise 35 Marks: 4 pts







(Rev. C4)

UNIT 2

Limits and the Derivative

Calculus is the area of mathematics that studies the dynamics of changing variables—that is, the rate ofchange. We have already considered the slope of a line as a rate of change. In this unit, we find the rate ofchange of other algebraic functions.


1. identify the largest set on which a function is continuous.

2. evaluate limits of algebraic functions.

3. interpret the derivative of a function as an instantaneous rate of change.

4. apply the rules of differentiation to differentiate a wide variety of algebraic functions.

5. apply the concept of marginal analysis to solve problems in economics.


• factor polynomials◦ Section A.3, pages 516–521 of the textbook◦ Section A.7, pages 543–544 of the textbook◦ Book V, chapter 2 of Language of Mathematics

• divide polynomials◦ Book V, Section 1.3 of Language of Mathematics◦ Book VI, Section 2.5 of Language of Mathematics

• rationalize expressions with radicals◦ Book VI, Section 2.8 of Language of Mathematics


(Rev. C4)

Introduction to LimitsIndications1. Read pages 92–97 of Section 2.1, “Introduction to Limits” in the textbook (to “Limits: An Algebraic

Approach”).



CommentsWe evaluate limits visually when we use the graph of a function to obtain those value. To evaluate the limit

limx→c

f(x)

we see the value at which eventually the dependent variable f will attain as the independent variable xapproaches a fixed value c from the left and from the right.

So, the more graphs of functions we know, the easier it is to evaluate limits. In Table 2.1 on the next page,we show the graphs of some basic functions. You should have memorized most of them as you workedthrough Unit 1.

From the graphs of these functions, we have the following limits. Note that the number c is any number inthe domain of the corresponding function.

1. For any constant function F (x) = b,

limx→c

b = b.

2. From the identity function I(x) = x,

limx→c

x = c.

3. From the quadratic function F (x) = x2,

limx→c

x2 = c2.

4. From the cubic function F (x) = x3,

limx→c

x3 = c3.

5. From the absolute value function F (x) = |x|,

limx→c|x| = |c|.


(Rev. C4)

Table 2.1 Graphs of basic functions

Function Graph

F (x) = c Constant function

F (x) = x Identity function

F (x) = x2 Basic quadratic function†

F (x) = x3 Basic cubic function

F (x) = |x| Absolute value function

F (x) =1

x

F (x) =√x Basic square root function

†We refer to the function F (x) = x2 as the “basic” quadratic function to distinguish it from other quadratic functions, such asG(x) = x2 + 4. We do the same for the cubic and square root functions.


(Rev. C4)

6. From the reciprocal function F (x) =1

xfor c 6= 0,

limx→c

1

x=

1

c.

7. From the square root function F (x) =√x,

limx→c

√x =√c

for c > 0 and

limx→0+

√x = 0.

Example 2.1.

a. limx→−2

x = −2

b. limx→6

√x =√

6

c. limx→2

1

x=

1

2

d. limx→1/3

x8 =1

38�

Pay attention to Theorem 1 (on page 95 of the textbook). This limit says that

limx→a

f(x) = L if and only if limx→a−

f(x) = L = limx→a+

f(x).

Definitions of limit and one-sided limits are on pages 94–95 of the textbook.

To conclude that a limit does not exist we must show that

limx→a−

f(x) = M 6= L = limx→a+

f(x).

Example 2.2. Let

f(x) =

3x− 1 if x < −1

1

x2 + 1if −1 < x < 3

x− 1

x2 + 3if x ≥ 3

This function is defined on the intervals (∞,−1), (−1, 3), [3,∞). We want to evaluate the limits at −1 and3 from the right and left. For x→ −1−, the function is a line 3x− 1. Thus,

limx→−1−

f(x) = limx→−1−

3x− 1 = 3(−1)− 1 = −4.


(Rev. C4)

For x→ −1+, the function is rational1

x2 + 1and

limx→−1+

f(x) = limx→−1+

1

x2 + 1=

1

2.

For x→ 3− the function is again the same rational function1

x+ 1. The only point where the graph of this

function is not defined is at −1. Hence,

limx→3−

f(x) = limx→3−

1

x2 + 1=

1

9 + 1=

1

10.

For x→ 3+ the function is the rational functionx− 1

x2 + 3. This rational function does not have vertical

asymptotes. [Why?] So its graph does not break and

limx→3+

f(x) = limx→3+

x− 1

x2 + 3=

3− 1

32 + 3=

2

12.

In summary, we have

limx→−1−

3x− 1 = −4 6= 1

2= lim

x→−1+

1

x+ 1

and therefore the limit at −1 does not exist.

limx→3−

1

x+ 1=

1

106= 1

6= lim

x→3+

x− 1

x2 + 3.

Again the limit at 3 does not exist. �


Indications1. Read the subsection “Limits: An Algebraic Approach” up to “Conceptual Insight” (pages 97–100) of

Section 2.1. Note that you will find it helpful to memorize the “Properties of Limits” given inTheorem 2 on page 97.



CommentsThe limits listed on page 35 of this manual, together with the properties of limits listed in Theorem 2 onpage 97 of the textbook, give the limits of many functions, among them polynomial and rational functions.


(Rev. C4)

If

P (x) = anxn + an−1x

n−1 + an−2xn−2 + ...+ a1x+ a0

is a polynomial of degree n, then by limits 3, 5, 2 and 1, applied in that order, for any number c,

limx→c

P (x) = limx→c

anxn + an−1x

n−1 + an−2xn−2 + ...+ a1x+ a0

= limx→c

anxn + lim

x→can−1x

n−1 + ...+ limx→c

a1x+ limx→c

a0

= an limx→c

xn + an−1 limx→c

xn−1 + ...+ a1 limx→c

x+ limx→c

a0

= ancn + an−1c

n−1 + an−2cn−2 + ...+ a1c+ a0

= P (c)

Hence,

limx→c

P (x) = P (c) (2.1)

Example 2.3.

limx→1/2

5x3 − x2 + 3x− 1 = 5

(1

2

)3

−(

1

2

)2

+ 3

(1

2

)− 1

= 5

(1

8

)−(

1

4

)+ 3

(1

2

)− 1

=7

8�

Remark:

1. Theorem 2 (on page 97 of the textbook) can only be applied if the limits of f(x) and g(x) at c are bothfinite.

2. Each limit in this theorem is also valid if x→ c is replaced by x→ c+ and x→ c−.

Theorem 3 (p. 98) and Limit 8 of Theorem 2 (p. 97) give the following limit for any odd number n, anypolynomial P and any number c:

limx→c

n√P (x) = n

√P (c).

For n even, we must have that P (c) ≥ 0, to conclude that

limx→c

n√P (x) = n

√P (c).

Example 2.4. From Example 2.3 we have

limx→1/2

4√

5x3 − x2 + 3x− 1 = 4

√limx→1/2

5x3 − x2 + 3x− 1 =4

√7

8�


(Rev. C4)

Limit 7 of Theorem 2 (p. 97) states that if

• R(x) =P (x)

Q(x)is a rational function, and

• limx→c

Q(x) = Q(c) 6= 0,

then

limx→c

R(x) =limx→c

P (x)

limx→c

Q(x)=P (c)

Q(c)= R(c) (2.2)

Example 2.5. To evaluate the limit

limx→−3

3x3 − 4x2 + 1

1− 2x2,

we note that the function is rational and that

limx→−3

1− 2x2 = 1− 2(−3)2 = −17 6= 0.

Hence, by Limit (2.2),

limx→−3

3x3 − 4x2 + 1

1− 2x2=

3(−3)3 − 4(−3)2 + 1

1− 2(−3)2=

116

17�

Limit (2.2) and Limit 8 give the following limit for any odd number n, any rational function R and anynumber c:

limx→c

n√R(x) = n

√R(c).

For n even, we must have that R(c) ≥ 0 to conclude that

limx→c

n√R(x) = n

√R(c).

Example 2.6. Similarly to Example 2.5

limx→−3

√3x3 − 4x2 + 1

−1 + 2x2=

√limx→−3

3x3 − 4x2 + 1

−1 + 2x2=

√116

17�

Example 2.7. By Equation (2.2)

limx→−4

3x3 − 4x2 + 1

−1 + 2x2=

3(−4)3 − 4(−4)2 + 1

−1 + 2(−4)2= −255

31.

Hence,

limx→−4

3

√3x3 − 4x2 + 1

−1 + 2x2=

3

√−255

31. �


(Rev. C4)

We apply Theorem 2 on page 97 of the textbook in the next example.


limx→1

ln(x6)− 3x2 + e2x

4e−x

We apply the properties of the exponential functions (Theorem 2 on page 64 of the textbook) and propertiesof the logarithmic functions (Theorem 1 on page 76 of the textbook)

ln(x6)− 3x2 + e2x

4e−x=

6 ln(x)− 3x2 + [ex]2

4e−x.

Thus,

limx→1

6 ln(x)− 3x2 + [ex]2

4e−x=

6 ln(1)− 3 + e2

4e−1=−3e+ e3

4. �


Indications1. Read Section 2.1, from “Conceptual: Insight” on page 100 up to Example 10 (pages 100–102).



CommentsIf limx→c

f(x) = 0 and limx→c

g(x) = 0, then the limit

limx→c

f(x)

g(x)

is called an indeterminate form of type 0/0.

To evaluate such limits, we factor or use any other algebraic property that allows us to replace the limit witha limit for which the limits in Theorem 2 (page 97 of the textbook) hold.

If the function

f(x)

g(x)

is rational, then the following application of the Fundamental Theorem of Algebra is very useful.


(Rev. C4)

If P (x) is a polynomial of degree n, and P (a) = 0, then a is a root of the polynomial. Thus theFundamental Theorem of Algebra says that

P (x) = (x− a)Q(x)

where Q(x) is a polynomial of degree n− 1. This statement says that it is possible to factor the polynomialP (x) by the factor x− a. To do so, we use known factorization properties or long division.

Example 2.9. Let

P (x) = −x2 + 5x− 2x3 + x4 − 18

be a polynomial of degree 4.

Since P (−2) = 0, we can factor P (x) by dividing by x− (−2) = x+ 2:

x+ 2)x4 − 2x3 − x2 + 5x− 18 arrange terms in the dividend and divisor in decreasing order

of their exponents

x+ 2

x3)x4 − 2x3 − x2 + 5x− 18 divide

x4

x= x3

x+ 2

x3)x4 − 2x3 − x2 + 5x− 18−x4 − 2x3

multiply x3(x+ 2) = x4 + 2x3

note that you will be subtracting, so−(x4 + 2x3) = −x4− 2x3

x+ 2

x3)x4 − 2x3 − x2 + 5x− 18−x4 − 2x3

−4x3 − x2 + 5x− 18

subtractx4 − 2x3 − x4 − 2x3 = −4x3;bring down −x2 + 5x− 18;the new dividend is −4x3 − x2 + 5x− 18

x+ 2

x3 − 4x2)x4 − 2x3 − x2 + 5x− 18−x4 − 2x3

−4x3 − x2 + 5x− 184x3 + 8x2

we repeat the process:

divide−4x3

x= −4x2,

then multiply; note that −(−4x3 − 8x2) = 4x3 + 8x2

x+ 2

x3 − 4x2)x4 − 2x3 − x2 + 5x− 18−x4 − 2x3

−4x3 − x2 + 5x− 184x3 + 8x2

7x2 + 5x− 18

the new dividend is 7x2 + 5x− 18


(Rev. C4)

x+ 2

x3 − 4x2 + 7x− 9)x4 − 2x3 − x2 + 5x− 18−x4 − 2x3

−4x3 − x2 + 5x− 184x3 + 8x2

7x2 + 5x− 18−7x2 − 14x

− 9x− 18− 9x− 18

0

repeating the process twice

The remainder must always be zero; hence,

P (x) = −x2 + 5x− 2x3 + x4 − 18 = (x+ 2)(x3 − 4x2 + 7x− 9). �

Example 2.10. You can verify that the limit

limx→−2

x4 − 2x3 − x2 + 5x− 18

x2 − 4

is an indeterminate form of type 0/0.

By Example 2.9 on the previous page, we have

x4 − 2x3 − x2 + 5x− 18

x2 − 4=

(x+ 2)(x3 − 4x2 + 7x− 9)

(x+ 2)(x− 2)

=x3 − 4x2 + 7x− 9

x− 2.

Thus, as in Example 2.6 on page 40, the given limit is equal to

limx→−2

x3 − 4x2 + 7x− 9

x− 2=

(−2)3 − 4(−2)2 + 7(−2)− 9

−2− 2

=47

4= 11.75. �

Another method we use to evaluate limits of indeterminate form 0/0 is rationalization based on the productof conjugate terms:

(a− b)(a+ b) = a2 + b2.

Example 2.11. The limit

limx→4

√x− 2

x− 4



(Rev. C4)

Observe that

(√x− 2)(

√x+ 2) = x− 4;

hence,√x− 2

x− 4

(√x+ 2√x+ 2

)=

x− 4

(√x+ 2)(x− 4)

=1√x+ 2

Thus,

limx→4

√x− 2

x− 4= lim

x→4

1√x+ 2

=1√

4 + 2=

1

4�

We now turn our attention to Theorem 4 on page 100 of the textbook. As stated, this theorem is not correct.What we have are the following results.

Theorem 2.1. If limx→c−

f(x) = 0, then

a. limx→c−

1

f(x)=∞ whenever f(x) > 0 for x→ c−.

b. limx→c−

1

f(x)= −∞ whenever f(x) < 0 for x→ c−.

Let us try to understand this theorem visually.

c c

lim→ −

= 0 lim→ −

1( )

= ∞

Figure 2.1 Limits of f(x) and its reciprocal1

f(x)at c from the left

From Figure 2.1 we see that if f(x) > 0 for x is close to c from the left, then the values of f(x) are positive

and very small. Hence, for this same x, the values of the reciprocal1

f(x)are positive and very large; they

increase without bound. Thus, the statement of part (a) of the theorem.


(Rev. C4)

c c c

c

lim𝑥→𝑐−

𝑓 𝑥 = 0

lim𝑥→𝑐−

1

𝑓(𝑥)= −∞

Figure 2.2 Limits of f(x) and its reciprocal1

f(x)at c from the left

From Figure 2.2 we see that if f(x) < 0 for x close to c from the left, then the values of f(x) are negative

and very small. Hence, for this same x, the values of the reciprocal1

f(x)are negative and very small; they

decrease without bound. Thus, the statement of part (b) of the theorem.

In the exercises of this section, you are are asked to explain visually the next theorem.

Theorem 2.2. If limx→c+

f(x) = 0, then

a. limx→c+

1

f(x)=∞ whenever f(x) > 0 for x→ c+.

b. limx→c+

1

f(x)= −∞ whenever f(x) < 0 for x→ c+.

Example 2.12. Let g(x) = 1− x3.

a. g(x) = 1− x3 > 0 for x→ 1− (values of x close to 1 from the left).

Thus, by Theorem 2.1 on the previous page

limx→1−

1

1− x3=∞.

b. g(x) = 1− x3 < 0 for x→ 1+ (values of x close to 1 from the right).

Thus, by Theorem 2.2

limx→1+

1

1− x3= −∞. �


(Rev. C4)

Example 2.13. The graph of the logarithmic function logb x is shown in Figure 3 on page 74 of thetextbook. We see that lim

x→1logb x = 0 and

logb x > 0 for c→ 1+.

So by Theorem 2.2 on the previous page

limx→1+

1

logb x=∞. �

In the following definition we establish the meaning of an infinity limit at a number.

Theorem 2.3.

a. If

limx→c−

f(x) =∞ and limx→c+

f(x) =∞,

then

limx→c

f(x) =∞.

b. If

limx→c−

f(x) = −∞ and limx→c+

f(x) = −∞,

then

limx→c

f(x) = −∞.

Example 2.14. Let g(x) = x2 − 2x+ 1.

a. For x→ 1−, we have g(x) > 0. Thus, by Theorem 2.1 on page 44

limx→1−

1

x2 − 2x+ 1=∞.

b. For x→ 1+, we have g(x) > 0. Thus, by Theorem 2.2 on the previous page

limx→1+

1

x2 − 2x+ 1=∞.

Hence, by Theorem 2.3

limx→1

1

x2 − 2x+ 1=∞. �


(Rev. C4)

See the difference of our conclusions in this last example. The limit from the right is equal to the limit formthe left. It is infinity. We cannot claim that the limit does not exist. That the limit is infinity is not contraryto the existence of a limit.

Theorem 2.1 on page 44 and Theorem 2.2 on page 45 imply the following theorem.

Theorem 2.4. If limx→c

f(x) = 0, then

a. limx→c

1

f(x)=∞ whenever f(x) > 0 for x→ c.

b. limx→c

1

f(x)= −∞ whenever f(x) < 0 for x→ c.

The conditions of Theorem 2.4 must be fulfilled for us to conclude that the limit of the reciprocal is infinityor negative infinity.

Moreover, consider the graph of the function h(x) in Figure 2.3.

Figure 2.3 Graph of a function h(x), which is positive and negative for x→ c

From this graph we see that limx→c

h(x) = 0 and h(x) = 0 for infinity many x close to c. Hence, the

reciprocal1

h(x)is undefined for infinitely many x close to c. So the limit lim

x→c

1

h(x)is undefined.

Therefore, from the limit

limx→c

f(x) = 0

alone, we cannot conclude that

limx→c

1

f(x)= ±∞ or does not exist.

By Theorem 2.4, the limit of the reciprocal may be infinity or negative infinity, depending whether thevalues of f(x) are positive or negative for x→ c.

Moreover, if the values of f(x) are zero for infinitely many x→ c, then the limit of its reciprocal at c isundefined.


(Rev. C4)

The negation of Theorem 1, on page 95 of the textbook, and Theorem 2.3 on page 46 establish the nextresult.

Theorem 2.5. A limit

limx→c

f(x) does not exist

if and only if

limx→c−

f(x) 6= limx→c=

f(x).

The limit at 1 of Example 2.12 on page 45 does not exist.

We replace Theorem 4 on page 100 of the textbook with the following corollary.

Corollary 2.6.

a. If limx→c

f(x) = L > 0 and limx→c

1

g(x)=∞, then

limx→c

f(x)

g(x)=∞.

b. If limx→c

f(x) = L > 0 and limx→c

1

g(x)= −∞, then

limx→c

f(x)

g(x)= −∞.

c. If limx→c

f(x) = L < 0 and limx→c

1

g(x)=∞, then

limx→c

f(x)

g(x)= −∞.

d. If limx→c

f(x) = L < 0 and limx→c

1

g(x)= −∞, then

limx→c

f(x)

g(x)=∞.

Go over the tutorials “Limit 1” and “Evaluation of Limits.”


limx→1

x2 + 3

1− x3


(Rev. C4)

we must see first that this limit is not and indeterminate form 0/0. In this case, we must consider the limits

from the right and left of1

1− x3. We evaluated these limits in Example 2.12 on page 45. Thus,

limx→1−

1

1− x3=∞ and lim

x→1+

1

1− x3= −∞.

On the other hand, limx→1

x3 + 3 = 1 + 3 = 4 > 0.

By part (a) of Corollary 2.6 on the previous page

limx→1−

x2 + 3

1− x3=∞.

By part (b) of the same Corollary

limx→1+

x2 + 3

1− x3= −∞.

The limit at 1 from the right and from the left are not equal. Therefore the limit at 1 does not exist. �


limx→1

x3 − 2

x2 − 2x+ 1

again we consider the limits form the left and right of 1. In Example 2.14 on page 46, we found that

limx→1

1

x2 − 2x+ 1=∞.

By Equation (2.1) on page 39

limx→1

x3 − 2 = −1 < 0.

By part (c) of Corollary 2.6 on the preceding page

limx→1

x3 − 2

x2 − 2x+ 1= −∞. �

Note: The “Limit of a Difference Quotient” of Example 10 on page 101 will always result in anindeterminate form 0/0. Thus, to evaluate them you must apply an algebraic simplification.

Exercises3. Explain visually Theorem 2.2 on page 54 of the textbook.

4. Explain visually Theorem 2.4 on page 56.

5. Do odd-numbered exercises 73–89 in Section 2.2.

6. Solve problems 91 and 95 in the same section.


(Rev. C4)

Infinite Limits and Limits at InfinityIndications1. Disregard the content of the subsection “Infinite Limits” in Section 2.2 of your textbook.



CommentsWe have established in Theorem 2.5 on page 48 what we mean by the statement, “the limit does not exist.”

The limit

limx→c

f(x) does not exist

if and only if

limx→c+

f(x) 6= limx→c−

f(x)

Example 2.17. If

f(x) =

2 + x2 if x < 3√x− 3 if 3 ≤ x ≤ 12

1

x− 12if x > 12

then we have

limx→3+

f(x) = limx→3+

√x− 3 =

√3− 3 = 0

and

limx→3−

f(x) = limx→3−

2 + x2 = 2 + 32 = 11.

