INSTRUCTOR
SOLUTIONS
MANUAL
Complete Solutions Manual
for
SINGLE VARIABLE CALCULUS
EARLY TRANSCENDENTALS
SEVENTH EDITION
DANIEL ANDERSON
University of Iowa
JEFFERY A. COLE
Anoka-Ramsey Community College
DANIEL DRUCKER
Wayne State University
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PREFACE
This Complete Solutions Manual contains solutions to all exercises in the text Single Variable
Calculus, Early Transcendentals, Seventh Edition, by James Stewart. A student version of this
manual is also available; it contains solutions to the odd-numbered exercises in each section, the
review sections, the True-False Quizzes, and the Problem Solving sections, as well as solutions to
all the exercises in the Concept Checks. No solutions to the projects appear in the student version.
It is our hope that by browsing through the solutions, professors will save time in determining
appropriate assignments for their particular class.
We use some nonstandard notation in order to save space. If you see a symbol that you don’t
recognize, refer to the Table of Abbreviations and Symbols on page iv.
We appreciate feedback concerning errors, solution correctness or style, and manual style. Any
comments may be sent directly to jeff.cole@anokaramsey.edu, or in care of the publisher:
Brooks/Cole, Cengage Learning, 20 Davis Drive, Belmont CA 94002-3098.
We would like to thank Stephanie Kuhns and Kathi Townes, of TECHarts, for their production
services; and Elizabeth Neustaetter, of Brooks/Cole, Cengage Learning, for her patience and support. All of these people have provided invaluable help in creating this manual.
Jeffery A. Cole
Anoka-Ramsey Community College
James Stewart
McMaster University
and University of Toronto
Daniel Drucker
Wayne State University
Daniel Anderson
University of Iowa
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ABBREVIATIONS AND SYMBOLS
CD
concave downward
CU
concave upward
D
the domain of i
FDT
First Derivative Test
HA
horizontal asymptote(s)
I
interval of convergence
IP
inflection point(s)
R
radius of convergence
VA
vertical asymptote(s)
CAS
=
indicates the use of a computer algebra system.
H
indicates the use of l’Hospital’s Rule.
m
indicates the use of Formula m in the Table of Integrals in the back endpapers.
s
indicates the use of the substitution {x = sin {> gx = cos { g{}.
=
=
=
c
=
indicates the use of the substitution {x = cos {> gx = 3 sin { g{}.
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CONTENTS
■
1
DIAGNOSTIC TESTS
■
1
FUNCTIONS AND MODELS
1.1
Four Ways to Represent a Function
1.2
Mathematical Models: A Catalog of Essential Functions
1.3
New Functions from Old Functions
1.4
Graphing Calculators and Computers
1.5
Exponential Functions
1.6
Inverse Functions and Logarithms
Review
■
■
27
38
51
69
75
2.1
The Tangent and Velocity Problems
2.2
The Limit of a Function
2.3
Calculating Limits Using the Limit Laws
2.4
The Precise Definition of a Limit
2.5
Continuity
2.6
Limits at Infinity; Horizontal Asymptotes
2.7
Derivatives and Rates of Change
2.8
The Derivative as a Function
Problems Plus
78
87
97
115
129
140
153
165
169
Derivatives of Polynomials and Exponential Functions
Applied Project
3.2
75
104
DIFFERENTIATION RULES
3.1
20
61
LIMITS AND DERIVATIVES
Review
3
9
46
Principles of Problem Solving
2
9
■
Building a Better Roller Coaster
The Product and Quotient Rules
182
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169
179
vi
■
CONTENTS
3.3
Derivatives of Trigonometric Functions
3.4
The Chain Rule
Applied Project
3.5
Where Should a Pilot Start Descent?
■
Implicit Differentiation
210
Laboratory Project
3.6
Derivatives of Logarithmic Functions
3.7
Rates of Change in the Natural and Social Sciences
3.8
Exponential Growth and Decay
3.9
Related Rates
3.10
Linear Approximations and Differentials
3.11
Problems Plus
262
285
297
297
The Calculus of Rainbows
■
4.2
The Mean Value Theorem
4.3
How Derivatives Affect the Shape of a Graph
4.4
Indeterminate Forms and l’Hospital’s Rule
4.5
Summary of Curve Sketching
4.6
Graphing with Calculus and Calculators
4.7
Optimization Problems
Applied Project
■
260
270
Applied Project
5
253
Taylor Polynomials
■
Maximum and Minimum Values
4.8
Newton’s Method
4.9
Antiderivatives
Review
309
314
338
350
374
394
The Shape of a Can
■
307
415
416
427
437
Problems Plus
461
INTEGRALS
471
5.1
Areas and Distances
5.2
The Definite Integral
Discovery Project
5.3
231
240
APPLICATIONS OF DIFFERENTIATION
4.1
224
225
245
Hyperbolic Functions
Review
■
Families of Implicit Curves
■
Laboratory Project
4
191
198
471
481
■
Area Functions
The Fundamental Theorem of Calculus
492
494
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210
CONTENTS
5.4
Indefinite Integrals and the Net Change Theorem
5.5
The Substitution Rule
Review
521
Problems Plus
6
■
531
APPLICATIONS OF INTEGRATION
6.1
Areas Between Curves
Applied Project
6.3
Volumes by Cylindrical Shells
6.4
Work
6.5
Average Value of a Function
566
576
582
Applied Project
■
Calculus and Baseball
Applied Project
■
Where To Sit at the Movies
585
586
587
595
TECHNIQUES OF INTEGRATION
603
7.1
Integration by Parts
7.2
Trigonometric Integrals
7.3
Trigonometric Substitution
7.4
Integration of Rational Functions by Partial Fractions
7.5
Strategy for Integration
7.6
Integration Using Tables and Computer Algebra Systems
603
615
624
■
Approximate Integration
7.8
Improper Integrals
Patterns in Integrals
664
673
675
689
703
Problems Plus
719
FURTHER APPLICATIONS OF INTEGRATION
8.1
635
652
7.7
Review
■
549
550
Discovery Project
8
The Gini Index
Volumes
Problems Plus
■
■
537
537
6.2
Review
7
505
511
Arc Length
727
Discovery Project
■
Arc Length Contest
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735
727
■
vii
viii
■
CONTENTS
8.2
Area of a Surface of Revolution
Discovery Project
8.3
Applications to Physics and Engineering
Discovery Project
Applications to Economics and Biology
8.5
Probability
Problems Plus
■
757
758
765
771
781
9.1
Modeling with Differential Equations
9.2
Direction Fields and Euler’s Method
9.3
Separable Equations
781
784
792
Applied Project
■
How Fast Does a Tank Drain?
Applied Project
■
Which Is Faster, Going Up or Coming Down?
9.4
Models for Population Growth
9.5
Linear Equations
9.6
Predator-Prey Systems
Problems Plus
■
744
761
DIFFERENTIAL EQUATIONS
Review
10
743
Complementary Coffee Cups
■
8.4
Review
9
735
Rotating on a Slant
■
805
807
817
824
829
837
PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.1
Curves Defined by Parametric Equations
Laboratory Project
10.2
■
Polar Coordinates
■
860
Bézier Curves
■
Families of Polar Curves
10.4
Areas and Lengths in Polar Coordinates
10.5
Conic Sections
10.6
Conic Sections in Polar Coordinates
Problems Plus
874
875
Laboratory Project
Review
843
Running Circles Around Circles
Calculus with Parametric Curves
Laboratory Project
10.3
806
893
905
916
922
935
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890
857
843
CONTENTS
11
■
11.1
Sequences
939
Laboratory Project
■
Logistic Sequences
952
11.2
Series
11.3
The Integral Test and Estimates of Sums
11.4
The Comparison Tests
11.5
Alternating Series
11.6
Absolute Convergence and the Ratio and Root Tests
11.7
Strategy for Testing Series
11.8
Power Series
11.9
Representations of Functions as Power Series
11.10
Taylor and Maclaurin Series
956
11.11
985
998
1002
■
Applied Project
Problems Plus
■
1011
1021
An Elusive Limit
Applications of Taylor Polynomials
Review
971
980
Laboratory Project
■
939
INFINITE SEQUENCES AND SERIES
1036
1037
Radiation from the Stars
1051
1052
1065
APPENDIXES
1077
A
Numbers, Inequalities, and Absolute Values
B
Coordinate Geometry and Lines
C
Graphs of Second-Degree Equations
D
Trigonometry
E
Sigma Notation
G
The Logarithm Defined as an Integral
H
Complex Numbers
1077
1082
1088
1092
1100
1104
1105
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■
ix
DIAGNOSTIC TESTS
Test A Algebra
1. (a) (33)4 = (33)(33)(33)(33) = 81
(c) 334 =
(e)
2 32
3
1
1
=
34
81
2
= 32 =
2. (a) Note that
(b) 334 = 3(3)(3)(3)(3) = 381
523
= 523321 = 52 = 25
521
1
1
1
1
(f ) 1633@4 = 3@4 = I
3 = 3 =
4
2
8
16
16
(d)
9
4
I
I
I
I
I
I
I
I
I
I
I
200 = 100 · 2 = 10 2 and 32 = 16 · 2 = 4 2. Thus 200 3 32 = 10 2 3 4 2 = 6 2.
(b) (3d3 e3 )(4de2 )2 = 3d3 e3 16d2 e4 = 48d5 e7
(c)
3{3@2 | 3
{2 | 31@2
32
=
{2 | 31@2
3{3@2 | 3
2
=
{
({2 | 31@2 )2
{4 | 31
{4
= 7
=
=
9{3 | 6
9{3 | 6 |
9|
(3{3@2 | 3 )2
3. (a) 3({ + 6) + 4(2{ 3 5) = 3{ + 18 + 8{ 3 20 = 11{ 3 2
(b) ({ + 3)(4{ 3 5) = 4{2 3 5{ + 12{ 3 15 = 4{2 + 7{ 3 15
(c)
I
I 2
I I
I I
I I 2 I I
d+ e
d3 e =
d 3 d e+ d e3
e =d3e
Or: Use the formula for the difference of two squares to see that
I
I I
I I 2 I 2
d+ e
d3 e =
d 3
e = d 3 e.
(d) (2{ + 3)2 = (2{ + 3)(2{ + 3) = 4{2 + 6{ + 6{ + 9 = 4{2 + 12{ + 9.
Note: A quicker way to expand this binomial is to use the formula (d + e)2 = d2 + 2de + e2 with d = 2{ and e = 3:
(2{ + 3)2 = (2{)2 + 2(2{)(3) + 32 = 4{2 + 12{ + 9
(e) See Reference Page 1 for the binomial formula (d + e)3 = d3 + 3d2 e + 3de2 + e3 . Using it, we get
({ + 2)3 = {3 + 3{2 (2) + 3{(22 ) + 23 = {3 + 6{2 + 12{ + 8.
4. (a) Using the difference of two squares formula, d2 3 e2 = (d + e)(d 3 e), we have
4{2 3 25 = (2{)2 3 52 = (2{ + 5)(2{ 3 5).
(b) Factoring by trial and error, we get 2{2 + 5{ 3 12 = (2{ 3 3)({ + 4).
(c) Using factoring by grouping and the difference of two squares formula, we have
{3 3 3{2 3 4{ + 12 = {2 ({ 3 3) 3 4({ 3 3) = ({2 3 4)({ 3 3) = ({ 3 2)({ + 2)({ 3 3).
(d) {4 + 27{ = {({3 + 27) = {({ + 3)({2 3 3{ + 9)
This last expression was obtained using the sum of two cubes formula, d3 + e3 = (d + e)(d2 3 de + e2 ) with d = {
and e = 3. [See Reference Page 1 in the textbook.]
(e) The smallest exponent on { is 3 12 , so we will factor out {31@2 .
3{3@2 3 9{1@2 + 6{31@2 = 3{31@2 ({2 3 3{ + 2) = 3{31@2 ({ 3 1)({ 3 2)
(f ) {3 | 3 4{| = {|({2 3 4) = {|({ 3 2)({ + 2)
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°
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1
2
¤
5. (a)
(b)
(c)
NOT FOR SALE
DIAGNOSTIC TESTS
({ + 1)({ + 2)
{+2
{2 + 3{ + 2
=
=
{2 3 { 3 2
({ + 1)({ 3 2)
{32
(2{ + 1)({ 3 1) { + 3
{31
2{2 3 { 3 1 { + 3
·
=
·
=
{2 3 9
2{ + 1
({ 3 3)({ + 3) 2{ + 1
{33
{2
{+1
{2
{+1
{2
{+1 {32
{2 3 ({ + 1)({ 3 2)
3
=
3
=
3
·
=
34
{+2
({ 3 2)({ + 2)
{+2
({ 3 2)({ + 2)
{+2 {32
({ 3 2)({ + 2)
{2
=
{2 3 ({2 3 { 3 2)
{+2
1
=
=
({ + 2)({ 3 2)
({ + 2)({ 3 2)
{32
|
|
{
{
3
3
| 2 3 {2
(| 3 {)(| + {)
|+{
{
|
{
| {|
=
·
=
=
=
= 3({ + |)
(d)
1
1
1
1 {|
{3|
3(| 3 {)
31
3
3
|
{
|
{
I
I
I
I
I
I
I
I
I
10
10
5+2
50 + 2 10
5 2 + 2 10
= 5 2 + 2 10
=
= I
·I
= I 2
2
5
3
4
532
532
5+2
5 32
6. (a) I
I
I
I
4+k32
4+k32
4+k+2
4+k34
k
1
= I
= I
= I
=
·I
(b)
k
k
4+k+2
k 4+k+2
k 4+k+2
4+k+2
7. (a) {2 + { + 1 = {2 + { +
1
4
+13
1
4
2
= { + 12 +
3
4
(b) 2{2 3 12{ + 11 = 2({2 3 6{) + 11 = 2({2 3 6{ + 9 3 9) + 11 = 2({2 3 6{ + 9) 3 18 + 11 = 2({ 3 3)2 3 7
8. (a) { + 5 = 14 3 12 {
(b)
2{
2{ 3 1
=
{+1
{
C { + 12 { = 14 3 5 C
3
{
2
=9 C {=
2
3
·9 C {=6
i 2{2 = (2{ 3 1)({ + 1) C 2{2 = 2{2 + { 3 1 C { = 1
(c) {2 3 { 3 12 = 0 C ({ + 3)({ 3 4) = 0 C { + 3 = 0 or { 3 4 = 0 C { = 33 or { = 4
(d) By the quadratic formula, 2{2 + 4{ + 1 = 0 C
I s
I
I
I
2 32 ± 2
34 ± 42 3 4(2)(1)
34 ± 8
34 ± 2 2
32 ± 2
=
=
=
=
= 31 ±
{=
2(2)
4
4
4
2
1
2
I
2.
(e) {4 3 3{2 + 2 = 0 C ({2 3 1)({2 3 2) = 0 C {2 3 1 = 0 or {2 3 2 = 0 C {2 = 1 or {2 = 2 C
I
{ = ±1 or { = ± 2
(f ) 3 |{ 3 4| = 10 C |{ 3 4| =
10
3
C { 3 4 = 3 10
3 or { 3 4 =
(g) Multiplying through 2{(4 3 {)31@2 3 3
C {=
2
3
or { =
22
3
I
4 3 { = 0 by (4 3 {)1@2 gives 2{ 3 3(4 3 {) = 0 C
2{ 3 12 + 3{ = 0 C 5{ 3 12 = 0 C 5{ = 12 C { =
9. (a) 34 ? 5 3 3{ $ 17
10
3
12
.
5
C 39 ? 33{ $ 12 C 3 A { D 34 or 34 $ { ? 3.
In interval notation, the answer is [34> 3).
(b) {2 ? 2{ + 8 C {2 3 2{ 3 8 ? 0 C ({ + 2)({ 3 4) ? 0. Now, ({ + 2)({ 3 4) will change sign at the critical
values { = 32 and { = 4. Thus the possible intervals of solution are (3"> 32), (32> 4), and (4> "). By choosing a
single test value from each interval, we see that (32> 4) is the only interval that satis¿es the inequality.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
TEST B ANALYTIC GEOMETRY
¤
3
(c) The inequality {({ 3 1)({ + 2) A 0 has critical values of 32> 0> and 1. The corresponding possible intervals of solution
are (3"> 32), (32> 0), (0> 1) and (1> "). By choosing a single test value from each interval, we see that both intervals
(32> 0) and (1> ") satisfy the inequality. Thus, the solution is the union of these two intervals: (32> 0) (1> ").
(d) |{ 3 4| ? 3 C 33 ? { 3 4 ? 3 C 1 ? { ? 7. In interval notation, the answer is (1> 7).
2{ 3 3
$1 C
{+1
2{ 3 3
31 $0 C
{+1
Now, the expression
{34
may change signs at the critical values { = 31 and { = 4, so the possible intervals of solution
{+1
(e)
2{ 3 3
{+1
3
$0 C
{+1
{+1
2{ 3 3 3 { 3 1
$0 C
{+1
{34
$ 0.
{+1
are (3"> 31), (31> 4], and [4> "). By choosing a single test value from each interval, we see that (31> 4] is the only
interval that satis¿es the inequality.
10. (a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick
s = 1 and t = 2 and observe that (1 + 2)2 6= 12 + 22 . In general, (s + t)2 = s2 + 2st + t 2 .
(b) True as long as d and e are nonnegative real numbers. To see this, think in terms of the laws of exponents:
I I
I
de = (de)1@2 = d1@2 e1@2 = d e.
(c) False. To see this, let s = 1 and t = 2, then
I
12 + 22 6= 1 + 2.
(d) False. To see this, let W = 1 and F = 2, then
(e) False. To see this, let { = 2 and | = 3, then
(f ) True since
1 + 1(2)
6= 1 + 1.
2
1
1
1
6= 3 .
233
2
3
{
1
1@{
· =
, as long as { 6= 0 and d 3 e 6= 0.
d@{ 3 e@{ {
d3e
Test B Analytic Geometry
1. (a) Using the point (2> 35) and p = 33 in the point-slope equation of a line, | 3 |1 = p({ 3 {1 ), we get
| 3 (35) = 33({ 3 2) i | + 5 = 33{ + 6 i | = 33{ + 1.
(b) A line parallel to the {-axis must be horizontal and thus have a slope of 0. Since the line passes through the point (2> 35),
the |-coordinate of every point on the line is 35, so the equation is | = 35.
(c) A line parallel to the |-axis is vertical with unde¿ned slope. So the {-coordinate of every point on the line is 2 and so the
equation is { = 2.
(d) Note that 2{ 3 4| = 3 i 34| = 32{ + 3 i | = 12 { 3 34 . Thus the slope of the given line is p = 12 . Hence, the
slope of the line we’re looking for is also
So the equation of the line is | 3 (35) =
1
2
(since the line we’re looking for is required to be parallel to the given line).
1
2 ({
3 2) i | + 5 = 12 { 3 1 i | = 12 { 3 6.
2. First we’ll ¿nd the distance between the two given points in order to obtain the radius, u, of the circle:
u=
s
s
I
[3 3 (31)]2 + (32 3 4)2 = 42 + (36)2 = 52. Next use the standard equation of a circle,
({ 3 k)2 + (| 3 n)2 = u2 , where (k> n) is the center, to get ({ + 1)2 + (| 3 4)2 = 52.
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°
licated, or posted to a publicly accessible website, in whole or in part.
par
© Cengage Learning. All Rights Reserved.
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¤
NOT FOR SALE
DIAGNOSTIC TESTS
3. We must rewrite the equation in standard form in order to identify the center and radius. Note that
{2 + | 2 3 6{ + 10| + 9 = 0 i {2 3 6{ + 9 + | 2 + 10| = 0. For the left-hand side of the latter equation, we
factor the ¿rst three terms and complete the square on the last two terms as follows: {2 3 6{ + 9 + | 2 + 10| = 0 i
({ 3 3)2 + | 2 + 10| + 25 = 25 i ({ 3 3)2 + (| + 5)2 = 25. Thus, the center of the circle is (3> 35) and the radius is 5.
4. (a) D(37> 4) and E(5> 312)
i pDE =
312 3 4
316
4
=
=3
5 3 (37)
12
3
(b) | 3 4 = 3 43 [{ 3 (37)] i | 3 4 = 3 43 { 3
28
3
i 3| 3 12 = 34{ 3 28 i 4{ + 3| + 16 = 0. Putting | = 0,
we get 4{ + 16 = 0, so the {-intercept is 34, and substituting 0 for { results in a |-intercept of 3 16
.
3
(c) The midpoint is obtained by averaging the corresponding coordinates of both points:
(d) g =
37+5 4+(312)
2 >
2
s
s
I
I
[5 3 (37)]2 + (312 3 4)2 = 122 + (316)2 = 144 + 256 = 400 = 20
= (31> 34).
(e) The perpendicular bisector is the line that intersects the line segment DE at a right angle through its midpoint. Thus the
perpendicular bisector passes through (31> 34) and has slope
3
4
[the slope is obtained by taking the negative reciprocal of
the answer from part (a)]. So the perpendicular bisector is given by | + 4 = 34 [{ 3 (31)] or 3{ 3 4| = 13.
(f ) The center of the required circle is the midpoint of DE, and the radius is half the length of DE, which is 10. Thus, the
equation is ({ + 1)2 + (| + 4)2 = 100.
5. (a) Graph the corresponding horizontal lines (given by the equations | = 31 and
| = 3) as solid lines. The inequality | D 31 describes the points ({> |) that lie
on or above the line | = 31. The inequality | $ 3 describes the points ({> |)
that lie on or below the line | = 3. So the pair of inequalities 31 $ | $ 3
describes the points that lie on or between the lines | = 31 and | = 3.
(b) Note that the given inequalities can be written as 34 ? { ? 4 and 32 ? | ? 2,
respectively. So the region lies between the vertical lines { = 34 and { = 4 and
between the horizontal lines | = 32 and | = 2. As shown in the graph, the
region common to both graphs is a rectangle (minus its edges) centered at the
origin.
(c) We ¿rst graph | = 1 3 12 { as a dotted line. Since | ? 1 3 12 {, the points in the
region lie below this line.
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TEST C FUNCTIONS
¤
5
(d) We ¿rst graph the parabola | = {2 3 1 using a solid curve. Since | D {2 3 1,
the points in the region lie on or above the parabola.
s
(e) We graph the circle {2 + | 2 = 4 using a dotted curve. Since {2 + | 2 ? 2, the
region consists of points whose distance from the origin is less than 2, that is,
the points that lie inside the circle.
(f ) The equation 9{2 + 16| 2 = 144 is an ellipse centered at (0> 0). We put it in
standard form by dividing by 144 and get
|2
{2
+
= 1. The {-intercepts are
16
9
I
located at a distance of 16 = 4 from the center while the |-intercepts are a
I
distance of 9 = 3 from the center (see the graph).
Test C Functions
1. (a) Locate 31 on the {-axis and then go down to the point on the graph with an {-coordinate of 31. The corresponding
|-coordinate is the value of the function at { = 31, which is 32. So, i (31) = 32.
(b) Using the same technique as in part (a), we get i (2) E 2=8.
(c) Locate 2 on the |-axis and then go left and right to ¿nd all points on the graph with a |-coordinate of 2. The corresponding
{-coordinates are the {-values we are searching for. So { = 33 and { = 1.
(d) Using the same technique as in part (c), we get { E 32=5 and { E 0=3.
(e) The domain is all the {-values for which the graph exists, and the range is all the |-values for which the graph exists.
Thus, the domain is [33> 3], and the range is [32> 3].
2. Note that i (2 + k) = (2 + k)3 and i(2) = 23 = 8. So the difference quotient becomes
(2 + k)3 3 8
8 + 12k + 6k2 + k3 3 8
12k + 6k2 + k3
k(12 + 6k + k2 )
i (2 + k) 3 i (2)
=
=
=
=
= 12 + 6k + k2 .
k
k
k
k
k
3. (a) Set the denominator equal to 0 and solve to ¿nd restrictions on the domain: {2 + { 3 2 = 0
i
({ 3 1)({ + 2) = 0 i { = 1 or { = 32. Thus, the domain is all real numbers except 1 or 32 or, in interval
notation, (3"> 32) (32> 1) (1> ").
(b) Note that the denominator is always greater than or equal to 1, and the numerator is de¿ned for all real numbers. Thus, the
domain is (3"> ").
(c) Note that the function k is the sum of two root functions. So k is de¿ned on the intersection of the domains of these two
root functions. The domain of a square root function is found by setting its radicand greater than or equal to 0. Now,
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¤
DIAGNOSTIC TESTS
4 3 { D 0 i { $ 4 and {2 3 1 D 0 i ({ 3 1)({ + 1) D 0 i { $ 31 or { D 1. Thus, the domain of
k is (3"> 31] [1> 4].
4. (a) ReÀect the graph of i about the {-axis.
(b) Stretch the graph of i vertically by a factor of 2, then shift 1 unit downward.
(c) Shift the graph of i right 3 units, then up 2 units.
5. (a) Make a table and then connect the points with a smooth curve:
{
32
31
0
1
2
|
38
31
0
1
8
(b) Shift the graph from part (a) left 1 unit.
(c) Shift the graph from part (a) right 2 units and up 3 units.
(d) First plot | = {2 . Next, to get the graph of i ({) = 4 3 {2 ,
reÀect i about the x-axis and then shift it upward 4 units.
(e) Make a table and then connect the points with a smooth curve:
{
0
1
4
9
|
0
1
2
3
(f ) Stretch the graph from part (e) vertically by a factor of two.
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TEST D TRIGONOMETRY
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(g) First plot | = 2{ . Next, get the graph of | = 32{ by reÀecting the graph of
| = 2{ about the x-axis.
(h) Note that | = 1 + {31 = 1 + 1@{. So ¿rst plot | = 1@{ and then shift it
upward 1 unit.
6. (a) i (32) = 1 3 (32)2 = 33 and i (1) = 2(1) + 1 = 3
(b) For { $ 0 plot i ({) = 1 3 {2 and, on the same plane, for { A 0 plot the graph
of i ({) = 2{ + 1.
7. (a) (i j)({) = i (j({)) = i(2{ 3 3) = (2{ 3 3)2 + 2(2{ 3 3) 3 1 = 4{2 3 12{ + 9 + 4{ 3 6 3 1 = 4{2 3 8{ + 2
(b) (j i)({) = j(i ({)) = j({2 + 2{ 3 1) = 2({2 + 2{ 3 1) 3 3 = 2{2 + 4{ 3 2 3 3 = 2{2 + 4{ 3 5
(c) (j j j)({) = j(j(j({))) = j(j(2{ 3 3)) = j(2(2{ 3 3) 3 3) = j(4{ 3 9) = 2(4{ 3 9) 3 3
= 8{ 3 18 3 3 = 8{ 3 21
Test D Trigonometry
300
5
=
=
180
180
3
5 180
5
=
= 150
2. (a)
6
6
18
=3
=3
180
180
10
360
180
=
E 114=6
(b) 2 = 2
1. (a) 300 = 300
(b) 318 = 318
3. We will use the arc length formula, v = u, where v is arc length, u is the radius of the circle, and is the measure of the
central angle in radians. First, note that 30 = 30
4. (a) tan(@3) =
= . So v = (12)
= 2 cm.
180
6
6
I I 3 You can read the value from a right triangle with sides 1, 2, and 3.
(b) Note that 7@6 can be thought of as an angle in the third quadrant with reference angle @6. Thus, sin(7@6) = 3 12 ,
since the sine function is negative in the third quadrant.
(c) Note that 5@3 can be thought of as an angle in the fourth quadrant with reference angle @3. Thus,
sec(5@3) =
1
1
=
= 2, since the cosine function is positive in the fourth quadrant.
cos(5@3)
1@2
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DIAGNOSTIC TESTS
5. sin = d@24
6. sin { =
1
3
i d = 24 sin and cos = e@24 i e = 24 cos and sin2 { + cos2 { = 1 i cos { =
So, using the sum identity for the sine, we have
t
13
1
9
=
2
I
2
. Also, cos | =
3
4
5
i sin | =
t
13
16
25
I
I
I 2 2 3
4+6 2
1
1 4
· =
=
4+6 2
sin({ + |) = sin { cos | + cos { sin | = · +
3 5
3
5
15
15
7. (a) tan sin + cos =
(b)
sin sin2 cos2 1
sin + cos =
+
=
= sec cos cos cos cos 2 sin {@(cos {)
sin {
2 tan {
=
=2
cos2 { = 2 sin { cos { = sin 2{
1 + tan2 {
sec2 {
cos {
8. sin 2{ = sin {
C 2 sin { cos { = sin { C 2 sin { cos { 3 sin { = 0 C sin { (2 cos { 3 1) = 0 C
sin { = 0 or cos { =
1
2
i { = 0,
,
3
,
5
,
3
2.
9. We ¿rst graph | = sin 2{ (by compressing the graph of sin {
by a factor of 2) and then shift it upward 1 unit.
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1
FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function
1. The functions i ({) = { +
I
I
2 3 { and j(x) = x + 2 3 x give exactly the same output values for every input value, so i
and j are equal.
2. i ({) =
{({ 3 1)
{2 3 {
=
= { for { 3 1 6= 0, so i and j [where j({) = {] are not equal because i (1) is unde¿ned and
{31
{31
j(1) = 1.
3. (a) The point (1> 3) is on the graph of i, so i (1) = 3.
(b) When { = 31, | is about 30=2, so i (31) E 30=2.
(c) i ({) = 1 is equivalent to | = 1= When | = 1, we have { = 0 and { = 3.
(d) A reasonable estimate for { when | = 0 is { = 30=8.
(e) The domain of i consists of all {-values on the graph of i . For this function, the domain is 32 $ { $ 4, or [32> 4].
The range of i consists of all |-values on the graph of i . For this function, the range is 31 $ | $ 3, or [31> 3].
(f ) As { increases from 32 to 1, | increases from 31 to 3. Thus, i is increasing on the interval [32> 1].
4. (a) The point (34> 32) is on the graph of i , so i (34) = 32. The point (3> 4) is on the graph of j, so j(3) = 4.
(b) We are looking for the values of { for which the |-values are equal. The |-values for i and j are equal at the points
(32> 1) and (2> 2), so the desired values of { are 32 and 2.
(c) i ({) = 31 is equivalent to | = 31. When | = 31, we have { = 33 and { = 4.
(d) As { increases from 0 to 4, | decreases from 3 to 31. Thus, i is decreasing on the interval [0> 4].
(e) The domain of i consists of all {-values on the graph of i . For this function, the domain is 34 $ { $ 4, or [34> 4].
The range of i consists of all |-values on the graph of i . For this function, the range is 32 $ | $ 3, or [32> 3].
(f ) The domain of j is [34> 3] and the range is [0=5> 4].
5. From Figure 1 in the text, the lowest point occurs at about (w> d) = (12> 385). The highest point occurs at about (17> 115).
Thus, the range of the vertical ground acceleration is 385 $ d $ 115. Written in interval notation, we get [385> 115].
6. Example 1: A car is driven at 60 mi@h for 2 hours. The distance g
traveled by the car is a function of the time w. The domain of the
function is {w | 0 $ w $ 2}, where w is measured in hours. The range
of the function is {g | 0 $ g $ 120}, where g is measured in miles.
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10
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CHAPTER 1
FUNCTIONS AND MODELS
Example 2: At a certain university, the number of students Q on
campus at any time on a particular day is a function of the time w after
midnight. The domain of the function is {w | 0 $ w $ 24}, where w is
measured in hours. The range of the function is {Q | 0 $ Q $ n},
where Q is an integer and n is the largest number of students on
campus at once.
Example 3: A certain employee is paid $8=00 per hour and works a
pay
maximum of 30 hours per week. The number of hours worked is
240
238
236
rounded down to the nearest quarter of an hour. This employee’s
gross weekly pay S is a function of the number of hours worked k.
The domain of the function is [0> 30] and the range of the function is
4
2
{0> 2=00> 4=00> = = = > 238=00> 240=00}.
0
0.25 0.50 0.75
29.50 29.75 30
hours
7. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails
the Vertical Line Test.
8. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [32> 2] and the range
is [31> 2].
9. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [33> 2] and the range
is [33> 32) [31> 3].
10. No, the curve is not the graph of a function since for { = 0, ±1, and ±2, there are in¿nitely many points on the curve.
11. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person’s weight
dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual
increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.
12. First, the tub was ¿lled with water to a height of 15 in. Then a person got into the tub, raising the water level to 20 in. At
around 12 minutes, the person stood up in the tub but then immediately sat down. Finally, at around 17 minutes, the person got
out of the tub, and then drained the water.
13. The water will cool down almost to freezing as the ice melts. Then, when
the ice has melted, the water will slowly warm up to room temperature.
14. Runner A won the race, reaching the ¿nish line at 100 meters in about 15 seconds, followed by runner B with a time of about
19 seconds, and then by runner C who ¿nished in around 23 seconds. B initially led the race, followed by C, and then A.
C then passed B to lead for a while. Then A passed ¿rst B, and then passed C to take the lead and ¿nish ¿rst. Finally,
B passed C to ¿nish in second place. All three runners completed the race.
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SECTION 1.1
FOUR WAYS TO REPRESENT A FUNCTION
¤
11
15. (a) The power consumption at 6 AM is 500 MW> which is obtained by reading the value of power S when w = 6 from the
graph. At 6 PM we read the value of S when w = 18> obtaining approximately 730 MW=
(b) The minimum power consumption is determined by ¿nding the time for the lowest point on the graph, w = 4> or 4 AM. The
maximum power consumption corresponds to the highest point on the graph, which occurs just before w = 12> or right
before noon. These times are reasonable, considering the power consumption schedules of most individuals and
businesses.
16. The summer solstice (the longest day of the year) is
17. Of course, this graph depends strongly on the
around June 21, and the winter solstice (the shortest day)
geographical location!
is around December 22. (Exchange the dates for the
southern hemisphere.)
18. The value of the car decreases fairly rapidly initially, then
19. As the price increases, the amount sold
somewhat less rapidly.
decreases.
20. The temperature of the pie would increase rapidly, level
21.
off to oven temperature, decrease rapidly, and then level
off to room temperature.
22. (a)
(b)
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CHAPTER 1
FUNCTIONS AND MODELS
(c)
(d)
23. (a)
(b) From the graph, we estimate the number of US cell-phone
subscribers to be about 126 million in 2001 and 207 million
in 2005.
24. (a)
(b) From the graph in part (a), we estimate the temperature at
9:00 AM to be about 87 F
25. i ({) = 3{2 3 { + 2=
i (2) = 3(2)2 3 2 + 2 = 12 3 2 + 2 = 12=
i (32) = 3(32)2 3 (32) + 2 = 12 + 2 + 2 = 16=
i (d) = 3d2 3 d + 2=
i (3d) = 3(3d)2 3 (3d) + 2 = 3d2 + d + 2=
i (d + 1) = 3(d + 1)2 3 (d + 1) + 2 = 3(d2 + 2d + 1) 3 d 3 1 + 2 = 3d2 + 6d + 3 3 d + 1 = 3d2 + 5d + 4=
2i (d) = 2 · i (d) = 2(3d2 3 d + 2) = 6d2 3 2d + 4=
i (2d) = 3(2d)2 3 (2d) + 2 = 3(4d2 ) 3 2d + 2 = 12d2 3 2d + 2=
i (d2 ) = 3(d2 )2 3 (d2 ) + 2 = 3(d4 ) 3 d2 + 2 = 3d4 3 d2 + 2=
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SECTION 1.1
FOUR WAYS TO REPRESENT A FUNCTION
¤
13
2 [i (d)]2 = 3d2 3 d + 2 = 3d2 3 d + 2 3d2 3 d + 2
= 9d4 3 3d3 + 6d2 3 3d3 + d2 3 2d + 6d2 3 2d + 4 = 9d4 3 6d3 + 13d2 3 4d + 4=
i (d + k) = 3(d + k)2 3 (d + k) + 2 = 3(d2 + 2dk + k2 ) 3 d 3 k + 2 = 3d2 + 6dk + 3k2 3 d 3 k + 2=
26. A spherical balloon with radius u + 1 has volume Y (u + 1) =
4
(u
3
+ 1)3 = 43 (u3 + 3u2 + 3u + 1). We wish to ¿nd the
amount of air needed to inÀate the balloon from a radius of u to u + 1. Hence, we need to ¿nd the difference
Y (u + 1) 3 Y (u) = 43 (u3 + 3u2 + 3u + 1) 3 43 u3 = 43 (3u2 + 3u + 1).
27. i ({) = 4 + 3{ 3 {2 , so i (3 + k) = 4 + 3(3 + k) 3 (3 + k)2 = 4 + 9 + 3k 3 (9 + 6k + k2 ) = 4 3 3k 3 k2 ,
and
(4 3 3k 3 k2 ) 3 4
k(33 3 k)
i (3 + k) 3 i (3)
=
=
= 33 3 k.
k
k
k
28. i ({) = {3 , so i (d + k) = (d + k)3 = d3 + 3d2 k + 3dk2 + k3 ,
and
29.
(d3 + 3d2 k + 3dk2 + k3 ) 3 d3
k(3d2 + 3dk + k2 )
i (d + k) 3 i (d)
=
=
= 3d2 + 3dk + k2 .
k
k
k
1
d3{
1
3
i ({) 3 i (d)
d = {d = d 3 { = 31({ 3 d) = 3 1
= {
{3d
{3d
{3d
{d({ 3 d)
{d({ 3 d)
d{
{+3
{ + 3 3 2({ + 1)
32
i ({) 3 i (1)
{ + 3 3 2{ 3 2
{
+
1
{+1
=
=
=
30.
{31
{31
{31
({ + 1)({ 3 1)
=
3({ 3 1)
1
3{ + 1
=
=3
({ + 1)({ 3 1)
({ + 1)({ 3 1)
{+1
31. i ({) = ({ + 4)@({2 3 9) is de¿ned for all { except when 0 = {2 3 9
C 0 = ({ + 3)({ 3 3) C { = 33 or 3, so the
domain is {{ M R | { 6= 33> 3} = (3"> 33) (33> 3) (3> ").
32. i ({) = (2{3 3 5)@({2 + { 3 6) is de¿ned for all { except when 0 = {2 + { 3 6
C 0 = ({ + 3)({ 3 2) C
{ = 33 or 2, so the domain is {{ M R | { 6= 33> 2} = (3"> 33) (33> 2) (2> ").
s
I
3
2w 3 1 is de¿ned for all real numbers. In fact 3 s(w), where s(w) is a polynomial, is de¿ned for all real numbers.
33. i (w) =
Thus, the domain is R> or (3"> ").
34. j(w) =
I
I
3 3 w 3 2 + w is de¿ned when 3 3 w D 0
C
w $ 3 and 2 + w D 0
C
w D 32. Thus, the domain is
C
{({ 3 5) A 0. Note that {2 3 5{ 6= 0 since that would result in
32 $ w $ 3, or [32> 3].
35. k({) = 1
I
4
{2 3 5{ is de¿ned when {2 3 5{ A 0
division by zero. The expression {({ 3 5) is positive if { ? 0 or { A 5. (See Appendix A for methods for solving
inequalities.) Thus, the domain is (3"> 0) (5> ").
36. i (x) =
1
1
x+1
is de¿ned when x + 1 6= 0 [x 6= 31] and 1 +
6= 0. Since 1 +
=0 i
1
x+1
x+1
1+
x+1
1
= 31 i 1 = 3x 3 1 i x = 32, the domain is {x | x 6= 32, x 6= 31} = (3"> 32) (32> 31) (31> ").
x+1
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NOT FOR SALE
CHAPTER 1
37. I (s) =
FUNCTIONS AND MODELS
s
I
I
I
I
I
2 3 s is de¿ned when s D 0 and 2 3 s D 0. Since 2 3 s D 0 i 2 D s i
s$2 i
0 $ s $ 4, the domain is [0> 4].
38. k({) =
I
I
4 3 {2 . Now | = 4 3 {2
i | 2 = 4 3 {2
C {2 + | 2 = 4, so
the graph is the top half of a circle of radius 2 with center at the origin. The domain
is { | 4 3 {2 D 0 = { | 4 D {2 = {{ | 2 D |{|} = [32> 2]. From the graph,
the range is 0 $ | $ 2, or [0> 2].
39. i ({) = 2 3 0=4{ is de¿ned for all real numbers, so the domain is R,
or (3"> ")= The graph of i is a line with slope 30=4 and |-intercept 2.
40. I ({) = {2 3 2{ + 1 = ({ 3 1)2 is de¿ned for all real numbers, so the
domain is R, or (3"> "). The graph of I is a parabola with vertex (1> 0).
41. i (w) = 2w + w2 is de¿ned for all real numbers, so the domain is R, or
(3"> "). The graph of i is a parabola opening upward since the
coef¿cient of w2 is positive. To ¿nd the w-intercepts, let | = 0 and solve
for w. 0 = 2w + w2 = w(2 + w) i w = 0 or w = 32. The w-coordinate of
the vertex is halfway between the w-intercepts, that is, at w = 31. Since
i (31) = 2(31) + (31)2 = 32 + 1 = 31, the vertex is (31> 31).
42. K(w) =
(2 + w)(2 3 w)
4 3 w2
=
, so for w 6= 2, K(w) = 2 + w. The domain
23w
23w
is {w | w 6= 2}. So the graph of K is the same as the graph of the function
i (w) = w + 2 (a line) except for the hole at (2> 4).
I
{ 3 5 is de¿ned when { 3 5 D 0 or { D 5, so the domain is [5> ").
I
Since | = { 3 5 i | 2 = { 3 5 i { = | 2 + 5, we see that j is the
43. j({) =
top half of a parabola.
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SECTION 1.1
44. I ({) = |2{ + 1| =
=
+
+
2{ + 1
if 2{ + 1 D 0
3(2{ + 1)
if 2{ + 1 ? 1
FOUR WAYS TO REPRESENT A FUNCTION
if { D 3 12
2{ + 1
if { ? 3 12
32{ 3 1
The domain is R, or (3"> ").
45. J({) =
3{ + |{|
. Since |{| =
{
; 3{ + {
A
?
{
J({) =
A
= 3{ 3 {
{
+
{
if { D 0
3{
if { ? 0
; 4{
A
?
{
=
A
2{
=
if { ? 0
{
if { A 0
, we have
if { A 0
=
if { ? 0
+
4
if { A 0
2
if { ? 0
Note that J is not de¿ned for { = 0. The domain is (3"> 0) (0> ").
46. j({) = |{| 3 { =
+
{3{
if { D 0
3{ 3 {
if { ? 0
=
+
0
if { D 0
32{
if { ? 0
.
The domain is R, or (3"> ").
47. i ({) =
+
{+2
if { ? 0
13{
if { D 0
The domain is R.
48. i ({) =
+
3 3 12 {
if { $ 2
2{ 3 5
if { A 2
The domain is R.
49. i ({) =
+
{ + 2 if { $ 31
{2
if { A 31
Note that for { = 31, both { + 2 and {2 are equal to 1. The domain is R.
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15
16
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NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
;
{+9
A
?
32{
50. i ({) =
A
=
36
if
{ ? 33
if |{| $ 3
if
{A 3
Note that for { = 33, both { + 9 and 32{ are equal to 6; and for { = 3, both 32{
and 36 are equal to 36. The domain is R.
51. Recall that the slope p of a line between the two points ({1 > |1 ) and ({2 > |2 ) is p =
|2 3 |1
and an equation of the line
{2 3 {1
connecting those two points is | 3 |1 = p({ 3 {1 ). The slope of the line segment joining the points (1> 33) and (5> 7) is
5
7 3 (33)
= , so an equation is | 3 (33) = 52 ({ 3 1). The function is i ({) = 52 { 3
531
2
52. The slope of the line segment joining the points (35> 10) and (7> 310) is
11
,
2
1 $ { $ 5.
5
310 3 10
= 3 , so an equation is
7 3 (35)
3
| 3 10 = 3 53 [{ 3 (35)]. The function is i ({) = 3 53 { + 53 , 35 $ { $ 7.
53. We need to solve the given equation for |.
| =1±
I
{ + (| 3 1)2 = 0 C (| 3 1)2 = 3{ C | 3 1 = ± 3{ C
I
3{. The expression with the positive radical represents the top half of the parabola, and the one with the negative
radical represents the bottom half. Hence, we want i ({) = 1 3
54. {2 + (| 3 2)2 = 4
C (| 3 2)2 = 4 3 {2
the function i ({) = 2 +
I
3{. Note that the domain is { $ 0.
I
C | 3 2 = ± 4 3 {2
I
4 3 {2 , 32 $ { $ 2.
C | =2±
I
4 3 {2 . The top half is given by
55. For 0 $ { $ 3, the graph is the line with slope 31 and |-intercept 3, that is, | = 3{ + 3. For 3 ? { $ 5, the graph is the line
with slope 2 passing through (3> 0); that is, | 3 0 = 2({ 3 3), or | = 2{ 3 6. So the function is
i ({) =
+
3{ + 3 if 0 $ { $ 3
2{ 3 6 if 3 ? { $ 5
56. For 34 $ { $ 32, the graph is the line with slope 3 32 passing through (32> 0); that is, | 3 0 = 3 32 [{ 3 (32)], or
| = 3 32 { 3 3. For 32 ? { ? 2, the graph is the top half of the circle with center (0> 0) and radius 2. An equation of the circle
is {2 + | 2 = 4, so an equation of the top half is | =
I
4 3 {2 . For 2 $ { $ 4, the graph is the line with slope
3
2
through (2> 0); that is, | 3 0 = 32 ({ 3 2), or | = 32 { 3 3. So the function is
; 3
3 { 3 3 if 34 $ { $ 32
A
A
?I 2
i ({) =
4 3 {2 if 32 ? { ? 2
A
A
=3
if 2 $ { $ 4
2{ 3 3
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SECTION 1.1
FOUR WAYS TO REPRESENT A FUNCTION
¤
17
57. Let the length and width of the rectangle be O and Z . Then the perimeter is 2O + 2Z = 20 and the area is D = OZ .
Solving the ¿rst equation for Z in terms of O gives Z =
20 3 2O
= 10 3 O. Thus, D(O) = O(10 3 O) = 10O 3 O2 . Since
2
lengths are positive, the domain of D is 0 ? O ? 10. If we further restrict O to be larger than Z , then 5 ? O ? 10 would be
the domain.
58. Let the length and width of the rectangle be O and Z . Then the area is OZ = 16, so that Z = 16@O. The perimeter is
S = 2O + 2Z , so S (O) = 2O + 2(16@O) = 2O + 32@O, and the domain of S is O A 0, since lengths must be positive
quantities. If we further restrict O to be larger than Z , then O A 4 would be the domain.
59. Let the length of a side of the equilateral triangle be {. Then by the Pythagorean Theorem, the height | of the triangle satis¿es
I
1 2
{ = {2 , so that | 2 = {2 3 14 {2 = 34 {2 and | = 23 {. Using the formula for the area D of a triangle,
2
I I
D = 12 (base)(height), we obtain D({) = 12 ({) 23 { = 43 {2 , with domain { A 0.
|2 +
60. Let the volume of the cube be Y and the length of an edge be O. Then Y = O3 so O =
I
3
Y , and the surface area is
I 2
= 6Y 2@3 , with domain Y A 0.
V(Y ) = 6O2 = 6 3 Y
61. Let each side of the base of the box have length {, and let the height of the box be k. Since the volume is 2, we know that
2 = k{2 , so that k = 2@{2 , and the surface area is V = {2 + 4{k. Thus, V({) = {2 + 4{(2@{2 ) = {2 + (8@{), with
domain { A 0.
62. The area of the window is D = {k + 12 1 2
{2
, where k is the height of the rectangular portion of the window.
= {k +
2{
8
The perimeter is S = 2k + { + 12 { = 30 C 2k = 30 3 { 3 12 { C k = 14 (60 3 2{ 3 {). Thus,
60 3 2{ 3 {
+4
{2
D({) = {
+
= 15{ 3 12 {2 3 4 {2 + 8 {2 = 15{ 3 48 {2 3 8 {2 = 15{ 3 {2
.
4
8
8
Since the lengths { and k must be positive quantities, we have { A 0 and k A 0. For k A 0, we have 2k A 0 C
30 3 { 3 12 { A 0 C 60 A 2{ + { C { ?
60
60
. Hence, the domain of D is 0 ? { ?
.
2+
2+
63. The height of the box is { and the length and width are O = 20 3 2{, Z = 12 3 2{. Then Y = OZ { and so
Y ({) = (20 3 2{)(12 3 2{)({) = 4(10 3 {)(6 3 {)({) = 4{(60 3 16{ + {2 ) = 4{3 3 64{2 + 240{.
The sides O, Z , and { must be positive. Thus, O A 0 C 20 3 2{ A 0 C { ? 10;
Z A 0 C 12 3 2{ A 0 C { ? 6; and { A 0. Combining these restrictions gives us the domain 0 ? { ? 6.
64. We can summarize the monthly cost with a piecewise
de¿ned function.
+
35
F({) =
35 + 0=10({ 3 400)
if 0 $ { $ 400
if { A 400
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CHAPTER 1
FUNCTIONS AND MODELS
65. We can summarize the amount of the ¿ne with a
piecewise de¿ned function.
;
15(40 3 {)
A
?
I ({) = 0
A
=
15({ 3 65)
if 0 $ { ? 40
if 40 $ { $ 65
if { A 65
66. For the ¿rst 1200 kWh, H({) = 10 + 0=06{.
For usage over 1200 kWh, the cost is
H({) = 10 + 0=06(1200) + 0=07({ 3 1200) = 82 + 0=07({ 3 1200).
Thus,
H({) =
+
10 + 0=06{
if 0 $ { $ 1200
82 + 0=07({ 3 1200) if { A 1200
67. (a)
(b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400.
On $26,000, tax is assessed on $16,000, and
10%($10,000) + 15%($6000) = $1000 + $900 = $1900.
(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so
the graph of W is a line segment from (10,000> 0) to (20,000> 1000).
The tax on $30,000 is $2500, so the graph of W for { A 20,000 is
the ray with initial point (20,000> 1000) that passes through
(30,000> 2500).
68. One example is the amount paid for cable or telephone system repair in the home, usually measured to the nearest quarter hour.
Another example is the amount paid by a student in tuition fees, if the fees vary according to the number of credits for which
the student has registered.
69. i is an odd function because its graph is symmetric about the origin. j is an even function because its graph is symmetric with
respect to the |-axis.
70. i is not an even function since it is not symmetric with respect to the |-axis. i is not an odd function since it is not symmetric
about the origin. Hence, i is neither even nor odd. j is an even function because its graph is symmetric with respect to the
|-axis.
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SECTION 1.1
FOUR WAYS TO REPRESENT A FUNCTION
¤
19
71. (a) Because an even function is symmetric with respect to the |-axis, and the point (5> 3) is on the graph of this even function,
the point (35> 3) must also be on its graph.
(b) Because an odd function is symmetric with respect to the origin, and the point (5> 3) is on the graph of this odd function,
the point (35> 33) must also be on its graph.
72. (a) If i is even, we get the rest of the graph by reÀecting
180 about the origin.
about the |-axis.
73. i ({) =
{2
i (3{) =
{
.
+1
74. i ({) =
3{
{
3{
= 2
=3 2
= 3i({).
(3{)2 + 1
{ +1
{ +1
So i is an odd function.
75. i ({) =
{2
.
+1
{4
i (3{) =
{2
(3{)2
= 4
= i ({).
4
(3{) + 1
{ +1
So i is an even function.
3{
{
{
, so i(3{) =
=
.
{+1
3{ + 1
{31
Since this is neither i ({) nor 3i ({), the function i is
neither even nor odd.
(b) If i is odd, we get the rest of the graph by rotating
76. i ({) = { |{|.
i (3{) = (3{) |3{| = (3{) |{| = 3({ |{|)
= 3i ({)
So i is an odd function.
