Calculus I - Cognella Academic PublishingFirst published in the United States of America in 2011 by Cognella, a division of University Readers, Inc. Trademark Notice: Product or corporate

Calculus IBy Tunc Geveci

Included in this preview:

• Copyright Page• Table of Contents• Sections:

1.3 Limits & Continuity: The Concepts1.5 The Calculation of Limits2.5 Local Linear Approximations and the Differential2.7 The Chain Rule3.1 Increasing/decreasing Behavior and Extrema3.5 Application of Maxima and Minima4.5 Orders of Magnitude5.3 The Fundamental Theorem of Calculus: Part 15.4 The Fundamental Theorem of Calculus: Part 2

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Tunc Geveci

Calculus I: First Edition

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Copyright © 2011 by Tunc Geveci. All rights reserved. No part of this publication may be reprinted, reproduced, transmitted, or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information retrieval system without the written permission of University Readers, Inc.

First published in the United States of America in 2011 by Cognella, a division of University Readers, Inc.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

15 14 13 12 11 1 2 3 4 5

Printed in the United States of America

ISBN: 978-1-935551-42-3

Contents

1 Functions, Limits and Continuity 11.1 Powers of x, Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Combinations of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.3 Limits and Continuity: The Concepts . . . . . . . . . . . . . . . . . . . . . . . . 311.4 The Precise De�nitions (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . 411.5 The Calculation of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.6 In�nite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581.7 Limits at In�nity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 681.8 The Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2 The Derivative 932.1 The Concept of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 932.2 The Derivatives of Powers and Linear Combinations . . . . . . . . . . . . . . . . 1072.3 The Derivatives of Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.4 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1312.5 Local Linear Approximations and the Di�erential . . . . . . . . . . . . . . . . . . 1382.6 The Product Rule and the Quotient Rule . . . . . . . . . . . . . . . . . . . . . . 1482.7 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1572.8 Related Rate Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1672.9 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1742.10 Implicit Di�erentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

3 Maxima and Minima 1933.1 Increasing/decreasing Behavior and Extrema . . . . . . . . . . . . . . . . . . . . 1933.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2053.3 Concavity and Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2143.4 Sketching the Graph of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . 2263.5 Applications of Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . 233

4 Special Functions 2494.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2494.2 The Derivative of an Inverse Function . . . . . . . . . . . . . . . . . . . . . . . . 2624.3 The Natural Exponential and Logarithm . . . . . . . . . . . . . . . . . . . . . . . 2724.4 Arbitrary Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2854.5 Orders of Magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2914.6 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3024.7 Hyperbolic and Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 3184.8 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

iii

iv CONTENTS

5 The Integral 3475.1 The Approximation of Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3475.2 The De�nition of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3575.3 The Fundamental Theorem of Calculus: Part 1 . . . . . . . . . . . . . . . . . . . 3715.4 The Fundamental Theorem of Calculus: Part 2 . . . . . . . . . . . . . . . . . . . 3855.5 Integration is a Linear Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . 4035.6 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4155.7 The Di�erential Equation y0 = f . . . . . . . . . . . . . . . . . . . . . . . . . . . 427

A Precalculus Review 435A.1 Solutions of Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . 435A.2 The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440A.3 The Number Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442A.4 Decimal Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451A.5 The Coordinate Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457A.6 Special Angles and Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . 471

B Some Theorems on Limits and Continuity 481

C The Continuity of an Inverse Function 489

D L’Hôpital’s Rule (A Proof) 491

E The Natural Logarithm as an Integral 493

F Answers to Some Problems 501

G Basic Derivatives and Integrals 535

1.3. LIMITS AND CONTINUITY: THE CONCEPTS 31

a) Determine the fundamental period p of f,b) Determine the part of the natural domain of f in the interval [�p/2, p/2] and whether f isan odd or even function,c) Sketch the graph of f on the interval [�p/2, p/2].

27. f(x) = sin(6x)28. f(x) = cos(x/3)29. f(x) = tan(�x)30. f(x) = sec(x/4).

31. f(x) =psin(x)

32. f(x) =pcos(x)

33. f(x) =ptan(x)

34. f (x) =psec(x)

1.3 Limits and Continuity: The ConceptsIn this section we will discuss the concept of the limit of a function at a point. This conceptprovides a general framework for problems such as the determination of the slope of a tangentline to the graph of a function or the velocity of an object in motion. We will also discuss therelated concept of continuity.

The Slope of a Tangent Line

How should we determine the slope of the graph of a function? Let’s begin with a case wherewe know the answer. Let f be a linear function so that f (x) = mx+ b, where m and b areconstants. The graph of f is a line with slope m. If a is an arbitrary point on the number lineand x 6= a, then

f (x)� f (a)x� a =

(mx+ b)� (ma+ b)x� a =

m (x� a)x� a = m.

x

y

a x

x � a

m�x � a�

Figure 1: The line y = mx+ b has slope m

The case of a nonlinear function is not that straightforward. Let’s consider a speci�c case.

Example 1 Let F (x) = x2. The graph of F is a familiar parabola, as shown in Figure 2.

�3 �2 2 3x

4

9

y

�2, 4�

�3, 9�

Figure 2

32 CHAPTER 1. FUNCTIONS, LIMITS AND CONTINUITY

A reasonable notion of the slope of the graph of F cannot yield a single number, unlike thecase of a linear function. For example, if we compare the behavior of the graph of F near(3, F (3)) = (3, 9) and (2, F (2)) = (2, 4), the graph appears to rise more steeply near (3, 9).

Let’s focus our attention on the behavior of the function near the point 3. If x 6= 3, we will referto the line that passes through the points (3, F (3)) and (x, F (x)) as a secant line. The slopeof such a secant line is

F (x)� F (3)x� 3 =

x2 � 9x� 3 .

x

y

x

F�x� � F�3�

x � 33

9

Figure 3: A secant line

Since the secant line is almost “tangential” to the graph of F at (3, F (3)) if x is close to 3,its slope should approximate the slope of the tangent line to the graph of F at (3, F (3)). Table1 displays the slope of the secant line that passes through the points (3, F (3)) and (x, F (x))for x = 3 + 10�n, where n = 1, 2, 3, 4, 5. The numbers indicate that the slope of the secant lineapproximates 6 if x is close to 3.

x F (x)�F (3)x�3

3 + 10�1 6.13 + 10�2 6.013 + 10�3 6.0013 + 10�4 6.00013 + 10�5 6.00001

Table 1

A little algebra clari�es the situation. We cannot merely replace x by 3 in the expression forthe slope of a secant line. This leads to the unde�ned or indeterminate expression 0/0. Onthe other hand, we can simplify the expression for any x 6= 3:

F (x)� F (3)x� 3 =

x2 � 9x� 3 =

(x� 3) (x+ 3)x� 3 = x+ 3.

If x is close to 3, then x+ 3 �= 3+ 3 = 6. It seems reasonable to declare that the slope of thetangent line to the graph of F at (3, F (3)) is 6. Since the tangent line should pass through(3, F (3)) and has slope 6, it is the graph of the equation

y = F (3) + 6 (x� 3) = 9 + 6 (x� 3)

in the xy-plane (this is the point-slope form of the equation of the tangent line with basepoint3). Figure 4 shows the graph of F and the tangent line to the graph of F at (3, F (3)). Thepicture is consistent with our intuitive notion of a tangent line. We will identify the slope of


the graph of F at (3, F (3)) with the slope of the tangent line to the graph of F at that point.¤

3x

9

y

�3, F�3��

Figure 4

The Informal De�nitions of Limits and Continuity

Let F (x) = x2, as in Example 1 The slope of the secant line that passes through (3, F (3)) and(x, F (x)) is a function of x in its own right. Let’s set

f (x) =F (x)� F (3)

x� 3 =x2 � 9x� 3 ,

so that f (x) is de�ned if x 6= 3. We have f (x) �= 6 if x 6= 3 and x is close to 3. Given anarbitrary function f and a point a on the number line, it will be useful to discuss whether f (x)approximates a certain number L if x 6= a and x is close to a, even if f (x) is not necessarily theslope of a secant line. This leads to the concept of the limit of a function at a point.

De�nition 1 ( THE LIMIT OF A FUNCTION AT A POINT) Assume that f (x) isde�ned for each x in some open interval that contains the point a, with the possible exceptionof a itself. The limit of f(x) at a is L if f (x) is as close to L as desired provided that x 6= aand x is su�ciently close to a. In this case we write

lim��

f(x) = L

(read “the limit of f(x) as x approaches a is L”).

x

y

a x

L

f�x� �x, f�x��

�a, L�

Figure 5: f (x) is close to the limit L at a if x 6= a and x is close to a

You can imagine that f (x) gets closer and closer to L as x approaches a, with the restriction thatx 6= a. We will refer to De�nition 1 has been dubbed as “the informal de�nition of the limit”,since the phrases “as close to L as desired ” and “su�ciently close” have not been quanti�ed.For almost all our purposes in calculus, an intuitive understanding of the concept of the limitwill be adequate. The precise de�nition of the limit of a function at a point will be discussed inthe next section.


Example 2 Let

f (x) =x2 � 9x� 3 .

Then f (x) is de�ned if x 6= 3. As in Example 1,

f (x) =x2 � 9x� 3 =

(x� 3) (x+ 3)x� 3 = x+ 3,

so that f (x) �= 6 if x �= 3 and x 6= 3. We have

|f(x)� 6| = |(x+ 3)� 6| = |x� 3| ,

so that f (x) is as close to 6 as desired if x is su�ciently close to 3. Therefore, the limit of f at3 is 6:

limx�3

f (x) = limx�3

x2 � 9x� 3 = 6.

¤

We can set x = a+h so that h represents the deviation of x from a. We have h 6= 0 correspondingto x 6= a and h approaches 0 if and only if x approaches a. Thus,

limx�a

f (x) = limh�0

f (a+ h) .

Example 3 Let

f (x) =

�x� 2x� 4 if x 6= 4.

a) Calculate f (4 + h) for h = ±10�n, n = 1, 2, 3, 4. Do the numbers lead to a conjecture aboutlimx�4 f (x)?

b) Plot the graph of with a graphing device. Does the picture support your conjecture?

Solution

a) We have

f (4 + h) =

�4 + h� 2h

.

Table 2 displays f (4 + h) for h = ±10�n, n = 1, 2, 3, 4. The numbers indicate that |f (4 + h)� 0.25|becomes smaller and smaller as |h| becomes small. Thus, we can expect that limx�4 f (x) = 1/4(|f (4 + h)� 0.25| is rounded to 2 signi�cant digits).

h f(4 + h) |f (4 + h)� 0.25|10�1 0.248 457 1. 5× 10�3�10�1 0.251 582 1. 6× 10�310�2 0.249 844 1. 6× 10�4�10�2 0.250 156 1. 6× 10�410�3 0.249 984 1. 6× 10�5�10�3 0.250 016 1. 6× 10�510�4 0.249 998 1. 6× 10�6�10�4 0.250 002 1. 6× 10�6

Table 2


b) Figure 6 supports the conjecture that limx�4f (x) = 0.25. In fact, the program that generatedthe picture seems to be unaware of the fact that f is not de�ned at 4. This is not surprising,since such a device samples some values of x, calculates the corresponding values of the function,and joins the resulting points by line segments. It is immaterial that f is not de�ned at 4, aslong as f (x) �= 4 when x is near 0. ¤

1 2 3 4 5x

0.25

0.5y

�4, 0.25�

Figure 6

A function may fail to have a limit at a point:

Example 4 Let

f (x) =

½x if x < 1,

x+ 2 if x � 1.a) Sketch the graph of f .b) Show that f does not have a limit at 1.

Solution

a) Figure 7 shows the graph of f .

�2 �1 1 2 3x

1

3

y

Figure 7

b) If x < 1 and x is close to 1, then f (x) = x �= 1. On the other hand, if x > 1 and x is near 1,f (x) = x + 2 �= 3. Thus, f (x) does not approximate a de�nite number if x 6= 1 and x is near1. Therefore f does not have a limit at 1. ¤The limit of a function at a point need not be the same as the value of the function at thatpoint, as in the following example.

Example 5 Let

f (x) =

½x if x 6= 12 if x = 1.

a) Sketch the graph of f .b) Show that limx�1 f (x) 6= f (1) .Solution


a) Figure 8 shows the graph of f . The fact that f (1) 6= 2 is indicated by a little hollow circle.The point (1, 2) belongs to the graph of f since f (1) = 2.

�2 �1 1 2x

�1

�2

2

1

�1, 2�

Figure 8

b) Since f (x) = x for each x 6= 1,

limx�1

f (x) = limx�1

x = 1.

On the other hand, f (1) = 2. Therefore, limx�1 f (x) 6= f (1). ¤We use special terminology that applies to a case where the limit of a function at a point is thevalue of the function at that point:

De�nition 2 (CONTINUITY) A function f is said to be continuous at a point a if f (x)is de�ned in some open interval that contains a and limx�a f (x) = f (a) .

Example 6 Let g (x) = x+ 3. Then g is continuous at 3. Indeed,

limx�3

g (x) = limx�3

(x+ 3) = 6 = g (3) .

¤

Since limx�a f (x) = limh�0 f (a+ h), a function f is continuous at a point a if and only if

limh�0

f (a+ h) = f (a) .

x

y

a

f�a�

f�x� � f�a�h� �x, f�x��

�a, f�a��

x � a�hh

Figure 9: f(x) is close to f (a) if x is close to a

Example 7 Calculate sin (�/6 + 10�n) for n = 2, 3, 4, 5. Do the numbers suggest that the sinefunction is continuous at �/6?


Solution

We have sin (�/6) = 0.5. The numbers in Table 3 indicate that |sin (�/6 + h)� sin (�/6)| shouldbe as small as desired provided that |h| is small enough, and suggest that sine is continuous at�/6. ¤

x sin (x) |sin (x)� 0.5|�6 + 10

�2 0.508 635 8. 6× 10�3�6 + 10

�3 0.500 866 8. 7× 10�4�6 + 10

�4 0.500 087 8. 7× 10�5�6 + 10

�5 0.500 009 8. 7× 10�6

Table 3

We will say that a function f is discontinuous at a if f is not continuous at a.

Example 8 Let

f (x) =x2 � 9x� 3 if x 6= 3,

as in Example 2. The function f is discontinuous at 3 since f is not de�ned at 3, even thoughlimx�3 f (x) exists. ¤

Example 9 Let

f (x) =

½x if x < 1,

x+ 2 if x � 1,as in Example 4. The function f is discontinuous at 1 since limx�1 f (x) does not exist. ¤

Example 10 Let

f (x) =

½x if x 6= 12 if x = 1,

as in Example 5. The function f is discontinuous at 1 since limx�1 f (x) = limx�1 x = 1 6=f (1) .¤

Let’s take another look at the function of Example 4:

f (x) =

½x if x < 1,

x+ 2 if x � 1.

�2 �1 1 2 3x

1

3

y

Figure 10

We observed that f does not have a limit at 1, since f (x) approaches 1 if x approaches 1from the left, and f (x) approaches 3 if x approaches 1 from the right. These are examples of“one-sided limits”:


De�nition 3

a) The right-limit of f at a is L+ if f (x) is as close to L+ as desired provided that x > aand x is su�ciently close to a. In this case we write

limx�a+

f (x) = L+

(read “the limit of f (x) as x approaches a from the right is L+”).b) The left-limit of f at a is L� if f (x) is as close to L� as desired provided that x < a andx is su�ciently close to a. In this case we write

limx�a� f (x) = L�

(read “the limit of f (x) as x approaches a from the left is L�”).

The language is suggestive: You can imagine that f (x) approaches L+ as x approaches a fromthe right on the number line, and that f (x) approaches L� as x approaches a from the left.

If f is the function of Example 4, then

limx�1+

f (x) = limx�1+

(x+ 2) = 3,

andlimx�1�

f (x) = limx�1�

x = 1.

De�nition 4 We say that f has a jump discontinuity at a if limx�a+ f (x) and limx�a� f (x)exist but are not equal.

Thus, the function of Example 9 has a jump discontinuity at 1.

Clearly, limx�a f (x) exists if and only if both one-sided limits of f at a exist and limx�a+ f (x) =limx�a� f (x). If this is the case,

limx�a

f (x) = limx�a� f (x) = lim

x�a+f (x) .

The notion of one-sided continuity is related to one-sided limits:

De�nition 5 A function f is said to be continuous at a from the right if f is de�ned at a andlimx�a+ f (x) = f (a). Similarly, f is continuous at a from the left if limx�a� f (x) = f (a) .

The function f of Example 4 is continuous at 1 from the right, since

limx�1+

f (x) = 3 = f (2) .

The function is discontinuous at 1 from the left, since

limx�1�

f (x) = 1 6= f (1) .

By the de�nition of continuity from the right and from the left, a function f is continuous at apoint a if and only if f is continuous at a from the right and from the left.

As discussed in Section A3 of Appendix A, a point is in the interior of an interval if itbelongs to the interval but it is not an endpoint of the interval.


De�nition 6 A function f is continuous on the interval J if f is continuous at each pointin the interior of J , and the appropriate one-sided continuity is valid at any endpoint of J thatis in J .

Note that there is a break in the graph of the function of Example 9, corresponding to thediscontinuity at 1. If f is the function of Example 10, then f is also discontinuous at 1. Thepoint (1, 2) which is on the graph of f seems to have left a hole at the point (1, 1). Such breaksor holes in the graph of a function indicate discontinuities. If f is continuous at each pointof an interval, the graph of f on that interval is a “continuous curve” withoutany breaks or holes. We will refer to such a portion of the graph of a function simply as acontinuous curve.

A word of caution: Figure 7 that displays the graph of the function f of Example 4 wasgenerated by a program that takes into account the discontinuities of a function. Figure 11shows another computer generated graph for the same function. The picture shows a continuouscurve as the graph of the function, even though f is discontinuous at 3. The picture includesa line segment that appears to be vertical and seems to connect the points (1, 1) and (1, 3).That is a spurious line segment and is not part of the graph of f (the graph of a functioncannot contain a vertical line segment!). The graphing utility that produced Figure 11 sampledvalues of x immediately to the left of 1 and immediately to the right of 1, and connected thecorresponding points on the graph of f with a line segment.

�2 �1 1 2 3x

1

3

y

Figure 11: Virtual continuity

Example 11 Let f (x) = 1/x.

Since f (x) attains arbitrarily large positive values as x approaches 0 from the right, the functiondoes not have a right-limit at 0. The function does not have a left-limit at 0 either, since f (x)attains negative values of arbitrarily large magnitude as x approaches 0 from the left. Forexample,

f

�1

10n

¶= 10n and f

�� 1

10n

¶= �10n,

where n is arbitrarily large. ¤

�3 3�2 2�1 1x

�10

�5

5

10y

Figure 12: Unbounded discontinuity at 0


De�nition 7 We will say that f has an unbounded discontinuity at a if f (x) attains valuesof arbitrarily large magnitude as x approaches a from the right or from the left.

Thus, the function f of Example 11 has an unbounded discontinuity at 0.

A function can have a discontinuity other than a jump discontinuity or an unbounded disconti-nuity:

Example 12 Let f(x) = sin (1/x). Figure 13 shows a computer generated graph of f .

�1 �0.5 0.5 1x

�1

1y

Figure 13: Discontinuity at 0 due to oscillations.

There appears to be a dark blob around the interval [�1, 1] on the vertical axis. In particular,the picture does not indicate the existence of a de�nite number to which f (x) approaches as xapproaches 0. Thus, the picture suggests that the function does not have a limit at 0. Indeed,there are points that are arbitrarily close to 0 at which the function has values 1 or �1 (determinesuch points as an exercise). You can imagine that the graph of f oscillates between 1 and �1“in�nitely often” near 0. ¤

Problems

In problems 1 - 8 the graph of a function f is displayed. Determine limx�a+ f (x), limx�a� f (x)and limx�a f (x), as indicated by the picture, provided that such values exists. Based on yourresponse, is f continuous at a?

�2 2 4 6x

1

2

3

4

5

6y

1 : a = 2

�1 1 2x

�5

�4

�3

�2

�1

1

2y

2 : a = 1

1.5. THE CALCULATION OF LIMITS 47

10.

f (x) =

��sin (x) if x < 1,

x� 1x2 � 1 if x > 1.

, limx�1+

f (x) =1

2

In problems 11 and 12, justify the indicated one-sided continuity of f in accordance with theprecise de�nition.

11. If

f (x) =

��

1

x+ 2if x < �2,

x2 + 4 if x � �2,then f is continuous at �2 from the right.

12. If

f (x) =

��x3 + 2x2 if x 1,

1

xif x > 1,

then f is continuous at 1 from the left.

1.5 The Calculation of LimitsIn this section we will provide guidelines for the determination of limits.

A Portfolio of Continuous Functions

Since the limit of a function f at a point a is simply the value of f at a, let’s begin by takingstock of a rich collection of continuous functions:

Polynomials, rational functions, sine, cosine, tangent and secant are continuous ontheir respective natural domains.

You can �nd the justi�cation of these facts in Appendix B.

In particular, a polynomial is continuous on the entire number line.

Example 1 Let

f (x) = 1� 12x2 +

1

24x4.

Evaluate limx��2 f (x).

Solution

Figure 1 shows the graph of f . The graph is a curve without any breaks or holes, consistentwith the continuity of f on the number line.

�4 �2 2 4x

1

2

3

y

Figure 1


By the continuity of f at�2,

limx��

2f (x) = f

³�2´= 1� 1

2

³�2´2+1

24

³�2´4=1

6.

¤

Example 2 Let

f (x) =x2 + 1

x2 � 1 .

a) Determine the points at which f is discontinuous.b) Evaluate limx�2 f (x) .

Solution

a) The rational function f is continuous at any point of its natural domain, i.e., at any pointwhere the denominator does not vanish, Since

x2 � 1 = 0 x = ±1,f is continuous at a if a 6= 1 and a 6= �1.b) By the continuity of f at 2,

limx�2

f (x) = f (2) =5

3.

Figure 2 shows the graph of f . Since f is continuous at each point of the intervals (��,�1),(�1, 1) and (1,+�), the corresponding parts of the graph of f are continuous curves. On theother hand, f is not de�ned at 1 or �1, so that f is discontinuous at these points, and the graphhas breaks at x = 1 and x = �1. ¤

�4 �2 2 4x

�6�6

2

4

6

y

�1 1

Figure 2: The function has discontinuities at ±1

The trigonometric functions sine and cosine are periodic functions that are de�ned on the entirenumber line. Figure 3 shows their graphs on the interval [�2�, 2�].

�2 Π �Π ΠΠ2

2 Πx

�1

1y

y � sin�x�

�2 Π �Π Π 2 Πx

�1

1y

y � cos�x�

Π

2

Figure 3


The graphs are “continuous waves”, consistent with the fact that the functions are periodicfunctions that are continuous on the number line.

Example 3 Determinelim

x��/6sin (x) and lim

x��/6cos (x) .

Solution

By the continuity of sine at �/6

limx��/6

sin (x) = sin³�6

´=1

2.

By the continuity of cosine at �/6,

limx��/6

cos (x) = cos³�6

´=

�3

2.