Hence, by Theorem 2.5 on page 48, we conclude that

limx→3

f(x) does not exist.

Also,

limx→12−

f(x) = limx→12−

2 + x2 = 146


(Rev. C4)

and

limx→12+

f(x) = limx→12+

1

x− 12=∞.

Again, by Theorem 2.5 on page 48

limx→12

f(x) does not exist. �

Unfortunately, it is customary to say that the limit does not exist when a limit is infinity or negative infinity,but this custom creates confusion with Theorem 2.3 on page 46.

So, if a limit is either infinity or negative infinity, we do not say that the limit does not exist; instead wesimply say that the limit is infinity (∞) or negative infinity (−∞), whichever may be appropriate.

It is important to learn to evaluate infinite limits properly. That is, applying theorems, proportions andcorollaries, i.e., proved results.

Infinite limits are important because they determine the existence of vertical asymptotes. The nextdefinition establishes the existence of vertical asymptotes together with their interpretation of theirdirection with respect to its graph.

Definition 2.7.

a. If limx→c−

f(x) =∞, then the vertical line x = c is a vertical asymptote of the function f(x) on the

left side of c and in the positive direction (the values of f(x) increase without bound as x→ c−.In term of the graph, we have

c

c

c

Figure 2.4 Vertical asymptote in the positive direction on the left side of c

b. If limx→c+

f(x) =∞, then the vertical line x = c is a vertical asymptote of the function f(x) on the

right side of c and in the positive direction (the values of f(x) increase without bound as x→ c+.In term of the graph, we have


(Rev. C4)

c

c

c

Figure 2.5 Vertical asymptote in the positive direction on the right side of c

c. If limx→c−

f(x) = −∞, then the vertical line x = c is a vertical asymptote of the function f(x) on

the left side of c and in the negative direction (the values of f(x) decrease without bound asx→ c−. In term of the graph, we have

c

c

c

Figure 2.6 Vertical asymptote in the negative direction on the left side of c

d. If limx→c+

f(x) = −∞, then the vertical line x = c is a vertical asymptote of the function f(x) on

the right side of c and in the negative direction (the values of f(x) decrease without bound asx→ c+. In term of the graph, we have

c

Figure 2.7 Vertical asymptote in the negative direction on the right side of c


(Rev. C4)

We leave as an exercise the definition and interpretation of the limit limx→c

f(x) = ±∞.

Example 2.18. From Figure 3 on page 108 of the textbook, we see that

limx→2+

x2 + 20

5(x− 2)2=∞ and lim

x→2−

x2 + 20

5(x− 2)2=∞.

So, by Theorem 2.3 on page 46

limx→2

x2 + 20

5(x− 2)2=∞.

The function has x = 2 as vertical asymptote in the positive direction on both sides of 2. �

Example 2.19. The graph of the logarithmic function logb x is shown in Figure 3 on page 74 of thetextbook. Hence,

limx→0+

logb x = −∞.

The y-axis is a vertical asymptote of logb x in the negative direction. �

The use of tables to evaluate limits, as it is done on page 106 and Example 1 on page 107 of the textbook, isnot how limits must be evaluated.

The evaluation of limits must be supported by the application of well established (proved) results ordefinitions. We can use the following theorem together with Theorem 2.1 on page 44, Theorem 2.2 onpage 45, and Corollary 2.6 on page 48 from the previous subsection to evaluate infinite limits.

Theorem 2.8. If

limx→c

f(x) = L 6= 0 and limx→c

g(x) =∞,

then

a. limx→c

f(x)g(x) =∞ if L > 0

b. limx→c

f(x)g(x) = −∞ if L < 0

If

limx→c


g(x) = −∞,

then

a. limx→c

f(x)g(x) = −∞ if L > 0

b. limx→c

f(x)g(x) =∞ if L < 0

Theorem 2.8 is also valid if x→ c is replaced by x→ c+ and x→ c−.


(Rev. C4)

See that Theorem 2.8 on the preceding page includes Corollary 2.6 on page 48 because

f(x)

g(x)= f(x)

(1

g(x)

).

Example 2.20. For the limit

limx→1

1− 3x

(x− 1)2,

we have

limx→1

1− 3x = −2 < 0,

and for x→ 1, we have (x− 1)2 > 0. Hence, by Theorem 2.1 on page 44

limx→1

1

(x− 1)2=∞

and by Theorem 2.8 on the previous page,

limx→1

1− 3x

(x− 1)2= −∞.

The line x = 1 is a vertical asymptote of1− 3x

(x− 1)2in the negative direction on both sides of 1. �


limx→1−

√1− x− 2

(x− 1)2,

we have

limx→1−

√1− x− 2 = −2 < 0,

and as in the previous example

limx→1

1

(x− 1)2=∞

Hence, by Theorem 2.8 on the preceding page,

limx→1−

√1− x− 2

(x− 1)2= −∞.

The line x = 1 is a vertical asymptote of√

1− x− 2

(x− 1)2in the negative direction on the left side of 1. �


(Rev. C4)


limx→1

x2 − 2

x− 1,

we have

limx→1

x2 − 2 = −1 < 0.

For x→ 1+, we have x− 1 > 0, and for x→ 1−, we have x− 1 < 0. Hence, by Theorem 2.1 on page 44and Theorem 2.2 on page 45

limx→1+

1

x− 1=∞ and lim

x→1−

1

x− 1= −∞.

By Theorem 2.8 on page 53,

limx→1+

x2 − 2

x− 1= −∞ and lim

x→1−

x2 − 2

x− 1=∞.

The line x = 1 is a vertical asymptote ofx2 − 2

x− 1in the negative direction on the right side of 1 and in the

positive direction on the left side of 1.

By Theorem 2.5 on page 48,

limx→1

x2 − 2

x− 1does not exist. �

Example 2.23. From the graphs of the exponential function (Figure 4 on page 65 of the textbook) andlogarithmic function (Figure 3 on page 74 of the textbook), we have

limx→0

ex = 1 and limx→0+

logb x = −∞.

Thus, by Theorem 2.8 on page 53

limx→0+

logb x

ex= −∞ �

Remark: You must know how to apply Theorems 2.1 to 2.4 on pages 44–47, Corollary 2.6 on page 48, andTheorem 2.8 on page 53. In Example 2.20 on the preceding page, we conclude that

limx→1

1− 3x

(x− 1)2= −∞

because

limx→1

1− 3x = −2 < 0 and limx→1

1

(x− 1)2=∞.


(Rev. C4)

Not because

limx→1

1− 3x

(x− 1)2= (−2)(∞) = −∞.

This is incorrect.

The symbol ∞ do not represent a quantity; hence, we cannot treat it in the same manner as we treat anumber. Expressions like

(2)(−∞) = −∞, 5

∞= 0, 4 +∞ =∞

do not make sense. They are mathematically incorrect.

Similarly, division by 0 is undefined—it has no meaning. Hence, expressions like

5

0=∞, −2

0= −∞

do not make sense either. They are equally incorrect.

From Theorem 2.8 on page 53, we see that if

limx→c


g(x) = ±∞,

then the vertical line x = c is a vertical asymptote of the function

F (x) = f(x)g(x).

Example 2.24. The function

F (x) =x

x+ 2− |2x− 4|

x2 + 1

is not defined at x = −2. To evaluate its limit at −2, we have

F (x) =x(x2 + 1)− (x+ 2)|2x− 4|

(x+ 2)(x2 + 1)

and

limx→−2+

x(x2 + 1)− (x+ 2)|2x− 4| = −2((−2)2 + 1) = −10 < 0.

For x→ −2+, we have (x+ 2)(x2 + 1) > 0; hence, by Theorem 2.1 on page 44 and Theorem 2.8 onpage 53,

limx→−2+

x(x2 + 1)− (x+ 2)|2x− 4|(x+ 2)(x2 + 1)

= −∞.


(Rev. C4)

So the line x = −2 is a vertical asymptote of the given function in the negative direction on the right sideof 2. �

On the other hand, if

limx→c

f(x) = 0 and limx→c

g(x) = 0

then the function

F (x) =f(x)

g(x)

may or may not have a vertical asymptote. As we said before, the limit limx→c

F (x) yields an indeterminate

form 0/0, and its factorization or rationalization may give a finite limit.

Example 2.25. In Example 2.10 on page 43, we indicated that the limit

limx→−2

x4 − 2x3 − x2 + 5x− 18

x2 − 4


Moreover, also by Example 2.10 on page 43,

limx→−2

x3 − 4x2 + 7x− 9

x− 2= 11.75.

Thus, the vertical line x = −2 is not an asymptote even when the function

F (x) =x4 − 2x3 − x2 + 5x− 18

x2 − 4

is not defined at −2.

Note also that 2 is not in the domain of F . To determine if x = 2 is a vertical asymptote, we evaluate thelimit lim

x→2+F (x), and see that

limx→2

x4 − 2x3 − x2 + 5x− 18 = −12 < 0 and limx→2+

x2 − 4 = 0,

also x2 − 4 > 0 for x→ 2+.

So, by Theorem 2.8 on page 53,

limx→2

x4 − 2x3 − x2 + 5x− 18

x2 − 4= −∞,

and the vertical line x = 2 is an asymptote in the negative direction on both sides of 2. �

In this course, if a function has a vertical asymptote, it is on a number not in its domain.


(Rev. C4)

Exercises7. Give the definition and interpretation (including a graph) of the limits lim

x→cf(x) =∞ and

limx→c

f(x) = −∞

8. Do odd-numbered exercises 17–23 in Section 2.2 of the textbook.

9. Do odd-numbered exercises 41–49 in Section 2.2.

Indications1. Read the subsection “Limits at Infinity,” on pages 109–112 of the textbook.



CommentsTo determine horizontal asymptotes we evaluate limits at infinity or at negative infinity.

Definition 2.9. A horizontal line y = b is the horizontal asymptote of the function f(x) if either

a. limx→∞

f(x) = b.

In terms of the graph of the function, the values of f(x) approached the value b as x increaseswithout bound.

b. limx→−∞

f(x) = b.

In terms of the graph of the function, the values of f(x) approached the value b as x decreaseswithout bound.

Figure 2.8 and Figure 2.9 on the following page show the graph of two functions with horizontalasymptotes.

b

b

Figure 2.8 Horizontal asymptote in the positive direction, given by limx→∞

f(x) = b

From the graph of the exponential functions ex and e−x (Figure 4 on page 65 of the textbook), we have

limx→−∞

ex = 0 and limx→∞

e−x = 0. (2.3)


(Rev. C4)

b

b

Figure 2.9 Horizontal asymptote in the negative direction, given by limx→−∞

f(x) = b

Note: Theorem 2: Properties of Limits on page 97 of the textbook, Corollary 2.6 on page 48, andTheorem 2.8 on page 53 are also valid if x→ c is replaced with either x→∞ or x→ −∞. We state themhere for reference proposes.

Theorem 2.10. If f and g are two functions and

limx→±∞

f(x) = L limx→±∞

g(x) = M

where L and M are real numbers. Then

1. limx→±∞

k = k for any constant k

2. limx→±∞

f(x)± g(x) = limx→±∞

f(x)± limx→±∞

g(x) = L+M

3. limx→±∞

kf(x) = k limx→±∞

f(x) = kL for any constant k

4. limx→±∞

f(x)g(x) =

[lim

x→±∞f(x)

] [lim

x→±∞g(x)

]= L(M)

5. limx→±∞

f(x)

g(x)=

limx→±∞

f(x)

limx→±∞

g(x)=

L

Mif M 6= 0

6. limx→±∞

n√f(x) = n

√lim

x→±∞f(x) =

n√L if L > 0 and n is odd.

Theorem 2.11. If

limx→±∞

f(x) = L 6= 0 and limx→±∞

g(x) =∞,

then

a. limx→±∞

f(x)g(x) =∞ if L > 0

b. limx→±∞

f(x)g(x) = −∞ if L < 0


(Rev. C4)

If

limx→±∞

f(x) = L 6= 0 and limx→±∞

g(x) = −∞,

then

a. limx→±∞

f(x)g(x) = −∞ if L > 0

b. limx→±∞

f(x)g(x) =∞ if L < 0

Corollary 2.12.

a. If limx→±∞

f(x) = L > 0 and limx→±∞

1

g(x)=∞, then

limx→±∞

f(x)

g(x)=∞.

b. If limx→±∞


1

g(x)= −∞, then

limx→±∞

f(x)

g(x)= −∞.

c. If limx→±∞

f(x) = L < 0 and limx→±∞

1

g(x)=∞, then

limx→±∞

f(x)

g(x)= −∞.

d. If limx→±∞

f(x) = L < 0 and limx→±∞

1

g(x)= −∞, then

limx→±∞

f(x)

g(x)=∞.

Corollary 2.12 follows from Theorem 2.11 on the preceding page because

f(x)

g(x)= f(x)

(1

g(x)

).

Limits 3 and 4 of Theorem 2 on page 110 of the textbook are particular cases of Theorem 2.11 on theprevious page, with f(x) = k for any constant k, and g(x) = xp for any real number p.

It is incorrect to conclude that for any function f

limx→∞

1

f(x)=∞

because

limx→∞

f(x) = 0.


(Rev. C4)

If the function f is similar to the graph of the function in Figure 2.8 on page 58 with b = 0, then there are

infinitely many values of x so that f(x) = 0. Hence, then its reciprocal1

f(x)is not defined for infinitely

many x. So the limit

limx→∞

1

f(x)is undefined.

However, we have limits 1 and 2 of Theorem 2 on page 110 of your textbook. Namely, for any positive realnumber p,

limx→∞

k

xp= 0 and lim

x→−∞

k

xp= 0. (2.4)

Remark: See that Theorem 2.11 on page 59 is saying that

limx→±∞

f(x)g(x) =∞

if

limx→±∞


g(x) =∞.

It is not saying that

limx→±∞

f(x)g(x) = L(∞) =∞.

We must justify our results properly.

Once we have Theorem 2.10 on page 59, we can be precise (and it is important to be precise) and removethe “or” from Theorem 3 on page 111 of the textbook.

Theorem 2.13. If

p(x) = anxn + an−1x

n−1 + · · ·+ ax+ a0, an 6= 0, n ≥ 1

then

a. limx→∞

p(x) = limx→∞

anxn =∞ if an > 0.

b. limx→∞

p(x) = limx→∞

anxn = −∞ if an < 0.

c. limx→−∞

p(x) = limx→−∞

anxn =∞ if an > 0 and n is even.

d. limx→−∞


anxn =∞ if an < 0 and n is odd.

e. limx→−∞


anxn = −∞ if an > 0 and n is odd.

f. limx→−∞


anxn = −∞ if an < 0 and n is even.


(Rev. C4)

Example 2.26. If

p(x) = anxn + an−1x

n−1 + · · ·+ ax+ a0, an 6= 0, n ≥ 1

then

p(x) = xn(an +

an−1x

+ · · ·+ a

xn−1+a0xn

).

By Theorem 2.10 on page 59 and Limits (2.4)

limx→±∞

an +an−1x

+ · · ·+ a

xn−1+a0xn

= an

and by Limits 3 and 4 of Theorem 2 on page 110 of the textbook

limx→±∞

xn =∞ if n is even.

Hence, by Theorem 2.11 on page 59

limx→±∞

p(x) =∞ if an > 0

and

limx→±∞

p(x) = −∞ if an < 0

Similarly, we have

limx→−∞

p(x) = −∞ if n is odd and an > 0

and

limx→−∞

p(x) =∞ if n is odd and an < 0 �

In general for any polynomial p(x)

limx→±∞

k

p(x)= 0 for any constant k (2.5)

Let us see why. As in Example 2.26 on the previous page

k

p(x)=

k

xn(an +

an−1x

+ · · ·+ a

xn−1+a0xn

) =

(k

xn

) 1

an +an−1x

+ · · ·+ a

xn−1+a0xn

.

By part 5 of Theorem 2.10 on page 59

limx→±∞

1

an +an−1x

+ · · ·+ a

xn−1+a0xn

=1

an


(Rev. C4)

and by Limits (2.4)

limx→±∞

k

xn= 0.

By part 4 of Theorem 2.10 on page 59

limx→±∞

k

p(x)=

[lim

x→±∞

k

xn

] limx→±∞

1

an +an−1x

+ · · ·+ a

xn−1+a0xn

1

an

= 0

(1

an

)= 0.

The Limit (2.5) has an interesting consequence for rational functions.

Example 2.27. Consider the rational function

h(x) =x2 − 2x− 2

2x− 2x4

We factor x2 in the numerator and x4 in the denominator.

x2

x4

1− 2

x− 2

x22

x3− 2

=1

x2

1− 2

x− 2

x22

x3− 2

By Equation (2.5) on the preceding page and Theorem 2.11 on page 59

limx→±∞

1

x2

1− 2

x− 2

x22

x3− 2

= 0

(−1

2

)= 0.

The x-axis is a horizontal asymptote of the function in both directions, positive and negative. �

If we apply the same argument to a rational function

f(x) =p(x)

q(x)=

anxn + an−1x

n−1 + · · ·+ ax+ a0bmxm + bm−1xm−1 + · · ·+ bx+ b0

with n < m and an, bm 6= 0, we have

p(x) = xn(an +

an−1x

+ · · ·+ a

xn−1+a0xn

)and

q(x) = xm(bm +

bm−1x

+ · · ·+ b

xm−1+a0xm

),


(Rev. C4)

then

p(x)

q(x)=xn

xm

an +an−1x

+ · · ·+ a

xn−1+a0xn

bm +bm−1x

+ · · ·+ b

xm−1+a0xm

=1

xm−n

an +an−1x

+ · · ·+ a

xn−1+a0xn

bm +bm−1x

+ · · ·+ b

xm−1+a0xm

since m− n > 0; by Limit (2.4) on page 61

limx→±∞

1

xm−n= 0

and

limx→±∞

an +an−1x

+ · · ·+ a

xn−1+a0xn

bm +bm−1x

+ · · ·+ b

xm−1+a0xm

=anbm.

By Theorem 2.10 on page 59

limx→±∞

p(x)

q(x)= 0

(anbm

)= 0.

This is part 1 of (B) of Theorem 4 on page 113 of the textbook.

Exercises10. Apply Theorem 2.13 on page 61 to do odd-numbered exercises 25–31 on page 115.

Indications1. Read the subsection “Finding Horizontal Asymptotes,” on pages 112–114 of the textbook.



CommentsExample 2.28. If we consider

h(x) =2x− 2x4

x2 − 2x− 2

We again factor x4 in the numerator and x2 in the denominator.

x4

x2

2

x3− 2

1− 2

x− 2

x2

= x2

2

x3− 2

1− 2

x− 2

x2


(Rev. C4)

By Equation (2.5) on page 62

limx→±∞

2

x3− 2

1− 2

x− 2

x2

= 2

and by Theorem 2.13 on page 61

limx→±∞

x2 =∞.

Hence, by Theorem 2.11 on page 59

limx→±∞

2x− 2x4

x2 − 2x− 2= lim

x→±∞x2

2

x3− 2

1− 2

x− 2

x2

=∞. �


f(x) =p(x)

q(x)=

anxn + an−1x

n−1 + · · ·+ ax+ a0bmxm + bm−1xm−1 + · · ·+ bx+ b0

with n > m and an, bm 6= 0, we have

p(x) = xn(an +

an−1x

+ · · ·+ a

xn−1+a0xn

)and

q(x) = xm(bm +

bm−1x

+ · · ·+ b

xm−1+a0xm

),

then

p(x)

q(x)=xn

xm

an +an−1x

+ · · ·+ a

xn−1+a0xn

bm +bm−1x

+ · · ·+ b

xm−1+a0xm

= xn−m

an +an−1x

+ · · ·+ a

xn−1+a0xn

bm +bm−1x

+ · · ·+ b

xm−1+a0xm

since n−m > 0; by Theorem 2.13 on page 61

limx→∞

xn−m =∞

limx→−∞

xn−m =∞ n−m is even

limx→−∞

xn−m = −∞ n−m is odd


(Rev. C4)

and by Limits (2.4)

limx→∞

an +an−1x

+ · · ·+ a

xn−1+a0xn

bm +bm−1x

+ · · ·+ b

xm−1+a0xm

=anbm.

By Theorem 2.11 on page 59

limx→∞

p(x)

q(x)=

∞ if

anbm

> 0

−∞ ifanbm

< 0

limx→−∞

p(x)

q(x)=

∞ ifanbm

> 0,m− n is even

−∞ ifanbm

> 0,m− n is odd

−∞ ifanbm

< 0,m− n is even

∞ ifanbm

< 0,m− n is odd < 0

This is part 3 of (B) of Theorem 4 on page 113 of the textbook without the “or.”

Example 2.29. If we consider the rational function

f(x) =x3 − 3x2 + 1

3x− 5x3

We factor x3 in both the numerator and denominator

x3 − 3x2 + 1 = x3(

1− 3

x+

1

x3

)and 3x− 5x3 = x3

(3

x2− 5

).