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CHAPTER 1 FUNCTIONS AND MODELS
77. i ({) = 1 + 3{2 3 {4 .
78. i ({) = 1 + 3{3 3 {5 , so
i (3{) = 1+3(3{)2 3(3{)4 = 1+3{2 3{4 = i ({).
So i is an even function.
i (3{) = 1 + 3(3{)3 3 (3{)5 = 1 + 3(3{3 ) 3 (3{5 )
= 1 3 3{3 + {5
Since this is neither i ({) nor 3i ({), the function i is
neither even nor odd.
79. (i) If i and j are both even functions, then i (3{) = i ({) and j(3{) = j({). Now
(i + j)(3{) = i(3{) + j(3{) = i ({) + j({) = (i + j)({), so i + j is an even function.
(ii) If i and j are both odd functions, then i (3{) = 3i ({) and j(3{) = 3j({). Now
(i + j)(3{) = i(3{) + j(3{) = 3i ({) + [3j({)] = 3[i ({) + j({)] = 3(i + j)({), so i + j is an odd function.
(iii) If i is an even function and j is an odd function, then (i + j)(3{) = i (3{) + j(3{) = i ({) + [3j({)] = i ({) 3 j({),
which is not (i + j)({) nor 3(i + j)({), so i + j is neither even nor odd. (Exception: if i is the zero function, then
i + j will be odd. If j is the zero function, then i + j will be even.)
80. (i) If i and j are both even functions, then i (3{) = i ({) and j(3{) = j({). Now
(i j)(3{) = i (3{)j(3{) = i ({)j({) = (i j)({), so ij is an even function.
(ii) If i and j are both odd functions, then i (3{) = 3i ({) and j(3{) = 3j({). Now
(i j)(3{) = i (3{)j(3{) = [3i ({)][3j({)] = i ({)j({) = (i j)({), so i j is an even function.
(iii) If i is an even function and j is an odd function, then
(i j)(3{) = i (3{)j(3{) = i ({)[3j({)] = 3[i ({)j({)] = 3(i j)({), so i j is an odd function.
1.2 Mathematical Models: A Catalog of Essential Functions
1. (a) i ({) = log2 { is a logarithmic function.
(b) j({) =
I
4
{ is a root function with q = 4.
(c) k({) =
2{3
is a rational function because it is a ratio of polynomials.
1 3 {2
(d) x(w) = 1 3 1=1w + 2=54w2 is a polynomial of degree 2 (also called a quadratic function).
(e) y(w) = 5w is an exponential function.
(f ) z() = sin cos2 is a trigonometric function.
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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS
¤
21
2. (a) | = { is an exponential function (notice that { is the exponent).
(b) | = { is a power function (notice that { is the base).
(c) | = {2 (2 3 {3 ) = 2{2 3 {5 is a polynomial of degree 5.
(d) | = tan w 3 cos w is a trigonometric function.
(e) | = v@(1 + v) is a rational function because it is a ratio of polynomials.
(f ) | =
I
I
{3 3 1@(1 + 3 {) is an algebraic function because it involves polynomials and roots of polynomials.
3. We notice from the ¿gure that j and k are even functions (symmetric with respect to the |-axis) and that i is an odd function
(symmetric with respect to the origin). So (b) | = {5 must be i . Since j is Àatter than k near the origin, we must have
(c) | = {8 matched with j and (a) | = {2 matched with k.
4. (a) The graph of | = 3{ is a line (choice J).
(b) | = 3{ is an exponential function (choice i ).
(c) | = {3 is an odd polynomial function or power function (choice I ).
(d) | =
I
3
{ = {1@3 is a root function (choice j).
5. (a) An equation for the family of linear functions with slope 2
is | = i ({) = 2{ + e, where e is the |-intercept.
(b) i (2) = 1 means that the point (2> 1) is on the graph of i . We can use the
point-slope form of a line to obtain an equation for the family of linear
functions through the point (2> 1). | 3 1 = p({ 3 2), which is equivalent
to | = p{ + (1 3 2p) in slope-intercept form.
(c) To belong to both families, an equation must have slope p = 2, so the equation in part (b), | = p{ + (1 3 2p),
becomes | = 2{ 3 3. It is the only function that belongs to both families.
6. All members of the family of linear functions i({) = 1 + p({ + 3) have
graphs that are lines passing through the point (33> 1).
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CHAPTER 1 FUNCTIONS AND MODELS
7. All members of the family of linear functions i({) = f 3 { have graphs
that are lines with slope 31. The |-intercept is f.
8. The vertex of the parabola on the left is (3> 0), so an equation is | = d({ 3 3)2 + 0. Since the point (4> 2) is on the
parabola, we’ll substitute 4 for { and 2 for | to ¿nd d. 2 = d(4 3 3)2
i d = 2, so an equation is i ({) = 2({ 3 3)2 .
The |-intercept of the parabola on the right is (0> 1), so an equation is | = d{2 + e{ + 1. Since the points (32> 2) and
(1> 32=5) are on the parabola, we’ll substitute 32 for { and 2 for | as well as 1 for { and 32=5 for | to obtain two equations
with the unknowns d and e.
(32> 2):
2 = 4d 3 2e + 1 i 4d 3 2e = 1
(1)
(1> 32=5):
32=5 = d + e + 1 i d + e = 33=5
(2)
2 · (2) + (1) gives us 6d = 36 i d = 31. From (2), 31 + e = 33=5 i e = 32=5, so an equation
is j({) = 3{2 3 2=5{ + 1.
9. Since i (31) = i (0) = i(2) = 0, i has zeros of 31, 0, and 2, so an equation for i is i ({) = d[{ 3 (31)]({ 3 0)({ 3 2),
or i ({) = d{({ + 1)({ 3 2). Because i (1) = 6, we’ll substitute 1 for { and 6 for i ({).
6 = d(1)(2)(31) i 32d = 6 i d = 33, so an equation for i is i ({) = 33{({ + 1)({ 3 2).
10. (a) For W = 0=02w + 8=50, the slope is 0=02, which means that the average surface temperature of the world is increasing at a
rate of 0=02 C per year. The W -intercept is 8=50, which represents the average surface temperature in C in the year 1900.
(b) w = 2100 3 1900 = 200 i W = 0=02(200) + 8=50 = 12=50 C
11. (a) G = 200, so f = 0=0417G(d + 1) = 0=0417(200)(d + 1) = 8=34d + 8=34. The slope is 8=34, which represents the
change in mg of the dosage for a child for each change of 1 year in age.
(b) For a newborn, d = 0, so f = 8=34 mg.
12. (a)
(b) The slope of 34 means that for each increase of 1 dollar for a
rental space, the number of spaces rented decreases by 4. The
|-intercept of 200 is the number of spaces that would be occupied
if there were no charge for each space. The {-intercept of 50 is the
smallest rental fee that results in no spaces rented.
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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS
13. (a)
(b) The slope of
9
5
means that I increases
9
5
¤
23
degrees for each increase
of 1 C. (Equivalently, I increases by 9 when F increases by 5
and I decreases by 9 when F decreases by 5.) The I -intercept of
32 is the Fahrenheit temperature corresponding to a Celsius
temperature of 0.
14. (a) Let g = distance traveled (in miles) and w = time elapsed (in hours). At
w = 0, g = 0 and at w = 50 minutes = 50 ·
have two points: (0> 0) and
5
1
60
=
5
6
(b)
h, g = 40. Thus we
40 3 0
= 48 and so g = 48w.
, so p = 5
30
6
6 > 40
(c) The slope is 48 and represents the car’s speed in mi@h.
15. (a) Using Q in place of { and W in place of |, we ¿nd the slope to be
equation is W 3 80 = 16 (Q 3 173) C W 3 80 = 16 Q 3
(b) The slope of
1
6
173
6
W2 3 W1
80 3 70
10
1
=
=
= . So a linear
Q2 3 Q1
173 3 113
60
6
307
C W = 16 Q + 307
6
6 = 51=16 .
means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket
chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 F.
(c) When Q = 150, the temperature is given approximately by W = 16 (150) +
307
6
= 76=16 F E 76 F.
16. (a) Let { denote the number of chairs produced in one day and | the associated
cost. Using the points (100> 2200) and (300> 4800), we get the slope
480032200
3003100
=
2600
200
= 13. So | 3 2200 = 13({ 3 100) C
| = 13{ + 900.
(b) The slope of the line in part (a) is 13 and it represents the cost (in dollars)
of producing each additional chair.
(c) The |-intercept is 900 and it represents the ¿xed daily costs of operating
the factory.
17. (a) We are given
change in pressure
4=34
=
= 0=434. Using S for pressure and g for depth with the point
10 feet change in depth
10
(g> S ) = (0> 15), we have the slope-intercept form of the line, S = 0=434g + 15.
(b) When S = 100, then 100 = 0=434g + 15 C 0=434g = 85 C g =
85
0=434
E 195=85 feet. Thus, the pressure is
100 lb@in2 at a depth of approximately 196 feet.
18. (a) Using g in place of { and F in place of |, we ¿nd the slope to be
So a linear equation is F 3 460 =
1
4
80
1
F2 3 F1
460 3 380
=
= .
=
g2 3 g1
800 3 480
320
4
(g 3 800) C F 3 460 = 14 g 3 200 C F = 14 g + 260.
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NOT FOR SALE
CHAPTER 1 FUNCTIONS AND MODELS
(b) Letting g = 1500 we get F =
1
4
(1500) + 260 = 635.
(c)
The cost of driving 1500 miles is $635.
(d) The |-intercept represents the ¿xed cost, $260.
The slope of the line represents the cost per
mile, $0=25.
(e) A linear function gives a suitable model in this situation because you have ¿xed monthly costs such as insurance and car
payments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these for each
additional mile driven is a constant.
19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form
i ({) = d cos(e{) + f seems appropriate.
(b) The data appear to be decreasing in a linear fashion. A model of the form i ({) = p{ + e seems appropriate.
20. (a) The data appear to be increasing exponentially. A model of the form i ({) = d · e{ or i({) = d · e{ + f seems appropriate.
(b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form i ({) = d@{ seems
appropriate.
Exercises 21 – 24: Some values are given to many decimal places. These are the results given by several computer algebra systems — rounding is left
to the reader.
21. (a)
(b) Using the points (4000> 14=1) and (60,000> 8=2), we obtain
8=2 3 14=1
({ 3 4000) or, equivalently,
60,000 3 4000
| E 30=000105357{ + 14=521429.
| 3 14=1 =
A linear model does seem appropriate.
(c) Using a computing device, we obtain the least squares regression line | = 30=0000997855{ + 13=950764.
The following commands and screens illustrate how to ¿nd the least squares regression line on a TI-84 Plus.
Enter the data into list one (L1) and list two (L2). Press
to enter the editor.
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SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS
Find the regession line and store it in Y1 . Press
.
Note from the last ¿gure that the regression line has been stored in Y1 and that Plot1 has been turned on (Plot1 is
highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing
pressing
or by
.
Now press
to produce a graph of the data and the regression
line. Note that choice 9 of the ZOOM menu automatically selects a window
that displays all of the data.
(d) When { = 25,000, | E 11=456; or about 11=5 per 100 population.
(e) When { = 80,000, | E 5=968; or about a 6% chance.
(f ) When { = 200,000, | is negative, so the model does not apply.
22. (a)
(b)
Using a computing device, we obtain the least squares
(c) When { = 100 F, | = 264=7 E 265 chirps@min.
regression line | = 4=856{ 3 220=96.
23. (a) A linear model seems appropriate over the time interval
considered.
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26
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CHAPTER 1 FUNCTIONS AND MODELS
(b) Using a computing device, we obtain the regression line | E 0=0265{ 3 46=8759. It is plotted in the graph in part (a).
(c) For { = 2008, the linear model predicts a winning height of 6.27 m, considerably higher than the actual winning height
of 5.96 m.
(d) It is not reasonable to use the model to predict the winning height at the 2100 Olympics since 2100 is too far from the
1896–2004 range on which the model is based.
24. By looking at the scatter plot of the data, we rule out
the power and logarithmic models.
We try various models:
Scatter plot
| = 30=430 545 454 5{ + 870=183 636 4
Linear
| = 0=004 893 939 4{2 3 19=786 075 76{ + 20 006=954 85
Quadratic:
| = 30=000 073 193 47{3 + 0=439 114 219 1{2 3 878=429 871 8{ + 585 960=983
Cubic:
| = 0=000 007 902 097 9{4 3 0=062 578 787 9{3 + 185=842 283 8{2 3 245 290=9304{ + 121 409 472=7
Quartic:
Exponential:
| = 2=618 230 2 × 1021 (0=976 789 309 4){
Linear model
Quadratic model
Quartic model
Exponential model
Cubic model
After examining the graphs of these models, we see that all the models are good and the quartic model is the best.
Using this model, we obtain estimates 13=6% and 10=2% for the rural percentages in 1988 and 2002 respectively.
25. If { is the original distance from the source, then the illumination is i ({) = n{32 = n@{2 . Moving halfway to the lamp gives
us an illumination of i
1 32
{ = n 12 {
= n(2@{)2 = 4(n@{2 ), so the light is 4 times as bright.
2
26. (a) If D = 60, then V = 0=7D0=3 r 2=39, so you would expect to ¿nd 2 species of bats in that cave.
(b) V = 4 i 4 = 0=7D0=3
to be 334 m2 .
i
40
7
= D3@10
i D=
40 10@3
7
r 333=6, so we estimate the surface area of the cave
27. (a) Using a computing device, we obtain a power function Q = fDe , where f r 3=1046 and e r 0=308.
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SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS
¤
27
(b) If D = 291, then Q = fDe r 17=8, so you would expect to ¿nd 18 species of reptiles and amphibians on Dominica.
28. (a) W = 1=000 431 227g 1=499 528 750
(b) The power model in part (a) is approximately W = g1=5 . Squaring both sides gives us W 2 = g3 , so the model matches
Kepler’s Third Law, W 2 = ng 3 .
1.3 New Functions from Old Functions
1. (a) If the graph of i is shifted 3 units upward, its equation becomes | = i ({) + 3.
(b) If the graph of i is shifted 3 units downward, its equation becomes | = i ({) 3 3.
(c) If the graph of i is shifted 3 units to the right, its equation becomes | = i ({ 3 3).
(d) If the graph of i is shifted 3 units to the left, its equation becomes | = i ({ + 3).
(e) If the graph of i is reÀected about the {-axis, its equation becomes | = 3i ({).
(f ) If the graph of i is reÀected about the |-axis, its equation becomes | = i (3{).
(g) If the graph of i is stretched vertically by a factor of 3, its equation becomes | = 3i ({).
(h) If the graph of i is shrunk vertically by a factor of 3, its equation becomes | = 13 i ({).
2. (a) To obtain the graph of | = i ({) + 8 from the graph of | = i ({), shift the graph 8 units upward.
(b) To obtain the graph of | = i ({ + 8) from the graph of | = i ({), shift the graph 8 units to the left.
(c) To obtain the graph of | = 8i({) from the graph of | = i ({), stretch the graph vertically by a factor of 8.
(d) To obtain the graph of | = i (8{) from the graph of | = i ({), shrink the graph horizontally by a factor of 8.
(e) To obtain the graph of | = 3i ({) 3 1 from the graph of | = i ({), ¿rst reÀect the graph about the {-axis, and then shift it
1 unit downward.
(f ) To obtain the graph of | = 8i ( 18 {) from the graph of | = i ({), stretch the graph horizontally and vertically by a factor
of 8.
3. (a) (graph 3) The graph of i is shifted 4 units to the right and has equation | = i ({ 3 4).
(b) (graph 1) The graph of i is shifted 3 units upward and has equation | = i ({) + 3.
(c) (graph 4) The graph of i is shrunk vertically by a factor of 3 and has equation | = 13 i({).
(d) (graph 5) The graph of i is shifted 4 units to the left and reÀected about the {-axis. Its equation is | = 3i ({ + 4).
(e) (graph 2) The graph of i is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is
| = 2i ({ + 6).
4. (a) To graph | = i ({) 3 2, we shift the graph of i , 2
(b) To graph | = i ({ 3 2), we shift the graph of i,
units downward.The point (1> 2) on the graph of i
2 units to the right.The point (1> 2) on the graph of i
corresponds to the point (1> 2 3 2) = (1> 0).
corresponds to the point (1 + 2> 2) = (3> 2).
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¤
CHAPTER 1 FUNCTIONS AND MODELS
(c) To graph | = 32i ({), we reÀect the graph about the
(d) To graph | = i ( 13 {) + 1, we stretch the graph
{-axis and stretch the graph vertically by a factor of 2.
horizontally by a factor of 3 and shift it 1 unit upward.
The point (1> 2) on the graph of i corresponds to the
The point (1> 2) on the graph of i corresponds to the
point (1> 32 · 2) = (1> 34).
point (1 · 3> 2 + 1) = (3> 3).
5. (a) To graph | = i (2{) we shrink the graph of i
horizontally by a factor of 2.
(b) To graph | = i
1 2 { we stretch the graph of i
horizontally by a factor of 2.
The point (4> 31) on the graph of i corresponds to the
The point (4> 31) on the graph of i corresponds to the
point 12 · 4> 31 = (2> 31).
(c) To graph | = i (3{) we reÀect the graph of i about
the |-axis.
(d) To graph | = 3i (3{) we reÀect the graph of i about
the |-axis, then about the {-axis.
The point (4> 31) on the graph of i corresponds to the
point (31 · 4> 31) = (34> 31).
6. The graph of | = i ({) =
point (2 · 4> 31) = (8> 31).
The point (4> 31) on the graph of i corresponds to the
point (31 · 4> 31 · 31) = (34> 1).
I
3{ 3 {2 has been shifted 2 units to the right and stretched vertically by a factor of 2.
Thus, a function describing the graph is
| = 2i ({ 3 2) = 2
7. The graph of | = i ({) =
s
s
I
3({ 3 2) 3 ({ 3 2)2 = 2 3{ 3 6 3 ({2 3 4{ + 4) = 2 3{2 + 7{ 3 10
I
3{ 3 {2 has been shifted 4 units to the left, reÀected about the {-axis, and shifted downward
1 unit. Thus, a function describing the graph is
|=
31 ·
 ~} €
reÀect
about {-axis
i ({ + 4)
 ~} €
shift
4 units left
3 1
 ~} €
shift
1 unit left
This function can be written as
s
s
I
| = 3i ({ + 4) 3 1 = 3 3({ + 4) 3 ({ + 4)2 3 1 = 3 3{ + 12 3 ({2 + 8{ + 16) 3 1 = 3 3{2 3 5{ 3 4 3 1
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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NOT FOR SALE
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS
8. (a) The graph of | = 2 sin { can be obtained from the graph
of | = sin { by stretching it vertically by a factor of 2.
9. | =
¤
29
I
(b) The graph of | = 1 + { can be obtained from
I
the graph of | = { by shifting it upward 1 unit.
1
: Start with the graph of the reciprocal function | = 1@{ and shift 2 units to the left.
{+2
10. | = ({ 3 1)3 : Start with the graph of | = {3 and shift 1 unit to the right.
I
11. | = 3 3 {: Start with the graph of | =
I
3
{ and reÀect about the {-axis.
12. | = {2 + 6{ + 4 = ({2 + 6{ + 9) 3 5 = ({ + 3)2 3 5: Start with the graph of | = {2 , shift 3 units to the left, and then shift
5 units downward.
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© Cengage Learning. All Rights Reserved.
30
¤
13. | =
NOT FOR SALE
CHAPTER 1 FUNCTIONS AND MODELS
I
I
{ 3 2 3 1: Start with the graph of | = {, shift 2 units to the right, and then shift 1 unit downward.
14. | = 4 sin 3{: Start with the graph of | = sin {, compress horizontally by a factor of 3, and then stretch vertically by a
factor of 4.
15. | = sin({@2): Start with the graph of | = sin { and stretch horizontally by a factor of 2.
16. | =
1
2
3 2: Start with the graph of | = , stretch vertically by a factor of 2, and then shift 2 units downward.
{
{
17. | =
1
(1
2
3 cos {): Start with the graph of | = cos {, reÀect about the {-axis, shift 1 unit upward, and then shrink vertically by
a factor of 2.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS
I
18. | = 1 3 2 { + 3: Start with the graph of | =
¤
31
I
{, shift 3 units to the left, stretch vertically by a factor of 2, reÀect about the
{-axis, and then shift 1 unit upward.
19. | = 1 3 2{ 3 {2 = 3({2 + 2{) + 1 = 3({2 + 2{ + 1) + 2 = 3({ + 1)2 + 2: Start with the graph of | = {2 , reÀect about
the {-axis, shift 1 unit to the left, and then shift 2 units upward.
20. | = |{| 3 2: Start with the graph of | = |{| and shift 2 units downward.
21. | = |{ 3 2|: Start with the graph of | = |{| and shift 2 units to the right.
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32
¤
22. | =
CHAPTER 1 FUNCTIONS AND MODELS
1
4
tan({ 3 4 ): Start with the graph of | = tan {, shift
I
23. | = | { 3 1|: Start with the graph of | =
4
units to the right, and then compress vertically by a factor of 4.
I
{, shift it 1 unit downward, and then reÀect the portion of the graph below the
{-axis about the {-axis.
24. | = |cos {|: Start with the graph of | = cos {, shrink it horizontally by a factor of , and reÀect all the parts of the graph
below the {-axis about the {-axis.
25. This is just like the solution to Example 4 except the amplitude of the curve (the 30 N curve in Figure 9 on June 21) is
2
14 3 12 = 2. So the function is O(w) = 12 + 2 sin 365
(w 3 80) . March 31 is the 90th day of the year, so the model gives
O(90) E 12=34 h. The daylight time (5:51 AM to 6:18 PM) is 12 hours and 27 minutes, or 12=45 h. The model value differs
from the actual value by
12=45312=34
12=45
E 0=009, less than 1%.
26. Using a sine function to model the brightness of Delta Cephei as a function of time, we take its period to be 5=4 days, its
amplitude to be 0=35 (on the scale of magnitude), and its average magnitude to be 4=0. If we take w = 0 at a time of average
brightness, then the magnitude (brightness) as a function of time w in days can be modeled by the formula
2 w .
P (w) = 4=0 + 0=35 sin 5=4
27. (a) To obtain | = i (|{|), the portion of the graph of | = i ({) to the right of the |-axis is reÀected about the |-axis.
(b) | = sin |{|
(c) | =
s
|{|
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© Cengage Learning. All Rights Reserved.
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SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS
¤
33
28. The most important features of the given graph are the {-intercepts and the maximum
and minimum points. The graph of | = 1@i({) has vertical asymptotes at the {-values
where there are {-intercepts on the graph of | = i({). The maximum of 1 on the graph
of | = i ({) corresponds to a minimum of 1@1 = 1 on | = 1@i ({). Similarly, the
minimum on the graph of | = i ({) corresponds to a maximum on the graph of
| = 1@i ({). As the values of | get large (positively or negatively) on the graph of
| = i ({), the values of | get close to zero on the graph of | = 1@i ({).
29. i ({) = {3 + 2{2 ; j({) = 3{2 3 1.
3
2
G = R for both i and j.
2
(a) (i + j)({) = ({ + 2{ ) + (3{ 3 1) = {3 + 5{2 3 1, G = R.
(b) (i 3 j)({) = ({3 + 2{2 ) 3 (3{2 3 1) = {3 3 {2 + 1, G = R.
(c) (i j)({) = ({3 + 2{2 )(3{2 3 1) = 3{5 + 6{4 3 {3 3 2{2 , G = R.
(d)
{3 + 2{2
1
i
I
since 3{2 3 1 6= 0.
({) =
,
G
=
{
|
{
=
6
±
j
3{2 3 1
3
I
I
3 3 {, G = (3"> 3]; j({) = {2 3 1, G = (3"> 31] [1> ").
I
I
(a) (i + j)({) = 3 3 { + {2 3 1, G = (3"> 31] [1> 3], which is the intersection of the domains of i and j.
I
I
(b) (i 3 j)({) = 3 3 { 3 {2 3 1, G = (3"> 31] [1> 3].
30. i ({) =
(c) (i j)({) =
(d)
I
I
3 3 { · {2 3 1, G = (3"> 31] [1> 3].
I
i
33{
i
, G = (3"> 31) (1> 3]. We must exclude { = ±1 since these values would make unde¿ned.
({) = I
j
j
{2 3 1
31. i ({) = {2 3 1, G = R;
j({) = 2{ + 1, G = R.
(a) (i j)({) = i (j({)) = i(2{ + 1) = (2{ + 1)2 3 1 = (4{2 + 4{ + 1) 3 1 = 4{2 + 4{, G = R.
(b) (j i)({) = j(i ({)) = j({2 3 1) = 2({2 3 1) + 1 = (2{2 3 2) + 1 = 2{2 3 1, G = R.
(c) (i i )({) = i(i ({)) = i ({2 3 1) = ({2 3 1)2 3 1 = ({4 3 2{2 + 1) 3 1 = {4 3 2{2 , G = R.
(d) (j j)({) = j(j({)) = j(2{ + 1) = 2(2{ + 1) + 1 = (4{ + 2) + 1 = 4{ + 3, G = R.
32. i ({) = { 3 2; j({) = {2 + 3{ + 4.
G = R for both i and j, and hence for their composites.
2
(a) (i j)({) = i (j({)) = i({ + 3{ + 4) = ({2 + 3{ + 4) 3 2 = {2 + 3{ + 2.
(b) (j i)({) = j(i ({)) = j({ 3 2) = ({ 3 2)2 + 3({ 3 2) + 4 = {2 3 4{ + 4 + 3{ 3 6 + 4 = {2 3 { + 2.
(c) (i i )({) = i(i ({)) = i ({ 3 2) = ({ 3 2) 3 2 = { 3 4.
(d) (j j)({) = j(j({)) = j({2 + 3{ + 4) = ({2 + 3{ + 4)2 + 3({2 + 3{ + 4) + 4
= ({4 + 9{2 + 16 + 6{3 + 8{2 + 24{) + 3{2 + 9{ + 12 + 4
= {4 + 6{3 + 20{2 + 33{ + 32
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°
licated, or posted to a publicly accessible website, in whole or in part.
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34
¤
NOT FOR SALE
CHAPTER 1 FUNCTIONS AND MODELS
33. i ({) = 1 3 3{; j({) = cos {.
G = R for both i and j, and hence for their composites.
(a) (i j)({) = i (j({)) = i(cos {) = 1 3 3 cos {.
(b) (j i)({) = j(i ({)) = j(1 3 3{) = cos(1 3 3{).
(c) (i i )({) = i(i ({)) = i (1 3 3{) = 1 3 3(1 3 3{) = 1 3 3 + 9{ = 9{ 3 2.
(d) (j j)({) = j(j({)) = j(cos {) = cos(cos {) [Note that this is not cos { · cos {.]
34. i ({) =
I
I
{, G = [0> "); j({) = 3 1 3 {, G = R.
(a) (i j)({) = i (j({)) = i
sI
I
I
3
3
13{ =
1 3 { = 6 1 3 {.
The domain of i j is {{ |
I
3
1 3 { D 0} = {{ | 1 3 { D 0} = {{ | { $ 1} = (3"> 1].
s
I
I
(b) (j i)({) = j(i ({)) = j( { ) = 3 1 3 {.
The domain of j i is {{ | { is in the domain of i and i ({) is in the domain of j}. This is the domain of i ,
that is, [0> ").
sI
I
I
I
{ = 4 {. The domain of i i is {{ | { D 0 and { D 0} = [0> ").
(c) (i i )({) = i(i ({)) = i ( { ) =
(d) (j j)({) = j(j({)) = j
s
I
I
3
1 3 { = 3 1 3 3 1 3 {, and the domain is (3"> ").
1
{+1
, G = {{ | { 6= 0}; j({) =
, G = {{ | { 6= 32}
{
{+2
{+1
{+1
{+1
1
{+2
=
(a) (i j)({) = i (j({)) = i
=
+
+
{+1
{+2
{+2
{+2
{+1
{+2
2
{ + 2{ + 1 + {2 + 4{ + 4
2{2 + 6{ + 5
({ + 1)({ + 1) + ({ + 2)({ + 2)
=
=
=
({ + 2)({ + 1)
({ + 2)({ + 1)
({ + 2)({ + 1)
35. i ({) = { +
Since j({) is not de¿ned for { = 32 and i (j({)) is not de¿ned for { = 32 and { = 31,
the domain of (i j)({) is G = {{ | { 6= 32> 31}.
1
{2 + 1 + {
{+
+1
{
{2 + { + 1
{2 + { + 1
1
= =
= 2
= 2 {
(b) (j i)({) = j(i ({)) = j { +
{
{ + 2{ + 1
({ + 1)2
1
{ + 1 + 2{
+2
{+
{
{
Since i({) is not de¿ned for { = 0 and j(i ({)) is not de¿ned for { = 31,
the domain of (j i )({) is G = {{ | { 6= 31> 0}.
1
1
1
1
1
1
{
={+ + 2
(c) (i i )({) = i (i ({)) = i { +
= {+
+
+ 2
1 = {+
{
+1
{
{
{
{
{
+1
{+ {
{
{({) {2 + 1 + 1 {2 + 1 + {({)
{4 + {2 + {2 + 1 + {2
=
=
2
{({ + 1)
{({2 + 1)
=
{4 + 3{2 + 1
>
{({2 + 1)
G = {{ | { 6= 0}
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS
{+1
(d) (j j)({) = j(j({)) = j
{+2
{ + 1 + 1({ + 2)
{+1
+1
{+1+{+2
2{ + 3
{
+
2
{+2
=
=
=
=
{+1
{ + 1 + 2({ + 2)
{ + 1 + 2{ + 4
3{ + 5
+2
{+2
{+2
Since j({) is not de¿ned for { = 32 and j(j({)) is not de¿ned for { = 3 53 ,
the domain of (j j)({) is G = { | { 6= 32> 3 53 .
{
, G = {{ | { 6= 31}; j({) = sin 2{, G = R.
1+{
sin 2{
(a) (i j)({) = i (j({)) = i(sin 2{) =
1 + sin 2{
3
3
+ 2q i { 6=
+ q [q an integer].
Domain: 1 + sin 2{ 6= 0 i sin 2{ 6= 31 i 2{ 6=
2
4
{
2{
(b) (j i)({) = j(i ({)) = j
= sin
.
1+{
1+{
36. i ({) =
Domain: {{ | { 6= 31}
(c) (i i )({) = i(i ({)) = i
{
1+{
{
{
· (1 + {)
{
{
{ = 1+{ =
= 1+{
=
1+{+{
2{ + 1
{
1+
1+
· (1 + {)
1+{
1+{
Since i({) is not de¿ned for { = 31, and i(i ({)) is not de¿ned for { = 3 12 ,
the domain of (i i )({) is G = {{ | { 6= 31> 3 12 }.
(d) (j j)(j) = j(j({)) = j(sin 2{) = sin(2 sin 2{).
Domain: R
37. (i j k)({) = i (j(k({))) = i (j({2 )) = i (sin({2 )) = 3 sin({2 ) 3 2
I
I
38. (i j k)({) = i (j(k({))) = i (j( {)) = i (2
{
I
) = 2 { 3 4
39. (i j k)({) = i (j(k({))) = i (j({3 + 2)) = i [({3 + 2)2 ]
= i ({6 + 4{3 + 4) =
s
I
({6 + 4{3 + 4) 3 3 = {6 + 4{3 + 1
I
40. (i j k)({) = i (j(k({))) = i (j( 3 { )) = i
I
I
3
3
{
{
I
I
=
tan
3
3
{31
{31
41. Let j({) = 2{ + {2 and i ({) = {4 . Then (i j)({) = i(j({)) = i (2{ + {2 ) = (2{ + {2 )4 = I ({).
42. Let j({) = cos { and i ({) = {2 . Then (i j)({) = i (j({)) = i ( cos {) = (cos {)2 = cos2 { = I ({).
I
3
I
{
{
I = I ({).
. Then (i j)({) = i (j({)) = i ( 3 { ) =
1+{
1+ 3{
43. Let j({) =
I
3
{ and i ({) =
44. Let j({) =
u
I
{
{
{
and i ({) = 3 {. Then (i j)({) = i (j({)) = i
= 3
= J({).
1+{
1+{
1+{
45. Let j(w) = w2 and i (w) = sec w tan w. Then (i j)(w) = i (j(w)) = i (w2 ) = sec(w2 ) tan(w2 ) = y(w).
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°
licated, or posted to a publicly accessible website, in whole or in part.
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¤
35
36
NOT FOR SALE
¤
CHAPTER 1 FUNCTIONS AND MODELS
46. Let j(w) = tan w and i (w) =
47. Let k({) =
tan w
w
. Then (i j)(w) = i(j(w)) = i (tan w) =
= x(w).
1+w
1 + tan w
I
I
{, j({) = { 3 1, and i({) = {. Then
sI
I
I
(i j k)({) = i (j(k({))) = i (j( {)) = i ( { 3 1) =
{ 3 1 = U({).
48. Let k({) = |{|, j({) = 2 + {, and i({) =
I
8
{. Then
(i j k)({) = i (j(k({))) = i (j(|{|)) = i (2 + |{|) =
49. Let k({) =
I
{, j({) = sec {, and i({) = {4 . Then
s
8
2 + |{| = K({).
I
I
I 4
I
(i j k)({) = i (j(k({))) = i (j( { )) = i (sec { ) = (sec { ) = sec4 ( { ) = K({).
50. (a) i (j(1)) = i (6) = 5
(b) j(i (1)) = j(3) = 2
(c) i (i (1)) = i (3) = 4
(d) j(j(1)) = j(6) = 3
(e) (j i)(3) = j(i(3)) = j(4) = 1
(f ) (i j)(6) = i (j(6)) = i (3) = 4
51. (a) j(2) = 5, because the point (2> 5) is on the graph of j. Thus, i(j(2)) = i (5) = 4, because the point (5> 4) is on the
graph of i .
(b) j(i(0)) = j(0) = 3
(c) (i j)(0) = i (j(0)) = i (3) = 0
(d) (j i )(6) = j(i (6)) = j(6). This value is not de¿ned, because there is no point on the graph of j that has
{-coordinate 6.
(e) (j j)(32) = j(j(32)) = j(1) = 4
(f ) (i i )(4) = i (i (4)) = i(2) = 32
52. To ¿nd a particular value of i (j({)), say for { = 0, we note from the graph that j(0) E 2=8 and i (2=8) E 30=5. Thus,
i (j (0)) E i(2=8) E 30=5. The other values listed in the table were obtained in a similar fashion.
{
j({)
i (j({))
{
j({)
35
30=2
34
33
2=2
31=7
31
3
30=2
4
34
32
1=2
2=8
0
2=8
33=3
1
2=2
2
1=2
30=5
3
30=2
5
31=9
34=1
i (j({))
30=5
31=7
33=3
34
32=2
1=9
53. (a) Using the relationship distance = rate · time with the radius u as the distance, we have u(w) = 60w.
(b) D = u2
i (D u)(w) = D(u(w)) = (60w)2 = 3600w2 . This formula gives us the extent of the rippled area
(in cm2 ) at any time w.
54. (a) The radius u of the balloon is increasing at a rate of 2 cm@s, so u(w) = (2 cm@s)(w s) = 2w (in cm).
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NOT FOR SALE
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS
(b) Using Y = 43 u3 , we get (Y u)(w) = Y (u(w)) = Y (2w) = 43 (2w)3 =
The result, Y =
32
w3 ,
3
¤
3
32
3 w .
gives the volume of the balloon (in cm3 ) as a function of time (in s).
55. (a) From the ¿gure, we have a right triangle with legs 6 and g, and hypotenuse v.
By the Pythagorean Theorem, g2 + 62 = v2
i v = i (g) =
I
g 2 + 36.
(b) Using g = uw, we get g = (30 km@h)(w hours) = 30w (in km). Thus,
g = j(w) = 30w.
(c) (i j)(w) = i (j(w)) = i (30w) =
s
I
(30w)2 + 36 = 900w2 + 36. This function represents the distance between the
lighthouse and the ship as a function of the time elapsed since noon.
56. (a) g = uw
i g(w) = 350w
(b) There is a Pythagorean relationship involving the legs with lengths g and 1 and the hypotenuse with length v:
I
g2 + 12 = v2 . Thus, v(g) = g2 + 1.
(c) (v g)(w) = v(g(w)) = v(350w) =
57. (a)
K(w) =
+
0
if w ? 0
1
if w D 0
s
(350w)2 + 1
(b)
Y (w) =
+
0
if w ? 0
so Y (w) = 120K(w).
120 if w D 0
Starting with the formula in part (b), we replace 120 with 240 to reÀect the
(c)
different voltage. Also, because we are starting 5 units to the right of w = 0,
we replace w with w 3 5. Thus, the formula is Y (w) = 240K(w 3 5).
58. (a) U(w) = wK(w)
=
+
0 if w ? 0
w
if w D 0
(b) Y (w) =
+
0
if w ? 0
2w if 0 $ w $ 60
(c) Y (w) =
so Y (w) = 2wK(w), w $ 60.
+
0
if w ? 7
4 (w 3 7) if 7 $ w $ 32
so Y (w) = 4(w 3 7)K(w 3 7), w $ 32.
59. If i({) = p1 { + e1 and j({) = p2 { + e2 , then
(i j)({) = i (j({)) = i (p2 { + e2 ) = p1 (p2 { + e2 ) + e1 = p1 p2 { + p1 e2 + e1 .
So i j is a linear function with slope p1 p2 .
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37
38
¤
NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
60. If D({) = 1=04{, then
(D D)({) = D(D({)) = D(1=04{) = 1=04(1=04{) = (1=04)2 {,
(D D D)({) = D((D D)({)) = D((1=04)2 {) = 1=04(1=04)2 { = (1=04)3 {, and
(D D D D)({) = D((D D D)({)) = D((1=04)3 {) = 1=04(1=04)3 {> = (1=04)4 {.
These compositions represent the amount of the investment after 2, 3, and 4 years.
Based on this pattern, when we compose q copies of D, we get the formula (D D · · · D)({) = (1=04)q {.

~}
€
q D0 s
61. (a) By examining the variable terms in j and k, we deduce that we must square j to get the terms 4{2 and 4{ in k. If we let
i ({) = {2 + f, then (i j)({) = i (j({)) = i (2{ + 1) = (2{ + 1)2 + f = 4{2 + 4{ + (1 + f). Since
k({) = 4{2 + 4{ + 7, we must have 1 + f = 7. So f = 6 and i ({) = {2 + 6.
(b) We need a function j so that i (j({)) = 3(j({)) + 5 = k({). But
k({) = 3{2 + 3{ + 2 = 3({2 + {) + 2 = 3({2 + { 3 1) + 5, so we see that j({) = {2 + { 3 1.
62. We need a function j so that j(i ({)) = j({ + 4) = k({) = 4{ 3 1 = 4({ + 4) 3 17. So we see that the function j must be
j({) = 4{ 3 17.
63. We need to examine k(3{).
k(3{) = (i j)(3{) = i(j(3{)) = i (j({)) [because j is even]
= k({)
Because k(3{) = k({), k is an even function.
64. k(3{) = i (j(3{)) = i (3j({)). At this point, we can’t simplify the expression, so we might try to ¿nd a counterexample to
show that k is not an odd function. Let j({) = {, an odd function, and i ({) = {2 + {. Then k({) = {2 + {> which is neither
even nor odd.
Now suppose i is an odd function. Then i (3j({)) = 3i (j({)) = 3k({). Hence, k(3{) = 3k({), and so k is odd if
both i and j are odd.
Now suppose i is an even function. Then i (3j({)) = i (j({)) = k({). Hence, k(3{) = k({), and so k is even if j is
odd and i is even.
1.4 Graphing Calculators and Computers
1. i ({) =
I
{3 3 5{2
(a) [35> 5] by [35> 5]
(b) [0> 10] by [0> 2]
(c) [0> 10] by [0> 10]
(There is no graph shown.)
The most appropriate graph is produced in viewing rectangle (c).
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NOT FOR SALE
SECTION 1.4
GRAPHING CALCULATORS AND COMPUTERS
2. i ({) = {4 3 16{2 + 20
(a) [33> 3] by [33> 3]
(b) [310> 10] by [310> 10]
(c) [350> 50] by [350> 50]
(d) [35> 5] by [350> 50]
The most appropriate graph is produced in viewing rectangle (d).
3. Since the graph of i ({) = {2 3 36{ + 32 is a parabola opening upward,
an appropriate viewing rectangle should include the minimum point.
Completing the square, we get i({) = ({ 3 18)2 3 292, and so the
minimum point is (18> 3292).
4. An appropriate viewing rectangle for i ({) = {3 + 15{2 + 65{ should
include the high and low points.
5. 50 3 0=2{ D 0
i 50 D 0=2{ i { $ 250, so the domain of the
I
root function i({) = 50 3 0=2{ is (3"> 250].
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39
40
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NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
6. 15{ 3 {2 D 0
i {(15 3 {) D 0 i 0 $ { $ 15, so the domain
I
of the root function i ({) = 15{ 3 {2 is [0> 15]. The graph is a
semicircle, so an appropriate viewing rectangle should make it look like a
semicircle.
7. The graph of i ({) = {3 3 225{ is symmetric with respect to the origin.
Since i ({) = {3 3 225{ = {({2 3 225) = {({ + 15)({ 3 15), there
are {-intercepts at 0, 315, and 15. i (20) = 3500.
8. The graph of i ({) = {@({2 + 100) is symmetric with respect to the
origin.
9. The period of j({) = sin(1000{) is
2
1000
E 0=0063 and its range is
[31> 1]. Since i ({) = sin2 (1000{) is the square of j, its range is
[0> 1] and a viewing rectangle of [30=01> 0=01] by [0> 1=1] seems
appropriate.
10. The period of i ({) = cos(0=001{) is
2
0=001
E 6300 and its range
is [31> 1], so a viewing rectangle of [310,000> 10,000] by
[31=5> 1=5] seems appropriate.
11. The domain of | =
I
I
{ is { D 0, so the domain of i({) = sin { is [0> ")
and the range is [31> 1]. With a little trial-and-error experimentation, we ¿nd
that an Xmax of 100 illustrates the general shape of i , so an appropriate
viewing rectangle is [0> 100] by [31=5> 1=5].
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NOT FOR SALE
SECTION 1.4
12. One period of | = sec { occurs on the interval 3 2 >
3 2 ? 20{ ?
3
2
1
i 3 40
?{?
3
,
40
2
or equivalently,
3
2> 2
GRAPHING CALCULATORS AND COMPUTERS
¤
.
30=025 ? { ? 0=075.
13. The ¿rst term, 10 sin {, has period 2 and range [310> 10]. It will be the dominant term in any “large” graph of
| = 10 sin { + sin 100{, as shown in the ¿rst ¿gure. The second term, sin 100{, has period
2
100
=
50
and range [31> 1].
It causes the bumps in the ¿rst ¿gure and will be the dominant term in any “small” graph, as shown in the view near the
origin in the second ¿gure.
14. | = {2 + 0=02 sin(50{)
15. (a) The ¿rst ¿gure shows the "big
picture" for i ({) = ({ 3 10)3 23{ .
The second ¿gure shows a maximum
near { = 10.
(b) You need more than one window because no single window can show what the function looks like globally
and the detail of the function near { = 10.
I
16. The function i ({) = {2 30 3 { has domain (3"> 30]. Its graph is very
steep near { = 30, so part of the graph may appear to be missing.
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42
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NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
17. We must solve the given equation for | to obtain equations for the upper and
lower halves of the ellipse.
4{2 + 2| 2 = 1 C 2| 2 = 1 3 4{2
u
1 3 4{2
|=±
2
18. | 2 3 9{2 = 1
C
| 2 = 1 + 9{2
C |2 =
1 3 4{2
2
C
I
C | = ± 1 + 9{2
19. From the graph of | = 3{2 3 6{ + 1 and | = 0=23{ 3 2=25 in the viewing rectangle [31> 3] by [32=5> 1=5], it is dif¿cult to
see if the graphs intersect. If we zoom in on the fourth quadrant, we see the graphs do not intersect.
20. From the graph of | = 6 3 4{ 3 {2 and | = 3{ + 18 in the viewing
rectangle [36> 2] by [35> 20], we see that the graphs intersect twice.
The points of intersection are (34> 6) and (33> 9).
21. We see that the graphs of i ({) = {4 3 { and j({) = 1 intersect twice.
The {-coordinates of these points (which are the solutions of the equations)
are approximately 30=72 and 1=22. Alternatively, we could ¿nd these
values by ¿nding the zeros of k({) = {4 3 { 3 1.
22. We see that the graphs of i ({) =
I
{ and j({) = {3 3 1 intersect once.
The {-coordinate of this point (which is the solution of the equation) is
approximately 1=29. Alternatively, we could ¿nd this value by ¿nding the
I
zero of k({) = { 3 {3 + 1.
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NOT FOR SALE
SECTION 1.4
23. We see that the graphs of i ({) = tan { and j({) =
GRAPHING CALCULATORS AND COMPUTERS
¤
I
1 3 {2 intersect
once. Using an intersect feature or zooming in, we ¿nd this value to be
approximately 0=65. Alternatively, we could ¿nd this value by ¿nding the
I
positive zero of k({) = tan { 3 1 3 {2 .
Note: After producing the graph on a TI-84 Plus, we can ¿nd the approximate value 0.65 by using the following keystrokes:
. The “.6” is just a guess for 0=65.
24. (a)
The {-coordinates of the three points of intersection are
{ E 33=29, 32=36 and 1=20.
(b) Using trial and error, we ¿nd that p E 0=3365. Note that p could also be negative.
25. j({) = {3 @10 is larger than i ({) = 10{2
whenever { A 100.
26. i ({) = {4 3 100{3 is larger than j({) = {3
whenever { A 101.
27. We see from the graphs of | = |tan { 3 {| and | = 0=01 that there are two
solutions to the equation | = |tan { 3 {| = 0=01 for 3@2 ? { ? @2:
{ r 30=31 and { r 0=31. The condition |tan { 3 {| ? 0=01 holds for any
{ lying between these two values, that is, 30=31 ? { ? 0=31.
28. S ({) = 3{5 3 5{3 + 2{, T({) = 3{5 . These graphs are signi¿cantly different only in the region close to the origin.
The larger a viewing rectangle one chooses, the more similar the two graphs look.
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44
¤
NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
29. (a) The root functions | =
|=
I
{,
I
I
4
{ and | = 6 {
(b) The root functions | = {,
I
I
| = 3 { and | = 5 {
(c) The root functions | =
I
I
| = 4 { and | = 5 {
I
I
{, | = 3 {,
(d) • For any q, the qth root of 0 is 0 and the qth root of 1 is 1; that is, all qth root functions pass through the points (0> 0)
and (1> 1).
• For odd q, the domain of the qth root function is R, while for even q, it is {{ M R | { D 0}.
I
I
• Graphs of even root functions look similar to that of {, while those of odd root functions resemble that of 3 {.
• As q increases, the graph of
30. (a) The functions | = 1@{ and
| = 1@{3
I
q
{ becomes steeper near 0 and Àatter for { A 1.
(b) The functions | = 1@{2 and
(c) The functions | = 1@{, | = 1@{2 ,
| = 1@{4
| = 1@{3 and | = 1@{4
(d) • The graphs of all functions of the form | = 1@{q pass through the point (1> 1).
• If q is even, the graph of the function is entirely above the {-axis. The graphs of 1@{q for q even are similar to
one another.
• If q is odd, the function is positive for positive { and negative for negative {. The graphs of 1@{q for q odd are
similar to one another.
• As q increases, the graphs of 1@{q approach 0 faster as { < ".
31. i ({) = {4 + f{2 + {. If f ? 31=5, there are three humps: two minimum points
and a maximum point. These humps get Àatter as f increases, until at f = 31=5
two of the humps disappear and there is only one minimum point. This single
hump then moves to the right and approaches the origin as f increases.
I
1 + f{2 . If f ? 0, the function is only de¿ned on
I
I
31
3f > 1
3f , and its graph is the top half of an ellipse. If f = 0, the
32. i ({) =
graph is the line | = 1. If f A 0, the graph is the top half of a hyperbola. As f
approaches 0, these curves become Àatter and approach the line | = 1.
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SECTION 1.4
GRAPHING CALCULATORS AND COMPUTERS
¤
45
33. | = {q 23{ . As q increases, the maximum of the
function moves further from the origin, and gets
larger. Note, however, that regardless of q, the
function approaches 0 as { < ".
|{|
. The “bullet” becomes broader as
f 3 {2
34. | = I
f increases.
35. | 2 = f{3 + {2 . If f ? 0, the loop is to the right of the origin, and if f is positive,
it is to the left. In both cases, the closer f is to 0, the larger the loop is. (In the
limiting case, f = 0, the loop is “in¿nite,” that is, it doesn’t close.) Also, the
larger |f| is, the steeper the slope is on the loopless side of the origin.
I
(b) | = sin({2 )
36. (a) | = sin( { )
This function is not periodic; it oscillates less
This function oscillates more frequently as |{| increases.
frequently as { increases.
Note also that this function is even, whereas sin { is odd.
37. The graphing window is 95 pixels wide and we want to start with { = 0 and end with { = 2. Since there are 94 “gaps”
Thus, the {-values that the calculator actually plots are { = 0 + 2
· q,
94
where q = 0, 1, 2, = = = , 93, 94. For | = sin 2{, the actual points plotted by the calculator are 2
· q> sin 2 · 2
· q for
94
94
· q> sin 96 · 2
· q for q = 0, 1, = = = , 94= But
q = 0, 1, = = = , 94= For | = sin 96{, the points plotted are 2
94
94
between pixels, the distance between pixels is
sin 96 ·
2
94
230
.
94
2
2
· q = sin 94 · 2
94 · q + 2 · 94 · q = sin 2q + 2 · 94 · q
[by periodicity of sine]> q = 0, 1, = = = , 94
= sin 2 · 2
94 · q
So the |-values, and hence the points, plotted for | = sin 96{ are identical to those plotted for | = sin 2{.
Note: Try graphing | = sin 94{. Can you see why all the |-values are zero?
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NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
38. As in Exercise 35, we know that the points being plotted for | = sin 45{ are
But
sin 45 ·
2
94
· q = sin 47 ·
2
94
· q> sin 45 ·
2
94
· q for q = 0, 1, = = = , 94.
· q 3 2 · 2
· q = sin q 3 2 · 2
·q
94
94
2
= sin(q) cos 2 · 2
[Subtraction formula for the sine]
94 · q 3 cos(q) sin 2 · 94 · q
2
2 = 0 · cos 2 · 94 · q 3 (±1) sin 2 · 94 · q
· q > q = 0, 1, = = = , 94
= ± sin 2 · 2
94
2
94
So the |-values, and hence the points, plotted for | = sin 45{ lie on either | = sin 2{ or | = 3 sin 2{.
1.5 Exponential Functions
1. (a)
433
28
28
28
=
=
=
= 2836 = 22 = 4
238
43
(22 )3
26
1
1
(b) I
= 4@3 = {34@3
3
4
{
{
2. (a) 84@3 = (81@3 )4 = 24 = 16
(b) {(3{2 )3 = { · 33 ({2 )3 = 27{ · {6 = 27{7
3. (a) e8 (2e)4 = e8 · 24 e4 = 16e12
(b)
4. (a)
(6| 3 )4
64 (| 3 )4
1296| 12
=
=
= 648| 7
2| 5
2| 5
2| 5
{2q · {3q31
{2q+3q31
{5q31
=
= q+2 = {4q33
q+2
q+2
{
{
{
s I
I sI
d e
d
e
d1@2 e1@4
I
= I
= 1@3 1@3 = d(1@231@3) e(1@431@3) = d1@6 e31@12
(b) I
3
3
3
d e
de
d e
5. (a) i ({) = d{ , d A 0
(c) (0> ")
(b) R
(d) See Figures 4(c), 4(b), and 4(a), respectively.