¤The natural domain of

tan (x) =sin (x)

cos (x)

consists of all x such that cos (x) 6= 0. Thus, tangent is continuous at each x that is not anodd multiple of ±�/2. Figure 4 shows the graph of y = tan (x) on (�3�/2, 3�/2). The parts ofthe graph on the intervals (�3�/2,��/2), (��/2, �/2) and (�/2, 3�/2) are continuous curves,consistent with the continuity of the function at each point of such an interval.

x

�20

�10

10

20y

�3Π�2 �Π�2 Π�2 3Π�2�Π Π

Figure 4: y = tan (x)

The natural domain ofsec (x) =

1

cos (x)

also consists of all x such that cos (x) 6= 0. Thus, secant is continuous at each x that is not anodd multiple of ±�/2. Figure 5 shows the graph of secant on the interval (�3�/2, 3�/2). Theparts of the graph of secant on the intervals (�3�/2,��/2), (��/2, �/2) and (�/2, 3�/2) arecontinuous curves, consistent with the continuity of secant on these intervals.

x

�20

�10

10

20y

�3Π�2 �Π�2 Π�2 3Π�2�Π Π

Figure 5: y = sec (x)


Example 4 Evaluatelim

x��/4tan (x) and lim

x��/4sec (x) .

Solution

By the continuity of tangent at �/4,

limx��/4

tan (x) = tan³�4

´= 1.

By the continuity of secant at �/4,

limx��/4

sec (x) = sec (�/4) =1

cos (�/4)=

11�2

=�2.

¤A rational power of x de�nes a function that is continuous at each point of its natural domain:

If r is a rational number the function de�ned by x� is continuous at each point ofis natural domain, with the understanding that continuity is from the right or fromthe left only, if appropriate.

You can �nd the proof of this fact in Appendix B.

Example 5 Let f (x) =�x = x1/2. The square-root function f is continuous on the

interval [0,+�). The continuity of f at 0 is only from the right. The graph of f on any intervalthat is contained in [0,+�) is a continuous curve, consistent with the continuity of f on anysuch interval. ¤

1 2 4 6 8x

1

2

y

y � x

Figure 6: y =�x

The cube-root function is a prototype of functions de�ned by x1/n, where n is an odd positiveinteger:

Example 6 Let f (x) = x1/3. Then, f is continuous on the entire number line. The graph ofthe cube-root function on any interval is a continuous curve. ¤

�8 �6 �4 �2 2 4 6 8x

�1

�2

1

2y

y � x1�3

Fiugre 7: y = x1/3


Example 7 Let f (x) = x3/4 =¡x1/4

¢3. Then, f is continuous on [0,+�). The continuity of

f at 0 is only from the right. ¤

2 4 6 8x

2

4

y

y � x3�4

Figure 8: y = x3/4

Example 8 Let f (x) = x2/3 =¡x1/3

¢2. Then, f is continuous on the entire number line. ¤

�8 �4 4 8x

1

2

3

4y

y � x2�3

Figure 9: y = x2/3

Limits and Removable Discontinuities

The following observation enables us to compute a limit by making use of our knowledge aboutcontinuous functions:

Assume that f(x) = g(x) for each x in an open interval J that contains the pointa, with the possible exception of a itself, and that g is continuous at a. Then,

lim��

f(x) = g(a).

Proof

Since g is continuous at a, we have limx�a g (x) = g (a). Since f (x) = g (x) for each x in J suchthat x 6= a, and the de�nition of the limit of f at a does not involve a,

limx�a

f (x) = limx�a

g (x) = g (a) .

¥

Example 9 Let

f (x) =4x� 16x2 � 16 .

Determine limx�4 f (x).


Solution

Sincex2 � 16 = 0 x = 4 or x = �4,

the rational function f is not de�ned at 4. We are led to the indeterminate form 0/0 if we tryto replace x in the expression for f (x) by 4. Let us simplify the expression for f (x):

f (x) =4x� 16x2 � 16 =

4 (x� 4)(x� 4) (x+ 4) =

4

x+ 4

if x 6= 4. Thus, if we set

g (x) =4

x+ 4,

then f (x) = g (x) for each x such that x 6= 4 and x 6= �4. The rational function g is de�nedand continuous at x = 4. Since f (x) = g (x) for each x near 4 such that x 6= 4. Therefore

limx�4

f (x) = limx�4

g (x) = g (4) =4

8=1

2.

1 2 3 4 5 6x

0.5

1y

�4, 0.5�

Figure 10

Figure 10 shows a computer generated graph of f on the interval [0, 6], as generated by agraphing utility. The computer does not seem to be aware of the fact that f is not de�ned at 4,and has produced a continuous curve. It would have produced the same picture if it had beenasked to plot the graph of the function g which is continuous at each point of [0, 6], including 4,since g (x) = f (x) for each x 6= 4 in that interval. The picture is consistent with the fact thatlimx�4 f (x) = 0.5. It appears that we can “remove ” the discontinuity of f at 4 by declaringthat its value at 4 is 0.5. ¤

De�nition 1 Assume that a function f is de�ned in an open interval J that contains the pointa, with the possible exception of a itself, and that f is discontinuous at a. We say that f hasa removable discontinuity at a if limx�a f (x) exists.

The terminology is appropriate since g is continuous at a if

g (x) =

½f (x) if x 6= a and x � J,

limx�a f (x) if x = a.

We can say that the discontinuity of f at a is removed by de�ning or rede�ning its value at aproperly.Note that the function f of Example 9 has a removable discontinuity at 4.


Limits and Continuity of Combinations of Functions

The following rules are relevant to the calculation of the limits of sums and products of functions:

LIMITS OF ARITHMETIC COMBINATIONS OF FUNCTIONS

Assume that lim�� f(x) and lim�� g(x) exist and that c is a constant.

1. The constant multiple rule for limits:

lim��

cf(x) = c lim��

f(x)

2. The sum rule for limits:

lim��

(f(x) + g(x)) = lim��

f(x) + lim��

g(x)

(the limit of a sum is the sum of the limits).3. The product rule for limits:

lim��

f(x)g(x) =³lim��

f(x)´³

lim��

g(x)´

(the limit of a product is the product of the limits).4. The quotient rule for limits: If lim��g(x) 6= 0,

lim��

f(x)

g(x)=lim��f(x)

lim��g(x)

(the limit of a quotient is the quotient of the limits).

The above rules are plausible: If limx�a f (x) = L1 and limx�a g (x) = L2, we have f (x) �= L1and g (x) �= L2 if x 6= a and x �= a. Therefore,

cf (x) �= cL1, f (x) + g (x) �= L1 + L2, f (x) g (x) �= L1L2,and

f (x)

g (x)�= L1L2,

if L2 6= 0. You can �nd the proofs of the above statements in Appendix B.

Since a function f is continuous at a point a if limx�a f (x) = f (a), the rules for the limitsof arithmetic combinations of functions lead to the continuity of arithmetic combinations ofcontinuous functions:

Assume that f and g are continuous at a and c is a constant. Then the constantmultiple cf , the sum f + g and the product fg are continuous at a. If g(a) 6= 0,the quotient f/g is also continuous at a.

Example 10 Evaluatelim

x��/4

�x cos (x) .

Solution

Since�x de�nes a continuous function on [0,�) and cosine is continuous on the entire number

line, the product�x cos (x) de�nes a function that is continuous at �/4. Therefore,

limx��/4

�x cos (x) =

r�

4cos³�4

´=

��

2

Ã�2

2

!=

�2�

4.

¤


Example 11 Determine limh�0 f (h) if

f (h) =

�4 + h� 2h

.

Solution

The function f is not de�ned at 0. The attempt to replace h by 0 leads to the indeterminateform 0/0. We will obtain another expression for f (h) by rationalizing the numerator. If h 6= 0,

�4 + h� 2h

=

��4 + h� 2h

¶��4 + h+ 2�4 + h+ 2

¶

=

¡�4 + h

¢2 � 22h¡�4 + h+ 2

¢ = (4 + h)� 4h¡�4 + h+ 2

¢ = h

h¡�4 + h+ 2

¢ = 1�4 + h+ 2

If we setg (h) =

1�4 + h+ 2

,

the function g is de�ned at 0. In fact, g is continuous at 0 since the square-root function iscontinuous at 4 and the denominator is nonzero at h = 0. Since f (h) = g (h) if h 6= 0 and |h|is small enough,

limh�0

f (h) = limh�0

g (h) = g (0) =1�4 + 2

=1

4.

Figure 11 shows the graph of f on the interval [�2, 2], as plotted by a graphing utility. Thegraphing utility would have produced the same picture if it had been asked to plot the graph ofg. The picture is consistent with the fact that limh�0 f (h) = 1/4. ¤

�2 �1 0 1 2h

0.5

�0, 0.25�

Figure 11

Many functions are formed by composing simpler functions. Therefore, we should be able tocalculate limits involving composite functions:

THE LIMIT OF A COMPOSITE FUNCTION

Assume that lim�� g(x) = L and f is continuous at L. Then,

lim��

(f � g)(x) = lim��

f(g(x)) = f(L).

It is easy to remember this fact in the following form:

lim��

f(g(x)) = f( lim��

g(x)).

You can �nd the proof of the above statement in Appendix B. The statement is plausible: Asx approaches a, g (x) approaches L. Therefore, f (g (x)) approaches f (L) by the continuity off at L.


If g is continuous at a, we have limx�a g (x) = g (a). Therefore, the above fact about the limitsof composite functions leads to the continuity of compositions of continuous functions:

Assume that g is continuous at a and f is continuous at g(a). Then, f � g is con-tinuous at a. We have

lim��

f(g(x)) = f(g(a)).

Example 12 Evaluate

limx�1

cos

Ã�¡x2 � 1¢

6 (x� 1)

!

Solution

We have

limx�1

�¡x2 � 1¢

6 (x� 1) = limx�1

� (x� 1) (x+ 1)6 (x� 1) = lim

x�1

� (x+ 1)

6=2�

6=�

3.

Since cosine is continuous at �/3,

limx�1

cos

Ã�¡x2 � 1¢

6 (x� 1)

!= cos

Ãlimx�1

�¡x2 � 1¢

6 (x� 1)

!= cos

³�3

´=1

2.

¤

Example 13 Let

F (x) = tan

�3�

4 (x2 � 1)¶.

Justify the continuity of F at 2. Determine limx�2 F (x).

Solution

If we set

u = g (x) =3�

4 (x2 � 1) and f (u) = tan (u) ,

then F (x) = f (g (x)) so that F = f � g. The rational function g is continuous at 2 andg (2) = �/4. The tangent function f is continuous at �/4. Therefore F = f � g is continuous at2. Thus,

limx�2

F (x) = F (2) = f (g (2)) = tan³�4

´= 1.

¤

We will often encounter functions of the form sin (�x) and cos (�x), where � is a constant. Sucha function is continuous at any point on the number line, since it can be expressed as f � g,where g (x) = �x and f (u) = sin (u) or f (u) = cos (u), and both f and g are continuous at anypoint. Recall that a trigonometric polynomial is a linear combination of functions of theform sin(nx) and cos (nx), where n is an integer (as we saw in Section 1.2). Since such functionsare continuous at each point on the number line, a trigonometric polynomial is continuous ateach x � R.

Example 14 Evaluate

limx��/2

�sin (x) +

1

3sin (3x)

¶


Solution

By the continuity of a trigonometric polynomial,

limx��/2

�sin (x) +

1

3sin (3x)

¶= sin

³�2

´+1

3sin

�3�

2

¶= 1 +

1

3(�1) = 2

3.

Figure 12 shows the graph of

f (x) = sin (x) +1

3sin (3x)

on the interval [�2�, 2�]. The graph is a continuous curve, consistent with the continuity of fon [�2�, 2�]. ¤

�2 Π �Π Π 2 Πx

�0.5

0.5

y

Figure 12 : A trigonometric polynomial is continuous

In some cases, the following theorem is helpful to determine the limit of a function:

THE SQUEEZE THEOREM Assume that h(x) � f(x) � g(x) for all x 6= a in anopen interval containing a, and

lim��

h(x) = lim��

g(x) = L.

Then lim�� f(x) = L as well.The squeeze theorem is intuitively plausible: If the values of f are squeezed between the corre-sponding values of h and g, and both h(x) and g(x) approach the same limit L as x approachesa, we should have limx�a f(x) = L. You can �nd the proof of the squeeze theorem in AppendixB.

x

y

f

h

g

a

L

Figure 13: The illustration of the squeeze theorem


Example 15 Determine

limx�0

x2 sin

�1

x

¶.

Solution

Since �1 sin(1/x) 1, we have

�x2 x2 sin(1/x) x2

for x 6= 0. We havelimx�0

¡�x2¢ = limx�0

¡x2¢= 0.

Therefore,

limx�0

x2 sin

�1

x

¶= 0,

by the Squeeze Theorem. ¤

�1 �0.5 0.5 1x

�0.5

�1

0.5

1y

y � x2

y��x2

y � x2sin�1�x�

Figure 14: �x2 x2 sin(1/x) x2

Problems

In problems 1 - 12,a) Determine whether f is continuous at a. Justify your response,b) Determine limx�a f (x) .

1.

f (x) =cos (x)

sin2 (x), a = �/6.

2.

f (x) =cos (x)

sin2 (x), a = �.

3.

f (x) =x4 � 16x� 2 , a = 3.

4.

f (x) =x3 + x2 + 1

x+ 3, a = 2.

5.f (x) =

px2 � 9, a = 2.

6.f (x) =

p4� x2, a = 3.

7.f (x) =

px2 � 9, a = 4.

8.f (x) =

¡x2 � 1¢3/4 , a = 5.

9.

f (x) =

½x2 � 4 if x 2,4� x2 if x > 2.

, a = 2.

10.

f (x) =

½x2 � 1 if x 2,4� x2 if x > 2.

, a = 2.

11.

f (x) =

½4x� 3 if x 6= 2,�3 if x = 2.

, a = 2.

12.

f (x) =

½x2 + 1 if x 6= 3,8 if x = 3.

, a = 3.


In problems 13 -20,a) Determine the function g that is continuous at a such that g (x) = f (x) if x 6= a and x isnear a.b) Evaluate the limit of f at a.

13.

f (x) =x2 � 4x3 � 8 , a = 2

14.

f (x) =2x2 + 5x� 3x2 + x� 6 , a = �3

15.

f (x) = tan (x) cos (x) , a = �/2

16.

f (x) =sec (x)

tan (x), a = 3�/2

17.

f (x) =x3 � 27x� 3 , a = 3

18.

f (x) =

1

x2� 19

x� 3 , a = 319.

f (x) =

�x� 4x� 16 , a = 16

20.

f (x) =x1/3 � 2x� 8 , a = 8

In problems 21-26, evaluate the indicated limit. Indicate the steps that lead to the �nal result:

21.lim

x��/4cos2 (x)

22.

limx��/3

sin (x)

cos2 (x)

23.

limx�4

x2 � x� 122x2 � 12x+ 16

24.

limh�0

�9 + h� 3h

25.

limx�3+

x2 � 4x+ 3�x� 3

26.

limh�0

(4 + h)3 � 64h

In problems 27 - 30, express F as f � g and evaluate limx�aF (x) .

27.

F (x) =

sx2 � 9x� 3 , a = 3

28.

F (x) =

�x2 � 16x� 4

¶1/3

, a = 4

29.F (x) =

psin (x), a = �/6

30.

F (x) = cos

��x2 � 4�3x� 6

¶, a = 2

1.6 In�nite LimitsA function may attain arbitrarily large values near a point. In this section we will discuss suchcases.

The De�nitions

Let’s begin with a speci�c case:

Example 1 Let

f (x) =1

x� 1 .

138 CHAPTER 2. THE DERIVATIVE

Π2

Π 3 Π2

2 Πt

�1

1

position

Π2

Π 3 Π2

2 Πt

�1

1

velocity

Π2

Π 3 Π2

2 Πt

�1

1

acceleration

Figure 10

Problems

In problems 1 - 4, f (t) is the position at time t of an object in one-dimensional motion.a) Determine v (t), the velocity of the object at time t, and a (t), the acceleration of the objectat time t.b) Calculate v (t0) and a (t0).

1.

f (t) = 200t� 5t2, t0 = 1

2.

f (t) = 5t2 + 100; t0 = 4

3.f (t) = 10 sin (t) , t0 = �/6

4.

f (t) = 3 sin (t) + 8 cos (t) , t0 = �/2

2.5 Local Linear Approximations and the Di�erentialThe derivative of a function f at a point a can be interpreted as the slope of the tangentline to the graph of f at (a, f(a)). The tangent line is the graph of a linear function that is thebest linear approximation to f near a in a sense that will be explained in this section.

Local Linear Approximations

Given a function f that is di�erentiable at the point a, the tangent line to the graph of f at(a, f (a)) is the graph of the equation

y = f(a) + f 0(a) (x� a) .We will give a name to the underlying linear function:

De�nition 1 The linear approximation to f based at a is

L�(x) = f(a) + f0(a)(x� a).

We refer to a as the basepoint.

2.5. LOCAL LINEAR APPROXIMATIONS AND THE DIFFERENTIAL 139

x

y

a

La

�a, f�a��

Figure 1: The graph of La is a tangent line

Example 1 Let f (x) = x2�2x+4, as in Example 1 of Section 2.1. We showed that f 0 (3) = 4and the tangent line to to graph of f at (3, f (3)) is the graph of the equation

y = f (3) + f 0 (3) (x� 3) = 7 + 4 (x� 3) .Thus, the linear approximation to f based at 3 is

L3 (x) = 7 + 4 (x� 3) .Figure 2 illustrates the e�ect of zooming in towards the point (3, f (3)) = (3, 7). Note that wecan hardly distinguish between the graphs of f and L3 in the third frame. This indicates thatL3 (x) approximates f (x) very well if x is close to the basepoint 3. On the other hand, we donot expect L3 (x) to approximate f (x) when x is far from 3. The linear function L3 is a "localapproximation" to f . ¤

3 6x

710

20

y

�3, f�3��

2.6 3.4

5

9

�3, f�3��

2.8 3.2

6.5

7.5

�3, f�3��

Figure 2

Let’s assess the error in the approximation of f (x) by L3 (x) algebraically. Since L3 (x) isexpected to be a good approximation to f when x is near 3, it is convenient to set x = 3 + h,so that h (= x� 3) represents the deviation of x from the basepoint 3. We have

L3 (3 + h) = 7 + 4 (x� 3)|x=3+h = 7 + 4h.

Therefore,

f (3 + h)� L3 (3 + h) = (3 + h)2 � 2 (3 + h) + 4� (7 + 4h)= 9 + 6h+ h2 � 6� 2h+ 4� 7� 4h= h2


Thus, the absolute error is|f (3 + h)� L3 (3 + h)| = h2.

Note that h2 is much smaller than |h| if |h| is small. For example,

¡10�2

¢2= 10�4 and

¡10�3

¢2= 10�6.

Thus, the absolute error in the approximation of f (x) by L3 (x) is much smaller than thedistance of x from the basepoint 3 if x is close to 3. This numerical fact is consistent with ourgraphical observation. ¤

Example 2 Let

f (x) =1

x.

a) Determine L2, the linear approximation to f based at 2.b) Calculate f (2 + h) and L2 (2 + h) for h = �10�n, n = 1, 2, 3. Compare |f (2 + h)� L2 (2 + h)|with |h|.Solution

a) By the power rule,

f 0 (x) =d

dx

�1

x

¶=d

dx

¡x�1

¢= �x�2 = � 1

x2.

Therefore, f 0 (2) = �1/4 and

L2 (x) = f (2) + f0 (2) (x� 2) = 1

2� 14(x� 2) .

Thus,

L2 (2 + h) =1

2� 14h.

Figure 3 shows the graphs of f and L2 (the dashed line) in a small viewing window that iscentered at (2, f (2)) = (2, 0.5).

1.8 1.9 2.1 2.2

0.46

0.48

0.52

0.54

Figure 3

b) Table 1 displays the required data. We see that |f (2 + h)� L2 (2 + h)| is much smallerthan |h| for the values of h that are considered. Indeed, f

¡2� 10�3¢ and L2

¡2� 10�3¢ are

represented by the same decimal, rounded to 6 signi�cant digits. Therefore, the numbers supportour analysis of the error in linear approximations. ¤

h f (2 + h) L2 (2 + h) |f (2 + h)� L2 (2 + h)|�10�1 0.526 316 0.525 1. 3× 10�3�10�2 0.502 513 0.502 5 1. 3× 10�5�10�3 0.500 25 0.500 25 1. 3× 10�7


Table 1

Remark We have identi�ed the rate of change of a function f at a point a with f 0 (a),and f 0 (a) is the rate of the linear function La. The fact that La (x) approximates f (x) verywell if x is close to a justi�es this identi�cation. After all, there is no question that the rate ofchange of the linear function

La (x) = f (a) + f0 (a) (x� a)

is f 0 (a) at any point. �

In particular, if f 0 (a) = 0 we declare that the rate of change of f at a is 0. This does not meanthat we have f (x) = f (a) for each x in some interval centered at a. On the other hand,

La (x) = f (a) + f0 (a) (x� a) = f (a) ,

and the rate of change of the constant function La is 0. Since

f (x) �= La (x) = f (a) ,

and the magnitude of the error can be expected to be much smaller than |x� a| if |x� a|is small, the restriction of f to a small interval centered at a is almost a constant function.Therefore, it is reasonable to declare that the rate of change of f at a is 0.

Example 3 As in Example 2 of Section 2.3, where we determined the tangent line to the graphof cosine at (0, 0), the linear approximation to cosine based at 0 is

L0 (x) = cos (0) +

�d

dxcos (x)

¯̄̄¯x=0

¶x = 1.

Thus, L0 is a constant function and its graph, i.e., the tangent line to the graph of cosine at(1, 0), is a horizontal line, as shown in Figure 4.

� Π2

Π2

x

�1

1

y

Figure 4

Obviously, the rate of change of L0 is 0. We declare that the rate of change of cosine at 0 isalso 0, even though cos (x) 6= 0 if x deviates from 0 slightly. This is justi�ed in view of the factthat cos (x) �= 1 if x �= 0, and the absolute error in the approximation is much smaller than |x|is |x| is small. For example, cos (0.01) �= 0.999 95, |cos (0.01)� 1| �= 5. 0× 10�5, and 5. 0× 10�5is much smaller than 10�2. ¤


The Di�erential

It is useful to consider all the local linear approximations to a given function at once by consid-ering the basepoint to be a variable. In this case it is convenient to work with di�erences and achange in the notation seems to be in order. We will denote an increment along the x-axis by�x. Thus,

f 0 (x) = limh�0

f(x+ h)� f(x)h

= lim�x�0

f (x+�x)� f (x)�x

.

Therefore,f (x+�x)� f (x)

�x�= f 0 (x)

if |�x| is small, so thatf (x+�x)� f (x) �= f 0 (x)�x.

We will give the expression f 0 (x)�x a special name:

De�nition 2 The di�erential of the function f is

d f = f 0(x)�x.

Thus, df is a function of two independent variables, the basepoint x and the increment�x. We can indicate this explicitly by writing

d f(x,�x) = f 0(x)�x.

We havef(x+�x)� f(x) �= d f(x,�x)

if |�x| is small. The idea behind the di�erential is the same as the idea of local linear ap-proximations. The di�erential merely keeps track of local linear approximations to afunction as the basepoint varies. Note that d f(x,�x) is the change correspondingto the increment �x along the tangent line to the graph of f at (x, f(x)), as illustrated inFigure 5.

x

y

x

�x, f�x��

�x��x, f�x��x��

x � �x

f�x � �x� � f�x�

�x

df�x, �x�

Figure 5

Example 4 Let f (x) =�x. Approximate

�4.1 via the di�erential of f .