Hence,

x3 − 3x2 + 1

3x− 5x3=

x3(

1− 3

x+

1

x3

)x3(

3

x2− 5

) =1− 3

x+

1

x33

x2− 5

.

By Limits (2.5) and Theorem 2.10 on page 59

limx→±∞

f(x) = −1

5. �


f(x) =p(x)

q(x)=anx

n + an−1xn−1 + · · ·+ ax+ a0

bnxn + bn−1xn−1 + · · ·+ bx+ b0


(Rev. C4)

with an, bn 6= 0, we have

p(x) = xn(an +

an−1x

+ · · ·+ a

xn−1+a0xn

)and

q(x) = xn(bn +

bn−1x

+ · · ·+ b

xn−1+a0xn

),

then

p(x)

q(x)=xn

xn

an +an−1x

+ · · ·+ a

xn−1+a0xn

bn +bn−1x

+ · · ·+ b

xn−1+a0xn

=an +

an−1x

+ · · ·+ a

xn−1+a0xn

bn +bn−1x

+ · · ·+ b

xn−1+a0xn

.

By Limits (2.5) and Theorem 2.10 on page 59

limx→±∞

p(x)

q(x)=anbn.

This is part 2 of (B) of Theorem 4 on page 113 of the textbook.

We use Theorem 4 on page 113 of the textbook to find horizontal asymptotes. In the next example, wepresent another application of this theorem for another type of functions.

Example 2.30. We want to know if the function

F (x) =

√3x4 + x2 − 6

8− 3x2

has a horizontal asymptote.

To apply Theorem 4 (p. 163), we have

|8− 3x2| =√

(8− 3x2)2

and 8− 3x2 < 0 for x→∞.

Therefore, |8− 3x2| = −(8− 3x2), and√(8− 3x2)2 = −(8− 3x2)

We conclude that√

3x4 + x2 − 6

8− 3x2= −

√3x4 + x2 − 6

(8− 3x2)2


(Rev. C4)

and

limx→∞

√3x4 + x2 − 6

8− 3x2= lim

x→∞−

√3x4 + x2 − 6

(8− 3x2)2= −

√limx→∞

3x4 + x2 − 6

(8− 3x2)2= −

√3

9= − 1√

3

You can use the same method to verify that

limx→−∞

√3x4 + x2 − 6

8− 3x2= lim

x→−∞

√3x4 + x2 − 6

(8− 3x2)2=

√limx→∞

3x4 + x2 − 6

(8− 3x2)2= − 1√

3�

In the next example, we show another method for evaluating this same limit.

Example 2.31. We factor the term of highest degree inside the square root:

√3x4 + x2 − 6 =

√x4(

3 +1

x2− 6

x4

)= x2

√3 +

1

x2− 6

x4.

Then,

limx→∞

√3x3 + x2 − 6

8− 3x2= lim

x→∞

x2√

3 +1

x2− 6

x4

x2(

8

x2− 3

) = limx→∞

√3 +

1

x2− 6

x4

8

x2− 3

=

√3

−3= − 1√

3�

We finish this section with an example in which we ask you to justify the steps.

Example 2.32. We want to know if the function

F (x) =4x− 6√8 + x2

has a horizontal asymptote in the negative direction.

For x→ −∞ we have√x2 = −x, thus

limx→−∞

4x− 6√8 + x2

= limx→−∞

x

(4− 6

x

)√x2(

8

x2+ 1

) = limx→−∞

x

(4− 6

x

)−x√

8

x2+ 1

= limx→−∞

4− 6

x

−√

8

x2+ 1

= −4

The line y = −4 is a horizontal asymptote in the negative direction. �

Exercises11. Do odd-numbered exercises 33–39 on page 115, 51–63 on page 116 of the textbook.

12. Solve odd-numbered problems 81, 83 on page 116.


(Rev. C4)

ContinuityIndications1. Read Section 2.3 of the textbook, “Continuity.” You may skip the paragraphs marked “Explore and

Discuss.”



CommentsFrom the definition of continuity on page 120 of the textbook, a function f is continuous at a number c ifand only if

limx→c

f(x) = f(c).

We also have the definition of continuity from the right and left.

Definition 2.14. A function f(x) is continuous at the point x = c

a. from the right if

1. limx→c+

f(x) exist

2. f(c) exist

3. limx→c+

f(x) = f(c).

b. from the left if

1. limx→c−

f(x) exist

2. f(c) exist

3. limx→c−

f(x) = f(c).

This statement tells us that the evaluation of limits of continuous functions is relatively simple. So it isimportant to know which functions are continuous on their domains. Theorem 1 on page 122 of thetextbook lists some of these functions. We can add the following item to this list.

(G) If a function f is continuous on its domain D, then the function |f(x)| is also continuous on thesame domain D.

From Figure 4 on page 65 of the textbook and Figure 3 on page 74 of the textbook we see that theexponential and logarithmic functions are continuous on their domains.


(Rev. C4)

In general, we can say that the algebraic functions we work with in this course are continuous on theirdomains. Indeed, the definition of continuity on page 120 of the textbook, together with the “Properties ofLimits” on page 97 of the textbook, allow us to compile the following list of continuous functions.

Theorem 2.15.

Let f and g be two continuous functions at c; thus,

limx→c

f(x) = f(c) and limx→c

g(x) = g(c).

Then, by Theorem 2 on page 97 of the textbook,

a. limx→c

f(x)± g(x) = limx→c

f(x)± limx→c

g(x) = f(c)± g(c);

thus, f ± g is continuous at c.

b. limx→c

kf(x) = k limx→c

f(x) = kf(c) for any constant k;

thus, the function kf is continuous at c.

c. limx→c

f(x)g(x) = limx→c

f(x) · limx→c

g(x) = f(c)g(c);

thus, the function fg is continuous at c.

d. limx→c

f(x)

g(x)=

limx→c

f(x)

limx→c

g(x)=f(c)

g(c)if g(c) 6= 0;

thus, the functionf

gis continuous on its domain.

Each property in Theorem 2.15 is also valid if the functions f and g are continuous from the right or theleft of c; that is, if x→ c is replaced by x→ c+ or x→ c−.

Example 2.33. Consider the function

F (x) =x

x+ 2− |2x− 4|

x2 + 1.

This function is the difference of two functions

f(x) =x

x+ 2and g(x) =

|2x− 4|x2 + 1

.

The function f is continuous on its domain (−∞,−2) ∪ (−2,∞) because it is rational. By part (d) ofTheorem 2.15, the function g(x) is continuous for all real numbers because it is the quotient of twocontinuous functions, namely |2x− 4| (see (G) on page 69) and x2 + 1. [Note that x2 + 1 6= 0 for any x.]Therefore, F is continuous on its domain. �

Observe that the domain of a function determines if a limit is well defined. For instance, the limit

limx→0

√x


(Rev. C4)

is not well defined because the domain of the function g(x) =√x is the interval [0,∞) and the given limit

indicates that x approaches 0 from both the left and the right. As the function is not defined for x on the leftof 0, the limit does not make sense. However, the limit

limx→0+

√x

is well defined because the function is defined for x to the right of 0.

Example 2.34. Let

f(x) =

{2 + x2 if x < 0√x− 3 if x ≥ 3

The domain of this function is (−∞, 0) ∪ [3,∞), since the function is not defined on the interval [0, 3).The limits

limx→0

f(x) limx→3

f(x) limx→0+

f(x) and limx→3−

f(x)

are not well defined, but the limits

limx→0−

f(x) and limx→3+

f(x)

are well defined. The function f is continuous at 3 from the right because

limx→3+

f(x) = limx→3+

√x− 3 = f(3) =

√3− 3 = 0.

The function f is not continuous from the left at 0 because f(0) is not defined. However,

limx→0−

f(x) = limx→0−

2 + x2 = 2.

Remember that the limit of a function f at c may well exist even when the function is not defined at c. �

We end this section with a summary on the evaluation of limits of the quotient of two functions.

Summary.

(A) Evaluation of a limit limx→c

f(x)

g(x).

1. If the functionf(x)

g(x)is continuous at c, then the limit is

f(c)

g(c).

2. If g(c) = 0, then we factor, simplify or rationalize the function to obtain either a continuous function at

c and apply point 1 above, or a functionp(x)

q(x)which is not an indeterminate form 0/0.

3. Ifp(x)

q(x)is not an indeterminate form 0/0 at c, then either


(Rev. C4)

i. limx→c

p(x) = 0 and limx→c

q(x) 6= 0, and therefore the limit is 0, or

ii. limx→c

p(x) = L 6= 0 and limx→c

q(x) = 0.

4. If limx→c

q(x) = 0, and Theorem 2.4 on page 47 applies, then

limx→c

1

q(x)=∞ or lim

x→c

1

q(x)= −∞.

and we apply Theorem 2.8 on page 53 to the limit limx→c

p(x)

q(x).

5. If limx→c

q(x) = 0, and Theorem 2.4 on page 47 does not apply, then

limx→c

1

q(x)does not exist or it is undefined.

In this case, the given limit ofp(x)

q(x)at c, neither exist nor is undefined.

The above steps are also applied for x→ c− and x→ c+.

(B) Evaluation of a limit limx→∞

f(x)

g(x).

1. If limx→∞

f(x) = L and limx→∞

g(x) = M 6= 0, then we apply Theorem 2.10 on page 59 and the limit isL

M.

2. Iff(x)

g(x)is rational or can be simplified or rationalized into a rational function, then we apply

Theorem 4 on page 113 of the textbook without the “or.”

3. If limx→∞

f(x) = L and limx→∞

g(x) = ±∞, then we apply Theorem 2.11 on page 59.

Exercises13. Do the exercises titled “Explore and Discuss 1,” on page 119 of the textbook.

14. Do odd-numbered exercises 9–37, 43, 49, 53, 69, 73 in Section 2.3 of the textbook.

15. Solve odd-numbered problems 89–95 in Section 2.3 of the textbook.

The DerivativeIndications1. Read Section 2.4 of the textbook, “The Derivative.” You may skip the paragraphs marked “Explore and

Discuss.”


(Rev. C4)



CommentsDefinition 2.16.

If f is a function defined on an interval [a, b], then its average rate of change is the quotient

f(b)− f(a)

b− a.

The units of the average rate of change areunits of funits of x

.

Observe that

f(b)− f(a)

b− a=f(a)− f(b)

a− b.

Example 2.35. Let

R(x) = 20x− 0.02x2 with domain 0 ≤ x ≤ 1000

be the revenue function of Example 1 on page 131 of the textbook.

The average rate of change on the interval [100, 400] is

R(400)−R(100)

400− 100= 10.

The interpretation is that, on the production from 100 planters to 400 planters, on average the revenueincreases at a rate of 10 dollars per planter. [What would happen if production was raised to 500 planters?— to 900 planters? — to 1000 planters?] �

Example 2.36. Let

f(x) = 16x2

be the falling distance function of Example 2 on pages 131–132 of the textbook.

On the interval [2, 3], the average rate of change is

f(3)− f(2)

3− 2= 80.

The interpretation is that from 2 to 3 seconds, on average the object is falling at a rate of 80 feet per second.�


(Rev. C4)

Geometrically, the average rate of change is the slope of the secant line joining the points (a, f(a)) and(b, f(b)) on the graph of f . [Recall, from your high school algebra course, that a secant is a line thatintersects two (or more) points on a curve; if the curve is a straight line, the slope of the secant is the slopeof the line.]

From the average rate of change, we pass to the instantaneous rate of change. As you consider thefollowing definition, examine Figure 2.10.

Definition 2.17.

Let f be a function defined on an interval (u, v) around a; that is, u < a < v.

The average rate of change of f on the interval [x, a], if u < x < a, or on the interval [a, x], ifa < x < v, is in either case

f(x)− f(a)

x− a.

The instantaneous rate of change of f at a is

limx→a

f(x)− f(a)

x− a.

The units of the instantaneous rate of change areunits of funits of x

.

Figure 2.10 Geometrical interpretation of h = x− a (top) or h = a− x (bottom) on the intervalu < a < v

Example 2.37. Let

f(x) = 16x2

be the function in Example 2.36 on the previous page.

The instantaneous rate of change of f at 3 is

limx→3

f(x)− f(3)

x− 3= lim

x→3

16x2 − 16(32)

x− 3

= limx→3

16

(x2 − 9

x− 3

)= lim

x→316

((x− 3)(x+ 3)

x− 3

)


(Rev. C4)

= limx→3

16(x+ 3)

= 16(6)

= 96.

The interpretation is that, at 3 seconds, the velocity of the falling object is 96 feet per second. �

Note that the derivative has two different interpretations: geometrical and physical. If f ′(a) = b, then thetwo interpretations are as follows.

GeometricalThe slope of the line tangent to the curve at the point (a, f(a)) is b. [Recall that a line is tangent to a curveat a point if the slope of the line is the best straight-line approximation to the slope of the curve at thatpoint.]

PhysicalThe variable f is increasing (if b > 0) or decreasing (if b < 0) at a rate of b units of f per unit of a.

In Example 2.37 on the preceding page, we gave the physical interpretation of an instantaneous rate ofchange. The geometrical interpretation is that the slope of the line tangent to the curve f(x) = 16x2 at thepoint (3, f(3)) is 96.

To find the equation of the line L tangent to the curve y = f(x) at the point (a, f(a)), we need the slope ofL.

By definition of the derivative, f ′(a) is the slope of the line L; hence,

L = f ′(a)x+ b.

Since the point (a, f(a)) is on the line, we have

f(a) = f ′(a)a+ b

so

b = f(a)− f ′(a)a

and

L = f ′(a)x+ f(a)− f ′(a)a = f ′(a)(x− a) + f(a) (2.6)

Example 2.38. The equation of the line tangent to the curve

f(x) = 16x2 at (3, f(3))

is

L = f ′(3)(x− 3) + f(3) = 96(x− 3) + 144 = 96x− 144. �

If h = x− a, then, h→ 0 as x→ a. See Figure 2.10 on the previous page.


(Rev. C4)

We conclude that

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

h→0

f(a+ h)− f(a)

h.

These limits give the specific value of the derivative of the function f at the point a.

Example 2.39. If f(x) =√x, then f ′(4) is obtained using one of the limits above. Thus, by

Example 2.11 on page 43, we have

f ′(4) = limx→4

f(x)− f(4)

x− 4= lim

x→4

√x−√

4

x− 4= lim

x→4

√x− 2

x− 4=

1

4

The slope of the tangent line L is1

4. (See Figure 2.11.)

L

2

4

Figure 2.11 Tangent line of the function f(x) =√x at the point (4, 2)

More generally, for any number a, you can verify that

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

√x−√a

x− a= lim

x→a

√x−√a

x− a= lim

x→a

1√x+√a

=1

2√a

Since a is any real number, we can conclude that

f ′(x) =1

2√x

for any real number x.

Compare this example with Example 7 on page 139 of the textbook. �


(Rev. C4)

The derivative

f ′(x) =1

2√x

of f(x) =√x

is not defined at 0; that is, the function f is not differentiable at 0.

If we look at the graph of the function f (see Table 2.1 on page 36 of this Study Guide), we clearlyunderstand why this is so: the function f does not have a tangent line at (0,0).

Example 2.40. To obtain the derivative function of f(x) = |x| we find f ′(a) for any number a, thus

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

|x| − |a|x− a

. (2.7)

Since x→ a, then x→ a+ and x→ a−. So to evaluate this limit, we must consider several cases.

Case 1x, a,> 0; hence,

limx→a

|x| − |a|x− a

= limx→a

x− ax− a

= 1

Case 2x, a < 0; hence,

limx→a

|x| − |a|x− a

= limx→a

−x+ a

x− a= −1

Case 3If a = 0,

f ′(0) = limx→0

|x|x

and we consider

limx→0+

x

x= 1 and lim

x→0−

−xx

= −1.

Therefore Limit (2.7) does not exist at 0, and the function is not differentiable at 0 (see Table 2.1 onpage 36).

In conclusion, since a is any real number, we have

f ′(x) =

{1 if x > 0

−1 if x < 0

See the graph of f in Table 2.1 on page 36 to see that, based on the slope of the function at any number x,this derivative is indeed correct. �


(Rev. C4)

Exercises16. Do the exercises titled “Explore and Discuss 1,” on page 138, “Explore and Discuss 2,” on page 140 of

the textbook.

17. Do odd-numbered exercises 1–55 and 69, 71, 75 in Section 2.4 of the textbook.

18. Solve odd-numbered problems 81–85 in Section 2.4.

Basic Differentiation PropertiesIndications1. Read Section 2.5 of the textbook, “Basic Differentiation Properties.” You may skip the paragraphs

marked “Explore and Discuss 1.”



CommentsIf y = f(x), then, for any x, the notation for the derivative is

f ′(x) = y′ =dy

dx=df

dx,

and the notation for the derivative at a number a is

f ′(a) = y′(a) =dy

dx

∣∣∣x=a

=df

dx

∣∣∣x=a

.

For f(x) = xn, with n a positive integer, we know that

xn − an = (x− a)(xn−1 + xn−2a+ ...+ xan−2 + an−1)

Thus,

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

xn − an

x− a

= limx→a

(x− a)(xn−1 + xn−2a+ ...+ xan−2 + an−1)

x− a= lim

x→axn−1 + xn−2a+ ...+ xan−2 + an−1

= an−1 + an−2a+ ...+ aan−2 + an−1

= nan−1


(Rev. C4)

Since this equation holds for any number a, we have f ′(x) = nxn−1, where n is any real number. Thus

d

dxxn = nxn−1.

Example 2.41. If f(x) = x−4, then

d

dxx−4 = −4x−5 = − 4

x5

and

d

dxx−4∣∣∣x=2

= −4x−5∣∣∣x=2

= − 4

x5

∣∣∣x=2

= − 4

25= −1

8.

Since the derivative function

f ′(x) = − 4

x5

is not defined at 0, the function f is not differentiable at 0. �

To differentiate the functions in this section, you must identify them with one or more of the rules given:power rule, constant multiple property, and sum and difference properties.

Example 2.42. To differentiate the function

f(x) =5

9√x,

we factor the constant

f(x) =5

9

(1√x

),

and we express the variable as a power

f(x) =5

9(x−1/2).

We then apply the constant multiple property and the power rule

f ′(x) =5

9

d

dxx−1/2 =

5

9

(−1

2x−3/2

)= − 5

18√x3. �

Example 2.43. To differentiate the function

g(x) =(x− 3)2 + 6x3√

x


(Rev. C4)

we must change it into a sum of power functions; so we multiply out and apply the properties of theexponents.

g(x) =x2 − 6x+ 9 + 6x3

x1/2=

6x3

x1/2+

x2

x1/2− 6x

x1/2+

9

x1/2

= 6(x5/2) + x3/2 − 6(x1/2) + 9(x−1/2).

We again apply the constant multiple property and the power rule.

g′(x) = 6d

dxx5/2 +

d

dxx3/2 − 6

d

dxx1/2 + 9

d

dxx−1/2

= 6

(5x3/2

2

)+

3x1/2

2− 6

(x−1/2

2

)+ 9

(−x

−3/2

2

)= 15

√x3 +

3√x

2− 3√

x− 9

2√x3

�

Exercises19. Do the exercises titled “Explore and Discuss 1,” on page 146 of the textbook.

20. Do odd-numbered exercises 1–73 and 77–81 in Section 2.5 of the textbook.

21. Solve problems 89 and 95 in Section 2.5.

Basic DifferentialsIndications1. Read the “Comments” section below.

2. Read Section 2.6 of the textbook, “Differentials.”


CommentsDifferentials are used to approximate the value of a function f at a number b such that

• the number b is near to a fixed number a (i.e., b ≈ a)

• the exact value of f(a) is known.

In this course, we use the linear approximation; that is we use the tangent line.

We know that√

9 = 3, and hence√

9.3 ≈ 3.

There are different ways to approximate the value of√

9.3.


(Rev. C4)

The linear approximation makes use of the tangent line to the curve y =√x at the point (9, 3). To

understand why we choose the tangent line, see Figure 2.12, below.

Figure 2.12 Curve y =√x showing the line tangent to the point (9, 3)

We want to approximate the value of√

9.3, and the closest exact value we know is√

9 = 3.

We see from Figure 2.12 that if L is the line tangent to the curve f(x) =√x at (9, 3), then

f(9.3) ≈ L(9.3).

So we need to find the equation of the line L.

Using Equation (2.6) on page 75, we have

L(x) = f ′(9)(x− 9) + f(9) =x− 9

2√

9+√

9 =x− 9

6+ 3.

Hence,

L(9.3) =9.3− 9

6+ 3 = 3 +

3

60= 3 +

1

20= 3.05

and√

9.3 = f(9.3) ≈ L(9.3) = 3.05.

In summary,

1. we want to approximate the value of√

9.3.

2. we find√

9 = 3 to be the closest exact value to√

9.3.


(Rev. C4)

3. we consider the function f(x) =√x, which gives f(9.3) =

√9.3 and f(9) =

√9 = 3.