6. (a) The number h is the value of d such that the slope of the tangent line at { = 0 on the graph of | = d{ is exactly 1.
(b) h E 2=71828
(c) i ({) = h{
7. All of these graphs approach 0 as { < 3", all of them pass through the point
(0> 1), and all of them are increasing and approach " as { < ". The larger the
base, the faster the function increases for { A 0, and the faster it approaches 0 as
{ < 3".
Note: The notation “{ < "” can be thought of as “{ becomes large” at this point.
More details on this notation are given in Chapter 2.
8. The graph of h3{ is the reÀection of the graph of h{ about the |-axis, and the
graph of 83{ is the reÀection of that of 8{ about the |-axis. The graph of 8{
increases more quickly than that of h{ for { A 0, and approaches 0 faster
as { < 3".
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 1.5
EXPONENTIAL FUNCTIONS
9. The functions with bases greater than 1 (3{ and 10{ ) are increasing, while those
with bases less than 1
1 {
3
and
1 {
10
are decreasing. The graph of
reÀection of that of 3{ about the |-axis, and the graph of
1
10
{
1 {
3
is the
is the reÀection of
that of 10{ about the |-axis. The graph of 10{ increases more quickly than that of
3{ for { A 0, and approaches 0 faster as { < 3".
10. Each of the graphs approaches " as { < 3", and each approaches 0 as
{ < ". The smaller the base, the faster the function grows as { < 3", and
the faster it approaches 0 as { < ".
11. We start with the graph of | = 10{
(Figure 3) and shift it 2 units to the left to
obtain the graph of | = 10{+2 .
12. We start with the graph of | = (0=5){
(Figure 3) and shift it 2 units downward
to obtain the graph of | = (0=5){ 3 2.
The horizontal asymptote of the ¿nal
graph is | = 32.
13. We start with the graph of | = 2{ (Figure 2),
reÀect it about the |-axis, and then about the
{-axis (or just rotate 180 to handle both
reÀections) to obtain the graph of | = 323{ .
In each graph, | = 0 is the horizontal
asymptote.
| = 2{
| = 23{
| = 323{
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48
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NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
14. We start with the graph of | = h{ (Figure 12) and
reÀect the portion of the graph in the ¿rst quadrant
about the |-axis to obtain the graph of | = h|{| .
15. We start with the graph of | = h{ (Figure 13) and reÀect about the |-axis to get the graph of | = h3{ . Then we compress
the graph vertically by a factor of 2 to obtain the graph of | = 12 h3{ and then reÀect about the {-axis to get the graph of
| = 3 12 h3{ . Finally, we shift the graph upward one unit to get the graph of | = 1 3 12 h3{ .
16. We start with the graph of | = h{ (Figure 13) and reÀect about the {-axis to get the graph of | = 3h{ . Then shift the graph
upward one unit to get the graph of | = 1 3 h{ . Finally, we stretch the graph vertically by a factor of 2 to obtain the graph of
| = 2(1 3 h{ ).
17. (a) To ¿nd the equation of the graph that results from shifting the graph of | = h{ 2 units downward, we subtract 2 from the
original function to get | = h{ 3 2.
(b) To ¿nd the equation of the graph that results from shifting the graph of | = h{ 2 units to the right, we replace { with { 3 2
in the original function to get | = h({32) .
(c) To ¿nd the equation of the graph that results from reÀecting the graph of | = h{ about the {-axis, we multiply the original
function by 31 to get | = 3h{ .
(d) To ¿nd the equation of the graph that results from reÀecting the graph of | = h{ about the |-axis, we replace { with 3{ in
the original function to get | = h3{ .
(e) To ¿nd the equation of the graph that results from reÀecting the graph of | = h{ about the {-axis and then about the
|-axis, we ¿rst multiply the original function by 31 (to get | = 3h{ ) and then replace { with 3{ in this equation to
get | = 3h3{ .
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SECTION 1.5
EXPONENTIAL FUNCTIONS
¤
49
18. (a) This reÀection consists of ¿rst reÀecting the graph about the {-axis (giving the graph with equation | = 3h{ )
and then shifting this graph 2 · 4 = 8 units upward. So the equation is | = 3h{ + 8.
(b) This reÀection consists of ¿rst reÀecting the graph about the |-axis (giving the graph with equation | = h3{ )
and then shifting this graph 2 · 2 = 4 units to the right. So the equation is | = h3({34) .
2
19. (a) The denominator is zero when 1 3 h13{ = 0
C
{2
2
h13{ = 1
C
1 3 {2 = 0
C
{ = ±1. Thus,
13h
the function i({) =
has domain {{ | { 6= ±1} = (3"> 31) (31> 1) (1> ").
1 3 h13{2
1+{
(b) The denominator is never equal to zero, so the function i ({) = cos { has domain R, or (3"> ").
h
20. (a) The sine and exponential functions have domain R, so j(w) = sin(h3w ) also has domain R.
(b) The function j(w) =
I
1 3 2w has domain {w | 1 3 2w D 0} = {w | 2w $ 1} = {w | w $ 0} = (3"> 0].
21. Use | = Fd{ with the points (1> 6) and (3> 24).
4 = d2
i d = 2 [since d A 0] and F =
22. Use | = Fd{ with the points (31> 3) and 1>
the point 1> 43 , we get
4
3
= Fd1
i
4
3
4
3
F = d6
and 24 = Fd3
6 = Fd1
6
2
i 24 =
6 3
d
d
i
= 3. The function is i ({) = 3 · 2{ .
. From the point (31> 3), we have 3 = Fd31 , hence F = 3d. Using this and
= (3d)d
i
4
9
= d2
i d=
2
3
[since d A 0] and F = 3( 23 ) = 2. The
function is i({) = 2( 23 ){ .
k
5{ 5k 3 1
5{+k 3 5{
5{ 5k 3 5{
i ({ + k) 3 i ({)
{ 5 31
=
=
=
=5
.
23. If i ({) = 5 , then
k
k
k
k
k
{
24. Suppose the month is February. Your payment on the 28th day would be 22831 = 227 = 134,217,728 cents, or
$1,342,177.28. Clearly, the second method of payment results in a larger amount for any month.
25. 2 ft = 24 in, i (24) = 242 in = 576 in = 48 ft.
j(24) = 224 in = 224 @(12 · 5280) mi E 265 mi
26. We see from the graphs that for { less than about 1=8, j({) = 5{ A i ({) = {5 , and then near the point (1=8> 17=1) the curves
intersect. Then i({) A j({) from { E 1=8 until { = 5. At (5> 3125) there is another point of intersection, and for { A 5 we
see that j({) A i({). In fact, j increases much more rapidly than i beyond that point.
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50
¤
NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
27. The graph of j ¿nally surpasses that of i at { E 35=8.
28. We graph | = h{ and | = 1,000,000,000 and determine where
h{ = 1 × 109 . This seems to be true at { E 20=723, so h{ A 1 × 109
for { A 20=723.
29. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 · 25 = 3200
(b) In w hours, there will be w@3 doubling periods. The initial population is 100,
so the population | at time w is | = 100 · 2w@3 .
(c) w = 20 i | = 100 · 220@3 E 10,159
(d) We graph |1 = 100 · 2{@3 and |2 = 50,000. The two curves intersect at
{ E 26=9, so the population reaches 50,000 in about 26=9 hours.
30. (a) Three hours represents 6 doubling periods (one doubling period is 30 minutes).
500 · 26 = 32,000
(b) In w hours, there will be 2w doubling periods. The initial population is 500,
so the population | at time w is | = 500 · 22w .
(c) w =
40
60
=
2
3
i | = 500 · 22(2@3) E 1260
(d) We graph |1 = 500 · 22w and |2 = 100,000. The two curves intersect at
w E 3=82, so the population reaches 100,000 in about 3=82 hours.
31. Let w = 0 correspond to 1950 to get the model S = dew , where d r 2614=086 and e r 1=01693. To estimate the population in
1993, let w = 43 to obtain S r 5381 million. To predict the population in 2020, let w = 70 to obtain S r 8466 million.
32. Let w = 0 correspond to 1900 to get the model S = dew , where d r 80=8498 and e r 1=01269. To estimate the population in
1925, let w = 25 to obtain S r 111 million. To predict the population in 2020, let w = 120 to obtain S r 367 million.
33.
From the graph, it appears that i is an odd function (i is unde¿ned for { = 0).
To prove this, we must show that i (3{) = 3i ({).
1
1 3 1@{ h1@{
1 3 h1@(3{)
1 3 h(31@{)
h1@{ 3 1
h
i (3{) =
=
=
· 1@{ = 1@{
1@(3{)
(31@{)
1
1+h
1+h
h
h
+1
1 + 1@{
h
1@{
13h
= 3i ({)
=3
1 + h1@{
so i is an odd function.
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SECTION 1.6
34. We’ll start with e = 31 and graph i ({) =
INVERSE FUNCTIONS AND LOGARITHMS
¤
1
for d = 0=1, 1, and 5.
1 + dhe{
From the graph, we see that there is a horizontal asymptote | = 0 as { < 3"
and a horizontal asymptote | = 1 as { < ". If d = 1, the y-intercept is 0> 12 .
As d gets smaller (close to 0), the graph of i moves left. As d gets larger, the graph
of i moves right.
As e changes from 31 to 0, the graph of i is stretched horizontally. As e
changes through large negative values, the graph of i is compressed horizontally.
(This takes care of negatives values of e.)
If e is positive, the graph of i is reÀected through the y-axis.
Last, if e = 0, the graph of i is the horizontal line | = 1@(1 + d).
1.6 Inverse Functions and Logarithms
1. (a) See De¿nition 1.
(b) It must pass the Horizontal Line Test.
2. (a) i 31 (|) = {
C i ({) = | for any | in E. The domain of i 31 is E and the range of i 31 is D.
(b) See the steps in (5).
(c) ReÀect the graph of i about the line | = {.
3. i is not one-to-one because 2 6= 6, but i (2) = 2=0 = i (6).
4. i is one-to-one because it never takes on the same value twice.
5. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the
function is not one-to-one.
6. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one-to-one.
7. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one-to-one.
8. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the
function is not one-to-one.
9. The graph of i ({) = {2 3 2{ is a parabola with axis of symmetry { = 3
32
e
=3
= 1. Pick any {-values equidistant
2d
2(1)
from 1 to ¿nd two equal function values. For example, i (0) = 0 and i (2) = 0, so i is not one-to-one.
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CHAPTER 1
FUNCTIONS AND MODELS
10. The graph of i ({) = 10 3 3{ is a line with slope 33. It passes the Horizontal Line Test, so i is one-to-one.
Algebraic solution: If {1 6= {2 , then 33{1 6= 33{2
11. j({) = 1@{.
{1 6= {2
i 1@{1 6= 1@{2
i 10 3 3{1 6= 10 3 3{2
i i ({1 ) 6= i({2 ), so i is one-to-one.
i j ({1 ) 6= j ({2 ), so j is one-to-one.
Geometric solution: The graph of j is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test,
so j is one-to-one.
12. j({) = cos {.
j(0) = 1 = j(2), so j is not one-to-one.
13. A football will attain every height k up to its maximum height twice: once on the way up, and again on the way down.
Thus, even if w1 does not equal w2 , i (w1 ) may equal i (w2 ), so i is not 1-1.
14. i is not 1-1 because eventually we all stop growing and therefore, there are two times at which we have the same height.
C i 31 (17) = 6.
15. (a) Since i is 1-1, i (6) = 17
(b) Since i is 1-1, i 31 (3) = 2 C i (2) = 3.
16. First, we must determine { such that i ({) = 3. By inspection, we see that if { = 1, then i (1) = 3. Since i is 1-1 (i is an
increasing function), it has an inverse, and i 31 (3) = 1. If i is a 1-1 function, then i (i 31 (d)) = d, so i (i 31 (2)) = 2.
17. First, we must determine { such that j({) = 4. By inspection, we see that if { = 0, then j({) = 4. Since j is 1-1 (j is an
increasing function), it has an inverse, and j31 (4) = 0.
18. (a) i is 1-1 because it passes the Horizontal Line Test.
(b) Domain of i = [33> 3] = Range of i 31 . Range of i = [31> 3] = Domain of i 31 .
(c) Since i(0) = 2, i 31 (2) = 0.
(d) Since i(31=7) E 0, i 31 (0) = 31=7.
19. We solve F =
5
(I
9
3 32) for I : 95 F = I 3 32 i I = 95 F + 32. This gives us a formula for the inverse function, that
is, the Fahrenheit temperature I as a function of the Celsius temperature F. I D 3459=67 i
9
F
5
9
F
5
+ 32 D 3459=67 i
D 3491=67 i F D 3273=15, the domain of the inverse function.
p0
1 3 y 2 @f2
20. p = s
i 13
y2
p2
= 02
2
f
p
i
y2
p2
= 1 3 02
2
f
p
u
p2
p2
i y = f 1 3 02 .
i y 2 = f2 1 3 02
p
p
This formula gives us the speed y of the particle in terms of its mass p, that is, y = i 31 (p).
21. | = i({) = 1 +
I
2 + 3{
(| D 1)
i |31=
I
2 + 3{
i (| 3 1)2 = 2 + 3{ i (| 3 1)2 3 2 = 3{ i
{ = 13 (| 3 1)2 3 23 . Interchange { and |: | = 13 ({ 3 1)2 3 23 . So i 31 ({) = 13 ({ 3 1)2 3 23 . Note that the domain of i 31
is { D 1.
22. | = i ({) =
4{ 3 1
2{ + 3
i |(2{ + 3) = 4{ 3 1 i 2{| + 3| = 4{ 3 1 i 3| + 1 = 4{ 3 2{|
3| + 1 = (4 3 2|){ i { =
i
3{ + 1
3| + 1
3{ + 1
. Interchange { and |: | =
. So i 31 ({) =
.
4 3 2|
4 3 2{
4 3 2{
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SECTION 1.6
23. | = i ({) = h2{31
i ln | = 2{ 3 1
INVERSE FUNCTIONS AND LOGARITHMS
¤
i 1 + ln | = 2{ i { = 12 (1 + ln |).
Interchange { and |: | = 12 (1 + ln {). So i 31 ({) = 12 (1 + ln {).
24. | = i ({) = {2 3 {
|+
1
4
= ({ 3 12 )2
i 31 ({) =
1
2
+
({ D 12 )
i
t
{ + 14 .
25. | = i ({) = ln ({ + 3)
{3
| = {2 3 { + 14 3
t
= | + 14 i { =
i
1
2
i { + 3 = h|
1
4
1
2
i | = ({ 3 12 )2 3 14 i
t
+ | + 14 . Interchange { and |: | =
1
2
+
t
{ + 14 . So
i { = h| 3 3. Interchange { and |: | = h{ 3 3. So i 31 ({) = h{ 3 3.
h{
|
i
i | + 2|h{ = h{ i | = h{ 3 2|h{ i | = h{ (1 3 2|) i h{ =
1 + 2h{
1 3 2|
{
{
|
31
. Interchange { and |: | = ln
. So i ({) = ln
. Note that the range of i and the
{ = ln
1 3 2|
1 3 2{
1 3 2{
26. | = i ({) =
domain of i 31 is (0> 12 ).
I
i | 3 1 = {4 i { = 4 | 3 1 [not ± since
I
I
{ D 0]. Interchange { and |: | = 4 { 3 1. So i 31 ({) = 4 { 3 1. The
I
I
graph of | = 4 { 3 1 is just the graph of | = 4 { shifted right one unit.
27. | = i ({) = {4 + 1
From the graph, we see that i and i 31 are reÀections about the line | = {.
28. | = i ({) = 2 3 h{
i h{ = 2 3 |
i { = ln(2 3 |). Interchange
{ and |: | = ln(2 3 {). So i 31 ({) = ln(2 3 {). From the graph, we see
that i and i 31 are reÀections about the line | = {.
29. ReÀect the graph of i about the line | = {. The points (31> 32), (1> 31),
(2> 2), and (3> 3) on i are reÀected to (32> 31), (31> 1), (2> 2), and (3> 3)
on i 31 .
30. ReÀect the graph of i about the line | = {.
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CHAPTER 1
31. (a) | = i ({) =
FUNCTIONS AND MODELS
I
1 3 {2
(0 $ { $ 1 and note that | D 0) i
s
i {2 = 1 3 | 2 i { = 1 3 | 2 . So
|2 = 1 3 {2
I
i 31 ({) = 1 3 {2 , 0 $ { $ 1. We see that i 31 and i are the same
function.
(b) The graph of i is the portion of the circle {2 + | 2 = 1 with 0 $ { $ 1 and
0 $ | $ 1 (quarter-circle in the ¿rst quadrant). The graph of i is symmetric
with respect to the line | = {, so its reÀection about | = { is itself, that is,
i 31 = i .
I
3
1 3 {3
| 3 = 1 3 {3 i {3 = 1 3 | 3 i
s
I
{ = 3 1 3 |3 . So j 31 ({) = 3 1 3 {3 . We see that j and j31 are the
32. (a) | = j({) =
i
same function.
(b) The graph of j is symmetric with respect to the line | = {, so its reÀection
about | = { is itself, that is, j 31 = j.
33. (a) It is de¿ned as the inverse of the exponential function with base d, that is, logd { = |
(b) (0> ")
(c) R
C d| = {.
(d) See Figure 11.
34. (a) The natural logarithm is the logarithm with base h, denoted ln {.
(b) The common logarithm is the logarithm with base 10, denoted log {.
(c) See Figure 13.
1
1
1
= 33 since 333 = 3 =
.
27
3
27
35. (a) log5 125 = 3 since 53 = 125.
(b) log3
36. (a) ln(1@h) = ln 1 3 ln h = 0 3 1 = 31
(b) log10
I
10 = log10 101@2 =
6
37. (a) log2 6 3 log2 15 + log2 20 = log2 ( 15
) + log2 20
1
2
by (7).
[by Law 2]
6
= log2 ( 15
· 20)
[by Law 1]
= log2 8, and log2 8 = 3 since 23 = 8.
100 3 log3 50 = log3 18·50
= log3 ( 19 ), and log3 19 = 32 since 332 = 19 =
(b) log3 100 3 log3 18 3 log3 50 = log3
38. (a) h32 ln 5 = hln 5
32
(9)
= 532 =
39. ln 5 + 5 ln 3 = ln 5 + ln 35
5
= ln(5 · 3 )
100 18
1
1
=
52
25
(9)
10
(9)
(b) ln ln hh
= ln(h10 ) = 10
[by Law 3]
[by Law 1]
= ln 1215
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NOT FOR SALE
SECTION 1.6
40. ln(d + e) + ln(d 3 e) 3 2 ln f = ln[(d + e)(d 3 e)] 3 ln f2
= ln
INVERSE FUNCTIONS AND LOGARITHMS
¤
[by Laws 1, 3]
(d + e)(d 3 e)
f2
[by Law 2]
d2 3 e2
f2
41. 13 ln({ + 2)3 + 12 ln { 3 ln({2 + 3{ + 2)2 = ln[({ + 2)3 ]1@3 +
or ln
= ln({ + 2) + ln
1
2
ln
{
({2 + 3{ + 2)2
I
{
{2 + 3{ + 2
[by Laws 3, 2]
[by Law 3]
I
({ + 2) {
({ + 1)({ + 2)
I
{
= ln
{+1
= ln
[by Law 1]
Note that since ln { is de¿ned for { A 0, we have { + 1, { + 2, and {2 + 3{ + 2 all positive, and hence their logarithms
are de¿ned.
42. (a) log12 10 =
ln 10
E 0=926628
ln 12
43. To graph these functions, we use log1=5 { =
(b) log2 8=4 =
ln 8=4
E 3=070389
ln 2
ln {
ln {
and log50 { =
.
ln 1=5
ln 50
These graphs all approach 3" as { < 0+ , and they all pass through the
point (1> 0). Also, they are all increasing, and all approach " as { < ".
The functions with larger bases increase extremely slowly, and the ones with
smaller bases do so somewhat more quickly. The functions with large bases
approach the |-axis more closely as { < 0+ .
44. We see that the graph of ln { is the reÀection of the graph of h{ about the
line | = {, and that the graph of log10 { is the reÀection of the graph of 10{
about the same line. The graph of 10{ increases more quickly than that
of h{ . Also note that log10 { < " as { < " more slowly than ln {.
45. 3 ft = 36 in, so we need { such that log2 { = 36
C
{ = 236 = 68,719,476,736. In miles, this is
68,719,476,736 in ·
1 mi
1 ft
·
E 1,084,587=7 mi.
12 in 5280 ft
46.
From the graphs, we see that i ({) = {0=1 A j({) = ln { for approximately 0 ? { ? 3=06, and then j({) A i({) for
3=06 ? { ? 3=43 × 1015 (approximately). At that point, the graph of i ¿nally surpasses the graph of j for good.
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55
56
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NOT FOR SALE
CHAPTER 1
FUNCTIONS AND MODELS
47. (a) Shift the graph of | = log10 { ¿ve units to the left to
(b) ReÀect the graph of | = ln { about the {-axis to obtain
the graph of | = 3 ln {.
obtain the graph of | = log10 ({ + 5). Note the vertical
asymptote of { = 35.
| = log10 {
| = log10 ({ + 5)
48. (a) ReÀect the graph of | = ln { about the |-axis to obtain
| = ln {
| = 3 ln {
(b) ReÀect the portion of the graph of | = ln { to the right
the graph of | = ln (3{).
of the |-axis about the |-axis. The graph of | = ln |{|
is that reÀection in addition to the original portion.
| = ln {
| = ln (3{)
| = ln {
| = ln |{|
49. (a) The domain of i ({) = ln { + 2 is { A 0 and the range is R.
(b) | = 0 i ln { + 2 = 0 i ln { = 32 i { = h32
(c) We shift the graph of | = ln { two units upward.
50. (a) The domain of i ({) = ln({ 3 1) 3 1 is { A 1 and the range is R.
(b) | = 0 i ln({ 3 1) 3 1 = 0 i ln({ 3 1) = 1 i
{ 3 1 = h1
i {=h+1
(c) We shift the graph of | = ln { one unit to the right and one unit downward.
51. (a) h734{ = 6
C
(b) ln(3{ 3 10) = 2
52. (a) ln({2 3 1) = 3
7 3 4{ = ln 6
C
C
C
7 3 ln 6 = 4{
3{ 3 10 = h2
{2 3 1 = h3
C
C
C
3{ = h2 + 10
{2 = 1 + h3
C
{ = 14 (7 3 ln 6)
C
{ = 13 (h2 + 10)
I
{ = ± 1 + h3 .
(b) h2{ 3 3h{ + 2 = 0 C (h{ 3 1)(h{ 3 2) = 0 C h{ = 1 or h{ = 2 C { = ln 1 or { = ln 2, so { = 0 or ln 2.
53. (a) 2{35 = 3
C log2 3 = { 3 5 C { = 5 + log2 3.
ln 3
Or: 2{35 = 3 C ln 2{35 = ln 3 C ({ 3 5) ln 2 = ln 3 C { 3 5 =
ln 2
C {=5+
ln 3
ln 2
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© Cengage Learning. All Rights Reserved.
SECTION 1.6
INVERSE FUNCTIONS AND LOGARITHMS
¤
57
(b) ln { + ln({ 3 1) = ln({({ 3 1)) = 1 C {({ 3 1) = h1 C {2 3 { 3 h = 0. The quadratic formula (with d = 1,
I
e = 31, and f = 3h) gives { = 12 1 ± 1 + 4h , but we reject the negative root since the natural logarithm is not
I
de¿ned for { ? 0. So { = 12 1 + 1 + 4h .
C hln(ln {) = h1
54. (a) ln(ln {) = 1
(b) hd{ = Fhe{
C ln hd{ = ln[F(he{ )] C d{ = ln F + ln he{
d{ 3 e{ = ln F
55. (a) ln { ? 0
C ln { = h1 = h C hln { = hh
C (d 3 e){ = ln F
i { ? h0
C {=
C { = hh
C d{ = ln F + e{ C
ln F
d3e
i { ? 1. Since the domain of i ({) = ln { is { A 0, the solution of the original inequality
is 0 ? { ? 1.
(b) h{ A 5 i ln h{ A ln 5 i { A ln 5
56. (a) 1 ? h3{31 ? 2
1
3
?{?
1
(1
3
i ln 1 ? 3{ 3 1 ? ln 2 i 0 ? 3{ 3 1 ? ln 2 i 1 ? 3{ ? 1 + ln 2 i
+ ln 2)
(b) 1 3 2 ln { ? 3 i 32 ln { ? 2 i ln { A 31 i { A h31
57. (a) We must have h{ 3 3 A 0
i h{ A 3 i { A ln 3. Thus, the domain of i ({) = ln(h{ 3 3) is (ln 3> ").
(b) | = ln(h{ 3 3) i h| = h{ 3 3 i h{ = h| + 3 i { = ln(h| + 3), so i 31 ({) = ln(h{ + 3).
Now h{ + 3 A 0 i h{ A 33, which is true for any real {, so the domain of i 31 is R.
58. (a) By (9), hln 300 = 300 and ln(h300 ) = 300.
(b) A calculator gives hln 300 = 300 and an error message for ln(h300 ) since h300 is larger than most calculators can evaluate.
59. We see that the graph of | = i ({) =
Enter { =
I
{3 + {2 + { + 1 is increasing, so i is 1-1.
s
| 3 + | 2 + | + 1 and use your CAS to solve the equation for |.
Using Derive, we get two (irrelevant) solutions involving imaginary expressions,
as well as one which can be simpli¿ed to the following:
I
I I
I
3
| = i 31 ({) = 3 64 3 G 3 27{2 + 20 3 3 G + 27{2 3 20 + 3 2
I I
where G = 3 3 27{4 3 40{2 + 16.
Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent
to that given by Derive. For example, Maple’s expression simpli¿es to
P = 108{2 + 12
1 P 2@3 3 8 3 2P 1@3
, where
6
2P 1@3
I
48 3 120{2 + 81{4 3 80.
60. (a) If we use Derive, then solving { = | 6 + | 4 for | gives us six solutions of the form | = ±
D
D
D
+
> 32 cos
+
32 sin > 2 sin
3
3
3
3
I I
D
3
+
({ D 0) is | = 3 E 3 1 with E = 2 sin
3
4
.
valid for { M 0> 27
EM
6
and D = sin31
I I
3
E
3
3 1, where
27{ 3 2
. The inverse for | = {6 + {4
2
, but because the domain of D is 0>
3
4
27
, this expression is only
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FUNCTIONS AND MODELS
Happily, Maple gives us the rest of the solution! We solve { = | 6 + | 4 for | to get the two real solutions
I s 1@3 2@3
I s
(F
3 2F 1@3 + 4)
6 F
, where F = 108{ + 12 3 { (27{ 3 4), and the inverse for | = {6 + {4 ({ D 0)
±
1@3
6
F
4
>" .
is the positive solution, whose domain is 27
(b)
Mathematica also gives two real solutions, equivalent to those of Maple.
I I
6 I
3
4G1@3 + 2 3 2G31@3 3 2 , where
The positive one is
6
I I I
G = 32 + 27{ + 3 3 { 27{ 3 4. Although this expression also has domain
4
> " , Mathematica is mysteriously able to plot the solution for all { D 0.
27
61. (a) q = i (w) = 100 · 2w@3
q
= 2w@3
100
i
write this as w = i 31 (q) = 3 ·
i log2
q w
=
100
3
i w = 3 log2
q . Using formula (10), we can
100
ln(q@100)
. This function tells us how long it will take to obtain q bacteria (given the
ln 2
number q).
(b) q = 50,000 i w = i
62. (a) T = T0 (1 3 h3w@d )
31
i
ln 50,000
ln 500
100
(50,000) = 3 ·
=3
E 26=9 hours
ln 2
ln 2
T
= 1 3 h3w@d
T0
i h3w@d = 1 3
T
T0
i 3
T
w
i
= ln 1 3
d
T0
w = 3d ln(1 3 T@T0 ). This gives us the time w necessary to obtain a given charge T.
(b) T = 0=9T0 and d = 2 i w = 32 ln (1 3 0=9T0 @T0 ) = 32 ln 0=1 E 4=6 seconds.
63. (a) sin31
I 3
2
3
=
since sin 3 =
I
3
2
and
is in 3 2 > 2 .
3
(b) cos31 (31) = since cos = 31 and is in [0> ].
64. (a) tan31
I1
3
(b) sec31 2 =
3
65. (a) arctan 1 =
(b) sin31
I1
2
=
=
6
since tan 6 =
since sec 3 = 2 and
4
3
since tan 4 = 1 and
4
I I1
3
since sin 4 =
66. (a) cot31 3 3 =
(b) arccos(3 12 ) =
5
6
2
3
I1
2
and
6
is in 3 2 > 2 .
is in [0> 2 ) [> 3
).
2
4
and
is in 3 2 > 2 .
4
is in 3 2 > 2 .
I
since cot 5
= 3 3 and
6
since cos 2
= 3 12 and
3
2
3
5
6
is in (0> ).
is in [0> ].
67. (a) In general, tan(arctan {) = { for any real number {. Thus, tan(arctan 10) = 10.
= sin31 sin 3 = sin31
(b) sin31 sin 7
3
[Recall that
7
3
=
3
I
3
2
=
3
since sin 3 =
I
3
2
and
3
+ 2 and the sine function is periodic with period 2.]
is in 3 2 > 2 .
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INVERSE FUNCTIONS AND LOGARITHMS
68. (a) Let denote the angle whose secant is 4, that is, = sec31 4,
as shown in the ¿gure.
Thus, tan(sec31 4) = tan =
I
15
1
=
I
15.
Alternative solution:
sec2 = 1 + tan2 i 42 = 1 + tan2 i tan =
I
15
[Note that sec = 4 A 0 i is in [0> 2 ) i tan A 0.]
(b) Let = sin31
3
5
[see the ¿gure].
sin 2 sin31 35 = sin(2) = 2 sin cos = 2 35 45 =
69. Let | = sin31 {. Then 3 2 $ | $
2
24
25
i cos | D 0, so cos(sin31 {) = cos | =
s
I
1 3 sin2 | = 1 3 {2 .
70. Let | = sin31 {. Then sin | = {, so from the triangle we see that
{
.
tan(sin31 {) = tan | = I
1 3 {2
71. Let | = tan31 {. Then tan | = {, so from the triangle we see that
{
.
sin(tan31 {) = sin | = I
1 + {2
72. Let | = tan31 {. Then tan | = {, so from the triangle we see that
cos(2 tan31 {) = cos 2| = cos2 | 3 sin2 |
2 2
1
{
1 3 {2
= I
3 I
=
2
2
1 + {2
1+{
1+{
73.
The graph of sin31 { is the reÀection of the graph of
sin { about the line | = {.
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60
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CHAPTER 1
FUNCTIONS AND MODELS
The graph of tan31 { is the reÀection of the graph of
74.
tan { about the line | = {.
75. j({) = sin31 (3{ + 1).
Domain (j) = {{ | 31 $ 3{ + 1 $ 1} = {{ | 32 $ 3{ $ 0} = { | 3 23 $ { $ 0 = 3 23 > 0 .
Range (j) = | | 3 2 $ | $ 2 = 3 2 > 2 .
76. (a) i ({) = sin sin31 {
Since one function undoes what the other one does, we get the
identity function, | = {, on the restricted domain 31 $ { $ 1.
(b) j({) = sin31 (sin {)
This is similar to part (a), but with domain R.
Equations for j on intervals of the form
3 2 + q> 2 + q , for any integer q, can be
found using j({) = (31)q { + (31)q+1 q.
The sine function is monotonic on each of these intervals, and hence, so is j (but in a linear fashion).
77. (a) If the point ({> |) is on the graph of | = i({), then the point ({ 3 f> |) is that point shifted f units to the left. Since i
is 1-1, the point (|> {) is on the graph of | = i 31 ({) and the point corresponding to ({ 3 f> |) on the graph of i is
(|> { 3 f) on the graph of i 31 . Thus, the curve’s reÀection is shifted down the same number of units as the curve itself is
shifted to the left. So an expression for the inverse function is j 31 ({) = i 31 ({) 3 f.
(b) If we compress (or stretch) a curve horizontally, the curve’s reÀection in the line | = { is compressed (or stretched)
vertically by the same factor. Using this geometric principle, we see that the inverse of k({) = i(f{) can be expressed as
k31 ({) = (1@f) i 31 ({).
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61
1 REVIEW
1. (a) A function i is a rule that assigns to each element { in a set D exactly one element, called i({), in a set E. The set D is
called the domain of the function. The range of i is the set of all possible values of i ({) as { varies throughout the
domain.
(b) If i is a function with domain D, then its graph is the set of ordered pairs {({> i ({)) | { M D}.
(c) Use the Vertical Line Test on page 15.
2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given
below.
Verbally: An assignment of students to chairs in a classroom (a description in words)
Numerically: A tax table that assigns an amount of tax to an income (a table of values)
Visually: A graphical history of the Dow Jones average (a graph)
Algebraically: A relationship between distance, rate, and time: g = uw (an explicit formula)
3. (a) If a function i satis¿es i (3{) = i ({) for every number { in its domain, then i is called an even function. If the graph of
a function is symmetric with respect to the |-axis, then i is even. Examples of an even function: i({) = {2 ,
i ({) = {4 + {2 , i ({) = |{|, i ({) = cos {.
(b) If a function i satis¿es i (3{) = 3i ({) for every number { in its domain, then i is called an odd function. If the graph
of a function is symmetric with respect to the origin, then i is odd. Examples of an odd function: i({) = {3 ,
I
i ({) = {3 + {5 , i ({) = 3 {, i ({) = sin {.
4. A function i is called increasing on an interval L if i ({1 ) ? i ({2 ) whenever {1 ? {2 in L.
5. A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world
phenomenon.
6. (a) Linear function: i({) = 2{ + 1, i ({) = d{ + e
7.
(b) Power function: i({) = {2 , i({) = {d
(c) Exponential function: i ({) = 2{ , i ({) = d{
(d) Quadratic function: i ({) = {2 + { + 1, i ({) = d{2 + e{ + f
(e) Polynomial of degree 5: i({) = {5 + 2
(f ) Rational function: i ({) =
{
S ({)
, i ({) =
where S ({) and
{+2
T({)
T({) are polynomials
8. (a)
(b)
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(c)
(d)
(e)
(f )
(g)
(h)
9. (a) The domain of i + j is the intersection of the domain of i and the domain of j; that is, D K E.
(b) The domain of i j is also D K E.
(c) The domain of i @j must exclude values of { that make j equal to 0; that is, {{ M D K E | j({) 6= 0}.
10. Given two functions i and j, the composite function i j is de¿ned by (i j) ({) = i (j ({)). The domain of i j is the
set of all { in the domain of j such that j({) is in the domain of i .
11. (a) If the graph of i is shifted 2 units upward, its equation becomes | = i ({) + 2.
(b) If the graph of i is shifted 2 units downward, its equation becomes | = i ({) 3 2.
(c) If the graph of i is shifted 2 units to the right, its equation becomes | = i ({ 3 2).
(d) If the graph of i is shifted 2 units to the left, its equation becomes | = i ({ + 2).
(e) If the graph of i is reÀected about the {-axis, its equation becomes | = 3i ({).
(f ) If the graph of i is reÀected about the |-axis, its equation becomes | = i (3{).
(g) If the graph of i is stretched vertically by a factor of 2, its equation becomes | = 2i ({).
(h) If the graph of i is shrunk vertically by a factor of 2, its equation becomes | = 12 i ({).
(i) If the graph of i is stretched horizontally by a factor of 2, its equation becomes | = i 12 { .
(j) If the graph of i is shrunk horizontally by a factor of 2, its equation becomes | = i (2{).
12. (a) A function i is called a one-to-one function if it never takes on the same value twice; that is, if i ({1 ) 6= i ({2 ) whenever
{1 6= {2 . (Or, i is 1-1 if each output corresponds to only one input.)
Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more
than once.
(b) If i is a one-to-one function with domain D and range E, then its inverse function i 31 has domain E and range D and is
de¿ned by
i 31 (|) = { C i ({) = |
for any | in E. The graph of i 31 is obtained by reÀecting the graph of i about the line | = {.
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63
13. (a) The inverse sine function i ({) = sin31 { is de¿ned as follows:
sin31 { = |
Its domain is 31 $ { $ 1 and its range is 3
C
sin | = {
and
3
$|$
2
2
$|$ .
2
2
(b) The inverse cosine function i ({) = cos31 { is de¿ned as follows:
cos31 { = |
C
cos | = {
0$|$
and
Its domain is 31 $ { $ 1 and its range is 0 $ | $ .
(c) The inverse tangent function i ({) = tan31 { is de¿ned as follows:
tan31 { = |
Its domain is R and its range is 3
1. False.
C
tan | = {
and
3
?|?
2
2
?|? .
2
2
Let i ({) = {2 , v = 31, and w = 1. Then i (v + w) = (31 + 1)2 = 02 = 0, but
i (v) + i(w) = (31)2 + 12 = 2 6= 0 = i (v + w).
2. False.
Let i ({) = {2 . Then i (32) = 4 = i (2), but 32 6= 2.
3. False.
Let i ({) = {2 . Then i (3{) = (3{)2 = 9{2 and 3i ({) = 3{2 . So i (3{) 6= 3i ({).
4. True.
If {1 ? {2 and i is a decreasing function, then the |-values get smaller as we move from left to right.
Thus, i ({1 ) A i ({2 ).
5. True.
See the Vertical Line Test.
6. False.
Let i ({) = {2 and j({) = 2{. Then (i j)({) = i(j({)) = i (2{) = (2{)2 = 4{2 and
(j i)({) = j(i({)) = j({2 ) = 2{2 . So i j 6= j i .
I
3
{. But 1@i ({) = 1@{3 , which is not equal to i 31 ({).
7. False.
Let i ({) = {3 . Then i is one-to-one and i 31 ({) =
8. True.
We can divide by h{ since h{ 6= 0 for every {.
9. True.
The function ln { is an increasing function on (0> ").
10. False.
Let { = h. Then (ln {)6 = (ln h)6 = 16 = 1, but 6 ln { = 6 ln h = 6 · 1 = 6 6= 1 = (ln {)6 . What is true, however,
is that ln({6 ) = 6 ln { for { A 0.
11. False.
12. False.
ln {
ln h2
2 ln h
{
h2
=
=
= 2 and ln = ln
= ln h = 1, so in general the statement
ln d
ln h
ln h
d
h
{
is false. What is true, however, is that ln = ln { 3 ln d.
d
3
It is true that tan 4 = 31, but since the range of tan31 is 3 2 > 2 , we must have tan31 (31) = 3 4 .
Let { = h2 and d = h. Then
13. False.
For example, tan31 20 is de¿ned; sin31 20 and cos31 20 are not.
14. False.
For example, if { = 33, then
s
I
(33)2 = 9 = 3, not 33.
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CHAPTER 1 FUNCTIONS AND MODELS
1. (a) When { = 2, | E 2=7. Thus, i (2) E 2=7.
(c) The domain of i is 36 $ { $ 6, or [36> 6].
(b) i ({) = 3 i { E 2=3, 5=6
(d) The range of i is 34 $ | $ 4, or [34> 4].
(e) i is increasing on [34> 4], that is, on 34 $ { $ 4.
(f ) i is not one-to-one since it fails the Horizontal Line Test.
(g) i is odd since its graph is symmetric about the origin.
2. (a) When { = 2, | = 3. Thus, j(2) = 3.
(b) j is one-to-one because it passes the Horizontal Line Test.
(c) When | = 2, { E 0=2. So j31 (2) E 0=2.
(d) The range of j is [31> 3=5], which is the same as the domain of j 31 .
(e) We reÀect the graph of j through the line | = { to obtain the graph of j 31 .
3. i ({) = {2 3 2{ + 3, so i (d + k) = (d + k)2 3 2(d + k) + 3 = d2 + 2dk + k2 3 2d 3 2k + 3, and
(d2 + 2dk + k2 3 2d 3 2k + 3) 3 (d2 3 2d + 3)
k(2d + k 3 2)
i (d + k) 3 i(d)
=
=
= 2d + k 3 2.
k
k
k
4. There will be some yield with no fertilizer, increasing yields with increasing
fertilizer use, a leveling-off of yields at some point, and disaster with too
much fertilizer use.
5. i ({) = 2@(3{ 3 1).
Domain: 3{ 3 1 6= 0 i 3{ 6= 1 i { 6= 13 . G = 3"> 13 13 > "
Range:
6. j({) =
I
16 3 {4 .
7. k({) = ln({ + 6).
all reals except 0 (| = 0 is the horizontal asymptote for i .) U = (3"> 0) (0> ")
I
Domain: 16 3 {4 D 0 i {4 $ 16 i |{| $ 4 16 i |{| $ 2. G = [32> 2]
I
Range: | D 0 and | $ 16 i 0 $ | $ 4. U = [0> 4]
Domain: { + 6 A 0 i { A 36. G = (36> ")
Range:
{ + 6 A 0, so ln({ + 6) takes on all real numbers and, hence, the range is R.
U = (3"> ")
8. | = I (w) = 3 + cos 2w. Domain: R.
Range:
G = (3"> ")
31 $ cos 2w $ 1 i 2 $ 3 + cos 2w $ 4 i 2 $ | $ 4. U = [2> 4]
9. (a) To obtain the graph of | = i ({) + 8, we shift the graph of | = i ({) up 8 units.
(b) To obtain the graph of | = i ({ + 8), we shift the graph of | = i ({) left 8 units.
(c) To obtain the graph of | = 1 + 2i ({), we stretch the graph of | = i ({) vertically by a factor of 2, and then shift the
resulting graph 1 unit upward.
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(d) To obtain the graph of | = i ({ 3 2) 3 2, we shift the graph of | = i ({) right 2 units (for the “32” inside the
parentheses), and then shift the resulting graph 2 units downward.
(e) To obtain the graph of | = 3i ({), we reÀect the graph of | = i ({) about the {-axis.
(f ) To obtain the graph of | = i 31 ({), we reÀect the graph of | = i({) about the line | = { (assuming i is one–to-one).
10. (a) To obtain the graph of | = i ({ 3 8), we shift the
graph of | = i ({) right 8 units.
(c) To obtain the graph of | = 2 3 i ({), we reÀect the
(b) To obtain the graph of | = 3i({), we reÀect the graph
of | = i ({) about the {-axis.
(d) To obtain the graph of | = 12 i ({) 3 1, we shrink the
graph of | = i ({) about the {-axis, and then shift the
graph of | = i ({) by a factor of 2, and then shift the
resulting graph 2 units upward.
resulting graph 1 unit downward.
(e) To obtain the graph of | = i 31 ({), we reÀect the
graph of | = i ({) about the line | = {.
(f) To obtain the graph of | = i 31 ({ + 3), we reÀect the
graph of | = i ({) about the line | = { [see part (e)],
and then shift the resulting graph left 3 units.
11. | = 3 sin 2{: Start with the graph of | = sin {, compress horizontally by a factor of 2, and reÀect about the {-axis.
12. | = 3 ln({ 3 2):
Start with the graph of | = ln {>
shift 2 units to the right, and
stretch vertically by a factor of 3.
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CHAPTER 1 FUNCTIONS AND MODELS
13. | =
1
2 (1
+ h{ ):
Start with the graph of | = h{ ,
shift 1 unit upward, and compress
vertically by a factor of 2.
14. | = 2 3
I
{:
Start with the graph of | =
I
{,
reÀect about the {-axis, and shift
2 units upward.
15. i ({) =
1
:
{+2
Start with the graph of i ({) = 1@{
and shift 2 units to the left.
16. i ({) =
+
3{
if { ? 0
{
h 3 1 if { D 0
On (3"> 0), graph | = 3{ (the line with slope 31 and |-intercept 0)
with open endpoint (0> 0).
On [0> "), graph | = h{ 3 1 (the graph of | = h{ shifted 1 unit downward)
with closed endpoint (0> 0).
17. (a) The terms of i are a mixture of odd and even powers of {, so i is neither even nor odd.
(b) The terms of i are all odd powers of {, so i is odd.
2
2
(c) i (3{) = h3(3{) = h3{ = i ({), so i is even.
(d) i (3{) = 1 + sin(3{) = 1 3 sin {. Now i (3{) 6= i ({) and i(3{) 6= 3i({), so i is neither even nor odd.
032
= 32, and an equation is | 3 0 = 32({ + 1) or,
31 + 2
I
equivalently, | = 32{ 3 2. The circle has equation {2 + | 2 = 1; the top half has equation | = 1 3 {2 (we have solved for
+
32{ 3 2 if 32 $ { $ 31
positive |). Thus, i ({) = I
1 3 {2 if 31 ? { $ 1
18. For the line segment from (32> 2) to (31> 0), the slope is
19. i ({) = ln {,
G = (0> "); j({) = {2 3 9, G = R.
(a) (i j)({) = i (j({)) = i({2 3 9) = ln({2 3 9).
Domain: {2 3 9 A 0 i {2 A 9 i |{| A 3 i { M (3"> 33) (3> ")
(b) (j i)({) = j(i ({)) = j(ln {) = (ln {)2 3 9. Domain: { A 0, or (0> ")
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(c) (i i )({) = i(i ({)) = i (ln {) = ln(ln {). Domain: ln { A 0 i { A h0 = 1, or (1> ")
(d) (j j)({) = j(j({)) = j({2 3 9) = ({2 3 9)2 3 9. Domain: { M R, or (3"> ")
20. Let k({) = { +
21.
I
I
1
{, j({) = {, and i ({) = 1@{. Then (i j k)({) = s
I = I ({).
{+ {
Many models appear to be plausible. Your choice depends on whether you
think medical advances will keep increasing life expectancy, or if there is
bound to be a natural leveling-off of life expectancy. A linear model,
| = 0=2493{ 3 423=4818, gives us an estimate of 77=6 years for the
year 2010.
22. (a) Let { denote the number of toaster ovens produced in one week and
| the associated cost. Using the points (1000> 9000) and
(1500> 12,000), we get an equation of a line:
| 3 9000 =
12,000 3 9000
({ 3 1000) i
1500 3 1000
| = 6 ({ 3 1000) + 9000 i | = 6{ + 3000.
(b) The slope of 6 means that each additional toaster oven produced adds $6 to the weekly production cost.
(c) The |-intercept of 3000 represents the overhead cost —the cost incurred without producing anything.
23. We need to know the value of { such that i ({) = 2{ + ln { = 2. Since { = 1 gives us | = 2, i 31 (2) = 1.
24. | =
{+1
|+1
. Interchanging { and | gives us { =
2{ + 1
2| + 1
|(2{ 3 1) = 1 3 { i | =
i 2{| + { = | + 1 i 2{| 3 | = 1 3 { i
13{
= i 31 ({).
2{ 3 1
25. (a) h2 ln 3 = (hln 3 )2 = 32 = 9
(b) log10 25 + log10 4 = log10 (25 · 4) = log10 100 = log10 102 = 2
(c) tan arcsin 12 = tan 6 =
1
I
3
t
I
2 t 9
(d) Let = cos31 45 , so cos = 45 . Then sin cos31 45 = sin = 1 3 cos2 = 1 3 45 = 25
= 35 .
26. (a) h{ = 5
i { = ln 5
(b) ln { = 2 i { = h2
{
(c) hh = 2 i h{ = ln 2 i { = ln(ln 2)
(d) tan31 { = 1 i tan tan31 { = tan 1 i { = tan 1 (E 1=5574)
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CHAPTER 1 FUNCTIONS AND MODELS
27. (a)
The population would reach 900 in about 4=4 years.
100,000
i 100S + 900S h3w = 100,000 i 900S h3w = 100,000 3 100S i
100 + 900h3w
1000 3 S
1000 3 S
9S
100,000 3 100S
i 3w = ln
i w = 3 ln
, or ln
; this is the time
=
900S
9S
9S
1000 3 S
(b) S =
h3w
required for the population to reach a given number S .
9 · 900
(c) S = 900 i w = ln
= ln 81 E 4=4 years, as in part (a).
1000 3 900
28.
For large values of {, | = d{ has the largest |-values and | = logd { has the smallest |-values. This makes sense because
they are inverses of each other.
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PRINCIPLES OF PROBLEM SOLVING
1.
(base) (height), in two ways, we see that
I
4|
1
(4) (|) = 12 (k) (d), so d =
. Since 42 + | 2 = k2 , | = k2 3 16, and
2
k
I
4 k2 3 16
.
d=
k
2.
Refer to Example 1, where we obtained k =
By using the area formula for a triangle,
1
2
S 2 3 100
. The 100 came from
2S
4 times the area of the triangle. In this case, the area of the triangle is
1
(k)(12)
2
= 6k. Thus, k =
2S k + 24k = S 2
3. |2{ 3 1| =
+
2{ 3 1 if { D
1 3 2{ if { ?
1
2
1
2
and
S 2 3 4 (6k)
2S
i 2S k = S 2 3 24k i
i k (2S + 24) = S 2
|{ + 5| =
+
{+5
i k=
S2
.
2S + 24
if { D 35
3{ 3 5 if { ? 35
Therefore, we consider the three cases { ? 35, 35 $ { ? 12 , and { D 12 .
If { ? 35, we must have 1 3 2{ 3 (3{ 3 5) = 3 C { = 3, which is false, since we are considering { ? 35.
If 35 $ { ? 12 , we must have 1 3 2{ 3 ({ + 5) = 3 C { = 3 73 .
If { D 12 , we must have 2{ 3 1 3 ({ + 5) = 3 C { = 9.
So the two solutions of the equation are { = 3 73 and { = 9.
4. |{ 3 1| =
+
{ 3 1 if { D 1
1 3 { if { ? 1
and
|{ 3 3| =
+
{ 3 3 if { D 3
3 3 { if { ? 3
Therefore, we consider the three cases { ? 1, 1 $ { ? 3, and { D 3.
If { ? 1, we must have 1 3 { 3 (3 3 { ) D 5 C 0 D 7, which is false.
If 1 $ { ? 3, we must have { 3 1 3 (3 3 {) D 5 C { D 92 , which is false because { ? 3.
If { D 3, we must have { 3 1 3 ({ 3 3) D 5 C 2 D 5, which is false.
All three cases lead to falsehoods, so the inequality has no solution.
5. i ({) = {2 3 4 |{| + 3. If { D 0, then i ({) = {2 3 4{ + 3 = |({ 3 1)({ 3 3)|.
Case (i):
If 0 ? { $ 1, then i ({) = {2 3 4{ + 3.
Case (ii): If 1 ? { $ 3, then i ({) = 3({2 3 4{ + 3) = 3{2 + 4{ 3 3.
Case (iii): If { A 3, then i ({) = {2 3 4{ + 3.
This enables us to sketch the graph for { D 0. Then we use the fact that i is an even
function to reÀect this part of the graph about the |-axis to obtain the entire graph. Or, we
could consider also the cases { ? 33, 33 $ { ? 31, and 31 $ { ? 0.
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CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING
6. j({) = {2 3 1 3 {2 3 4.
2
{ 3 1 =
+ 2
{ 3 1 if |{| D 1
1 3 {2
if |{| ? 1
and {2 3 4 =
+ 2
{ 3 4 if |{| D 2
4 3 {2
if |{| ? 2
So for 0 $ |{| ? 1, j({) = 1 3 {2 3 (4 3 {2 ) = 33, for
1 $ |{| ? 2, j({) = {2 3 1 3 (4 3 {2 ) = 2 {2 3 5, and for
|{| D 2, j({) = {2 3 1 3 ({2 3 4) = 3=
7. Remember that |d| = d if d D 0 and that |d| = 3d if d ? 0. Thus,
{ + |{| =
+
2{ if { D 0
0
and
if { ? 0
| + ||| =
+
2|
0
if | D 0
if | ? 0
We will consider the equation { + |{| = | + ||| in four cases.