Solution

Since

f 0 (x) =d

dx

�x =

1

2�x

,


The di�erential of f is

df (x,�x) = f 0 (x)�x =1

2�x(�x) =

�x

2�x.

It is natural to set x = 4 and �x = 0.1 for the approximation of�4.1 = f (4.1) since f (4) =�

4 = 2. Thus,

�4.1� 2 = f (4.1)� f (4) �= df (4, 0.1) = 0.1

2�4=0.1

4= 0.025.

Therefore, �4.1 = 2 +

³�4.1� 2

´ �= 2 + 0.025 = 2.025.We have �

4.1 �= 2. 024 85,rounded to 6 signi�cant digits, and¯̄̄�

4.1� 2.025¯̄̄�= 1.5× 10 �4.

Note that the absolute error in the approximation of�4.1 via the di�erential is much smaller

than �x = 0.1. ¤

Remark 1 As we saw in Section 2.4, the rate of change of the position f (t) of an object inone-dimensional motion at time t is the instantaneous velocity v (t). If the time increment is�t > 0 is small then

f (t+�t)� f (t) �= df (t,�t) = f 0 (t)�t = v (t)�t.

Thus, the displacement over the time time interval [t, t+�t] is approximately v (t)�t if �t issmall.For example, if f (t) = cos (t) then v (t) = � sin (t) so that

f (t+�t)� f (t) �= � sin (t)�t.

In particular,

f³�6+ 0.1

´� f

³�6

´ �= � sin³�6

´(0.1) = �0.1

2= �0.05.

The (�) sign indicates that the motion is in the negative direction. �

The Traditional Notation for the Di�erential

We wrotedf (x,�x) = f 0(x)�x.

Traditionally, the increment �x is denoted by dx within the context of di�erentials. Thus,

df (x, dx) = f 0(x)dx.

If we use the Leibniz notation for f 0 (x), we have

df (x, dx) =df

dx(x) dx.


x

y

x

�x, f�x��

�x � dx, f�x � dx��

x � dx

f�x � dx� � f�x�

dx

df

Figure 6: The geometric meaning of the di�erential

We usually do not bother to indicate that the di�erential depends on x and dx, and write

df =df

dxdx.

This is convenient and traditional notation, but you should keep in mind that the “fraction”

df

dx

is a symbolic fraction, and that the symbol dx that appears as the denominator does not havethe same meaning as dx that stands for the increment in the value of the independent variable.The expression

df =df

dxdx

is analogous to the expression

�f =�f

�x�x,

where �x 6= 0 and �f = f (x+�x)� f (x).If we refer to the function as y = y(x), we can write

dy =dy

dxdx

The above expression is analogous to the expression

�y =�y

�x�x,

where �x 6= 0 and �y = y (x+�x)� y (x).

Example 5 Let f (x) = x1/3

a) Determine the di�erential df.b) Make use of the di�erential df to approximate (8.01)1/3. Determine the absolute error in theapproximation by treating the value that is obtained from your calculator as the exact value.Compare with the deviation from the basepoint that you have chosen.

Solution

a)

df =df

dxdx =

�d

dx

³x1/3

´¶dx =

�1

3x�2/3

¶dx =

1

3x2/3dx.


b) Since 8.01 is close to 8, and f (8) = 81/3 = 2, the natural choice for the basepoint is 8. Thus,dx = 8.01� 8 = 0.01. The value of the di�erential corresponding to x = 8 and dx = 0.01 isÃ

1

3¡82/3

¢!(0.01) =

0.01

3 (4)=0.01

12.

Therefore,

(8.01)1/3 � 2 = f (8.01)� f (8) �= 0.01

12

so that(8.01)

1/3 �= 2 + 0.0112

�= 2. 000 83

A calculator tells us that (8.01)1/3 �= 2. 000 83, rounded to 6 signi�cant digits. Thus, theapproximation via the di�erential gave us the same decimal, rounded to 6 signi�cant digits.There is a nonzero of course. Indeed,¯̄̄

¯�2 +

0.01

12

¶� (8.01)1/3

¯̄̄¯ �= 3. 5× 10�7.

Thus, the absolute error in the approximation is much smaller than 0.01, the deviation of 8.01from the basepoint 8. ¤

Example 6 The volume of a spherical ball of radius r is

V =4

3�r3.

a) Determine the di�erential dV .b) Use the di�erential to approximate the change in the volume of the ball if the ball is in�atedand its radius increases from 20 centimeters to 20.1 centimeters.

Solution

a) We havedV

dr=d

dr

�4

3�r3

¶=4

3�d

dr

¡r3¢=4

3�¡3r2¢= 4�r2.

Therefore,

dV =dV

drdr = 4�r2dr.

Note that 4�r2 is the surface area of sphere of radius r. Therefore, the change in the volume ofa spherical ball that corresponds to a small change in the radius can be approximated by theproduct of the area of its boundary and the increment of the radius.

b) In particular,V (20.1)� V (20) �= 4� ¡202¢ (0.1) �= 502.655

(cm3). The actual change in the volume is

V (20.1)� V (20) = 4

3� (20.1)3 � 4

3� (20)3 �= 505.172

(cm3). Therefore, the error in the approximation of the change in the volume via the di�erentialis approximately 2.517

¡cm3

¢. This may not be considered to be a small number. On the other

hand, the relative error is usually more appropriate in assessing error. Thus,

(V (20.1)� V (20))� 4� ¡202¢ (0.1)V (20)

�= 2.517

33510.3�= 7.5× 10�5,


and this number is small.

We can also approximate the relative change in the volume, i.e.,

V (20.1)� V (20)V (20)

,

via the di�erential by calculating

dV (20, 0.1)

V (20)=4�¡202¢(0.1)

V (20)�= 1.5× 10�2.

This approximatesV (20.1)� V (20)

V (20)�= 1.507 51× 10�2

with an error that is approximately 7× 10�5. ¤

The Accuracy of Local Linear Approximations

Theorem 1 Assume that f is di�erentiable at a, and that L� is the linear approxi-mation to f based at a. We have

f(a+ h) = L�(a+ h) + hq (h) ,

wherelim�0

q(h) = 0.

Thus, hq (h) represents the error in the approximation of f by La at x = a+ h. Since the erroris the product of h and q (h), and q (h) � 0 as h � 0, its magnitude is much smaller than|h| = |x� a| if x is close to the basepoint a.

With reference to Example 1, q (h) = h2.

Proof

As in Example 1, we will set x = a+ h, so that h = x� a represents the deviation of x from aand has small magnitude if x is near a. We have

La (a+ h) = f (a) + f 0 (a) (x� a)|x�a=h = f (a) + f 0 (a)h.Therefore,

f (a+ h)� La (a+ h) = f (a+ h)� (f (a) + f 0 (a)h)= (f (a+ h)� f (a))� f 0 (a)h= h

�f (a+ h)� f (a)

h� f 0 (a)

¶

Let’s set

q (h) =f (a+ h)� f (a)

h� f 0 (a) ,

so that q (h) is the di�erence between the di�erence quotient and the derivative. We have

limh�0

q (h) = limh�0

�f (a+ h)� f (a)

h� f 0 (a)

¶= 0,

since the di�erence quotient approaches the derivative as h� 0.


Thus,f (a+ h)� La (a+ h) = hq (h) ,

so thatf (a+ h) = La (a+ h) + hq (h) ,

where limh�0 q (h) = 0. ¥The analysis of the error in the approximation of di�erences via the di�erential is along similarlines:

Theorem 2 Assume that f is di�erentiable at x. Then,

f(x+�x)� f(x) = d f(x,�x) + �x q (�x) ,

wherelim

��0q (�x) = 0.

Proof

We have

f (x+�x)� f (x)� df (x,�x) = f (x+�x)� f (x)� f 0 (x)�x= �x

�f (x+�x)� f (x)

�x� f 0 (x)

¶.

If we set

q (�x) =f (x+�x)� f (x)

�x� f 0 (x) ,

thenf (x+�x)� f (x)� df (x,�x) = �xq (�x) .

We have

lim�x�0

q (�x) = lim�x�0

�f (x+�x)� f (x)

�x� f 0 (x)

¶= 0,

since

lim�x�0

f (x+�x)� f (x)�x

� .f 0 (x) .Thus,

f (x+�x)� f (x)� df (x,�x) = �xq (�x) ,where lim�x�0 q (�x) = 0.

Problems

In problems 1 - 6,a) Determine La, the linear approximation to f based at a,b) Make use of La (b) to approximate f (b) if such a point b is indicated,c) [C] Calculate the absolute error in the approximation of f (b) by La (b) and compare with|b� a|.d) [C] Plot the graphs of f and La in a su�ciently small viewing window centered at (a, f (a))that demonstrates the accuracy of the linear approximation near a.

2.7. THE CHAIN RULE 157

2.7 The Chain Rule

In the previous sections of this chapter we discussed the rules for the di�erentiation of thesums, products and quotients of functions. In this section you will learn how to di�erentiatea function that can be expressed as a composition of functions with known derivatives. Therelevant di�erentiation rule is the chain rule. For example, if F (x) = sin

¡x2¢, the rules that

you have learned until now do not lead to the derivative of F , at least not immediately. On theother hand, we can express F as f � g, where f (u) = sin (u) and g (x) = x2, and we know howto di�erentiate both f and g. The chain rule will enable you to determine F 0 easily.

Introduction to the Chain Rule

THE CHAIN RULE Assume that g is di�erentiable at x and f is di�erentiableat g(x). Then f � g is di�erentiable at x and we have

(f � g)0(x) = f0(g(x))g

0(x).

Example 1 Let F (x) = sin¡x2¢. Determine F 0 (x).

Solution

If we set g (x) = x2 and f (u) = sin (u), then f (g (x)) = f¡x2¢= sin

¡x2¢. Therefore, F = f �g.

We have

f 0 (u) =d

dusin (u) = cos (u) , g0 (x) =

d

dx

¡x2¢= 2x.

By the chain rule,

F 0 (x) = (f � g)0 (x) = f 0 (g (x)) g0 (x) = cos ¡x2¢ (2x) = 2x cos ¡x2¢ .¤

A Plausibility Argument for the Chain Rule

The di�erence quotient that is relevant to the di�erentiation of f � g is

(f � g) (x+�x)� (f � g) (x)�x

=f (g (x+�x))� f (g (x))

�x.

Let’s set g (x) = u and g (x+�x) = u+�u so that

�u = g (x+�x)� g (x) .

Thus,f (g (x+�x))� f (g (x))

�x=f (u+�u)� f (u)

�x

Assume that |�x| is small. Since g is di�erentiable at x it is continuous at x. Therefore |�u| isalso small. As we have seen in Section 2.5,

f (u+�u)� f (u) �= df (u,�u) = f 0 (u)�u.

Thus,f (g (x+�x))� f (g (x))

�x=f (u+�u)� f (u)

�x�= f 0 (u)�u

�x


Threfore we should have

(f � g)0 (x) = lim�x�0

f (g (x+�x))� f (g (x))�x

= lim�x�0

f 0 (u)�u�x

= f 0 (u) lim�x�0

�u

�x

= f 0 (g (x)) lim�x�0

g (x+�x)� g (x)�x

= f 0 (g (x)) g0 (x) ,

as claimed.

You can �nd the proof of the chain rule at the end of this section. The proof is along the linesof the above plausibility argument.

Remark 1 (Caution) In order to determine the derivative of the composite function f � g atx, we must evaluate g0 at x and f 0 at g (x). The chain rule does not say that

(f � g)0 (x) = f 0 (x) g0 (x) .

For example, if f (x) = g (x) = x2, then (f � g) (x) = f (g (x)) = ¡x2¢2 = x4, so that(f � g)0 (x) = 4x3, by the power rule. On the other hand, f 0 (x) g0 (x) = (2x) (2x) = 4x2. �

We can visualize the composite function f � g schematically, where the functions are viewed asinput-output mechanisms. The input for “the outer function” f is the output g (x) of the “innerfunction” g:

xg� g (x)

f� f (g (x))

Thus, it should be easy to remember to evaluate f 0 at g (x) in the evaluation of (f � g)0 (x). �

Example 2 LetF (x) =

px2 + 1.

Determine F 0.

Solution

If we setu = g (x) = x2 + 1 and f (u) =

�u,

then F (x) = f (g (x)), so that F = f � g. We have

f 0 (u) =d

du

�u =

1

2�u,

so thatf 0 (g (x)) = f 0

¡x2 + 1

¢=

1

2�x2 + 1

.

We also have

g0 (x) =d

dx

¡x2 + 1

¢= 2x.

By the chain rule,

F 0 (x) = f 0 (g (x)) g0 (x) =�

1

2�x2 + 1

¶(2x) =

x�x2 + 1

.


The above expression is valid for each x � R since x2 + 1 > 0. Figure 1 shows the graphs ofF and F 0. Note that the graph of F 0 has the horizontal asymptote y = �1 at �� and thehorizontal asymptote y = 1 at +� (con�rm by evaluating the relevant limits). ¤

�4 �2 2 4x

1

2

3

4

y

F

�4 �2 2 4x

�1

1y

F'

Figure 1

The Chain Rule in the Leibniz Notation

As in the implementation of the other rules for di�erentiation, it is usually more practical touse the Leibniz notation when we apply the chain rule. Assume that F (x) = f (u (x)). By thechain rule,

F 0 (x) = f 0 (u (x))u0 (x) .

The above relationship can be expressed in the Leibniz notation as follows:

d

dxf(u(x)) =

Ãdf

du

¯̄̄¯�=�(�)

!du

dx=df

du(u(x))

du

dx.

Example 3 Determined

dxtan

¡x3¢.

Solution

If we set u (x) = x3 then tan¡x3¢= tan (u (x)). Therefore,

d

dxtan

¡x3¢=

�d

dutan (u)

¯̄̄¯u=x3

¶�d

dx

¡x3¢¶=¡sec2 (u)

¯̄u=x3

¢ ¡3x2

¢=¡sec2

¡x3¢¢ ¡

3x2¢

= 3x2 sec2¡x3¢.

¤

The chain rule enables us to evaluate the derivative of a translation of a function easily: Ifc is a constant,

d

dxg(x� c) = dg

du(x� c) .


Indeed, if we set f (x) = g (x� c) and u (x) = x� c,df

dx(x) =

d

dxg (u (x)) =

�dg

du(u (x))

¶�d

dx(x� c)

¶

=

�dg

du(x� c)

¶(1)

=dg

du(x� c) .

It is practical to implement the chain rule directly in a speci�c case, as in the following example.


dx(x� 4)2/3 .

Solution

If we set u (x) = x� 4,

d

dx(x� 4)2/3 = d

dx(u (x))2/3 =

Ãd

du

³u2/3

´¯̄̄¯u=x�4

!�d

dx(x� 4)

¶

=

Ã2

3u�1/3

¯̄̄¯u=x�4

!(1)

=2

3 (x� 4)1/3.

¤We will come across many functions of the form g (�x), where � is a constant. If we setu (x) = �x,

d

dxg (�x) =

d

dxg (u (x)) =

�dg

du

¯̄̄¯u=�x

¶�d

dx(�x)

¶= g0 (�x) (�) = �g0 (�x) .

Again, it is practical to implement the chain rule directly in a speci�c case, as in the followingexample.

Example 5 Let � be an arbitrary constant. then

d

dxsin (�x) = � cos (�x) and

d

dxcos (�x) = �� sin (�x) .

We can derive these formulas with the help of the chain rule:

d

dxsin (�x) =

�d

dusin (u)

¯̄̄¯u=�x

¶�d

dx(�x)

¶= (cos (u)|u=�x) (�)

= � cos (�x) .

Similarly,

d

dxcos (�x) =

�d

ducos (u)

¯̄̄¯u=�x

¶�d

dx(�x)

¶= (� sin (u)|u=�x) (�)= �� sin (�x) .


¤If y is the dependent variable of f , and we refer to f (u) as y (u), then the expression

d

dxf (u (x)) =

Ãdf

du

¯̄̄¯u=u(x)

!du

dx

readsd

dxy (u (x)) =

Ãdy

du

¯̄̄¯u=u(x)

!du

dx.

We can simply writedy

dx=dy

du

du

dx,

with the understanding that the letter y on the left-hand side refers to y (u (x)), and dy/du isevaluated at u (x). This somewhat imprecise expression for the chain rule is appealing due toits “symbolic correctness”: If we pretend that we are dealing with genuine fractions, and notjust symbolic fractions, the cancellation of du on the right-hand side of the expression yieldsdy/dx. Aside from its “symbolic correctness”, an appealing feature of the above expression isits interpretation in terms of rates of change. Indeed, dy/dx is the rate of change of y withrespect to x, dy/du is the rate of change of y with respect to u (at u (x)), and du/dx is the rateof change of u with respect to x. Therefore, we can read the chain rule as follows:

The rate of change of y with respect to x= (the rate of change of y with respect to u)× (the rate of change of u with respect to x) .

Remark 2 (Another Plausibility Argument for the Chain Rule)

Let’s set u = u (x), �u = u (x+�x) � u (x), and �y = y (u (x+�x)) � y (u (x)) so that�y = y (u+�u)� y (u). If we assume that �x 6= 0 and �u 6= 0,

�y

�x=�y

�u

�u

�x.

We can read the above equality as follows:

The average rate of change of y with respect to x

= (the average rate of change of y with respect to u)

× (the average rate of change of u with respect to x) .

We havedy

dx= lim

�x�0

�y

�x= lim

�x�0

��y

�u

�u

�x

¶=

�lim

�x�0

�y

�u

¶�lim

�x�0

�u

�x

¶,

assuming that �u 6= 0 if �x 6= 0. Since �u = u (x+�x)�u (x) approaches 0 as �x approaches0 (di�erentiability implies continuity),

dy

dx=

�lim

�x�0

�y

�u

¶�lim

�x�0

�u

�x

¶=

�lim

�u�0

�y

�u

¶�lim

�x�0

�u

�x

¶=dy

du

du

dx.

Thus, we can consider the chain rule to be the limiting case of an obvious fact about averagerates of change. This plausibility argument does not lead to a rigorous proof, as in the case of theplausibility argument that relied on di�erentials, since we may have �u = u (x+�x)�u (x) = 0even if �x 6= 0. �


Example 6 Let f (x) = sin2/3 (x) .Determine f 0 (x).

Solution

We set f (x) = y (x) = (sin (x))2/3 and u = sin (x), so that y (u) = u2/3. By the chain rule,

f 0 (x) =dy

dx=dy

du

du

dx=

�d

duu2/3

¶�d

dxsin (x)

¶

=

�2

3u�1/3

¶cos (x) =

2

3(sin (x))�1/3 cos (x) =

2 cos (x)

3 sin1/3 (x).

Therefore,

f 0 (x) =2 cos (x)

3 sin1/3 (x)

if sin (x) 6= 0.

�Π� 3 Π2

� Π2

0 Π 3 Π2

Π2

x

1y

f

�Π� 3 Π2

� Π2

Π 3 Π2

Π2

x

1

y

f'

Figure 2

Figure 2 shows the graphs of f and f 0 on the interval [�3�/2, 3�/2]. Note that the graph of fhas cusps at ��, 0 and � (De�nition 2 of Section 2.2) and the graph of f 0 has vertical asymptotesat these points. For example,

limx�0�

2

3cos (x) =

2

3> 0,

andlimx�0�

1

sin1/3 (x)= ��

since sin1/3 (x) < 0 if ��/2 < x < 0 and

limx�0

sin1/3 (x) = 0.

Therefore,

limx�0�

f 0 (x) = limx�0�

�2

3cos (x)

¶Ã1

sin1/3 (x)

!= ��.

Similarly,limx�0+

f 0 (x) = +�.¤


Remark 3 As in Example 6, if a function is of the form ur (x), where r is a rational exponent,we can apply the chain rule to evaluate its derivative. Indeed, if we set y = ur (x) = (u (x))

r

and u = u (x), then y = ur. By the chain rule and the power rule,

d

dxur (x) =

dy

dx=dy

du

du

dx

=

�d

duur¶du

dx=¡rur�1

¢ dudx= rur (x)

du

dx.

Thus,d

dxu�(x) = ru��1

du

dx.

Since the above expression reduces to the power rule if u (x) = x, it may be referred to asthe function-power rule. The implementation of the function-power rule is slightly fasterthan the direct implementation of the chain rule, and the rule is easy to remember due to thesimilarity with the ordinary power rule (don’t neglect du/dx, though). �


dxcos10 (x) .

Solution

By the function-power rule:

d

dxcos10 (x) = 10 cos9 (x)

�d

dxcos (x)

¶= 10 cos9 (x) (� sin (x)) = �10 cos9 (x) sin (x) .

The direct implementation of the chain rule is not much slower: Set y (x) = (cos (x))10 andu = cos (x) so that y (u) = u10. By the chain rule and the power rule,

d

dxcos10 (x) =

dy

dx=dy

du

du

dx

=

�d

dxu10¶�

d

dxcos (x)

¶=¡10u9

¢(� sin (x)) = �10 cos9 (x) sin (x) .

¤

The Chain Rule for more than two Functions

The chain rule can be extended to cover cases that involve the composition of more than twofunctions: For example, if F = f � g � h, then

F (x) = (f � g) (h(x)) ,

so thatF 0(x) = (f � g)0 (h(x))h0(x) = f 0 (g(h (x))) g0(h(x))h0 (x) .

The following schematic description of the composition should make it easier to remember whereto evaluate the derivatives:

x� h(x)� g(h(x))� f(g(h(x)))

The expression of the chain rule in “the prime notation” is somewhat unwieldy when the com-position of more than two functions is involved. We may refer to the functions with the symbols


that denote their dependent variables, and use the Leibniz notation: If we set y = y(u(v(x)),then

dy

dx=dy

du

du

dx=dy

du

�du

dv

dv

dx

¶,

so thatdy

dx=dy

du

du

dv

dv

dx.

The symbolic cancellations are helpful in checking that we are on the right track. Note thatdy/du is evaluated at u(v(x)) and du/dv is evaluated at v(x).

Example 8 Determine

d

dx

scos

�1

x

¶.

Solution

We set

y =

scos

�1

x

¶, u = cos

�1

x

¶and v =

1

x,

so thaty =

�u and u = cos (v) .

By the chain rule,

dy

dx=dy

du

du

dv

dv

dx=

�d

du

�u

¶�d

dvcos (v)

¶�d

dx

¡x�1

¢¶

=

�1

2�u

¶(� sin (v)) ¡�x�2¢ = sin

�1

x

¶

2x2

scos

�1

x

¶

The expression is valid if x 6= 0 and cos (1/x) > 0. ¤

The Proof of the Chain Rule

We set u = g (x) and �u = g(x+�x)� g(x), so that g(x+�x) = u+�u. Then,

(f � g) (x+�x)� (f � g) (x) = f (g(x+�x))� f(g(x)) = f(u+�u)� f(u)).As in Theorem 2 of Section 2.5,

f(u+�u)� f(u) = f 0 (u)�u+�uq (�u) ,where

lim�u�0

q (�u) = 0.

Therefore,

f (g (x+�x))� f (g (x))�x

=f(u+�u)� f(u)

�x

=f 0 (u)�u+�uq (�u)

�x

=f 0 (g(x))�u+�uq (�u)

�x

= f 0 (g(x))�u

�x+�u

�xq (�u) ,


where �x 6= 0. We have

lim�x�0

�u

�x= lim

�x�0

g (x+�x)� g (x)�x

= g0 (x) .