4. we find the line L tangent to the function f(x) =√x at the point (9, 3).

5. we have f(9.3) ≈ L(9.3).

In mathematics, a differential in x is a small increment of x, and it is denoted as ∆x (read “delta x”).Hence, if f(a) = b, then

f(a+ ∆x) ≈ b = f(a).

That is, the linear approximation applies when we know the exact value of f(a) and we want toapproximate the value of f(a+ ∆x).

The linear approximation works because, if L is the line tangent to the curve f(x) at the point (a, b), then

f(a+ ∆x) ≈ L(a+ ∆x).

Thus, by Equation (2.6) on page 75, we have

L = f ′(a)(x− a) + f(a)

and

L(a+ ∆x) = f ′(a)(a+ ∆x− a) + f(a) = f ′(a)∆x+ f(a).

Therefore,

f(a+ ∆x) ≈ f ′(a)∆x+ f(a).

Example 2.44. We want to approximate the value of 3√

3. The closest exact value is 3√

1 = 1.

So we consider the function f(x) = 3√x, and we have 3 = 1 + 2 = 1 + ∆x, where ∆x = 2.

The linear approximation gives

3√

3 =3√

1 + ∆x = f(1 + ∆x) ≈ f ′(1)∆x+ f(1) = 1 + f ′(1)(2)

sinced

dx3√x∣∣∣x=1

=d

dxx1/3

∣∣∣x=1

=1

3x2/3

∣∣∣x=1

=1

3we have

3√

3 ≈ 1 +2

3. �

Let us consider the geometric interpretation of the approximation

f(a+ ∆x) ≈ f ′(a)∆x+ f(a).

The equation of the line tangent to the curve y = f(x) at (a, f(a)) is

L = f ′(a)(x− a) + f(a);


(Rev. C4)

therefore,

L(a+ ∆x) = f ′(a)∆x+ f(a) and L(a+ ∆x)− f(a) = f ′(a)∆x.

Hence,

f(a+ ∆x)− f(a) ≈ f ′(a)∆x = L(a+ ∆x)− f(a).

We can see, in Figure 2.13, that this is indeed the case

Figure 2.13 Geometrical interpretation of f(a+ ∆x) ≈ f ′(a)∆x+ f(a)

The increment from f(a) to f(a+ ∆x) is in the direction of the variable y; therefore, we define

∆y = f(a+ ∆x)− f(a).

The quantity f ′(a)∆x also has a distinct notation and name: it is called the “differential of f at a,” and it isdenoted by dy. So,

dy = f ′(a)∆x.

What we did here applies to any number a, so we can generalize it to any number x. Thus, dy is thedifferential of f and

dy = f ′(x)∆x.

We have

∆y = f(x+ ∆x)− f(x)

and

f(x+ ∆x) ≈ f ′(x)∆x+ f(x).


(Rev. C4)

Therefore,

∆y = f(x+ ∆x)− f(x) ≈ f ′(x)∆x = dy. (2.8)

The approximation

∆y ≈ dy

is also used to approximate the difference f(x+ ∆x)− f(x), because it is easier to evaluate f ′(x)∆x.

Example 2.45. Let

x =√

5, 000− 800p

be the price-demand function for the production of x glass beads.

What is the extra demand if the current price of $1.80 per bead is decreased to $1.63?

The answer to this question is the difference x(1.63)− x(1.80) (remember that x is a decreasing function).Using differentials, we approximate this difference with the differential of x, with a = 1.63 and∆x = −0.17. [Why?]

Hence,

d

dx

√5, 000− 800p

∣∣∣x=1.63

= − 800

2√

5000− 800p

∣∣∣x=1.63

= − 400√3696

= − 100√231

,

and

x(1.63)− x(1.80) ≈ − 100√231

(−0.17) =17√231

.

The exact value of the extra demand is

x(1.63)− x(1.80) =√

3696−√

3560 = 4√

231− 2√

890

thus

4√

231− 2√

890 ≈ 17√231

.

You can check with your calculator how good the approximation we found is. If one does not have acalculator, it is easier to evaluate the differential than the difference on the left. �

From Equation (2.8), if ∆x = 1, then

∆y = f(x+ 1)− f(x) ≈ f ′(x),

which gives us another interpretation of the derivative:

f ′(x) is approximately equal to the change of f caused by increasing the variable x by one unit tox+ 1.


(Rev. C4)

Example 2.46. Let

D = 1000− 40x2 for 1 ≤ x ≤ 5

be the demand function of Exercise 46 on page 161 of the textbook. Then,

D′(x) = −80x and D′(3) = −240.

Thus,

D(3 + 1)−D(3) ≈ D′(3).

Therefore, if the price increases from $3.00 to $4.00, the demand for chocolates decreases byapproximately 240 pounds.

You can evaluate D(3)−D(4) and see that the approximation is off by 40 pounds. Depending of thevolume of production, this may be an acceptable approximation. If it is not, we must use another method ofapproximation or evaluate the exact value. �

Exercises22. Do odd-numbered exercises 1–35 and 41, 43 in Section 2.6 of the textbook.


Marginal Analysis in Business and EconomicsIndications1. Read Section 2.7 of the textbook, “Marginal Analysis in Business and Economics.”



CommentsWe have

d

dx

g(x)

x=g′(x)x− g(x)

x2=g′(x)

x− g(x)

x2.

Hence, the marginal average cost is

d

dxC(x) =

C ′(x)

x− C(x)

x2=C ′(x)

x− C(x)

x=C ′(x)− C(x)

x.


(Rev. C4)

The marginal average revenue is

d

dxR(x) =

R′(x)−R(x)

x,

and the marginal average profit is

d

dxP (x) =

P ′(x)− P (x)

x.

In the next example, we consider the different interpretations of the concepts presented in this section.

Example 2.47. The total profit (in dollars) from the sale of x charcoal grills is

P (x) = 20x− 0.02x2 − 320 for 0 ≤ x ≤ 1000.

The marginal profit, if 40 grills are produced, is

P ′(x) = 20− 0.04x.

Thus,

P ′(40) = 20− 0.04(40) = 18.4.

The two interpretation of this result are

• at a production level of 40, the profit increases at a rate of $18.40 per grill.

• the profit gained by producing the 41st grill is approximately equal to $18.40.

The average profit per grill, if 40 are produced, is

P (x) =20x− 0.02x2 − 320

x= 20− 0.02x− 320

x.

Hence,

P (40) = 11.2;

that is, if 40 grills are produced on average the profit is $11.20 per grill.

The marginal average profit at a level of 40 is

d

dxP (40) =

P ′(40)− P (40)

40=

18.40− 11.20

40= 0.18.

The two interpretation of this statement are

• when the production level is 40 grills, on average, the profit increases at a rate of $0.18 per grill.

• the increase in average profit, if 41 grills are produced, is approximately equal to $0.18 per grill. �


(Rev. C4)


21. Solve the odd-numbered problems 33–47 in Section 2.7 of the textbook.

Finishing this Unit1. Review the objectives of this unit and make sure you are able to meet each of them.

2. Study the section of the “Chapter 2 Review” titled “Important Terms, Symbols, and Concepts.”


4. You may want to do odd-numbered problems 1–35, 43–63 and 87–91 from the “Review Exercise”section in Chapter 2.

5. Complete the “Practice Examination” provided for this unit. Evaluate yourself, first checking youranswers against those provided in the “Answers” section at the end of the textbook, and thencomparing your solutions with those provided in Chapter 2 “Review” on pages 2–51, 2–68 of theStudent’s Solutions Manual. The number of points in a question may indicate the number of steps inthe solution. Give yourself full credit if your answer is correct and you give a complete solution, evenif your solution differs from that shown in the Student’s Solutions Manual.


(Rev. C4)


Do the following exercises from the Chapter 2 “Review Exercises.”














(Rev. C4)

UNIT 3

Additional Derivative Topics

In this unit, we differentiate the exponential and logarithmic functions, as well as the product, quotient andcomposition of functions. We apply implicit differentiation to solve related rate problems, and discusselasticity of demand.


1. differentiate the exponential and logarithmic functions.

2. solve problems of compound interest.

3. apply the Product, Quotient and Chain rules to differentiation.

4. use implicit differentiation.

5. solve problems of related rates.

6. solve problems of elasticity of demand.


• apply Pythagoras’s Theorem◦ Book VII, Triangles and Quadrilaterals, on page 51 of Language of Mathematics

The Constant e and Continuous Compound InterestIndications1. Read Section 3.1 of the textbook, “The Constant e and Continuous Compound Interest.”




(Rev. C4)

CommentsExponential growth problems are in general of two types: compound interest and growth. Themathematical model

A = Pert,

is the same in either case; only the interpretation of A, P , r and t is different.

Compound InterestP = initial amount or principal,r = interest,t = time of investment,A = total amount at time t.

GrowthP = initial population,r = rate of growth,t = time of growth,A = amount of growth.

Exponential decay is a problem of growth; the only difference in the mathematical model is that the rate isnegative. The interpretation is the same.

The doubling time in a problem of exponential growth is the time that it takes for an initial amount P todouble; hence,

2P = Pert thus 2 = ert

The corresponding problem in decay is the half-life; that is, the time that it takes for an initial amount todecrease to one-half. Hence,

P

2= Pert thus

1

2= ert

Example 3.1. The half-life of radium is 95 years. What is its rate of decay?

From

1

2= ert,

we solve for r and find that

r = − ln 2

t.

In this case, t = 95 and we have

r = − ln 2

95≈ −0.0073. �


(Rev. C4)


2. Solve odd-numbered problems 27–37, 43, 45 in Section 3.1.

Derivatives of Exponential and Logarithmic FunctionsIndications1. Read Section 3.2 of the textbook, “Derivatives of Exponential and Logarithmic Functions.”



CommentsDerivatives of exponential functions:

d

dxex = ex

d

dxax = ln a(ax)

Derivatives of logarithmic functions:

d

dxlnx =

1

x

d

dxloga x =

1

ln a(x)

It is often convenient to use the properties of logarithmic and exponential functions before we differentiate.

Example 3.2.

a.d

dxlog4

√x =

d

dxlog4(x)1/2 =

d

dx

1

2log4 x =

1

2(ln 4)x

b.d

dxlog2

(5

x2

)=

d

dxlog2 5− log2(x

2) =d

dxlog2 5− 2 log2 x = − 2

ln 2(x)

c.d

dx5x − ln(6x) =

d

dx5x − ln(6)− lnx = ln 5(5x)− 1

x�


4. Solve odd-numbered problems 67–75 on page 195.


(Rev. C4)

Derivatives of Products and QuotientsIndications1. Read Section 3.3 of the textbook, “Derivatives of Products and Quotients.”



CommentsThe Product Rule can be applied to more than two functions, using the strategy shown below.

f(x)g(x)h(x) = [f(x)g(x)]h(x).

Thus

d

dxf(x)g(x)h(x) =

d

dx[f(x)g(x)]h(x)

=

[d

dx[f(x)g(x)]

]h(x) + [f(x)g(x)]

d

dxh(x)

= [f ′(x)g(x) + f(x)g′(x)]h(x) + [f(x)g(x)]h′(x)

= f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x)

Example 3.3. We apply the previous differentiation

d

dxe2x lnx =

d

dx(ex)2 lnx

=d

dx(ex)(ex) lnx

=

(d

dxex)

(ex lnx) + ex(d

dxex)

(lnx)

= (ex)(ex)

(d

dxlnx

)= e2x lnx+ e2x lnx+

e2x

x

= 2e2x lnx+e2x

x= e2x

(ln(x2) +

1

x

)�

Example 3.4.

d

dx

√xex log3(x

2) =

(d

dxx1/2

)(ex log3(x

2)) +√x

(d

dxex)

log3(x2) +√xex

(d

dx2 log3 x

)=ex log3(x

2)

2√x

+√xex log3(x

2) +2√xex

(ln 3)x�


(Rev. C4)

Example 3.5.

d

dx[f(x)]3 =

d

dx[f(x)][f(x)][f(x)]

=

(d

dxf(x)

)[f(x)f(x)] + f(x)

(d

dxf(x)

)f(x) + f(x)f(x)

(d

dxf(x)

)= f ′(x)[f(x)]2 + f(x)f ′(x)f(x) + [f(x)]2f ′(x)

= 3[f(x)]2f ′(x) �

We use the Quotient Rule to differentiate the reciprocal of a function f(x). That is,

d

dx

1

f(x)=

0[f ′(x)]− f ′(x)

[f ′(x)]2= − f ′(x)

[f(x)]2

Example 3.6.

d

dxe−x =

d

dx

1

ex= − ex

e2x= −e−x. �

Example 3.7.

d

dx

1

lnx= −

1

x(lnx)2

= − 1

x(lnx)2. �

When applying the product and quotient rules, we need the derivatives of the functions involved. To preventmistakes, we recommend you to find these derivatives first, simplify them if it is possible, and then applythese rules.

Example 3.8. In the function

f(x) = e2x(x2 − 3 lnx)

we have F (x) = e2x and S(x) = x2 − 3 lnx. See Theorem 1 on page 196 of the textbook. thus,

F ′(x) = 2e2x and S ′(x) = 2x− 3

x.

We apply the product rule

f ′(x) = e2x(

2x− 3

x

)+ 2e2x(x2 − 3 lnx).

We can factor e2x.

f ′(x) = e2x[2x− 3

x+ 2x2 − 6 lnx

]. �


(Rev. C4)

Example 3.9. In the function

f(x) =3x3 + 2x− 1

x4ex

we have T (x) = 3x3 + 2x− 1 and B(x) = x4ex. See Theorem 2 on page 199 of the textbook. Then,

T ′(x) = 9x2 + 2

and for the derivative of B, we apply the product rule

B′(x) = 4x3ex + x4ex = x3ex(4 + x).

We apply the quotient rule

f ′(x) =(9x2 + 2)x4ex − (3x3 + 2x− 1)(x3ex(4 + x))

x8e2x

=x3ex[(9x2 + 2)x− (3x3 + 2x− 1)(4 + x)]

x8e2x

=−3x4 − 3x3 − 2x2 − 5x+ 4

x5ex�



The Chain RuleIndications1. Read Section 3.4 of the textbook, “The Chain Rule.”



CommentsCertain cases of the general power rule introduced on page 207 of the textbook are worth noticing, becausethey appear frequently in the solution of problems.

Case 1If n =

1

2, then [g(x)]1/2 =

√g(x).


(Rev. C4)

Therefore,

d

dx

√g(x) =

d

dx[g(x)]1/2 =

1

2[g(x)]−1/2g′(x) =

g′(x)

2√g(x)

Case 2If n = −1

2, then [g(x)]−1/2 =

1√g(x)

.

Therefore,

d

dx

1√g(x)

=d

dx[g(x)]−1/2 = −1

2[g(x)]−3/2g′(x) = − g′(x)

2[g(x)]3/2

Case 3If n = −2, then [g(x)]−2 =

1

[g(x)]2.

Therefore,

d

dx

1

[g(x)]2=

d

dx[g(x)]−2 = (−2)[g(x)]−3g′(x) = − 2g′(x)

[g(x)]3

Example 3.10.

a.d

dx

√ex − x4 + x =

ex − 4x3 + 1

2√ex − x4 + x

b.d

dx

6√x5 − 4ex

= − 6(5x4 − 4ex)

2(x5 − 4ex)3/2

c.d

dx

1

(lnx− ex)2= −

2

(1

x− ex

)(lnx− ex)3

�

The composition of the function g(x) with the exponential and logarithmic functions gives

eg(x) and ln(g(x)).

respectively.

By the chain rule, we have

d

dxeg(x) = g′(x)eg(x)

and

d

dxln(g(x)) =

g′(x)

g(x)


(Rev. C4)

Example 3.11.

a.d

dxln(√

4x3 − 6x2 − 1) =d

dx

1

2ln(4x3 − 6x2 − 1) =

12x2 − 12x

2(4x3 − 6x2 − 1)

b.d

dxekx = kekx

c.d

dx

√x2 − e2x − log2 x =

2x2 − 2e2x − 1

(lnx)(ln 2)

2√x2 − e2x − log2 x

�

The next example is interesting.

Example 3.12. Let h(x) = ln |x|. This function is defined for all x except 0.

f(x) =

{ln(x) x > 0 ifln(−x) x < 0 if

Its derivative is

f ′(x) =

1

xx > 0 if

−1

−xx < 0 if

=

1

xx > 0 if

1

xx < 0 if

=1

x

That is,d

dxln |x| = 1

x. �

As in the previous section we recommend you to identify the inside and outside functions, differentiatethem and then apply the chain rule.

Example 3.13. For the function

f(x) =√

ln(x2 − ex + 2)

the inside function is I(x) = ln(x2 − ex + 2) and the outside function is E(I) =√I . Thus,

dE

dI=

1

2√I

The derivative of the inside function also requires the chain rule

I ′(x) =2x− ex

x2 − ex + 2.

The chain rule is

df

dx=dE

dI

dI

dx=

(1

2√I

)2x− ex

x2 − ex + 2=

2x− ex

2√

ln(x2 − ex + 2)(x2 − ex + 2). �


(Rev. C4)

Go over the tutorial “Differentiation.”



Implicit DifferentiationIndications1. Read Section 3.5 of the textbook, “Implicit Differentiation.”


3. Complete the exercises assigned below. If you have difficulty, consult your tutor to discuss theproblem.

An equation with variables x and y represents a curve on the Cartesian plane. It may be not the curve of afunction. The equation x2 + y2 = 16 represents a circle with center at (0, 0) and radius 4. This is not thecurve of a function. It does not satisfy the Vertical Line Test. However, in small parts of the curve, it is afunction. See the red portion of the circle in the figure below.

Figure 3.1 Curve x2 + y2 = 16 showing in red a portion which is a function.

This means that, for this portion, we do have a function, which is implicit in y. That is, y = f(x). So theequation can be considered to be x2 + (f(x))2 + 16.

When we take the derivative of this equation with respect to x, we have by the general power rule

2x+ 2(f(x))f ′(x) = 0 2x+ 2yy′ = 0. (3.1)


(Rev. C4)

If we have a curve z = f(x, y) on the Cartesian plane, then the curve passes through the point (a, b) if forx = a, y = b satisfies the equation z.

Example 3.14. Consider the curve 5x3 − y − 1 = 0 of Exercise 18 on page 220 of the textbook. Thepoint (1, 4) is on the curve, or the curve possess though this point because x = 1, y = 4 satisfy the equationof the curve. Indeed,

5(1)3 − 4− 1 = 5− 4− 1 = 0.

However, the point (2,−4) is not on the curve because for x = 2 and y = −4 we have

5(2)3 − (−4)− 1 = 40 + 5− 1 6= 0. �

Recall that for a function y = f(x), the equation of the tangent line to the curve at (a, f(a)) is

y = f ′(a)(x− a) + f(a).

Similarly the equation of a tangent line to the curve z = f(x, y) at the point (a, b) is

(y − b) = y′(a)(x− a).

Example 3.15. To find the equation of the tangent line to the curve 5x3 − y − 1 = 0, in the previousexample, at (1, 4), we first find the slope of the line y′(1) for x = 1, y = 4.

15x2 − y′ = 0 and 15(1)2 − y′(1) = 0.

So y′(1) = 15 and the equation of the tangent line is

(y − 4) = 15(x− 1) or y = 15x− 11.

Let us see which points on the curve have slope 0. From the equation above, we have y′ = 15x2 and y′ = 0if x = 0. For this value of x on the curve, we have that y = −1. Hence the curve has one point (0,−1) witha horizontal tangent line. �

Example 3.16. Consider the circle in Figure 3.1 on the previous page. To find all points where thetangent line is vertical, we need to find the points on the circle where y′ is undefined. From Equation (3.1)on the preceding page we have

y′ =−2x

2y= −x

y.

Thus, y′ is undefined if y = 0. For this value of y we have x2 + 0 = 16. thus, x = ±4. Therefore, the circlehas two vertical tangent lines at the points (0, 4) and (0,−4). This is clear from Figure 3.1. �




(Rev. C4)

Related RatesIndications1. Read Section 3.6 of the textbook, “Related Rates.”



CommentsObserve that in a related rate problem, we have constants and variables that change with respect to time.The challenge is to find a rate of change when certain conditions hold.

To find the correct mathematical model, we must identify the given rates and the rate we are looking for.Whenever possible, we draw a figure that shows the constants and the variables.

In the examples below, study the strategy we follow to solve such problems.

Example 3.17. See Example 2 in Section 3.6 of the textbook.

Suppose that two motorboats leave from the same point at the same time. If one travels north at 15miles per hour and the other travels east at 20 miles per hour, how fast will the distance between thembe changing after 2 hours?

Solution Strategy

Step 1We identify the constants and the variables that change with respect to time.

There are no constants.

x and y: the speed of the two boats

z: the distance between the two boats

Step 2We draw the figure showing the variables we found (see Figure 2 in Section 3.6 of the textbook).