(1) { D 0> | D 0
2{ = 2|
(2) { D 0, | ? 0
2{ = 0
{=|
{=0
(3) { ? 0, | D 0
0 = 2|
(4) { ? 0> | ? 0
0=0
0=|
Case 1 gives us the line | = { with nonnegative { and |.
Case 2 gives us the portion of the |-axis with | negative.
Case 3 gives us the portion of the {-axis with { negative.
Case 4 gives us the entire third quadrant.
8. |{ 3 || + |{| 3 ||| $ 2
[call this inequality (B)]
Case (i):
{ D | D 0.
Then (B)
C
{3|+{3| $ 2
C
{3| $1
C
| D { 3 1.
Case (ii):
| D { D 0.
Then (B)
C
|3{+{3| $ 2
C
0 $ 2 (true).
Case (iii):
{ D 0 and | $ 0.
Then (B)
C
{3|+{+| $ 2
C
2{ $ 2
C
{ $ 1.
Case (iv):
{ $ 0 and | D 0.
Then (B)
C
|3{3{3| $ 2
Case (v):
| $ { $ 0.
Then (B)
C
{3|3{+| $ 2
C
32{ $ 2
C
{ D 31.
C
0 $ 2 (true).
Case (vi):
{ $ | $ 0.
Then (B)
C
|3{3{+| $ 2
C
|3{$1
C
| $ { + 1.
Note: Instead of considering cases (iv), (v), and (vi), we could have noted that
the region is unchanged if { and | are replaced by 3{ and 3|, so the region is
symmetric about the origin. Therefore, we need only draw cases (i), (ii), and
(iii), and rotate through 180 about the origin.
9. (a) To sketch the graph of
i ({) = max {{> 1@{}, we ¿rst graph
j({) = { and k({) = 1@{ on the same
coordinate axes. Then create the graph of
i by plotting the largest |-value of j and k
for every value of {.
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CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING
(b)
(c)
On the TI-84 Plus, max is found under LIST, then under MATH. To graph i ({) = max {2 > 2 + {> 2 3 { , use
Y = max({2 > max(2 + {> 2 3 {)).
10. (a) If max {{> 2|} = 1, then either { = 1 and 2| $ 1
ru { $ 1 and 2| = 1. Thus, we obtain the set of
points such that { = 1 and | $ 12 [a vertical line
with highest point (1> 12 ) ru { $ 1 and | = 12
a horizontal line with rightmost point (1> 12 ) .
(b) The graph of max{{> 2|} = 1 is shown in part (a), and
the graph of max{{> 2|} = 31 can be found in a
similar manner. The inequalities in
31 $ max{{> 2|} $ 1 give us all the points on or
inside the boundaries.
(c) max{{> | 2 } = 1 C
{ = 1 and |2 $ 1 [31 $ | $ 1]
ru { $ 1 and |2 = 1 [| = ±1].
11. (log2 3)(log3 4)(log4 5) · · · (log31 32) =
ln 3
ln 2
ln 4
ln 3
ln 5
ln 4
···
ln 32
ln 31
=
ln 32
ln 25
5 ln 2
=
=
=5
ln 2
ln 2
ln 2
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CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING
I
t
I
3{ 3 {2 + 1
I
(3{)2 + 1 = ln 3{ + {2 + 1 ·
3{ 3 {2 + 1
#
$
{2 3 {2 + 1
31
1
I
I
I
= ln
= ln
= ln
3{ 3 {2 + 1
3{ 3 {2 + 1
{ + {2 + 1
I
I
= ln 1 3 ln { + {2 + 1 = 3 ln { + {2 3 1 = 3i ({)
s
s
I
i h{ = | + | 2 + 1 i
(b) | = ln { + {2 + 1 . Interchanging { and |, we get { = ln | + | 2 + 1
12. (a) i (3{) = ln 3{ +
h{ 3 | =
s
| 2 + 1 i h2{ 3 2|h{ + | 2 = | 2 + 1 i h2{ 3 1 = 2|h{
13. ln {2 3 2{ 3 2 $ 0
i |=
h2{ 3 1
= i 31 ({)
2h{
i {2 3 2{ 3 2 $ h0 = 1 i {2 3 2{ 3 3 $ 0 i ({ 3 3)({ + 1) $ 0 i { M [31> 3].
Since the argument must be positive, {2 3 2{ 3 2 A 0 i
I I {3 13 3 {3 1+ 3 A 0 i
I
I I I { M 3"> 1 3 3 1 + 3> " . The intersection of these intervals is 31> 1 3 3 1 + 3> 3 .
14. Assume that log2 5 is rational. Then log2 5 = p@q for natural numbers p and q. Changing to exponential form gives us
2p@q = 5 and then raising both sides to the qth power gives 2p = 5q . But 2p is even and 5q is odd. We have arrived at a
contradiction, so we conclude that our hypothesis, that log2 5 is rational, is false. Thus, log2 5 is irrational.
15. Let g be the distance traveled on each half of the trip. Let w1 and w2 be the times taken for the ¿rst and second halves of the trip.
For the ¿rst half of the trip we have w1 = g@30 and for the second half we have w2 = g@60. Thus, the average speed for the
entire trip is
60
2g
total distance
2g
120g
120g
·
=
=
=
=
= 40. The average speed for the entire trip
g
g 60
total time
w1 + w2
2g + g
3g
+
30
60
is 40 mi@h.
16. Let i({) = sin {, j({) = {, and k({) = {. Then the left-hand side of the equation is
[i (j + k)]({) = sin({ + {) = sin 2{ = 2 sin { cos {; and the right-hand side is
(i j)({) + (i k)({) = sin { + sin { = 2 sin {. The two sides are not equal, so the given statement is false.
17. Let Vq be the statement that 7q 3 1 is divisible by 6=
• V1 is true because 71 3 1 = 6 is divisible by 6.
• Assume Vn is true, that is, 7n 3 1 is divisible by 6. In other words, 7n 3 1 = 6p for some positive integer p. Then
7n+1 3 1 = 7n · 7 3 1 = (6p + 1) · 7 3 1 = 42p + 6 = 6(7p + 1), which is divisible by 6, so Vn+1 is true.
• Therefore, by mathematical induction, 7q 3 1 is divisible by 6 for every positive integer q.
18. Let Vq be the statement that 1 + 3 + 5 + · · · + (2q 3 1) = q2 .
• V1 is true because [2(1) 3 1] = 1 = 12 .
• Assume Vn is true, that is, 1 + 3 + 5 + · · · + (2n 3 1) = n2 . Then
1 + 3 + 5 + · · · + (2n 3 1) + [2(n + 1) 3 1] = 1 + 3 + 5 + · · · + (2n 3 1) + (2n + 1) = n2 + (2n + 1) = (n + 1)2
which shows that Vn+1 is true.
• Therefore, by mathematical induction, 1 + 3 + 5 + · · · + (2q 3 1) = q2 for every positive integer q.
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CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING
¤
73
19. i0 ({) = {2 and iq+1 ({) = i0 (iq ({)) for q = 0> 1> 2> = = =.
2
i1 ({) = i0 (i0 ({)) = i0 {2 = {2 = {4 , i2 ({) = i0 (i1 ({)) = i0 ({4 ) = ({4 )2 = {8 ,
q+1
i3 ({) = i0 (i2 ({)) = i0 ({8 ) = ({8 )2 = {16 , = = =. Thus, a general formula is iq ({) = {2
.
20. (a) i0 ({) = 1@(2 3 {) and iq+1 = i0 iq for q = 0> 1> 2> = = =.
i1 ({) = i0
1
23{
i2 ({) = i0
23{
3 3 2{
3 3 2{
4 3 3{
=
1
1
23
23{
=
=
=
23{
23{
=
,
2(2 3 {) 3 1
3 3 2{
3 3 2{
1
3 3 2{
=
=
,
23{
2(3 3 2{) 3 (2 3 {)
4 3 3{
23
3 3 2{
4 3 3{
1
4 3 3{
=
=
>===
3 3 2{
2(4 3 3{) 3 (3 3 2{)
5 3 4{
23
4 3 3{
q + 1 3 q{
.
Thus, we conjecture that the general formula is iq ({) =
q + 2 3 (q + 1){
i3 ({) = i0
To prove this, we use the Principle of Mathematical Induction. We have already veri¿ed that iq is true for q = 1.
n + 1 3 n{
. Then
Assume that the formula is true for q = n; that is, in ({) =
n + 2 3 (n + 1){
n + 1 3 n{
1
in+1 ({) = (i0 in )({) = i0 (in ({)) = i0
=
n + 1 3 n{
n + 2 3 (n + 1){
23
n + 2 3 (n + 1){
=
n + 2 3 (n + 1){
n + 2 3 (n + 1){
=
2 [n + 2 3 (n + 1){] 3 (n + 1 3 n{)
n + 3 3 (n + 2){
This shows that the formula for iq is true for q = n + 1. Therefore, by mathematical induction, the formula is true for all
positive integers q.
(b) From the graph, we can make several observations:
• The values at each ¿xed { = d keep increasing as q increases.
• The vertical asymptote gets closer to { = 1 as q increases.
• The horizontal asymptote gets closer to | = 1
as q increases.
• The {-intercept for iq+1 is the value of the
vertical asymptote for iq .
• The |-intercept for iq is the value of the
horizontal asymptote for iq+1 .
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2
LIMITS AND DERIVATIVES
2.1 The Tangent and Velocity Problems
1. (a) Using S (15> 250), we construct the following table:
(b) Using the values of w that correspond to the points
closest to S (w = 10 and w = 20), we have
w
T
slope = pS T
5
(5> 694)
6943250
5315
= 3 444
= 344=4
10
10
(10> 444)
4443250
10315
= 3 194
= 338=8
5
20
(20> 111)
1113250
20315
= 3 139
= 327=8
5
25
(25> 28)
283250
25315
30
(30> 0)
03250
30315
338=8 + (327=8)
= 333=3
2
= 3 222
10 = 322=2
= 3 250
15 = 316=6
(c) From the graph, we can estimate the slope of the
tangent line at S to be
2. (a) Slope =
(c) Slope =
2948 3 2530
42 3 36
=
2948 3 2806
42 3 40
=
3300
9
418
6
142
2
= 333=3.
E 69=67
(b) Slope =
= 71
(d) Slope =
2948 3 2661
42 3 38
3080 3 2948
44 3 42
=
287
4
= 71=75
=
132
2
= 66
From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeats@minute after 42 minutes.
After being stable for a while, the patient’s heart rate is dropping.
3. (a) | =
1
, S (2> 31)
13{
{
(i)
1=5
(ii)
1=9
(iii)
1=99
(iv)
1=999
(v)
2=5
(vi)
2=1
(vii)
2=01
(viii)
2=001
(b) The slope appears to be 1.
(c) Using p = 1, an equation of the tangent line to the
T({> 1@(1 3 {))
pS T
(1=5> 32)
2
(1=99> 31=010 101)
1=010 101
(2=5> 30=666 667)
0=666 667
(2=01> 30=990 099)
0=990 099
(1=9> 31=111 111)
1=111 111
(1=999> 31=001 001)
1=001 001
(2=1> 30=909 091)
0=909 091
(2=001> 30=999 001)
0=999 001
curve at S (2> 31) is | 3 (31) = 1({ 3 2), or
| = { 3 3.
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CHAPTER 2
LIMITS AND DERIVATIVES
4. (a) | = cos {, S (0=5> 0)
{
(b) The slope appears to be 3.
T
pS T
(i)
0
(0> 1)
(ii)
0=4
(0=4> 0=309017)
(iii)
0=49
(0=49> 0=031411)
(iv)
0=499
(0=499> 0=003142)
(1> 31)
(v)
1
(vi)
0=6
(vii)
0=51
(viii)
0=501
32
(0=6> 30=309017)
(0=51> 30=031411)
(0=501> 30=003142)
(c) | 3 0 = 3({ 3 0=5) or | = 3{ + 12 .
(d)
33=090170
33=141076
33=141587
32
33=090170
33=141076
33=141587
5. (a) | = |(w) = 40w 3 16w2 . At w = 2, | = 40(2) 3 16(2)2 = 16. The average velocity between times 2 and 2 + k is
yave =
40(2 + k) 3 16(2 + k)2 3 16
|(2 + k) 3 |(2)
324k 3 16k2
=
=
= 324 3 16k, if k 6= 0.
(2 + k) 3 2
k
k
(i) [2> 2=5]: k = 0=5, yave = 332 ft@s
(iii) [2> 2=05]: k = 0=05, yave = 324=8 ft@s
(ii) [2> 2=1]: k = 0=1, yave = 325=6 ft@s
(iv) [2> 2=01]: k = 0=01, yave = 324=16 ft@s
(b) The instantaneous velocity when w = 2 (k approaches 0) is 324 ft@s.
6. (a) | = |(w) = 10w 3 1=86w2 . At w = 1, | = 10(1) 3 1=86(1)2 = 8=14. The average velocity between times 1 and 1 + k is
yave =
10(1 + k) 3 1=86(1 + k)2 3 8=14
6=28k 3 1=86k2
|(1 + k) 3 |(1)
=
=
= 6=28 3 1=86k, if k 6= 0.
(1 + k) 3 1
k
k
(i) [1> 2]: k = 1, yave = 4=42 m@s
(ii) [1> 1=5]: k = 0=5, yave = 5=35 m@s
(iii) [1> 1=1]: k = 0=1, yave = 6=094 m@s
(iv) [1> 1=01]: k = 0=01, yave = 6=2614 m@s
(v) [1> 1=001]: k = 0=001, yave = 6=27814 m@s
(b) The instantaneous velocity when w = 1 (k approaches 0) is 6=28 m@s.
7. (a) (i) On the interval [1> 3], yave =
v(3) 3 v(1)
10=7 3 1=4
9=3
=
=
= 4=65 m@s.
331
2
2
(ii) On the interval [2> 3], yave =
10=7 3 5=1
v(3) 3 v(2)
=
= 5=6 m@s.
332
1
(iii) On the interval [3> 5], yave =
25=8 3 10=7
15=1
v(5) 3 v(3)
=
=
= 7=55 m@s.
533
2
2
(iv) On the interval [3> 4], yave =
17=7 3 10=7
v(4) 3 v(3)
=
= 7 m@s.
433
1
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SECTION 2.1
THE TANGENT AND VELOCITY PROBLEMS
¤
Using the points (2> 4) and (5> 23) from the approximate tangent
(b)
line, the instantaneous velocity at w = 3 is about
8. (a) (i) v = v(w) = 2 sin w + 3 cos w. On the interval [1> 2], yave =
(ii) On the interval [1> 1=1], yave =
23 3 4
E 6=3 m@s.
532
3 3 (33)
v(2) 3 v(1)
=
= 6 cm@s.
231
1
v(1=1) 3 v(1)
33=471 3 (33)
E
= 34=71 cm@s.
1=1 3 1
0=1
(iii) On the interval [1> 1=01], yave =
v(1=01) 3 v(1)
33=0613 3 (33)
E
= 36=13 cm@s.
1=01 3 1
0=01
(iv) On the interval [1> 1=001], yave =
33=00627 3 (33)
v(1=001) 3 v(1)
E
= 36=27 cm@s.
1=001 3 1
1=001 3 1
(b) The instantaneous velocity of the particle when w = 1 appears to be about 36=3 cm@s.
9. (a) For the curve | = sin(10@{) and the point S (1> 0):
{
T
pS T
{
T
2
(2> 0)
0
0=5
(0=5> 0)
1=5
(1=5> 0=8660)
1=7321
0=6
(0=6> 0=8660)
1=4
(1=4> 30=4339)
31=0847
0=7
(0=7> 0=7818)
0=8
(0=8> 1)
4=3301
0=9
(0=9> 30=3420)
1=3
1=2
1=1
(1=3> 30=8230)
(1=2> 0=8660)
(1=1> 30=2817)
32=7433
32=8173
pS T
0
32=1651
32=6061
35
3=4202
As { approaches 1, the slopes do not appear to be approaching any particular value.
(b)
We see that problems with estimation are caused by the frequent
oscillations of the graph. The tangent is so steep at S that we need to
take {-values much closer to 1 in order to get accurate estimates of
its slope.
(c) If we choose { = 1=001, then the point T is (1=001> 30=0314) and pS T E 331=3794. If { = 0=999, then T is
(0=999> 0=0314) and pS T = 331=4422. The average of these slopes is 331=4108. So we estimate that the slope of the
tangent line at S is about 331=4.
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NOT FOR SALE
CHAPTER 2
LIMITS AND DERIVATIVES
2.2 The Limit of a Function
1. As { approaches 2, i ({) approaches 5. [Or, the values of i ({) can be made as close to 5 as we like by taking { suf¿ciently
close to 2 (but { 6= 2).] Yes, the graph could have a hole at (2> 5) and be de¿ned such that i (2) = 3.
2. As { approaches 1 from the left, i ({) approaches 3; and as { approaches 1 from the right, i({) approaches 7. No, the limit
does not exist because the left- and right-hand limits are different.
3. (a) lim i ({) = " means that the values of i ({) can be made arbitrarily large (as large as we please) by taking {
{<33
suf¿ciently close to 33 (but not equal to 33).
(b) lim i({) = 3" means that the values of i ({) can be made arbitrarily large negative by taking { suf¿ciently close to 4
{<4+
through values larger than 4.
4. (a) As { approaches 2 from the left, the values of i({) approach 3, so lim i ({) = 3.
{<2
(b) As { approaches 2 from the right, the values of i ({) approach 1, so lim i({) = 1.
{<2+
(c) lim i ({) does not exist since the left-hand limit does not equal the right-hand limit.
{<2
(d) When { = 2, | = 3, so i (2) = 3.
(e) As { approaches 4, the values of i ({) approach 4, so lim i ({) = 4.
{<4
(f ) There is no value of i ({) when { = 4, so i (4) does not exist.
5. (a) As { approaches 1, the values of i ({) approach 2, so lim i ({) = 2.
{<1
(b) As { approaches 3 from the left, the values of i({) approach 1, so lim i ({) = 1.
{<3
(c) As { approaches 3 from the right, the values of i ({) approach 4, so lim i({) = 4.
{<3+
(d) lim i ({) does not exist since the left-hand limit does not equal the right-hand limit.
{<3
(e) When { = 3, | = 3, so i (3) = 3.
6. (a) k({) approaches 4 as { approaches 33 from the left, so
lim k({) = 4.
{<33
(b) k({) approaches 4 as { approaches 33 from the right, so lim k({) = 4.
{<33+
(c) lim k({) = 4 because the limits in part (a) and part (b) are equal.
{<33
(d) k(33) is not de¿ned, so it doesn’t exist.
(e) k({) approaches 1 as { approaches 0 from the left, so lim k({) = 1.
{<0
(f ) k({) approaches 31 as { approaches 0 from the right, so lim k({) = 31.
{<0+
(g) lim k({) does not exist because the limits in part (e) and part (f ) are not equal.
{<0
(h) k(0) = 1 since the point (0> 1) is on the graph of k.
(i) Since lim k({) = 2 and lim k({) = 2, we have lim k({) = 2.
{<2
{<2+
{<2
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NOT FOR SALE
SECTION 2.2
THE LIMIT OF A FUNCTION
¤
(j) k(2) is not de¿ned, so it doesn’t exist.
(k) k({) approaches 3 as { approaches 5 from the right, so lim k({) = 3.
{<5+
(l) k({) does not approach any one number as { approaches 5 from the left, so lim k({) does not exist.
{<5
7. (a) lim j(w) = 31
(b) lim j(w) = 32
w<0
w<0+
(c) lim j(w) does not exist because the limits in part (a) and part (b) are not equal.
w<0
(d) lim j(w) = 2
(e) lim j(w) = 0
w<2
w<2+
(f ) lim j(w) does not exist because the limits in part (d) and part (e) are not equal.
w<2
(h) lim j(w) = 3
(g) j(2) = 1
w<4
8. (a) lim U({) = 3"
(b) lim U({) = "
{<2
(c)
{<5
lim U({) = 3"
(d)
{<33
lim U({) = "
{<33+
(e) The equations of the vertical asymptotes are { = 33, { = 2, and { = 5.
9. (a) lim i ({) = 3"
(b) lim i ({) = "
(d) lim i ({) = 3"
(e) lim i ({) = "
{<37
{<33
{<6
(c) lim i ({) = "
{<0
{<6+
(f ) The equations of the vertical asymptotes are { = 37, { = 33, { = 0, and { = 6.
10.
lim i (w) = 150 mg and lim i (w) = 300 mg. These limits show that there is an abrupt change in the amount of drug in
w<12
+
w<12
the patient’s bloodstream at w = 12 h. The left-hand limit represents the amount of the drug just before the fourth injection.
The right-hand limit represents the amount of the drug just after the fourth injection.
11. From the graph of
;
1 + { if { ? 31
A
?
if 31 $ { ? 1 ,
i ({) = {2
A
=
2 3 { if { D 1
we see that lim i({) exists for all d except d = 31. Notice that the
{<d
right and left limits are different at d = 31.
12. From the graph of
;
1 + sin { if { ? 0
A
?
if 0 $ { $ ,
i ({) = cos {
A
=
sin {
if { A we see that lim i({) exists for all d except d = . Notice that the
{<d
right and left limits are different at d = .
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NOT FOR SALE
CHAPTER 2
LIMITS AND DERIVATIVES
13. (a) lim i ({) = 1
{<0
(b) lim i({) = 0
{<0+
(c) lim i ({) does not exist because the limits
{<0
in part (a) and part (b) are not equal.
14. (a) lim i ({) = 31
{<0
(b) lim i({) = 1
{<0+
(c) lim i ({) does not exist because the limits
{<0
in part (a) and part (b) are not equal.
15. lim i ({) = 31,
{<0
lim i({) = 2, i (0) = 1
{<0+
16. lim i ({) = 1, lim i ({) = 32, lim i ({) = 2,
{<0
{<3
{<3+
i (0) = 31, i (3) = 1
17. lim i ({) = 4,
{<3+
lim i ({) = 2, lim i({) = 2,
{<3
i (3) = 3, i(32) = 1
{<32
18. lim i ({) = 2, lim i ({) = 0, lim i ({) = 3,
{<0
{<0+
{<4
lim i ({) = 0, i (0) = 2, i (4) = 1
{<4+
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NOT FOR SALE
SECTION 2.2
{2 3 2{
:
3{32
19. For i ({) =
20. For i ({) =
{2
{
i ({)
{
i ({)
{
i ({)
1=9
0=655172
0
2=1
0=677419
1=95
0=661017
2=05
0=672131
1=99
0=665552
30=5
31
2=01
0=667774
1=995
0=666110
2=005
2=001
0=667221
0=666778
1=999
0=666556
30=95
319
30=999
3999
{2 3 2{
= 0=6̄ = 23 .
3{32
81
{2 3 2{
:
3{32
0=714286
0
30=9
39
30=99
399
It appears that lim
{<2 {2
¤
{2
2=5
It appears that lim
THE LIMIT OF A FUNCTION
{<31
{
32
i ({)
2
31=5
31=1
31=01
31=001
3
11
101
1001
{2 3 2{
does not exist since
{2 3 { 3 2
i ({) < " as { < 313 and i ({) < 3" as { < 31+ .
21. For i (w) =
w
h5w 3 1
:
w
22. For i (k) =
i (w)
0=5
22=364988
0=1
6=487213
0=01
5=127110
0=001
5=012521
0=0001
5=001250
It appears that lim
w<0
w
i (w)
1
0=236068
0=5
0=242641
0=1
0=248457
0=05
0=249224
0=01
0=249844
It appears that lim
{<0
i (k)
k
1=835830
0=5
131=312500
30=1
3=934693
0=1
88=410100
30=01
4=877058
0=01
80=804010
30=001
4=987521
0=001
80=080040
30=0001
4=998750
0=0001
80=008000
h5w 3 1
= 5.
w
i ({)
k
30=5
i (k)
30=5
48=812500
30=01
79=203990
30=0001
79=992000
30=1
72=390100
30=001
79=920040
(2 + k)5 3 32
= 80.
k<0
k
It appears that lim
I
{+432
:
23. For i ({) =
{
{
(2 + k)5 3 32
:
k
24. For i ({) =
{
tan 3{
:
tan 5{
i ({)
31
0=267949
30=1
0=251582
30=01
0=250156
30=5
0=258343
30=05
0=250786
I
{+432
= 0=25 = 14 .
{
{
It appears that lim
{<0
i({)
±0=2
0=439279
±0=1
0=566236
±0=05
0=591893
±0=01
±0=001
0=599680
0=599997
tan 3{
= 0=6 = 35 .
tan 5{
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NOT FOR SALE
CHAPTER 2
25. For i ({) =
{
0=5
LIMITS AND DERIVATIVES
{6 3 1
:
{10 3 1
26. For i ({) =
i ({)
0=985337
{
{
i ({)
1=5
9{ 3 5{
:
{
i ({)
0=183369
0=5
1=527864
0=711120
0=9
0=719397
1=1
0=484119
0=1
0=95
0=660186
1=05
0=540783
0=05
0=646496
0=99
0=612018
1=01
0=588022
0=01
0=599082
0=999
0=601200
1=001
0=598800
0=001
0=588906
It appears that lim
{<1
{6 3 1
= 0=6 = 35 .
{10 3 1
27. (a) From the graphs, it seems that lim
{<0
It appears that lim
{<0
{
i ({)
30=5
0=227761
30=05
0=534447
30=001
0=586669
30=1
0=485984
30=01
0=576706
9{ 3 5{
= 0=59. Later we will be able
{
to show that the exact value is ln(9@5).
cos 2{ 3 cos {
= 31=5.
{2
(b)
{
±0=1
31=493759
±0=001
31=499999
±0=01
±0=0001
28. (a) From the graphs, it seems that lim
{<0
sin {
= 0=32.
sin {
i ({)
31=499938
31=500000
(b)
{
i ({)
±0=1
0=323068
±0=001
±0=0001
0=318310
0=318310
±0=01
0=318357
Later we will be able to show that
the exact value is
29.
30.
1
.
lim
{+2
= 3" since the numerator is negative and the denominator approaches 0 from the positive side as { < 33+ .
{+3
lim
{+2
= " since the numerator is negative and the denominator approaches 0 from the negative side as { < 333 .
{+3
{<33+
{<33
31. lim
{<1
23{
= " since the numerator is positive and the denominator approaches 0 through positive values as { < 1.
({ 3 1)2
32. lim
{<5
h{
= 3" since the numerator is positive and the denominator approaches 0 from the negative side as { < 53 .
({ 3 5)3
33. Let w = {2 3 9. Then as { < 3+ , w < 0+ , and lim ln({2 3 9) = lim ln w = 3" by (5).
{<3+
34. lim cot { = lim
{< {< w<0+
cos {
= 3" since the numerator is negative and the denominator approaches 0 through positive values
sin {
as { < 3 .
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NOT FOR SALE
SECTION 2.2
35.
lim { csc { = lim
{<2 {<2 THE LIMIT OF A FUNCTION
¤
83
{
= 3" since the numerator is positive and the denominator approaches 0 through negative
sin {
values as { < 23 .
36. lim
{<2
{2 3 2{
{({ 3 2)
{
= lim
= lim
= 3" since the numerator is positive and the denominator
3 4{ + 4
{<2 ({ 3 2)2
{<2 { 3 2
{2
approaches 0 through negative values as { < 23 .
37. lim
{<2+
({ 3 4)({ + 2)
{2 3 2{ 3 8
= lim
= " since the numerator is negative and the denominator approaches 0 through
{2 3 5{ + 6 {<2+ ({ 3 3)({ 3 2)
negative values as { < 2+ .
38. (a) The denominator of | =
{ = 0 and { =
3
2
{2 + 1
{2 + 1
is equal to zero when
=
3{ 3 2{2
{(3 3 2{)
(b)
(and the numerator is not), so { = 0 and { = 1=5 are
vertical asymptotes of the function.
39. (a) i ({) =
1
.
{3 3 1
{
0=5
From these calculations, it seems that
lim i ({) = 3" and lim i ({) = ".
{<1
0=9
0=99
{<1+
0=999
0=9999
0=99999
i ({)
{
i ({)
31=14
1=5
0=42
33=69
1=1
3=02
333=7
1=01
33=0
3333=7
1=001
333=0
33333=7
1=0001
3333=0
333,333=7
1=00001
33,333=3
(b) If { is slightly smaller than 1, then {3 3 1 will be a negative number close to 0, and the reciprocal of {3 3 1, that is, i ({),
will be a negative number with large absolute value. So lim i({) = 3".
{<1
If { is slightly larger than 1, then {3 3 1 will be a small positive number, and its reciprocal, i ({), will be a large positive
number. So lim i ({) = ".
{<1+
(c) It appears from the graph of i that
lim i ({) = 3" and lim i ({) = ".
{<1
{<1+
40. (a) From the graphs, it seems that lim
{<0
tan 4{
= 4.
{
(b)
{
±0=1
±0=01
±0=001
±0=0001
i ({)
4=227932
4=002135
4=000021
4=000000
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CHAPTER 2
LIMITS AND DERIVATIVES
1@{
41. (a) Let k({) = (1 + {)
{
30=001
30=0001
30=00001
30=000001
0=000001
0=00001
0=0001
0=001
.
(b)
k({)
2=71964
2=71842
2=71830
2=71828
2=71828
2=71827
2=71815
2=71692
It appears that lim (1 + {)1@{ E 2=71828, which is approximately h.
{<0
In Section 3.6 we will see that the value of the limit is exactly h.
42. (a)
No, because the calculator-produced graph of i ({) = h{ + ln |{ 3 4| looks like an exponential function, but the graph of i
has an in¿nite discontinuity at { = 4. A second graph, obtained by increasing the numpoints option in Maple, begins to
reveal the discontinuity at { = 4.
(b) There isn’t a single graph that shows all the features of i . Several graphs are needed since i looks like ln |{ 3 4| for large
negative values of { and like h{ for { A 5, but yet has the in¿nite discontiuity at { = 4.
A hand-drawn graph, though distorted, might be better at revealing the main
features of this function.
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NOT FOR SALE
SECTION 2.2
43. For i ({) = {2 3 (2{@1000):
(a)
THE LIMIT OF A FUNCTION
(b)
{
i ({)
{
i({)
1
0=8
0=6
0=4
0=2
0=1
0=998000
0=638259
0=358484
0=158680
0=038851
0=008928
0=04
0=02
0=01
0=005
0=003
0=000572
30=000614
30=000907
30=000978
30=000993
0=05
0=001465
0=001
It appears that lim i ({) = 0.
30=001000
It appears that lim i({) = 30=001.
{<0
{<0
tan { 3 {
:
{3
44. For k({) =
(b) It seems that lim k({) = 13 .
(a)
{<0
{
k({)
1=0
0=5
0=1
0=05
0=01
0=005
0=55740773
0=37041992
0=33467209
0=33366700
0=33334667
0=33333667
(c)
Here the values will vary from one
{
k({)
0=001
0=0005
0=0001
0=00005
0=00001
0=000001
0=33333350
0=33333344
0=33333000
0=33333600
0=33300000
0=00000000
calculator to another. Every calculator will
eventually give false values.
(d) As in part (c), when we take a small enough viewing rectangle we get incorrect output.
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85
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NOT FOR SALE
CHAPTER 2
LIMITS AND DERIVATIVES
45. No matter how many times we zoom in toward the origin, the graphs of i ({) = sin(@{) appear to consist of almost-vertical
lines. This indicates more and more frequent oscillations as { < 0.
s
p0
. As y < f3 , 1 3 y 2@f2 < 0+ , and p < ".
2
2
1 3 y @f
46. lim p = lim s
y<f
y<f
47.
There appear to be vertical asymptotes of the curve | = tan(2 sin {) at { E ±0=90
and { E ±2=24. To ¿nd the exact equations of these asymptotes, we note that the
graph of the tangent function has vertical asymptotes at { =
must have 2 sin { =
2
+ q, or equivalently, sin { =
4
2
+ q. Thus, we
+ 2 q. Since
31 $ sin { $ 1, we must have sin { = ± 4 and so { = ± sin31
4
(corresponding
to { E ±0=90). Just as 150 is the reference angle for 30 , 3 sin31 4 is the
reference angle for sin31 4 . So { = ± 3 sin31 4 are also equations of
vertical asymptotes (corresponding to { E ±2=24).
{3 3 1
.
{31
48. (a) Let | = I
From the table and the graph, we guess
that the limit of | as { approaches 1 is 6.
{
|
0=99
0=999
0=9999
1=01
1=001
1=0001
5=92531
5=99250
5=99925
6=07531
6=00750
6=00075
{3 3 1
(b) We need to have 5=5 ? I
? 6=5. From the graph we obtain the approximate points of intersection S (0=9314> 5=5)
{31
and T(1=0649> 6=5). Now 1 3 0=9314 = 0=0686 and 1=0649 3 1 = 0=0649, so by requiring that { be within 0=0649 of 1,
we ensure that | is within 0=5 of 6.
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SECTION 2.3
CALCULATING LIMITS USING THE LIMIT LAWS
¤
87
2.3 Calculating Limits Using the Limit Laws
1. (a) lim [i ({) + 5j({)] = lim i ({) + lim [5j({)]
{<2
{<2
[Limit Law 1]
{<2
= lim i ({) + 5 lim j({)
{<2
(b) lim [j({)]3 =
{<2
l3
lim j({)
{<2
[Limit Law 6]
= ( 32)3 = 38
[Limit Law 3]
{<2
k
= 4 + 5(32) = 36
(c) lim
{<2
t
s
i ({) = lim i ({)
{<2
=
(d) lim
[Limit Law 11]
{<2
lim [3i ({)]
3i ({)
{<2
=
j({)
lim j({)
[Limit Law 5]
{<2
I
4=2
3 lim i ({)
{<2
=
lim j({)
[Limit Law 3]
{<2
=
lim [j({) k({)]
j({) k({)
{<2
=
{<2
i({)
lim i ({)
(e) Because the limit of the denominator is 0, we can’t
use Limit Law 5. The given limit, lim
{<2
3(4)
= 36
32
(f) lim
[Limit Law 5]
{<2
j({)
, does
k({)
=
not exist because the denominator approaches 0
lim j({) · lim k({)
{<2
{<2
lim i ({)
[Limit Law 4]
{<2
while the numerator approaches a nonzero number.
=
32 · 0
=0
4
2. (a) lim [i({) + j({)] = lim i ({) + lim j({) = 2 + 0 = 2
{<2
{<2
{<2
(b) lim j({) does not exist since its left- and right-hand limits are not equal, so the given limit does not exist.
{<1
(c) lim [i({)j({)] = lim i ({) · lim j({) = 0 · 1=3 = 0
{<0
{<0
{<0
(d) Since lim j({) = 0 and j is in the denominator, but lim i ({) = 31 6= 0, the given limit does not exist.
{<31
{<31
k
l
lk
(e) lim {3 i ({) = lim {3 lim i ({) = 23 · 2 = 16
{<2
(f ) lim
{<1
{<2
{<2
t
s
I
3 + i ({) = 3 + lim i ({) = 3 + 1 = 2
{<1
3. lim (5{3 3 3{2 + { 3 6) = lim (5{3 ) 3 lim (3{2 ) + lim { 3 lim 6
{<3
{<3
{<3
{<3
{<3
[Limit Laws 2 and 1]
= 5 lim {3 3 3 lim {2 + lim { 3 lim 6
[3]
= 5(33 ) 3 3(32 ) + 3 3 6
[9, 8, and 7]
{<3
{<3
{<3
{<3
= 105
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4.
NOT FOR SALE
¤
CHAPTER 2
LIMITS AND DERIVATIVES
lim ({4 3 3{)({2 + 5{ + 3) = lim ({4 3 3{) lim ({2 + 5{ + 3)
{<31
{<31
=
=
[Limit Law 4]
{<31
lim {2 + lim 5{ + lim 3
lim {4 3 lim 3{
{<31
{<31
{<31
{<31
{<31
lim {4 3 3 lim {
lim {2 + 5 lim { + lim 3
{<31
{<31
{<31
{<31
{<31
= (1 + 3)(1 3 5 + 3)
[2, 1]
[3]
[9, 8, and 7]
= 4(31) = 34
lim (w4 3 2)
w4 3 2
w<32
=
w<32 2w2 3 3w + 2
lim (2w2 3 3w + 2)
5. lim
[Limit Law 5]
w<32
=
=
lim w4 3 lim 2
w<32
2 lim w2 3
w<32
w<32
[1, 2, and 3]
3 lim w + lim 2
w<32
w<32
16 3 2
2(4) 3 3(32) + 2
[9, 7, and 8]
7
14
=
16
8
t
I
6. lim
x4 + 3x + 6 =
lim (x4 + 3x + 6)
=
x<32
[11]
x<32
=
t
lim x4 + 3 lim x + lim 6
x<32
x<32
[1, 2, and 3]
x<32
t
(32)4 + 3 (32) + 6
I
I
= 16 3 6 + 6 = 16 = 4
=
7. lim (1 +
{<8
[9, 8, and 7]
I
I
3
{ ) (2 3 6{2 + {3 ) = lim (1 + 3 { ) · lim (2 3 6{2 + {3 )
{<8
=
[Limit Law 4]
{<8
lim 1 + lim
{<8
{<8
I
3
{ · lim 2 3 6 lim {2 + lim {3
{<8
{<8
{<8
I = 1 + 3 8 · 2 3 6 · 82 + 83
[1, 2, and 3]
[7, 10, 9]
= (3)(130) = 390
8. lim
w<2
w2 3 2
3
w 3 3w + 5
2
2
w2 3 2
w<2 w3 3 3w + 5
42
3
lim (w2 3 2)
w<2
D
=C
lim (w3 3 3w + 5)
=
lim
[Limit Law 6]
[5]
w<2
3
=C
lim w2 3 lim 2
lim
w<2
w<2
w3 3
w<2
3 lim w + lim 5
w<2
432
=
8 3 3(2) + 5
2
4
2
=
=
7
49
2
w<2
42
D
[1, 2, and 3]
[9, 7, and 8]
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
SECTION 2.3
u
9. lim
{<2
2{2 + 1
=
3{ 3 2
u
2{2 + 1
{<2 3{ 3 2
y
x lim (2{2 + 1)
x {<2
=w
lim (3{ 3 2)
lim
CALCULATING LIMITS USING THE LIMIT LAWS
¤
89
[Limit Law 11]
[5]
{<2
y
x 2 lim {2 + lim 1
x {<2
{<2
=w
3 lim { 3 lim 2
[1, 2, and 3]
2(2)2 + 1
=
3(2) 3 2
[9, 8, and 7]
{<2
=
v
{<2
u
9
3
=
4
2
10. (a) The left-hand side of the equation is not de¿ned for { = 2, but the right-hand side is.
(b) Since the equation holds for all { 6= 2, it follows that both sides of the equation approach the same limit as { < 2, just as
in Example 3. Remember that in ¿nding lim i ({), we never consider { = d.
{<d
{2 3 6{ + 5
({ 3 5)({ 3 1)
= lim
= lim ({ 3 1) = 5 3 1 = 4
{<5
{<5
{35
{35
11. lim
{<5
{2 3 4{
{({ 3 4)
{
4
4
= lim
= lim
=
=
3 3{ 3 4 {<4 ({ 3 4)({ + 1) {<4 { + 1
4+1
5
12. lim
{2
{<4
{2 3 5{ + 6
does not exist since { 3 5 < 0, but {2 3 5{ + 6 < 6 as { < 5.
{35
13. lim
{<5
14. lim
{<31
{2 3 4{
does not exist since {2 3 3{ 3 4 < 0 but {2 3 4{ < 5 as { < 31.
3 3{ 3 4
{2
15. lim
w2 3 9
(w + 3)(w 3 3)
w33
33 3 3
36
6
= lim
= lim
=
=
=
+ 7w + 3 w<33 (2w + 1)(w + 3) w<33 2w + 1
2(33) + 1
35
5
16. lim
2{2 + 3{ + 1
(2{ + 1)({ + 1)
2{ + 1
2(31) + 1
31
1
= lim
= lim
=
=
=
{<31 ({ 3 3)({ + 1)
{<31 { 3 3
{2 3 2{ 3 3
31 3 3
34
4
w<33 2w2
{<31
(35 + k)2 3 25
(25 3 10k + k2 ) 3 25
310k + k2
k(310 + k)
= lim
= lim
= lim
= lim (310 + k) = 310
k<0
k<0
k<0
k<0
k
k
k
k
17. lim
k<0
8 + 12k + 6k2 + k3 3 8
(2 + k)3 3 8
12k + 6k2 + k3
= lim
= lim
k<0
k<0
k
k
k
2
= lim 12 + 6k + k = 12 + 0 + 0 = 12
18. lim
k<0
k<0
19. By the formula for the sum of cubes, we have
lim
{<32
{+2
{+2
1
1
1
= lim
= lim
=
=
.
{3 + 8 {<32 ({ + 2)({2 3 2{ + 4) {<32 {2 3 2{ + 4
4+4+4
12
20. We use the difference of squares in the numerator and the difference of cubes in the denominator.
lim
w<1
w4 3 1
(w2 3 1)(w2 + 1)
(w 3 1)(w + 1)(w2 + 1)
(w + 1)(w2 + 1)
2(2)
4
= lim
= lim
= lim
=
=
3
2
2
w<1
w 3 1 w<1 (w 3 1)(w + w + 1) w<1 (w 3 1)(w + w + 1)
w2 + w + 1
3
3
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90
NOT FOR SALE
¤
CHAPTER 2
LIMITS AND DERIVATIVES
I
2
I
I
I
9 + k 3 32
9+k33
9+k33
9+k+3
(9 + k) 3 9
= lim I
= lim I
= lim
·I
21. lim
k<0
k<0
k<0
k<0
k
k
9+k+3
k 9+k+3
k 9+k+3
= lim
k<0
22. lim
x<2
k
k
1
1
1
I
= lim I
=
=
k<0
3+3
6
9+k+3
9+k+3
2
I
I
I
I
4x + 1 3 32
4x + 1 3 3
4x + 1 3 3
4x + 1 + 3
I
= lim
= lim
·I
x<2
x32
x32
4x + 1 + 3 x<2 (x 3 2) 4x + 1 + 3
= lim
x<2
4(x 3 2)
4x + 1 3 9
I
= lim
I
x<2 (x 3 2)
(x 3 2) 4x + 1 + 3
4x + 1 + 3
4
4
2
= I
= lim I
=
x<2
3
4x + 1 + 3
9+3
1
{+4
1
+
{+4
1
1
1
4
{
= lim 4{ = lim
= lim
=
=3
23. lim
{<34 4 + {
{<34 4 + {
{<34 4{(4 + {)
{<34 4{
4(34)
16
24.
{2 + 2{ + 1
({ + 1)2
({ + 1)2
= lim
= lim
4
2
2
2
{<31
{<31 ({ + 1)({ 3 1)
{<31 ({ + 1)({ + 1)({ 3 1)
{ 31
lim
= lim
{<31
{+1
0
=
=0
({2 + 1)({ 3 1)
2(32)
I
2 I
2
I
I
I
I
I
I
1+w 3
13w
1+w3 13w
1+w3 13w
1+w+ 13w
I
I
= lim I
25. lim
= lim
·I
w<0
w<0
w
w
1 + w + 1 3 w w<0 w 1 + w + 1 3 w
(1 + w) 3 (1 3 w)
2w
2
= lim I
= lim I
I
I
I
= lim I
w<0 w
w<0 w
w<0
1+w+ 13w
1+w+ 13w
1+w+ 13w
2
2
I = =1
= I
2
1+ 1
26. lim
w<0
1
1
3 2
w
w +w
= lim
w<0
1
1
3
w
w(w + 1)
= lim
w<0
w+131
1
1
= lim
=
=1
w<0 w + 1
w(w + 1)
0+1
I
I
I
43 {
(4 3 { )(4 + { )
16 3 {
I
I
=
lim
= lim
{<16 16{ 3 {2
{<16 (16{ 3 {2 )(4 +
{ ) {<16 {(16 3 {)(4 + { )
27. lim
= lim
{<16
1
1
1
1
I =
I
=
=
16(8)
128
{(4 + { )
16 4 + 16
1
1
3
(3 + k)31 3 331
3k
3
+
k
3 = lim 3 3 (3 + k) = lim
= lim
28. lim
k<0
k<0
k<0 k(3 + k)3
k<0 k(3 + k)3
k
k
1
1
1
1
=3
=3
=3
= lim 3
k<0
3(3 + k)
lim [3(3 + k)]
3(3 + 0)
9
k<0
29. lim
w<0
w
1
1
I
3
w
1+w
I
I
I
13 1+w 1+ 1+w
13 1+w
3w
= lim I
I
I
I
= lim I
= lim
w<0 w
w<0
w<0 w
1+w
w w+1 1+ 1+w
1+w 1+ 1+w
31
1
31
= I
=3
I
I
= lim I
w<0
2
1+w 1+ 1+w
1+0 1+ 1+0
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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NOT FOR SALE
SECTION 2.3
CALCULATING LIMITS USING THE LIMIT LAWS
I
I
I
{2 + 9 3 5
{2 + 9 + 5
{2 + 9 3 5
({2 + 9) 3 25
I
I
= lim
30. lim
= lim
{<34
{<34
{<34 ({ + 4)
{+4
({ + 4) {2 + 9 + 5
{2 + 9 + 5
= lim
{<34
({ + 4)({ 3 4)
{2 3 16
I
= lim
I
2
{<34
({ + 4) { + 9 + 5
({ + 4) {2 + 9 + 5
{34
4
38
34 3 4
=3
=
= I
= lim I
2
{<34
5
+
5
5
16 + 9 + 5
{ +9+5
31. lim
k<0
({ + k)3 3 {3
({3 + 3{2 k + 3{k2 + k3 ) 3 {3
3{2 k + 3{k2 + k3
= lim
= lim
k<0
k<0
k
k
k
= lim
k<0
k(3{2 + 3{k + k2 )
= lim (3{2 + 3{k + k2 ) = 3{2
k<0
k
1
{2 3 ({ + k)2
1
3 2
2
({ + k)
{
({ + k)2 {2
{2 3 ({2 + 2{k + k2 )
3k(2{ + k)
= lim
= lim
32. lim
= lim
k<0
k<0
k<0
k<0 k{2 ({ + k)2
k
k
k{2 ({ + k)2
= lim
k<0
3(2{ + k)
32{
2
= 2 2 =3 3
{2 ({ + k)2
{ ·{
{
33. (a)
(b)
2
{
E
lim I
3
1 + 3{ 3 1
{<0
(c) lim
{<0
{
i ({)
30=001
30=0001
30=00001
30=000001
0=000001
0=00001
0=0001
0=001
0=6661663
0=6666167
0=6666617
0=6666662
0=6666672
0=6666717
0=6667167
0=6671663
The limit appears to be
I
I
I
{ 1 + 3{ + 1
{ 1 + 3{ + 1
1 + 3{ + 1
{
I
·I
= lim
= lim
{<0
{<0
(1 + 3{) 3 1
3{
1 + 3{ 3 1
1 + 3{ + 1
I
1
lim
1 + 3{ + 1
3 {<0
1 t
lim (1 + 3{) + lim 1
=
{<0
{<0
3
1 t
=
lim 1 + 3 lim { + 1
{<0
{<0
3
=
=
=
1 I
1+3·0+1
3
[Limit Law 3]
[1 and 11]
[1, 3, and 7]
[7 and 8]
1
2
(1 + 1) =
3
3
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.
3
¤
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92
¤
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CHAPTER 2
LIMITS AND DERIVATIVES
34. (a)
(b)
I
I
3+{3 3
E 0=29
{<0
{
lim
{
i ({)
30=001
30=0001
30=00001
30=000001
0=000001
0=00001
0=0001
0=001
0=2886992
0=2886775
0=2886754
0=2886752
0=2886751
0=2886749
0=2886727
0=2886511
The limit appears to be approximately 0=2887.
I I
I I
(3 + {) 3 3
3+{3 3
3+{+ 3
1
I
I = lim I
I
= lim I
(c) lim
·I
{<0
{<0 {
{<0
{
3+{+ 3
3+{+ 3
3+{+ 3
=
lim 1
{<0
I
I
lim 3 + { + lim 3
{<0
= t
[Limit Laws 5 and 1]
{<0
1
lim (3 + {) +
{<0
I
3
1
I
= I
3+0+ 3
1
= I
2 3
[7 and 11]
[1, 7, and 8]
35. Let i({) = 3{2 , j({) = {2 cos 20{ and k({) = {2 . Then
31 $ cos 20{ $ 1 i 3{2 $ {2 cos 20{ $ {2
i i ({) $ j({) $ k({).
So since lim i ({) = lim k({) = 0, by the Squeeze Theorem we have
{<0
{<0
lim j({) = 0.
{<0
I
I
I
{3 + {2 sin(@{), and k({) = {3 + {2 . Then
I
I
I
31 $ sin(@{) $ 1 i 3 {3 + {2 $ {3 + {2 sin(@{) $ {3 + {2 i
36. Let i({) = 3 {3 + {2 , j({) =
i ({) $ j({) $ k({). So since lim i ({) = lim k({) = 0, by the Squeeze Theorem
{<0
{<0
we have lim j({) = 0.
{<0
37. We have lim (4{ 3 9) = 4(4) 3 9 = 7 and lim {2 3 4{ + 7 = 42 3 4(4) + 7 = 7. Since 4{ 3 9 $ i ({) $ {2 3 4{ + 7
{<4
{<4
for { D 0, lim i ({) = 7 by the Squeeze Theorem.
{<4
38. We have lim (2{) = 2(1) = 2 and lim ({4 3 {2 + 2) = 14 3 12 + 2 = 2. Since 2{ $ j({) $ {4 3 {2 + 2 for all {,
{<1
{<1
lim j({) = 2 by the Squeeze Theorem.
{<1
39. 31 $ cos(2@{) $ 1
i 3{4 $ {4 cos(2@{) $ {4 . Since lim 3{4 = 0 and lim {4 = 0, we have
{<0
{<0
lim {4 cos(2@{) = 0 by the Squeeze Theorem.
{<0
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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NOT FOR SALE
SECTION 2.3
40. 31 $ sin(@{) $ 1
i h31 $ hsin(@{) $ h1
CALCULATING LIMITS USING THE LIMIT LAWS
¤
I
I
I
I
{/h $ { hsin(@{) $ { h. Since lim ( {/h) = 0 and
i
{<0+
kI
l
I
lim ( { h) = 0, we have lim
{ hsin(@{) = 0 by the Squeeze Theorem.
{<0+
{<0+
41. |{ 3 3| =
+
{33
if { 3 3 D 0
3({ 3 3)
if { 3 3 ? 0
+
=
{33
if { D 3
33{
if { ? 3
Thus, lim (2{ + |{ 3 3|) = lim (2{ + { 3 3) = lim (3{ 3 3) = 3(3) 3 3 = 6 and
{<3+
{<3+
{<3+
lim (2{ + |{ 3 3|) = lim (2{ + 3 3 {) = lim ({ + 3) = 3 + 3 = 6. Since the left and right limits are equal,
{<3
{<3
{<3
lim (2{ + |{ 3 3|) = 6.