Since g is di�erentiable at x, it is continuous at x. Thus,

lim�x�0

�u = lim�x�0

(g (x+�x)� g (x)) = 0.

Therefore,lim

�x�0q (�u) = 0.

Thus,

(f � g)0 (x) = lim�x�0

f (g (x+�x))� f (g (x))�x

= lim�x�0

�f 0 (g(x))

�u

�x+�u

�xq (�u)

¶

= f 0 (g (x)) lim�x�0

�u

�x+

�lim

�x�0

�u

�x

¶³lim

�x�0q (�u)

´= f 0 (g(x)) g0(x) + g0 (x) (0)= f 0 (g (x)) g0 (x) .

¥

Problems

In problems 1-23, compute f 0(x) ( It will be practical to use the Leibniz notation):

1.f (x) =

px2 � 2x+ 5

2.f(x) = x+

px2 + 4

3.f(x) =

¡x2 � 16¢2/3

4.f(x) =

1�x4 + 9

.

5.

f(x) =

r4 + x2

4� x2 .6.

f (x) =¡x2 � 4x+ 8¢2/3

7.f (x) = sin (10x)

8.f (x) = cos

³x4

´9.

f (x) = sin (�x)

10.

f (x) = cos(x) +1

9cos(3x) +

1

25cos(5x)

11.

f (x) = sin(�x)� 12sin(2�x) +

1

3sin(3�x)

12.f (x) = 10 cos

³x4� 1´

13.

f (x) =1

4sin

�1

2x+

�

6

¶14.

f(x) = tan(2x)

15.f(x) = cos(x2).

16.f(x) = cos(1/x)

17.f(x) = sin(

�x)


18.f(x) = sin2(3x)

19.f(x) =

psin(x/2)

20.f(x) = cos

³px2 + 1

´

21.f(x) = sin2(

1

x)

22.f(x) =

p4� cos3(2x)

23.f(x) =

ptan(x2)

In problems 24-26, compute f 0(x) and f 00(x):

24.f(x) = sin(4x)

25.f(x) = cos (1/x) .

26.f(x) = sin2(6x)

27. Letf(x) =

1�x2 + 16

.

a) Determine L3, the linear approximation to f based at 3,b) Make use of L3 to approximate f (2.8).

28. Letf (x) = sin2

¡x2¢.

a) Determine the di�erential of f .

b) Make use of the di�erential of f in order to approximate f³p

�/4 + 0.1´

In problems 29 and 30,a) Compute f 0 (x), determine the fundamental period p of f , and specify the part of the domainsof f and f 0 in the interval [�p/2, p/2],b) Determine whether the graph of f has vertical tangents or cusps on the interval [�p/2, p/2],c) [C] Make use of your graphing utility to plot the graphs of f and f 0 on [�p/2, p/2] Are thepictures consistent with your response to part b)?

29.f (x) = cos2/3 (x)

30.f (x) =

ptan (x)

The motion of an oscillating object such as a mass that is attached to a spring can be expresedby a position function of the form

y(t) = A cos (�t� �) ,where t represents time, A > 0 and � are constants (friction forces are neglected). We say thatthe object is in simple harmonic motion. The motion has period

T =2�

�.

The frequency of the motion is the reciprocal of its period, i.e.,

1

T=�

2�.

Since |A cos (�t� �)| = A |cos (�t� �)| A, the maximum distance of the object from theequilibrium position is A. We refer to A as the amplitude of the simple harmonic motion.

Chapter 3

Maxima and Minima

The sign of the derivative of a function provides us with valuable information about the in-creasing/decreasing behavior of the function and its maximum and minimum values.The sign of the second derivative of a function provides information about the increasing ordecreasing behavior of the derivative of the function. We will make use of such informationto determine the solutions of practical optimization problems.

3.1 Increasing/decreasing Behavior and ExtremaThe sign of the derivative of a function provides information about the intervals on which thefunction is increasing or decreasing and the points at which it attains its maximum and minimumvalues.

Some Terminology

Let’s begin by recalling some terminology. Let J denote an interval that can be a boundedopen interval such as (1, 3), a half-open interval such as [1, 3), or an unbounded interval such as(��, 2]. Recall that x is in the interior of an interval J if x � J , but x is not an endpoint ofJ . Thus, the interior of [1, 3) is the open interval (1, 3), and the interior of (��, 2] is the openinterval (��, 2). Also recall that a function f is said to be increasing on an interval J if, forany pair of points x1 and x2 in J such that x1 < x2, we have f (x1) < f(x2). A function f issaid to be decreasing on J if, for x1 and x2 in J such that x1 < x2, we have f (x1) < f(x2).A function f is said to be decreasing on J if, for x1 and x2 in J such that x1 < x2, we havef (x1) > f (x2). We will say that f is monotone on J if f is increasing, decreasing or constanton J .

A function f is said to have a local maximum or a local minimum at a point a if f (a)is its maximum or minimum value, respectively, relative to some interval that contains a inits interior. The absolute maximum of f on a set D is the maximum value of f on D,the absolute minimum of f on D is the minimum value of f on D. Here are the preciseexpressions:

De�nition 1 A function f has a local maximum at a if there exists an open interval J thatcontains a such that f (a) � f (x) for each x � J . The function f has a local minimum at a ifthere exists an open interval J that contains a such that f (a) f (x) for each x � J . In eithercase, we say that f has a local extremum at a. The absolute maximum of f on a set D isM if there exists cM � D such that f(cM ) =M and M � f (x) for each x � D. The absoluteminimum of f on a set D is m if there exists cm � D such that f(cm) = m and m f (x) foreach x � D. We may refer to absolute maxima and minima as absolute extrema.

193

194 CHAPTER 3. MAXIMA AND MINIMA

x

y

b

�b, f�b��

c�c, f�c��

a d

�d, f�d��

e

y � f�x�

Figure 1

It appears that the function of Figure 1 has local maxima at b and d, and a local minimum atc. The picture indicates that f (b) is the absolute maximum of f on [a, e] and f (a) (= 0) is theabsolute minimum of f on [a, e]. Note that f is constant on the interval [d, e].In this section we will focus on �nding the local extrema of a function. We will take up the issueof absolute extrema in the next section

Remark 1 According to our de�nition of a local extremum, if a function f has a local extremumat a point a, f must attain a maximum or minimum at a relative to the points in some openinterval that contains a. Thus, f (x) =

�x does not have a local minimum at 0, even though

f (x) � f (0) = 0 for each x � [0,+�). We have to single out the endpoints of an interval forspecial consideration when we investigate the absolute maximum and the absolute minimum ofa function on the entire interval. �

2 4 6 8x

1

2

y

Figure 2: The square-root function does not have a local minimum at 0

The Derivative Test for Monotonicity and Extrema

Recall that the tangent line to the graph of f at (a, f (a)) is the graph of the linear function

La (x) = f (a) + f0 (a) (x� a) .

x

y

a

La�a, f�a��

Figure 3

3.1. INCREASING/DECREASING BEHAVIOR AND EXTREMA 195

We have seen that La (x) approximates f (x) very well if x is close to a. If f 0 (a) > 0, thelinear function La is an increasing function. Therefore it is reasonable to expect that f isalso increasing in a small interval containing the point a. Similarly, if f 0 (a) < 0 then La is adecreasing function, and we would expect that f is decreasing in a su�ciently small intervalcontaining a. Actually, the sign of the derivative of a function gives us information about theincreasing/decreasing behavior of the function on any interval, irrespective of its size.

THE DERIVATIVE TEST FOR MONOTONICITY Assume that f is continuouson the interval J and that f is di�erentiable at each x in the interior of J . Iff 0(x) > 0 for each x in the interior of J , then f is increasing on J . If f 0(x) < 0 foreach x in the interior of J , then f is decreasing on J .

We will discuss the theoretical basis of the derivative test for monotonicity in the next section.An immediate corollary of the derivative test for monotonicity is that a function has a localmaximum or minimum at a point where its derivative changes sign:

THE DERIVATIVE TEST FOR LOCAL EXTREMA Assume that f is continu-ous at a, and a is contained in open interval (c, d) such that f 0(x) > 0 if c < x < a,and f 0(x) < 0 if a < x < d. Then f has a local maximum at a. Similarly, if f 0(x) < 0if c < x < a, and f 0(x) > 0 if a < x < d, then f has a local minimum at a.

Proof

a)

xac d

f' 0 f' 0f'�a� � 0

Figure 4: f has a local maximum at a

Since f 0 (x) > 0 if x � (c, a), f is increasing on (c, a]. Since f 0 (x) < 0 if x � (a, d), f isdecreasing on [a, d). Therefore, f (a) � f (x) for each x � (c, d), so that f has a local maximumat a.

b)

xac d

f' 0 f' 0f'�a� � 0

Figure 5: f has a local minimum at a

Since f 0 (x) < 0 if x � (c, a), f is decreasing on (c, a]. Since f 0 (x) > 0 if x � (a, d), f isincreasing on [a, d). Therefore, f (a) f (x) for each x � (c, d), so that f has a local minimumat a. ¥


Example 1 Let

f (x) =1

3x3 � 9x.

a) Determine the points at which f has a local maximum or minimum, and the correspondingvalues of f .b) Determine the absolute maximum and minimum values of f on the intervals (��, 0] and[0,+�), provided that such values exist. Justify your responses if you claim that such valuesdo not exist.

Solution

Figure 6 shows the graph of f . The picture indicates that f has a local maximum near �3 anda local,minimum near 3.

�6 �3 3 6x

�18

18

y

f�x� �x3

3� 9x

Figure 6

We will determine the exact values with the help of the �rst derivative test. We have

f 0 (x) =d

dx

�1

3x3 � 9x

¶=

1

3

¡3x2

¢� 9= x2 � 9 = (x+ 3) (x� 3) .

Figure 7 shows the graph of f 0.

�6 �3 3 6x

20

40y

f'

Figure 7

We have f 0 (x) = 0 if x = ±3, f 0 (x) > 0 if x < �3 or x > 3, and f 0 (x) < 0 if �3 < x < 3.By the derivative test for monotonicity, f is increasing on the intervals (��,�3] and [3,+�),and decreasing on [�3, 3]. Since f is increasing on the interval (��,�3] and decreasing on[�3, 3], f (3) � f (x) for each x � (��, 3). Therefore, f has a local maximum at �3. We havef (�3) = 18. Similarly, since f is decreasing on [�3, 3] and increasing on [3,+�), f has a localminimum at 3. We have f (�3) = �18. The pictures are consistent with our conclusions.

b) Since f is increasing on (��,�3] and decreasing on [�3, 0], the function attains its absolutemaximum on the interval (��, 0] at �3. We have f (�3) = 18. On the other hand, f doesnot have an absolute minimum on (��, 0] since f attains negative values of arbitrarily largemagnitute. Indeed,

limx�� f (x) = lim

x��

�1

3x3 � 9x

¶= limx��x

3

�1

3� 9

x2

¶= ��


since

limx��x

3 = �� and limx��

�1

3� 9

x2

¶=1

3> 0.

¤We saw that the derivative of the function of Example 1 vanishes at the points at which it has alocal maximum or minimum. Thus, the tangent lines at the corresponding points on its graphare horizontal. This is not an accident. Assume that a function f has a local maximum or alocal minimum at a point a, and that f 0 is continuous at a. If f 0 (a) > 0, then f 0 (x) �= f 0 (a) > 0if x is close to a, by the continuity of f 0 at a. Therefore, f 0 (x) > 0 for each x in some openinterval that contains a. The derivative test for monotonicity implies that f is increasing onthat interval. This contradicts the assumption that f has a local extremum at a. Similarly, iff 0 (a) < 0, then f is decreasing on an open interval containing a so that f cannot have a localextremum at a. Therefore, we must have f 0 (a) = 0. This is actually valid even if f 0 is notcontinuous at a:

FERMAT’S THEOREM If f has a local maximum or minimum at a and f isdi�erentiable at a we have f 0(a) = 0.

You can �nd the proof of Fermat’s Theorem at the end of this section.

De�nition 2 We say that the point a is a stationary point of f if f 0 (a) = 0.

Thus, a is a stationary point of f if the tangent line to the graph of f at (a, f (a)) is horizontal.Fermat’s Theorem says that a must be a stationary point of f if f is di�erentiable at a andhas a local extremum at a. The term “stationary point” has the following meaning within thecontext of one-dimensional motion: Assume that f (t) is the position at time t of an object inone-dimensional motion. The instantaneous velocity v (t) at the instant t is the rate of changeof f at t, i.e., v (t) = f 0 (t). Thus, we have v (t0) = 0 if t0 is a stationary point of f . Since theinstantaneous velocity of the object at t0 is 0, we may imagine that the object is “instantaneouslystationary” at the instant t0.

A word of caution: You should not misread Fermat’s Theorem. The condition, f 0 (a) = 0, isnecessary for f to have a local maximum or minimum at a. The theorem does not say thatthe condition f 0 (a) = 0 is su�cient for f to have a local extremum at a.

Example 2 Let f (x) = x3. Then f 0 (x) = 3x2 = 0 if x = 0, but f does not have a localmaximum or minimum at 0. The function is increasing on (��,+�). ¤

�4 �2 2 4x

�40

�20

20

40

y

Figure 8: y = x3

A function may have a local extremum at a point even though it is not di�erentiable at thatpoint.

Example 3 Let f (x) = x2/3.


The function is not di�erentiable at 0, as we discussed in Section 2.3. The graph of f has a cuspat (0, 0). The function has a local minimum at 0 (f (0) is actually the absolute minimum of fon R). ¤

�8 �4 4 8x

1

2

3

4

y

Figure 9: y = x2/3

De�nition 3 A point a is a critical point of the function f if a is a stationary point of f , i.e.,f 0 (a) = 0, or f is de�ned at a but not di�erentiable at a.

Thus, 0 is the only critical point of the function of Example 3.

By Fermat’s Theorem, if a function f is di�erentiable at a point a and f is di�erentiable at a,we must have f 0 (a) = 0. We just saw that a function can have a local extremum at a pointwhere it is not di�erentiable. Therefore, we have the following necessary condition for alocal extremum:

Assume that f has a local maximum or minimum at a. Then a is a critical point off .

Remark 2 We have seen that a function does not have to have a local extremum at a pointwhere its derivative is 0. Let’s also note that a function need not have a local extremum ata critical point where it is not di�erentiable. For example, if f (x) = x1/3, then f is notdi�erentiable at 0, as we saw in Section 2.3, so that 0 is a critical point of f . Figure 10 showsthe graph of f . The function does not have a local maximum or minimum at 0. Note that thereis a vertical tangent to the graph of f at (0, 0). �

�8 �4 4 8x

�2

�1

1

2

y

Figure 10

Our observations lead to the following strategy for the determination of the local extrema of afunction:

1. Determine the domains of f and f 0.

2. Determine the critical points of f .

3. Make use of the derivative test for monotonicity to determine whether f isincreasing or decreasing on the intervals that are separated from each otherby the critical points of f or points at which f is not de�ned.


4. Determine the local extrema of f based on the results of item 3.

Example 4 Let

f(x) =2

3x3 +

1

4x4.

a) Determine the local maxima and minima of f .b) Determine the absolute maximum and minimum values of f on the intervals (��, 0] and[0,+�), provided that such values exist. Justify your responses if you claim that such valuesdo not exist.

Solution

Figure 11 indicates that f has a local minimum near �2. The picture also inductees that thegraph of f has a horizontal tangent line at the origin, even though it does not have a localextremum there. We can con�rm all this with the help of the derivative test.

�3 �2 �1 1 2x

�1

2

4

6

8

y

f

Figure 11: f(x) = 23x

3 + 14x

4

The polynomial f is di�erentiable on the entire number line. Therefore, the only critical pointsof f are its stationary points. We have

f 0 (x) =d

dx

�2

3x3 +

1

4x4¶

=2

3

¡3x2

¢+1

4

¡4x3

¢= 2x2 + x3 = x2 (2 + x) .

Therefore, f 0 (x) = 0 if x = 0 or x = �2. Thus, the stationary points of f are �2 and 0. Figure12 displays the graph of f 0.

�3 �2 �1 1 2x

�10

�5

5

10

15

y

f'

Figure 12: f 0 (x) = x2 (2 + x)

Since x2 > 0 if x 6= 0, we see that the sign of f 0(x) = x2 (x+ 2) is determined by the sign ofthe factor x+ 2 if x 6= 0. Thus, f 0(x) < 0 if x < �2, f 0(x) > 0 if �2 < x < 0, and f 0 (x) > 0 ifx > 0. Therefore, f is decreasing on the interval (��,�2] and increasing on the entire interval[�2,+�). The conclusion is that f has a local minimum at �2. The function f does not havea local extremum at 0, even though 0 is a stationary point of f .


Table 1 summarizes the relationship between the sign of f 0 and the increasing/decreasing be-havior of f , and indicates the local minimum of f ..

x �2 0f 0 (x) � 0 + 0 +f decreasing min increasing increasing

Table 1

b) Since f is decreasing on (��,�2] and increasing on [�2, 0], the function attains its absoluteminimum on (��, 0] at �2. We have f (�2) = �4/3.Since f is increasing on [0,+�), the function attains its absolute minimum on [0,+�) at 0. Wehave f (0) = 0. The function does not attain an absolute maximum on [0,+�) since it attainsarbitrarily large values. Indeed,

limx�+� f (x) = lim

x�+�

�2

3x3 +

1

4x4¶= limx�+�x

4

�2

3x+1

4

¶= +�

since

limx�+�x

4 = +� and limx�+�

�2

3x+1

4

¶=1

4> 0.

¤A useful observation: In order to determine the sign of the derivative of a function f on aninterval that does not contain a critical point or a point of discontinuity of f , it is su�cient tosample a single point in the interval and determine the sign of f 0 at that point, provided thatf 0 is continuous on the interval. This is usually easier than working with inequalities. Indeed,assume that f 0 is continuous on the interval (a, b) and f 0 (x) 6= 0 for each x � (a, b). If x1and x2 are in (a, b) and the sign of f 0 (x1) and the sign of f 0 (x2) are di�erent, there must bea point c between x1 and x2 such that f 0 (c) = 0. This is a consequence of the IntermediateValue Theorem that was discussed in Section 2.10. Therefore, the sign of f 0 (x) is the same forall x � (a, b).

Example 5 Let f(x) = x(x� 3)2/3.a) Determine the critical points of f . Does the graph of f have vertical tangents or cusps atany of the critical points?b) Determine the points at which f has a local maximum or minimum.

Solution

Figure 13 displays the graph of f . The picture suggests that the graph of f has a cusp at 3where it has a local minimum The picture also indicates that f has a local maximum near 2.

�2 3 5x

�6

�4

�2

2

4

6

8

y

9

5

f

Figure 13: f(x) = x (x� 3)2/3


a) With the help of the product rule and the chain rule,

f 0 (x) =d

dx

³x(x� 3)2/3

´=

�d

dx(x)

¶(x� 3)2/3 + x

�d

dx(x� 3)2/3

¶

= (1) (x� 3)2/3 + x�2

3(x� 3)�1/3

¶

= (x� 3)2/3 + 2x

3 (x� 3)1/3

=3 (x� 3) + 2x3 (x� 3)1/3

=5x� 9

3(x� 3)1/3if x 6= 3. Therefore,

f 0 (x) = 0 5x� 9 = 0 x =9

5.

Therefore 9/5 is the stationary point of f . The function is not di�erentiable at 3, as suggestedby the expression for f 0(x) (you should be able to con�rm this), although f is de�ned at 3.Therefore, the critical points of f are 9/5 and 3. We have

limx�3�

f 0 (x) = limx�3�

�5x� 9

3(x� 3)1/3¶= ��,

and

limx�3+

f 0 (x) = limx�3+

�5x� 9

3(x� 3)1/3¶= +�

(con�rm with the help of Proposition 1 and Proposition 2 of Section 1.6). Therefore, the graphof f has a cusp at (3, f (3)) = (3, 0). Figure 14 shows the graph of f 0. The picture is consistentwith our conclusions.

�2 3 5x

�4

�2

2

4

6

8

y

9

5

f'

Figure 14

b) The function is de�ned everywhere. The intervals that are separated by the critical points off are

(��, 95], [9

5, 3] and [3,+�)

Since f 0 is continuous in the interior of each interval, and the only zero of f 0 is 9/5, f 0 has a con-stant sign in the interior of each interval. It is practical to sample a point in the interior of eachinterval. Table 2 summarizes the relationship between the sign of f 0, the increasing/decreasingbehavior of f , and the local extrema of f . By the derivative test for monotonicity, f is increasingon the interval (��, 9/5], decreasing on [9/5, 3] and increasing on [9/5,+�). Thus, f has alocal maximum at 9/5 and a local minimum at 3. ¤

x 9/5 3f 0 (x) + 0 � unde�ned +f (x) increasing max decreasing min increasing


Table 2

Example 6 Let

f(x) = x+1

x� 1 .

a) Determine the stationary points of f .b) Determine whether f has a local maximum or minimum at each stationary point.c) Determine the absolute maximum and minimum values of f on (1,+�), provided that suchvalues exist. Justify your responses if you claim that such values do not exist.

Solution

Figure 15 displays the graph of f . The picture suggests that the graph of f has a verticalasymptote at x = 1 and that f has a local maximum near 0 and a local minimum near 2.

�4 �2 2 4

�10

�5

5

10

f

1

Figure 15: f(x) = x+1

x� 1

a) We have

f 0 (x) =d

dx

�x+

1

x� 1¶=d

dx(x) +

d

dx

�1

x� 1¶

= 1 +� d

dx(x� 1)

(x� 1)2 = 1� 1

(x� 1)2

=(x� 1)2 � 1(x� 1)2 =

x2 � 2x(x� 1)2 =

x (x� 2)(x� 1)2 .

Therefore,

f 0 (x) = 0 x = 0 or x = 2.

Thus, the stationary points of f are 0 and 2.

b) We will apply the derivative test for monotonicity. We have to take into account the fact thatf and f 0 are not de�ned at x = 1 (you can show that the line x = 1 is a vertical asymptote forthe graphs of f and f 0). The intervals that are separated from each other by the discontinuitiesor the stationary points of f are

(��, 0], [0, 1), (1, 2] and [2,+�).

Notice that we have excluded the point 1 from the relevant intervals, since f is not de�ned at 1.You can determine the sign of f 0 in the interior of each interval by making use of your knowledgeabout inequalities or by sampling appropriate values of f 0. Figure 16 shows the graph of f 0.


�4 �2 2 4

�4

�2

1

2

f'

1

Table 3 summarizes the relationship between the sign of f 0, the increasing/decreasing behaviorof f , and the local extrema of f . The function is increasing on (��, 0], decreasing on [0, 1),decreasing on (1, 2], and increasing on [2,+�). Thus, f has a local maximum at 0 and a localminimum at 2. ¤

x 0 1 2f 0 (x) + 0 � unde�ned � 0 +f increasing max decreasing unde�ned decreasing min increasing

Table 3

c) Since f is decreasing on (1, 2] and increasing on [2,+�), it attains its absolute value on(1,+�) at 2. We have f (2) = 3.The function does not have an absolute maximum on (1,+�) since it attains arbitrarily largevalues. We have

limx�1+

f (x) = limx�+� f (x) = +�

(con�rm). ¤

The Proof of Fermat’s Theorem

Assume that f attains a local maximum at a. Then f(a + h) f(a), if |h| is small enough.Therefore, f(a+ h)� f(a) 0 if h > 0 and h is su�ciently small. Thus.

f(a+ h)� f(a)h

0

under these conditions. Therefore,

f 0(a) = limh�0

f(a+ h)� f(a)h

= limh�0+

f(a+ h)� f(a)h

0.