Step 3We identify the known rates and the rate to be found, along with the conditions required.

dx

dt= 15mi/h and

dy

dt= 20mi/h.

dz

dt=? after 2 hours.


(Rev. C4)

Step 4We must find a relation between the variables in the figure. Here we use Pythagoras’s theorem, because thevariables are the sides of a right triangle. Thus,

z2 = x2 + y2.

Step 5We differentiate this equation with respect to time:

dz

dt(2z) =

dx

dt(2x) +

dy

dt(2y);

thus,

dz

dt(z) =

dx

dt(x) +

dy

dt(y)

Step 6We solve the rate we want to find:

dz

dt=dx

dt

x

z+dy

dt

y

z.

Step 7

To finddz

dt, we need the values of x, y and z, and the related rates

dx

dtand

dy

dt.

The rates are known (Step 3), and we need to find x, y and z.

Here is where we use the conditions of the problem (i.e., after two hours x = 30 miles and y = 40 miles) tofind z. We use the equation from Step 4:

z =√x2 + y2 =

√302 + 402 = 50

Step 8We have all we need to find what we want:

dz

dt=dx

dt

x

z+dy

dt

y

z= (15)

30

50+ (20)

40

50= 25. �

Example 3.18. See Exercise 16 in Section 3.6 of the textbook.

A point is moving on the graph of 4x2 + 9y2 = 36. When the point is at (3, 0), its y coordinate isdecreasing by 2 units per second. How fast is its x coordinate changing at that moment?

Solution Strategy

Step 1The coordinates x and y are changing with respect to time.

There are no constants.


(Rev. C4)

Step 2dy

dt= −2. [Note that decreasing rates are indicated with a minus sign.]

We want to knowdx

dtwhen x = 3 and y = 0.

Step 3We can draw the given graph 4x2 + 9y2 = 36, but it is not necessary; this equation already gives therelation between the two variables.

Step 4By implicit differentiation,

dx

dt(8x) +

dy

dt(18y) = 0.

Step 5

We solve fordx

dt:

dx

dt= −dy

dt

18y

8x= −dy

dt

9y

4x.

Step 6When x = 3 and y = 0, we have

dx

dt= −(−2)

0

24= 0. �

Exercises11. Solve odd-numbered problems 15–25, 35–45 in Section 3.6 of the textbook.

Elasticity of DemandIndications1. Read Section 3.7 of the textbook, “Elasticity of Demand.”



CommentsIn Definition of “Elasticity of Demand” E(p) in Section 3.6 of the textbook, the demand is x and the priceis p. To find E(p), we must have x in terms of p; that is, x(p). The demand depends on the price. StudyFigure 2 on page 231 for the interpretation of x(p).


(Rev. C4)

The value of E(p) indicates the change of the demand according to the price.

The interpretation of E(p) = a is: if the price $p changes by 10%, then the demand changes byapproximately a(10)%.

In terms of revenue, the interpretation of E(p) is referred as inelastic demand, elastic demand and unitelasticity. It is important to understand these interpretations, found in Table 2 on page 232 of the textbook.

Example 3.19. Consider the price demand x = 1875− p2 of Exercise 48 on page 234 of the textbook.The elasticity of demand E(p) is

E(p) = − p(−2p)

1875− p2=

2p2

1875− p2.

Thus, E(15) =450

1875− 225=

3

11≈ 0.272 indicates that if the price $15.00 changes by 10%, then the

demand changes by 2.72%. For this price the demand is inelastic, and the price increase will increaserevenue.

The price demand is elastic for a price p, so that E(p) > 1. That is,

E(p) =2p2

1875− p2> 1.

For p2 < 1875, the elasticity of demand is positive and for 3p2 − 1875 > 0, it is greater than 1. Since p > 0

p <√

1875 = 25√

3 and p > 25.

For a price of 25 < p < 25√

3, the price demand is elastic. Hence, an increase in price between 25 and25√

3, the revenue decreases.

The price demand is unit elasticity if E(p) = 1 (no change). Thus,

E(p) =2p2

1875− p2= 1 p = 25.

A change in price of $25.00 produces the same percentage change in demand. �



Finishing This Unit1. Review the objectives of this unit and make sure you are able to meet each of them.

2. Study the section of the “Chapter 3 Summary and Review” titled “Important Terms, Symbols andConcepts.”


(Rev. C4)


4. You may want to do odd-numbered problems 1–33 from the “Review Exercises” at the end ofChapter 3.



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Practice ExaminationTime: 2 hoursTotal Points: 50Passing grade: 55%











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UNIT 4

Graphing and Optimization

Optimization problems aim to find the optimum (maximum or minimum) of a mathematical model, in ourcase, a function. “Optimum” could mean the maximum revenue, the minimum cost, the maximum area orthe minimum distance traveled to reach a specific point. In terms of the graph of a function, the optimumvalue corresponds to an absolute maximum or minimum. In this unit, we discuss how to graph functions soas to find these absolute values.


1. find the interval on which a function increases and decreases.

2. determine the critical points of a function.

3. determine the local maximum and minimum points of a function.

4. find the interval of concavity of a function.

5. determine the inflection points of a function.

6. apply L’Hôpital’s Rule to evaluate limits.

7. identify the vertical and horizontal asymptotes of a function.

8. sketch the graph of a function.

9. determine the absolute maximum and minimum of a function.

10. solve optimization problems.

PrerequisitesBefore beginning this unit, you must be able to

• find areas and perimeters of triangles and quadrilaterals◦ Book VII section Triangles and Quadrilaterals page 42 of Language of Mathematics

• find surface areas and volumes of regular of regular geometric bodies◦ Book VII section Triangles and Quadrilaterals pages 47–49 of Language of Mathematics


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First Derivative and GraphsIndications1. Read Section 4.1 of the textbook, “The First Derivative and Graphs.”



CommentsIf a function has a local maximum or minimum, that maximum or minimum occurs at a “critical point.”These points determine the intervals of increase and decrease of a function.

To find critical points, we must determine for which points the derivative is zero, which may be challengingfor certain functions. For example, the derivative of the function

f(x) = x4 − 3x3 + 2x

is

f ′(x) = 4x3 − 9x2 + 2,

but solving

4x3 − 9x2 + 2 = 0

requires that we use techniques other than a formula.

In some of the textbook exercises, you are asked to use your calculator to approximate the values of thesecritical points. We leave these exercises as optional. Other exercises are such that you can find the criticalpoints by factoring or applying the quadratic formula.

Example 4.1. To find the critical points of the polynomial function

P (x) =2x5

5− 5x3

3− 3x

We find its derivative P ′(x) = 2x4 − 5x2 − 3 and solve

2x4 − 5x2 − 3 = 0.

If we set u = x2, then we obtain a quadratic equation in terms of u.

2u2 − 5u− 3 = 0


(Rev. C4)

which can be solved applying the quadratic formula:

u =5±

√25− 4(2)(−3)

4=

5± 1

4.

Thus, u = x2 = 1 and u = x2 =3

2. The critical points are

c1 = 1 c2 = −1 c3 = −√

3

2c4 =

√3

2. �

To find the “sign” of a function, we must apply the following algebraic properties of a product and quotient.

ab > 0 anda

b> 0 if and only if either

a > 0 and b > 0 or a < 0 and b < 0.

ab < 0 anda

b< 0 if and only if either

a > 0 and b < 0 or a < 0 and b > 0.

To apply these properties, we often need to factor the function. The next examples present a method youmay want to apply to find the “sign” of a derivative function.

Example 4.2. Let

f(x) = (x2 − 9)2/3.

The derivative of this function is

f ′(x) =2

3(x2 − 9)−1/3(2x) =

4x

3(x2 − 9)1/3.

The critical point is the solution of

4x

3(x2 − 9)1/3= 0

or the points where the derivative is undefined.

Hence, either

4x = 0 or x2 − 9 = 0,

and x = 0, x = 3, or x = −3.

These three points divide the real line into four intervals:

(−∞,−3), (−3, 0), (0, 3), (3,∞).


(Rev. C4)

To determine the “sign” of the derivative in these intervals, take any number in the interval and test thederivative function. For instance −4 is in the interval (−∞,−3), and

f ′(−4) =4(−4)

3((−4)2 − 9)1/3< 0.

As you can see, the derivative is negative on the interval (−∞,−3).

The table below shows how we determine the “sign” of the derivative function.

Interval Number a in the Interval f ′(a) Conclusion

(−∞,−3) −4 f ′(−4) < 0 −

(−3, 0) −2 f ′(−2) > 0 +

(0, 3) 1 f ′(1) < 0 −

(3,∞) 4 f ′(4) > 0 +

We conclude from this table that the function has local minima at −3 and 3, and a local maximum at 0. �

Example 4.3. Let

f(x) = x3(x− 5)2.

The derivative of this function is

f ′(x) = 3x2(x− 5)2 + 2x3(x− 5) = x2(x− 5)(3(x− 5) + 2x) = 5x2(x− 5)(x− 3).

The critical points are x = 0, x = 5 and x = 3. [Why?]

We have four intervals,

(−∞, 0), (0, 3), (3, 5), (5,∞).

The “sign” of the derivative is

Interval Number a in the Interval (x− 5) (x− 3) f ′(x) is

(−∞, 0) −1 1− 5 1− 3 +

(0, 3) 1 1− 5 1− 3 +

(3, 5) 4 4− 5 4− 3 −

(5,∞) 6 6− 5 6− 3 +

The local maximum is at 3 and a local minimum is at 5. �


(Rev. C4)

The interpretation of the “sign” of the derivative gives the intervals where the function is increasing anddecreasing.

In Example 4.2 on page 107, we see that the function is increasing on the intervals (−3, 0) and (3,∞), andit is decreasing on the intervals (−∞,−3) and (0, 3).

In Example 4.3 on the preceding page, the function is increasing on the intervals (−∞, 3) and (5,∞), andit is decreasing on the interval (3, 5).

Exercises1. Do odd-numbered exercises 1–55 and 61–87 in Section 4.1 of the textbook.


Second Derivative and GraphsIndications1. Read Section 4.2 of the textbook, “Second Derivative and Graphs.”



CommentsThe second derivative determines the concavity of a function, and changes in concavity, from upwards todownwards, or downwards to upwards, may indicate inflection points.

Observe that the concavity is given by the “sign” of the second derivative function. We can use the methodwe used in the previous section to find the “sign” of the second derivative function.

To determine the concavity of a function f , we find the critical points of f ′; that is, the numbers c such thatf ′′(c) = 0 or f ′′(c) is undefined. These critical numbers give the intervals of concavity, and we make atable to determine the “sign” of f ′.

A quadratic function Q(x) = ax2 + bx+ c is either concave up or concave down. Since Q′′(x) = 2a theparabola Q is concave up if a > 0 and concave down if a < 0.

Example 4.4. Let

f(x) = e3x − 9ex.

Verify that the second derivative of this function is

f ′′(x) = 9e3x − 9ex = 9ex(e2x − 1).


(Rev. C4)

The critical point is x = 0. [Why?]

We have ex > 0 for all x; hence, the “sign” of f ′ is determined by e2x − 1.

Interval Number a in the Interval e2x − 1 f ′′(x) is

(−∞, 0) −1 e−2 − 1 −

(0,∞) 1 e2 − 1 +

The function is concave upwards on (0,∞) and concave downwards on (−∞, 0). �

For a function f to have an inflection point at c

• f must be continuous at c,

• f ′′(c) = 0 or f ′′(c) is undefined, and

• the concavity of f must change at c from up to down or vise versa.

In the next example, we show you why is important to check these three facts to conclude that a functionhas an inflection point.

Example 4.5. Let

f(x) =

{x3 + 1 if x ≥ 0

−x2 if x < 0

Its first and second derivatives are

f ′(x) =

{3x2 if x ≥ 0

−2x if x < 0and f ′′(x) =

{3x if x ≥ 0

−2 if x < 0

Thus, f ′′(0) = 0 and the concavity changes from down to up at 0, since f ′′(x) < 0 for x < 0 andf ′′(x) > 0 for x > 0. However, this function does not have an inflection point at 0. Its graph is shown inFigure 4.1 on the next page.

In “Conceptual: Insight” on pages 262–263 of the textbook, we have two functions which satisfied the firsttwo conditions for an inflection point at 0, but not the third.

Therefore, all three conditions must be satisfied for a graph to have an inflection point.

Example 4.6. Let

f(x) = ln(x2 + 6x+ 13).

First we observe that x2 + 6x+ 13 is a concave up parabola and b2 − 4ac = 36− 4(13) < 0 the paraboladoes not have roots. Hence, x2 + 6x+ 13 > 0 for all x and the domain of the function is all real numbers.


(Rev. C4)

1

Figure 4.1 The function f(x) does not have an inflection point at 0 �

The second derivative of this function is

f ′′(x) =−2x2 − 12x− 10

(x2 + 6x+ 13)2=−2(x+ 5)(x+ 1)

(x2 + 6x+ 13)2. [Verify this.]

The critical points for f ′ are −5, and −1.

Since (x2 + 6x+ 13)2 > 0 for any x, the “sign” of f ′ is determined by −2(x+ 5) and x+ 1.

Interval −2(x+ 5) (x+ 1) f ′′(x) is

(−∞,−5) + − −

(−5,−1) − − +

(−1,∞) − + −

The function is concave upwards on (−5,−1) and concave downwards on (−∞,−5) and (−1,∞). Theinflection points are at −5 and −1 because the function is continuous for at these points. �

To do Exercises 41–52 in Section 4.2 of the textbook, we must interpret the information given in terms ofthe graph of the function.

Example 4.7. See Exercise 50 in Section 4.2.

a. f(0) = −2, f(1) = 0, f(2) = 4; thus, (0,−2), (1, 0), (2, 4) are on the graph of f .

b. f ′(0) = 0, f ′(2) = 0, f ′(1) not defined; thus, 0, 2 and 1 are critical points.

c. f ′(x) > 0, on (0, 1) and (1, 2); thus, f is increasing on the intervals (0, 1) and (1, 2).

d. f ′(x) < 0 on (−∞, 0) and (2,∞); thus, f is decreasing on the intervals (−∞, 0) and (2,∞), and fhas a local minimum at 0, and a local maximum at 2.


(Rev. C4)

e. f ′′(1) undefined; thus, 1 is the critical point of f ′ and f may have an inflection point at 1.

f. f ′′(x) > 0 on (−∞, 1); thus, f is concave up on the interval (−∞, 1).

g. f ′′(x) < 0 on (1,∞); thus, f is concave down on the interval (1,∞), and f has an inflection point at 1.

From this information we sketch the function, plotting the points (0,−2), (1, 0), (2, 4) from point a, above.

The critical points of f and f ′ are 0, 2, 1. The intervals to be considered are (−∞, 0), (0, 1), (1, 2), (2,∞),by points b and e, above.

• On the interval (−∞, 0), the function is decreasing and concave up by points d and f, above.

• On the interval (0, 1), the function is increasing and concave up by points c and f, above.

• On the interval (1, 2), the function is increasing and concave down by points c and g, above.

• On the interval (2,∞), the function is decreasing and concave down by points d and g, above.

We join the points plotted, following the shape of the function, to obtain the graph shown in Figure 4.2. �

Figure 4.2 Graph of function for Example 4.7

The interpretation of the “sign” of the second derivative gives the intervals on which the function isconcave up and concave down. If the function is concave up, then it moves fast; if it is concave down, thenit moves slowly. Hence, if a function is increasing and concave up, it is growing at a fast rate, as it is thecase with the exponential function. If a function is increasing and concave down, then it is increasing at aslow rate, as it is the case with the logarithmic function.

The inflection point indicates where the rate of the function changes from fast (concave up) to slow(concave down), or from slow (concave down) to fast (concave up).


(Rev. C4)

Sign of f ′ Sign of f ′′ Graph of f Interpretation of f

+ + increasing & concave up increasing fast

+ − increasing & concave down increasing slowly

− + decreasing & concave up decreasing fast

− − decreasing & concave down decreasing slowly

Example 4.8. The function f(x) = x3 is increasing and has an inflection point at 0. Moreover, it isconcave down for x < 0, and concave up for x > 0. Hence, before 0, the function is increasing at a slowrate, and after 0, the function is increasing at a fast rate. �


4. Solve odd-numbered problems 75–91 on pages 271–272.

L’Hôpital’s RuleIndications1. Read Section 4.3 of the textbook, “L’Hôpital’s Rule.”



CommentsIn working with limits, you must interpret the symbols x→∞ and x→ −∞ correctly. The formerindicates that x is positive and very large, the latter says that x is negative and very small. Thus, thenotations for the limits

limx→∞

f(x) =∞ limx→∞

f(x) = −∞ limx→∞

f(x) = L

indicate that, for x positive and very large, the values of the function f are positive and very large, negativeand very small, and very close to the number L, respectively. Likewise,

limx→−∞

f(x) =∞ limx→−∞

f(x) = −∞ limx→−∞

f(x) = L

indicate that, for x negative and very small, the values of the function f are positive and very large,negative and very small, and very close to the number L, respectively.


(Rev. C4)

There are several theorems we can apply to evaluate limits, and we state them here to complement thediscussion in the textbook. Their proofs are beyond the scope of this course.

Theorem 4.1.

If

limx→±∞

h(x) = 0 and h(x) > 0 for x→ ±∞,

then

limx→±∞

1

h(x)=∞.

To understand this statement, note that, if limx→±∞

h(x) = 0, then, for x→ ±∞, the values of h(x) are very

close to 0.

Then, if h(x) > 0 for x→ ±∞,

h(x) is close to zero from the right (i.e., from the positive side), and

the reciprocal1

h(x)is positive and very large.

Theorem 4.1 holds if x→ ±∞ is replaced by x→ a, x→ a+ or x→ a−, for any number a.

Example 4.9. We know that

limx→2

(x− 2)2 = 0

and (x− 2)2 > 0 for any x 6= 2; hence,

limx→2

1

(x− 2)2=∞. �


limx→1−

x3 − x2 − x+ 1 = 0

and for x→ 1−, we take any number close to 1 on the left, say x = 0.9. Thus,

(0.9)3 − (0.9)2 − 0.9 + 1 = 0.019 > 0,

and we have

x3 − x2 − x+ 1 > 0 for x→ 1−.


(Rev. C4)

Therefore,

limx→1−

1

x3 − x2 − x+ 1=∞. �

If limx→±∞

h(x) = 0, then for x→ ±∞, the values of h(x) are very close to 0.

If h(x) < 0 for x→ ±∞, then

h(x) is close to zero from the left (i.e., the negative side), and

the reciprocal1

h(x)is negative and very small.

The latter fact gives us the next theorem.

Theorem 4.2.

If limx→±∞

h(x) = 0 and h(x) < 0 for x→ ±∞, then

limx→±∞

1

h(x)= −∞

Theorem 4.2 holds if x→ ±∞ is replaced by x→ a, x→ a+ or x→ a−, for any number a.


limx→2−

x3 − 2x2 + 8x− 16 = 0,

and for x→ 2−, we take any number close to 2 on the left, say x = 1.9. Thus,

(1.9)3 − 2(1.9)2 + 8(1.9)− 16 = −1.161 < 0,

and we have

x3 − 2x2 + 8x− 16 < 0 for x→ 2−.

Therefore,

limx→2−

1

x3 − 2x2 + 8x− 16= −∞ �


(Rev. C4)

Theorem 4.3.

If limx→±∞

f(x) = L and limx→±∞

g(x) =∞, then

limx→±∞

f(x)g(x) =

{∞ if L > 0

−∞ if L < 0

Intuitively we see that if, for x→ ±∞, the values of f(x) are close to L, and g(x) are positive and verylarge, then the product f(x)g(x) is positive and very large if L > 0, or negative and very small if L < 0.

You can use the same argument to convince yourself that the following theorem is true.

Theorem 4.4.

If limx→±∞

f(x) = L and limx→±∞

g(x) = −∞, then

limx→±∞

f(x)g(x) =

{−∞ if L > 0

∞ if L < 0

Theorems 4.3 and 4.4 hold if x→ ±∞ is replaced by x→ a, x→ a+ or x→ a−, for any number a.

Compare Theorem 4.3 and Theorem 4.4 with Theorem 2.8 on page 53.

We can apply these theorems in the evaluation of the limits in Example 1 on page 275 of the textbook.

A. limx→2

5 = 5 > 0 and limx→2

1

(x− 2)4=∞.

Then, by Theorem 4.3,

limx→2

5

(x− 2)4= lim

x→25

(1

(x− 2)4

)=∞.

B. limx→−1−

4 = 4 > 0 and limx→−1−

1

(x+ 1)3= −∞.

Then, by Theorem 4.3

limx→−1−

4

(x+ 1)3= lim

x→−1−4

(1

(x+ 1)3

)= −∞.

Example 4.12. Let us evaluate the limit

limx→1−

x2 + 1

x3 − x2 − x+ 1.


(Rev. C4)

We see that

limx→1−

x2 + 1 = 2 > 0 and limx→1−

x3 − x2 − x+ 1 = 0.