{<3
42. |{ + 6| =
+
{+6
if { + 6 D 0
3({ + 6)
if { + 6 ? 0
+
=
{+6
if { D 36
3({ + 6)
if { ? 36
We’ll look at the one-sided limits.
lim
{<36+
2{ + 12
2({ + 6)
= lim
= 2 and
|{ + 6|
{+6
{<36+
The left and right limits are different, so lim
{<36
lim
{<36
2{ + 12
2({ + 6)
= lim
= 32
|{ + 6|
{<36 3({ + 6)
2{ + 12
does not exist.
|{ + 6|
43. 2{3 3 {2 = {2 (2{ 3 1) = {2 · |2{ 3 1| = {2 |2{ 3 1|
|2{ 3 1| =
+
2{ 3 1
3(2{ 3 1)
if 2{ 3 1 D 0
if 2{ 3 1 ? 0
=
+
2{ 3 1
if { D 0=5
3(2{ 3 1)
if { ? 0=5
So 2{3 3 {2 = {2 [3(2{ 3 1)] for { ? 0=5.
Thus,
lim
{<0=5
31
31
2{ 3 1
2{ 3 1
31
=
=
= lim
= lim
= 34.
|2{3 3 {2 | {<0=5 {2 [3(2{ 3 1)] {<0=5 {2
(0=5)2
0=25
2 3 |{|
2 3 (3{)
2+{
= lim
= lim
= lim 1 = 1.
{<32
{<32 2 + {
{<32
2+{
2+{
2
1
1
1
1
3
= lim
3
= lim , which does not exist since the
45. Since |{| = 3{ for { ? 0, we have lim
{
|{|
{
3{
{
{<0
{<0
{<0
44. Since |{| = 3{ for { ? 0, we have lim
{<32
denominator approaches 0 and the numerator does not.
46. Since |{| = { for { A 0, we have lim
{<0+
47. (a)
1
1
3
{
|{|
= lim
{<0+
1
1
3
{
{
= lim 0 = 0.
{<0+
(b) (i) Since sgn { = 1 for { A 0, lim sgn { = lim 1 = 1.
{<0+
{<0+
(ii) Since sgn { = 31 for { ? 0, lim sgn { = lim 31 = 31.
{<0
{<0
(iii) Since lim sgn { 6= lim sgn {, lim sgn { does not exist.
{<0
{<0
{<0+
(iv) Since |sgn {| = 1 for { 6= 0, lim |sgn {| = lim 1 = 1.
{<0
{<0
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94
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CHAPTER 2
+
48. (a) i ({) =
LIMITS AND DERIVATIVES
{2 + 1
if { ? 1
2
({ 3 2)
if { D 1
lim i ({) = lim ({2 + 1) = 12 + 1 = 2,
{<1
{<1
lim i ({) = lim ({ 3 2)2 = (31)2 = 1
{<1+
{<1+
(b) Since the right-hand and left-hand limits of i at { = 1
(c)
are not equal, lim i({) does not exist.
{<1
49. (a) (i) lim j({) = lim
{<2+
{<2+
{2 + { 3 6
({ + 3)({ 3 2)
= lim
|{ 3 2|
|{ 3 2|
{<2+
= lim
{<2+
({ + 3)({ 3 2)
{32
[since { 3 2 A 0 if { < 2+ ]
= lim ({ + 3) = 5
{<2+
(ii) The solution is similar to the solution in part (i), but now |{ 3 2| = 2 3 { since { 3 2 ? 0 if { < 23 .
Thus, lim j({) = lim 3({ + 3) = 35.
{<2
{<2
(b) Since the right-hand and left-hand limits of j at { = 2
(c)
are not equal, lim j({) does not exist.
{<2
50. (a) (i) lim j({) = lim { = 1
{<1
{<1
(ii) lim j({) = lim (2 3 {2 ) = 2 3 12 = 1. Since lim j({) = 1 and lim j({) = 1, we have lim j({) = 1.
{<1+
{<1
{<1+
{<1+
{<1
Note that the fact j(1) = 3 does not affect the value of the limit.
(iii) When { = 1, j({) = 3, so j(1) = 3.
(iv) lim j({) = lim (2 3 {2 ) = 2 3 22 = 2 3 4 = 32
{<2
{<2
(v) lim j({) = lim ({ 3 3) = 2 3 3 = 31
{<2+
{<2+
(vi) lim j({) does not exist since lim j({) 6= lim j({).
{<2
(b)
;
{
A
A
A
A
?3
j({) =
A
2 3 {2
A
A
A
=
{33
{<2
{<2+
if { ? 1
if { = 1
if 1 ? { $ 2
if { A 2
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 2.3
51. (a) (i) [[{]] = 32 for 32 $ { ? 31, so
{<32+
(ii) [[{]] = 33 for 33 $ { ? 32, so
{<32
lim [[{]] =
CALCULATING LIMITS USING THE LIMIT LAWS
¤
lim (32) = 32
{<32+
lim [[{]] =
lim (33) = 33.
{<32
The right and left limits are different, so lim [[{]] does not exist.
{<32
(iii) [[{]] = 33 for 33 $ { ? 32, so
lim [[{]] =
{<32=4
lim (33) = 33.
{<32=4
(b) (i) [[{]] = q 3 1 for q 3 1 $ { ? q, so lim [[{]] = lim (q 3 1) = q 3 1.
{<q
{<q
(ii) [[{]] = q for q $ { ? q + 1, so lim [[{]] = lim q = q.
{<q+
{<q+
(c) lim [[{]] exists C d is not an integer.
{<d
52. (a) See the graph of | = cos {.
Since 31 $ cos { ? 0 on [3> 3@2), we have | = i({) = [[cos {]] = 31
on [3> 3@2).
Since 0 $ cos { ? 1 on [3@2> 0) (0> @2], we have i({) = 0
on [3@2> 0) (0> @2].
Since 31 $ cos { ? 0 on (@2> ], we have i ({) = 31 on (@2> ].
Note that i(0) = 1.
(b) (i) lim i ({) = 0 and lim i ({) = 0, so lim i ({) = 0.
{<0
{<0
{<0+
3
(ii) As { < (@2) , i ({) < 0, so
lim
{<(@2)
(iii) As { < (@2)+ , i ({) < 31, so
lim
i ({) = 0.
{<(@2)+
i ({) = 31.
(iv) Since the answers in parts (ii) and (iii) are not equal, lim i ({) does not exist.
{<@2
(c) lim i ({) exists for all d in the open interval (3> ) except d = 3@2 and d = @2.
{<d
53. The graph of i ({) = [[{]] + [[3{]] is the same as the graph of j({) = 31 with holes at each integer, since i (d) = 0 for any
integer d. Thus, lim i ({) = 31 and lim i ({) = 31, so lim i ({) = 31. However,
{<2
{<2
{<2+
i (2) = [[2]] + [[32]] = 2 + (32) = 0, so lim i ({) 6= i (2).
{<2
54. lim
y<f
#
O0
u
13
y2
f2
$
I
= O0 1 3 1 = 0. As the velocity approaches the speed of light, the length approaches 0.
A left-hand limit is necessary since O is not de¿ned for y A f.
55. Since s({) is a polynomial, s({) = d0 + d1 { + d2 {2 + · · · + dq {q . Thus, by the Limit Laws,
lim s({) = lim d0 + d1 { + d2 {2 + · · · + dq {q = d0 + d1 lim { + d2 lim {2 + · · · + dq lim {q
{<d
{<d
{<d
2
{<d
{<d
q
= d0 + d1 d + d2 d + · · · + dq d = s(d)
Thus, for any polynomial s, lim s({) = s(d).
{<d
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95
96
NOT FOR SALE
¤
CHAPTER 2
56. Let u({) =
LIMITS AND DERIVATIVES
s({)
where s({) and t({) are any polynomials, and suppose that t(d) 6= 0. Then
t({)
lim u({) = lim
{<d
{<d
lim s({)
s({)
{<d
=
t({)
lim t ({)
[Limit Law 5] =
{<d
57. lim [i({) 3 8] = lim
{<1
{<1
s(d)
t(d)
[Exercise 55] = u(d).
i ({) 3 8
i({) 3 8
· ({ 3 1) = lim
· lim ({ 3 1) = 10 · 0 = 0.
{<1
{<1
{31
{31
Thus, lim i ({) = lim {[i ({) 3 8] + 8} = lim [i ({) 3 8] + lim 8 = 0 + 8 = 8.
{<1
{<1
Note: The value of lim
{<1
lim
{<1
i ({) 3 8
exists.
{31
58. (a) lim i ({) = lim
{<0
{<0
{<1
{<1
i ({) 3 8
does not affect the answer since it’s multiplied by 0. What’s important is that
{31
i ({)
i ({) 2
= lim 2 · lim {2 = 5 · 0 = 0
·
{
{<0 {
{<0
{2
i ({)
i ({)
i({)
= lim
·
{
= lim 2 · lim { = 5 · 0 = 0
{<0 {
{<0
{<0 {
{<0
{2
(b) lim
59. Observe that 0 $ i ({) $ {2 for all {, and lim 0 = 0 = lim {2 . So, by the Squeeze Theorem, lim i ({) = 0.
{<0
{<0
{<0
60. Let i({) = [[{]] and j({) = 3[[{]]. Then lim i({) and lim j({) do not exist
{<3
{<3
[Example 10]
but lim [i ({) + j({)] = lim ([[{]] 3 [[{]]) = lim 0 = 0.
{<3
{<3
{<3
61. Let i({) = K({) and j({) = 1 3 K({), where K is the Heaviside function de¿ned in Exercise 1.3.57.
Thus, either i or j is 0 for any value of {. Then lim i ({) and lim j({) do not exist, but lim [i ({)j({)] = lim 0 = 0.
{<0
{<0
{<0
{<0
I
I
I
I
63{32
63{32
63{+2
33{+1
= lim I
·I
·I
62. lim I
{<2
{<2
33{31
33{31
63{+2
33{+1
&
% I
2
I
I
6 3 { 3 22
63{34
33{+1
33{+1
I
I
= lim I
·
=
lim
·
2
{<2
{<2
33{31
63{+2
63{+2
3 3 { 3 12
I
I
(2 3 {) 3 3 { + 1
33{+1
1
I
= lim I
=
{<2 (2 3 {)
{<2
2
63{+2
63{+2
= lim
63. Since the denominator approaches 0 as { < 32, the limit will exist only if the numerator also approaches
0 as { < 32. In order for this to happen, we need lim
{<32
2
3{ + d{ + d + 3 = 0 C
3(32)2 + d(32) + d + 3 = 0 C 12 3 2d + d + 3 = 0 C d = 15. With d = 15, the limit becomes
lim
{<32
3{2 + 15{ + 18
3({ + 2)({ + 3)
3({ + 3)
3(32 + 3)
3
= lim
= lim
=
=
= 31.
{<32 ({ 3 1)({ + 2)
{<32 { 3 1
{2 + { 3 2
32 3 1
33
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SECTION 2.4
THE PRECISE DEFINITION OF A LIMIT
¤
97
64. Solution 1: First, we ¿nd the coordinates of S and T as functions of u. Then we can ¿nd the equation of the line determined
by these two points, and thus ¿nd the {-intercept (the point U), and take the limit as u < 0. The coordinates of S are (0> u).
The point T is the point of intersection of the two circles {2 + | 2 = u2 and ({ 3 1)2 + | 2 = 1. Eliminating | from these
equations, we get u2 3 {2 = 1 3 ({ 3 1)2
u2 = 1 + 2{ 3 1 C { = 12 u2 . Substituting back into the equation of the
t
2
C | = u 1 3 14 u2
shrinking circle to ¿nd the |-coordinate, we get 12 u2 + | 2 = u2 C | 2 = u2 1 3 14 u2
C
(the positive |-value). So the coordinates of T are
u
|3u =
t
1 3 14 u2 3 u
1 2
2u
30
1 2
2u > u
t
1 3 14 u2 . The equation of the line joining S and T is thus
({ 3 0). We set | = 0 in order to ¿nd the {-intercept, and get
1 2
u
2
{ = 3u t
=
u
1 3 14 u2 3 1
3 12 u2
t
1 3 14 u2 + 1
13
1 2
4u
31
=2
t
1 3 14 u2 + 1
t
I
1 3 14 u2 + 1 = lim 2 1 + 1 = 4.
Now we take the limit as u < 0+ : lim { = lim 2
u<0+
u<0+
u<0+
So the limiting position of U is the point (4> 0).
Solution 2: We add a few lines to the diagram, as shown. Note that
_S TV = 90 (subtended by diameter S V). So _VTU = 90 = _RTW
(subtended by diameter RW ). It follows that _RTV = _W TU. Also
_S VT = 90 3 _VS T = _RUS . Since 4TRV is isosceles, so is
4TW U, implying that TW = W U. As the circle F2 shrinks, the point T
plainly approaches the origin, so the point U must approach a point twice
as far from the origin as W , that is, the point (4> 0), as above.
2.4 The Precise Definition of a Limit
1. If |i({) 3 1| ? 0=2, then 30=2 ? i ({) 3 1 ? 0=2
i 0=8 ? i ({) ? 1=2. From the graph, we see that the last inequality is
true if 0=7 ? { ? 1=1, so we can choose = min {1 3 0=7> 1=1 3 1} = min {0=3> 0=1} = 0=1 (or any smaller positive
number).
2. If |i({) 3 2| ? 0=5, then 30=5 ? i ({) 3 2 ? 0=5
i 1=5 ? i ({) ? 2=5. From the graph, we see that the last inequality is
true if 2=6 ? { ? 3=8, so we can take = min {3 3 2=6> 3=8 3 3} = min {0=4> 0=8} = 0=4 (or any smaller positive number).
Note that { 6= 3.
3. The leftmost question mark is the solution of
I
I
{ = 1=6 and the rightmost, { = 2=4. So the values are 1=62 = 2=56 and
2=42 = 5=76. On the left side, we need |{ 3 4| ? |2=56 3 4| = 1=44. On the right side, we need |{ 3 4| ? |5=76 3 4| = 1=76.
To satisfy both conditions, we need the more restrictive condition to hold — namely, |{ 3 4| ? 1=44. Thus, we can choose
= 1=44, or any smaller positive number.
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¤
NOT FOR SALE
CHAPTER 2
LIMITS AND DERIVATIVES
4. The leftmost question mark is the positive solution of {2 =
solution of {2 = 32 , that is, { =
t
1
2,
that is, { =
1
I
,
2
and the rightmost question mark is the positive
I1
3
2 . On the left side, we need |{ 3 1| ? 2 3 1 E 0=292 (rounding down to be safe). On
t
the right side, we need |{ 3 1| ? 32 3 1 E 0=224. The more restrictive of these two conditions must apply, so we choose
= 0=224 (or any smaller positive number).
5.
From the graph, we ¿nd that | = tan { = 0=8 when { E 0=675, so
4
3 1 E 0=675 i 1 E
when { E 0=876, so
4
4
3 0=675 E 0=1106. Also, | = tan { = 1=2
+ 2 E 0=876 i 2 = 0=876 3
4
E 0=0906.
Thus, we choose = 0=0906 (or any smaller positive number) since this is
the smaller of 1 and 2 .
6.
From the graph, we ¿nd that | = 2{@({2 + 4) = 0=3 when { = 23 , so
1 3 1 =
2
3
i 1 = 13 . Also, | = 2{@({2 + 4) = 0=5 when { = 2, so
1 + 2 = 2 i 2 = 1. Thus, we choose =
1
3
(or any smaller positive
number) since this is the smaller of 1 and 2 .
7.
From the graph with % = 0=2, we ¿nd that | = {3 3 3{ + 4 = 5=8 when
{ r 1=9774, so 2 3 1 r 1=9774 i 1 r 0=0226. Also,
| = {3 3 3{ + 4 = 6=2 when { r 2=022, so 2 + 2 r 2=0219 i
2 r 0=0219. Thus, we choose = 0=0219 (or any smaller positive
number) since this is the smaller of 1 and 2 .
For % = 0=1, we get 1 r 0=0112 and 2 r 0=0110, so we choose
= 0=011 (or any smaller positive number).
8.
From the graph with % = 0=5, we ¿nd that | = (h2{ 3 1)@{ = 1=5 when
{ r 30=303, so 1 r 0=303. Also, | = (h2{ 3 1)@{ = 2=5 when
{ r 0=215, so 2 r 0=215. Thus, we choose = 0=215 (or any smaller
positive number) since this is the smaller of 1 and 2 .
For % = 0=1, we get 1 r 0=052 and 2 r 0=048, so we choose
= 0=048 (or any smaller positive number).
9. (a)
From the graph, we ¿nd that | = tan2 { = 1000 when { E 1=539 and
{ E 1=602 for { near
.
2
Thus, we get E 1=602 3
2
E 0=031 for
P = 1000.
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NOT FOR SALE
SECTION 2.4
THE PRECISE DEFINITION OF A LIMIT
¤
99
From the graph, we ¿nd that | = tan2 { = 10,000 when { E 1=561 and
(b)
{ E 1=581 for { near
.
2
Thus, we get E 1=581 3
2
E 0=010 for
P = 10,000.
I
From the graph, we ¿nd that {2 @ { 3 5 = 100 i { E 5=066.
10.
Thus, 5 + E 5=0659 and E 0=065.
11. (a) D = u2 and D = 1000 cm2
i u2 = 1000 i u2 =
1000
i u=
t
1000
(u A 0)
E 17=8412 cm.
(b) |D 3 1000| $ 5 i 35 $ u2 3 1000 $ 5 i 1000 3 5 $ u2 $ 1000 + 5 i
t
t
t
t
t
t
995
1005
1000
995
1005
1000
$
u
$
i
17=7966
$
u
$
17=8858.
3
E
0=04466
and
3
E 0=04455. So
if the machinist gets the radius within 0=0445 cm of 17=8412, the area will be within 5 cm2 of 1000.
(c) { is the radius, i ({) is the area, d is the target radius given in part (a), O is the target area (1000), % is the tolerance in the
area (5), and is the tolerance in the radius given in part (b).
12. (a) W = 0=1z2 + 2=155z + 20 and W = 200
i
0=1z2 + 2=155z + 20 = 200 i [by the quadratic formula or
from the graph] z E 33=0 watts (z A 0)
(b) From the graph, 199 $ W $ 201 i 32=89 ? z ? 33=11.
(c) { is the input power, i ({) is the temperature, d is the target input power given in part (a), O is the target temperature (200),
% is the tolerance in the temperature (1), and is the tolerance in the power input in watts indicated in part (b) (0=11 watts).
13. (a) |4{ 3 8| = 4 |{ 3 2| ? 0=1
C |{ 3 2| ?
0=1
0=1
, so =
= 0=025.
4
4
(b) |4{ 3 8| = 4 |{ 3 2| ? 0=01 C |{ 3 2| ?
0=01
0=01
, so =
= 0=0025.
4
4
14. |(5{ 3 7) 3 3| = |5{ 3 10| = |5({ 3 2)| = 5 |{ 3 2|. We must have |i({) 3 O| ? %, so 5 |{ 3 2| ? %
C
|{ 3 2| ? %@5. Thus, choose = %@5. For % = 0=1, = 0=02; for % = 0=05, = 0=01; for % = 0=01, = 0=002.
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100
¤
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CHAPTER 2
LIMITS AND DERIVATIVES
15. Given % A 0, we need A 0 such that if 0 ? |{ 3 3| ? , then
(1 + 1 {) 3 2 ? %. But (1 + 1 {) 3 2 ? % C 1 { 3 1 ? % C
3
3
3
1
|{ 3 3| ? % C |{ 3 3| ? 3%. So if we choose = 3%, then
3
0 ? |{ 3 3| ? i (1 + 13 {) 3 2 ? %. Thus, lim (1 + 13 {) = 2 by
{<3
the de¿nition of a limit.
16. Given % A 0, we need A 0 such that if 0 ? |{ 3 4| ? , then
|(2{ 3 5) 3 3| ? %. But |(2{ 3 5) 3 3| ? % C |2{ 3 8| ? % C
|2| |{ 3 4| ? % C |{ 3 4| ? %@2. So if we choose = %@2, then
0 ? |{ 3 4| ? i |(2{ 3 5) 3 3| ? %. Thus, lim (2{ 3 5) = 3 by the
{<4
de¿nition of a limit.
17. Given % A 0, we need A 0 such that if 0 ? |{ 3 (33)| ? , then
|(1 3 4{) 3 13| ? %. But |(1 3 4{) 3 13| ? % C
|34{ 3 12| ? % C |34| |{ + 3| ? % C |{ 3 (33)| ? %@4. So if
we choose = %@4, then 0 ? |{ 3 (33)| ? i |(1 3 4{) 3 13| ? %.
Thus, lim (1 3 4{) = 13 by the de¿nition of a limit.
{<33
x
18. Given % A 0, we need A 0 such that if 0 ? |{ 3 (32)| ? , then
|(3{ + 5) 3 (31)| ? %. But |(3{ + 5) 3 (31)| ? % C
|3{ + 6| ? % C |3| |{ + 2| ? % C |{ + 2| ? %@3. So if we choose
= %@3, then 0 ? |{ + 2| ? i |(3{ + 5) 3 (31)| ? %. Thus,
lim (3{ + 5) = 31 by the de¿nition of a limit.
{<32
2 + 4{
19. Given % A 0, we need A 0 such that if 0 ? |{ 3 1| ? , then 2 + 4{
3 2 ? %. But 3 2 ? % C
3
3
4{ 3 4 4
3
3
3 ? % C 3 |{ 3 1| ? % C |{ 3 1| ? 4 %. So if we choose = 4 %, then 0 ? |{ 3 1| ? 2 + 4{
? %. Thus, lim 2 + 4{ = 2 by the de¿nition of a limit.
3
2
3
{<1
3
i
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 2.4
THE PRECISE DEFINITION OF A LIMIT
20. Given % A 0, we need A 0 such that if 0 ? |{ 3 10| ? , then 3 3 45 { 3 (35) ? %. But 3 3 45 { 3 (35) ? %
8 3 4 { ? % C 3 4 |{ 3 10| ? % C |{ 3 10| ? 5 %. So if we choose = 5 %, then 0 ? |{ 3 10| ? 5
5
4
4
4
4
3 3 { 3 (35) ? %. Thus, lim (3 3 {) = 35 by the de¿nition of a limit.
5
5
¤
101
C
i
{<10
2
{ + { 3 6
21. Given % A 0, we need A 0 such that if 0 ? |{ 3 2| ? , then ({ + 3)({ 3 2)
3 5 ? %
{32
Then 0 ? |{ 3 2| ? C
{32
3 5 ? % C
|{ + 3 3 5| ? % [{ 6= 2] C |{ 3 2| ? %. So choose = %.
({ + 3)({ 3 2)
i |{ 3 2| ? % i |{ + 3 3 5| ? % i 3 5 ? % [{ 6= 2] i
{32
2
{ + { 3 6
{2 + { 3 6
lim
= 5.
{ 3 2 3 5 ? %. By the de¿nition of a limit, {<2
{32
9 3 4{2
22. Given % A 0, we need A 0 such that if 0 ? |{ + 1=5| ? , then 3 6 ? % C
3 + 2{
(3 + 2{)(3 3 2{)
3 6 ? % C |3 3 2{ 3 6| ? % [{ 6= 31=5] C |32{ 3 3| ? % C |32| |{ + 1=5| ? % C
3 + 2{
|{ + 1=5| ? %@2. So choose = %@2. Then 0 ? |{ + 1=5| ? i |{ + 1=5| ? %@2 i |32| |{ + 1=5| ? % i
9 3 4{2
(3 + 2{)(3 3 2{)
|32{ 3 3| ? % i |3 3 2{ 3 6| ? % i 3 6 ? % [{ 6= 31=5] i 3 6 ? %.
3 + 2{
3 + 2{
By the de¿nition of a limit,
9 3 4{2
= 6.
{<31=5 3 + 2{
lim
23. Given % A 0, we need A 0 such that if 0 ? |{ 3 d| ? , then |{ 3 d| ? %. So = % will work.
24. Given % A 0, we need A 0 such that if 0 ? |{ 3 d| ? , then |f 3 f| ? %. But |f 3 f| = 0, so this will be true no matter
what we pick.
C {2 ? %
C |{|3 ? % C |{| ?
25. Given % A 0, we need A 0 such that if 0 ? |{ 3 0| ? , then {2 3 0 ? %
Then 0 ? |{ 3 0| ? i {2 3 0 ? %. Thus, lim {2 = 0 by the de¿nition of a limit.
I
I
%. Take = %.
{<0
26. Given % A 0, we need A 0 such that if 0 ? |{ 3 0| ? , then {3 3 0 ? %
Then 0 ? |{ 3 0| ? C |{| ?
I
I
3
%. Take = 3 %.
i {3 3 0 ? 3 = %. Thus, lim {3 = 0 by the de¿nition of a limit.
{<0
27. Given % A 0, we need A 0 such that if 0 ? |{ 3 0| ? , then |{| 3 0 ? %. But |{| = |{|. So this is true if we pick = %.
Thus, lim |{| = 0 by the de¿nition of a limit.
{<0
I
I
28. Given % A 0, we need A 0 such that if 0 ? { 3 (36) ? , then 8 6 + { 3 0 ? %. But 8 6 + { 3 0 ? %
I
8
6 + { ? % C 6 + { ? %8 C { 3 (36) ? %8 . So if we choose = %8 , then 0 ? { 3 (36) ? I
I
8 6 + { 3 0 ? %. Thus, lim 8 6 + { = 0 by the de¿nition of a right-hand limit.
i
{<36+
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C
102
NOT FOR SALE
¤
CHAPTER 2
LIMITS AND DERIVATIVES
C {2 3 4{ + 4 ? % C
I
C |{ 3 2| ? % C ({ 3 2)2 ? %. Thus,
29. Given % A 0, we need A 0 such that if 0 ? |{ 3 2| ? , then {2 3 4{ + 5 3 1 ? %
({ 3 2)2 ? %. So take = I%. Then 0 ? |{ 3 2| ? lim {2 3 4{ + 5 = 1 by the de¿nition of a limit.
{<2
30. Given % A 0, we need A 0 such that if 0 ? |{ 3 2| ? , then ({2 + 2{ 3 7) 3 1 ? %. But ({2 + 2{ 3 7) 3 1 ? %
C
2
{ + 2{ 3 8 ? % C |{ + 4| |{ 3 2| ? %. Thus our goal is to make |{ 3 2| small enough so that its product with |{ + 4|
is less than %. Suppose we ¿rst require that |{ 3 2| ? 1. Then 31 ? { 3 2 ? 1 i 1 ? { ? 3 i 5 ? { + 4 ? 7 i
|{ + 4| ? 7, and this gives us 7 |{ 3 2| ? % i |{ 3 2| ? %@7. Choose = min {1> %@7}. Then if 0 ? |{ 3 2| ? , we
have |{ 3 2| ? %@7 and |{ + 4| ? 7, so ({2 + 2{ 3 7) 3 1 = |({ + 4)({ 3 2)| = |{ + 4| |{ 3 2| ? 7(%@7) = %, as
desired. Thus, lim ({2 + 2{ 3 7) = 1 by the de¿nition of a limit.
{<2
31. Given % A 0, we need A 0 such that if 0 ? |{ 3 (32)| ? , then {2 3 1 3 3 ? % or upon simplifying we need
2
{ 3 4 ? % whenever 0 ? |{ + 2| ? . Notice that if |{ + 2| ? 1, then 31 ? { + 2 ? 1 i 35 ? { 3 2 ? 33 i
|{ 3 2| ? 5. So take = min {%@5> 1}. Then 0 ? |{ + 2| ? i |{ 3 2| ? 5 and |{ + 2| ? %@5, so
2
{ 3 1 3 3 = |({ + 2)({ 3 2)| = |{ + 2| |{ 3 2| ? (%@5)(5) = %. Thus, by the de¿nition of a limit, lim ({2 3 1) = 3.
{<32
32. Given % A 0, we need A 0 such that if 0 ? |{ 3 2| ? , then {3 3 8 ? %. Now {3 3 8 = ({ 3 2) {2 + 2{ + 4 .
If |{ 3 2| ? 1, that is, 1 ? { ? 3, then {2 + 2{ + 4 ? 32 + 2(3) + 4 = 19 and so
3
{ 3 8 = |{ 3 2| {2 + 2{ + 4 ? 19 |{ 3 2|. So if we take = min 1> % , then 0 ? |{ 3 2| ? 19
3
{ 3 8 = |{ 3 2| {2 + 2{ + 4 ? % · 19 = %. Thus, by the de¿nition of a limit, lim {3 = 8.
19
i
{<2
33. Given % A 0, we let = min 2>
%
8
. If 0 ? |{ 3 3| ? , then |{ 3 3| ? 2 i 32 ? { 3 3 ? 2 i
4 ? { + 3 ? 8 i |{ + 3| ? 8. Also |{ 3 3| ? 8% , so {2 3 9 = |{ + 3| |{ 3 3| ? 8 · 8% = %. Thus, lim {2 = 9.
{<3
34. From the ¿gure, our choices for are 1 = 3 3
2 =
I
9 3 % and
I
9 + % 3 3. The largest possible choice for is the minimum
value of { 1 > 2 }; that is, = min{ 1 > 2 } = 2 =
I
9 + % 3 3.
35. (a) The points of intersection in the graph are ({1 > 2=6) and ({2 > 3=4)
with {1 E 0=891 and {2 E 1=093. Thus, we can take to be the
smaller of 1 3 {1 and {2 3 1. So = {2 3 1 E 0=093.
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SECTION 2.4
THE PRECISE DEFINITION OF A LIMIT
¤
103
(b) Solving {3 + { + 1 = 3 + % gives us two nonreal complex roots and one real root, which is
I
2@3
3 12
216 + 108% + 12 336 + 324% + 81%2
{(%) =
I
1@3 . Thus, = {(%) 3 1.
2
6 216 + 108% + 12 336 + 324% + 81%
(c) If % = 0=4, then {(%) E 1=093 272 342 and = {(%) 3 1 E 0=093, which agrees with our answer in part (a).
1
1 3
? % whenever
36. 1. Guessing a value for Let % A 0 be given. We have to ¿nd a number A 0 such that {
2
1
|{ 3 2|
1 2 3 { 1
=
0 ? |{ 3 2| ? . But 3 = ? %. We ¿nd a positive constant F such that
?F i
{ 2
2{ |2{|
|2{|
|{ 3 2|
%
? F |{ 3 2| and we can make F |{ 3 2| ? % by taking |{ 3 2| ?
= . We restrict { to lie in the interval
|2{|
F
|{ 3 2| ? 1 i 1 ? { ? 3 so 1 A
1
1
A
{
3
i
1
1
1
?
?
6
2{
2
choose = min {1> 2%}.
i
1
1
1
? . So F = is suitable. Thus, we should
|2{|
2
2
2. Showing that works
Given % A 0 we let = min {1> 2%}. If 0 ? |{ 3 2| ? , then |{ 3 2| ? 1 i 1 ? { ? 3 i
1
1
1 |{ 3 2|
1
1
? (as in part 1). Also |{ 3 2| ? 2%, so 3 =
? · 2% = %. This shows that lim (1@{) = 12 .
{<2
|2{|
2
{
2
|2{|
2
37. 1. Guessing a value for I
I
Given % A 0, we must ¿nd A 0 such that | { 3 d| ? % whenever 0 ? |{ 3 d| ? . But
I
I
I
I
|{ 3 d|
I ? % (from the hint). Now if we can ¿nd a positive constant F such that { + d A F then
| { 3 d| = I
{+ d
|{ 3 d|
|{ 3 d|
I
I ?
? %, and we take |{ 3 d| ? F%. We can ¿nd this number by restricting { to lie in some interval
F
{+ d
t
I
I
I
{ + d A 12 d + d, and so
centered at d. If |{ 3 d| ? 12 d, then 3 12 d ? { 3 d ? 12 d i 12 d ? { ? 32 d i
t
t
I
I 1
F = 12 d + d is a suitable choice for the constant. So |{ 3 d| ?
d + d %. This suggests that we let
2
t
q
I r
1
d+ d % .
= min 12 d>
2
t
q
I r
1
d
+
d % . If 0 ? |{ 3 d| ? , then
2. Showing that works Given % A 0, we let = min 12 d>
2
t
t
I
I
I
I 1
{ + d A 12 d + d (as in part 1). Also |{ 3 d| ?
d + d %, so
|{ 3 d| ? 12 d i
2
s
I d@2 + d %
I
I
I
I
|{ 3 d|
I ? s
| { 3 d| = I
lim { = d by the de¿nition of a limit.
I = %. Therefore, {<d
{+ d
d@2 + d
38. Suppose that lim K(w) = O. Given % =
w<0
O3
1
2
so O 3
1
,
2
there exists A 0 such that 0 ? |w| ? ? K(w) ? O + 12 . For 0 ? w ? , K(w) = 1, so 1 ? O +
1
2
1
2
i |K(w) 3 O| ?
1
2
C
i O A 12 . For 3 ? w ? 0, K(w) = 0,
? 0 i O ? 12 . This contradicts O A 12 . Therefore, lim K(w) does not exist.
w<0
39. Suppose that lim i ({) = O. Given % =
{<0
1
,
2
there exists A 0 such that 0 ? |{| ? i |i({) 3 O| ? 12 . Take any rational
number u with 0 ? |u| ? . Then i (u) = 0, so |0 3 O| ? 12 , so O $ |O| ? 12 . Now take any irrational number v with
0 ? |v| ? . Then i (v) = 1, so |1 3 O| ? 12 . Hence, 1 3 O ? 12 , so O A 12 . This contradicts O ? 12 , so lim i({) does not
{<0
exist.
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¤
CHAPTER 2
LIMITS AND DERIVATIVES
40. First suppose that lim i ({) = O. Then, given % A 0 there exists A 0 so that 0 ? |{ 3 d| ? {<d
i |i({) 3 O| ? %.
Then d 3 ? { ? d i 0 ? |{ 3 d| ? so |i ({) 3 O| ? %. Thus, lim i ({) = O. Also d ? { ? d + {<d
i
0 ? |{ 3 d| ? so |i ({) 3 O| ? %. Hence, lim i ({) = O.
{<d+
Now suppose lim i ({) = O = lim i({). Let % A 0 be given. Since lim i ({) = O, there exists 1 A 0 so that
{<d
{<d
{<d+
d 3 1 ? { ? d i |i ({) 3 O| ? %. Since lim i ({) = O, there exists 2 A 0 so that d ? { ? d + 2
{<d+
|i ({) 3 O| ? %. Let be the smaller of 1 and 2 . Then 0 ? |{ 3 d| ? i d 3 1 ? { ? d or d ? { ? d + 2 so
|i ({) 3 O| ? %. Hence, lim i ({) = O. So we have proved that lim i ({) = O C
{<d
41.
1
C |{ + 3| ? I
4
10,000
42. Given P A 0, we need A 0 such that 0 ? |{ + 3| ? ({ + 3)4 ?
lim
{<33
1
P
lim i({) = O = lim i ({).
{<d
1
1
A 10,000 C ({ + 3)4 ?
({ + 3)4
10,000
i
C
{<d
{<d+
|{ 3 (33)| ?
i 1@({ + 3)4 A P. Now
1
10
1
AP
({ + 3)4
1
1
1
C |{ + 3| ? I
. So take = I
. Then 0 ? |{ + 3| ? = I
4
4
4
P
P
P
i
C
1
A P, so
({ + 3)4
1
= ".
({ + 3)4
43. Given P ? 0 we need A 0 so that ln { ? P whenever 0 ? { ? ; that is, { = hln { ? hP whenever 0 ? { ? . This
suggests that we take = hP . If 0 ? { ? hP , then ln { ? ln hP = P . By the de¿nition of a limit, lim ln { = 3".
{<0+
44. (a) Let P be given. Since lim i ({) = ", there exists 1 A 0 such that 0 ? |{ 3 d| ? 1
{<d
lim j({) = f, there exists 2 A 0 such that 0 ? |{ 3 d| ? 2
{<d
smaller of 1 and 2 . Then 0 ? |{ 3 d| ? i i ({) A P + 1 3 f. Since
i |j({) 3 f| ? 1 i j({) A f 3 1. Let be the
i i ({) + j({) A (P + 1 3 f) + (f 3 1) = P . Thus,
lim [i ({) + j({)] = ".
{<d
(b) Let P A 0 be given. Since lim j({) = f A 0, there exists 1 A 0 such that 0 ? |{ 3 d| ? 1
{<d
i
|j({) 3 f| ? f@2 i j({) A f@2. Since lim i ({) = ", there exists 2 A 0 such that 0 ? |{ 3 d| ? 2
{<d
i ({) A 2P@f. Let = min {1 > 2 }. Then 0 ? |{ 3 d| ? i i ({) j({) A
i
2P f
= P, so lim i ({) j({) = ".
{<d
f 2
(c) Let Q ? 0 be given. Since lim j({) = f ? 0, there exists 1 A 0 such that 0 ? |{ 3 d| ? 1
{<d
i
|j({) 3 f| ? 3f@2 i j({) ? f@2. Since lim i ({) = ", there exists 2 A 0 such that 0 ? |{ 3 d| ? 2
i
i ({) A 2Q@f. (Note that f ? 0 and Q ? 0 i 2Q@f A 0.) Let = min {1 > 2 }. Then 0 ? |{ 3 d| ? i
{<d
i ({) A 2Q@f i i ({) j({) ?
2Q f
· = Q, so lim i ({) j({) = 3".
{<d
f 2
2.5 Continuity
1. From De¿nition 1, lim i ({) = i (4).
{<4
2. The graph of i has no hole, jump, or vertical asymptote.
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SECTION 2.5
CONTINUITY
¤
3. (a) i is discontinuous at 34 since i (34) is not de¿ned and at 32, 2, and 4 since the limit does not exist (the left and right
limits are not the same).
(b) i is continuous from the left at 32 since
lim i ({) = i (32). i is continuous from the right at 2 and 4 since
{<32
lim i({) = i (2) and lim i ({) = i (4). It is continuous from neither side at 34 since i (34) is unde¿ned.
{<2+
{<4+
4. j is continuous on [34> 32), (32> 2), [2> 4), (4> 6), and (6> 8).
5. The graph of | = i ({) must have a discontinuity at
{ = 2 and must show that lim i ({) = i (2).
{<2+
6. The graph of | = i ({) must have discontinuities
at { = 31 and { = 4. It must show that
lim i ({) = i (31) and lim i({) = i(4).
{<31
7. The graph of | = i ({) must have a removable
discontinuity (a hole) at { = 3 and a jump discontinuity
at { = 5.
{<4+
8. The graph of | = i ({) must have a discontinuity
at { = 32 with
lim i ({) 6= i (32) and
{<32
lim i ({) 6= i(32). It must also show that
{<32+
lim i ({) = i (2) and lim i ({) 6= i (2).
{<2
{<2+
9. (a) The toll is $7 between 7:00 AM and 10:00 AM and between 4:00 PM and 7:00 PM .
(b) The function W has jump discontinuities at w = 7, 10, 16, and 19. Their
signi¿cance to someone who uses the road is that, because of the sudden jumps in
the toll, they may want to avoid the higher rates between w = 7 and w = 10 and
between w = 16 and w = 19 if feasible.
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CHAPTER 2
LIMITS AND DERIVATIVES
10. (a) Continuous; at the location in question, the temperature changes smoothly as time passes, without any instantaneous jumps
from one temperature to another.
(b) Continuous; the temperature at a speci¿c time changes smoothly as the distance due west from New York City increases,
without any instantaneous jumps.
(c) Discontinuous; as the distance due west from New York City increases, the altitude above sea level may jump from one
height to another without going through all of the intermediate values — at a cliff, for example.
(d) Discontinuous; as the distance traveled increases, the cost of the ride jumps in small increments.
(e) Discontinuous; when the lights are switched on (or off ), the current suddenly changes between 0 and some nonzero value,
without passing through all of the intermediate values. This is debatable, though, depending on your de¿nition of current.
11. If i and j are continuous and j(2) = 6, then lim [3i ({) + i ({) j({)] = 36
{<2
i
3 lim i ({) + lim i ({) · lim j({) = 36 i 3i (2) + i(2) · 6 = 36 i 9i (2) = 36 i i (2) = 4.
{<2
{<2
{<2
12. lim i ({) = lim 3{4 3 5{ +
{<2
{<2
t
I
3
{2 + 4 = 3 lim {4 3 5 lim { + 3 lim ({2 + 4)
{<2
{<2
{<2
I
= 3(2)4 3 5(2) + 3 22 + 4 = 48 3 10 + 2 = 40 = i(2)
By the de¿nition of continuity, i is continuous at d = 2.
13. lim i ({) = lim
{<31
{<31
4
{ + 2{3 =
lim { + 2 lim {3
{<31
{<31
4
By the de¿nition of continuity, i is continuous at d = 31.
4
= 31 + 2(31)3 = (33)4 = 81 = i(31).
lim (2w 3 3w2 )
2 lim w 3 3 lim w2
2(1) 3 3(1)2
31
2w 3 3w2
w<1
w<1
w<1
14. lim k(w) = lim
=
=
=
=
= k(1).
w<1
w<1 1 + w3
lim (1 + w3 )
lim 1 + lim w3
1 + (1)3
2
w<1
w<1
w<1
By the de¿nition of continuity, k is continuous at d = 1.
15. For d A 2, we have
lim (2{ + 3)
2{ + 3
{<d
=
{<d { 3 2
lim ({ 3 2)
lim i ({) = lim
{<d
[Limit Law 5]
{<d
2 lim { + lim 3
=
{<d
{<d
=
{<d
lim { 3 lim 2
2d + 3
d32
[1, 2, and 3]
{<d
[7 and 8]
= i (d)
Thus, i is continuous at { = d for every d in (2> "); that is, i is continuous on (2> ").
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SECTION 2.5
CONTINUITY
¤
16. For d ? 3, we have
lim j({) = lim 2
{<d
{<d
= 2 lim
{<d
=2
=2
=2
t
I
33{
I
33{
lim (3 3 {)
[11]
lim 3 3 lim {
[2]
{<d
t
[Limit Law 3]
{<d
{<d
I
33d
[7 and 8]
= j(d)
So j is continuous at { = d for every d in (3"> 3). Also, lim j({) = 0 = j(3), so j is continuous from the left at 3.
{<3
Thus, j is continuous on (3"> 3].
17. i ({) =
1
is discontinuous at d = 32 because i (32) is unde¿ned.
{+2
18. i ({) =
;
?
1
{+2
=
1
if { 6= 32
if { = 32
Here i (32) = 1, but
lim i ({) = 3" and
{<32
lim i ({) = ",
{<32+
so lim i ({) does not exist and i is discontinuous at 32.
{<32
19. i ({) =
+
h{
if { ? 0
2
if { D 0
{
The left-hand limit of i at d = 0 is lim i ({) = lim h{ = 1. The
{<0
{<0
right-hand limit of i at d = 0 is lim i ({) = lim {2 = 0. Since these
{<0+
{<0+
limits are not equal, lim i ({) does not exist and i is discontinuous at 0=
{<0
; 2
?{ 3{
{2 3 1
20. i ({) =
=
1
lim i ({) = lim
{<1
{<1
if { 6= 1
if { = 1
{2 3 {
{({ 3 1)
{
1
= lim
= lim
= ,
{<1 ({ + 1)({ 3 1)
{<1 { + 1
{2 3 1
2
but i (1) = 1, so i is discontinous at 1=
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108
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NOT FOR SALE
CHAPTER 2
;
cos {
A
?
21. i ({) = 0
A
=
1 3 {2
LIMITS AND DERIVATIVES
if { ? 0
if { = 0
if { A 0
lim i ({) = 1, but i (0) = 0 6= 1, so i is discontinuous at 0.
{<0
; 2
? 2{ 3 5{ 3 3
{33
22. i ({) =
=
6
lim i ({) = lim
{<3
{<3
if { 6= 3
if { = 3
(2{ + 1)({ 3 3)
2{2 3 5{ 3 3
= lim
= lim (2{ + 1) = 7,
{<3
{<3
{33
{33
but i (3) = 6, so i is discontinuous at 3.
23. i ({) =
({ 3 2)({ + 1)
{2 3 { 3 2
=
= { + 1 for { 6= 2. Since lim i ({) = 2 + 1 = 3, de¿ne i (2) = 3. Then i is
{<2
{32
{32
continuous at 2.
24. i ({) =
({ 3 2)({2 + 2{ + 4)
{2 + 2{ + 4
{3 3 8
4+4+4
=
=
for { 6= 2. Since lim i ({) =
= 3, de¿ne i (2) = 3.
{<2
{2 3 4
({ 3 2)({ + 2)
{+2
2+2
Then i is continuous at 2.
25. I ({) =
2{2 3 { 3 1
is a rational function, so it is continuous on its domain, (3"> "), by Theorem 5(b).
{2 + 1
{2 + 1
{2 + 1
=
is a rational function, so it is continuous on its domain,
3{31
(2{ + 1)({ 3 1)
3"> 3 12 3 12 > 1 (1> "), by Theorem 5(b).
26. J({) =
2{2
I
3
27. { 3 2 = 0 i { = 2 i { = 2, so T({) =
3
3
continuous everywhere by Theorem 5(a) and
I
3
I I
{32
has domain 3"> 3 2 3 2> " . Now {3 3 2 is
{3 3 2
I
3
{ 3 2 is continuous everywhere by Theorems 5(a), 7, and 9. Thus, T is
continuous on its domain by part 5 of Theorem 4.
28. The domain of U(w) =
hsin w
is (3"> ") since the denominator is never 0 [cos w D 31 i 2 + cos w D 1]. By
2 + cos w
Theorems 7 and 9, hsin w and cos w are continuous on R. By part 1 of Theorem 4, 2 + cos w is continuous on R and by part 5
of Theorem 4, U is continuous on R.
29. By Theorem 5(a), the polynomial 1 + 2w is continuous on R. By Theorem 7, the inverse trigonometric function arcsin { is
continuous on its domain, [31> 1]. By Theorem 9, D(w) = arcsin(1 + 2w) is continuous on its domain, which is
{w | 31 $ 1 + 2w $ 1} = {w | 32 $ 2w $ 0} = {w | 31 $ w $ 0} = [31> 0].
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NOT FOR SALE
SECTION 2.5
30. By Theorem 7, the trigonometric function tan { is continuous on its domain, { | { 6=
the composite function
2
CONTINUITY
¤
109
+ q . By Theorems 5(a), 7, and 9,
I
tan {
4 3 {2 is continuous on its domain [32> 2]. By part 5 of Theorem 4, E({) = I
is
4 3 {2
continuous on its domain, (32> 3@2) (3@2> @2) (@2> 2).
31. P ({) =
u
u
1
{+1
{+1
1+ =
is de¿ned when
D 0 i { + 1 D 0 and { A 0 or { + 1 $ 0 and { ? 0 i { A 0
{
{
{
or { $ 31, so P has domain (3"> 31] (0> "). P is the composite of a root function and a rational function, so it is
continuous at every number in its domain by Theorems 7 and 9.
2
2
32. By Theorems 7 and 9, the composite function h3u is continuous on R. By part 1 of Theorem 4, 1 + h3u is continuous on R.
By Theorem 7, the inverse trigonometric function tan31 is continuous on its domain, R. By Theorem 9, the composite
2
function Q(u) = tan31 1 + h3u is continuous on its domain, R.
33. The function | =
1
is discontinuous at { = 0 because the
1 + h1@{
left- and right-hand limits at { = 0 are different.
34. The function | = tan2 { is discontinuous at { =
2
+ n, where n is
2 any integer. The function | = ln tan { is also discontinuous
where tan2 { is 0, that is, at { = n. So | = ln tan2 { is
discontinuous at { =
2 q,
q any integer.
35. Because we are dealing with root functions, 5 +
I
I
{ is continuous on [0> "), { + 5 is continuous on [35> "), so the
I
5+ {
is continuous on [0> "). Since i is continuous at { = 4, lim i ({) = i (4) = 73 .
quotient i ({) = I
{<4
5+{
36. Because { is continuous on R, sin { is continuous on R, and { + sin { is continuous on R, the composite function
i ({) = sin({ + sin {) is continuous on R, so lim i ({) = i () = sin( + sin ) = sin = 0.
{<
37. Because {2 3 { is continuous on R, the composite function i ({) = h{
2 3{
is continuous on R, so
lim i ({) = i (1) = h1 3 1 = h0 = 1.
{<1
38. Because arctan is a continuous function, we can apply Theorem 8.
lim arctan
{<2
{2 3 4
3{2 3 6{
= arctan
({ + 2)({ 3 2)
{<2
3{({ 3 2)
lim
= arctan
{+2
{<2 3{
lim
= arctan 23 E 0=588
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NOT FOR SALE
CHAPTER 2
39. i ({) =
+
{2
I
{
LIMITS AND DERIVATIVES
if { ? 1
if { D 1
By Theorem 5, since i ({) equals the polynomial {2 on (3"> 1), i is continuous on (3"> 1). By Theorem 7, since i ({)
I
equals the root function { on (1> ")> i is continuous on (1> "). At { = 1, lim i({) = lim {2 = 1 and
{<1
{<1
I
I
lim i ({) = lim { = 1. Thus, lim i({) exists and equals 1. Also, i (1) = 1 = 1. Thus, i is continuous at { = 1.
{<1+
{<1
{<1+
We conclude that i is continuous on (3"> ").
40. i ({) =
+
sin {
if { ? @4
cos {
if { D @4
By Theorem 7, the trigonometric functions are continuous. Since i({) = sin { on (3"> @4) and i ({) = cos { on
I
lim sin { = sin 4 = 1@ 2 since the sine
(@4> "), i is continuous on (3"> @4) (@4> ")= lim i ({) =
{<(@4)
function is continuous at @4= Similarly,
at @4. Thus,
lim
{<(@4)
lim
{<(@4)+
i ({) =
lim
{<(@4)
{<(@4)+
I
cos { = 1@ 2 by continuity of the cosine function
I
i ({) exists and equals 1@ 2, which agrees with the value i (@4). Therefore, i is continuous at @4,
so i is continuous on (3"> ").
;
A
1 + {2
A
?
41. i ({) = 2 3 {
A
A
= ({ 3 2)2
if { $ 0
if 0 ? { $ 2
if { A 2
i is continuous on (3"> 0), (0> 2), and (2> ") since it is a polynomial on
each of these intervals. Now lim i ({) = lim (1 + {2 ) = 1 and lim i({) = lim (2 3 {) = 2> so i is
{<0
{<0
{<0+
{<0+
discontinuous at 0. Since i (0) = 1, i is continuous from the left at 0= Also, lim i ({) = lim (2 3 {) = 0>
{<2
{<2
lim i ({) = lim ({ 3 2)2 = 0, and i (2) = 0, so i is continuous at 2= The only number at which i is discontinuous is 0.
{<2+
{<2+
;
{+1
A
?
42. i ({) = 1@{
A
=I
{33
if { $ 1
if 1 ? { ? 3
if { D 3
i is continuous on (3"> 1), (1> 3), and (3> "), where it is a polynomial,
a rational function, and a composite of a root function with a polynomial,
respectively. Now lim i({) = lim ({ + 1) = 2 and lim i ({) = lim (1@{) = 1, so i is discontinuous at 1.
{<1
{<1
{<1+
{<1+
Since i (1) = 2, i is continuous from the left at 1. Also, lim i ({) = lim (1@{) = 1@3, and
{<3
{<3
I
lim i ({) = lim { 3 3 = 0 = i (3), so i is discontinuous at 3, but it is continuous from the right at 3=
{<3+
{<3+
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NOT FOR SALE
SECTION 2.5
;
{+2
A
A
?
43. i ({) = h{
A
A
=
23{
CONTINUITY
¤
111
if { ? 0
if 0 $ { $ 1
if { A 1
i is continuous on (3"> 0) and (1> ") since on each of these intervals
it is a polynomial; it is continuous on (0> 1) since it is an exponential.
Now lim i ({) = lim ({ + 2) = 2 and lim i ({) = lim h{ = 1, so i is discontinuous at 0. Since i (0) = 1, i is
{<0
{<0
{<0+
{<0+
continuous from the right at 0. Also lim i ({) = lim h{ = h and lim i ({) = lim (2 3 {) = 1, so i is discontinuous
{<1
{<1
{<1+
{<1+
at 1. Since i (1) = h, i is continuous from the left at 1.