Similarly, if we consider h < 0 such that |h| is su�ciently small, f(a + h) f(a) so thatf(a+ h)� f(a) 0. Since h < 0, we have

f(a+ h)� f(a)h

� 0,

so that

f 0(a) = limh�0

f(a+ h)� f(a)h

= limh�0�

f(a+ h)� f(a)h

� 0.

Since we deduced that f 0(a) 0 and f 0(a) � 0, we must have f 0(a) = 0, as claimed.The proof of the fact that f 0(a) = 0 at a point a at which f attains a local minimum is similar.¥


Remark 3 In the above proof we have used the fact that limh�0+ g (h) 0 if g (h) 0 whenh is su�ciently small and positive (and a similar fact for limh�0� g (h)). Indeed, if we assumethat limh�0+ g (h) = L > 0, then g (h) must also be positive if h > 0 and h is su�ciently small,since g (h) is as close to L as desired if h > 0 and h is small enough. This contradicts the factthat g (h) 0 when h is su�ciently small and positive. �

Problems

In problems 1 - 6, a function f is given and its graph is displayed. Find the critical points andthe points at which f has a local maximum or minimum.

1.

x

y

f (x) = x2 � 4x+ 3

2.

x

y

f (x) = 13x

3 + 12x

2 � 6x

3.

x

y

f (x) = �13x

3 + 3x

x

y

4. f (x) = (x� 2)2 (x+ 3)2/3

5.

x

y

f (x) = (x+ 3)2 (x� 2)1/3

6.

2 Πx

y

f (x) = sin (x)� cos (x) (only on the interval[0, 2�])

In problems 7 - 18, given the function f ,a) Find the critical points of f .b) Make use of the derivative test for monotonicity to determine intervals on which f is increasingor decreasing, and the points at which f has a local maximum or minimum.

7.

f(x) = x2 + 4x

8.

f (x) = �x2 + 3x� 2

3.5. APPLICATIONS OF MAXIMA AND MINIMA 233

7.f (x) = 4x+

1

x� 2In problems 8-10,a) Determine the domain of f ,b) Determine limx�� f (x) and limx�+� f (x), if applicable,c) Determine the domain of f 0 and the points at which the graph of f has a vertical tangent ora cusp,d) Determine the intervals on which f is increasing/decreasing, and the points at which f hasa local maximum or minimum,e) Sketch the graph of f . If you expect the graph to be symmetric with respect to the verticalaxis or with respect to the origin, your sketch should re�ect the relevant symmetry.

8. f (x) = x2(x� 3)1/5

9. f (x) = x(x+ 3)3/4

10. f (x) = x2 (x� 2)4/5

3.5 Applications of Maxima and Minima

In the previous sections of this chapter we developed powerful tools in order to �nd the maximaand minima of functions. In this section we will discuss some applications.

Optimization

Let us begin with some geometric applications.

Example 1 Determine the dimensions of the rectangle that has the greatest area among allrectangles which are inscribed in a circle of radius r.

Solution

We will consider the circle of radius r whose center is the origin of the xy-plane. Thus, the circleis the graph of the equation x2 + y2 = r2. Due to the symmetries, it is su�cient to considerinscribed rectangles whose sides are parallel to the coordinate axes, and we can assume that thevertices of the rectangles are on the unit circle. After all, we wish to maximize the area, and ifthe rectangle is strictly within the unit disk, we can enlarge the rectangle to one which has itsvertices on the circle.

x

y

x

r

P � �x, y�

y

Figure 1


With reference to Figure 1, the inscribed rectangle is completely determined by the point P =(x, y). We have y =

�r2 � x2. Therefore, the area of the rectangle is

(2x) (2y) = 4xy = 4xpr2 � x2.

We will maximize the square of the area in order to maximize the area, since the expressionswill be easier to work with, Thus, let’s set

f (x) =³4xpr2 � x2

´2= 16x2

¡r2 � x2¢ = 16r2x2 � 16x4.

Since 0 < x < r, we would like to determine the absolute maximum of f on the interval (0, r).We have f (0) = f (r) = 0. The cases x = 0 and x = r lead to the degenerate cases where the“rectangles” are intervals with 0 area. In any case, the search procedure for the determinationof the absolute extrema of a continuous function on a closed and bounded interval is applicable,as we discussed in Section 3.2. Since f is di�erentiable at any x � R, the only critical points off are its stationary points, i.e., points x such that f 0 (x) = 0. We have

f 0 (x) =d

dx

¡16r2x2 � 16x4¢ = 32r2x� 64x3 = 32x ¡r2 � 2x2¢ .

Therefore,f 0 (x) = 0 x = 0 or x = ± r�

2.

Thus, the only critical point of f in the interior of the interval [0, r] is r/�2.

Figure 2 shows the graph of f that corresponds to r = 2.

�2 2r

20

40

60

y

y � f�r�

Figure 2

We have

f

�r�2

¶= 16r2x2 � 16x4 ¯̄

x=r/�2= 16

�r�2

¶2Ãr2 �

�r�2

¶2!= 4r4.

Therefore, f¡r/�2¢> 0 = f (0) = f (r). Thus, the absolute maximum of f on [0, r] is 4r4, and

f attains this value at x = r/�2. Therefore, the maximum area of a rectangle that is inscribed

in a circle of radius r is sf

�r�2

¶=�4r2 = 2r.

The dimensions of a rectangle with maximum area are

2x = 2

�r�2

¶=�2r,


and

2y = 2

sr2 �

�r�2

¶2

= 2

rr2 � 1

2r2 = 2

�r�2

¶=�2r.

Therefore, an inscribed rectangle with maximum area is a square whose sides have length�2r.

Note that the ratio of the area of such a square and the area of the disc of radius r is

2r2

�r2=2

��= 0.64.

¤

Example 2 Find the points on the parabola y = x2 which are closest to the point (0, 4).

Solution

Let P = (x, y) be an arbitrary point on the parabola so that y = x2, as shown in Figure 3.

�2 2x

10

20

30

y

P � �x, x2�

�0,4�

Figure 3

The distance of P from (0, 4) ispx2 + (x2 � 4)2 =

px4 � 7x2 + 16.

We will minimize the distance if we minimize the square of the distance. Thus, let’s set f (x) =x4 � 7x2 + 16, so that f represents the square of the distance of an arbitrary point on theparabola from the point (0, 4). Note that f (�x) = f(x), so that f is even. We would like tominimize f (x) as x varies on the entire number line. Unlike Example 1, we are not able tocon�ne x to a closed and bounded interval that can be determined immediately, and implementthe search procedure for the absolute extrema of a continuous function on a closed and boundedinterval. We will make use of the derivative test for monotonicity.

We have

f 0 (x) =d

dx

¡x4 � 7x2 + 16¢ = 4x3 � 14x = 2x ¡2x2 � 7¢ .

Therefore,

f 0 (x) = 0 2x¡2x2 � 7¢ = 0 x = 0 or x = ±

r7

2.

Table 1 summarizes the relationship between the sign of f 0 and the increasing/decreasing be-havior of f .

x �q

72 0

q72

sign of f 0 � 0 + � 0 +f decreasing local min. increasing local max. decreasing local min. increasing


Table 1

Note that

limx�±� f (x) = lim

x�±�¡x4 � 7x2 + 16¢ = lim

x�±�x4

�1� 7

x2+16

x4

¶= +�,

since

limx�±�x

4 = +�and limx�±�

�1� 7

x2+16

x4

¶= 1 > 0.

The data displayed in Table 1 shows that f attains its absolute minimum on the entire numberline at ±p7/2 (we noted that f is even). Therefore, the points on the parabola at a minimumdistance from the point (0, 4) are

(�r7

2,7

2) and (

r7

2,7

2).

These points are at a distancesf(±

r7

2) =

r15

4=

�15

2�= 1.936 49

from (0, 4).

Figure 4 displays the graph of f , and Figure 5 indicates the points on the parabola y = x2 at aminimum distance from (0, 4). ¤

�1 1x

10

4

20

30

y

y � f�x�

7 �2� 7 �2

Figure 4

�2 2x

10

y

� 7 �2 , 7�2��0,4�

�� 7 �2 , 7�2�

Figure 5

Let us look at an example that involves the minimization of the cost of producing a certainitem:


Example 3 Assume that a manufacturer must produce cans in the shape of right circularcylinders. The volume of each can is required to be 250 cubic centimeters. Otherwise, themanufacturer is free to choose the dimensions of the can. The cost per square centimeter ofthe material for the top and the bottom of the can is twice as much as the cost per squarecentimeter of the material for the lateral surface. How should the manufacturer determine thedimensions of the can in order to minimize the cost?

Figure 6

Solution

Let’s assume that the material for the lateral surface of the can costs k cents per cm2, and thatthe material for the top and the bottom of the can costs 2k cents per cm2. If the cross sectionof the can is a circle of radius r (centimeters), and the height of the can is h (centimeters), thetotal area of the top and the bottom is 2�r2, so that the cost of the material for the top and thebottom is 2�r2 × 2k cents. The area of the lateral surface of the can is 2�rh (You can imaginethat a vertical cut is made, and the lateral surface is laid out on a �at surface: The shape is arectangle of base length 2�r, the perimeter of the cross section, and height h). Therefore, thecontribution of the lateral surface to the cost is 2�rh× k cents. Thus, the total cost is

2�r2 × 2k + 2�rh× k = 4�r2k + 2�rhk = ¡4�r2 + 2�rh¢ k.cents. The cost is minimized if 4�r2 + 2�rh is minimized. This involves two variables r and h.We will eliminate one of the variables by making use of the requirement that the volume of thecan must be 250 cm3. Therefore,

Volume = �r2h = 250,

so thath =

250

�r2.

Now we can express 4�r2 + 2�rh as a function of the radius r:

4�r2 + 2�rh = 4�r2 + 2�r

�250

�r2

¶= 4�r2 +

500

r.

Let us setf(r) = 4�r2 +

500

r.

We must determine r0 > 0 such that f(r0) f(r) for each r > 0. Thus, we must determinethe point r0 at which f attains its absolute minimum on the interval (0,+�). There are noadditional restrictions on r (A small r corresponds to a tall and skinny can, and a large r


corresponds to a short and fat can: Certain practical considerations impose restrictions on rand h, but these have not been stipulated). Therefore, we will implement the derivative test formonotonicity on the interval (0,+�). We have

f 0 (r) =d

dr

�4�r2 +

500

r

¶= 8�r � 500

r2=8�r3 � 500

r2.

Therefore,

f 0 (r) = 0 8�r3 � 500 = 0 r =

�500

8�

¶1/3

.

Table 2 summarizes the relationship between the sign of f 0 and the increasing/decreasing be-havior of f .

r 0�500

8�

¶1/3

sign of f 0 unde�ned � 0 +f unde�ned decreasing minimum increasing

Table 2

Note that

limr�0+

f (r) = limr�0+

�4�r2 +

500

r

¶= +�,

sincelimr�0+

4�r2 = 0 and limr�0+

500

r= +�.

We also have limr�+� f (r) = +�,since

limr�+� 4�r

2 = +� and limr�+�

500

r= 0.

Thus, f attains its absolute minimum on the interval (0,+�) at

r0 =

�500

8�

¶1/3

�= 2.709 63

Figure 7 shows the graph of f on the interval (0, 8). The picture is consistent with our analysis.

2 4 6 8r

300

600f

r0

Figure 7

The height of the “optimal can” is

h0 =250

�r20=

250

�

�500

8�

¶2/3= 250

�8

500

¶2/31

�1/3�= 10.838 5


In practice, it cannot be expected that the manufacturer will manufacture a can of radius 2.70963and height 10.8385. The precision is up to the negotiations between the manufacturer and thecustomer with regard to the dimensions of a “suboptimal can". ¤

Example 4 Assume that two hallways meet at a right angle, as shown in Figure 8. One hallwayis 3 meters wide, and the other is 5 meters wide. Determine whether it is possible to carry aladder which is 10 meters long around the corner horizontally.

A

B

Θ

Θ

3

5

Figure 8

Solution

With reference to Figure 8, the line segment AB must be longer than 10 for any value of theangle � between 0 and �/2, so that it is possible to carry the ladder around the corner. Thelength of AB is

f(�) =3

sin (�)+

5

cos (�).

Thus, we must determine the minimum value of f (�), where 0 < � < �/2, and see whether thatvalue is greater than 10.Figure 9 shows the graph of f on the interval (0, �/2).

0.7 Π2

Θ

10

20

30

40

Figure 9

We have

lim��0+

f (�) = lim��/2�

f (�) = +�


(con�rm). Figure 8 indicates that f attains its absolute minimum on the interval (0, �/2) at itsstationary point that seems to be near 0.7. Let’s di�erentiate f :

f 0 (�) =d

d�

�3

sin (�)+

5

cos (�)

¶

= 3d

d�(sin (�))�1 + 5

d

d�(cos (�))�1

= 3³� (sin (�))�2 cos (�)

´+ 5

³� (cos (�))�2 (� sin (�))

´= �3 cos (�)

sin2 (�)+5 sin (�)

cos2 (�).

You can check that f 00 (�) > 0 for each � � (0, �/2), so that the graph of f is concave upon (0, �/2) and f attains its absolute minimum on (0, �/2) at its stationary point in (0, �/2)(Theorem 2 of Section 3.3). You can determine an approximation to �0 such that f 0 (�0) = 0 withthe help of the approximate equation solver of your calculator, and con�rm that �0 �= .700 669.Therefore, the absolute minimum of f on the interval (0, �/2) is f(�0) �= 11.194 1. Since theladder in question is 10 meters long, it can be carried around the corner. ¤

Example 5 (Snell’s Law of Refraction) Assume that the speed of light in medium 1 is c1kilometers/second, and the speed of light in medium 2 is c2 kilometers/second.

A

C

B

d1

d2

EDx a � xΑ

Β

Figure 10

With reference to Figure 10, the point A is in medium 1 and the point B is in medium 2. Letthe distance from D to E be a. The horizontal line represents the demarcation between themedia. We will assume that light travels from one point to another so that the time of travel isminimized. In particular, light takes the shortest path to travel from A to C and from C to B.Thus, the required path from A to B consists of the line segments AC and CB. Snell’s law ofrefraction says that

sin (�)

sin (�)=c1c2.

The angle � is referred to as the angle of incidence, and the angle � is referred to as theangle of refraction. Thus, Snell’s law says that the ratio of the sine of the angle of incidenceto the sine of the angle of refraction is equal to the ratio of the speed of light in the �rst mediumto the speed of light in the second medium.

Let us establish Snell’s law. Our task is to determine the point C, i.e., the value of x, so thatthe time of travel from A to B is minimized.The time needed for the light to travel from A to C is

length of ACspeed of light in medium 1

=

pd21 + x

2

c1.


The time needed for the light to travel from C to B is

length of CBspeed of light in medium 2

=

pd22 + (a� x)2

c2.

The only variable is x. Let us set

f(x) =

pd21 + x

2

c1+

pd22 + (a� x)2

c2.

Thus, f(x) is the total time that is needed for light to travel from A to C and then from C toB. We need to minimize f (x). We have

f 0(x) =d

dx

Ãpd21 + x

2

c1+

pd22 + (a� x)2

c2

!=1

c1

xpd21 + x

2� 1

c2

a� xpd22 + (a� x)2

.

You can verify that

f 00(x) =1

c1

d21q(d21 + x

2)3+1

c2

d22q(d22 + (a� x)2)3

> 0

for each x � R. Therefore, the graph of f is concave up on [0, a]. We have

f 0 (0) =1

c1

xpd21 + x

2� 1

c2

a� xpd22 + (a� x)2

¯̄̄¯̄x=0

= � a

c2pd22 + a

2< 0,

and

f 0 (a) =1

c1

xpd21 + x

2� 1

c2

a� xpd22 + (a� x)2

¯̄̄¯̄x=a

=a

c1pd21 + a

2> 0.

Since f 0 is continuous on [0, a], there exists a point x0 � (0, a) such that f 0 (x0) = 0, by theIntermediate Value Theorem. Since f 00 (x) > 0 for each x, x0 is the only stationary point of fin the interval [0, a], and f attains its absolute minimum on [0, a] at x0 (Theorem 2 of Section3.6). We have

f 0 (x0) = 0 1

c1

xpd21 + x

2� 1

c2

a� xpd22 + (a� x)2

= 0

1

c1

x0pd21 + x

20

=1

c2

a� x0pd22 + (a� x0)2

.

With reference to Figure 10, � is the angle of incidence and � is the angle of refraction. Wehave

sin (�) =x0pd21 + x

20

, and sin (�) =a� x0p

d22 + (a� x0)2.

Therefore, f attains its minimum value if

1

c1sin (�) =

1

c2sin (�) ,

i.e.,sin (�)

sin (�)=c1c2.

Thus, we have established Snell’s law of refraction. ¤


Applications to Economics

Let us look at some applications to economics. We will assume that a company produces andsells a single product. We will denote the quantity that is produced by x, and assume thatx can be any real number. This is a modeling assumption. After all, in many cases x canonly attain positive integer values. For example, a company may produce calculators, in whichcase x may refer to the number of calculators produced by the company over a period of threemonths. Therefore, if the result of a calculation is that x =

�50000 �= 223.607, the decimal can

be rounded to 224.

Let us �rst examine the cost aspects. The cost function C is de�ned so that C(x) denotes thecost of producing quantity x of the product. If there is a �xed cost c per item, the total costof producing the quantity x is simply cx, so that C(x) = cx. In this case C is a simple linearfunction. More realistic models involve nonlinear cost functions. Cost functions of the form

a0 + a1x� a2x2 + a3x3,where a0, a1, a2 and a3 are certain positive constants are quite common. The term a0 re�ects thecost of maintaining the infrastructure of the company, even if there is no production. Obviously,the reliability of the predictions of a particular model depend on the construction of a realisticcost function. Such an e�ort involves many practical and theoretical considerations, and some ofyou may study such issues in other courses. Our discussions will be limited to certain conclusionsthat may be reached with the help of calculus, given a cost function.

The rate of change of the total cost C(x) with respect to the change in the level of the productionmay be of interest. Since we identify the rate of change of a function with its derivative, all wehave to do is to compute the derivative of the given cost function.

Remark 1 Let �x > 0 represent an increase in the production level. Recall our discussion ofthe di�erential in Section 3.2. We have

C(x+�x)� C(x) �= C0(x)�x,if �x is small. But, how small is “small”? Let’s assume that �x = 1 is small relative to xwithin the context of the production of many items. Then,

C(x+ 1)� C(x) �= C 0(x).Thus, C 0(x) approximates the change in the total cost corresponding to the increase in theproduction by a single item. For this reason, economists like to refer to the derivative of thecost function as marginal cost. As far as they are concerned, the derivative at x is “themarginal di�erence” in the cost due to the production of a single extra item. �

Example 6 Let the cost function be

C(x) = 100 + 10x� 0.1x2 + 0.001x3.You may imagine that a company produces a single type of computer, and that C(x) × 100 isthe total cost (in dollars) of producing x computers over a period of three months. The constantterm re�ects the fact that there is a cost of maintaining the infrastructure even if there is noproduction.

a) Plot the graph of C on the interval [0, 100] with the help of your graphing utility.b) Determine level of production at which marginal cost has its minimum value on the interval[0, 100].

Solution


a) Figure 11 shows the graph of the cost function.

20 40 60 80 100x

200

400

600

800

1000

y

y � C�x�

Figure 11: A cost function

b) Notice that the cost function C is an increasing function, as it should be. Since marginalcost is the derivative of the cost function, we have to compute the point at which C0 attains itsminimum value on the interval [0, 100]. We have

C0(x) =d

dx

¡100 + 10x� 0.1x2 + 0.001x3¢ = 10� 0.2x+ 0.003x2.

Figure 12 shows the graph of C 0.

20 40 60 80 100x

5

10

15

20y

a

y � C'�x�

Figure 12: A marginal cost function

Figure 11 indicates that marginal cost decreases up to a certain production level, and then startsto increase. We must �nd the value of x that minimizes C 0 (x). We will apply the derivativetest for monotonicity to C0. We have

(C0)0 (x) = C00 (x) =d

dx

¡10� 0.2x+ 0.003x2¢ = �0.2 + 0.006x.

Therefore,

C00 (x) = 0 �0.2 + 0.006x = 0 x = a =1

3× 102 �= 33.3,

We have C 00 (x) < 0 if 0 x < a, and C00(x) > 0 if a < x 100. Therefore, C 0 decreaseson the interval [0, a] and increases on the interval on the interval [a, 100]. Thus, C0 attains itsminimum value on the interval [0, 100] at a. Note that the graph of the cost function is concavedown on [0, a], and concave up on [a, 100]. The graph of C has an in�ection point at

(a,C (a)) =

�1

3× 102, C

�1

3× 102

¶¶�= (33.3, 359.3) .

¤Now let us look at the revenue side. If quantity x of the product of the company is sold at a�xed price p per unit, the total revenue is px. But it is not realistic to assume that the product


has the same price at all production levels. It is to be expected that the price will be lower ifthe supply is higher. Thus, the revenue function R is of the form

R (x) = xp(x),

where the function p (x) is not a constant, in general. The function p(x) is referred to as theprice function, or the demand function.

Remark 2 Economists refer to the derivative of the revenue function as marginal revenue.As in the case of marginal cost, the approximation

R(x+ 1)�R(x) �= R0 (x)

appears to be justi�ed in many cases of practical interest.�

Example 7 Let us consider the speci�c price function

p (x) = 15� 0.05x.

The corresponding revenue function is

R(x) = xp(x) = 15x� 0.05x2.

Figure 13 shows the graph of R. ¤

50 100 150 200x

500

1000

y

y � R�x�

Figure 13: A revenue function

In general, the ultimate goal of a company is to make pro�t. The pro�t function P is thedi�erence between the revenue function R and the cost function C:

P (x) = R(x)� C(x).

Proposition 1 If the pro�t function P attains a local maximum at a then the mar-ginal revenue at a is the same as the marginal cost at the production level a.

Proof

If P attains a local maximum at a, we must have P 0 (a) = 0. We have

P 0 (x) =d

dx(R(x)� C(x)) = R0 (x)� C0(x),

Therefore, R0(a) = C0(a), i.e., the marginal revenue is the same as marginal cost at a.¥


Remark 3 Proposition 1 can be interpreted as follows: The optimal production level is reachedwhen the additional revenue that is obtained by producing one additional item is the same asthe additional cost of producing one additional item. Graphically, the line that is tangent tothe graph of the revenue function at (a,R(a)) is parallel to the line that is tangent to the graphof the cost function at (a,C(x)), as illustrated in Figure 14. �

x

y

y � R�x�

y � C�x�

�a,R�a��

�a,C�a��

Figure 14

Example 8 LetC(x) = 100 + 10x� 0.1x2 + 0.001x3,

as in Example 6, andR(x) = 15x� 0.05x2,

as in Example 7. Determine the production level between 0 and 100 so that the pro�t ismaximized.