By Example 4.10 on page 114,

limx→1−

1

x3 − x2 − x+ 1=∞,

and by Theorem 4.3 on the previous page, we have

limx→1−

x2 + 1

x3 − x2 − x+ 1= lim

x→1−(x2 + 1)

(1

x3 − x2 − x+ 1

)=∞ �

Example 4.13. Let us evaluate the limit

limx→2+

e2x

4− x2.

We see that

limx→2+

e2x = e4 > 0 and limx→2+

4− x2 = 0.

Would we apply Theorem 4.3 or Theorem 4.4?

We find that 4− x2 < 0 for x→ 2+. Hence, by Theorem 4.2 on page 115,

limx→2+

1

4− x2= −∞.

We now apply Theorem 4.4 on the previous page, and we conclude that

limx→2+

e2x

4− x2= lim

x→2+e2x(

1

4− x2

)= −∞. �

Note that,

• if x→∞, then xn →∞ for any positive integer n.

• if x→ −∞, then xm →∞ for any positive even integer m, and

• xk → −∞ for any positive odd integer k.

In limit notation:

• limx→∞

xn =∞ for any positive integer n.

• limx→−∞

xm =∞ for any positive even integer m,

• limx→−∞

xk = −∞ for any positive odd integer m.


(Rev. C4)

From parts (C) and (D) of Figure 1 on page 274, we can see that for any constant k,

limx→±∞

k

xn= 0 for any positive integer n.

These results help us to see that the limit of a polynomial of degree n is the same as the limit of the term ofthe biggest power n. Indeed

anxn + an−1x

n−1 + an−2xn−2 + ...+ a1x+ a0

= xn(an +

an−1x

+an−2x2

+ ...+a1xn−1

+a0xn

).

Since

limx→±∞

1 +an−1x

+an−2x2

+ ...+a1xn−1

+a0xn

= an + limx→±∞

an−1x

+ limx→±∞

an−2x2

+ ...+ limx→±∞

a1xn−1

+ limx→±∞

a0xn

= an,

we have

limx→±∞

anxn + an−1x

n−1 + ...+ a1x+ a0 = limx→±∞

anxn.

Example 4.14. The polynomial

5x7 − 5x3 − x2

4−√

2x+ 1

is of degree 7. Thus, by Theorem 4.3 on page 116,

limx→∞

5x7 − 5x3 − x2

4−√

2x+ 1 = limx→∞

5x7 =∞,

and by Theorem 4.4 on page 116,

limx→−∞

5x7 − 5x3 − x2

4−√

2x+ 1 = limx→−∞

5x7 = −∞. �

L’Hôpital’s Rule is applied when the limits are of type 0/0 or∞/∞. It is important that you use thetheorems given in this section to determine correctly whether L’Hôpital’s Rule applies.


Curve-sketching TechniquesIndications1. Read Section 4.4 of the textbook, “Curve-sketching Techniques.”


(Rev. C4)



CommentsIn this section, the procedure to sketch graphs of function includes asymptotes: vertical, horizontal, andoblique or slant.

To determine asymptotes we must apply limits.

Horizontal Asymptotes The horizontal line y = L is a horizontal asymptote of a function f if

limx→±∞

f(x) = L.

Vertical Asymptotes The vertical line x = c is a vertical asymptote of a function f if

limx→c+

f(x) = ±∞ or limx→c−

f(x) = ±∞.

Oblique Asymptotes A line L(x) = mx+ b is an oblique or slant asymptote of a function f if

limx→±∞

f(x)− L(x) = 0.

Example 4.15. The rational function

f(x) =x2 − x− 20

x2 + 5x+ 4

has two possible vertical asymptotes where the denominator is zero.

x2 + 5x+ 4 = (x+ 4)(x+ 1) = 0.

These are: x = −4 and x = −1. We take the limits at these two numbers from either the right or left sides.

limx→−4+

x2 − x− 20

x2 + 5x+ 4= lim

x→−4+

(x+ 4)(x− 5)

(x+ 4)(x+ 1)= lim

x→−4+

x− 5

x+ 1= 3.

Hence, x = −4 is not a vertical asymptote.

limx→−1+

x2 − x− 20

x2 + 5x+ 4= lim

x→−1+

x− 5

x+ 1= −∞

because x+ 1 > 0 for x→ 1+, and

limx→−1+

x− 5 = −9 < 0 and limx→−1+

x+ 1 = 0.


(Rev. C4)

Hence, x = −1 is a vertical asymptote. Since,

limx→±∞

x2 − x− 20

x2 + 5x+ 4= 1

The horizontal line y = 1 is a horizontal asymptote. �

From the list of exercises in Section 4.4 of the textbook, we see that the functions of numbers 45, 46, 61, 62have oblique asymptotes.

Example 4.16. By long division the function

f(x) =x2

x+ 2

is equal to

x2

x+ 2= x− 2 +

4

x+ 2.

See also that

x2

x+ 2=x2 − 4 + 4

x+ 2=x2 − 4

x+ 2+

4

x+ 2

=(x− 2)(x+ 2)

x+ 2+

4

x+ 2= x− 2 +

4

x+ 2.

The line L(x) = x− 2 is an oblique asymptote because

limx→∞

f(x)− L(x) = limx→∞

4

x+ 1= 0.

Since the degree of the numerator polynomial is greater than the degree of the denominator polynomial,this rational function does not have horizontal asymptotes.

A possible vertical asymptote is x = −2. Since x+ 2 > 0 for x→ −2+, and

limx→−2+

x+ 2 = 0

we have that

limx→−2+

4

x+ 2=∞.

Therefore,

limx→−2+

x− 2 +4

x+ 2=∞

since

limx→−2+

x− 2 = −4.


(Rev. C4)

So, the vertical line x = −2 is indeed a vertical asymptote. �

Example 4.17. Consider the function

f(x) =x2 + 3xex + 4ex

ex.

If we factor ex in the numerator we have ex(3x+ 4) and the line L(x) = 3x+ 4 may be a slant asymptoteof f .

x2 + 3xex + 4ex

ex=x2 + ex(3x+ 4)

ex=x2

ex+ 3x+ 4.

Hence, f(x)− L(x) =x2

exand by L’Hôpital’s Rule twice

limx→∞

x2

ex= lim

x→∞

2

ex= 0.

Indeed the line L(x) is a slant asymptote. �



Absolute Maxima and MinimaIndications1. Read Section 4.5 of the textbook, “Absolute Maxima and Minima.”



CommentsIn this section, you are given two methods of finding the absolute maxima and minima: the procedure usedon page 298 of the textbook, known as the “closed-interval method”; and the second-derivative test (seeTheorem 3 on page 301).

The procedure on page 298 is applied only to continuous functions on a closed interval. If these conditionsdo not hold, we apply the second-derivative test. If this test is inconclusive, we apply the first-derivative test.


(Rev. C4)

Let Q(x) = ax2 + bx+ c be a quadratic function.

Q′(x) = 2ax+ b = 0 for x = − b

2a.

This is the only critical point of Q. And

Q′′(x) = 2a.

Theretofore, by the second derivative test, a quadratic function has

• a maximum at x = − b

2aif a < 0, and

• a minimum at x = − b

2aif a > 0.

Example 4.18. Consider the quadratic function

Q(x) = −3x2 +2x

5− 1

3.

Since a = −3 < 0 this function has a maximum at

− b

2a= − 2/5

2(−3)=

1

15.

The coordinates of this maximum point are(

1

15,−24

75

); since

Q

(1

15

)= −3

(1

15

)2

+2

5

(1

15

)− 1

3= −24

75.

The absolute maximum and minimum values of Q on the interval [0, 1] are given by the Closed Interval

Method. At the endpoints we have Q(0) = −1

3, Q(1) = −44

15. The critical point

1

15is in the interval.

Hence,

Values for x Values of Q(x)

0 −1

3

1

15−24

75

1 −44

15

The absolute minimum value of Q is −44

15at 1, and the absolute maximum value is −24

75at

1

15. �

The closed-interval method does not apply in exercises 27–66; it does apply in exercises 19–26, and 67–72(pp. 303–304) of the textbook.


(Rev. C4)

Exercises8. Do odd-numbered exercises 1–63, 67–77 in Section 4.5 of the textbook.

OptimizationIndications1. Read Section 4.6 of the textbook, “Optimization.”



CommentsTo solve optimization problems, we must determined which quantity is to be maximized or minimized.Once this is established, this quantity becomes the dependent variable of a function (say, f ). We must thensolve the problem (f(x)). There is no step-by-step procedure for constructing this function: we must applyour knowledge to make it up.

The optimization of a quantity is subject to some limitations, which give the constraint equation orequations associated with the problem.

The construction of the function to be maximized or minimized, and the establishment of the constraintequations, are the challenging part of solving optimization problems.

A procedure for solving optimization problems is given on page 305 of the textbook; we show you anotherin Example 4.19.

Example 4.19. Find the dimensions of a rectangle of area 225 square centimetres that has the smallestperimeter. What is the perimeter?

Step 1Determine the variable to be maximized or minimized. We must minimize the perimeter of a rectangle. Wecall this variable P .

Step 2Determine other variables. The perimeter depends on the size of the sides. We call these variables x and y.Thus,

P = 2x+ 2y.

Step 3Establish the constraint equation. The limitation is the area of the rectangle—it must be 225 squarecentimetres. Thus,

xy = 225.


(Rev. C4)

Step 4Construct the function to be optimized. We solve one variable from the constraint equation, say y; hence,

y =225

x.

The function is

P (x) = 2x+ 2

(225

x

)= 2x+

450

x.

Step 5Determine the interval of the independent variable. Since xy = 225 and x, y > 0, we have 0 < x < 225.This value gives the interval (0, 225), which is an open interval.

Step 6Apply a method to find the optimum value required. We need to minimize the function

P (x) = 2x+450

xon the interval (0, 225).

We apply the second-derivative test:

P ′(x) = 2− 450

x2= 0 if x =

√225 = 15;

P ′′(x) =900

x3and P ′′(15) =

900

153> 0.

Hence, the function has a minimum at x = 15.

Step 7Answer the question or questions required. For x = 15, the other dimension is

y =225

15= 15,

and the dimensions of the rectangle are 15× 15 centimetres. The perimeter is 4(15) = 60 centimetres. �

A quadratic function Q(x) = ax2 + bx+ c has a maximum if a < 0 and a minimum if a > 0. This extreme

is at x = − b

2a.

Example 4.20. Exercise 10 on page 313 of the textbook. Find two numbers whose sum is 19 and whoseproduct is maximum.

Step 1The variable to be maximized is the product of the two numbers. Say P .

Step 2The product is of two numbers, say x, y. Hence, P = xy.

Step 3


(Rev. C4)

The limitation is that the sum of x and y must be 19. Hence, x+ y = 19. We solve for y = 19− x.

Step 4The function to be maximized is

P (x) = x(19− x) = 19x− x2.

Step 5The range for the variable x is 0 < x < 19 because x, y cannot be zero.

Step 6It is a quadratic with a maximum at x = − 19

−2= 9.5.

Step 7The numbers are y = 19− 9.5 = 9.5. The product (not required) is 9.52 = 90.25. �

If the range of the variable x of the function to be maximized or minimized f is not a closed interval, thenthe Closed Interval Method (on page 298 of the textbook) does not apply.

Example 4.21. Exercise 14 on page 313 of the textbook. Find two positive numbers whose product is 19and whose sum is a minimum.

Step 1The variable to be minimized is the sum of the two numbers, say, S.

Step 2The sum is of two numbers, say x, y. Hence, S = x+ y.

Step 3The limitation is that the product of x and y must be 19. Hence, xy = 19. We solve for y =

19

x.

Step 4The function to be minimized is

S(x) = x+19

x.

Step 5The range for the variable x is 0 < x ≤ 19 because x, y cannot be zero. Not a closed interval.

Step 6We apply the second derivative test.

S ′(x) = 1− 19

x2= 0 x2 = 19 |x| =

√19.

Since x > 0. We have one critical number√

19.

S ′′(x) =2(19)

x3=

38

x3S ′′(√

19) =38

193/2=

2√19

> 0.


(Rev. C4)

So the function S has a minimum at x =2√19

.

Step 7

The numbers are x =2√19

and y = 19

(19√19

)= 2√

19. �

Note that in Exercises 44, 47, 48, 49, and 50, the function to be maximized or minimized is given. Theconstraint equation must be established.


10. solve the odd numbered problems 19–45 in Section 4.6 of the textbook.


2. Study the section of the “Chapter 4 Summary and Review” titled “Important Terms, Symbols andConcepts,” page 317 of the textbook.


4. You may want to do odd-numbered problems 1–51 from the “Review Exercise” section onpages 318–319.

5. Complete the “Practice Examination” provided for this unit. Evaluate yourself, first checking youranswers against those provided in the “Answers” section at the end of the textbook, and thencomparing your solutions with those provided on pages 4–101, 4–133 of the Student’s SolutionsManual. The number of points in a question may indicate the number of steps in the solution. Giveyourself full credit if your answer is correct and you give a complete solution, even if your solutiondiffers from that shown in the Student’s Solutions Manual.


(Rev. C4)

Practice ExaminationTime: 2 hoursTotal Points: 55Passing grade: 55%










(Rev. C4)

UNIT 5

Integration

In Unit 2, you learned how to differentiate. In this unit, you will learn to reconstruct a function from itsderivative, a process (known as “integration”) that is the reverse or dual operation of differentiation.


1. find the antiderivative of algebraic functions.

2. evaluate indefinite integrals.

3. use the method of substitution to integrate.

4. apply integration to solve problems in economics and life sciences.

5. use differential equation to solve problems of growth, decay, and learning.

6. approximate the area under a curve using Riemann sums.

7. integrate definite integrals.

8. state and apply the Mean Value Theorem to justify the Fundamental Theorem of Calculus.

Antiderivatives and Indefinite IntegralsIndications1. Read Section 5.1 of the textbook, “Antiderivatives and Indefinite Integrals.”




(Rev. C4)

CommentsIn the notation of the antiderivative or indefinite integral∫

f(x) dx,

the symbol∫

is called the integral, the function f(x) is called the integrand function, and dx is the

differential.

By the definition of the indefinite integral,∫d

dxf(x) dx = f(x).

The other operation,

d

dx

∫(taking the derivative of the indefinite integral), introduced on page 325 of the textbook is explained in thenext unit.

Observe that we do not have formulas or rules for integration as we do for differentiation; instead, we mustappeal to our knowledge of basic functions as listed on page 325.

In the following examples, we use algebraic properties to fit the integrand function into one or more ofthese basic functions.

Example 5.1. The integrand function in the integral∫ (x2 − 5 +

1

x

)2

dx

does not fit any of the basic functions on page 325. But if we multiply it out, we get

f(x) =

(x2 − 5 +

1

x

)(x2 − 5 +

1

x

)= x4 − 10x2 + 2x+ 25− 10

x+

1

x2

= x4 − 10x2 + 2x+ 25− 10

x+ x−2.

[Verify this.]

Each of the terms in this sum fits one of the basic formulas on page 362 or the properties on page 325, andwe can also apply the properties of Indefinite Integrals on page 326. Hence, as in the previous example


(Rev. C4)

∫ (x2 − 5 +

1

x

)2

dx =

∫x4 − 10x2 + 2x+ 25− 10

x+ x−2 dx

=x5

5− 10x3

3+ x2 + 25x− 10 ln |x| − 1

x+ C. �

Example 5.2. For the indefinite integral∫(2x3 − 3x)2√

xdx,

we have

(2x3 − 3x)2√x

=4x6 − 12x4 + 9x2√

x

= 4x11/2 − 12x7/2 + 9x3/2.

Hence,∫(2x3 − 3x)2√

xdx =

∫4x11/2 − 12x7/2 + 9x3/2 dx

=4x13/2

13/2− 12x9/2

9/2+

9x5/2

5/2

=8x13/2

13− 24x9/2

9+

18x5/2

5

=8x13/2

13− 8x9/2

3+

18x5/2

5+ C. �

To apply Formula 1 on page 325 of the textbook, we must have the integrand function in the right formxn. Apply algebraic operations to achieve this if necessary.

Integration can be checked out applying differentiation. See that∫f(x) dx = F (x) + C

if and only if

F ′(x) = f(x).

Example 5.3. The integrals of the basic functions on page 325 of the textbook are correct because

d

dx

(xn+1

n+ 1

)=

(n+ 1)xn

n+ 1= xn.

Also, by Example 3.12 on page 96

d

dxln |x| = 1

xand

d

dxex = ex. �


(Rev. C4)

Example 5.4. Exercise 28 (page 332) of the textbook. Is F (x) = x lnx− x+ e an antiderivative off(x) = ln x?

To answer this question we differentiate F .

d

dx(x lnx− x+ e) = x

(d

dxlnx

)+ lnx− 1 = 1 + lnx− 1 = ln x.

The answer is Yes, it is. �



Integration by SubstitutionIndications1. Read Section 5.2 of the textbook, “Integration by Substitution.”



CommentsThe three formulas listed on page 337 result from the Chain Rule.

For Formula 1, we have

d

dx(f(x))n+1 = (n+ 1)[f(x)]nf ′(x);

hence,

d

dx

(f(x))n+1

n+ 1= [f(x)]nf ′(x).

For Formula 2,

d

dxef(x) = ef(x)f ′(x).

For Formula 3,

d

dxln(f(x)) =

f ′(x)

f(x).


(Rev. C4)

Thus, the corresponding integrals are as shown below.

Integral 1∫[f(x)]nf ′(x) dx =

[f(x)]n+1

n+ 1+ C

Integral 2∫ef(x)f ′(x) dx = ef(x) + C

Integral 3∫f ′(x)

f(x)dx = ln |f(x)|+ C

To integrate properly, we need to develop the skill to fit the integrand function into one of the forms listedin these integrals. You can see that all three of them have f ′(x) dx as part of the integrand function, and asyou know, f ′(x) dx is the differential of f . Thus, we should be able to identify the function f and itsderivative f ′.

Example 5.5. If a is a constant, then in the integral∫eax dx,

we identify f as f(x) = ax to apply the Integral 2 above.

Since f ′(x) = a, we need the constant a in the integrand. We include it by multiplying the integral by

1

a(a),

and applying Property 4 on page 326 of the textbook. Thus,∫eax dx =

1

a

∫eax(a) dx =

eax

a+ C. �

Example 5.6. If a and b are constants, then in the integral∫1

ax+ bdx,

we identify f as f(x) = ax+ b and n = −1 to apply the Integral 1 above. Since f ′(x) = a, as in theprevious example, we include this constant in the integrand, and∫

1

ax+ bdx =

1

a

∫a

ax+ bdx =

1

a

∫(ax+ b)−1(a) dx =

ln |ax+ b|a

+ C. �

As a general rule, when integrating a quotient

f(x)

g(x),


(Rev. C4)

the first thing we should see is if g′(x) = f(x) or g′(x) = kf(x) for some constant k, in order to applyIntegral 3, above.

Example 5.7. To apply Integral 3 to the integral∫x

5x2 − 6dx,

we must recognize that

d

dx5x2 − 6 = 10x.

Thus, we apply Integral 3 with f(x) = 5x2 − 6∫x

5x2 − 6dx =

1

10

∫10x

5x2 − 6dx =

ln |5x2 − 6|10

+ C. �

Example 5.8. In the integral∫1

5x2 − 6dx,

we see that

d

dx5x2 − 6 = 10x.

Therefore, the integrand function cannot be fitted into Integral 3, because we cannot include a variable. It isnecessary to use another method (one that is beyond the scope of this course) to evaluate this integral. �

The substitution method considered on page 339 of the textbook is in fact a u-identification. We identifythe function f as we indicated above, and we set u = f ; hence, the differential is du = f ′(x)dx andIntegrals 1–3, listed above, become as listed on page 339 of the textbook.

This method is not always necessary, if the differential in the integrand function can be obtained bymultiplying it by a constant, as we saw in the previous two examples. Compare also Example 5.6 on thepreceding page with Example 5 on page 340 of the textbook.

The four steps of the substitution method on page 339 of the textbook is absolutely necessary when theintegrand function does not fit any of Integrals 1–3 listed on page 337 of the textbook (see Example 6 onpage 342).

Example 5.9. We need to use the substitution method to evaluate the integral∫x2√x+ 1 dx.

The substitution we choose is u = x+ 1; hence, the differential is

du = (1)dx = dx,


(Rev. C4)

and we need to replace every other term in the integrand function by an expression in terms of u. In thiscase, the only other expression is x2; hence,

u = x+ 1 and x2 = (u− 1)2.

The integral in terms of u is∫(u− 1)2

√u du =

∫(u2 − 2u+ 1)

√u du =

∫u5/2 − 2u3/2 + u1/2 du.

We apply the basic integral 1 (pp. 325) of the textbook, and obtain

∫u5/2 − 2u3/2 + u1/2 du =

2u7/2

7− 4u5/2

5+

2u3/2

3.