44. By Theorem 5, each piece of I is continuous on its domain. We need to check for continuity at u = U.
lim I (u) = lim
u<U
u<U
JPu
JP
JP
JP
JP
JP
=
and lim I (u) = lim
=
, so lim I (u) =
. Since I (U) =
,
2
u<U
U3
U2
U2
U2
U2
u<U+
u<U+ u
I is continuous at U. Therefore, I is a continuous function of u.
45. i ({) =
+
f{2 + 2{
if { ? 2
3
{ 3 f{
if { D 2
i is continuous on (3"> 2) and (2> "). Now lim i ({) = lim
{<2
lim i ({) = lim
{<2+
{<2+
{<2
2
f{ + 2{ = 4f + 4 and
3
{ 3 f{ = 8 3 2f. So i is continuous C 4f + 4 = 8 3 2f C 6f = 4 C f = 23 . Thus, for i
to be continuous on (3"> "), f = 23 .
46. i ({) =
; 2
{ 34
A
A
A
? {32
if { ? 2
d{2 3 e{ + 3
A
A
A
=
2{ 3 d + e
At { = 2:
if 2 $ { ? 3
if { D 3
lim i ({) = lim
{<2
{<2
{2 3 4
({ + 2)({ 3 2)
= lim
= lim ({ + 2) = 2 + 2 = 4
{32
{32
{<2
{<2
lim i ({) = lim (d{2 3 e{ + 3) = 4d 3 2e + 3
{<2+
{<2+
We must have 4d 3 2e + 3 = 4, or 4d 2e = 1 (1).
At { = 3:
lim i ({) = lim (d{2 3 e{ + 3) = 9d 3 3e + 3
{<3
{<3
lim i ({) = lim (2{ 3 d + e) = 6 3 d + e
{<3+
{<3+
We must have 9d 3 3e + 3 = 6 3 d + e, or 10d 4e = 3 (2).
Now solve the system of equations by adding 32 times equation (1) to equation (2).
38d + 4e = 32
10d 3 4e =
2d
=
So d = 12 . Substituting
1
2
for d in (1) gives us 32e = 31, so e =
1
2
3
1
as well. Thus, for i to be continuous on (3"> "),
d = e = 12 .
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¤
NOT FOR SALE
CHAPTER 2
47. (a) i ({) =
LIMITS AND DERIVATIVES
({2 + 1)({2 3 1)
({2 + 1)({ + 1)({ 3 1)
{4 3 1
=
=
= ({2 + 1)({ + 1) [or {3 + {2 + { + 1]
{31
{31
{31
for { 6= 1. The discontinuity is removable and j({) = {3 + {2 + { + 1 agrees with i for { 6= 1 and is continuous on R.
(b) i ({) =
{3 3 {2 3 2{
{({2 3 { 3 2)
{({ 3 2)({ + 1)
=
=
= {({ + 1) [or {2 + {] for { 6= 2. The discontinuity
{32
{32
{32
is removable and j({) = {2 + { agrees with i for { 6= 2 and is continuous on R.
(c) lim i ({) = lim [[sin {]] = lim 0 = 0 and lim i ({) = lim [[sin {]] = lim (31) = 31, so lim i ({) does not
{< {< {< {< +
{< +
{< +
{<
exist. The discontinuity at { = is a jump discontinuity.
48.
i does not satisfy the conclusion of the
i does satisfy the conclusion of the
Intermediate Value Theorem.
Intermediate Value Theorem.
49. i ({) = {2 + 10 sin { is continuous on the interval [31> 32], i (31) E 957, and i(32) E 1030. Since 957 ? 1000 ? 1030,
there is a number c in (31> 32) such that i (f) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in
(332> 331) such that i (f) = 1000=
50. Suppose that i(3) ? 6. By the Intermediate Value Theorem applied to the continuous function i on the closed interval [2> 3],
the fact that i (2) = 8 A 6 and i(3) ? 6 implies that there is a number f in (2> 3) such that i (f) = 6. This contradicts the fact
that the only solutions of the equation i ({) = 6 are { = 1 and { = 4. Hence, our supposition that i (3) ? 6 was incorrect. It
follows that i(3) D 6. But i (3) 6= 6 because the only solutions of i ({) = 6 are { = 1 and { = 4. Therefore, i (3) A 6.
51. i ({) = {4 + { 3 3 is continuous on the interval [1> 2]> i (1) = 31, and i(2) = 15. Since 31 ? 0 ? 15, there is a number f
in (1> 2) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation {4 + { 3 3 = 0 in the
interval (1> 2)=
I
3
{ + { 3 1 is continuous on the interval [0> 1]> i (0) = 31, and i (1) = 1. Since 31 ? 0 ? 1, there is a number f in
I
3
(0> 1) such that i(f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation { + { 3 1 = 0, or
I
3
{ = 1 3 {, in the interval (0> 1)=
52. i ({) =
53. The equation h{ = 3 3 2{ is equivalent to the equation h{ + 2{ 3 3 = 0. i ({) = h{ + 2{ 3 3 is continuous on the interval
[0> 1], i (0) = 32, and i (1) = h 3 1 E 1=72. Since 32 ? 0 ? h 3 1, there is a number f in (0> 1) such that i (f) = 0 by the
Intermediate Value Theorem. Thus, there is a root of the equation h{ + 2{ 3 3 = 0, or h{ = 3 3 2{, in the interval (0> 1).
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NOT FOR SALE
SECTION 2.5
CONTINUITY
¤
113
54. The equation sin { = {2 3 { is equivalent to the equation sin { 3 {2 + { = 0. i ({) = sin { 3 {2 + { is continuous on the
interval [1> 2]> i (1) = sin 1 E 0=84, and i (2) = sin 2 3 2 E 31=09. Since sin 1 A 0 A sin 2 3 2, there is a number f in
(1> 2) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation sin { 3 {2 + { = 0, or
sin { = {2 3 {, in the interval (1> 2).
55. (a) i ({) = cos { 3 {3 is continuous on the interval [0> 1], i (0) = 1 A 0, and i (1) = cos 1 3 1 E 30=46 ? 0. Since
1 A 0 A 30=46, there is a number f in (0> 1) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root
of the equation cos { 3 {3 = 0, or cos { = {3 , in the interval (0> 1).
(b) i (0=86) E 0=016 A 0 and i (0=87) E 30=014 ? 0, so there is a root between 0=86 and 0=87, that is, in the interval
(0=86> 0=87).
56. (a) i ({) = ln { 3 3 + 2{ is continuous on the interval [1> 2], i (1) = 31 ? 0, and i (2) = ln 2 + 1 E 1=7 A 0. Since
31 ? 0 ? 1=7, there is a number f in (1> 2) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of
the equation ln { 3 3 + 2{ = 0, or ln { = 3 3 2{, in the interval (1> 2).
(b) i (1=34) E 30=03 ? 0 and i (1=35) E 0=0001 A 0, so there is a root between 1=34 and 1=35> that is, in the
interval (1=34> 1=35).
57. (a) Let i ({) = 100h3{@100 3 0=01{2 = Then i (0) = 100 A 0 and
i (100) = 100h31 3 100 E 363=2 ? 0. So by the Intermediate
Value Theorem, there is a number f in (0> 100) such that i (f) = 0.
This implies that 100h3f@100 = 0=01f2 .
(b) Using the intersect feature of the graphing device, we ¿nd that the
root of the equation is { = 70=347, correct to three decimal places.
58. (a) Let i ({) = arctan { + { 3 1. Then i (0) = 31 ? 0 and
i (1) =
4
A 0. So by the Intermediate Value Theorem, there is a
number f in (0> 1) such that i (f) = 0. This implies that
arctan f = 1 3 f.
(b) Using the intersect feature of the graphing device, we ¿nd that the
root of the equation is { = 0=520, correct to three decimal places.
59. (i) If i is continuous at d, then by Theorem 8 with j(k) = d + k, we have
lim i (d + k) = i lim (d + k) = i (d).
k<0
k<0
(U) Let % A 0. Since lim i(d + k) = i (d), there exists A 0 such that 0 ? |k| ? k<0
i
|i (d + k) 3 i (d)| ? %. So if 0 ? |{ 3 d| ? , then |i ({) 3 i (d)| = |i (d + ({ 3 d)) 3 i (d)| ? %.
Thus, lim i ({) = i (d) and so i is continuous at d.
{<d
60. lim sin(d + k) = lim (sin d cos k + cos d sin k) = lim (sin d cos k) + lim (cos d sin k)
k<0
k<0
=
k<0
k<0
lim sin d lim cos k + lim cos d lim sin k = (sin d)(1) + (cos d)(0) = sin d
k<0
k<0
k<0
k<0
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114
NOT FOR SALE
¤
CHAPTER 2
LIMITS AND DERIVATIVES
61. As in the previous exercise, we must show that lim cos(d + k) = cos d to prove that the cosine function is continuous.
k<0
lim cos(d + k) = lim (cos d cos k 3 sin d sin k) = lim (cos d cos k) 3 lim (sin d sin k)
k<0
k<0
=
k<0
k<0
lim cos d lim cos k 3 lim sin d lim sin k = (cos d)(1) 3 (sin d)(0) = cos d
k<0
k<0
k<0
k<0
62. (a) Since i is continuous at d, lim i({) = i (d). Thus, using the Constant Multiple Law of Limits, we have
{<d
lim (fi )({) = lim fi ({) = f lim i ({) = fi (d) = (fi )(d). Therefore, fi is continuous at d.
{<d
{<d
{<d
(b) Since i and j are continuous at d, lim i ({) = i (d) and lim j({) = j(d). Since j(d) 6= 0, we can use the Quotient Law
{<d
{<d
lim i({)
i({)
i
i(d)
i
i
{<d
({) = lim
=
=
=
(d). Thus, is continuous at d.
{<d
{<d j({)
j
lim j({)
j(d)
j
j
of Limits: lim
{<d
63. i ({) =
+
0 if { is rational
1 if { is irrational
is continuous nowhere. For, given any number d and any A 0, the interval (d 3 > d + )
contains both in¿nitely many rational and in¿nitely many irrational numbers. Since i (d) = 0 or 1, there are in¿nitely many
numbers { with 0 ? |{ 3 d| ? and |i ({) 3 i (d)| = 1. Thus, lim i({) 6= i (d). [In fact, lim i ({) does not even exist.]
{<d
64. j({) =
+
0 if { is rational
{ if { is irrational
{<d
is continuous at 0. To see why, note that 3 |{| $ j({) $ |{|, so by the Squeeze Theorem
lim j({) = 0 = j(0). But j is continuous nowhere else. For if d 6= 0 and A 0, the interval (d 3 > d + ) contains both
{<0
in¿nitely many rational and in¿nitely many irrational numbers. Since j(d) = 0 or d, there are in¿nitely many numbers { with
0 ? |{ 3 d| ? and |j({) 3 j(d)| A |d| @2. Thus, lim j({) 6= j(d).
{<d
65. If there is such a number, it satis¿es the equation {3 + 1 = {
C {3 3 { + 1 = 0. Let the left-hand side of this equation be
called i ({). Now i (32) = 35 ? 0, and i (31) = 1 A 0. Note also that i ({) is a polynomial, and thus continuous. So by the
Intermediate Value Theorem, there is a number f between 32 and 31 such that i(f) = 0, so that f = f3 + 1.
66.
d
e
+ 3
= 0 i d({3 + { 3 2) + e({3 + 2{2 3 1) = 0. Let s({) denote the left side of the last
{3 + 2{2 3 1
{ +{32
equation. Since s is continuous on [31> 1], s(31) = 34d ? 0, and s(1) = 2e A 0, there exists a f in (31> 1) such that
s(f) = 0 by the Intermediate Value Theorem. Note that the only root of either denominator that is in (31> 1) is
I
I
(31 + 5 )@2 = u, but s(u) = (3 5 3 9)d@2 6= 0. Thus, f is not a root of either denominator, so s(f) = 0 i
{ = f is a root of the given equation.
67. i ({) = {4 sin(1@{) is continuous on (3"> 0) (0> ") since it is the product of a polynomial and a composite of a
trigonometric function and a rational function. Now since 31 $ sin(1@{) $ 1, we have 3{4 $ {4 sin(1@{) $ {4 . Because
lim (3{4 ) = 0 and lim {4 = 0, the Squeeze Theorem gives us lim ({4 sin(1@{)) = 0, which equals i(0). Thus, i is
{<0
{<0
{<0
continuous at 0 and, hence, on (3"> ").
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NOT FOR SALE
SECTION 2.6
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
¤
115
68. (a) lim I ({) = 0 and lim I ({) = 0, so lim I ({) = 0, which is I (0), and hence I is continuous at { = d if d = 0. For
{<0
{<0+
{<0
d A 0, lim I ({) = lim { = d = I (d). For d ? 0, lim I ({) = lim (3{) = 3d = I (d). Thus, I is continuous at
{<d
{<d
{<d
{<d
{ = d; that is, continuous everywhere.
(b) Assume that i is continuous on the interval L. Then for d M L, lim |i ({)| = lim i ({) = |i (d)| by Theorem 8. (If d is
{<d
{<d
an endpoint of L, use the appropriate one-sided limit.) So |i | is continuous on L.
+
1 if { D 0
(c) No, the converse is false. For example, the function i({) =
is not continuous at { = 0, but |i({)| = 1 is
31 if { ? 0
continuous on R.
69. De¿ne x(w) to be the monk’s distance from the monastery, as a function of time w (in hours), on the ¿rst day, and de¿ne g(w)
to be his distance from the monastery, as a function of time, on the second day. Let G be the distance from the monastery to
the top of the mountain. From the given information we know that x(0) = 0, x(12) = G, g(0) = G and g(12) = 0. Now
consider the function x 3 g, which is clearly continuous. We calculate that (x 3 g)(0) = 3G and (x 3 g)(12) = G.
So by the Intermediate Value Theorem, there must be some time w0 between 0 and 12 such that (x 3 g)(w0 ) = 0 C
x(w0 ) = g(w0 ). So at time w0 after 7:00 AM, the monk will be at the same place on both days.
2.6 Limits at Infinity; Horizontal Asymptotes
1. (a) As { becomes large, the values of i ({) approach 5.
(b) As { becomes large negative, the values of i ({) approach 3.
2. (a) The graph of a function can intersect a
The graph of a function can intersect a horizontal asymptote.
vertical asymptote in the sense that it can
It can even intersect its horizontal asymptote an in¿nite
meet but not cross it.
number of times.
(b) The graph of a function can have 0, 1, or 2 horizontal asymptotes. Representative examples are shown.
No horizontal asymptote
One horizontal asymptote
Two horizontal asymptotes
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NOT FOR SALE
¤
CHAPTER 2
LIMITS AND DERIVATIVES
3. (a) lim i ({) = 32
{<"
(d) lim i ({) = 3"
{<3
4. (a) lim j({) = 2
{<"
(d) lim j({) = 3"
{<2
(b) lim i ({) = 2
(c) lim i ({) = "
{<3"
{<1
(e) Vertical: { = 1, { = 3; horizontal: | = 32, | = 2
(b) lim j({) = 31
(c) lim j({) = 3"
(e) lim j({) = "
(f ) Vertical: { = 0, { = 2;
{<3"
{<0
{<2+
horizontal: | = 31, | = 2
5. lim i ({) = 3",
{<0
lim i({) = 5,
{<3"
lim i ({) = 35
{<"
8. lim i ({) = 3,
{<"
6. lim i ({) = ",
{<2
lim i ({) = 3",
{<32
lim i ({) = 0,
{<"
9. i (0) = 3,
lim i ({) = ",
{<0+
lim i ({) = 3",
{<3"
{<2
{<2+
i is odd
lim i ({) = ",
{<32+
lim i({) = 0,
{<3"
i (0) = 0
lim i ({) = 4,
{<0
lim i({) = 3",
lim i ({) = ",
{<2
lim i ({) = 0,
{<3"
lim i ({) = ",
{<"
lim i ({) = ",
{<0+
lim i ({) = 3"
{<0
10. lim i ({) = 3",
{<3
lim i ({) = 2,
{<"
i (0) = 0, i is even
lim i ({) = 2,
{<4+
7. lim i ({) = 3",
lim i({) = 3",
{<4
lim i({) = 3
{<"
11. If i ({) = {2@2{ , then a calculator gives i(0) = 0, i (1) = 0=5, i(2) = 1, i (3) = 1=125, i (4) = 1, i (5) = 0=78125,
i (6) = 0=5625, i (7) = 0=3828125, i (8) = 0=25, i (9) = 0=158203125, i(10) = 0=09765625, i (20) E 0=00038147,
i (50) E 2=2204 × 10312 , i (100) E 7=8886 × 10327 .
It appears that lim {2@2{ = 0.
{<"
12. (a) From a graph of i({) = (1 3 2@{){ in a window of [0> 10,000] by [0> 0=2], we estimate that lim i ({) = 0=14
{<"
(to two decimal places.)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 2.6
{<"
i ({)
10,000
100,000
1,000,000
0=135308
0=135333
0=135335
3{2 3 { + 4
(3{2 3 { + 4)@{2
= lim
2
{<"
2{ + 5{ 3 8
(2{2 + 5{ 3 8)@{2
[divide both the numerator and denominator by {2
(the highest power of { that appears in the denominator)]
lim (3 3 1@{ + 4@{2 )
{<"
[Limit Law 5]
lim (2 + 5@{ 3 8@{2 )
{<"
=
lim 3 3 lim (1@{) + lim (4@{2 )
{<"
{<"
=
{<"
[Limit Laws 1 and 2]
{<"
3 3 lim (1@{) + 4 lim (1@{2 )
{<"
{<"
[Limit Laws 7 and 3]
2 + 5 lim (1@{) 3 8 lim (1@{2 )
{<"
u
{<"
lim 2 + lim (5@{) 3 lim (8@{2 )
{<"
{<"
117
{<"
{
=
14. lim
¤
From the table, we estimate that lim i ({) = 0=1353 (to four decimal places.)
(b)
13. lim
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
{<"
=
3 3 0 + 4(0)
2 + 5(0) 3 8(0)
=
3
2
12{3 3 5{ + 2
=
1 + 4{2 + 3{3
=
u
lim
{<"
v
[Theorem 5 of Section 2.5]
12{3 3 5{ + 2
1 + 4{2 + 3{3
[Limit Law 11]
12 3 5@{2 + 2@{3
{<" 1@{3 + 4@{ + 3
[divide by {3 ]
lim
y
x lim (12 3 5@{2 + 2@{3 )
x {<"
=w
lim (1@{3 + 4@{ + 3)
[Limit Law 5]
{<"
y
x lim 12 3 lim (5@{2 ) + lim (2@{3 )
x {<"
{<"
{<"
=w
lim (1@{3 ) + lim (4@{) + lim 3
{<"
{<"
y
x 12 3 5 lim (1@{2 ) + 2 lim (1@{3 )
x
{<"
{<"
=w
lim (1@{3 ) + 4 lim (1@{) + 3
{<"
=
v
12 3 5(0) + 2(0)
0 + 4(0) + 3
=
u
12 I
= 4=2
3
[Limit Laws 1 and 2]
{<"
[Limit Laws 7 and 3]
{<"
[Theorem 5 of Section 2.5]
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¤
15. lim
{<"
NOT FOR SALE
CHAPTER 2
LIMITS AND DERIVATIVES
lim 3 3 2 lim 1@{
(3{ 3 2)@{
3 3 2@{
3 3 2(0)
3{ 3 2
3
{<"
{<"
= lim
= lim
=
=
=
2{ + 1 {<" (2{ + 1)@{ {<" 2 + 1@{
lim 2 + lim 1@{
2+0
2
{<"
{<"
1 3 {2
(1 3 {2 )@{3
1@{3 3 1@{
= lim
= lim
3
3
{<" ({ 3 { + 1)@{
{<" 1 3 1@{2 + 1@{3
3{+1
16. lim
{<" {3
lim 1@{3 3 lim 1@{
{<"
=
{<"
lim 1 3 lim 1@{2 + lim 1@{3
{<"
{<"
{<"
=
030
=0
130+0
{32
({ 3 2)@{2
1@{ 3 2@{2
=
lim
17. lim
=
lim
=
{<3" {2 + 1
{<3" ({2 + 1)@{2
{<3" 1 + 1@{2
18.
lim
{<3"
19. lim
w<"
lim 1@{ 3 2 lim 1@{2
{<3"
{<3"
lim 1 + lim 1@{2
{<3"
=
{<3"
0 3 2(0)
=0
1+0
4{3 + 6{2 3 2
(4{3 + 6{2 3 2)@{3
4 + 6@{ 3 2@{3
4+030
= lim
=2
= lim
=
3
3
3
{<3" (2{ 3 4{ + 5)@{
{<3" 2 3 4@{2 + 5@{3
2{ 3 4{ + 5
230+0
I
I
( w + w2 )@w2
1@w3@2 + 1
w + w2
0+1
=
lim
= lim
=
= 31
w<" (2w 3 w2 )@w2
w<" 2@w 3 1
2w 3 w2
031
I
I
w 3 w w @w3@2
w3w w
1@w1@2 3 1
1
031
=3
20. lim
=
lim
=
=
lim
w<" 2w3@2 + 3w 3 5
w<" (2w3@2 + 3w 3 5) @w3@2
w<" 2 + 3@w1@2 3 5@w3@2
2+030
2
(2{2 + 1)2
(2{2 + 1)2 @{4
[(2{2 + 1)@{2 ]2
= lim
= lim
2
2
2
2
4
2
{<" ({ 3 1) ({ + {)
{<" [({ 3 1) ({ + {)]@{
{<" [({ 3 2{ + 1)@{2 ][({2 + {)@{2 ]
21. lim
(2 + 1@{2 )2
(2 + 0)2
=
=4
{<" (1 3 2@{ + 1@{2 )(1 + 1@{)
(1 3 0 + 0)(1 + 0)
= lim
{2
{2 @{2
1
= lim I
= lim s
4
{<"
{<"
{ +1
{4 + 1@{2
({4 + 1)@{4
22. lim I
{<"
[since {2 =
I
{4 for { A 0]
1
1
= lim s
=1
= I
{<"
1+0
1 + 1@{4
s
I
I
(9{6 3 {)@{6
lim
9{6 3 {
9{6 3 { @{3
{<"
23. lim
=
=
lim
{<"
{<" ({3 + 1)@{3
{3 + 1
lim (1 + 1@{3 )
[since {3 =
I
{6 for { A 0]
{<"
s
9 3 1@{5
lim
{<"
=
lim 1 + lim (1@{3 )
{<"
{<"
=
t
I
I
9{6 3 {
9{6 3 { @{3
=
lim
24. lim
=
{<3"
{<3" ({3 + 1)@{3
{3 + 1
=
lim 9 3 lim (1@{5 )
{<"
{<"
1+0
=
I
930 =3
s
lim 3 (9{6 3 {)@{6
I
[since {3 = 3 {6 for { ? 0]
lim (1 + 1@{3 )
{<3"
t
s
5
3
lim 9 3 lim (1@{5 )
lim 3 9 3 1@{
I
{<3"
{<3"
{<3"
=
= 3 9 3 0 = 33
lim 1 + lim (1@{3 )
1+0
{<3"
{<3"
{<3"
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LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
¤
119
I
2
I
I
I
9{2 + { 3 3{
9{2 + { + 3{
9{2 + { 3 (3{)2
2
I
I
= lim
25. lim
9{ + { 3 3{ = lim
{<"
{<"
{<"
9{2 + { + 3{
9{2 + { + 3{
2
9{ + { 3 9{2
{
1@{
= lim I
·
= lim I
{<"
{<"
9{2 + { + 3{
9{2 + { + 3{ 1@{
{@{
1
1
1
1
= lim s
= lim s
= I
=
=
2
2
2
{<"
{<"
3
+
3
6
9+3
9{ @{ + {@{ + 3{@{
9 + 1@{ + 3
I
I
I
{ 3 {2 + 2{
{2 3 ({2 + 2{)
2
2
I
I
26. lim { + { + 2{ = lim { + { + 2{
= lim
{<3"
{<3"
{<3" { 3
{ 3 {2 + 2{
{2 + 2{
32{
32
32
s
s
I
=
= 31
= lim
{2 + 2{ {<3" 1 + 1 + 2@{
1 + 1 + 2 (0)
I
Note: In dividing numerator and denominator by {, we used the fact that for { ? 0, { = 3 {2 .
= lim
{<3"
{3
I
I
27. lim
{2 + d{ 3 {2 + e{ = lim
{<"
{<"
I
I
I
I
{2 + d{ 3 {2 + e{
{2 + d{ + {2 + e{
I
I
{2 + d{ + {2 + e{
({2 + d{) 3 ({2 + e{)
[(d 3 e){]@{
I
= lim I
= lim I
I
I
{<"
{2 + d{ + {2 + e{ {<"
{2 + d{ + {2 + e{ @ {2
28. For { A 0,
29. lim
{<"
d3e
d3e
d3e
s
I
=
= lim s
= I
{<"
2
1
+
0
+
1
+
0
1 + d@{ + 1 + e@{
I
I
I
I
{2 + 1 A {2 = {. So as { < ", we have {2 + 1 < ", that is, lim {2 + 1 = ".
{<"
({4 3 3{2 + {)@{3
{4 3 3{2 + {
=
lim
{<" ({3 3 { + 2)@{3
{3 3 { + 2
divide by the highest power
of { in the denominator
{ 3 3@{ + 1@{2
="
{<" 1 3 1@{2 + 2@{3
= lim
since the numerator increases without bound and the denominator approaches 1 as { < ".
30. lim (h3{ + 2 cos 3{) does not exist. lim h3{ = 0, but lim (2 cos 3{) does not exist because the values of 2 cos 3{
{<"
{<"
{<"
oscillate between the values of 32 and 2 in¿nitely often, so the given limit does not exist.
31.
lim ({4 + {5 ) = lim {5 ( {1 + 1) [factor out the largest power of {] = 3" because {5 < 3" and 1@{ + 1 < 1
{<3"
{<3"
as { < 3".
Or:
32.
lim
{<3"
lim
{<3"
4
{ + {5 = lim {4 (1 + {) = 3".
{<3"
1 + {6
(1 + {6 )@{4
=
lim
{4 + 1 {<3" ({4 + 1)@{4
divide by the highest power
of { in the denominator
= lim
{<3"
1@{4 + {2
="
1 + 1@{4
since the numerator increases without bound and the denominator approaches 1 as { < 3".
33. Let w = h{ . As { < ", w < ". lim arctan(h{ ) = lim arctan w =
{<"
w<"
34. Divide numerator and denominator by h3{ : lim
{<"
2
by (3).
h3{ 3 h33{
1 3 h36{
130
=1
= lim
=
3{
33{
{<" 1 + h36{
h +h
1+0
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CHAPTER 2
35. lim
{<"
LIMITS AND DERIVATIVES
1 3 h{
(1 3 h{ )@h{
1@h{ 3 1
031
1
= lim
= lim
=
=3
{<" (1 + 2h{ )@h{
{<" 1@h{ + 2
1 + 2h{
0+2
2
36. Since 0 $ sin2 { $ 1, we have 0 $
1
1
sin2 {
$ 2
. We know that lim 0 = 0 and lim 2
= 0, so by the Squeeze
{<"
{<" { + 1
{2 + 1
{ +1
sin2 {
= 0.
{<" {2 + 1
Theorem, lim
37. Since 31 $ cos { $ 1 and h32{ A 0, we have 3h32{ $ h32{ cos { $ h32{ . We know that lim (3h32{ ) = 0 and
{<"
lim h32{ = 0, so by the Squeeze Theorem, lim (h32{ cos {) = 0.
{<"
{<"
38. Let w = ln {. As { < 0+ , w < 3". lim tan31 (ln {) = lim tan31 w = 3 2 by (4).
{<0+
w<3"
39. (a)
(b)
{
i({)
310,000
30=4999625
31,000,000
30=4999996
3100,000
From the graph of i ({) =
I
{2 + { + 1 + {, we
30=4999962
From the table, we estimate the limit to be 30=5.
estimate the value of lim i ({) to be 30=5.
{<3"
2
I 2
I
I
{ + { + 1 3 {2
{ +{+13{
{2 + { + 1 + { = lim
{2 + { + 1 + { I
= lim I
{<3"
{<3"
{<3"
{2 + { + 1 3 {
{2 + { + 1 3 {
(c) lim
({ + 1)(1@{)
1 + (1@{)
s
= lim I
= lim
{<3"
{2 + { + 1 3 { (1@{) {<3" 3 1 + (1@{) + (1@{2 ) 3 1
=
Note that for { ? 0, we have
40. (a)
1
1+0
I
=3
2
3 1+0+031
I
{2 = |{| = 3{, so when we divide the radical by {, with { ? 0, we get
s
1 I 2
1I 2
{ + { + 1 = 3I
{ + { + 1 = 3 1 + (1@{) + (1@{2 ).
2
{
{
(b)
{
i({)
10,000
1=44339
100,000
1=44338
1,000,000
1=44338
From the table, we estimate (to four decimal
From the graph of
I
I
i ({) = 3{2 + 8{ + 6 3 3{2 + 3{ + 1, we estimate
places) the limit to be 1=4434.
(to one decimal place) the value of lim i ({) to be 1=4.
{<"
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LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
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121
I
I
I
I
3{2 + 8{ + 6 3 3{2 + 3{ + 1
3{2 + 8{ + 6 + 3{2 + 3{ + 1
I
I
(c) lim i ({) = lim
{<"
{<"
3{2 + 8{ + 6 + 3{2 + 3{ + 1
2
2
3{ + 8{ + 6 3 3{ + 3{ + 1
(5{ + 5)(1@{)
I
I
= lim I
= lim I
{<"
{<"
3{2 + 8{ + 6 + 3{2 + 3{ + 1
3{2 + 8{ + 6 + 3{2 + 3{ + 1 (1@{)
I
5
5 + 5@{
5
5 3
I = I =
s
= lim s
= I
E 1=443376
{<"
6
3+ 3
2 3
3 + 8@{ + 6@{2 + 3 + 3@{ + 1@{2
2{ + 1
2+
2{ + 1
{
= lim
= lim
41. lim
{<" { 3 2
{<" { 3 2
{<"
13
{
=
1
lim 2 +
{ = {<" 2
lim 1 3
{
{<"
1
lim 2 + lim
{
{<"
= {<"
2
lim 1 3 lim
{<"
{<"
{
1
{
2
{
2+0
= 2, so | = 2 is a horizontal asymptote.
130
The denominator { 3 2 is zero when { = 2 and the numerator is not zero, so we
investigate | = i ({) =
2{ + 1
as { approaches 2.
{32
lim i ({) = 3" because as
{<2
{ < 23 the numerator is positive and the denominator approaches 0 through
negative values. Similarly, lim i ({) = ". Thus, { = 2 is a vertical asymptote.
{<2+
The graph con¿rms our work.
1
{2 + 1
1
lim 1 + 2
1
+
{<"
{
{ +1
{2
{2
= lim
42. lim
= lim
=
2
3
{<" 2{2 3 3{ 3 2
{<" 2{2 3 3{ 3 2
{<"
3
2
23 3 2
lim
2
3
3
{
{
{<"
{ {2
{2
2
lim 1 + lim
=
| = i ({) =
{<"
{<"
1
{2
3
2
3 lim 2
lim 2 3 lim
{<"
{<" {
{<" {
=
1
1+0
= , so | =
23030
2
1
2
is a horizontal asymptote.
{2 + 1
{2 + 1
=
, so
lim
i ({) = "
2{2 3 3{ 3 2
(2{ + 1)({ 3 2)
{<(31@2)
because as { < (31@2)3 the numerator is positive while the denominator
approaches 0 through positive values. Similarly,
lim
{<(31@2)+
i({) = 3",
lim i ({) = 3", and lim i({) = ". Thus, { = 3 12 and { = 2 are vertical
{<2
{<2+
asymptotes. The graph con¿rms our work.
1
1
2{2 + { 3 1
1
1
lim
2
+
3
2+ 3 2
{
{2
2{2 + { 3 1
{2
{
{ = {<" = lim
43. lim
= lim
2
1
{<" {2 + { 3 2
{<" {2 + { 3 2
{<"
1
2
1+ 3 2
lim
1
+
3
{
{
{<"
{
{2
{2
1
1
lim 2 + lim
3 lim 2
2+030
{<"
{<" {
{<" {
= 2, so | = 2 is a horizontal asymptote.
=
=
1
1
1 + 0 3 2(0)
lim 1 + lim
3 2 lim 2
{<"
{<" {
{<" {
[continued]
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CHAPTER 2
| = i ({) =
LIMITS AND DERIVATIVES
2{2 + { 3 1
(2{ 3 1)({ + 1)
=
, so lim i({) = ",
{2 + { 3 2
({ + 2)({ 3 1)
{<32
lim i ({) = 3", lim i ({) = 3", and lim i ({) = ". Thus, { = 32
{<1
{<32+
{<1+
and { = 1 are vertical asymptotes. The graph con¿rms our work.
1 + {4
4
1 + {4
44. lim 2
= lim 2 { 4 = lim
{<" { 3 {4
{<" { 3 {
{<"
{4
=
| = i ({) =
1
1
1
lim
+1
+ lim 1
lim
+1
{<"
{4
{<" {4
{<"
{4
=
=
1
1
1
31
lim
3 lim 1
lim
31
{<" {2
{<"
{2
{<"
{2
0+1
= 31, so | = 31 is a horizontal asymptote.
031
1 + {4
1 + {4
1 + {4
=
. The denominator is
=
{2 3 {4
{2 (1 3 {2 )
{2 (1 + {)(1 3 {)
zero when { = 0, 31, and 1, but the numerator is nonzero, so { = 0, { = 31, and
{ = 1 are vertical asymptotes. Notice that as { < 0, the numerator and
denominator are both positive, so lim i ({) = ". The graph con¿rms our work.
{<0
45. | = i ({) =
{({2 3 1)
{({ + 1)({ 3 1)
{({ + 1)
{3 3 {
=
=
=
= j({) for { 6= 1.
{2 3 6{ + 5
({ 3 1)({ 3 5)
({ 3 1)({ 3 5)
{35
The graph of j is the same as the graph of i with the exception of a hole in the
graph of i at { = 1. By long division, j({) =
30
{2 + {
={+6+
.
{35
{35
As { < ±", j({) < ±", so there is no horizontal asymptote. The denominator
of j is zero when { = 5. lim j({) = 3" and lim j({) = ", so { = 5 is a
{<5
{<5+
vertical asymptote. The graph con¿rms our work.
46. lim
{<"
2h{
2h{
2
1@h{
2
= lim {
·
=
= 2, so | = 2 is a horizontal asymptote.
= lim
{<"
{<" 1 3 (5@h{ )
35
h 3 5 1@h{
130
h{
2h{
2(0)
=
= 0, so | = 0 is a horizontal asymptote. The denominator is zero (and the numerator isn’t)
35
035
when h 3 5 = 0 i h{ = 5 i { = ln 5.
lim
{<3" h{
{
2h{
= " since the numerator approaches 10 and the denominator
35
lim
{<(ln 5)+ h{
approaches 0 through positive values as { < (ln 5)+ . Similarly,
lim
{<(ln 5)
2h{
= 3". Thus, { = ln 5 is a vertical asymptote. The graph
35
h{
con¿rms our work.
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LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
¤
123
47. From the graph, it appears | = 1 is a horizontal asymptote.
3{3 + 500{2
3{ + 500{
3 + (500@{)
{3
lim
= lim 3
= lim
{<" {3 + 500{2 + 100{ + 2000
{<" { + 500{2 + 100{ + 2000
{<" 1 + (500@{) + (100@{2 ) + (2000@{3 )
{3
3+0
=
= 3, so | = 3 is a horizontal asymptote.
1+0+0+0
3
2
The discrepancy can be explained by the choice of the viewing window. Try
[3100,000> 100,000] by [31> 4] to get a graph that lends credibility to our
calculation that | = 3 is a horizontal asymptote.
48. (a)
From the graph, it appears at ¿rst that there is only one horizontal asymptote, at | E 0> and a vertical asymptote at
{ E 1=7. However, if we graph the function with a wider and shorter viewing rectangle, we see that in fact there seem to be
two horizontal asymptotes: one at | E 0=5 and one at | E 30=5. So we estimate that
I
2{2 + 1
E 30=5
{<"
{<3" 3{ 3 5
I
2{2 + 1
(b) i (1000) E 0=4722 and i(10,000) E 0=4715, so we estimate that lim
E 0=47.
{<" 3{ 3 5
I
2{2 + 1
i (31000) E 30=4706 and i(310,000) E 30=4713, so we estimate that lim
E 30=47.
{<3" 3{ 3 5
s
I
I
I
2 + 1@{2
2{2 + 1
2
2
= lim
[since { = { for { A 0] =
E 0=471404.
(c) lim
{<" 3{ 3 5
{<"
3 3 5@{
3
I
For { ? 0, we have {2 = |{| = 3{, so when we divide the numerator by {, with { ? 0, we
lim
I
2{2 + 1
E 0=5
3{ 3 5
and
lim
s
1 I 2
1I 2
2{ + 1 = 3 I
2{ + 1 = 3 2 + 1@{2 . Therefore,
{
{2
s
I
I
3 2 + 1@{2
2{2 + 1
2
= lim
=3
E 30=471404.
lim
{<3" 3{ 3 5
{<3"
3 3 5@{
3
get
49. Let’s look for a rational function.
(1)
lim i({) = 0 i degree of numerator ? degree of denominator
{<±"
(2) lim i ({) = 3" i there is a factor of {2 in the denominator (not just {, since that would produce a sign
{<0
change at { = 0), and the function is negative near { = 0.
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CHAPTER 2
LIMITS AND DERIVATIVES
(3) lim i ({) = " and lim i({) = 3" i vertical asymptote at { = 3; there is a factor of ({ 3 3) in the
{<3
{<3+
denominator.
(4) i (2) = 0 i 2 is an {-intercept; there is at least one factor of ({ 3 2) in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us
i ({) =
23{
as one possibility.
{2 ({ 3 3)
50. Since the function has vertical asymptotes { = 1 and { = 3, the denominator of the rational function we are looking for must
have factors ({ 3 1) and ({ 3 3). Because the horizontal asymptote is | = 1, the degree of the numerator must equal the
degree of the denominator, and the ratio of the leading coef¿cients must be 1. One possibility is i ({) =
{2
.
({ 3 1)({ 3 3)
51. (a) We must ¿rst ¿nd the function i . Since i has a vertical asymptote { = 4 and {-intercept { = 1, { 3 4 is a factor of the
denominator and { 3 1 is a factor of the numerator. There is a removable discontinuity at { = 31, so { 3 (31) = { + 1 is
a factor of both the numerator and denominator. Thus, i now looks like this: i ({) =
be determined. Then lim i ({) = lim
{<31
d = 5. Thus i ({) =
i (0) =
d({ 3 1)({ + 1)
d({ 3 1)
d(31 3 1)
2
2
= lim
=
= d, so d = 2, and
{<31 { 3 4
({ 3 4)({ + 1)
(31 3 4)
5
5
5({ 3 1)({ + 1)
is a ratio of quadratic functions satisfying all the given conditions and
({ 3 4)({ + 1)
5(31)(1)
5
= .
(34)(1)
4
(b) lim i ({) = 5 lim
{<"
{<31
d({ 3 1)({ + 1)
, where d is still to
({ 3 4)({ + 1)
{<"
{2
{2 3 1
({2 @{2 ) 3 (1@{2 )
130
= 5 lim
=5
= 5(1) = 5
2
{<"
3 3{ 3 4
({ @{2 ) 3 (3{@{2 ) 3 (4@{2 )
13030
52. | = i ({) = 2{3 3 {4 = {3 (2 3 {).
The |-intercept is i(0) = 0. The
{-intercepts are 0 and 2. There are sign changes at 0 and 2 (odd exponents on {
and 2 3 {). As { < ", i ({) < 3" because {3 < " and 2 3 { < 3". As
{ < 3", i ({) < 3" because {3 < 3" and 2 3 { < ". Note that the graph
of i near { = 0 Àattens out (looks like | = {3 ).
53. | = i ({) = {4 3 {6 = {4 (1 3 {2 ) = {4 (1 + {)(1 3 {).
The |-intercept is
i (0) = 0. The {-intercepts are 0, 31, and 1 [found by solving i ({) = 0 for {].
Since {4 A 0 for { 6= 0, i doesn’t change sign at { = 0. The function does change
sign at { = 31 and { = 1. As { < ±", i ({) = {4 (1 3 {2 ) approaches 3"
because {4 < " and (1 3 {2 ) < 3".
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SECTION 2.6
54. | = i ({) = {3 ({ + 2)2 ({ 3 1).
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
¤
125
The |-intercept is i (0) = 0. The {-intercepts
are 0, 32, and 1. There are sign changes at 0 and 1 (odd exponents on { and
{ 3 1). There is no sign change at 32. Also, i ({) < " as { < " because all
three factors are large. And i ({) < " as { < 3" because {3 < 3",
({ + 2)2 < ", and ({ 3 1) < 3". Note that the graph of i at { = 0 Àattens out
(looks like | = 3{3 ).
55. | = i ({) = (3 3 {)(1 + {)2 (1 3 {)4 .
The |-intercept is i (0) = 3(1)2 (1)4 = 3.
The {-intercepts are 3, 31, and 1. There is a sign change at 3, but not at 31 and 1.
When { is large positive, 3 3 { is negative and the other factors are positive, so
lim i ({) = 3". When { is large negative, 3 3 { is positive, so
{<"
lim i({) = ".
{<3"
56. | = i ({) = {2 ({2 3 1)2 ({ + 2) = {2 ({ + 1)2 ({ 3 1)2 ({ + 2).
The
|-intercept is i (0) = 0. The {-intercepts are 0, 31, 1> and 32. There is a sign
change at 32, but not at 0, 31, and 1. When { is large positive, all the factors are
positive, so lim i ({) = ". When { is large negative, only { + 2 is negative, so
{<"
lim i({) = 3".
{<3"
sin {
1
1
$
$ for { A 0. As { < ", 31@{ < 0 and 1@{ < 0, so by the Squeeze
{
{
{
sin {
Theorem, (sin {)@{ < 0. Thus, lim
= 0.
{<"
{
57. (a) Since 31 $ sin { $ 1 for all {> 3
(b) From part (a), the horizontal asymptote is | = 0. The function
| = (sin {)@{ crosses the horizontal asymptote whenever sin { = 0;
that is, at { = q for every integer q. Thus, the graph crosses the
asymptote an in¿nite number of times.
58. (a) In both viewing rectangles,
lim S ({) = lim T({) = " and
{<"
{<"
lim S ({) = lim T({) = 3".
{<3"
{<3"
In the larger viewing rectangle, S and T
become less distinguishable.
S ({)
3{5 3 5{3 + 2{
5 1
2 1
1
3
= 1 3 53 (0) + 23 (0) = 1 i
= lim
·
·
=
lim
+
{<" T({)
{<"
{<"
3{5
3 {2
3 {4
(b) lim
S and T have the same end behavior.
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126
NOT FOR SALE
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CHAPTER 2
LIMITS AND DERIVATIVES
59. (a) Divide the numerator and the denominator by the highest power of { in T({).
(a) If deg S ? deg T, then the numerator < 0 but the denominator doesn’t. So lim [S ({)@T({)] = 0.
{<"
(b) If deg S A deg T, then the numerator < ±" but the denominator doesn’t, so lim [S ({)@T({)] = ±"
{<"
(depending on the ratio of the leading coef¿cients of S and T).
60.
(i) q = 0
(ii) q A 0 (q odd)
(iii) q A 0 (q even)
From these sketches we see that
;
1 if q = 0
A
?
q
(a) lim { = 0 if q A 0
A
{<0+
=
" if q ? 0
(b) lim {q =
{<"
(v) q ? 0 (q even)
1 if q = 0
0 if q A 0
if q ? 0, q odd
if q ? 0, q even
if q = 0
if q A 0, q odd
if q A 0, q even
if q ? 0
I
I
5 {
5
5
1@ {
= I
· I = lim s
= 5 and
{<"
1@
{
{31
130
1 3 (1@{)
61. lim I
lim
;
A
A
A
A
?
A
3"
A
A
A
=
"
;
1
A
A
A
A
?3"
(d) lim {q =
{<3"
A
"
A
A
A
=
0
{<0
;
1 if q = 0
A
?
q
(c) lim { = " if q A 0
{<"
A
=
0 if q ? 0
{<"
(iv) q ? 0 (q odd)
I
10h{ 3 21 1@h{
10 3 (21@h{ )
10 3 0
10h{ 3 21
5 {
=
= 5. Since
,
·
= lim
? i ({) ? I
{
{
{
{<"
2h
1@h
2
2
2h
{31
we have lim i({) = 5 by the Squeeze Theorem.
{<"
62. (a) After w minutes, 25w liters of brine with 30 g of salt per liter has been pumped into the tank, so it contains
(5000 + 25w) liters of water and 25w · 30 = 750w grams of salt. Therefore, the salt concentration at time w will be
F(w) =
750w
30w g
=
.
5000 + 25w
200 + w L
(b) lim F(w) = lim
w<"
w<"
30w
30w@w
30
= lim
=
= 30. So the salt concentration approaches that of the brine
200 + w w<" 200@w + w@w
0+1
being pumped into the tank.
= y W (1 3 0) = y W
63. (a) lim y(w) = lim y W 1 3 h3jw@y
w<"
w<"
(b) We graph y(w) = 1 3 h39=8w and y(w) = 0=99y W , or in this case,
y(w) = 0=99. Using an intersect feature or zooming in on the point of
intersection, we ¿nd that w E 0=47 s.
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SECTION 2.6
LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES
¤
127
64. (a) | = h3{@10 and | = 0=1 intersect at {1 E 23=03.
If { A {1 , then h3{@10 ? 0=1.
(b) h3{@10 ? 0=1 i 3{@10 ? ln 0=1 i
1
= 310 ln 1031 = 10 ln 10 E 23=03
{ A 310 ln 10
65. Let j({) =
3{2 + 1
and i ({) = |j({) 3 1=5|. Note that
2{2 + { + 1
lim j({) =
{<"
3
2
and lim i({) = 0. We are interested in ¿nding the
{<"
{-value at which i ({) ? 0=05. From the graph, we ¿nd that { E 14=804,
so we choose Q = 15 (or any larger number).
I
4{2 + 1
66. For % = 0=5, we must ¿nd Q such that whenever { D Q, we have {+1
I
4{2 + 1
3 2 ? 0=5 C 1=5 ?
? 2=5.
{+1
We graph the three parts of this inequality on the same screen, and ¿nd that it holds whenever { A 2=82. So we choose Q = 3
I
4{2 + 1
? 2=1, and the graphs show that this holds whenever
(or any larger number). For % = 0=1, we must have 1=9 ?
{+1
{ A 18=9. So we choose Q = 19 (or any larger number).
I
4{2 + 1
67. For % = 0=5, we need to ¿nd Q such that 3 (32) ? 0=5 C
{+1
I
4{2 + 1
32=5 ?
? 31=5 whenever { $ Q. We graph the three parts of this
{+1
inequality on the same screen, and see that the inequality holds for { $ 36= So we
choose Q = 36 (or any smaller number).
I
4{2 + 1
? 31=9 whenever { $ Q= From the
For % = 0=1, we need 32=1 ?
{+1
graph, it seems that this inequality holds for { $ 322. So we choose Q = 322
(or any smaller number).
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CHAPTER 2
LIMITS AND DERIVATIVES
2{ + 1
A 100 whenever { D Q= From the graph, we
{+1
68. We need Q such that I
see that this inequality holds for { D 2500. So we choose Q = 2500 (or any
larger number).
69. (a) 1@{2 ? 0=0001
C {2 A 1@0=0001 = 10 000 C { A 100 ({ A 0)
I
I
(b) If % A 0 is given, then 1@{2 ? % C {2 A 1@% C { A 1@ %. Let Q = 1@ %.
1
1
1
1
Then { A Q i { A I
i 2 3 0 = 2 ? %, so lim 2 = 0.
{<" {
{
{
%
I
I
70. (a) 1@ { ? 0=0001 C
{ A 1@0=0001 = 104 C { A 108
I
I
(b) If % A 0 is given, then 1@ { ? % C
{ A 1@% C { A 1@%2 . Let Q = 1@%2 .
1
1
1
1
Then { A Q i { A 2 i I 3 0 = I ? %, so lim I = 0.
{<"
%
{
{
{
71. For { ? 0, |1@{ 3 0| = 31@{. If % A 0 is given, then 31@{ ? %
Take Q = 31@%. Then { ? Q
i { ? 31@% i |(1@{) 3 0| = 31@{ ? %, so lim (1@{) = 0.
{<3"
72. Given P A 0, we need Q A 0 such that { A Q
{AQ =
I
3
P
C { ? 31@%.
i {3 A P, so lim {3 = ".
I
I
3
P, so take Q = 3 P . Then
i {3 A P. Now {3 A P
C {A
i h{ A P. Now h{ A P
C { A ln P , so take
{<"
73. Given P A 0, we need Q A 0 such that { A Q
Q = max(1> ln P ). (This ensures that Q A 0.) Then { A Q = max(1> ln P) i h{ A max(h> P) D P ,
so lim h{ = ".
{<"
74. De¿nition
Let i be a function de¿ned on some interval (3"> d). Then lim i({) = 3" means that for every negative
{<3"
number P there is a corresponding negative number Q such that i ({) ? P whenever { ? Q. Now we use the de¿nition to
prove that lim 1 + {3 = 3". Given a negative number P , we need a negative number Q such that { ? Q i
{<3"
3
I
I
C {3 ? P 3 1 C { ? 3 P 3 1. Thus, we take Q = 3 P 3 1 and ¿nd that
i 1 + {3 ? P. This proves that lim 1 + {3 = 3".
1 + { ? P. Now 1 + {3 ? P
{?Q
{<3"
75. Suppose that lim i ({) = O. Then for every % A 0 there is a corresponding positive number Q such that |i ({) 3 O| ? %
{<"
whenever { A Q. If w = 1@{, then { A Q
C 0 ? 1@{ ? 1@Q
C
0 ? w ? 1@Q. Thus, for every % A 0 there is a
corresponding A 0 (namely 1@Q) such that |i(1@w) 3 O| ? % whenever 0 ? w ? . This proves that
lim i (1@w) = O = lim i({).
w<0+
{<"
Now suppose that lim i ({) = O. Then for every % A 0 there is a corresponding negative number Q such that
{<3"
|i ({) 3 O| ? % whenever { ? Q. If w = 1@{, then { ? Q
C 1@Q ? 1@{ ? 0 C 1@Q ? w ? 0. Thus, for every
% A 0 there is a corresponding A 0 (namely 31@Q) such that |i (1@w) 3 O| ? % whenever 3 ? w ? 0. This proves that
lim i (1@w) = O = lim i ({).
w<0
{<3"
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SECTION 2.7
DERIVATIVES AND RATES OF CHANGE
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129
2.7 Derivatives and Rates of Change
1. (a) This is just the slope of the line through two points: pS T =
i ({) 3 i (3)
{|
=
.
{{
{33
(b) This is the limit of the slope of the secant line S T as T approaches S : p = lim
{<3
2. The curve looks more like a line as the viewing rectangle gets smaller.
i({) 3 i (3)
.