Solution

We have

P (x) = R(x)� C(x)=¡15x� 0.05x2¢� ¡100 + 10x� 0.1x2 + 0.001x3¢ .

Figure 15 shows the graph of the pro�t function. Figure 15 indicates that the pro�t functionattains its absolute maximum on [0, 100] at its stationary point a in that interval. You cancon�rm this with the help of the derivative test for monotonicity. We have

P 0 (x) =d

dx

¡¡15x� 0.05x2¢� ¡100 + 10x� 0.1x2 + 0.001x3¢¢

= 5.0 + . 1x� .003x2.Therefore, the stationary point of P that is in the interval [0, 100] is 60.762 5, rounded to 6signi�cant digits. Thus, the optimal production level a is approximately 61. ¤

20 40 60 80 100x

�100

�50

50

100

150

y

y � P�x�

a

Figure 15: A pro�t function


Problems

1. Determine two positive numbers whose sum is the minimum among all pairs of positivenumbers whose product is 400.

2. Determine the dimensions of a rectangle whose area is the maximum among all rectangleswith perimeter 200 meters.

3. Determine the dimensions of a rectangle whose perimeter is the minimum among all rectangleswith area 800 m2.

4. Determine the dimensions of the rectangle that has the largest area among all rectanglesthat can be inscribed in a semicircle of radius 4.

5. Determine the dimensions of the rectangle of maximum area among all rectangles that canbe inscribed in an equilateral triangle of side length 10 such that one side lies on the base of thetriangle.

6. Determine the dimensions of the cylinder of largest volume among all right circular cylindersthat can be inscribed in a sphere of radius 10.

7. Determine the points on the ellipse

x2

4+ y2 = 1

that is closest to the point (1, 0).

8. Assume that a manufacturer must produce cans in the shape of right circular cylinders. Thevolume of each can is required to be 400 cubic centimeters. Otherwise, the manufacturer is freeto choose the dimensions of the can. The cost per square centimeter of the material for the topand the bottom of the can is the same as the cost per square centimeter of the material for thelateral surface. How should the manufacturer determine the dimensions of the can in order tominimize the cost?

9. A box with a square base and open top is required to have a volume of 1000 cubic centimeters.Determine the dimensions of the box so that the amount of the material that is used in theconstruction of the box is minimized.

10. A water trough has length 10 meters and a cross section in the shape of an isosceles trianglewith sides that are 50 centimetres. Use calculus to determine the length of the top of the triangleso that the volume of the trough is maximized.

11. Assume that a racket launcher is �red at an angle � from the horizotal ground. The ranges is the horizontal distance traveled by a projectile �red by the rocket launcher and is given bythe expression

s =v20 sin (2�)

g,

where v0 is the initial speed of the projectile and g is the constant gravitational acceleration (canbe assumed to be 9.8 meters/sec/sec.). Use calculus to determine the value of � that maximizesthe range s.

12 [C] Assume that the total cost C (x) of producing x items of a certain product is

C (x) = 0.002x3 � 0.1x2 + 4x.a) Plot the graph of C (x) and the marginal cost function C 0 (x).b) Determine level of production at which marginal cost has its minimum value. Intepret yourresponce graphically.

13 [C] Assume that the total cost C (x) of producing x items of a certain product is

C (x) = 0.002x3 � 0.1x2 + 4x


and the corresponding revenue is

R (x) = 10x� 0.04x2.

a) Show the graphs of C (x) and R (x) in the same picture.b) Determine production level so that the pro�t is maximized. Intepret your response in thelanguage of econmics and graphically.

4.5. ORDERS OF MAGNITUDE 291

Problems

In problems 1-6, simplify the given expression.

1. log10¡104¢

2. log2¡2�3

¢3. log5

¡53/4

¢4. 10log10(7)

5. 10� log10(4)

6. 2log2(81/4)

In problems 7-11, determine the solutions of the given equation.

7. 3 log10 (x) = 2

8. 2 log10 (x) = �59. 6 log2 (x) = 4

10. 102x+1 = 6

11. 2x2

= 14

In problems 12-24, determine the derivative:

12.d

dx10x

2+1

13.d

dx101/x

14.d

dx2�x

15.d

dx3�x

2

16.d

dx2sin(x)

17.d

dxlog10

¡x2 + 1

¢18.

d

dxlog2

¡�x¢

19.d

dxlog10

�x� 1x+ 4

¶

20.d

dxlog2

¡sin2 (x)

¢21.

d

dxx�3

22.d

dx�x

23.d

dxx�

24.d

dxx1/x

25 [C] Plot the graphs of f (x) = x�, g (x) = x4 and h (x) = x2 on the interval [0, 2] with thehelp of your graphing utility. What can you say about the relative size of their values?

26. [C] Plot the graphs of f (x) = xx, g (x) = x and h (x) = ex on the interval [1, 4]. with thehelp of your graphing utility. What can you say about the relative size of their values?

4.5 Orders of Magnitude

In this section we will see that an exponential function of the form a�x, where a > 1 and � > 0,increases much faster than any power of x, and a logarithmic function increases more slowlythan any positive power of x as x tends to �. Since exponential and logarithmic functions witharbitrary bases can be expressed in terms of the natural exponential function and the naturallogarithm, our emphasis will be on the “natural” functions.

292 CHAPTER 4. SPECIAL FUNCTIONS

Exponentials vs. Powers of x

In Section 4.3 we noted thatlim

x�+� ex = +�,

since e > 1. In fact, ex grows very rapidly as x increases. Table 1 displays en (rounded to 6signi�cant digits, as usual) for n = 10, 20, 30, 40. We see that en is much larger than n for thesampled values of n.

n en

10 22026.520 4.85165× 10830 1.06865× 101340 2.353 85× 1017

Table 1

We will see that ex grows faster than any power of x as x � +�. Let’s begin by examining afunction that involves ex and x2.

Example 1 Let

f (x) =ex

x2.

a) Determine limx�0 f (x) and the absolute minimum of f on (0,+�).b) Plot the graph of f on the interval (0, 8] with the help of your graphing utility. Does thepicture suggest that limx�+� f (x) = +�?c) Compute f (x) for x = 10, 20, 30, 40. Do the numbers suggest that limx�+� f (x) = +�?

Solution

a) We have

limx�0

ex = e0 = 1 > 0 and limx�0

1

x2= +�.

Therefore,

limx�0

f (x) = limx�0

�ex�1

x2

¶¶= +�.

Thus, f does not have an absolute maximum on (0,+�). In order to determine the absoluteminimum of f on (0,+�) we will apply the derivative test for monotonicity. By the quotientrule,

f 0 (x) =d

dx

�ex

x2

¶=ex¡x2¢� ex (2x)x4

=exx (x� 2)

x4=

�ex

x3

¶(x� 2)

if x 6= 0. We have f 0 (x) = 0 if x = 2, so that 2 is the only stationary point of f . Since ex > 0for each x and x3 > 0 if x > 0, the sign of f 0 (x) is determined by the sign of x � 2 if x > 0.Thus, f 0 (x) < 0 if 0 < x < 2 and f 0 (x) > 0 if x > 2. Therefore, f is decreasing on the interval(0, 2] and f is increasing on [0,+�). Thus, f attains its absolute minimum on (0,+�) at 2.We have

f (2) =e2

22=e2

4.

b) Figure 1 shows the graph of f . The picture indicates that limx�+� f (x) = +�.


2 4 6 8x

10

20

30

40

y

Figure 1: y =ex

x2

c) Table 2 displays f (x) for x = 10, 20, 30, 40. The numbers de�nitely support the expectationthat limx�+�f (x) = +�. ¤

x f (x)10 2202.6520 2.425 83× 10730 3.562 16× 101140 5.884 63× 1015

Table 2

The function of Example 1 represents the family of functions fn, where

fn (x) =ex

xn,

and n is a positive integer. Note that limx�+� ex = +� and limx�+� xn = +�. An attemptto evaluate limx�� f (x) by setting

limx�+�

ex

xn=limx�+� ex

limx�+� xn

leads to the indeterminate expression �/�. Actually, ex increases faster than xn as x� +�,so that the ratio ex/xn ��:

Proposition 1 We have

lim��+�

e�

xn= +�

for any integer n.

Proof

We have limx�+� ex = +�, so that statement of the proposition is valid if n = 0. If n < 0,then �n > 0, so that limx�+� x�n = +�. Therefore,

limx�+�

ex

xn= lim

x�+�¡exx�n

¢= +�,

as well.

The statement of the proposition is nontrivial if n is a positive integer. Let’s set

fn (x) =ex

xn, n = 1, 2, 3, . . .


As in Example 1, we will make use of the derivative test for monotonicity in order to determinethe absolute minimum of fn on the interval (0,+�). By the quotient rule,

f 0n (x) =d

dx

�ex

xn

¶=exxn � nxn�1ex

x2n=exxn�1 (x� n)

x2n=

�ex

xn+1

¶(x� n) .

Since ex > 0 for each x and xn+1 > 0 if x > 0, the sign of f 0n (x) is determined by the sign ofx� n. We have f 0n (x) = 0 if x = n. We also have

f 0n (x) < 0 if 0 < x < n, and f 0n (x) > 0 if x > n.

Therefore, fn is decreasing on (0, n], and increasing on [n,+�). Thus, fn attains its absoluteminimum on (0,+�) at x = n. We have

fn (n) =en

nn=³ en

ń.

Therefore,

fn (x) =ex

xn�³ en

ńfor each x > 0.

We can express fn (x) in terms of fn+1 (x):

fn(x) =ex

xn= (x)

�ex

xn+1

¶= xfn+1(x).

Since

fn+1 (x) ��

e

n+ 1

¶n+1

,

we have

fn(x) � x�

en+1

(n+ 1)n+1

¶for each x > 0. Since

limx�+�

�en+1

(n+ 1)n+1

¶x = +�,

limx�� fn(x) = +�, as well. ¥

Example 2 Let

f (x) =2x/4

x.

Determine limx�+� f (x).

Solution

We have

f (x) =2x/4

x=e(x/4) ln(2)

x.

Let’s set

z =³x4

´ln (2) x =

�4

ln (2)

¶z.

Since ln (2) > 0 we have z � +� as x� +�. Thus,


x�+�e(x/4) ln(2)

x= lim

z�+�ez³4

ln(2)

´z= lim

z�+�

�ln (2)

4

¶�ez

z

¶= +�,


thanks to Proposition 1. Figure 2 shows the graph of f on the interval [20, 60] (The axes arecentered at (20, 0)). The picture is consistent with the fact that

limx�+�

2x/4

x= +�.

¤

30 40 50 60x

100

200

300

y

Figure 2: y =2x/4

x

Example 3 Let f (x) = x2e�x.

a) Determine limx�+� f (x).b) Determine the absolute maximum of f on [0,+�).c) Plot the graph of f on [0, 8] with the help of your graphing utility. Does the picture supportyour responses to part a) and part b)?

Solution

a)


x�+�x2e�x = lim

x�+�x2

ex= limx�+�

1ex

x2

= 0,

since

limx�+�

ex

x2= +�,

by Proposition 1.

b) By the product rule and the chain rule,

f 0 (x) =d

dx

¡x2e�x

¢= (2x)

¡e�x

¢+ x2

¡�e�x¢ = xe�x (2� x) .Since e�x > 0, the sign of f 0 (x) is determined by the sign of 2�x if x > 0. Therefore, f 0 (x) = 0if x = 2 and

f 0 (x) > 0 if 0 < x < 2 and f 0 (x) < 0 if x > 2.

By the derivative test for monotonicity, f is increasing on [0, 2] and decreasing on [2,+�) (notethat f (0) = 0). Therefore, f attains its absolute maximum on the interval [0,+�) at 2. Wehave

f (2) = 22e�2 =4

e2�= 0.541341

c) Figure 3 shows the graph of f on the interval [0, 8]. The picture is consistent with the factthat limx�+� f (x) = 0 and our calculation of the absolute maximum of f on [0,+�). ¤


2 4 6 8x

0.5

y

Figure 3: y = x2e�x

Note that an attempt to evaluate limx�+� x2e�x as�lim

x�+�x2

¶�lim

x�+� e�x¶=

�lim

x�+�x2

¶�lim

x�+�1

ex

¶

leads to the indeterminate form (+�) (0). ¤

Logarithmic Growth

We havelim

x�+� ln (x) = +�,and

d

dxln (x) =

1

x> 0,

consistent with the fact that the natural logarithm is an increasing function on the interval(0,+�).

5 10 15 20x

�1

1

2

3

y

Figure 4: The natural logarithm increases slowly if x is large

But the rate of growth of ln(x) at x (i.e., 1/x) tends to 0 as x becomes large. In fact, ln (x)increases more slowly than any positive power of x as x increases. The following propositionmakes this statement more precise:

Proposition 2 Let a > 0, a 6= 1, and r > 0. Then

lim��+�

ln (x)

xr= 0.

Proof

We set y = ln (x) so that x = ey, and y � +� as x� +�. Therefore,

limx�+�

ln (x)

xr= limy�+�

y

(ey)r= limy�+�

y

ery.


If we set z = ry then z � +� as y � +�, since r > 0. Thus,

limx�+�

ln (x)

xr= limy�+�

y

ery= lim

z�+�z/r

ez= lim

z�+�1

r

�� 1ez

z

�� = 0,

since

limz�+�

ez

z= +�

by Proposition 1. ¥Note that an attempt to evaluate

limx�+�

ln (x)

xr

aslimx�+� ln (x)limx�+� xr

leads to the indeterminate form �/�.

Example 4 Let

f (x) =ln (x)�x.

a) Determine limx�+� f (x) and limx�0+ f (x) .b) Determine the absolute maximum of f on the interval (0,+�).c) Plot the graph of on the interval [1, 100] with the help of your graphing utility. Is the pictureconsistent with your response to part a) and part b)?d) Compute f (x) for x = 10k, k = 2, 3, 4, 5. Do the numbers support your statement concerninglimx�+� f (x)?

Solution

a) We have

f (x) =ln (x)�x=ln (x)

x1/2.

By Proposition 2 (with r = 1/2) limx��f (x) = 0.As for limx�0+f (x), we have limx�0+ ln (x) = �� and limx�0+1/

�x = +�. Therefore,

limx�0+

f (x) = limx�0+

�ln (x)

�1�x

¶¶= ��.

b) By the quotient rule,

f 0 (x) =d

dx

�ln (x)

x1/2

¶=

�1

x

¶x1/2 � ln (x)

�1

2x�1/2

¶¡x1/2

¢2 =

1�x� ln (x)2�x

x=2� ln (x)2x�x.

Since x�x > 0 if x > 0, the sign of f 0 (x) is determined by the sign of 2� ln (x). We have

2� ln (x) = 0 2 = ln (x) e2 = x.

Furthermore,2� ln (x) > 0 if 0 < x < e2 and 2� ln (x) < 0 if x > e2.


By the derivative test for monotonicity, f is increasing on (0, e2], and f is decreasing on [e2,+�).Therefore, f attains its absolute maximum on the interval (0,+�) at e2 �= 7.389 06. We have

f¡e2¢=ln (x)�x

¯̄̄¯x=e2

=ln¡e2¢

�e2

=2

e�= 0.735759

c) Figure 5 shows the graph of f on the interval. The picture is consistent with our response topart a) and part b), but does not provide strong support for the statements about limx�0+f (x)and limx�+� f (x).

20 40x

0.75

y

e2

�e2, 2�e2�

Figure 5: y =ln (x)�x

d) Table 4 displays f (x) for x = 10k, k = 2, 3, 4, 5. The numbers in Table 4 support thestatement that limx�+� f (x) = 0. ¤

x f (x)102 .460 517103 .218 442104 9.210 34× 10�2105 3.640 71× 10�2

Table 4

Example 5 Let f (x) = x ln (x).

a) Determine limx�0+ f (x) and limx�+� f (x) .b) Determine the absolute minimum of f on the interval (0,+�).c) Sketch the graph of f .

Solution

a) We set z = 1/x so that z � +� as x approaches 0 from the right. Thus,

limx�0+

f (x) = limx�0+

x ln (x) = limz�+�

�1

z

¶ln

�1

z

¶= limz�+�

� ln (z)z

= 0

by Proposition 2.Note that an attempt to evaluate limx�0+ x ln (x) as�

limx�0+

x

¶�limx�0+

ln (x)

¶

leads to the indeterminate form 0× (��).As for limx�+� f (x), limx�+� x ln (x) = +� since limx�+� x = +� and limx�+� ln (x) =+�.


b) By the product rule,

f 0 (x) =d

dx(x ln (x)) = ln (x) + x

�1

x

¶= ln (x) + 1.

Therefore,

f 0 (x) = 0 ln (x) + 1 = 0 ln (x) = �1 x = e�1 =1

e�= 0.367 879.

Furthermore,

f 0 (x) < 0 if 0 < x <1

e, and f 0 (x) > 0 if x >

1

e.

By the derivative test for monotonicity,

f is decreasing on (0,1

e] and increasing on [

1

e,+�).

Therefore, f attains its absolute minimum on (0,+�) at 1/e. The corresponding value of f is

1

eln

�1

e

¶= �1

eln (e) = �1

e�= �0.367 879.

c) Figure 6 displays the graph of f on the interval [0, 4]. Even though f is not de�ned at 0, thegraph is consistent with the fact that limx�0+ f (x) = 0. ¤

1 2 3 4x

2

4

y

�1�e, �1�e�

Figure 6: y = x ln (x)

The Natural Exponential Function as a Limit of Polynomials

The natural exponential function can be approximated by polynomials. We will examine such afamily of polynomials. In Chapter 9 we will discuss another family of approximating polynomials.Let

pn (x) =³1 +

x

n

ń,

where n is a positive integer. Each pn (x) is a polynomial, and the degree of pn (x) is n. The �rstmember of this family of polynomials is p1 (x) = 1+x. Note that p1 is the linear approximationto the natural exponential function based at 0 (Example 2 of Section 5.3). Figure 7 displays thegraphs of the natural exponential function and the polynomials

p1 (x) = 1 + x, p3 (x) =³1 +

x

3

´3and p20 (x) =

³1 +

x

20

´2.


�1 1 2 3x

5

10

y

p1

p3

p20

y � ex

Figure 7

Figure 7 indicates that pn (x) approximates ex with increasing accuracy as n increases. This isindeed the case.

Proposition 3 We have

lim��(1+

x

n)�= e� for each x � R.

In particular,

lim��(1+

1

n)� = e

Since limn�� 1/n = 0, Proposition 3 follows from the following fact:

Proposition 4

lim�0

(1 + xh)1�

= e� .

Proof

We have

(1 + xh)1/h

=³eln(1+xh)

´1/h= eln(1+xh)/h.

By the continuity of the natural exponential function,

limh�0

eln(1+xh)/h = elimh�0 ln(1+xh)/h.

Therefore, in order to show that

limh�0

(1 + xh)1/h

= ex,

it is su�cient to prove that

limn��

�1

hln (1 + xh)

¶= x.


This is indeed the case: If x = 0, the equality is obvious since ln (1) = 0. Let’s assume thatx 6= 0 and set z = xh. Then z 6= 0 if h 6= 0 and z � 0 as h� 0. Therefore,

limh�0

ln (1 + xh)

h= limh�0

�ln (1 + xh)

xh(x)

¶= x lim

h�0

ln (1 + xh)

xh

= x limz�0

ln (1 + z)

z

= x limz�0

ln (1 + z)� ln (1)z

= x

�d

duln (u)

¯̄̄¯u=1

¶= x

�1

u

¯̄̄¯u=1

¶= x.

¥Note that an attempt the evaluate limn�� (1 + 1/n)

n as

�limn��

�1 +

1

n

¶¶limn�� n

leads to the indeterminate form 1�. One may be tempted to say that 1� = 1, since 1n = 1for each n, but the actual limit need not be 1, as Proposition 3 shows.

Problems

In problems 1-4,a) Determine the indicated limits and the asymptotes for the graph of f .b) Use the derivative test to determine the intervals on which f is increasing/decreasing, andthe points at which f has a local extremum.c) Sketch the graph of f.

1.. f (x) =ex/2

x, limx�0±

f (x) , limx�±�f (x)

2.. f (x) = x2e�x, limx�±�f (x)

3. f (x) =ln (x)

x1/3, limx�0+

f (x) , limx�+�f (x)

4. f (x) =�x ln (x) , lim

x�0+f (x) , lim

x�+�f (x)

In problems 6-7,a) Determine the indicated limits and the asymptotes for the graph of f .b) Use the derivative test to determine the intervals on which f is increasing/decreasing, andthe points at which f has a local extremum.c) Use the second derivative test to determine the intervals on which the graph of f is concaveup/concave down, and the x-coordinates of the in�ection points of the graph of f .d) Sketch the graph of f.

5.. f (x) = xe�x2/4, lim

x�±�f (x) .

6. f (x) = e�x2/9, lim

x�±�f (x)

7. f (x) =10ex

2 + ex, limx�±�f (x)

In problems 8-11, determine the absolute maximum and the absolute minimum of f on theinterval I, provided that such values exist. Justify your response if you claim that such a valuedoes not exist.

5.3. THE FUNDAMENTAL THEOREM OF CALCULUS: PART 1 371

5.3 The Fundamental Theorem of Calculus: Part 1

In sections 5.1 and 5.2 we introduced the concept of the integral. The Fundamental Theoremof Calculus establishes the link between the two fundamental concepts of calculus, namely, thederivative and the integral. We will discuss the �rst part of the theorem in this section and thesecond part of the theorem in Section 5.5.

The Fundamental Theorem of Calculus (Part 1)

The �rst part of the Fundamental Theorem of Calculus states that the integral of the deriv-ative of a function on an interval is equal to the di�erence between the values ofthe function at the endpoints of the interval:

Theorem 1 (THE FUNDAMENTAL THEOREM OF CALCULUS (Part 1)) As-sume that F 0 is continuous on [a, b] Then

Z �

�

F 0(x)dx = F (b)� F (a).

F 0 (a) and F 0 (b) can be interpreted as the one sided derivatives F 0+ (a) and F 0� (b), respectively.

The Proof of Theorem 1

Let P = {x0, x1, . . . , xk�1, xk, . . . , xn�1, xn} be a partition of [a, b], so that x0 = a and xn = b.We can express the change in the value of F over the interval [a, b] as the sum of the changes inthe value of F over the subintervals determined by P :

F (b)� F (a) = F (xn)� F (x0)= [F (xn)� F (xn�1)] + [F (xn�1)� F (xn�2)] + · · ·+ [F (x2)� F (x1)] + [F (x1)� F (x0)]

=nXk=1

[F (xk)� F (xk�1)] .

By the Mean Value Theorem (Theorem 3 of Section 3.2), there exists x�k � (xk�1, xk) such that

F (xk)� F (xk�1) = F 0 (x�k) (xk � xk�1) = F 0 (x�k)�xk.

Therefore,

F (b)� F (a) =nXk=1

F 0 (x�k)�xk.

We havenXk=1

F 0 (x�k)�xk �=Z b

a

F 0 (x) dx

if ||P || = maxk�xk is small, and the approximation is as accurate as desired if ||P || is smallenough. Therefore,

F (b)� F (a) �=Z b

a

F 0 (x) dx,

and ¯̄̄¯̄(F (b)� F (a))� Z b

a

F 0 (x) dx

¯̄̄¯̄

372 CHAPTER 5. THE INTEGRAL

is as small as desired. This means that the numbers

F (b)� F (a) andZ b

a

F 0 (x) dx

are equal.¥We may refer to the �rst part of the Fundamental Theorem of Calculus simply as “the Fun-damental Theorem of Calculus” or “the Fundamental Theorem”, until we introduce the secondpart of the Fundamental Theorem and a distinction is necessary.