We then substitute u = x+ 1 to give our answer in terms of x. Thus,

2(x+ 1)7/2

7− 4(x+ 1)5/2

5+

2(x+ 1)3/2

3+ C. �

Note: If the integral is given in terms of a variable, say x, your answer must be in terms of the samevariable. Do not leave it in terms of the variable used in the substitution.



Differential Equations; Growth and DecayIndications1. Read Section 5.3 of the textbook, “Differentia Equations; Growth and Decay.”



CommentsDifferential equation are used to solve problems, so keep in mind Table 1 on page 354, which tells youwhich differential equation you must apply to solve a particular problem.


(Rev. C4)

Observe that problems of unlimited growth and exponential decay refer to functions each of which has arate of change that is proportional to the function itself. See, for example, like Exercises 48 and 49 onpage 355 of the textbook.

When reading a differential equation we must identify the dependent and independent variables. In theequation (5) on page 349 of the textbook, we have

dA

dt= rA A(0) = P.

The function A is a function of t. Thus, A(t) and its derivative with respect tot t is equal to r times itself.Moreover, A(0) = P is a condition of A.

Example 5.10. The equation

dy

dt= −3y

indicates that y is a function of t and its derivative is equal to −3 times itself. �



The Definite IntegralIndications1. Read Section 5.4, “The Definite Integral” (up to “The Definite Integrals as a Limit of Sums”).


CommentsIn calculus the area “under the curve f on the interval [a, b]” is the area between the curve of the function fon the interval [a, b] and the x-axis.

If the curve f consists of straight lines, the area under the curve consists of sums of areas of rectangles andtriangles.

Example 5.11. The area under the curve of

f(x) =

{x+ 2 if x < 0

2− x if x ≥ 0

on the interval [−2, 3] consists of two triangles. See Figure 5.1 on the next page.


(Rev. C4)

Figure 5.1 Area under the curve of f(x) on the interval [−2, 3], Example 5.11

The total area is

4(2)

2+

1

2=

5

2. �

Example 5.12. The area under the curve of

f(x) =

4x

3+

5

3if x < 1

−x4

+13

4if x ≥ 1

on the interval [−2, 3] consists of three triangles and one rectangle. See Figure 5.2.

Figure 5.2 Area under the curve of f(x) on the interval [−2, 3], Example 5.12

The total area is(3

4

)(1)

2+

(5

4

)(5

2

)2

+

3

(1

2

)2

+ 3

(5

2

)=

163

16. �


(Rev. C4)

However, when the curve f is a smooth arc, the exact value of the area under the curve poses a moredifficult problem, one that requires the use of calculus.

If A is the area under the curve f on the interval [a, b], and Ln, Rn are the left and right sums, as stated onFigures 2 and 3 page 359 of the textbook, then

Ln ≤ A ≤ Rn.

If Sn is the Riemann sum then

A ≈ Sn.

Note that the use of rectangles gives only an approximation of A. Therefore, the larger the n the better theapproximation.

To find the exact value of the area A, we use the Mean Value Theorem.

Theorem 5.1.

If F is a differentiable function on the interval (a, b) and continuous on the interval [a, b], then thereexists a < c < b such that

F (b)− F (a)

b− a= F ′(c)

or

F (b)− F (a) = F ′(c)(b− a).

To make up the Riemann sum, we take n rectangles and divide the interval (a, b) into n parts:

a = x1 < x2 < x3 < · · · < xn < xn+1 = b.

The bases of the n rectangles are

(x2 − x1), (x3 − x2), (x4 − x3), . . . , (xn+1 − xn).

We take arbitrary numbers ck so that

x1 < c1 < x2 < c2 < x3 < c3 < x4 < .... < cn < xn+1;

then, the heights of the rectangles are

f(c1), f(c2), f(c3), . . . , f(cn).

For these n rectangles,

A ≈ f(c1)(x2 − x1) + f(c2)(x3 − x2) + f(c3)(x4 − x3) + · · ·+ f(cn)(xn+1 − xn).


(Rev. C4)

In sigma notation,

A ≈n∑i=1

f(ci)(xi+1 − xi). (5.1)

Let us use the Mean Value Theorem (Theorem 5.1 on the previous page) to see what the sum inEquation (5.1) is.

Let F be the antiderivative of the function f ; that is, F ′ = f .

Hence, F is continuous in the interval [a, b] and differentiable in (a, b), and for any 1 ≤ i ≤ n, the functionF is continuous in the interval [xi, xi+1] and differentiable in (xi, xi+1) for i = 1, 2, 3....n. By the MeanValue Theorem, then,

F (xi+1)− F (xi) = F ′(ci)(xi+1 − xi) = f(ci)(xi+1 − xi).

for some xi < ci < xi+1.

Hence,

n∑i=1

f(ci)(xi+1 − xi) =n∑i=1

F (xi+1)− F (xi).

The sumn∑i=1

F (xi+1)− F (xi)

is equal to F (b)− F (a) because

n∑i=1

F (xi+1)− F (xi) = [F (x2)− F (x1)] + [F (x3)− F (x2)] + . . .

· · ·+ [F (xn)− F (xn−1)] + [F (xn+1)− F (xn)]

= F (xn+1)− F (x1) = F (b)− F (a)

Therefore,

A ≈n∑i=1

f(ci)(xi+1 − xi) = F (b)− F (a).

As we said earlier, the more rectangles (the larger n) the better the approximation; furthermore, the morerectangles, the smaller (xi+1− xi). Since (xi+1− xi) is a small quantity of the variable x, it is a differential

∆xi = (xi+1 − xi).

Hence, as n→∞ the sum

n∑i=1

f(ci)∆xi


(Rev. C4)

approaches the area A. This is precisely a limit, and we have found that

A = limn→∞

n∑i=1

f(ci)∆xi = limn→∞

F (b)− F (a) = F (b)− F (a).

This equation is the Fundamental Theorem of Calculus. It is fundamental because a simple geometricalconcept (area) is given by the antiderivative of a function, which is not a geometrical concept. That thesetwo concepts are so intimately related is what makes this theorem so outstanding. Moreover, to arrive tothis theorem was not easy—we needed to learn about limits, a concept which took humankind a long timeto understand fully—and we needed to apply the Mean Value Theorem.

We write F (x)∣∣∣ba

to indicate the difference F (b)− F (a); that is

F (x)∣∣∣ba

= F (b)− F (a).

What we have shown so far is that, if f is a continuous positive function on the interval [a, b] withantiderivative F , then the area under the curve on the interval is the limit.

limn→∞

n∑i=1

f(ci)∆xi = F (x)∣∣∣ba

= F (b)− F (a).

The notation for this limit has to indicate the function f , the interval [a, b] and the fact that it is the limit ofa sum; hence, instead of Σ we use the integral symbol

∫and we indicate the interval [a, b] by writing

b as an upper limit of integration, and

a as a lower limit of integration, as follows:∫ b

a

.

We indicate the integrand function as f and the differential as dx; therefore, we have the definite integral ofthe function f on the interval [a, b]:∫ b

a

f(x) dx = limn→∞

n∑i=1


= F (b)− F (a).

In summary, if f is a continuous positive function on the interval [a, b] with antiderivative F , then the areaunder the curve is∫ b

a

f(x) dx = limn→∞

n∑i=1


= F (b)− F (a).

Example 5.13. The function f(x) =√

3x+ 5 is positive on the interval [0, 6]; hence, the area under thecurve is∫ 6

0

(3x+ 5)1/2 dx =1

3

∫ 6

0

(3x+ 5)1/2(3) dx =2(3x+ 5)3/2

9

∣∣∣60

=2(3(6) + 5)3/2

9− (2)53/2

9≈ 22.03. �


(Rev. C4)

Example 5.14. The function f(x) = x2 − 4 is negative on the interval [0, 2] and positive in the interval[2, 5].

If we consider the definite integral on the interval [0, 2], we have∫ 2

0

x2 − 4 dx =x3

3− 4x

∣∣∣20

=8

3− 8 = −16

3< 0.

On the other hand, the definite integral on the interval [2, 5] is∫ 5

2

x2 − 4 dx =x3

3− 4x

∣∣∣52

=

(125

3− 40

)−(

8

3− 8

)= 7 > 0.

Since an area is positive, it is clear that −16

3is not the area under the curve on the interval [0, 2]. The

definite integral is negative because the function is negative on this interval. �

If the function f(x) is negative on the interval [a, b], then the graph of f(x) +K, with K a positiveconstant, is equal to the translation K units up of the graph of f(x). Thus, the area under the curve of f(x)is equal to the area of the rectangle with base (b− a) and height K, minus the area under the curve off(x) +K. See Figure 5.3.

Figure 5.3 Graph of f(x) translated K units up

Hence, the area under the curve f(x) +K on the interval [a, b] is∫ b

a

f(x) +K dx = F (x) +Kx∣∣∣ba

= F (b) +Kb− (F (a)−Ka) = (F (b)− F (a)) +K(b− a),


(Rev. C4)

and the area under the curve f(x) on the interval [a, b] is

K(b− a)− [F (b)− F (a)) +K(b− a)] = −[F (b)− F (a)].

Therefore, the area under the curve f(x) on the interval [a, b] is

−∫ b

a

f(x) dx = −[F (b)− F (a)].

Example 5.15. The area under the curve f(x) = x2 − 4 on the interval [0, 5] is, by Example 5.14 onpage 139, equal to

−∫ 2

0

x2 − 4 dx+

∫ 5

2

x2 − 4 dx =16

3+ 7 =

37

3.

On the other hand,∫ 5

0

x2 − 4 dx =x3

3− 4x

∣∣∣50

=125

3− 4(5) =

65

3. �

This last example illustrates the difference between the area below the curve and the evaluation of thedefinite integral. Only when the integrand function is positive on the interval of integration do the definiteintegrals coincide with the area below the curve.

Indications1. Read Section 5.4 from “The Definite Integral as a Limit of Sums” to the end of the section.


Exercises7. Do odd-numbered exercises 31–53, in Section 5.4 the textbook.

8. Do odd-numbered problems 73–77 in Section 5.4.

The Fundamental Theorem of CalculusIndications1. Read Section 5.5 of the textbook, “The Fundamental Theorem of Calculus.”




(Rev. C4)

CommentsIf the function f(x) is continuous on an interval I , and a is in this interval, then for any a < x, the integral∫ x

a

f(x) dx

is a function and∫ x

a

f(x) dx = F (x)− F (a).

Thus, its derivative is

d

dx

∫ x

a

f(x) dx =d

dxF (x)− F (a) = F ′(x) = f(x);

that is,

d

dx

∫ x

a

f(x) dx = f(x).

This result confirms that differentiation and integration are opposite (dual) operations of each other.


10. Do odd-numbered problems 69–85 in Section 5.5.


2. Study “Chapter 5 Review” titled “Important Terms, Symbols and Concepts.”


4. You may want to do odd-numbered problems 1–31, 39–55 and 67–75 from the “Review Exercises”section at the end of Chapter 5.



(Rev. C4)


Do the following exercises from the “Review Exercises” at the end of Chapter 5.











(Rev. C4)

UNIT 6

Additional Integration Topics

In this unit, we consider a geometric application of the definite integral: the evaluation of areas betweencurves. We will learn to apply the integration technique of parts and to use tables of integration to integratea larger type of functions.


1. find the area between several curves.

2. determine if a function is a probability density function.

3. determine the probability of a variable selected at random.

4. find the total income produced by a rate of flow of a continuous income stream.

5. define consumer’s surplus and producer’s surplus in terms of a definite integral.

6. apply the technique of parts to integrate.

7. Apply the Trapezoidal Rule to approximate the value of a definite integral.

8. Apply the Simpson’s Rule to approximate the value of a definite integral.

9. integrate using tables of integration.

Area between CurvesIndications1. Read Section 6.1 of the textbook, “Area between Curves.” You may skip Example 6 and Example 7.




(Rev. C4)

CommentsIf

g(x) ≤ f(x) for all x in the interval [a, b],

then f(x) is above the function g(x) and

f(x)− g(x) ≥ 0 on the interval [a, b].

Hence,∫ b

a

f(x)− g(x) dx is the area below the curve f(x)− g(x).

On the other hand,∫ b

a

f(x) dx−∫ b

a

g(x) dx

is the area below f minus the area below g on the interval [a, b].

Since the definite integral is a limit∫ b

a

f(x)− g(x) dx =

∫ b

a

f(x) dx−∫ b

a

g(x) dx.

Hence, the area below the curve f(x)− g(x) is equal to the area between the curves f and g on the interval[a, b].

When finding areas between curves, you will find it helpful to sketch the curves on the same coordinateplane to determine the order of the functions; that is, whether f(x) ≤ g(x) or f(x) ≥ g(x) on the interval[a, b].

If we are required to find the area between two curves f and g, and the interval is not given, we assume thatthe curves intersect. We must find the points of intersection. If those points are, say, (a, c) and (b, d), thenthe interval of integration is [a, b].

Example 6.1. Find the area between the curves f(x) = x2 and g(x) = x3. A sketch of the functionsindicates that the functions intersect at two points that we must determine. Thus, we determine that

x2 = x3, x3 − x2 = 0 and x2(x− 1) = 0,

which give x = 0 and x = 1.

For these values of x, the values of y are 0 and 1; hence, the points of intersection are (0, 0) and (1, 1). Wealso see that f(x) ≥ g(x) on the interval [0, 1].

Thus, the area between the curves is∫ 1

0

x2 − x3 dx =x3

3− x4

4

∣∣∣10

=1

3− 1

4=

1

12. �


(Rev. C4)

Example 6.2. Find the area between the curves

f(x) = 3− 2x2 and g(x) = x+ 3.

Figure 6.1 shows the two curves.

Figure 6.1 Curves f(x) = 3− 2x2 and g(x) = x+ 3, for Example 6.2

The intersection of the two curves is given by

3− 2x2 = x+ 3, x+ 2x2 = 0 and x(1 + 2x) = 0.

Thus, x = 0 and x = −1

2.

For these values of x the values of y are 0 and5

2, respectively.

Hence, the points of intersection are (0, 0) and(−1

2,5

2

). The interval of intersection is

[−1

2, 0

].

We have f(x) ≥ g(x) on this interval; hence, the area is∫ 0

−1/2(3− 2x2)− (x+ 3) dx =

∫ 0

−1/2−2x2 − x dx

= −2x3

3− x2

2

∣∣∣0−1/2

= −(

2

3(8)− 1

2(4)

)=

1

24. �

Example 6.3. Find the area bounded by the curves

f(x) = x2 − 2x+ 7, g(x) = 4x+ 2 and x = −1.

A sketch of these three curves gives the area shown in Figure 6.2 on the next page.


(Rev. C4)

Figure 6.2 Curves f(x) = x2 − 2x+ 7, g(x) = 4x+ 2 and x = −1, for Example 6.3

We must find the points of intersection of f and g. Thus,

x2 − 2x+ 7 = 4x+ 2

x2 − 6x+ 5 = 0

x =6±

√36− 4(5)

2

x =6± 4

2x = 1; x = 5.

The points of intersection are (1, 6) and (5, 22). The area between these curves consists of two parts, in thefirst, f(x) ≥ g(x) on the interval [−1, 1], and in the second, g(x) ≥ f(x) on the interval [1, 5]. Thus,∫ 1

−1x2 − 2x+ 7− (4x+ 2) dx+

∫ 5

1

4x+ 2− (x2 − 2x+ 7) dx

=

∫ 1

−1x2 − 6x+ 5 dx+

∫ 5

1

−x2 + 6x− 5 dx

=x3

3− 3x2 + 5x

∣∣∣1−1− x3

3+ 3x2 − 5x

∣∣∣51

=

(1

3+ 2

)−(−1

3− 8

)+

(−125

3+ 50

)−(−1

3− 2

)=

64

3�


2. Solve odd-numbered problems 79–85 and 89–91 from Section 6.1.


(Rev. C4)

Applications in Business and EconomicsIndications1. Read Section 6.2 of the textbook, “Applications in Business and Economics.”


Exercises3. Do odd numbered exercises 5–13 in Section 6.2 of the textbook.


Integration by PartsIndications1. Read Section 6.3 of the textbook, “Integration by Parts.”



CommentsBasically, there are only two techniques of integration: substitution and parts. To decide which one we mustuse, we need to observe the form of the integrand function. Before parts can be used, the integrand functionmust be the product of two functions, f and g, and we must know the antiderivative of one of them.

Example 6.4. Let us consider the integrals listed below, and determine which method we would use toevaluate each of them.

a.∫xe3x dx

b.∫xe3x

2

dx

c.∫

lnx

xdx

d.∫

lnx

x2dx

e.∫x(x+ 4)6 dx


(Rev. C4)

For (a), let us see if the integral fits one of the basic integrals. In this case, the closest one is Integral 2 onpage 337 of the textbook. But in this case, f(x) = 3x and f ′(x) = 3; hence, the integral does not fit thisbasic integral.

We try parts instead. The integrand function is the product of two functions, f(x) = x and g(x) = e3x. Weknow the antiderivatives of f and g, so we can apply parts, and we have two choices: either u = e3x anddv = x dx or u = x and dv = e3x dx.

For (b), we again try to fit Integral 2 on page 337 of the textbook. In this case, f(x) = 3x2, and f ′(x) = 6x.This example fits the basic integral, with the exception of a constant that we can include. We apply thisbasic integral and find∫

xe3x2

dx =1

6

∫e3x

2

(6x) dx.

In (c), the integrand function is the product of two functions, f(x) = ln x and g(x) =1

x.

Being observant, we notice that f ′(x) = g(x); hence, the integral has the form∫f(x)f ′(x) dx =

∫lnx

(1

x

)dx,

which fits Integral 1 on page 337, with n = 1. We apply this basic integral and find∫lnx

xdx =

(lnx)2

2+ C.

In (d), the integrand function is the product of two functions, f(x) = lnx and g(x) =1

x2.

There is no relation between f and g, but we know the antiderivative of g; hence, we apply parts, withu = lnx and dv = x−2 dx.

In (e), the integrand function is the product of two functions, f(x) = x and g(x) = (x+ 4)6. It does not fitthe Integral 1 on page 337. [Why?] We know the antiderivative of f , and we apply parts with u = (x+ 4)6

and dv = x dx.

Another method is substitution, with u = x+ 4; thus, du = dx, and the integrand function becomes(u− 4)(u6). Which method is easier? �

Example 6.5. In Example 6.4 (a), we apply parts to integrate∫xe3x dx.

If we choose u = e3x and dv = x, then du = 3e3xdx and v =x2

2, and

∫xe3x dx =

x2e3x

2− 3

2

∫x2e3x dx.


(Rev. C4)

The integral∫x2e3x dx is more difficult that the original one, and our choice is not good.

So we set u = x and dv = e3x dx, thus du = dx and v = frace3x3, and we have∫xe3x

2

dx =xe3x

3− 1

3

∫e3x dx =

xe3x

3− e3x

9+ C. �

Example 6.6. In Example 6.4 (d), we use parts to integrate∫lnx

x2dx

with u = lnx and dv = x−2 dx, we find du =dx

xand v = −1

x.

Hence,∫lnx

x2dx = − lnx

x−∫− 1

x2dx = − lnx

x− 1

x+ C. �

Integration is more challenging than differentiation, and to do it properly, you must practice and practice.Experience will be your guide to finding the right approach.

Example 6.7. We will try different approaches with the integral from Example 6.4 (b),∫x3ex

2

dx.

The integrand function is the product of two functions, and there are several possibilities

f(x) = x3 and g(x) = ex2

In this case, we know the antiderivative of f but not of g; hence u = ex2 and dv = x3 dx. Therefore

du = 2xex2dx and v =

x4

4.

By parts,∫x3ex

2

dx =x4ex

2

4− 1

2

∫x5ex

2

dx.

This is not a good choice since the integral∫x5ex

2dx is more difficult than the original one.

f(x) = x2 and g(x) = xex2

In this case, we know the antiderivatives of both f and g. You can verify that the choice of u = xex2 and

dv = x2dx is not good.

So we take u = x2 and dv = xex2dx; then, du = 2x dx and v =

ex2

2.


(Rev. C4)

By parts,∫x3ex

2

dx =xex

2ex

2

2−∫xex

2

dx =e2x

2

2− 1

2

∫ex

2

(2x) dx =xe2x

2

2− ex

2

2+ C �

Exercises5. Do odd numbered exercises 1–65 in Section 6.3 of the textbook.


Other Integration MethodsPrerequisitesTo complete this section, you must know how to complete perfect squares.

Indications1. Read Section 6.4 of the textbook, “Other Integration Methods.”



CommentsWe have two methods of integration: substitution and parts. We also have tables of integrals. However,these are not enough to integrate all possible functions. When none of these are applicable, we resort tomethods of approximation.

In this course we cover two of these methods to approximate the value of a definite integral: TrapezoidRule (page 418 of the textbook) and Simpson’s Rule (page 420 of the textbook).