{33
3. (a) (i) Using De¿nition 1 with i ({) = 4{ 3 {2 and S (1> 3),
i ({) 3 i (d)
(4{ 3 {2 ) 3 3
3({2 3 4{ + 3)
3({ 3 1)({ 3 3)
= lim
= lim
= lim
{<d
{<1
{<1
{<1
{3d
{31
{31
{31
p = lim
= lim (3 3 {) = 3 3 1 = 2
{<1
(ii) Using Equation 2 with i ({) = 4{ 3 {2 and S (1> 3),
4(1 + k) 3 (1 + k)2 3 3
i (d + k) 3 i(d)
i (1 + k) 3 i (1)
= lim
= lim
k<0
k<0
k<0
k
k
k
p = lim
4 + 4k 3 1 3 2k 3 k2 3 3
3k2 + 2k
k(3k + 2)
= lim
= lim
= lim (3k + 2) = 2
k<0
k<0
k<0
k<0
k
k
k
= lim
(b) An equation of the tangent line is | 3 i(d) = i 0 (d)({ 3 d) i | 3 i (1) = i 0 (1)({ 3 1) i | 3 3 = 2({ 3 1),
or | = 2{ + 1.
The graph of | = 2{ + 1 is tangent to the graph of | = 4{ 3 {2 at the
(c)
point (1> 3). Now zoom in toward the point (1> 3) until the parabola and
the tangent line are indistiguishable.
4. (a) (i) Using De¿nition 1 with i ({) = { 3 {3 and S (1> 0),
p = lim
{<1
i ({) 3 0
{ 3 {3
{(1 3 {2 )
{(1 + {)(1 3 {)
= lim
= lim
= lim
{<1
{<1
{<1
{31
{31
{31
{31
= lim [3{(1 + {)] = 31(2) = 32
{<1
(ii) Using Equation 2 with i ({) = { 3 {3 and S (1> 0),
p = lim
k<0
= lim
k<0
(1 + k) 3 (1 + k)3 3 0
i(d + k) 3 i (d)
i (1 + k) 3 i (1)
= lim
= lim
k<0
k<0
k
k
k
1 + k 3 (1 + 3k + 3k2 + k3 )
3k3 3 3k2 3 2k
k(3k2 3 3k 3 2)
= lim
= lim
k<0
k<0
k
k
k
= lim (3k2 3 3k 3 2) = 32
k<0
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LIMITS AND DERIVATIVES
(b) An equation of the tangent line is | 3 i (d) = i 0 (d)({ 3 d) i | 3 i (1) = i 0 (1)({ 3 1) i | 3 0 = 32({ 3 1),
or | = 32{ + 2.
The graph of | = 32{ + 2 is tangent to the graph of | = { 3 {3 at the
(c)
point (1> 0). Now zoom in toward the point (1> 0) until the cubic and the
tangent line are indistinguishable.
5. Using (1) with i ({) = 4{ 3 3{2 and S (2> 34) [we could also use (2)],
4{ 3 3{2 3 (34)
i ({) 3 i (d)
33{2 + 4{ + 4
p = lim
= lim
= lim
{<d
{<2
{<2
{3d
{32
{32
= lim
{<2
(33{ 3 2)({ 3 2)
= lim (33{ 3 2) = 33(2) 3 2 = 38
{<2
{32
Tangent line: | 3 (34) = 38({ 3 2) C | + 4 = 38{ + 16 C | = 38{ + 12.
6. Using (2) with i ({) = {3 3 3{ + 1 and S (2> 3),
p = lim
k<0
= lim
k<0
i (d + k) 3 i (d)
i (2 + k) 3 i (2)
(2 + k)3 3 3(2 + k) + 1 3 3
= lim
= lim
k<0
k<0
k
k
k
8 + 12k + 6k2 + k3 3 6 3 3k 3 2
9k + 6k2 + k3
k(9 + 6k + k2 )
= lim
= lim
k<0
k<0
k
k
k
= lim (9 + 6k + k2 ) = 9
k<0
Tangent line: | 3 3 = 9({ 3 2) C | 3 3 = 9{ 3 18
C
| = 9{ 3 15
I
I
I
I
( { 3 1)( { + 1)
{3 1
{31
1
1
I
I
= lim
= lim I
= .
= lim
{<1
{<1 ({ 3 1)( { + 1)
{<1 ({ 3 1)( { + 1)
{<1
{31
2
{+1
7. Using (1), p = lim
Tangent line: | 3 1 = 12 ({ 3 1) C
8. Using (1) with i ({) =
| = 12 { +
1
2
2{ + 1
and S (1> 1),
{+2
2{ + 1
2{ + 1 3 ({ + 2)
31
i ({) 3 i (d)
{31
{
+
2
{+2
p = lim
= lim
= lim
= lim
{<d
{<1
{<1
{<1 ({ 3 1)({ + 2)
{3d
{31
{31
= lim
{<1
1
1
1
=
=
{+2
1+2
3
Tangent line: | 3 1 = 13 ({ 3 1) C
| 3 1 = 13 { 3
1
3
C
| = 13 { +
2
3
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SECTION 2.7
DERIVATIVES AND RATES OF CHANGE
¤
131
9. (a) Using (2) with | = i ({) = 3 + 4{2 3 2{3 ,
i (d + k) 3 i (d)
3 + 4(d + k)2 3 2(d + k)3 3 (3 + 4d2 3 2d3 )
= lim
k<0
k<0
k
k
p = lim
3 + 4(d2 + 2dk + k2 ) 3 2(d3 + 3d2 k + 3dk2 + k3 ) 3 3 3 4d2 + 2d3
k<0
k
= lim
3 + 4d2 + 8dk + 4k2 3 2d3 3 6d2 k 3 6dk2 3 2k3 3 3 3 4d2 + 2d3
k<0
k
= lim
8dk + 4k2 3 6d2 k 3 6dk2 3 2k3
k(8d + 4k 3 6d2 3 6dk 3 2k2 )
= lim
k<0
k<0
k
k
= lim
= lim (8d + 4k 3 6d2 3 6dk 3 2k2 ) = 8d 3 6d2
k<0
(b) At (1> 5): p = 8(1) 3 6(1)2 = 2, so an equation of the tangent line
(c)
is | 3 5 = 2({ 3 1) C | = 2{ + 3.
At (2> 3): p = 8(2) 3 6(2)2 = 38, so an equation of the tangent
line is | 3 3 = 38({ 3 2) C | = 38{ + 19.
10. (a) Using (1),
1
1
I 3I
{
d
p = lim
= lim
{<d
{<d
{3d
I
I
d3 {
I
I
I
I
I
( d 3 {)( d + {)
d3{
d{
I
I
I
I
= lim I
= lim I
{<d
{3d
d{ ({ 3 d) ( d + { ) {<d d{ ({ 3 d) ( d + { )
1
1
31
31
I
I
= lim I
= 3 3@2 or 3 d33@2 [d A 0]
= I
I
{<d
2
2d
d{ ( d + { )
d2 (2 d )
(b) At (1> 1): p = 3 12 , so an equation of the tangent line
is | 3 1 =
3 12 ({
3 1) C | =
3 12 {
+
(c)
3
.
2
1
At 4> 12 : p = 3 16
, so an equation of the tangent line
is | 3
1
2
1
1
= 3 16
({ 3 4) C | = 3 16
{ + 34 .
11. (a) The particle is moving to the right when v is increasing; that is, on the intervals (0> 1) and (4> 6). The particle is moving to
the left when v is decreasing; that is, on the interval (2> 3). The particle is standing still when v is constant; that is, on the
intervals (1> 2) and (3> 4).
(b) The velocity of the particle is equal to the slope of the tangent line of the
graph. Note that there is no slope at the corner points on the graph. On the
interval (0> 1)> the slope is
330
= 3. On the interval (2> 3), the slope is
130
133
331
= 32. On the interval (4> 6), the slope is
= 1.
332
634
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LIMITS AND DERIVATIVES
12. (a) Runner A runs the entire 100-meter race at the same velocity since the slope of the position function is constant.
Runner B starts the race at a slower velocity than runner A, but ¿nishes the race at a faster velocity.
(b) The distance between the runners is the greatest at the time when the largest vertical line segment ¿ts between the two
graphs—this appears to be somewhere between 9 and 10 seconds.
(c) The runners had the same velocity when the slopes of their respective position functions are equal—this also appears to be
at about 9=5 s. Note that the answers for parts (b) and (c) must be the same for these graphs because as soon as the velocity
for runner B overtakes the velocity for runner A, the distance between the runners starts to decrease.
13. Let v(w) = 40w 3 16w2 .
40w 3 16w2 3 16
38 2w2 3 5w + 2
v(w) 3 v(2)
316w2 + 40w 3 16
= lim
= lim
= lim
w<2
w<2
w<2
w<2
w32
w32
w32
w32
y(2) = lim
= lim
w<2
38(w 3 2)(2w 3 1)
= 38 lim (2w 3 1) = 38(3) = 324
w<2
w32
Thus, the instantaneous velocity when w = 2 is 324 ft@s.
14. (a) Let K(w) = 10w 3 1=86w2 .
10(1 + k) 3 1=86(1 + k)2 3 (10 3 1=86)
K(1 + k) 3 K(1)
= lim
k<0
k<0
k
k
y(1) = lim
10 + 10k 3 1=86(1 + 2k + k2 ) 3 10 + 1=86
k<0
k
= lim
10 + 10k 3 1=86 3 3=72k 3 1=86k2 3 10 + 1=86
k<0
k
= lim
6=28k 3 1=86k2
= lim (6=28 3 1=86k) = 6=28
k<0
k<0
k
= lim
The velocity of the rock after one second is 6=28 m@s.
10(d + k) 3 1=86(d + k)2 3 (10d 3 1=86d2 )
K(d + k) 3 K(d)
= lim
(b) y(d) = lim
k<0
k<0
k
k
10d + 10k 3 1=86(d2 + 2dk + k2 ) 3 10d + 1=86d2
k<0
k
= lim
10d + 10k 3 1=86d2 3 3=72dk 3 1=86k2 3 10d + 1=86d2
10k 3 3=72dk 3 1=86k2
= lim
k<0
k<0
k
k
= lim
= lim
k<0
k(10 3 3=72d 3 1=86k)
= lim (10 3 3=72d 3 1=86k) = 10 3 3=72d
k<0
k
The velocity of the rock when w = d is (10 3 3=72d) m@s=
(c) The rock will hit the surface when K = 0 C 10w 3 1=86w2 = 0 C w(10 3 1=86w) = 0 C w = 0 or 1=86w = 10.
The rock hits the surface when w = 10@1=86 E 5=4 s.
(d) The velocity of the rock when it hits the surface is y
10
1=86
10 = 10 3 3=72 1=86
= 10 3 20 = 310 m@s.
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NOT FOR SALE
SECTION 2.7
DERIVATIVES AND RATES OF CHANGE
¤
1
d2 3 (d + k)2
1
3 2
2
v(d + k) 3 v(d)
(d + k)
d
d2 (d + k)2
d2 3 (d2 + 2dk + k2 )
15. y(d) = lim
= lim
= lim
= lim
k<0
k<0
k<0
k<0
k
k
k
kd2 (d + k)2
3(2dk + k2 )
3k(2d + k)
3(2d + k)
32d
32
= lim
= lim 2
= 2 2 = 3 m@s
k<0 kd2 (d + k)2
k<0 kd2 (d + k)2
k<0 d (d + k)2
d ·d
d
= lim
So y (1) =
32
32
1
2
32
m@s.
= 32 m@s, y(2) = 3 = 3 m@s, and y(3) = 3 = 3
13
2
4
3
27
16. (a) The average velocity between times w and w + k is
(w + k)2 3 8(w + k) + 18 3 w2 3 8w + 18
v(w + k) 3 v(w)
w2 + 2wk + k2 3 8w 3 8k + 18 3 w2 + 8w 3 18
=
=
(w + k) 3 w
k
k
=
2wk + k2 3 8k
= (2w + k 3 8) m@s.
k
(i) [3> 4]: w = 3, k = 4 3 3 = 1, so the average
(ii) [3=5> 4]: w = 3=5, k = 0=5, so the average velocity
velocity is 2(3) + 1 3 8 = 31 m@s.
(iii) [4> 5]: w = 4, k = 1, so the average velocity
is 2(3=5) + 0=5 3 8 = 30=5 m@s.
(iv) [4> 4=5]: w = 4, k = 0=5, so the average velocity
is 2(4) + 1 3 8 = 1 m@s.
(b) y(w) = lim
k<0
v(w + k) 3 v(w)
= lim (2w + k 3 8) = 2w 3 8,
k<0
k
is 2(4) + 0=5 3 8 = 0=5 m@s.
(c)
so y (4) = 0.
17. j 0 (0) is the only negative value. The slope at { = 4 is smaller than the slope at { = 2 and both are smaller than the slope
at { = 32. Thus, j0 (0) ? 0 ? j0 (4) ? j 0 (2) ? j 0 (32).
18. Since j(5) = 33, the point (5> 33) is on the graph of j. Since j0 (5) = 4, the slope of the tangent line at { = 5 is 4.
Using the point-slope form of a line gives us | 3 (33) = 4({ 3 5), or | = 4{ 3 23.
19. For the tangent line | = 4{ 3 5: when { = 2, | = 4(2) 3 5 = 3 and its slope is 4 (the coef¿cient of {). At the point of
tangency, these values are shared with the curve | = i ({); that is, i (2) = 3 and i 0 (2) = 4.
20. Since (4> 3) is on | = i ({), i (4) = 3. The slope of the tangent line between (0> 2) and (4> 3) is 14 , so i 0 (4) =
1
4.
21. We begin by drawing a curve through the origin with a
slope of 3 to satisfy i (0) = 0 and i 0 (0) = 3. Since
i 0 (1) = 0, we will round off our ¿gure so that there is
a horizontal tangent directly over { = 1. Last, we
make sure that the curve has a slope of 31 as we pass
over { = 2. Two of the many possibilities are shown.
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134
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CHAPTER 2
LIMITS AND DERIVATIVES
22. We begin by drawing a curve through the origin with a slope of 1 to satisfy
j(0) = 0 and j 0 (0) = 1. We round off our ¿gure at { = 1 to satisfy j0 (1) = 0,
and then pass through (2> 0) with slope 31 to satisfy j(2) = 0 and j 0 (2) = 31.
We round the ¿gure at { = 3 to satisfy j 0 (3) = 0, and then pass through (4> 0)
with slope 1 to satisfy j(4) = 0 and j 0 (4) = 1= Finally we extend the curve on
both ends to satisfy lim j({) = " and lim j({) = 3".
{<"
{<3"
23. Using (4) with i ({) = 3{2 3 {3 and d = 1,
i (1 + k) 3 i (1)
[3(1 + k)2 3 (1 + k)3 ] 3 2
= lim
k<0
k<0
k
k
i 0 (1) = lim
(3 + 6k + 3k2 ) 3 (1 + 3k + 3k2 + k3 ) 3 2
3k 3 k3
k(3 3 k2 )
= lim
= lim
k<0
k<0
k<0
k
k
k
= lim
= lim (3 3 k2 ) = 3 3 0 = 3
k<0
Tangent line: | 3 2 = 3({ 3 1) C | 3 2 = 3{ 3 3 C | = 3{ 3 1
24. Using (5) with j({) = {4 3 2 and d = 1,
j({) 3 j(1)
({4 3 2) 3 (31)
{4 3 1
({2 + 1)({2 3 1)
= lim
= lim
= lim
{<1
{<1
{<1 { 3 1
{<1
{31
{31
{31
j 0 (1) = lim
({2 + 1)({ + 1)({ 3 1)
= lim [({2 + 1)({ + 1)] = 2(2) = 4
{<1
{<1
{31
= lim
Tangent line: | 3 (31) = 4({ 3 1) C | + 1 = 4{ 3 4 C | = 4{ 3 5
25. (a) Using (4) with I ({) = 5{@(1 + {2 ) and the point (2> 2), we have
(b)
5(2 + k)
32
I (2 + k) 3 I (2)
1 + (2 + k)2
I (2) = lim
= lim
k<0
k<0
k
k
0
= lim
k<0
= lim
k<0
5k + 10 3 2(k2 + 4k + 5)
5k + 10
32
+ 4k + 5
k2 + 4k + 5
= lim
k<0
k
k
k2
32k2 3 3k
k(32k 3 3)
32k 3 3
33
= lim
= lim
=
k(k2 + 4k + 5) k<0 k(k2 + 4k + 5) k<0 k2 + 4k + 5
5
So an equation of the tangent line at (2> 2) is | 3 2 = 3 35 ({ 3 2) or | = 3 35 { +
16
5 .
26. (a) Using (4) with J({) = 4{2 3 {3 , we have
J(d + k) 3 J(d)
[4(d + k)2 3 (d + k)3 ] 3 (4d2 3 d3 )
= lim
k<0
k<0
k
k
J0 (d) = lim
= lim
k<0
4d2 + 8dk + 4k2 3 (d3 + 3d2 k + 3dk2 + k3 ) 3 4d2 + d3
8dk + 4k2 3 3d2 k 3 3dk2 3 k3
= lim
k<0
k
k
k(8d + 4k 3 3d2 3 3dk 3 k2 )
= lim (8d + 4k 3 3d2 3 3dk 3 k2 ) = 8d 3 3d2
k<0
k<0
k
= lim
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SECTION 2.7
At the point (2> 8), J0 (2) = 16 3 12 = 4, and an equation of the
DERIVATIVES AND RATES OF CHANGE
¤
(b)
tangent line is | 3 8 = 4({ 3 2), or | = 4{. At the point (3> 9),
J0 (3) = 24 3 27 = 33, and an equation of the tangent line is
| 3 9 = 33({ 3 3), or | = 33{ + 18=
27. Use (4) with i ({) = 3{2 3 4{ + 1.
i 0 (d) = lim
k<0
i (d + k) 3 i(d)
[3(d + k)2 3 4(d + k) + 1] 3 (3d2 3 4d + 1)]
= lim
k<0
k
k
= lim
3d2 + 6dk + 3k2 3 4d 3 4k + 1 3 3d2 + 4d 3 1
6dk + 3k2 3 4k
= lim
k<0
k
k
= lim
k(6d + 3k 3 4)
= lim (6d + 3k 3 4) = 6d 3 4
k<0
k
k<0
k<0
28. Use (4) with i (w) = 2w3 + w.
i (d + k) 3 i (d)
[2(d + k)3 + (d + k)] 3 (2d3 + d)
= lim
k<0
k<0
k
k
i 0 (d) = lim
2d3 + 6d2 k + 6dk2 + 2k3 + d + k 3 2d3 3 d
6d2 k + 6dk2 + 2k3 + k
= lim
k<0
k<0
k
k
= lim
k(6d2 + 6dk + 2k2 + 1)
= lim (6d2 + 6dk + 2k2 + 1) = 6d2 + 1
k<0
k<0
k
= lim
29. Use (4) with i (w) = (2w + 1)@(w + 3).
2(d + k) + 1 2d + 1
3
i
(d
+
k)
3
i
(d)
(d + k) + 3
d+3
(2d + 2k + 1)(d + 3) 3 (2d + 1)(d + k + 3)
i 0 (d) = lim
= lim
= lim
k<0
k<0
k<0
k
k
k(d + k + 3)(d + 3)
(2d2 + 6d + 2dk + 6k + d + 3) 3 (2d2 + 2dk + 6d + d + k + 3)
k<0
k(d + k + 3)(d + 3)
5k
5
5
= lim
=
= lim
k<0 k(d + k + 3)(d + 3)
k<0 (d + k + 3)(d + 3)
(d + 3)2
= lim
30. Use (4) with i ({) = {32 = 1@{2 .
1
d2 3 (d + k)2
1
3 2
2
i (d + k) 3 i (d)
(d + k)
d
d2 (d + k)2
d2 3 (d2 + 2dk + k2 )
= lim
= lim
= lim
i 0 (d) = lim
k<0
k<0
k<0
k<0
k
k
k
kd2 (d + k)2
= lim
k<0
32dk 3 k2
k(32d 3 k)
32d 3 k
32
32d
= lim
= lim 2
= 2 2 = 3
k<0 kd2 (d + k)2
k<0 d (d + k)2
kd2 (d + k)2
d (d )
d
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CHAPTER 2
LIMITS AND DERIVATIVES
31. Use (4) with i ({) =
I
1 3 2{.
i (d + k) 3 i (d)
= lim
k<0
k<0
k
i 0 (d) = lim
s
I
1 3 2(d + k) 3 1 3 2d
k
2 I
s
2
s
s
I
I
1
3
2(d
+
k)
3
1 3 2d
1 3 2(d + k) 3 1 3 2d
1 3 2(d + k) + 1 3 2d
s
= lim
·s
= lim
I
I
k<0
k
1 3 2(d + k) + 1 3 2d k<0 k
1 3 2(d + k) + 1 3 2d
= lim
k<0
(1 3 2d 3 2k) 3 (1 3 2d)
32k
s
= lim s
I
I
k<0
k
1 3 2(d + k) + 1 3 2d
k
1 3 2(d + k) + 1 3 2d
32
32
32
31
I
= lim s
= I
= I
= I
I
k<0
1 3 2d + 1 3 2d
2 1 3 2d
1 3 2d
1 3 2(d + k) + 1 3 2d
4
.
13{
32. Use (4) with i ({) = I
4
4
s
3I
1
3d
1
3
(d
+
k)
i
(d
+
k)
3
i
(d)
= lim
i 0 (d) = lim
k<0
k<0
k
k
I
I
13d3 13d3k
I
I
I
I
13d3 13d3k
13d3k 13d
= 4 lim
= 4 lim I
I
k<0
k<0 k 1 3 d 3 k
k
13d
I
I
I
I
I
I
13d3 13d3k
13d+ 13d3k
( 1 3 d)2 3 ( 1 3 d 3 k)2
I
I
= 4 lim I
·I
= 4 lim I
I
I
I
k<0 k 1 3 d 3 k
k<0 k 1 3 d 3 k
13d
13d+ 13d3k
1 3 d( 1 3 d + 1 3 d 3 k)
= 4 lim
k<0
(1 3 d) 3 (1 3 d 3 k)
k
I
I
I
= 4 lim I
I
I
I
I
k<0 k 1 3 d 3 k
k 1 3 d 3 k 1 3 d( 1 3 d + 1 3 d 3 k)
1 3 d( 1 3 d + 1 3 d 3 k)
1
1
I
I
I
I
= 4 lim I
=4· I
I
I
k<0
1 3 d 1 3 d( 1 3 d + 1 3 d)
1 3 d 3 k 1 3 d( 1 3 d + 1 3 d 3 k)
=
2
2
4
I
=
=
(1 3 d)1 (1 3 d)1@2
(1 3 d)3@2
(1 3 d)(2 1 3 d)
Note that the answers to Exercises 33 – 38 are not unique.
(1 + k)10 3 1
= i 0 (1), where i ({) = {10 and d = 1.
k<0
k
33. By (4), lim
Or: By (4), lim
k<0
(1 + k)10 3 1
= i 0 (0), where i ({) = (1 + {)10 and d = 0.
k
I
4
I
16 + k 3 2
= i 0 (16), where i ({) = 4 { and d = 16.
34. By (4), lim
k<0
k
I
4
I
16 + k 3 2
Or: By (4), lim
= i 0 (0), where i ({) = 4 16 + { and d = 0.
k<0
k
35. By Equation 5, lim
{<5
2{ 3 32
= i 0 (5), where i({) = 2{ and d = 5.
{35
36. By Equation 5, lim
{<@4
tan { 3 1
= i 0 (@4), where i ({) = tan { and d = @4.
{ 3 @4
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SECTION 2.7
DERIVATIVES AND RATES OF CHANGE
cos( + k) + 1
= i 0 (), where i ({) = cos { and d = .
k
37. By (4), lim
k<0
Or: By (4), lim
k<0
cos( + k) + 1
= i 0 (0), where i ({) = cos( + {) and d = 0.
k
w4 + w 3 2
= i 0 (1), where i (w) = w4 + w and d = 1.
w<1
w31
38. By Equation 5, lim
i (5 + k) 3 i (5)
[100 + 50(5 + k) 3 4=9(5 + k)2 ] 3 [100 + 50(5) 3 4=9(5)2 ]
= lim
k<0
k
k
(100 + 250 + 50k 3 4=9k2 3 49k 3 122=5) 3 (100 + 250 3 122=5)
34=9k2 + k
= lim
= lim
k<0
k<0
k
k
k(34=9k + 1)
= lim
= lim (34=9k + 1) = 1 m@s
k<0
k<0
k
39. y(5) = i 0 (5) = lim
k<0
The speed when w = 5 is |1| = 1 m@s.
i (5 + k) 3 i (5)
[(5 + k)31 3 (5 + k)] 3 (531 3 5)
= lim
k<0
k<0
k
k
40. y(5) = i 0 (5) = lim
5 3 5k(5 + k) 3 (5 + k)
1
1
1
1
353k3 +5
3k3
5(5 + k)
5
5 = lim
= lim 5 + k
= lim 5 + k
k<0
k<0
k<0
k
k
k
= lim
k<0
5 3 25k 3 5k2 3 5 3 k
35k2 3 26k
k(35k 3 26)
35k 3 26
326
= lim
= lim
= lim
=
m@s
k<0 5k(5 + k)
k<0 5k(5 + k)
k<0 5(5 + k)
5k(5 + k)
25
=
The speed when w = 5 is 3 26
25
26
25
= 1=04 m@s.
41. The sketch shows the graph for a room temperature of 72 and a refrigerator
temperature of 38 . The initial rate of change is greater in magnitude than the
rate of change after an hour.
42. The slope of the tangent (that is, the rate of change of temperature with respect
to time) at w = 1 h seems to be about
43. (a) (i) [2002> 2006]:
75 3 168
E 30=7 F@min.
132 3 0
233 3 141
92
Q(2006) 3 Q(2002)
=
=
= 23 millions of cell phone subscribers per year
2006 3 2002
4
4
(ii) [2002> 2004]:
Q(2004) 3 Q(2002)
182 3 141
41
=
=
= 20=5 millions of cell phone subscribers per year
2004 3 2002
2
2
(iii) [2000> 2002]:
Q(2002) 3 Q(2000)
141 3 109
32
=
=
= 16 millions of cell phone subscribers per year
2002 3 2000
2
2
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¤
CHAPTER 2
LIMITS AND DERIVATIVES
(b) Using the values from (ii) and (iii), we have
20=5 + 16
= 18=25 millions of cell phone subscribers per year.
2
(c) Estimating A as (2000> 107) and B as (2004> 175), the slope at 2002
is
68
175 3 107
=
= 17 millions of cell phone subscribers per
2004 3 2000
4
year.
16,680 3 12,440
4240
Q(2008) 3 Q(2006)
=
=
= 2120 locations per year
2008 3 2006
2
2
Q(2007) 3 Q(2006)
15,011 3 12,440
(ii) [2006> 2007]:
=
= 2571 locations per year
2007 3 2006
1
Q(2006) 3 Q(2005)
12,440 3 10,241
(iii) [2005> 2006]:
=
= 2199 locations per year
2006 3 2005
1
2571 + 2199
= 2385 locations per year.
(b) Using the values from (ii) and (iii), we have
2
44. (a) (i) [2006> 2008]:
(c) Estimating A as (2005> 10,060) and B as (2007> 14,800), the slope at
2005 is
14,800 3 10,060
4740
=
= 2370 locations per year.
2007 3 2005
2
(d) Estimating C as (2006> 12,860) and D as (2008> 17,100), the slope
at 2007 is
4240
17,100 3 12,860
=
= 2120 locations per year.
2008 3 2006
2
We conclude that the rate of growth is decreasing.
45. (a) (i)
F(105) 3 F(100)
6601=25 3 6500
{F
=
=
= $20=25@unit.
{{
105 3 100
5
F(101) 3 F(100)
6520=05 3 6500
{F
=
=
= $20=05@unit.
{{
101 3 100
1
5000 + 10(100 + k) + 0=05(100 + k)2 3 6500
F(100 + k) 3 F(100)
20k + 0=05k2
(b)
=
=
k
k
k
= 20 + 0=05k, k 6= 0
(ii)
So the instantaneous rate of change is lim
k<0
F(100 + k) 3 F(100)
= lim (20 + 0=05k) = $20@unit.
k<0
k
2
2
w+k
w
3 100,000 1 3
60
60
2
2 (w + k)
w
w
k
2wk
k2
w+k
+
3 13
+
= 100,000 3 +
+
= 100,000 1 3
30
3600
30
3600
30
3600
3600
46. {Y = Y (w + k) 3 Y (w) = 100,000 1 3
=
100,000
250
k (3120 + 2w + k) =
k (3120 + 2w + k)
3600
9
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NOT FOR SALE
SECTION 2.7
DERIVATIVES AND RATES OF CHANGE
Dividing {Y by k and then letting k < 0, we see that the instantaneous rate of change is
w
Flow rate (gal@min)
0
33333=3
10
20
30
40
50
60
500
9
¤
139
(w 3 60) gal@min.
Water remaining Y (w) (gal)
100> 000
32777=7
69> 444=4
31666=6
25> 000
32222=2
44> 444=4
31111=1
11> 111=1
3 555=5
0
2> 777=7
0
The magnitude of the Àow rate is greatest at the beginning and gradually decreases to 0.
47. (a) i 0 ({) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are
dollars per ounce.
(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17@ounce. So the cost
of producing the 800th (or 801st) ounce is about $17.
(c) In the short term, the values of i 0 ({) will decrease because more ef¿cient use is made of start-up costs as { increases. But
eventually i 0 ({) might increase due to large-scale operations.
48. (a) i 0 (5) is the rate of growth of the bacteria population when w = 5 hours. Its units are bacteria per hour.
(b) With unlimited space and nutrients, i 0 should increase as w increases; so i 0 (5) ? i 0 (10). If the supply of nutrients is
limited, the growth rate slows down at some point in time, and the opposite may be true.
49. W 0 (8) is the rate at which the temperature is changing at 8:00 AM . To estimate the value of W 0 (8), we will average the
difference quotients obtained using the times w = 6 and w = 10.
Let D =
W (6) 3 W (8)
75 3 84
W (10) 3 W (8)
90 3 84
=
= 4=5 and E =
=
= 3. Then
638
32
10 3 8
2
W 0 (8) = lim
w<8
W (w) 3 W (8)
D+E
4=5 + 3
r
=
= 3=75 F@h.
w38
2
2
50. (a) i 0 (8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is $8 per pound.
The units for i 0 (8) are pounds@(dollars@pound).
(b) i 0 (8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally
less willing to buy a product when its price increases.
51. (a) V 0 (W ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg@L)@ C.
(b) For W = 16 C, it appears that the tangent line to the curve goes through the points (0> 14) and (32> 6). So
V 0 (16) E
8
6 3 14
=3
= 30=25 (mg@L)@ C. This means that as the temperature increases past 16 C, the oxygen
32 3 0
32
solubility is decreasing at a rate of 0=25 (mg@L)@ C.
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CHAPTER 2
LIMITS AND DERIVATIVES
52. (a) V 0 (W ) is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature. Its units
are (cm@s)@ C.
(b) For W = 15 C, it appears the tangent line to the curve goes through the points (10> 25) and (20> 32). So
32 3 25
= 0=7 (cm@s)@ C. This tells us that at W = 15 C, the maximum sustainable speed of Coho salmon is
20 3 10
V 0 (15) E
changing at a rate of 0.7 (cm@s)@ C. In a similar fashion for W = 25 C, we can use the points (20> 35) and (25> 25) to
obtain V 0 (25) E
rapidly.
25 3 35
= 32 (cm@s)@ C. As it gets warmer than 20 C, the maximum sustainable speed decreases
25 3 20
53. Since i ({) = { sin(1@{) when { 6= 0 and i (0) = 0, we have
i 0 (0) = lim
k<0
i (0 + k) 3 i (0)
k sin(1@k) 3 0
= lim
= lim sin(1@k). This limit does not exist since sin(1@k) takes the
k<0
k<0
k
k
values 31 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.)
54. Since i({) = {2 sin(1@{) when { 6= 0 and i (0) = 0, we have
i (0 + k) 3 i (0)
k2 sin(1@k) 3 0
1
= lim
= lim k sin(1@k). Since 31 $ sin $ 1, we have
k<0
k<0
k<0
k
k
k
i 0 (0) = lim
1
1
$ |k| i 3 |k| $ k sin $ |k|. Because lim (3 |k|) = 0 and lim |k| = 0, we know that
k<0
k<0
k
k
1
lim k sin
= 0 by the Squeeze Theorem. Thus, i 0 (0) = 0.
k<0
k
3 |k| $ |k| sin
2.8 The Derivative as a Function
1. It appears that i is an odd function, so i 0 will be an even function—that
is, i 0 (3d) = i 0 (d).
(a) i 0 (33) E 30=2
(b) i 0 (32) E 0
(c) i 0 (31) E 1
(d) i 0 (0) E 2
(e) i 0 (1) E 1
(f) i 0 (2) E 0
(g) i 0 (3) E 30=2
2. Your answers may vary depending on your estimates.
(a) Note: By estimating the slopes of tangent lines on the
graph of i , it appears that i 0 (0) E 6.
(b) i 0 (1) E 0
(c) i 0 (2) E 31=5
(d) i 0 (3) E 31=3
(e) i 0 (4) E 30=8
(f) i 0 (5) E 30=3
(g) i 0 (6) E 0
(h) i 0 (7) E 0=2
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SECTION 2.8
THE DERIVATIVE AS A FUNCTION
¤
141
3. (a)0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then
negative again. The actual function values in graph II follow the same pattern.
(b)0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a ¿xed positive quantity, then suddenly
become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)0 = I, since the slopes of the tangents to graph (c) are negative for { ? 0 and positive for { A 0, as are the function values of
graph I.
(d)0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then
positive, then 0, then negative again, and the function values in graph III follow the same pattern.
Hints for Exercises 4 –11: First plot {-intercepts on the graph of i 0 for any horizontal tangents on the graph of i . Look for any corners on the graph
of i — there will be a discontinuity on the graph of i 0 . On any interval where i has a tangent with positive (or negative) slope, the graph of i 0 will be
positive (or negative). If the graph of the function is linear, the graph of i 0 will be a horizontal line.
4.
5.
6.
7.
8.
9.
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10.
11.
12. The slopes of the tangent lines on the graph of | = S (w) are always
positive, so the |-values of | = S 0(w) are always positive. These values start
out relatively small and keep increasing, reaching a maximum at about
w = 6. Then the |-values of | = S 0(w) decrease and get close to zero. The
graph of S 0 tells us that the yeast culture grows most rapidly after 6 hours
and then the growth rate declines.
13. (a) F 0 (w) is the instantaneous rate of change of percentage
of full capacity with respect to elapsed time in hours.
(b) The graph of F 0 (w) tells us that the rate of change of
percentage of full capacity is decreasing and
approaching 0.
14. (a) I 0 (y) is the instantaneous rate of change of fuel
economy with respect to speed.
(b) Graphs will vary depending on estimates of I 0 , but
will change from positive to negative at about y = 50.
(c) To save on gas, drive at the speed where I is a
maximum and I 0 is 0, which is about 50 mi@ h.
15. It appears that there are horizontal tangents on the graph of P for w = 1963
and w = 1971. Thus, there are zeros for those values of w on the graph of
P 0 . The derivative is negative for the years 1963 to 1971.
16. See Figure 1 in Section 3.3.
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SECTION 2.8
17.
THE DERIVATIVE AS A FUNCTION
¤
143
18.
The slope at 0 appears to be 1 and the slope at 1
As { increases toward 1, i 0 ({) decreases from very large
appears to be 2=7. As { decreases, the slope gets
numbers to 1. As { becomes large, i 0 ({) gets closer to 0.
closer to 0. Since the graphs are so similar, we might
As a guess, i 0 ({) = 1@{2 or i 0 ({) = 1@{ makes sense.
guess that i 0 ({) = h{ .
19. (a) By zooming in, we estimate that i 0 (0) = 0, i 0
and i 0 (2) = 4.
1
2
= 1, i 0 (1) = 2,
(b) By symmetry, i 0 (3{) = 3i 0 ({). So i 0 3 12 = 31, i 0 (31) = 32,
and i 0 (32) = 34.
(c) It appears that i 0 ({) is twice the value of {, so we guess that i 0 ({) = 2{.
i({ + k) 3 i ({)
({ + k)2 3 {2
= lim
k<0
k<0
k
k
2
{ + 2k{ + k2 3 {2
2k{ + k2
k(2{ + k)
= lim
= lim
= lim (2{ + k) = 2{
= lim
k<0
k<0
k<0
k<0
k
k
k
(d) i 0 ({) = lim
20. (a) By zooming in, we estimate that i 0 (0) = 0, i 0
i 0 (1) E 3, i 0 (2) E 12, and i 0 (3) E 27.
1
2
E 0=75,
(b) By symmetry, i 0 (3{) = i 0 ({). So i 0 3 12 E 0=75,
i 0 (31) E 3, i 0 (32) E 12, and i 0 (33) E 27.
(d) Since i 0 (0) = 0, it appears that i 0 may have the
(c)
form i 0 ({) = d{2 . Using i 0 (1) = 3, we have d = 3,
so i 0 ({) = 3{2 .
i({ + k) 3 i ({)
({ + k)3 3 {3
({3 + 3{2 k + 3{k2 + k3 ) 3 {3
= lim
= lim
k<0
k<0
k
k
k
3{2 k + 3{k2 + k3
k(3{2 + 3{k + k2 )
2
= lim
= lim
= lim (3{ + 3{k + k2 ) = 3{2
k<0
k<0
k<0
k
k
(e) i 0 ({) = lim
k<0
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i ({ + k) 3 i({)
= lim
k<0
k<0
k
21. i 0 ({) = lim
= lim
k<0
1
k
2
k
1
k<0 2
= lim
=
1
2 ({
+ k) 3
1
3
k
3
1
2{
3
1
3
= lim
1
{
2
k<0
+ 12 k 3 13 3 12 { +
k
1
3
1
2
Domain of i = domain of i 0 = R.
22. i 0 ({) = lim
k<0
= lim
k<0
i({ + k) 3 i ({)
[p({ + k) + e] 3 (p{ + e)
p{ + pk + e 3 p{ 3 e
= lim
= lim
k<0
k<0
k
k
k
pk
= lim p = p
k<0
k
Domain of i = domain of i 0 = R.
5(w + k) 3 9(w + k)2 3 (5w 3 9w2 )
i (w + k) 3 i(w)
= lim
23. i (w) = lim
k<0
k<0
k
k
0
= lim
5w + 5k 3 9(w2 + 2wk + k2 ) 3 5w + 9w2
5w + 5k 3 9w2 3 18wk 3 9k2 3 5w + 9w2
= lim
k<0
k
k
= lim
5k 3 18wk 3 9k2
k(5 3 18w 3 9k)
= lim
= lim (5 3 18w 3 9k) = 5 3 18w
k<0
k<0
k
k
k<0
k<0
Domain of i = domain of i 0 = R.
1=5({ + k)2 3 ({ + k) + 3=7 3 1=5{2 3 { + 3=7
i({ + k) 3 i ({)
= lim
24. i ({) = lim
k<0
k<0
k
k
0
= lim
k<0
1=5{2 + 3{k + 1=5k2 3 { 3 k + 3=7 3 1=5{2 + { 3 3=7
3{k + 1=5k2 3 k
= lim
k<0
k
k
= lim (3{ + 1=5k 3 1) = 3{ 3 1
k<0
Domain of i = domain of i 0 = R.
25. i 0 ({) = lim
k<0
= lim
k<0
i({ + k) 3 i ({)
[({ + k)2 3 2({ + k)3 ] 3 ({2 3 2{3 )
= lim
k<0
k
k
{2 + 2{k + k2 3 2{3 3 6{2 k 3 6{k2 3 2k3 3 {2 + 2{3
k
k(2{ + k 3 6{2 3 6{k 3 2k2 )
2{k + k2 3 6{2 k 3 6{k2 3 2k3
= lim
k<0
k<0
k
k
= lim (2{ + k 3 6{2 3 6{k 3 2k2 ) = 2{ 3 6{2
= lim
k<0
Domain of i = domain of i 0 = R.
I
I
w3 w+k
1
1
I
I
I
I
I
I
I
3I
j(w + k) 3 j(w)
w3 w+k
w+ w+k
w+k
w+k w
w
I ·I
I
I
= lim
= lim
= lim
26. j0 (w) = lim
k<0
k<0
k<0
k<0
k
k
k
k w+k w
w+ w+k
= lim
k<0
k
w 3 (w + k)
3k
31
I I
I I
I I
I
I
= lim I
I
= lim I
I
k<0 k
k<0
w+k w w+ w+k
w+k w w+ w+k
w+k w w+ w+k
31
1
31
I = I = 3 3@2
= I I I
2w
w w w+ w
w 2 w
Domain of j = domain of j0 = (0> ").
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THE DERIVATIVE AS A FUNCTION
¤
145
&
%s
s
I
I
9 3 ({ + k) 3 9 3 {
9 3 ({ + k) + 9 3 {
j({ + k) 3 j({)
0
s
= lim
27. j ({) = lim
I
k<0
k<0
k
k
9 3 ({ + k) + 9 3 {
= lim
k<0
[9 3 ({ + k)] 3 (9 3 {)
3k
ks
l = lim ks
l
I
I
k<0
k
9 3 ({ + k) + 9 3 {
k
9 3 ({ + k) + 9 3 {
31
31
= lim s
= I
I
k<0
2 93{
9 3 ({ + k) + 9 3 {
Domain of j = (3"> 9], domain of j0 = (3"> 9).
({ + k)2 3 1
{2 3 1
3
i({ + k) 3 i ({)
2({ + k) 3 3
2{ 3 3
28. i 0 ({) = lim
= lim
k<0
k<0
k
k
[({ + k)2 3 1](2{ 3 3) 3 [2({ + k) 3 3]({2 3 1)
[2({ + k) 3 3](2{ 3 3)
= lim
k<0
k
= lim
({2 + 2{k + k2 3 1)(2{ 3 3) 3 (2{ + 2k 3 3)({2 3 1)
k[2({ + k) 3 3](2{ 3 3)
= lim
(2{3 + 4{2 k + 2{k2 3 2{ 3 3{2 3 6{k 3 3k2 + 3) 3 (2{3 + 2{2 k 3 3{2 3 2{ 3 2k + 3)
k(2{ + 2k 3 3)(2{ 3 3)
= lim
4{2 k + 2{k2 3 6{k 3 3k2 3 2{2 k + 2k
k(2{2 + 2{k 3 6{ 3 3k + 2)
= lim
k<0
k(2{ + 2k 3 3)(2{ 3 3)
k(2{ + 2k 3 3)(2{ 3 3)
= lim
2{2 + 2{k 3 6{ 3 3k + 2
2{2 3 6{ + 2
=
(2{ + 2k 3 3)(2{ 3 3)
(2{ 3 3)2
k<0
k<0
k<0
k<0
Domain of i = domain of i 0 = (3"> 32 ) ( 32 > ").
1 3 2(w + k)
1 3 2w
3
J(w + k) 3 J(w)
3 + (w + k)
3+w
= lim
29. J (w) = lim
k<0
k<0
k
k
0
[1 3 2(w + k)](3 + w) 3 [3 + (w + k)](1 3 2w)
[3 + (w + k)](3 + w)
= lim
k<0
k
= lim
3 + w 3 6w 3 2w2 3 6k 3 2kw 3 (3 3 6w + w 3 2w2 + k 3 2kw)
36k 3 k
= lim
k<0 k(3 + w + k)(3 + w)
k[3 + (w + k)](3 + w)
= lim
37k
37
37
= lim
=
k(3 + w + k)(3 + w) k<0 (3 + w + k)(3 + w)
(3 + w)2
k<0
k<0
Domain of J = domain of J0 = (3"> 33) (33> ").
i({ + k) 3 i ({)
({ + k)3@2 3 {3@2
[({ + k)3@2 3 {3@2 ][({ + k)3@2 + {3@2 ]
= lim
= lim
k<0
k<0
k
k
k [({ + k)3@2 + {3@2 ]
k 3{2 + 3{k + k2
({ + k)3 3 {3
{3 + 3{2 k + 3{k2 + k3 3 {3
= lim
= lim
= lim
k<0 k[({ + k)3@2 + {3@2 ]
k<0
k<0 k[({ + k)3@2 + {3@2 ]
k[({ + k)3@2 + {3@2 ]
30. i 0 ({) = lim
k<0
= lim
k<0
3{2 + 3{k + k2
3{2
= 3@2 = 32 {1@2
3@2
3@2
({ + k)
+{
2{
Domain of i = domain of i 0 = [0> "). Strictly speaking, the domain of i 0 is (0> ") because the limit that de¿nes i 0 (0) does
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LIMITS AND DERIVATIVES
not exist (as a two-sided limit). But the right-hand derivative (in the sense of Exercise 56) does exist at 0, so in that sense one
could regard the domain of i 0 to be [0> ").
4
{ + 4{3 k + 6{2 k2 + 4{k3 + k4 3 {4
i({ + k) 3 i ({)
({ + k)4 3 {4
= lim
= lim
31. i ({) = lim
k<0
k<0
k<0
k
k
k
3
4{3 k + 6{2 k2 + 4{k3 + k4
= lim
= lim 4{ + 6{2 k + 4{k2 + k3 = 4{3
k<0
k<0
k
0
Domain of i = domain of i 0 = R.
32. (a)
(b) Note that the third graph in part (a) has small negative values for its slope, i 0 ; but as { < 63 , i 0 < 3".
See the graph in part (d).
i ({ + k) 3 i ({)
k
&
%s
s
I
I
6 3 ({ + k) 3 6 3 {
6 3 ({ + k) + 6 3 {
s
= lim
I
k<0
k
6 3 ({ + k) + 6 3 {
(c) i 0 ({) = lim
k<0
= lim
k<0
(d)
[6 3 ({ + k)] 3 (6 3 {)
3k
ks
l = lim I
I
I
k<0 k
6
3
{
3
k+ 63{
k
6 3 ({ + k) + 6 3 {
31
31
= lim I
= I
I
k<0
2 63{
63{3k+ 63{
Domain of i = (3"> 6], domain of i 0 = (3"> 6).
33. (a) i 0 ({) = lim
k<0
i ({ + k) 3 i ({)
[({ + k)4 + 2({ + k)] 3 ({4 + 2{)
= lim
k<0
k
k
= lim
{4 + 4{3 k + 6{2 k2 + 4{k3 + k4 + 2{ + 2k 3 {4 3 2{
k
= lim
4{3 k + 6{2 k2 + 4{k3 + k4 + 2k
k(4{3 + 6{2 k + 4{k2 + k3 + 2)
= lim
k<0
k
k
k<0
k<0
= lim (4{3 + 6{2 k + 4{k2 + k3 + 2) = 4{3 + 2
k<0
(b) Notice that i 0 ({) = 0 when i has a horizontal tangent, i 0 ({) is
positive when the tangents have positive slope, and i 0 ({) is
negative when the tangents have negative slope.
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SECTION 2.8
THE DERIVATIVE AS A FUNCTION
i({ + k) 3 i ({)
[({ + k) + 1@({ + k)] 3 ({ + 1@{)
= lim
= lim
34. (a) i 0 ({) = lim
k<0
k<0
k<0
k
k
{[({ + k)2 + 1] 3 ({ + k)({2 + 1)
({3 + 2k{2 + {k2 + {) 3 ({3 + { + k{2 + k)
= lim
k<0
k({ + k){
k({ + k){
= lim
k{2 + {k2 3 k
k({2 + {k 3 1)
{2 + {k 3 1
{2 3 1
1
= lim
= lim
=
, or 1 3 2
k<0
k<0
k({ + k){
k({ + k){
({ + k){
{2
{
k<0
147
({ + k)2 + 1
{2 + 1
3
{+k
{
k
= lim
k<0
¤
(b) Notice that i 0 ({) = 0 when i has a horizontal tangent, i 0 ({) is
positive when the tangents have positive slope, and i 0 ({) is
negative when the tangents have negative slope. Both functions
are discontinuous at { = 0.
35. (a) X 0 (w) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.
X (w + k) 3 X(w)
X(w + k) 3 X (w)
E
for small values of k.
k
k
X (2000) 3 X(1999)
4=0 3 4=2
For 1999: X 0 (1999) E
=
= 30=2
2000 3 1999
1
0
For 2000: We estimate X (2000) by using k = 31 and k = 1, and then average the two results to obtain a ¿nal estimate.
4=2 3 4=0
X (1999) 3 X (2000)
=
= 30=2;
k = 31 i X 0 (2000) E
1999 3 2000
31
4=7 3 4=0
X (2001) 3 X (2000)
=
= 0=7.
k = 1 i X 0 (2000) E
2001 3 2000
1
0
1
So we estimate that X (2000) E 2 [(30=2) + 0=7] = 0=25.
(b) To ¿nd X 0 (w), we use lim
k<0
w
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
0
30=2
0=25
0=9
0=65
30=15
30=45
30=45
30=25
0=6
1=2
X (w)
36. (a) S 0 (w) is the rate at which the percentage of Americans under the age of 18 is changing with respect to time. Its units are
percent per year (%@yr).
(b) To ¿nd S 0 (w), we use lim
k<0
For 1950: S 0 (1950) E
S (w + k) 3 S (w) S (w + k) 3 S (w)
E
for small values of k.
k
k
35=7 3 31=1
S (1960) 3 S (1950)
=
= 0=46
1960 3 1950
10
For 1960: We estimate S 0 (1960) by using k = 310 and k = 10, and then average the two results to obtain a
¿nal estimate.
31=1 3 35=7
S (1950) 3 S (1960)
=
= 0=46
1950 3 1960
310
k = 310 i S 0 (1960) E
k = 10 i S 0 (1960) E
S (1970) 3 S (1960)
34=0 3 35=7
=
= 30=17
1970 3 1960
10
So we estimate that S 0 (1960) E 12 [0=46 + (30=17)] = 0=145.
w
1950
1960
1970
1980
1990
2000
0
0=460
0=145
30=385
30=415
30=115
0=000
S (w)
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LIMITS AND DERIVATIVES
(c)
(d) We could get more accurate values for S 0 (w) by obtaining data for the mid-decade years 1955, 1965, 1975, 1985, and 1995.
37. i is not differentiable at { = 34, because the graph has a corner there, and at { = 0, because there is a discontinuity there.
38. i is not differentiable at { = 0, because there is a discontinuity there, and at { = 3, because the graph has a vertical tangent
there.
39. i is not differentiable at { = 31, because the graph has a vertical tangent there, and at { = 4, because the graph has a corner
there.
40. i is not differentiable at { = 31, because there is a discontinuity there, and at { = 2, because the graph has a corner there.
41. As we zoom in toward (31> 0), the curve appears more and more like a straight
line, so i ({) = { +
s
|{| is differentiable at { = 31. But no matter how much
we zoom in toward the origin, the curve doesn’t straighten out—we can’t
eliminate the sharp point (a cusp). So i is not differentiable at { = 0.
42. As we zoom in toward (0> 1), the curve appears more and more like a straight
line, so i is differentiable at { = 0. But no matter how much we zoom in toward
(1> 0) or (31> 0), the curve doesn’t straighten out—we can’t eliminate the sharp
point (a cusp). So i is not differentiable at { = ±1.
43. d = i , e = i 0 , f = i 00 . We can see this because where d has a horizontal tangent, e = 0, and where e has a horizontal tangent,
f = 0. We can immediately see that f can be neither i nor i 0 , since at the points where f has a horizontal tangent, neither d
nor e is equal to 0.
44. Where g has horizontal tangents, only f is 0, so g0 = f. f has negative tangents for { ? 0 and e is the only graph that is
negative for { ? 0, so f0 = e. e has positive tangents on R (except at { = 0), and the only graph that is positive on the same
domain is d, so e0 = d. We conclude that g = i , f = i 0 , e = i 00 , and d = i 000 .