Example 1 Let

F (x) =2

3x3/2.

By the power rule,

F 0 (x) =d

dx

�2

3x3/2

¶=2

3

d

dxx3/2 =

2

3

�3

2x1/2

¶=�x

if x � 0 (we have to interpret F 0 (0) as F 0+ (0)). Thus, F 0 is continuous on [0, 1], so that theFundamental Theorem of Calculus is applicable on [0, 1]. Therefore,

Z 1

0

�xdx =

Z 1

0

F 0 (x) dx = F (1)� F (0) = 2

3.

Thus, the area of the region between the graph of y =�x and the interval [0, 1] is 2/3. The

region is illustrated in Figure 1. ¤

1 2x

1

y

Figure 1:Z 1

0

�x =

2

3

Example 2 Evaluate Z ��/40

d

dxcos¡x2¢dx

Solution

If we set F (x) = cos¡x2¢, we have

Z ��/40

d

dxcos¡x2¢dx =

Z ��/40

d

dxF (x) dx = F

�r�

4

¶� F (0)

= cos³�4

´� cos (0) =

�2

2� 1,

by Theorem 1.


Note that

f (x) =d

dxcos¡x2¢= �2x sin ¡x2¢ ,

and we have f (x) 0 if 0 x p�/4. Thus, the area of the region G between the graph of f

and the intervalh0,p�/4

iis

�Z ��/40

f (x) dx = �Z ��/40

d

dxcos¡x2¢dx = 1�

�2

2.

Figure 2 shows the region G. ¤

�2 2x

22

y

Π �4

Figure 2

We were able to compute the integrals in the above examples by expressing the integrand as thederivative of a familiar function. We will use this procedure to compute many integrals:

Corollary (Corollary to the Fundamental Theorem of Calculus) Assume that f iscontinuous on [a, b] and that F 0(x) = f(x) for each x [a, b]. Then

Z �

�

f(x)dx = F (b)� F (a).

Proof

By the Fundamental Theorem of Calculus (Part 1),

Z b

a

f (x) dx =

Z b

a

F 0 (x) dx = F (b)� F (a)

¥

As in Theorem 1, F 0 (a) and F 0 (b) can be interpreted as the one sided derivatives F 0+ (a) andF 0� (b), respectively.

We may refer to the corollary to the Fundamental Theorem of Calculus simply as “the Funda-mental Theorem of Calculus”.

De�nition 1 A function F is an antiderivative of f on an interval J if F 0(x) = f(x) foreach x in J .

The derivative should be interpreted as the appropriate one-sided derivative at an endpoint ofthe relevant interval.


We will denote F (b)� F (a) asF (x)|ba .

Thus, we can express the Corollary to the Fundamental Theorem of Calculus as follows:

Z �

�

f(x)dx = F (x)|��if F is an antiderivative of f on [a, b].

Example 3 Evaluate Z 9

4

�xdx.

Solution

With reference to Example 1, if

f(x) =�x and F (x) =

2

3x3/2,

then F is an antiderivative of f on the interval [0,+�), since

F 0 (x) = f (x)

for each x � (0,+�), and F 0+ (0) = f (0).Therefore, Z 9

4

�xdx =

2

3x3/2

¯̄̄¯9

4

=2

3

³93/2 � 43/2

´=2

3(27� 8) = 38

3.

¤We have been referring to “an antiderivative of a function”. Indeed, a function has in�nitelymany antiderivatives. On the other hand, any two antiderivatives of the same function can di�erat most by an additive constant:

Proposition 1 Let F be an antiderivative of f on the interval J .

a) If C is a constant, then F + C is also an antiderivative of f on J.b) If G is any antiderivative of f on the interval J , there exists a constant C suchthat G(x) = F (x) + C for each x in J .

Proof

a) Since F is an antiderivative of f on J , we have

d

dxF (x) = f(x) for each x � J.

If C is an arbitrary constant,

d

dx(F (x) + C) =

d

dxF (x) +

d

dx(C) = f (x) + 0 = f(x)

for each x in J . Therefore, F + C is also an antiderivative of f on the interval J .b) Since F and G are antiderivatives of f on the interval J , we have

d

dxF (x) = f(x) and

d

dxG(x) = f(x)


for each x � J . Therefore, there exists a constant C such that G(x) = F (x) + C for all x in J(Corollary to Theorem 5 of Section 3.2). ¥By Proposition 1, if F is an antiderivative of f , we can express any antiderivative of f as F +C,where C is a constant. We will use the notationZ

f(x)dx

to denote any antiderivative of f and refer toZf(x)dx

as the inde�nite integral of f . Thus,Zf (x) dx = F (x) + C.

Example 4 If

F (x) =1

3x3

and f (x) = x2, then F is an antiderivative of f (on the entire number line), since

d

dx

�1

3x3¶=1

3

¡3x2

¢= x2

for each x � R. Therefore, we can express the inde�nite integral of f asZx2dx =

1

3x3 + C,

where C is an arbitrary constant. ¤

Remark 1 (Caution) We may refer to an integralZ b

a

f (x) dx

as a de�nite integral, if we feel the need to make a distinction between an integral and aninde�nite integral. In spite of the similarities between the terminology and the notation, theinde�nite integral of f and the integral of f on an interval [a, b] are distinct entities.The (de�nite) integral Z b

a

f (x) dx

is a number that can be approximated with arbitrary accuracy by Riemann sums, whereas, theinde�nite integral Z

f (x) dx

represents any function whose derivative is equal to the function f . In either case, we will referto f as the integrand. The Fundamental Theorem establishes a link between a de�nite integraland an inde�nite integral: Z b

a

f (x) dx =

Zf (x)

¯̄̄¯x=b

x=a

.

�


Example 5 Let C denote an arbitrary constant. Show that the statementsZ2 sin (x) cos (x) dx = sin2 (x) + C

and Z2 sin (x) cos (x) dx = � cos2 (x) + C

are both correct.

Solution

We haved

dxsin2(x) = 2 sin(x) cos(x)

andd

dx

¡� cos2(x)¢ = �2 cos (x) (� sin(x)) = 2 sin(x) cos(x)for each x � R. Therefore, both sin2 (x) and � cos2 (x) are antiderivatives for 2 sin (x) cos (x).Therefore, we can express the inde�nite integral of 2 sin (x) cos (x) asZ

2 sin (x) cos (x) dx = sin2 (x) + C

or Z2 sin (x) cos (x) dx = � cos2 (x) + C,

where C denotes an arbitrary constant.Since sin2 (x) and � cos2 (x) are antiderivatives of the same function, they must di�er by aconstant. Indeed,

sin2(x)� ¡� cos2 (x)¢ = sin2 (x) + cos2 (x) = 1for all x � R. ¤

Remark 2 We should be able to use any antiderivative of the integrand in order to evaluatean integral. Indeed, if

d

dxF (x) = f (x) and

d

dxG (x) = f (x)

for each x in some interval J , there exists a constant C such that G(x) � F (x) = C for eachx � J . Therefore, Z b

a

f (x) dx = F (x)|x=bx=a = F (b)� F (a),and Z b

a

f(x)dx = G (x)|x=bx=a = G(b)�G(a) = (F (b) + C)� (F (a) + C) = F (b)� F (a).

Therefore, we do not have to include an arbitrary constant in the expression for an inde�niteintegral when we use the inde�nite integral to evaluate a de�nite integral. �

Example 6 With reference to Example 5,Z �/2

�/4

2 sin (x) cos (x) dx = sin2 (x)¯̄�/2�/2

= 1��1�2

¶2

= 1� 12=1

2.

We also have Z �/2

�/4

2 sin (x) cos (x) dx = � cos2 (x)¯̄�/2�/4

= (0) +

�1�2

¶2

=1

2.

¤


The Inde�nite Integrals of Basic Functions

We will refer to the determination of the antiderivatives of functions as antidi�erentiation.Traditionally, the term “integration” is also used instead of the term “antidi�erentiation”, eventhough we should make a distinction between integrals and antiderivatives. The particularcontext in which the terms are used should clarify the intended meaning. Antidi�erentiation isnot as straightforward as di�erentiation. Computer algebra systems are very helpful in �ndingthe inde�nite integrals of many functions. On the other hand, it is convenient to have theinde�nite integrals of frequently encountered functions at your �ngertips. Let’s begin with ashort list of inde�nite integrals. You will learn about some rules of antidi�erentiation in the restof this chapter and in the next chapter. These rules will enable you to expand the scope of thisshort list considerably. The letter C denotes an arbitrary constant.

A Short List of Antiderivatives

1.

Zx� dx =

x�+1

r + 1+C, r 6= �1 (if x� is de�ned)

2.

Z1

xdx = ln(|x|) + C(on any interval that that does not contain 0)

3.

Zsin(�x) dx = � 1

�cos(�x) + C (� is a nonzero constant)

4.

Zcos(�x) dx =

1

�sin(x) + C (� is a nonzero constant)

5.

Ze�dx = e�+C

6.Ra� dx =

1

ln(a)ax +C, where a > 0

7.

Zsinh(x)dx = cosh(x) + C

8.

Zcosh(x)dx = sinh(x) + C

9.

Z1

x2+1dx = arctan(x) + C

By the de�nition of the inde�nite integral, each formula is con�rmed by di�erentiation.

1. Let J be an interval that is contained in the natural domain of xr. By the power rule,

d

dx

�xr+1

r + 1

¶=

1

r + 1

d

dx

¡xr+1

¢=

1

r + 1(r + 1)xr = xr

for each x in J (the derivative may have to be interpreted as a one-sided derivative at 0).Therefore, Z

xrdx =xr+1

r + 1+ C

on the interval J . We will refer to the above antidi�erentiation rule as the reverse powerrule since it is a consequence of the power rule for di�erentiation.

2. If x > 0,d

dxln (|x|) = d

dxln (x) =

1

x.

If x < 0,

d

dxln (|x|) = d

dxln (�x) =

Ãd

duln (u)

¯̄̄¯u=�x

!�d

dx(�x)

¶=

�1

�x¶(�1) = 1

x,


with the help of the chain rule.Therefore, Z

1

xdx = ln(|x|) + C

on any interval that does not contain 0.Figure 3 shows the graph of y = ln (|x|). Note that ln (|x|) de�nes an even function, so that thegraph of the function is symmetric with respect to the vertical axis. Also note that

limx�0�

ln (|x|) = limx�0+

ln (|x|) = ��.

�10 �5 5 10x

�1

1

2

y

y � ln��x��

Figure 3: y = ln (|x|)

Formulas 3 - 9 are equivalent to the following di�erentiation formulas, respectively:

d

dx

�� 1�cos (�x)

¶= sin (�x) ,

d

dx

�1

�sin (�x)

¶= cos (�x) ,

d

dxex = ex,

d

dx

�1

ln (a)ax¶=

1

ln (a)

d

dxax =

1

ln (a)(ln (a) ax) = ax,

d

dxcosh (x) = sinh (x) ,

d

dxsinh (x) = cosh (x) ,

d

dxarctan (x) =

1

x2 + 1,

Example 7

a) Determine Zx2/3dx

by the reverse power rule. Con�rm the result by di�erentiation.b) Compute Z 27

�8x2/3dx.

Interpret the integral as signed area.

Solution


a) By the reverse power rule,

Zx2/3dx =

x2/3+1

2/3 + 1=x5/3

5/3=3

5x5/3 + C,

where C is an arbitrary constant.We have

d

dx

�3

5x5/3 + C

¶=3

5

�5

3x2/3

¶= x2/3

for each x � R. Therefore, the statementZx2/3dx =

3

5x5/3 + C

is valid on R.

b) By the Fundamental Theorem of Calculus,

Z 27

�8x2/3dx =

3

5x5/3

¯̄̄¯ = 3

5

³275/3 � (�8)5/3

´=3

5

¡35 + 25

¢= 165.

Note that f is continuous on R, even though f is not di�erentiable at 0, so that there is noproblem about the existence of an integral of f , or the application of the Fundamental Theorem.Since x2/3 � 0 for each x, the area of the region between the graph of f (x) = x2/3 and theinterval [�8, 27] is 165. ¤

�8 27x

4

9

y

Figure 4: The region between the graph of y = x2/3 and [�8, 27]

Example 8

a) Determine Z1

x2dx,

and the intervals on which the expression is valid.b) Compute Z �1

�2

1

x2dx.

Interpret the integral as signed area.

Solution

a) By the reverse power rule,Z1

x2dx =

Zx�2dx =

x�1

�1 = �1

x+ C,


where C is an arbitrary constant. The expressionZ1

x2dx = �1

x+ C

is valid on the interval (��, 0) and on the interval (0,+�).b) By the Corollary to the Fundamental Theorem of Calculus,Z �1

�2

1

x2dx = �1

x

¯̄̄¯�1

�2=

�� 1

(�1)¶�� 1

(�2)¶= 1� 1

2=1

2.

Since 1/x2 > 0, the area of the region between the graph of f(x) = 1/x2 and the interval[�2,�1] is 0.5. Figure 5 illustrates the region. ¤

�2 �1 1 2x

5

10y

Figure 5

Example 9 Since x�2 > 0, the following claim cannot be valid:Z 2

�1

1

x2dx = �1

x

¯̄̄¯2

�1=

��12

¶� (1) = �3

2.

Why is the above line incorrect?

SolutionWe have

d

dx

�1

x2

¶= �1

x

if and only if x 6= 0. But 0 is in the interval [�1, 2], so that the Corollary to the FundamentalTheorem of Calculus (Corollary )cannot be applied as indicated above. ¤

Example 10 Evaluate Z �2

�4

1

xdx.

Solution

Since Z1

xdx = ln (|x|) + C,

on any interval contained in (��, 0) or (0,+�), and [�4,�2] is contained in (��, 0), we canuse the above inde�nite integral to evaluate the given de�nite integral. By the FundamentalTheorem of Calculus,Z �2

�4

1

xdx = ln (|x|)|�2�4 = ln (|�2|)� ln (|�4|) = ln (2)� ln(4) �= �0.693147


Thus, the signed area of the region between the graph of the function de�ned by 1/x and theinterval [�4,�2] is ln (2)� ln(4). The region is illustrated in Figure 6. ¤

�4 �2 2 4x

�4

�2

2

4

y

Figure 6

Remark 3 (Caution) We must be careful with the use of the antidi�erentiation formula,Z1

xdx = ln (|x|) + C.

For example, we might be tempted to write,Z 3

�2

1

xdx = ln (|x|)|3�2 = ln(3)� ln (2) .

The above statement is not valid since it is not true thatd

dxln (|x|) = 1

x

at 0, and 0 � (�2, 3). �Example 11 Evaluate Z ln(10)

0

exdx.

Solution

We have Zexdx = ex + C,

where C is an arbitrary constant. By the Fundamental Theorem of Calculus,Z ln(10)

0

exdx = ex|ln(10)x=0 = eln(10) � e0 = 10� 1 = 9.

Thus, the area of the region between the graph of the natural exponential function and theinterval [0, ln(10)] is 9. Figure 7 illustrates the region. ¤

�1 1x

10

20

y

ln�10�

Figure 7


Example 12 Con�rm the following claims that were made in Example 4 of Section 5.2:Z �

0

sin(x)dx = 2 andZ 4�/3

�

sin (x) dx = �12.

Solution

We have Zsin (x) dx = � cos (x) + C,

where C denotes an arbitrary constant, as usual. By the Fundamental Theorem of Calculus ,Z �

0

sin (x) dx = � cos (x)|�0 = � cos (�)� (� cos (0)) = 1 + 1 = 2,

and Z 4�/3

�

sin (x) dx = � cos (x)|4�/3� = � cos�4�

3

¶� (� cos (�)) = �

��12

¶� 1 = �1

2.

¤

The Fundamental Theorem of Calculus and One-Dimensional Motion

Let’s interpret the Fundamental Theorem of Calculus within the context of one-dimensionalmotion. Assume that f (t) is the position at time t of an object in one dimensional motion,and let v (t) be its instantaneous velocity at time t, so that v (t) = f 0 (t). Also assume that vis continuous on [a, b]. By the Fundamental Theorem of Calculus,

Z �

�

v(t)dt =

Z �

�

f 0(t)dt = f(b)� f(a).

We will refer to the change in the position of the object over the time time interval [a, b] as thedisplacement of the object over that time interval. Thus, the displacement of the objectover the time interval [a, b] is equal to the integral of the velocity function on [a, b].Even though the above fact is a direct consequence of the Fundamental Theorem of Calculus,it is helpful to interpret the proof of the theorem within the context of one-dimensional motion.If P = {t0, t1, t2, . . . , tn�1, tn} is a partition of [a, b], so that t0 = a and tn = b, we can expressthe displacement over [a, b] as the sum of the displacements over the subintervals. Thus,

f (b)� f (a) = f(tn)� f (t0)= [f(tn)� f(tn�1)] + [f (tn�1)� f (tn�2))] + · · ·+ [f (t2)� f (t1)] + [f (t1)� f (t0)]

=nXk=1

[f (tk)� f (tk�1)] .

By the Mean Value Theorem, there exists t�k � (tk�1, tk) such that

f (tk)� f (tk�1) = f 0 (t�k) (tk � tk�1) = v (t�k)�tk.Therefore,

Displacement over [a, b] =nXk=1

[f (tk)� f (tk�1)] =nXk=1

v (t�k)�tk.


SincenXk=1

v (t�k)�tk �=Z b

a

v (t) dt

if ||P || is small, and the approximation is as accurate as desired provided that ||P || is smallenough, ¯̄̄

¯̄Displacement over [a, b]�Z b

a

v (t) dt

¯̄̄¯̄

is arbitrarily small. This is the case if and only if

Displacement over [a, b] =Z b

a

v (t) dt.

In particular the units match. For example, if distance is measured in centimeters and time ismeasured in seconds, velocity is expressed in terms of centimeters per second. This is consistentwith the fact that

Displacement over [a, b] =Z b

a

v (t) dt �=nXk=1

v (t�k)�tk.

Indeed, the unit of v (t�k)�tk is

centimetersecond

× second = centimeter.

Graphically, the displacement of the object over the time interval [a, b] is the signed area ofthe region between the graph of the velocity function and the interval [a, b]. We must distinguishbetween the displacement of an object over a time interval and the distance traveled by theobject over the same time interval. If v (t) 0 for each t � [a, b], the object is moving in thenegative direction over the time interval [a, b]. Therefore, the distance traveled is

�Z b

a

v (t) dt.

More generally, if we wish to calculate the distance traveled by an object over the time interval[a, b], we need to determine the subintervals of [a, b] on which the velocity has constant sign.If the velocity is negative over a subinterval, the relevant integral must be multiplied by (�1).Graphically, the distance traveled over the time interval [a, b] is the area between the graph ofthe velocity function and the interval [a, b].

Example 13 With the above notation, assume that an object that is attached to a spring hasvelocity v (t) = cos (2t) .

a) Sketch the graph of the velocity function on [0, �].b) Determine the displacement of the object over the time interval [0, 3�/4].c) Determine the distance traveled by the object over the time interval [0, 3�/4].

Solution

a) Figure 8 shows the graph of the velocity function on [0, �].


Πx

�1

1y

Π

4

3 Π

4

Figure 8

b) Since Zcos (�t) dt =

1

�sin (�t) + C,

for any � 6= 0, we have Zcos (2t) dt =

1

2sin (2t) + C.

Therefore, the displacement of the object over the time interval [0, 3�/4] is

Z 3�/4

0

v (t) dt =

Z 3�/4

0

cos (2t) dt =1

2sin (2t)

¯̄̄¯3�/4

0

=1

2sin

�3�

2

¶� 12sin (0) = �1

2

(centimeters).

c) We see that v (t) > 0 if 0 < t < �/4 and v (t) < 0 if �/4 < t < 3�/4. Thus, the object ismoving in the positive direction over the time interval [0, �/4] and in the negative direction overthe time interval [�/4, 3�/4]. We have

Z �/4

0

v (t) dt =1

2sin (2t)

¯̄̄¯�/4

0

=1

2sin³�2

´� 12sin (0) =

1

2,

and Z 3�/4

�/4

v (t) dt =1

2sin (2t)

¯̄̄¯3�/4

�/4

=1

2sin

�3�

2

¶� 12sin³�2

´= �1

2� 12= �1.

Therefore, total distance traveled is

Z �/4

0

v (t) dt�Z 3�/4

�/4

v (t) dt =1

2� (�1) = 1

2

(centimeters). Graphically, the distance traveled is the area of the region between the velocityfunction and the interval [0, 3�/4]. ¤

Problems

In problems 1-4, evaluate the integral (make use the Fundamental Theorem of Calculus):


1. Z ��/2��/4

d

dxsin¡x2¢dx

2. Z 2

1

d

dx

p9� x2dx

3. Z �/2

�/3

d

dx

rcos³x2

´dx

4. Z 2

�3

d

dx

�1

x2 + 4

¶dx

In problems 5-22, use the Fundamental Theorem of Calculus to evaluate the integral, if applica-ble, otherwise state why the theorem does not apply.

5. Z 2

�3x2dx

6. Z 3

�2x�2dx

7. Z 9

4

�xdx

8. Z 27

0

x2/3dx

9. Z 9

1

1�xdx

10. Z 1

0

1

xdx

11. Z �/2

�/3

sin (x) dx

12. Z ��/6

��cos (x) dx

13. Z �/2

�/3

sin³x2

´dx

14. Z �/9

�/18

cos (9x) dx

15. Z �2

�4

1

xdx

16. Z 2

�1

1

xdx

17 Z e2

e�1

1

xdx

18. Z ln(10)

ln(3)

exdx

19. Z �ln(2)

0

d

dxe�x

2

dx

20. Z log10(4)

log10(2)

10xdx

21. Z 0

�1

d

dxarctan (x) dx

22. Z 1/�3

��3

1

1 + x2dx

23. Assume that the velocity of an object in one-dimensional motion is t at time t. Calculatethe displacement of the object over the time interval [2, 5].

24. Assume that the velocity of an object in one-dimensional motion is cos (4t) at time t.Calculate the displacement of the object over the time interval [�/6, �/4].

5.4 The Fundamental Theorem of Calculus: Part 2The second part of the Fundamental Theorem of Calculus shows that every continuous functionhas an antiderivative, even though such an antiderivative may not be expressible in terms offamiliar functions. The theorem leads to the de�nition of new special functions.


Some Properties of the Integral

In preparation for the second part of the Fundamental Theorem of Calculus, we will discusssome general facts about the integral that will be useful in other contexts as well.

Proposition 1 Assume that f and g are continuous on the interval [a, b] and f(x) � g(x)for each x [a, b]. Then Z �

�

f(x)dx �Z �

�

g(x)dx.