The formula for the area of a trapezoid is given on page 418 of the textbook. From Figure 6.3 on the

following page, the area of the trapezoid is the sum of the areas of a rectangle ah and a trianglech

2. If

b = a+ c, then the sum of the areas is

ah+ch

2=

2ah+ ch

2=

(a+ a+ c)h

2=

(a+ b)h

2.

To apply the Trapezoid Rule correctly, we must understand the equality for Tn. The interval of integrationis [a, b] and this interval is divided into n equal parts. The greater n the better the approximation. Theformula is

Tn = [f(a) + 2f(x1) + 2f(x2) + · · ·+ 2f(xn−1 + f(b)]

[b− a2n

]


(Rev. C4)

c

a

h

Figure 6.3 Trapezoid

and ∫ b

a

f(x) ≈ Tn.

Be careful with the identification of the xi, 1 ≤ i ≤ n. If ∆x =b− an

, then

x1 = a+ ∆x, x2 = x1 + ∆x, . . . xn−1 = xn−2 + ∆x.

For the Simpson’s rule the interval [a, b] is divided into 2n equal parts (always an even number of parts).See that there are 2n− 1 (odd) divisions xi, 1 ≤ i ≤ 2n− 1 of the interval and the coefficients alternatebetween 4 and 2, starting and ending with 4. Hence,

S2n = [f(a) + 4f(x1) + 2f(x2) + · · ·+ 4f(x2n−1 + f(b)]

[b− a6n

].

and ∫ b

a

f(x) ≈ S2n.

In this case if ∆x =b− a2n

, then

x1 = a+ ∆x, x2 = x1 + ∆x, . . . x2n−1 = x2n−2 + ∆x.

Remark: The application of these rules for large n may be tedious, so you have to take it slowly and surely.

When applying either of these two rules, we recommend you follow these steps.

Step 1 Identify n and the interval [a, b].

Step 2 Evaluate ∆x. Use decimals only when the fraction ∆x is an exact decimal i.e.1

4= 0.25.


(Rev. C4)

Step 3 Identify the xi in the interval [a, b]. See that xi = xi−1 + ∆x.

Step 4 Make a table with the values of xi and f(xi. Starting with a and f(a) and ending with b and f(b).See Example 1 on page 419 of the textbook.

Step 5 Apply the rule.

To use tables of integration, we must decide on the function u. Observe that du is the differential of u; thatis, if we decide on u = f(x), then du = f ′(x)dx must be in the integrand function as well.

In the following examples, the numbers of the integrals refer to the integrals in Table 1 of Appendix C,pages 547–549 of the textbook.

Example 6.8. For the integral∫dx

1 + 2ex,

we may want to set u = 2ex so that the integrand function is1

1 + uand its

antiderivative is ln |1 + u|; however, du = ex dx, and we cannot fit the differential into the integral.

We then look further and see that Integral 48—with u = x, du = dx, c = 1, d = 2 and a = 1—applies, and∫dx

1 + 2ex= x− ln |1 + 2ex|+ C. �

Example 6.9. For the integral∫ex

3 + 2exdx,

we see that Integral 48 does not apply, because if u = ex then du = ex dx, and these values do not fit inthis integral. However, for this u and du, the integral becomes∫

du

3 + 2u=

1

2

∫2

3 + 2udu =

ln |3 + 2u|2

+ C. �

Example 6.10. In the integral∫lnx

x√

4 + ln xdx,

we have the function lnx, and its derivative is1

x—both are in the integrand function.

Hence, we set u = lnx, then du =dx

x, and the integral becomes∫

u√4 + u

du.


(Rev. C4)

We see that Integral 25 applies, with a = 4, b = 1; and∫u√

4 + udu =

2(u− 8)

3

√4 + u =

2(lnx− 8)

3

√4 + ln x+ C. �

The note on page 134 of this Study Guide also applies here.

We can also complete perfect squares to fit one of the integrals in the tables.

Example 6.11. Let us integrate∫1√

x2 + 4xdx.

We see that the integrand function does not fit any of the integrals in Appendix C, but if we complete theperfect square, we have

x2 + 4x = x2 + 4x+ 4− 4 = (x+ 2)2 − 4.

If we set u = x+ 2, then du = dx, the integral in terms of u is∫1√

u2 − 4du.

From Integral 43, with a = 2, we have∫1√

u2 − 4du = ln |u+

√u2 − 4|,

and the answer in terms of x is

ln |x+ 2 +√x2 + 4x|+ C. �

Example 6.12. To integrate∫x√

2x2 + 6x+ 1/2dx,

we also complete the perfect square. Thus,

2x2 + 6x+1

2= 2(x2 + 3x) +

1

2

= 2(x2 + 3x+

9

4

)− 18

4+

1

2= 2(x+

3

2

)2− 4.

If we set u = x+3

2, then du = dx and x = u− 3

2. The given integral becomes

∫ u− 3

2√2u2 − 4

du =1

2√

2

∫2u− 3√u2 − 2

du,


(Rev. C4)

and by the properties of the integrals,

1

2√

2

∫2u− 3√2u2 − 4

du =1

2√

2

[ ∫ 2u√u2 − 2

du− 3

∫1√

u2 − 2du].

The first integral fits Integral 1 on page 132; thus, you can verify that∫2u√u2 − 2

du =

∫(2u)(u2 − 2)−1/2 du = 2(u2 − 2)1/2.

The second integral follows by Integral 43 with a =√

2:∫1√

u2 − 2du = ln |u+

√u2 − 2|.

The answer in terms of x is

1

2√

2

2

√(x+

3

2

)2

− 2− 3

ln

∣∣∣∣∣∣x+3

2+

√(x+

3

2

)2

− 2

∣∣∣∣∣∣+ C �

Tables of integrals save us work and offer us a practical way to integrate. However, you should not neglectthe substitution and parts methods you learned earlier in this unit. Unfortunately, not all integrals fit into atable of integrals.

Note: If you use a table of integrals in your assignments, you must identify the number of the integral youare applying and state the corresponding u and du.


8. Solve odd-numbered problems 71, 73, and 77–85 in Section 6.4.


2. Study “Chapter 6: Summary and Review” titled “Important Terms, Symbols and Concepts.”


4. You may want to do odd-numbered problems 1–29, 35–41 from the “Review Exercises” section at theend of Chapter 6 in the textbook.

5. Complete the “Practice Examination” provided for this unit. Evaluate yourself, first checking youranswers against those provided in the “Answers” section at the end of the textbook, and thencomparing your solutions with those provided on pages 6–37, 6–50 of the Student’s Solutions Manual.


(Rev. C4)

The number of points in a question may indicate the number of steps in the solution. Give yourself fullcredit if your answer is correct and you give a complete solution, even if your solution differs from thatshown in the Student’s Solutions Manual.


(Rev. C4)

Practice ExaminationTime: 1.5 hoursTotal points: 32Passing grade: 55%


To obtain full credit, you must justify all your answers and show your work.

1. Refer to Exercise 14 on page 431.

Sketch the given curves and show the area bounded by them. Marks: 5 pts








(Rev. C4)

UNIT 7

Multivariable Calculus

We finish this course by extending the concepts of maxima and minima for functions of one variable thatwe considered in Unit 4 to functions of several variables.


1. evaluate functions of several variables.

2. locate points in a three-dimensional coordinate system.

3. identify the curve levels of functions of several 4 maximum and minimum of functions of severalvariables.

4. solve optimization problems with mathematical models consisting of functions of several variables.

Functions of Several VariablesPrerequisitesTo complete this section, you must be able to identify conic curves (i.e., parabolas, ellipses, hyperbolas).

Indications1. Read Section 7.1 of the textbook, “Functions of Several Variables.”



CommentsIn previous sections, we worked with functions of one variable, where the notation f(x) indicates that thevariable f depends on the variable x. Functions of several variables have one dependent variable f and two


(Rev. C4)

or more independent variables x, y, z, . . . . However, the notation f(x, y) has a similar meaning: f dependson the variables x, y.

Example 7.1. The cost of production of a product C depends on the cost of materials x and the cost oflabour y; hence C(x, y). �

Example 7.2. The surface area S of a box depends on the length `, the width w, and the height h; thusS(`, w, h). �

More examples of functions of several variables can be found on page 437 of the textbook.

The graph of a two-variable function f(x, y) is visually represented in a three-dimensional coordinatesystem. The variables x, y are located on the “floor,” and the corresponding value of f is located withrespect to the z-axis. Functions of more than two variables do not have a visual representation.

In general, it is not easy to visualize functions of two variables, although computers are helpful and areoften used. One method we can use to get an idea of the graph of a function is to obtain the “curve levels”of the function.

Definition 7.1. Let f(x, y) be a two-variable function. Then, for each constant k, the curvef(x, y) = k is the curve of level k.

The curve of level k is the curve we obtain after slicing the graph of f with a plane parallel to the floor atthe height k (see Figure 7.1, below).

Figure 7.1 Curve of level k

If f(x, y) is a function, then we draw the curves f(x, y) = k, for some constants k, on the sametwo-dimensional coordinate plane.

Example 7.3. If the level curves are concentric circles in increasing order, as shown in Figure 7.2 on thefollowing page, then the surface is a paraboloid.

To visualize the surface, look at the circles and visually “lift” them, to get the image of the surface in yourhead.

Example 7.4. Let us consider level curves that are parabolas, as shown in Figure 7.3 on the next page.


(Rev. C4)

Figure 7.2 Level curves forming concentric circles of increasing order

Figure 7.3 Level curves forming parabolas of increasing order

If we “lift” the curves, we “see” the surface shown in Figure 7.4.

Figure 7.4 Curved surface formed of the parabolas shown in Figure 7.3

Cross sections are similar to curve levels, but in this case, the surfaces are “sliced” by planes parallel to the“walls” framed by the x and z axes and the y and z axes. For example, the plane x = a is perpendicular tothe x-axis and parallel to the plane yz, and passes through the point (a, 0, 0). Similarly, the plane y = b isperpendicular to the y-axis and parallel to the plane xz, and passes through the point (0, b, 0).


(Rev. C4)

To identify level curves and cross sections, you must know the standard equations of conic curves. In thiscourse, we do not require a precise sketch of a surface, but it is important that you learn to visualize someeasy functions.

Example 7.5. To obtain the standard equations of conics, we complete perfect squares. Hence, the levelcurves k for the function

f(x, y) = 31 + 10x+ 24y − x2 − y2

are obtained as follows.

31 + 10x+ 24y − x2 − y2 = k

31− k = x2 − 10x+ y2 − 24y

31− k + 25 + 144 = x2 − 10x+ 25 + y2 − 24y + 144

200− k = (x− 5)2 + (y − 12)2

The equation corresponds to a circle.

For k = 0, 200 = (x− 5)2 + (y − 12)2.

The equation describes circle with centre (5, 12) and radius√

200 = 10√

2.

For k = 25 175 = (x− 5)2 + (y − 12)2.

The equation describes a circle with centre (5, 12) and radius√

175.

For different values of k, we obtain concentric circles in descending order. Since 200− k > 0, we knowthat k < 200. Thus, the surface is an upside-down cone of height of 200 units.

The curve on the intersection of the surface f(x, y) and the plane x = a, is

z = f(a, y) = 31 + 10a+ 24y − a2 − y2

= (31 + 10a− a2 + 144)− (y2 − 24y + 144)

= (175 + 10a− a2)− (y − 12)2.

This equation describes a upside-down parabola with vertex (a, 12, 175 + 10a− a2). For instance if x = 0,then the parabola is z = 175− (y − 12)2 on the plane yz.

The curve on the intersection of the surface f(x, y) and the plane y = b, is

z = f(x, b) = 31 + 10x+ 24b− x2 − b2

= (31 + 24b− b2 + 25)− (x2 − 10x+ 25)

= (56 + 24b− b2)− (x− 5)2.

This equation describes a upside-down parabola with vertex (5, b, 56 + 24b− b2). If b = 0, thenz = 56− (x− 5)2, and we have parabola on the xz plane.

You should convince yourself that, if we slice an upside-down cone with planes perpendicular to its base,then the curve on the plane is an concave up parabola. �


(Rev. C4)

Exercises1. Do odd-numbered exercises 1–25, and 31–47 in Section 7.1 of the textbook.

2. Solve odd-numbered exercises 65–71 in Section 7.1.

Partial DerivativesIndications1. Read Section 7.2 of the textbook, “Partial Derivatives.”



CommentsTo differentiate functions of several variables, we do not use the limits on page 445 of the textbook. Instead,we use what we have learned about differentiation of functions with one variable. We differentiate withrespect to one variable, considering (mentally) the others as constants.

The notation is very important. If z = f(x, y), then

∂z

∂x=∂f

∂x= fx(x, y)

denotes the partial derivative of f with respect to x; that is, we consider y as constant, and differentiate fwith respect to x.

Similarly,

∂z

∂y=∂f

∂y= fy(x, y)

denotes the partial derivative of f with respect to y. Mentally, we consider x as a constant, and wedifferentiate f with respect to y.

The notation

∂z

∂x

∣∣∣x=a

=∂f

∂x

∣∣∣x=a

= fx(a, y)

is the value of the partial derivative of f at a. We find the partial derivative fx(x, y), and then we substitutex = a, to get fx(a, x). We interpret fy(x, b) in a similar way. Thus

fx(a, b) =∂f

∂x

∣∣∣x=a,y=b


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and

fy(a, b) =∂f

∂y

∣∣∣x=a,y=b

.

The interpretation of the partial derivatives is similar to the interpretation of single variables. So, iffx(a, b) = c, then

a. if we slice the surface z = f(x, y) with the plane y = b, we obtain a curve on this plane, and the linetangent to this curve at the point (a, b, f(a, b)) has slope c.

b. if y = b is fixed, then when x = a, the variable f is increasing (if c > 0) or decreasing (if c < 0), at arate of c units of f per unit of x.

c. if y = b is fixed, then by increasing the variable x by one unit, the variable f increases (if c > 0) ordecreases (if c < 0) by approximately c units of f .

We leave the interpretation of fy(a, b) as an exercise.

The notations for the second-order partial derivatives are listed on page 448 of the textbook.

Observe that the notation fxy gives the order of differentiation, first with respect to x, then with respect to y,from left to right on the subindex xy. This notation corresponds to

∂z2

∂y∂x;

the terms ∂y∂x being read from right to left, first with respect to x, then with respect to y.



Maxima and MinimaIndications1. Read Section 7.3 of the textbook, “Maxima and Minima.”



CommentsThe second derivative test is inconclusive when AC −B2 (Case 4 of Theorem 2 on page 454). In this case,we evaluate the function f at points close to (a, b) to try to gain an idea of the behaviour of the function.


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We may also use cross sections close to the point. As well, using a computer to graph the function aroundthe point is very helpful. Note, however, that in this course, our examples and exercises involve onlyconclusive results.

Finding the critical points requires algebra and ingenuity. Study the examples in this section, and do notgive up easily. See how the constraint equation is used to find the critical points.

Example 7.6. Exercise 26 on page 459 of the textbook.

f(x, y) = −x2 + 2xy − 2y2 − 20x+ 34y + 40.

The partial derivatives set to 0 are:

fx(x, y) = −2x+ 2y − 20 = 0 fy(x, y) = 2x− 4y + 34 = 0.

We have two equations with two unknowns. We can solve them by adding fx and fy. By doing this weeliminate x from both equations. Thus, fx(x, y) + fy(x, y) = −2y + 14 = 0 and y = 7. From eitherequation we get 0 = −2x+ 14− 20 = −2x− 6 and x = −3. The only critical point is (−3, 7).

fxx(x, y) = −2 fyy(x, y) = −4 fxy(x, y) = 2.

Hence,

fxxfyy − fxy = (−2)(−4)− (2) = 6 > 0.

By Theorem 2 (pp. 454) of the textbook, the function has a local maximum at (−3, 7). �



Maxima and Minima Using Lagrange MultipliersIndications1. Read Section 7.4 of the textbook, “Maxima and Minima Using Lagrange Multipliers.”



CommentsWe have two methods for finding the maxima and minima of a function of two variables: thesecond derivative test (see page 454 of the textbook), and Lagrange multipliers (see page 462 of the


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textbook). The example in the “Insight” on page 468 shows that the latter is more efficient than the formerin some cases.

The application of Lagrange multipliers requires the solution of the system

Fx(x, y, z, λ) = 0

Fy(x, y, z, λ) = 0

Fλ(x, y, z, λ) = 0.

This is the difficult part, and there is no specific strategy for it. Note that we need the value of x and y; thevalue of λ is not necessary, although we may find it if it helps us to find x and y (see the fence problem onpage 461).

Example 7.7. See Exercise 10 on page 468.

Let

f(x, y) = 25− x2 − y2

with the constraint

g(x, y) = 2x+ y − 10.

Hence,

F (x, y, λ) = 25− x2 − y2 + λ(2x+ y − 10).

The system of equations is

−2x+ 2λ = 0

−2y + λ = 0

2x+ y − 10 = 0.

From the first equation, x = λ, and replacing in the second equation, we obtain −2y + x = 0. Hence,x = 2y, and from the third equation, we find

2(2y) + y − 10 = 5y − 10 = 0.

Therefore y = 2 and x = 4. The maximum is at (4, 2) and the maximum value is f(4, 2) = 5. We did notneed the value of λ to find the solution. �

Lagrange multipliers can be extended to functions of more than two variables, and more than oneconstraint. In this course, we consider only Theorem 2 on page 466. Again we have four equations, andthere is no one strategy to solve them.

Example 7.8. Let

f(x, y, z) = xyz


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with the constraint

g(x, y, z) = 2x+ y + 2z − 120.

Hence.

F (x, y, z, λ) = xyz + λ(2x+ y + 2z − 120).

The system of equations is

yz + 2λ = 0 (7.1)xz + λ = 0 (7.2)xy + 2λ = 0 (7.3)

2x+ y + 2z − 120 = 0 (7.4)

We must observe first that we are looking for the maximum of f , and f(1, 1, 1, ) = 1 > 0. Hence, if eitherx, y, or z is zero, then f(x, y, z) = 0 and this is not the maximum of f .

Now, we multiply Equation (7.2) by 2, and we get 2xz + 2λ = 0, and this result, with Equation (7.1), givesyz = −2λ = 2xz. Since z 6= 0, we can cancel z, and we have y = 2x.

From 2xz = −2λ and Equation (7.3), we have 2xz = xy, and we cancel x because it is nonzero. Thus,2z = y.

From Equation (7.4), we have

y + y + y = 3y = 120,

and we conclude that y = 40, x = y/2 = 20 and z = y/2 = 20.

The function is at its maximum at (20, 40, 20), the maximum value being f(20, 40, 20) = 16000. �


8. Solve odd-numbered problems 31, 33, and 35 in Section 7.4.


2. Study the “Chapter 7 Review” section titled “Important Terms, Symbols and Concepts,”pages 499–500 of the textbook (Sections 7.1 to 7.4 only).



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4. You may want to do odd-numbered problems 1, 7, 9, 13, 15, 17, 23, and 29 from the “ReviewExercises” section on pages 501–502.

5. Complete the “Practice Examination” provided for this unit. Evaluate yourself, first checking youranswers against those provided in the “Answers” section at the end of the textbook, and thencomparing your solutions with those provided on pages 7–49 to 7–54 of the Student’s SolutionsManual. The number of points in a question may indicate the number of steps in the solution. Giveyourself full credit if your answer is correct and you give a complete solution, even if your solutiondiffers from that shown in the Student’s Solutions Manual.


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Practice ExaminationTime: 1.5 hoursTotal points: 30Passing grade: 55%

Do the following exercises from the Chapter 7 “Review Exercises,” pages 501–502 of the textbook.








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Index

area under the curve, 135asympote

horizontal, 67vertical, 57

closed interval method, 121conjugate terms, 43contraction

of a function, 9critical point, 106cross sections, 160curve levels, 159

definite integral, 139derivative

interpretation of, 108notation of, 78

exponential decay, 90exponential growth, 90

fractioncomparation of, 18equality of, 16

functionabsolute value graph, 35cubic graph, 35identity

graph, 35quadratic

graph, 35Fundamental Theorem of Algebra, 41Fundamental Theorem of Calculus, 139

half-life, 90

infinite limitat infinity, 113

inflection point, 109interpretation of, 112

integral, 129integrand function, 129

integrationlower limit, 139upper limit, 139

limitdoes not exist, 50

linesketching of, 20

linear approximation, 80

Mean Value Theorem, 137

one-to-one function, 30

piecewise function, 14

rate of changeaverage, 73instantaneous, 74interpretations of, 75

rationalization, 43reflection

of a function, 11Riemann sum, 137

second derivativeinterpretation of, 112

second-derivative test, 121shift

of a function, 8sign of a function, 107Simpson’s Rule, 152stretch

of a function, 9

Trapezoid Rule, 151

variabledependent, 2independent, 2


Mathematics 260 Calculus for Social Sciences and Economics

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