Figure 1 illustrates the graphical meaning of Proposition 1 if 0 f(x) < g(x) for each x � [a, b]:The area of the region between the graph of f and the interval [a, b] is less than the area of theregion between the graph of g and [a, b].

x

y

f

g

a b

Figure 1

We will leave the rigorous proof of Proposition 1 to a course in advanced calculus. Let’s providea plausibility argument:

Let P = {x0, x1, . . . , xn�1, xn} be a partition of [a, b] and x�k � [xk�1, xk], k = 1, 2, . . . , n. Wehave

nXk=1

f(x�k)�xk nXk=1

g (x�k)�xk,

since f(x) g(x) for each x � [a, b]. Since

nXk=1

f(x�k)�xk �=Z b

a

f(x)dx andnXk=1

g (x�k)�xk �=Z b

a

g(x)dx,

if ||P || = maxk�xk is small, it is plausible that

Z b

a

f(x)dx Z b

a

g(x)dx.

¥

Corollary 1 (The Triangle Inequality for Integrals) Assume that f is continuous on[a, b]. Then ¯̄̄

¯̄Z �

�

f(x)dx

¯̄̄¯̄ � Z �

�

|f(x)|dx.


Proof

It can be shown that |f | is continuous on [a, b] if f is continuous on [a, b]. We have

� |f (x)| f (x) |f (x)|for each x � [a, b]. By Proposition 1,Z b

a

�|f (x)| dx Z b

a

f (x) dx Z b

a

|f (x)| dx.

By the constant multiple rule for integrals,Z b

a

� |f (x)| dx = �Z b

a

|f (x)| dx.

Therefore,

�Z b

a

|f (x)| dx Z b

a

f (x) dx Z b

a

|f (x)| dx.The above inequalities imply that¯̄̄

¯̄Z b

a

f (x) dx

¯̄̄¯̄ Z b

a

|f (x)| dx.

¥We have dubbed the Corollary 1 as “the triangle inequality for integrals”, since we canview the inequality ¯̄̄

¯̄Z b

a

f (x) dx

¯̄̄¯̄ Z b

a

|f (x)| dx

as a generalization of the triangle inequality for numbers. Indeed, if P = {x0, x1, . . . , xn�1, xn}is a partition of [a, b] and x�k � [xk�1, xk] for k = 1, 2, . . . , n, we have¯̄̄

¯̄ nXk=1

f(x�k)�xk

¯̄̄¯̄ nX

k=1

|f(x�k)|�xk

by the triangle inequality for numbers. If ||P || = maxk�xk is small,¯̄̄¯̄ nXk=1

f(x�k)�xk

¯̄̄¯̄ �=

¯̄̄¯̄Z b

a

f(x)dx

¯̄̄¯̄

andnXk=1

|f(x�k)|�xk �=Z b

a

|f(x)| dx.

Therefore, the inequality ¯̄̄¯̄Z b

a

f(x)dx

¯̄̄¯̄ Z b

a

|f(x)| dx

is not surprising.

De�nition 1 The mean value (or the average value) of a continuous function f on theinterval [a, b] is

1

b� a

Z �

�

f(x)dx.


Thus, the mean value of f on [a, b] is the ratio of the integral of f on [a, b] and the length of theinterval [a, b].

The terminology of De�nition 1 is reasonable. Indeed, if

�x =b� an

, xk = a+ k�x, k = 1, 2, . . . , n,

thennXk=1

f (xk)�x �=Z b

a

f (x) dx

if �x is small, i.e., n is large. Therefore,

1

b� anXk=1

f (xk)�x �= 1

b� aZ b

a

f (x) dx.

We have1

b� anXk=1

f (xk)�x =1

b� anXk=1

f (xk)

�b� an

¶=1

n

nXk=1

f (xk) .

Therefore,1

n

nXk=1

f (xk) �= 1

b� aZ b

a

f (x) dx

if n is large. The quantity1

n

nXk=1

f (xk)

is the mean of the values of the function at the points xk, k = 1, 2, . . . , n.

A Continuous function attains its mean value on an interval:

Theorem 1 (THE MEAN VALUE THEOREM FOR INTEGRALS) Assume thatf is continuous on [a, b]. There exists c [a, b] such that

f(c) =1

b� a

Z �

�

f(x)dx.

Proof

Let m and M be the minimum and the maximum value of f on [a, b], respectively. Since

m f(x) M

for each x � [a, b], we have

Z b

a

mdx Z b

a

f(x)dx Z b

a

Mdx,

by Proposition 1. Therefore,

m (b� a) Z b

a

f(x)dx M(b� a).


Thus,

m 1

b� aZ b

a

f(x)dx M.

By the Intermediate Value Theorem for continuous functions (Theorem 1 of Section 2.9), thereexists c � [a, b] such that

f(c) =1

b� aZ b

a

f(x)dx.

¥

Since

f(c) =1

b� aZ b

a

f(x)dx f (c) (b� a) =Z b

a

f(x)dx,

we can interpret the Mean Value Theorem for Integrals in the case of a positive-valued functionf graphically: The area of the region between the graph of f and the interval [a, b] is the sameas the area of a rectangle that has as its base the interval [a, b] and has height equal to the valueof f at some c in [a, b], as illustrated in Figure 2.

x

y

a bc

f

y � f�c�

Figure 2

An integral is multiplied by (�1) if the upper and lower limits are interchanged:

De�nition 2 Assume that a < b. We de�neZ �

�

f(x)dx = �Z �

�

f(x)dx.

Remark 1 If F is an antiderivative of f , we have

Z a

b

f(x)dx = �Z b

a

f(x)dx = � (F (b)� F (a)) = F (a)� F (b).

Therefore, Z a

b

f(x)dx = F (x)|ab ,

just as Z b

a

f(x)dx = F (x)|ba .

Thus, we need not pay attention to the positions of a and b on the number line relative to eachother, when we make use of the Fundamental Theorem to evaluate the integral. �


Remark 2 By De�nition 2, if v(t) is the velocity at time t of an object in one-dimensionalmotion, and f is the corresponding position function, we haveZ a

b

v(t)dt = �Z b

a

v(t)dt = �Z b

a

f 0 (t) dt = � (f(b)� f(a)) = f(a)� f(b).

Thus, if we imagine that time �ows backwards from b to a, the integral of the velocity functionfrom b to a is still the change in the position function. �

Example 1 Determine Z 0

�/2

cos (x) dx.

Solution

By De�nition 2,

Z 0

�/2

cos (x) dx = �Z �/2

0

cos (x) dx = �ÃZ

cos (x) dx

¯̄̄¯x=�/2

x=0

!

= �³sin (x)|�/20

´= �

³sin³�2

´� sin (0)

´= �1.

We can obtain the same result as follows:Z 0

�/2

cos (x) dx = sin (x)|0�/2 = sin (0)� sin³�2

´= �1.

¤We de�ne an integral that has the same lower and upper limits to be 0:

De�nition 3 Z �

�

f(x)dx = 0.

The following argument suggests that the above de�nition is reasonable:Assume that f is continuous in some open interval that contains the point a and that |f (x)| Mfor each x in that interval. If the positive integer n is large enough,¯̄̄

¯̄Z a+1/n

a�1/nf (x) dx

¯̄̄¯̄ Z a+1/n

a�1/n|f(x)| dx

Z a+1/n

a�1/nMdx =M

�2

n

¶,

with the help of the triangle inequality for integrals. Therefore,

limn��

Z a+1/n

a�1/nf (x) dx = 0.

Thus, it is natural to set Z a

a

f(x)dx = limn��

Z a+1/n

a�1/nf (x) dx = 0.

¥The above de�nitions enable us to express the generalized version of the additivity of theintegral with respect to intervals:


Theorem 2 If f is continuous on an interval that contains the points a, b and c, wehave Z �

�

f(x)dx+

Z �

�

f(x)dx =

Z �

�

f(x)dx.

Proof

We know that the statement of Thorem 2 is valid if a < b < c. Assume that a < c < b. Then,Z c

a

f(x)dx+

Z b

c

f(x)dx =

Z b

a

f(x)dx.

Therefore, Z c

a

f(x)dx =

Z b

a

f(x)dx�Z b

c

f(x)dx =

Z b

a

f(x)dx��Z c

b

f(x)dx

¶

=

Z b

a

f(x)dx+

Z c

b

f(x)dx,

as claimed.

Let’s consider the case a = b < c. Then,Z b

a

f(x)dx+

Z c

b

f(x)dx =

Z a

a

f(x)dx+

Z c

a

f(x)dx = 0 +

Z c

a

f(x)dx =

Z c

a

f(x)dx.

Other cases are handled in a similar fashion. ¥

The Second Part of the Fundamental Theorem

We will de�ne functions via integrals. Assume that f is continuous on an interval J that containsthe point a. Let us set

F (x) =

Z x

a

f(t)dt

for each x � J . Note that the upper limit of the integral is the variable x, and we used the lettert to denote the “dummy” integration variable (we could have used any letter other than x). Ifx > a, then F (x) is the signed area of the region between the graph of f and the interval [a, x],as illustrated in Figure 3.

y

a x

y � f�t�

t

Figure 3: F (x) =R xaf(t)dt

If x < a, we have

F (x) = �Z a

x

f(x)dx,


so that F (x) is (�1) times the signed area of the region between the graph of f and the interval[a, x], as illustrated in Figure 4.

y

ax

y � f�t�

t

Figure 4: F (x) = � R axf(x)dx

Note that

F (a) =

Z a

a

f(t)dt = 0.

Example 2 Set

F (x) =

Z x

2

t2dt.

a) Determine F (x), F (3) and F (1).b) Interpret the meaning of F (x) graphically. Sketch the graph of F .c) Determine F 0 (x).

Solution

a) By the reverse power rule, Zt2dt =

t3

3+ C,

where C is an arbitrary constant. Therefore,

F (x) =

Z x

2

t2dt =t3

3

¯̄̄¯x

2

=x3

3� 2

3

3=1

3x3 � 8

3.

In particular,

F (3) =

Z 3

2

t2dt =19

3and F (1) =

Z 1

2

t2dt = �73.

b) We have

F (2) =

Z 2

2

t2dt = 0.

If x > 2, then F (x) is the area between the graph of f(t) = t2 and the interval [2, x], asillustrated in Figure 5.

�4 �2 2 4t

4

8

12

y

y � t2

x

Figure 5: F (x) =R x2t2dt


If x < 2, then

F (x) = �Z 2

x

t2dt.

Therefore, F (x) is (�1) times the area of the region between the graph of f (t) = t2 and theinterval [x, 2], as illustrated in Figure 6.

2t

4

8

y

y � t2

x

Figure 6: F (x) = � R 2xt2dt if x < 2

Figure 7 shows the graph of F .

�2 �1 1 2 3x

�6

�4

�2

2

4

6

y

y � F�x�

Figure 7: y = F (x) =R x2t2dt

c) We have

F 0 (x) =d

dx

Z x

2

t2dt =d

dx

�1

3x3 � 8

3

¶=1

3

¡3x2

¢= x2.

Note that x2 is the value of the integrand t2 at t = x. ¤

Example 3 Set

F (x) =

Z x

�

sin (t) dt.

a) Determine F (x), F (3�/2) and F (�/3).b) Interpret the meaning of F (x) graphically. Sketch the graph of F on the interval [0, 3�].c) Determine F 0 (x).

Solution

a) We have

F (x) =

Z x

�

sin (t) dt = � cos (t)|x� = � cos (x) + cos (�) = � cos (x)� 1.

In particular,

F (3�/2) =

Z 3�/2

�

sin (t) dt = �1 and F (�/3) =Z �/3

�

sin (t) dt = �32.


b) We have

F (�) =

Z �

�

sin (t) dt = 0.

If x > �, F (x) is the signed area of the region between the graph of sine and the interval [�, x],as illustrated in Figure 8.

3 Πt

�1

1y

2Π xΠ

Figure 8: F (x) =R x�sin (t) dt

If x < �, we have

F (x) =

Z x

�

sin (t) dt = �Z �

x

sin (t) dt.

Therefore, F (x) is (�1) times the signed area of the region between the graph of sine and theinterval [x, �], as illustrated in Figure 9.

�Π 2 Πt

�1

1y

xΠ

Figure 9: F (x) = � R �xsin (t) dt

Figure 10 shows the graph of y = F (x) = � cos (x)� 1 on the interval [�2�, 2�].

�2 Π �Π Π 2 Πx

�2

�1

1y

Figure 10: y = F (x) = � cos(x)� 1

c) We have

F 0 (x) =d

dx

Z x

�

sin (t) dt =d

dx(� cos (x)� 1) = sin (x) .


Note that sin (x) is the value of the integrand sin (t) at t = x. ¤In examples 2 and 3, it turned out that

d

dx

Z x

a

f(t)dt = f(x).

That is a general fact:

Theorem 3 (The Fundamental Theorem of Calculus, Part 2) Assume that f is con-tinuous on the interval J , and a is a given point in J . If

F (x) =

Z �

�

f(t)dt,

then F 0(x) = f(x) for each x J .

The derivative should be interpreted as the appropriate one-sided derivative at an endpoint ofJ .

Remark 3 The second part of the Fundamental Theorem of Calculus asserts that

d

dx

Z x

a

f (t) dt = f(x)

for each x � J (provided that f is continuous on J). Therefore,

F (x) =

Z �

�

f(t)dt

de�nes an antiderivative of f on J. �

A Plausibility Argument for Theorem 3:

We have

F (x+�x)� F (x) =Z x+�x

a

f(t)dt�Z x

a

f(t)dt =

Z x+�x

x

f(t)dt.

a x x � �x t�x

f�x�

Figure 11:R x+�x

xf (t) dt �= f (x)�x

With reference to Figure 11, if �x > 0 and small, this quantity is approximately the area of therectangle that has as its base the interval [x, x+�x] and has height f(x). Therefore

F (x+�x)� F (x) �= f(x)�x,so that

F (x+�x)� F (x)�x

�= f(x)


if �x is small. Thus, it is plausible that

F 0(x) = lim�x�0

F (x+�x)� F (x)�x

= f(x).

¥The Proof of Theorem 3

We will show that F 0 (x) = f (x) at a point x in the interior of J . If x is an endpoint of J ,the equality of the appropriate one-sided derivative of F and f (x) is established in a similarmanner.Let �x > 0. As in our plausibility argument,

F (x+�x)� F (x) =Z x+�x

x

f(t)dt.

Therefore,F (x+�x)� F (x)

�x=

1

�x

Z x+�x

x

f(t)dt.

Thus, the di�erence quotient is the mean value of f on the interval [x, x +�x]. By the MeanValue Theorem for Integrals (Theorem 1), there exists a point c (x,�x) in the interval [x, x+�x]such that

1

�x

Z x+�x

x

f(t)dt = f (c (x,�x))

(we have used the notation “c (x,�x)” in order to indicate that c depends on x and �x).Therefore,

F 0+ (x) = lim�x�0+

F (x+�x)� F (x)�x

= lim�x�0+

f (c (x,�x)) .

Since c (x,�x) is between x and x+�x, we have lim�x�0 c (x,�x) = x. Since f is continuousat x,

lim�x�0+

f (c (x,�x)) = lim�x�0

f (c (x,�x)) = f³lim

�x�0c (x,�x)

´= f (x) .

Therefore,F 0+ (x) = lim

�x�0+f (c (x,�x)) = f (x) .

If �x < 0

F (x+�x)� F (x)�x

=1

�x

Z x+�x

x

f(t)dt =1

(��x)

Ã�Z x+�x

x

f(t)dt

!=

1

(��x)Z x

x+�x

f(t)dt.

The �nal expression is the mean value of f on the interval [x + �x, x]. By the Mean ValueTheorem for Integrals, there exists c (x,�x) � [x+�x, x] such that

1

(��x)Z x

x+�x

f(t)dt = f (c (x,�x)) .

Therefore,

F 0� (x) = lim�x�0�

F (x+�x)� F (x)�x

= lim�x�0�

f (c (x,�x)) = f

�lim

�x�0�c (x,�x)

¶= f(x),

since c (x,�x) is between x+�x and x, and f is continuous at x.Thus,

F 0 (x) = F 0+(x) = F0�(x) = f(x).

¥


Example 4 (The function erf) Set

F (x) =

Z x

0

2��e�t

2

dt

a) Determine F 0 (x).b) Interpret F (x) in terms of area.

Solution

a) By the second part of the Fundamental Theorem of Calculus,

F 0 (x) =d

dx

Z x

0

2��e�t

2

dt =2��e�x

2

at each x � R, since

f (t) =2��e�t

2

is continuous on R.b) If x > 0, F (x) is the area between the graph of f and the interval [0, x], as illustrated inFigure 12.

�2 2t

1

y

x

Figure 12: F (x) =R x0

2��e�t

2

dt

If x < 0,

F (x) =

Z x

0

f (t) dt = �Z 0

x

f (t) dt,

so that F (x) is (�1) × (the area between the graph of f and the interval [x, 0]), as illustratedin Figure 13.

�2 2t

1

y

x

Figure 13: F (x) = � R 0x

2��e�t

2

dt

The function F is a built-in function in computer algebra systems such as Maple or Mathematica,since it occurs in many statistical applications, and is referred to as the error function erf.Figure 14 shows the graph of F . ¤


�4 �2 2 4x

�1

1

y

y � erf�x�

Figure 14: y = erf (x) =R x0

2��e�t

2

dt

Example 5 (The natural logarithm de�ned as an integral)

If x > 0, we have Z x

1

1

tdt = ln(t)|x1 = ln (x)� ln (1) = ln (x) ,

sinced

dtln (t) =

1

t, t > 0.

Thus, ln(x) is the area between the graph of y = 1/t and the interval [1, x] if x > 1, as illustratedin Figure 15.

2

2

4

y

x1 t

y �1

t

Figure 15: ln (x) =R x1

1t dt

If 0 < x < 1,

ln (x) = �Z 1

x

1

tdt,

so that ln (x) is (�1)× (the area between the graph of y = 1/t and the interval [x, 1]), as illus-trated in Figure 16.

2t

2

4

y

x t1

y �1

t

Figure 16: ln(x) = � R 1x

1t dt


If we didn’t know about the natural logarithm, and needed an antiderivative if 1/x, we couldhave set

F (x) =

Z x

1

1

tdt, x > 0.

By the second part of the Fundamental Theorem of Calculus, we have

F 0 (x) =d

dx

Z x

1

1

tdt =

1

x

for each x > 0, so that F is an antiderivative of the function de�ned by 1/x on (0,+�). Thus,we can introduce the natural logarithm as the function F and de�ne the natural exponentialfunction as its inverse. This approach enables us to derive all the properties of the naturallogarithm and the natural exponential function rigorously. This program is carried out inAppendix E. ¤

Example 6 The sine integral function Si is de�ned by the expression

Si (x) =

Z x

0

sin (t)

tdt

Determine Si0 (x).

Solution

Since

limt�0

sin (t)

t= 1,

if we set

f (t) =

��sin (t)

tif t 6= 0,

1 if t = 0,

then f is continuous on the entire number line. We can interpret the integralZ x

0

sin (t)

tdt

as Z x

0

f (t) dt.

�4 Π �2 Π 2 Π 4 Πt

0.5

1

y

Figure 17: y =sin (t)

t


Thus, the second part of the Fundamental Theorem of Calculus is applicable:

d

dxSi (x) =

d

dx

Z x

0

f (t) dt =

��

sin (x)

xif x = 0,

1 if x = 0.

Figure 17 shows the graph of f and Figure 18 shows the graph of the sine integral function Si.¤

�4 Π �2 Π 2 Π 4 Πx

�1.5

1.5

y

Figure 18: The sine integral function

Example 7 Set

F (x) =

Z x

0

t2 � 4t2 + 1

dt.

a) Determine F 0 (x).b) Determine the intervals on which F is increasing/decreasing,

Solution

a) The integrand is continuous on the entire number line, since it is a rational function andt2 + 1 6= 0 for any t � R. Figure 19 shows the graph of the integrand.

�6 �4 �2 2 4 6t

�4

�3

�2

�1

1y

Figure 19: y =t2 � 4t2 + 1

By the second part of the Fundamental Theorem of Calculus,

F 0 (x) =d

dx

Z x

0

t2 � 4t2 + 1

dt =x2 � 4x2 + 1

for each x � R.

b) We will make of the derivative test for monotonicity. We have

F 0 (x) = 0 x2 � 4x2 + 1

= 0 x = ±2.


We also have F 0 (x) > 0 if x < �2, F 0 (x) < 0 if �2 < x < 2 and F 0 (x) > 0 if x > 2. Therefore,F is increasing on (��, 2], decreasing on [�2, 2] and increasing on [2,+�).In the next chapter we will introduce new special functions and we will be able to express F (x)in terms of one of these functions. In the mean time, you can make use of your computationalutility to obtain approximates values for F (for example, you can use midpoint sums), and plota graph of F . Figure 20 shows such a graph.

�8 �4 4 8x

�4

�2

2

4

y

Figure 20: F (x) =R x0

t2 � 4t2 + 1

dt

Incidentally,

F (2) =

Z 2

0

t2 � 4t2 + 1

dt �= �3.535 74 and F (�2) =Z �2

0

t2 � 4t2 + 1

dt �= 3. 535 74

¤

Now that we have established the second part of the Fundamental Theorem of Calculus, let usdisplay both parts of the Theorem in a symmetric fashion (the restrictions on the functions havebeen stated earlier):

THE FUNDAMENTAL THEOREM OF CALCULUS

1. Z �

�

d f(t)

dtdt = f(x)� f(a).

2.d

dx

Z �

�

f(t)dt = f(x).

It is worthwhile to repeat the meaning of the Fundamental Theorem: The �rst part of thetheorem states that the integral of the derivative of a function on an interval is thedi�erence between the values of the function at the endpoints. The second part ofthe theorem states that the derivative of the functionZ x

a

f(t)dt

is the value of the integrand at the upper limit. We can say that di�erentiation andintegration are reverse operations in the precise sense of the Fundamental Theorem.

We may refer to either part of the Fundamental Theorem of Calculus simply as “the FundamentalTheorem of Calculus”.


Problems

In problems 1-8 , compute the indicated derivative.

1.d

dx

Z x

1

1

t4 + 1dt

2.d

dx

Z x

�

q4 + sin2 (t)dt

3.d

dx

Z 4

x

1�4 + u2

du

4.d

dx

Z sin(x)

2

�9� t2dt

5.d

dx

Z x

0

cos

��t2

2

¶dt

6.d

dx

Z �x

0

1q(1� t2) ¡1� 1

4 t2¢dt

7.d

dx

Z x2

x

t2�t2 + 1dt

8.d

dx

Z ln(x)

0

e�tdt

9. Let

F (x) =

Z x

0

cos¡t2¢dt.

Use the second derivative test for local extrema to determine the smallest a > 0 such that Fhas a local maximum at a, and the smallest b > 0 such that F has a local minimum at b.

10. Let

F (x) =

Z x

1

1

(t� 1)2 + 1dt.

a) Show that F is increasing on (��,+�).b) Determine the intervals on which the graph of F is concave up/concave down, and the pointof in�ection

11. Let

F (x) =

Z x

�/2

r1� 1

2sin2 (t)dt.

Express Z 5�/6

�/6

r1� 1

2sin2 (x)dx

in terms of F.

12. Let

F (x) =

Z x

1

sin (t)

tdt.

Express Z �

�/2

sin (x)

xdx.

in terms of F .

13. Let

F (x) =

Z x

0

1q(1� t2) ¡1� 1

4 t2¢dt.

Express Z 1/2

1/3

1q(1� x2) ¡1� 1

4x2¢dx

in terms of F .

Documents

Calculus I - Cognella Academic PublishingFirst published in the United States of America in 2011 by Cognella, a division of University Readers, Inc. Trademark Notice: Product or corporate