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THE DERIVATIVE AS A FUNCTION
¤
149
45. We can immediately see that d is the graph of the acceleration function, since at the points where d has a horizontal tangent,
neither f nor e is equal to 0. Next, we note that d = 0 at the point where e has a horizontal tangent, so e must be the graph of
the velocity function, and hence, e0 = d. We conclude that f is the graph of the position function.
46. d must be the jerk since none of the graphs are 0 at its high and low points. d is 0 where e has a maximum, so e0 = d. e is 0
where f has a maximum, so f0 = e. We conclude that g is the position function, f is the velocity, e is the acceleration, and d is
the jerk.
47. i 0 ({) = lim
k<0
i({ + k) 3 i ({)
[3({ + k)2 + 2({ + k) + 1] 3 (3{2 + 2{ + 1)
= lim
k<0
k
k
= lim
(3{2 + 6{k + 3k2 + 2{ + 2k + 1) 3 (3{2 + 2{ + 1)
6{k + 3k2 + 2k
= lim
k<0
k
k
= lim
k(6{ + 3k + 2)
= lim (6{ + 3k + 2) = 6{ + 2
k<0
k
k<0
k<0
i 0 ({ + k) 3 i 0 ({)
[6({ + k) + 2] 3 (6{ + 2)
(6{ + 6k + 2) 3 (6{ + 2)
= lim
= lim
k<0
k<0
k<0
k
k
k
i 00 ({) = lim
= lim
k<0
6k
= lim 6 = 6
k<0
k
We see from the graph that our answers are reasonable because the graph of
i 0 is that of a linear function and the graph of i 00 is that of a constant
function.
i({ + k) 3 i ({)
[({ + k)3 3 3({ + k)] 3 ({3 3 3{)
= lim
k<0
k<0
k
k
({3 + 3{2 k + 3{k2 + k3 3 3{ 3 3k) 3 ({3 3 3{)
3{2 k + 3{k2 + k3 3 3k
= lim
= lim
k<0
k<0
k
k
48. i 0 ({) = lim
= lim
k<0
k(3{2 + 3{k + k2 3 3)
= lim (3{2 + 3{k + k2 3 3) = 3{2 3 3
k<0
k
i 0 ({ + k) 3 i 0 ({)
[3({ + k)2 3 3] 3 (3{2 3 3)
(3{2 + 6{k + 3k2 3 3) 3 (3{2 3 3)
= lim
= lim
k<0
k<0
k<0
k
k
k
i 00 ({) = lim
6{k + 3k2
k(6{ + 3k)
= lim
= lim (6{ + 3k) = 6{
k<0
k<0
k<0
k
k
= lim
We see from the graph that our answers are reasonable because the graph of
i 0 is that of an even function (i is an odd function) and the graph of i 00 is
that of an odd function. Furthermore, i 0 = 0 when i has a horizontal
tangent and i 00 = 0 when i 0 has a horizontal tangent.
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LIMITS AND DERIVATIVES
2({ + k)2 3 ({ + k)3 3 (2{2 3 {3 )
i ({ + k) 3 i ({)
= lim
k<0
k<0
k
k
49. i 0 ({) = lim
k(4{ + 2k 3 3{2 3 3{k 3 k2 )
= lim (4{ + 2k 3 3{2 3 3{k 3 k2 ) = 4{ 3 3{2
k<0
k<0
k
= lim
i 00 ({) = lim
k<0
4({ + k) 3 3({ + k)2 3 (4{ 3 3{2 )
i 0 ({ + k) 3 i 0 ({)
k(4 3 6{ 3 3k)
= lim
= lim
k<0
k<0
k
k
k
= lim (4 3 6{ 3 3k) = 4 3 6{
k<0
i 000 ({) = lim
k<0
i (4) ({) = lim
i 00 ({ + k) 3 i 00 ({)
[4 3 6({ + k)] 3 (4 3 6{)
36k
= lim
= lim
= lim (36) = 36
k<0
k<0 k
k<0
k
k
k<0
i 000 ({ + k) 3 i 000 ({)
36 3 (36)
0
= lim
= lim = lim (0) = 0
k<0
k<0 k
k<0
k
k
The graphs are consistent with the geometric interpretations of the
derivatives because i 0 has zeros where i has a local minimum and a local
maximum, i 00 has a zero where i 0 has a local maximum, and i 000 is a
constant function equal to the slope of i 00 .
50. (a) Since we estimate the velocity to be a maximum at w = 10, the acceleration is 0 at w = 10.
(b) Drawing a tangent line at w = 10 on the graph of d, d appears to decrease by 10 ft@s2 over a period of 20 s.
So at w = 10 s, the jerk is approximately 310@20 = 30=5 (ft@s2 )@s or ft@s3 .
51. (a) Note that we have factored { 3 d as the difference of two cubes in the third step.
i 0(d) = lim
{<d
= lim
{<d
(b) i 0(0) = lim
k<0
i ({) 3 i (d)
{1@3 3 d1@3
{1@3 3 d1@3
= lim
= lim 1@3
{<d
{<d ({
{3d
{3d
3 d1@3 )({2@3 + {1@3 d1@3 + d2@3 )
1
1
= 2@3 or 13 d32@3
{2@3 + {1@3 d1@3 + d2@3
3d
i(0 + k) 3 i(0)
= lim
k<0
k
I
3
k30
1
= lim 2@3 . This function increases without bound, so the limit does not
k<0 k
k
exist, and therefore i 0(0) does not exist.
(c) lim |i 0 ({)| = lim
{<0
{<0
1
= " and i is continuous at { = 0 (root function), so i has a vertical tangent at { = 0.
3{2@3
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SECTION 2.8
52. (a) j 0(0) = lim
{<0
(b) j 0(d) = lim
{<d
¤
j({) 3 j(0)
{2@3 3 0
1
= lim
= lim 1@3 , which does not exist.
{<0
{<0 {
{30
{
j({) 3 j(d)
{2@3 3 d2@3
({1@3 3 d1@3 )({1@3 + d1@3 )
= lim
= lim 1@3
{<d
{<d ({
{3d
{3d
3 d1@3 )({2@3 + {1@3 d1@3 + d2@3 )
= lim
{<d {2@3
2d1@3
2
{1@3 + d1@3
= 2@3 = 1@3 or
1@3
1@3
2@3
+{ d
+d
3d
3d
2 31@3
d
3
(c) j({) = {2@3 is continuous at { = 0 and
lim |j0({)| = lim
{<0
THE DERIVATIVE AS A FUNCTION
{<0
2
3 |{|1@3
(d)
= ". This shows that
j has a vertical tangent line at { = 0.
53. i ({) = |{ 3 6| =
+
{36
if { 3 6 D 6
3({ 3 6) if { 3 6 ? 0
So the right-hand limit is lim
{<6+
is lim
{<6
=
+
{ 3 6 if { D 6
6 3 { if { ? 6
i ({) 3 i (6)
|{ 3 6| 3 0
{36
= lim
= lim
= lim 1 = 1, and the left-hand limit
{36
{36
{<6+
{<6+ { 3 6
{<6+
i({) 3 i(6)
|{ 3 6| 3 0
63{
= lim
= lim
= lim (31) = 31. Since these limits are not equal,
{36
{36
{<6
{<6 { 3 6
{<6
i ({) 3 i (6)
does not exist and i is not differentiable at 6.
{36
+
1
if { A 6
0
0
However, a formula for i is i ({) =
31 if { ? 6
i 0 (6) = lim
{<6
Another way of writing the formula is i 0 ({) =
{36
.
|{ 3 6|
54. i ({) = [[{]] is not continuous at any integer q, so i is not differentiable
at q by the contrapositive of Theorem 4. If d is not an integer, then i
is constant on an open interval containing d, so i 0(d) = 0. Thus,
i 0({) = 0, { not an integer.
+ 2
if { D 0
{
55. (a) i ({) = { |{| =
2
3{ if { ? 0
(b) Since i ({) = {2 for { D 0, we have i 0 ({) = 2{ for { A 0.
[See Exercise 19(d).] Similarly, since i ({) = 3{2 for { ? 0,
we have i 0 ({) = 32{ for { ? 0. At { = 0, we have
i 0 (0) = lim
{<0
i ({) 3 i (0)
{ |{|
= lim
= lim |{| = 0=
{<0 {
{<0
{30
So i is differentiable at 0. Thus, i is differentiable for all {.
0
(c) From part (b), we have i ({) =
+
2{ if { D 0
32{ if { ? 0
,
= 2 |{|.
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LIMITS AND DERIVATIVES
i (4 + k) 3 i (4)
5 3 (4 + k) 3 1
= lim
k
k
k<0
3k
= lim
= 31
k<0 k
0
56. (a) i3
(4) = lim
(b)
k<0
and
0
(4)
i+
1
31
i (4 + k) 3 i (4)
5 3 (4 + k)
= lim
= lim
k
k
k<0+
k<0+
1 3 (1 3 k)
1
= lim
=1
= lim
k<0+ k(1 3 k)
k<0+ 1 3 k
;
0
if { $ 0
A
?
53{
if 0 ? { ? 4
(c) i ({) =
A
=
1@(5 3 {) if { D 4
At 4 we have lim i({) = lim (5 3 {) = 1 and lim i({) = lim
{<4
{<4
{<4+
{<4+
1
= 1, so lim i ({) = 1 = i (4) and i is
{<4
53{
continuous at 4. Since i (5) is not de¿ned, i is discontinuous at 5. These expressions show that i is continuous on the
intervals (3"> 0), (0> 4), (4> 5) and (5> "). Since lim i ({) = lim (5 3 {) = 5 6= 0 = lim i ({), lim i ({) does
{<0+
{<0
{<0+
{<0
not exist, so i is discontinuous (and therefore not differentiable) at 0.
0
0
(d) From (a), i is not differentiable at 4 since i3
(4) 6= i+
(4), and from (c), i is not differentiable at 0 or 5.
57. (a) If i is even, then
i 0 (3{) = lim
k<0
= lim
k<0
i (3{ + k) 3 i (3{)
i [3({ 3 k)] 3 i (3{)
= lim
k<0
k
k
i ({ 3 k) 3 i({)
i ({ 3 k) 3 i ({)
= 3 lim
k<0
k
3k
= 3 lim
{{<0
[let {{ = 3k]
i ({ + {{) 3 i ({)
= 3i 0 ({)
{{
Therefore, i 0 is odd.
(b) If i is odd, then
i 0 (3{) = lim
k<0
= lim
k<0
= lim
i (3{ + k) 3 i (3{)
i [3({ 3 k)] 3 i (3{)
= lim
k<0
k
k
3i({ 3 k) + i ({)
i ({ 3 k) 3 i({)
= lim
k<0
k
3k
{{<0
[let {{ = 3k]
i ({ + {{) 3 i ({)
= i 0 ({)
{{
Therefore, i 0 is even.
58. (a)
(b) The initial temperature of the water is close to room temperature because of the water that was in the pipes. When the
water from the hot water tank starts coming out, gW @gw is large and positive as W increases to the temperature of the water
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¤
153
in the tank. In the next phase, gW @gw = 0 as the water comes out at a constant, high temperature. After some time, gW @gw
becomes small and negative as the contents of the hot water tank are exhausted. Finally, when the hot water has run out,
gW @gw is once again 0 as the water maintains its (cold) temperature.
(c)
59.
In the right triangle in the diagram, let {| be the side opposite angle ! and {{
the side adjacent to angle !. Then the slope of the tangent line c
is p = {|@{{ = tan !. Note that 0 ? ! ?
2
.
2
We know (see Exercise 19)
0
that the derivative of i({) = { is i ({) = 2{. So the slope of the tangent to
the curve at the point (1> 1) is 2. Thus, ! is the angle between 0 and
2
whose
tangent is 2; that is, ! = tan31 2 E 63 .
2 Review
1. (a) lim i ({) = O: See De¿nition 2.2.1 and Figures 1 and 2 in Section 2.2.
{<d
(b) lim i ({) = O: See the paragraph after De¿nition 2.2.2 and Figure 9(b) in Section 2.2.
{<d+
(c) lim i({) = O: See De¿nition 2.2.2 and Figure 9(a) in Section 2.2.
{<d
(d) lim i ({) = ": See De¿nition 2.2.4 and Figure 12 in Section 2.2.
{<d
(e) lim i ({) = O: See De¿nition 2.6.1 and Figure 2 in Section 2.6.
{<"
2. In general, the limit of a function fails to exist when the function does not approach a ¿xed number. For each of the following
functions, the limit fails to exist at { = 2.
The left- and right-hand
There is an
There are an in¿nite
limits are not equal.
in¿nite discontinuity.
number of oscillations.
3. (a) – (g) See the statements of Limit Laws 1– 6 and 11 in Section 2.3.
4. See Theorem 3 in Section 2.3.
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CHAPTER 2 LIMITS AND DERIVATIVES
5. (a) See De¿nition 2.2.6 and Figures 12–14 in Section 2.2.
(b) See De¿nition 2.6.3 and Figures 3 and 4 in Section 2.6.
6. (a) | = {4 : No asymptote
(b) | = sin {: No asymptote
(c) | = tan {: Vertical asymptotes { =
2
+ q, q an integer
(d) | = tan31 {: Horizontal asymptotes | = ± 2
(e) | = h{ : Horizontal asymptote | = 0
lim h{ = 0
{<3"
(f ) | = ln {: Vertical asymptote { = 0
lim ln { = 3"
{<0+
(g) | = 1@{: Vertical asymptote { = 0, horizontal asymptote | = 0
I
(h) | = {: No asymptote
7. (a) A function i is continuous at a number d if i({) approaches i (d) as { approaches d; that is, lim i ({) = i (d).
{<d
(b) A function i is continuous on the interval (3"> ") if i is continuous at every real number d. The graph of such a
function has no breaks and every vertical line crosses it.
8. See Theorem 2.5.10.
9. See De¿nition 2.7.1.
10. See the paragraph containing Formula 3 in Section 2.7.
11. (a) The average rate of change of | with respect to { over the interval [{1 > {2 ] is
(b) The instantaneous rate of change of | with respect to { at { = {1 is lim
{2 <{1
i ({2 ) 3 i({1 )
.
{2 3 {1
i ({2 ) 3 i ({1 )
.
{2 3 {1
12. See De¿nition 2.7.4. The pages following the de¿nition discuss interpretations of i 0 (d) as the slope of a tangent line to the
graph of i at { = d and as an instantaneous rate of change of i ({) with respect to { when { = d.
13. See the paragraphs before and after Example 7 in Section 2.8.
14. (a) A function i is differentiable at a number d if its derivative i 0 exists
(b)
at { = d; that is, if i 0 (d) exists.
(c) See Theorem 2.8.4. This theorem also tells us that if i is not
continuous at d, then i is not differentiable at d.
15. See the discussion and Figure 7 on page 159.
1. False.
Limit Law 2 applies only if the individual limits exist (these don’t).
2. False.
Limit Law 5 cannot be applied if the limit of the denominator is 0 (it is).
3. True.
Limit Law 5 applies.
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CHAPTER 2 REVIEW
4. True.
¤
155
The limit doesn’t exist since i ({)@j({) doesn’t approach any real number as { approaches 5.
(The denominator approaches 0 and the numerator doesn’t.)
5. False.
Consider lim
{<5
{({ 3 5)
sin({ 3 5)
or lim
. The ¿rst limit exists and is equal to 5. By Example 3 in Section 2.2,
{<5
{35
{35
we know that the latter limit exists (and it is equal to 1).
6. False.
If i ({) = 1@{, j({) = 31@{, and d = 0, then lim i({) does not exist, lim j({) does not exist, but
{<0
{<0
lim [i({) + j({)] = lim 0 = 0 exists.
{<0
7. True.
{<0
Suppose that lim [i ({) + j({)] exists. Now lim i({) exists and lim j({) does not exist, but
{<d
{<d
{<d
lim j({) = lim {[i ({) + j({)] 3 i ({)} = lim [i ({) + j({)] 3 lim i({) [by Limit Law 2], which exists, and
{<d
{<d
{<d
{<d
we have a contradiction. Thus, lim [i ({) + j({)] does not exist.
{<d
8. False.
Consider lim [i({)j({)] = lim ({ 3 6)
{<6
{<6
so i (6)j(6) 6= 1.
9. True.
10. False.
1
. It exists (its value is 1) but i (6) = 0 and j(6) does not exist,
{36
A polynomial is continuous everywhere, so lim s({) exists and is equal to s(e).
{<e
Consider lim [i({) 3 j({)] = lim
{<0
{<0
approaches ".
1
1
3 4 . This limit is 3" (not 0), but each of the individual functions
{2
{
11. True.
See Figure 8 in Section 2.6.
12. False.
Consider i ({) = sin { for { D 0. lim i ({) 6= ±" and i has no horizontal asymptote.
{<"
+
1@({ 3 1) if { 6= 1
13. False.
Consider i ({) =
14. False.
The function i must be continuous in order to use the Intermediate Value Theorem. For example, let
+
1
if 0 $ { ? 3
i ({) =
There is no number f M [0> 3] with i (f) = 0.
31 if { = 3
15. True.
Use Theorem 2.5.8 with d = 2, e = 5, and j({) = 4{2 3 11. Note that i (4) = 3 is not needed.
16. True.
Use the Intermediate Value Theorem with d = 31, e = 1, and Q = , since 3 ? ? 4.
2
if { = 1
17. True, by the de¿nition of a limit with % = 1.
18. False.
For example, let i ({) =
+2
{ + 1 if { 6= 0
2
if { = 0
Then i({) A 1 for all {, but lim i ({) = lim {2 + 1 = 1.
{<0
19. False.
{<0
See the note after Theorem 4 in Section 2.8.
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CHAPTER 2 LIMITS AND DERIVATIVES
i 0(u) exists i i is differentiable at u
20. True.
g 2|
is the second derivative while
g{2
2
g|
g 2|
then
=
0,
but
= 1.
g{2
g{
21. False.
g|
g{
i i is continuous at u
2
i
lim i ({) = i (u).
{<u
is the ¿rst derivative squared. For example, if | = {,
i ({) = {10 3 10{2 + 5 is continuous on the interval [0> 2], i (0) = 5, i (1) = 34, and i (2) = 989. Since
22. True.
34 ? 0 ? 5, there is a number f in (0> 1) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a
root of the equation {10 3 10{2 + 5 = 0 in the interval (0> 1). Similarly, there is a root in (1> 2).
23. True.
See Exercise 68(b) in Section 2.5.
24. False
See Exercise 68(c) in Section 2.5.
1. (a) (i) lim i ({) = 3
(ii)
{<2+
lim i ({) = 0
{<33+
(iii) lim i ({) does not exist since the left and right limits are not equal. (The left limit is 32.)
{<33
(iv) lim i ({) = 2
{<4
(v) lim i ({) = "
(vi) lim i ({) = 3"
(vii) lim i ({) = 4
(viii) lim i ({) = 31
{<0
{<2
{<"
{<3"
(b) The equations of the horizontal asymptotes are | = 31 and | = 4.
(c) The equations of the vertical asymptotes are { = 0 and { = 2.
(d) i is discontinuous at { = 33, 0, 2, and 4. The discontinuities are jump, in¿nite, in¿nite, and removable, respectively.
2.
lim i({) = 32,
{<"
lim i ({) = 3",
{<3+
{<3"
{<3
lim i ({) = 0,
lim i ({) = ",
{<33
lim i ({) = 2,
i is continuous from the right at 3
3. Since the exponential function is continuous, lim h{
{<1
4. Since rational functions are continuous, lim
{<3 {2
5. lim
{<33
6. lim
{<1+
3 3{
= h131 = h0 = 1.
{2 3 9
32 3 9
0
= 2
=
= 0.
+ 2{ 3 3
3 + 2(3) 3 3
12
{2
{2 3 9
({ + 3)({ 3 3)
{33
33 3 3
36
3
= lim
= lim
=
=
=
+ 2{ 3 3 {<33 ({ + 3)({ 3 1) {<33 { 3 1
33 3 1
34
2
{2
{2 3 9
{2 3 9
= 3" since {2 + 2{ 3 3 < 0+ as { < 1+ and 2
? 0 for 1 ? { ? 3.
+ 2{ 3 3
{ + 2{ 3 3
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CHAPTER 2 REVIEW
¤
3
k 3 3k2 + 3k 3 1 + 1
(k 3 1)3 + 1
k3 3 3k2 + 3k
= lim
= lim
= lim k2 3 3k + 3 = 3
7. lim
k<0
k<0
k<0
k<0
k
k
k
Another solution: Factor the numerator as a sum of two cubes and then simplify.
[(k 3 1) + 1] (k 3 1)2 3 1(k 3 1) + 12
(k 3 1)3 + 1
(k 3 1)3 + 13
lim
= lim
= lim
k<0
k<0
k<0
k
k
k
2
= lim (k 3 1) 3 k + 2 = 1 3 0 + 2 = 3
k<0
8. lim
w<2
(w + 2)(w 3 2)
w2 3 4
w+2
2+2
4
1
= lim
= lim
=
=
=
w3 3 8 w<2 (w 3 2)(w2 + 2w + 4) w<2 w2 + 2w + 4
4+4+4
12
3
I
I
u
u
4
+
=
"
since
(u
3
9)
<
0
as
u
<
9
and
A 0 for u 6= 9.
u<9 (u 3 9)4
(u 3 9)4
9. lim
10. lim
y<4+
43y
43y
1
= lim
= lim
= 31
|4 3 y| y<4+ 3(4 3 y) y<4+ 31
11. lim
x<1 x3
x4 3 1
(x2 + 1)(x2 3 1)
(x2 + 1)(x + 1)(x 3 1)
(x2 + 1)(x + 1)
2(2)
4
= lim
= lim
= lim
=
=
2
2
x<1
x<1
+ 5x 3 6x x<1 x(x + 5x 3 6)
x(x + 6)(x 3 1)
x(x + 6)
1(7)
7
I
I
I
I
{+63{
{+63{
{+6+{
( { + 6 )2 3 {2
I
I
=
lim
=
lim
·
{<3 {3 3 3{2
{<3
{<3 {2 ({ 3 3)
{2 ({ 3 3)
{+6+{
{+6+{
12. lim
= lim
{<3
= lim
{<3
13. Since { is positive,
{ + 6 3 {2
3({2 3 { 3 6)
3({ 3 3)({ + 2)
I
= lim
I
= lim
I
{<3 {2 ({ 3 3)
{<3 {2 ({ 3 3)
{2 ({ 3 3) { + 6 + {
{+6+{
{+6+{
{2
3({ + 2)
5
5
I
=3
=3
9(3 + 3)
54
{+6+{
I
{2 = |{| = {. Thus,
s
I
I
I
I
1 3 9@{2
{2 3 9@ {2
{2 3 9
130
1
= lim
= lim
=
=
{<" 2{ 3 6
{<" (2{ 3 6)@{
{<"
2 3 6@{
230
2
lim
14. Since { is negative,
I
{2 = |{| = 3{. Thus,
lim
{<3"
s
I
I
I
I
1 3 9@{2
{2 3 9
{2 3 9@ {2
130
1
= lim
= lim
=
=3
{<3" (2{ 3 6)@(3{)
{<3" 32 + 6@{
2{ 3 6
32 + 0
2
15. Let w = sin {. Then as { < 3 , sin { < 0+ , so w < 0+ . Thus, lim ln(sin {) = lim ln w = 3".
{< 16.
lim
{<3"
w<0+
1 3 2{2 3 {4
(1 3 2{2 3 {4 )@{4
1@{4 3 2@{2 3 1
03031
31
1
=
=
=
= lim
= lim
4
4
4
{<3" (5 + { 3 3{ )@{
{<3" 5@{4 + 1@{3 3 3
5 + { 3 3{
0+033
33
3
I
I 2
{ + 4{ + 1 3 {
{2 + 4{ + 1 + {
({2 + 4{ + 1) 3 {2
·I
= lim I
{<"
{<"
1
{2 + 4{ + 1 + {
{2 + 4{ + 1 + {
l
k
I
(4{ + 1)@{
divide by { = {2 for { A 0
= lim I
{<" (
{2 + 4{ + 1 + {)@{
I
17. lim
{2 + 4{ + 1 3 { = lim
{<"
4
4 + 1@{
4+0
= lim s
= =2
= I
{<"
2
1+0+0+1
1 + 4@{ + 1@{2 + 1
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158
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CHAPTER 2 LIMITS AND DERIVATIVES
2
18. Let w = { 3 {2 = {(1 3 {). Then as { < ", w < 3", and lim h{3{ = lim hw = 0.
{<"
w<3"
19. Let w = 1@{. Then as { < 0+ , w < " , and lim tan31 (1@{) = lim tan31 w =
w<"
{<0+
20. lim
{<1
1
1
+ 2
{31
{ 3 3{ + 2
.
2
1
{32
1
1
+
= lim
+
{<1 { 3 1
{<1 ({ 3 1)({ 3 2)
({ 3 1)({ 3 2)
({ 3 1)({ 3 2)
1
{31
1
= lim
=
= 31
= lim
{<1 ({ 3 1)({ 3 2)
{<1 { 3 2
132
= lim
21. From the graph of | = cos2 { @{2 , it appears that | = 0 is the horizontal
asymptote and { = 0 is the vertical asymptote. Now 0 $ (cos {)2 $ 1 i
cos2 {
1
0
$
$ 2
2
{
{2
{
lim
{<±"
i 0$
cos2 {
1
$ 2 . But lim 0 = 0 and
{<±"
{2
{
1
cos2 {
= 0, so by the Squeeze Theorem, lim
= 0.
2
{<±"
{
{2
cos2 {
= " because cos2 { < 1 and {2 < 0+ as { < 0, so { = 0 is the
{<0
{2
Thus, | = 0 is the horizontal asymptote. lim
vertical asymptote.
22. From the graph of | = i ({) =
I
I
{2 + { + 1 3 {2 3 {, it appears that there are 2 horizontal asymptotes and possibly 2
vertical asymptotes. To obtain a different form for i , let’s multiply and divide it by its conjugate.
I
I
I
I
{2 + { + 1 + {2 3 {
({2 + { + 1) 3 ({2 3 {)
I
I
{2 + { + 1 3 {2 3 { I
= I
i1 ({) =
{2 + { + 1 + {2 3 {
{2 + { + 1 + {2 3 {
2{ + 1
I
= I
{2 + { + 1 + {2 3 {
Now
2{ + 1
I
lim i1 ({) = lim I
{<"
{2 + { + 1 + {2 3 {
{<"
2 + (1@{)
s
= lim s
{<"
1 + (1@{) + (1@{2 ) + 1 3 (1@{)
=
[since
I
{2 = { for { A 0]
2
= 1,
1+1
so | = 1 is a horizontal asymptote. For { ? 0, we have
I
{2 = |{| = 3{, so when we divide the denominator by {,
with { ? 0, we get
&
%u
u
I
I
I
I
{2 + { + 1 + {2 3 {
{2 + { + 1 + {2 3 {
1
1
1
I
=3
=3
1+ + 2 + 13
{
{
{
{
{2
Therefore,
2{ + 1
2 + (1@{)
ks
l
I
= lim
lim i1 ({) = lim I
s
2
2
{<3"
{<"
{ +{+1+ { 3{
3
1 + (1@{) + (1@{2 ) + 1 3 (1@{)
{<3"
=
2
= 31>
3(1 + 1)
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CHAPTER 2 REVIEW
159
so | = 31 is a horizontal asymptote.
The domain of i is (3"> 0] [1> "). As { < 03 , i ({) < 1, so
I
{ = 0 is not a vertical asymptote. As { < 1+ , i ({) < 3, so { = 1
is not a vertical asymptote and hence there are no vertical asymptotes.
23. Since 2{ 3 1 $ i({) $ {2 for 0 ? { ? 3 and lim (2{ 3 1) = 1 = lim {2 , we have lim i ({) = 1 by the Squeeze Theorem.
{<1
{<1
{<1
24. Let i({) = 3{2 , j({) = {2 cos 1@{2 and k({) = {2 . Then since cos 1@{2 $ 1 for { 6= 0, we have
i ({) $ j({) $ k({) for { 6= 0, and so lim i ({) = lim k({) = 0 i
{<0
{<0
lim j({) = 0 by the Squeeze Theorem.
{<0
25. Given % A 0, we need A 0 such that if 0 ? |{ 3 2| ? , then |(14 3 5{) 3 4| ? %. But |(14 3 5{) 3 4| ? %
|35{ + 10| ? %
C
C
|35| |{ 3 2| ? % C |{ 3 2| ? %@5. So if we choose = %@5, then 0 ? |{ 3 2| ? i
|(14 3 5{) 3 4| ? %. Thus, lim (14 3 5{) = 4 by the de¿nition of a limit.
{<2
I
I
I
3
{ 3 0| ? %. Now | 3 { 3 0| = | 3 {| ? % i
I
s
I 3
I
I
3
|{| = | 3 {| ? %3 . So take = %3 . Then 0 ? |{ 3 0| = |{| ? %3 i | 3 { 3 0| = | 3 {| = 3 |{| ? %3 = %.
I
Therefore, by the de¿nition of a limit, lim 3 { = 0.
26. Given % A 0 we must ¿nd A 0 so that if 0 ? |{ 3 0| ? , then |
{<0
27. Given % A 0, we need A 0 so that if 0 ? |{ 3 2| ? , then {2 3 3{ 3 (32) ? %. First, note that if |{ 3 2| ? 1, then
31 ? { 3 2 ? 1, so 0 ? { 3 1 ? 2 i |{ 3 1| ? 2. Now let = min {%@2> 1}. Then 0 ? |{ 3 2| ? 2
{ 3 3{ 3 (32) = |({ 3 2)({ 3 1)| = |{ 3 2| |{ 3 1| ? (%@2)(2) = %.
i
Thus, lim ({2 3 3{) = 32 by the de¿nition of a limit.
{<2
I
28. Given P A 0, we need A 0 such that if 0 ? { 3 4 ? , then 2@ { 3 4 A P. This is true
{ 3 4 ? 4@P 2 . So if we choose = 4@P 2 , then 0 ? { 3 4 ? I
lim 2@ { 3 4 = ".
C
I
{ 3 4 ? 2@P
I
i 2@ { 3 4 A P. So by the de¿nition of a limit,
{<4+
29. (a) i ({) =
I
3{ if { ? 0, i ({) = 3 3 { if 0 $ { ? 3, i ({) = ({ 3 3)2 if { A 3.
(i) lim i ({) = lim (3 3 {) = 3
{<0+
{<0+
(iii) Because of (i) and (ii), lim i ({) does not exist.
{<0
2
(v) lim i ({) = lim ({ 3 3) = 0
{<3+
{<3+
(b) i is discontinuous at 0 since lim i ({) does not exist.
{<0
C
(ii) lim i ({) = lim
{<0
{<0
I
3{ = 0
(iv) lim i ({) = lim (3 3 {) = 0
{<3
{<3
(vi) Because of (iv) and (v), lim i ({) = 0.
{<3
(c)
i is discontinuous at 3 since i (3) does not exist.
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CHAPTER 2 LIMITS AND DERIVATIVES
30. (a) j({) = 2{ 3 {2 if 0 $ { $ 2, j({) = 2 3 { if 2 ? { $ 3, j({) = { 3 4 if 3 ? { ? 4, j({) = if { D 4.
Therefore, lim j({) = lim
{<2
{<2
2{ 3 {2 = 0 and lim j({) = lim (2 3 {) = 0. Thus, lim j({) = 0 = j (2),
{<2+
{<2
{<2+
so j is continuous at 2. lim j({) = lim (2 3 {) = 31 and lim j({) = lim ({ 3 4) = 31. Thus,
{<3
{<3
{<3+
{<3+
lim j({) = 31 = j(3), so j is continuous at 3. lim j({) = lim ({ 3 4) = 0 and lim j({) = lim = .
{<3
{<4
{<4
{<4+
{<4+
Thus, lim j({) does not exist, so j is discontinuous at 4. But lim j({) = = j(4), so j is continuous from the
{<4
{<4+
right at 4.
(b)
31. sin { and h{ are continuous on R by Theorem 7 in Section 2.5. Since h{ is continuous on R, hsin { is continuous on R by
Theorem 9 in Section 2.5. Lastly, { is continuous on R since it’s a polynomial and the product {hsin { is continuous on its
domain R by Theorem 4 in Section 2.5.
32. {2 3 9 is continuous on R since it is a polynomial and
composition
I
{ is continuous on [0> ") by Theorem 7 in Section 2.5, so the
I
{2 3 9 is continuous on { | {2 3 9 D 0 = (3"> 33] [3> ") by Theorem 9. Note that {2 3 2 6= 0 on this
I
{2 3 9
set and so the quotient function j({) = 2
is continuous on its domain, (3"> 33] [3> ") by Theorem 4.
{ 32
33. i ({) = {5 3 {3 + 3{ 3 5 is continuous on the interval [1> 2], i (1) = 32, and i(2) = 25. Since 32 ? 0 ? 25, there is a
number f in (1> 2) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation
{5 3 {3 + 3{ 3 5 = 0 in the interval (1> 2).
34. i ({) = cos
I
{ 3 h{ + 2 is continuous on the interval [0> 1], i(0) = 2, and i (1) r 30=2. Since 30=2 ? 0 ? 2, there is a
number f in (0> 1) such that i (f) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation
I
I
cos { 3 h{ + 2 = 0, or cos { = h{ 3 2, in the interval (0> 1).
35. (a) The slope of the tangent line at (2> 1) is
i ({) 3 i (2)
32({2 3 4)
32({ 3 2)({ + 2)
9 3 2{2 3 1
8 3 2{2
= lim
= lim
= lim
= lim
{<2
{<2
{<2 { 3 2
{<2
{<2
{32
{32
{32
{32
lim
= lim [32({ + 2)] = 32 · 4 = 38
{<2
(b) An equation of this tangent line is | 3 1 = 38({ 3 2) or | = 38{ + 17.
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161
36. For a general point with {-coordinate d, we have
2@(1 3 3{) 3 2@(1 3 3d)
2(1 3 3d) 3 2(1 3 3{)
6({ 3 d)
= lim
= lim
{<d (1 3 3d)(1 3 3{)({ 3 d)
{<d (1 3 3d)(1 3 3{)({ 3 d)
{3d
p = lim
{<d
6
6
=
(1 3 3d)(1 3 3{)
(1 3 3d)2
= lim
{<d
For d = 0, p = 6 and i (0) = 2, so an equation of the tangent line is | 3 2 = 6({ 3 0) or | = 6{ + 2= For d = 31, p =
and i (31) = 12 , so an equation of the tangent line is | 3
1
2
3
8
= 38 ({ + 1) or | = 38 { + 78 .
37. (a) v = v(w) = 1 + 2w + w2 @4. The average velocity over the time interval [1> 1 + k] is
yave =
1 + 2(1 + k) + (1 + k)2 4 3 13@4
10k + k2
10 + k
v(1 + k) 3 v(1)
=
=
=
(1 + k) 3 1
k
4k
4
So for the following intervals the average velocities are:
(i) [1> 3]: k = 2, yave = (10 + 2)@4 = 3 m@s
(iii) [1> 1=5]: k = 0=5, yave = (10 + 0=5)@4 = 2=625 m@s
(b) When w = 1, the instantaneous velocity is lim
k<0
(ii) [1> 2]: k = 1, yave = (10 + 1)@4 = 2=75 m@s
(iv) [1> 1=1]: k = 0=1, yave = (10 + 0=1)@4 = 2=525 m@s
v(1 + k) 3 v(1)
10 + k
10
= lim
=
= 2=5 m@s.
k<0
k
4
4
38. (a) When Y increases from 200 in3 to 250 in3 , we have {Y = 250 3 200 = 50 in3 , and since S = 800@Y ,
{S = S (250) 3 S (200) =
is
800
800
3
= 3=2 3 4 = 30=8 lb@in2 . So the average rate of change
250
200
{S
30=8
lb@in2
.
=
= 30=016
{Y
50
in3
(b) Since Y = 800@S , the instantaneous rate of change of Y with respect to S is
lim
k<0
{Y
Y (S + k) 3 Y (S )
800@(S + k) 3 800@S
800 [S 3 (S + k)]
= lim
= lim
= lim
k<0
k<0
k<0
{S
k
k
k(S + k)S
= lim
k<0
3800
800
=3 2
(S + k)S
S
which is inversely proportional to the square of S .
39. (a) i 0 (2) = lim
{<2
= lim
{<2
i({) 3 i (2)
{3 3 2{ 3 4
= lim
{<2
{32
{32
(c)
({ 3 2)({2 + 2{ + 2)
= lim ({2 + 2{ + 2) = 10
{<2
{32
(b) | 3 4 = 10({ 3 2) or | = 10{ 3 16
40. 26 = 64, so i ({) = {6 and d = 2.
41. (a) i 0 (u) is the rate at which the total cost changes with respect to the interest rate. Its units are dollars@(percent per year).
(b) The total cost of paying off the loan is increasing by $1200@(percent per year) as the interest rate reaches 10%. So if the
interest rate goes up from 10% to 11%, the cost goes up approximately $1200.
(c) As u increases, F increases. So i 0 (u) will always be positive.
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CHAPTER 2 LIMITS AND DERIVATIVES
42.
43.
45. (a) i 0 ({) = lim
k<0
i ({ + k) 3 i ({)
= lim
k<0
k
= lim
k<0
k
44.
s
s
I
I
3 3 5({ + k) 3 3 3 5{ 3 3 5({ + k) + 3 3 5{
s
I
k
3 3 5({ + k) + 3 3 5{
[3 3 5({ + k)] 3 (3 3 5{)
35
35
s
= lim s
= I
I
I
k<0
2 3 3 5{
3 3 5({ + k) + 3 3 5{
3 3 5({ + k) + 3 3 5{
(b) Domain of i : (the radicand must be nonnegative) 3 3 5{ D 0 i
5{ $ 3 i { M 3"> 35
Domain of i 0 : exclude
{ M 3"> 35
3
5
because it makes the denominator zero;
(c) Our answer to part (a) is reasonable because i 0 ({) is always negative and
i is always decreasing.
46. (a) As { < ±", i ({) = (4 3 {)@(3 + {) < 31, so there is a horizontal
asymptote at | = 31. As { < 33+ , i ({) < ", and as { < 333 ,
i ({) < 3". Thus, there is a vertical asymptote at { = 33.
(b) Note that i is decreasing on (3"> 33) and (33> "), so i 0 is negative on
those intervals. As { < ±", i 0 < 0. As { < 333 and as { < 33+ ,
i 0 < 3".
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163
4 3 ({ + k)
43{
3
i({
+
k)
3
i
({)
3
+
({
+
k)
3+{
(3 + {) [4 3 ({ + k)] 3 (4 3 {) [3 + ({ + k)]
= lim
= lim
(c) i 0 ({) = lim
k<0
k<0
k<0
k
k
k [3 + ({ + k)] (3 + {)
= lim
(12 3 3{ 3 3k + 4{ 3 {2 3 k{) 3 (12 + 4{ + 4k 3 3{ 3 {2 3 k{)
k[3 + ({ + k)](3 + {)
= lim
37k
37
7
= lim
=3
k [3 + ({ + k)] (3 + {) k<0 [3 + ({ + k)] (3 + {)
(3 + {)2
k<0
k<0
(d) The graphing device con¿rms our graph in part (b).
47. i is not differentiable: at { = 34 because i is not continuous, at { = 31 because i has a corner, at { = 2 because i is not
continuous, and at { = 5 because i has a vertical tangent.
48. The graph of d has tangent lines with positive slope for { ? 0 and negative slope for { A 0, and the values of f ¿t this pattern,
so f must be the graph of the derivative of the function for d. The graph of f has horizontal tangent lines to the left and right of
the {-axis and e has zeros at these points. Hence, e is the graph of the derivative of the function for f. Therefore, d is the graph
of i , f is the graph of i 0 , and e is the graph of i 00 .
49. F 0 (1990) is the rate at which the total value of US currency in circulation is changing in billions of dollars per year. To
estimate the value of F 0 (1990), we will average the difference quotients obtained using the times w = 1985 and w = 1995.
Let D =
E=
187=3 3 271=9
384=6
F(1985) 3 F(1990)
=
=
= 16=92 and
1985 3 1990
35
35
F(1995) 3 F(1990)
409=3 3 271=9
137=4
=
=
= 27=48. Then
1995 3 1990
5
5
F 0 (1990) = lim
w<1990
F(w) 3 F(1990)
D+E
16=92 + 27=48
44=4
E
=
=
= 22=2 billion dollars@year.
w 3 1990
2
2
2
50. (a) Drawing slope triangles, we obtain the following estimates: I 0 (1950) E
and I 0 (1987) E
0=2
10
1=1
10
= 0=11, I 0 (1965) E
31=6
10
= 30=16,
= 0=02.
(b) The rate of change of the average number of children born to each woman was increasing by 0=11 in 1950, decreasing
by 0=16 in 1965, and increasing by 0=02 in 1987.
(c) There are many possible reasons:
• In the baby-boom era (post-WWII), there was optimism about the economy and family size was rising.
• In the baby-bust era, there was less economic optimism, and it was considered less socially responsible to have a
large family.
• In the baby-boomlet era, there was increased economic optimism and a return to more conservative attitudes.
51. |i ({)| $ j({)
C 3j({) $ i ({) $ j({) and lim j({) = 0 = lim 3j({).
{<d
{<d
Thus, by the Squeeze Theorem, lim i ({) = 0.
{<d
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CHAPTER 2 LIMITS AND DERIVATIVES
52. (a) Note that i is an even function since i({) = i (3{). Now for any integer q,
[[q]] + [[3q]] = q 3 q = 0, and for any real number n which is not an integer,
[[n]] + [[3n]] = [[n]] + (3 [[n]] 3 1) = 31. So lim i ({) exists (and is equal to 31)
{<d
for all values of d.
(b) i is discontinuous at all integers.
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PROBLEMS PLUS
1. Let w =
I
6
{, so { = w6 . Then w < 1 as { < 1, so
I
3
{31
w2 3 1
(w 3 1)(w + 1)
w+1
1+1
2
lim I
= lim 3
= lim
= lim 2
= 2
= .
{<1
w<1 w 3 1
w<1 (w 3 1) (w2 + w + 1)
w<1 w + w + 1
1 +1+1
3
{31
I
I
I
3
{2 + 3 { + 1 .
Another method: Multiply both the numerator and the denominator by ( { + 1)
I
I
d{ + e 3 2
d{ + e + 2
d{ + e 3 4
. Now since the denominator
2. First rationalize the numerator: lim
= lim I
·I
{<0
{
d{ + e + 2 {<0 { d{ + e + 2
approaches 0 as { < 0, the limit will exist only if the numerator also approaches 0 as { < 0. So we require that
d
=1 i
d(0) + e 3 4 = 0 i e = 4. So the equation becomes lim I
{<0
d{ + 4 + 2
d
I
= 1 i d = 4.
4+2
Therefore, d = e = 4.
3. For 3 12 ? { ?
Therefore, lim
{<0
1
,
2
we have 2{ 3 1 ? 0 and 2{ + 1 A 0, so |2{ 3 1| = 3(2{ 3 1) and |2{ + 1| = 2{ + 1.
|2{ 3 1| 3 |2{ + 1|
3(2{ 3 1) 3 (2{ + 1)
34{
= lim
= lim
= lim (34) = 34.
{<0
{<0
{<0
{
{
{
4. Let U be the midpoint of RS , so the coordinates of U are
Since the slope pRS =
1
2
{> 12 {2 since the coordinates of S are {> {2 . Let T = (0> d).
{2
1
= {, pTU = 3 (negative reciprocal). But pTU =
{
{
1 2
{ 3d
2
1
2{ 3 0
=
{2 3 2d
, so we conclude that
{
31 = {2 3 2d i 2d = {2 + 1 i d = 12 {2 + 12 . As { < 0, d < 12 > and the limiting position of T is 0> 12 .
5. (a) For 0 ? { ? 1, [[{]] = 0, so
lim
{<0
[[{]]
= lim
{
{<0
31
{
[[{]]
[[{]]
31
[[{]]
= 0, and lim
= 0. For 31 ? { ? 0, [[{]] = 31, so
=
, and
{
{
{
{<0+ {
= ". Since the one-sided limits are not equal, lim
{<0
[[{]]
does not exist.
{
(b) For { A 0, 1@{ 3 1 $ [[1@{]] $ 1@{ i {(1@{ 3 1) $ {[[1@{]] $ {(1@{) i 1 3 { $ {[[1@{]] $ 1.
As { < 0+ , 1 3 { < 1, so by the Squeeze Theorem, lim {[[1@{]] = 1.
{<0+
For { ? 0, 1@{ 3 1 $ [[1@{]] $ 1@{ i {(1@{ 3 1) D {[[1@{]] D {(1@{) i 1 3 { D {[[1@{]] D 1.
As { < 03 , 1 3 { < 1, so by the Squeeze Theorem, lim {[[1@{]] = 1.
{<0
Since the one-sided limits are equal, lim {[[1@{]] = 1.
{<0
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CHAPTER 2 PROBLEMS PLUS
2
2
2
6. (a) [[{]] + [[|]] = 1. Since [[{]] and [[|]]2 are positive integers or 0, there are
only 4 cases:
Case (i): [[{]] = 1, [[|]] = 0
i1 $ { ? 2 and 0 $ | ? 1
Case (ii): [[{]] = 31, [[|]] = 0 i31 $ { ? 0 and 0 $ | ? 1
Case (iii):[[{]] = 0, [[|]] = 1
i0 $ { ? 1 and 1 $ | ? 2
Case (iv): [[{]] = 0, [[|]] = 31 i0 $ { ? 1 and 31 $ | ? 0
(b) [[{]]2 3 [[|]]2 = 3. The only integral solution of q2 3 p2 = 3 is q = ±2
and p = ±1. So the graph is
{({> |) | [[{]] = ±2, [[|]] = ±1} =
,
2 $ { $ 3 or 32 $ { ? 1>
.
({> |) 1 $ | ? 2 or 31 $ | ? 0
+
(c) [[{ + |]]2 = 1 i [[{ + |]] = ±1 i 1 $ { + | ? 2
or 31 $ { + | ? 0
(d) For q $ { ? q + 1, [[{]] = q. Then [[{]] + [[|]] = 1 i [[|]] = 1 3 q i
1 3 q $ | ? 2 3 q. Choosing integer values for q produces the graph.
7. i is continuous on (3"> d) and (d> "). To make i continuous on R, we must have continuity at d. Thus,
lim i ({) = lim i ({) i
{<d+
{<d
lim {2 = lim ({ + 1) i d2 = d + 1 i d2 3 d 3 1 = 0 i
{<d+
{<d
I [by the quadratic formula] d = 1 ± 5 2 E 1=618 or 30=618.
8. (a) Here are a few possibilities:
(b) The “obstacle” is the line { = | (see diagram). Any intersection of the graph of i with the line | = { constitutes a ¿xed
point, and if the graph of the function does not cross the line somewhere in (0> 1), then it must either start at (0> 0)
(in which case 0 is a ¿xed point) or ¿nish at (1> 1) (in which case 1 is a ¿xed point).
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CHAPTER 2 PROBLEMS PLUS
¤
167
(c) Consider the function I ({) = i({) 3 {, where i is any continuous function with domain [0> 1] and range in [0> 1]. We
shall prove that i has a ¿xed point. Now if i (0) = 0 then we are done: i has a ¿xed point (the number 0), which is what
we are trying to prove. So assume i (0) 6= 0. For the same reason we can assume that i (1) 6= 1. Then I (0) = i (0) A 0
and I (1) = i(1) 3 1 ? 0. So by the Intermediate Value Theorem, there exists some number f in the interval (0> 1) such
that I (f) = i (f) 3 f = 0. So i (f) = f, and therefore i has a ¿xed point.
9. lim i ({) = lim
{<d
{<d
=
1
2
1
2
·2+
and lim j({) = lim
{<d
[i ({) + j({)] +
{<d
1
2
1
2
· 1 = 32 ,
[i({) 3 j({)] =
+ j({)] +
1
lim [i ({)
2 {<d
[i ({) + j({)] 3 i ({) = lim [i ({) + j({)] 3 lim i({) = 2 3
{<d
k
lk
l
So lim [i ({)j({)] = lim i({) lim j({) =
{<d
1
lim [i ({)
2 {<d
{<d
{<d
3
2
·
1
2
{<d
3
2
3 j({)]
= 12 .
= 34 .
Another solution: Since lim [i ({) + j({)] and lim [i ({) 3 j({)] exist, we must have
{<d
lim [i ({) + j({)]2 =
{<d
lim [i ({) j({)] =
{<d
=
{<d
lim 1
{<d 4
1
4
{<d
2
2
lim [i ({) + j({)] and lim [i ({) 3 j({)]2 = lim [i ({) 3 j({)] , so
{<d
{<d
[i ({) + j({)]2 3 [i ({) 3 j({)]2
[because all of the i 2 and j2 cancel]
lim [i ({) + j({)]2 3 lim [i ({) 3 j({)]2 = 14 22 3 12 = 34 .
{<d
{<d
10. (a) Solution 1: We introduce a coordinate system and drop a perpendicular
from S , as shown. We see from _QFS that tan 2 =
|
, and from
13{
_QES that tan = |@{ . Using the double-angle formula for tangents,
we get
2 tan 2( |@{)
|
= tan 2 =
=
. After a bit of
13{
1 3 tan2 1 3 ( |@{)2
simpli¿cation, this becomes
2{
1
= 2
13{
{ 3 |2
C | 2 = { (3{ 3 2).
As the altitude DP decreases in length, the point S will approach the {-axis, that is, | < 0, so the limiting location of S
must be one of the roots of the equation {(3{ 3 2) = 0. Obviously it is not { = 0 (the point S can never be to the left of
the altitude DP, which it would have to be in order to approach 0) so it must be 3{ 3 2 = 0, that is, { = 23 .
Solution 2: We add a few lines to the original diagram, as shown. Now note
that _ES T = _S EF (alternate angles; TS k EF by symmetry) and
similarly _FTS = _TFE. So {ES T and {FTS are isosceles, and
the line segments ET, TS and S F are all of equal length. As |DP| < 0,
S and T approach points on the base, and the point S is seen to approach a
position two-thirds of the way between E and F, as above.
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CHAPTER 2 PROBLEMS PLUS
(b) The equation | 2 = {(3{ 3 2) calculated in part (a) is the equation of
the curve traced out by S . Now as |DP| < ", 2 <
,
2
<
,
4
{ < 1, and since tan = |@{, | < 1. Thus, S only traces out the
part of the curve with 0 $ | ? 1.
11. (a) Consider J({) = W ({ + 180 ) 3 W ({). Fix any number d. If J(d) = 0, we are done: Temperature at d = Temperature
at d + 180 . If J(d) A 0, then J(d + 180 ) = W (d + 360 ) 3 W (d + 180 ) = W (d) 3 W (d + 180 ) = 3J(d) ? 0.
Also, J is continuous since temperature varies continuously. So, by the Intermediate Value Theorem, J has a zero on the
interval [d> d + 180 ]. If J(d) ? 0, then a similar argument applies.
(b) Yes. The same argument applies.
(c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that
altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity.
j({ + k) 3 j({)
({ + k)i ({ + k) 3 {i ({)
{i ({ + k) 3 {i ({)
ki({ + k)
= lim
= lim
+
k<0
k<0
k<0
k
k
k
k
12. j 0 ({) = lim
= { lim
k<0
i ({ + k) 3 i ({)
+ lim i ({ + k) = {i 0 ({) + i ({)
k<0
k
because i is differentiable and therefore continuous.
13. (a) Put { = 0 and | = 0 in the equation: i (0 + 0) = i(0) + i (0) + 02 · 0 + 0 · 02
i i (0) = 2i (0).
Subtracting i(0) from each side of this equation gives i (0) = 0.
i(0) + i (k) + 02 k + 0k2 3 i (0)
i (0 + k) 3 i (0)
i (k)
i({)
(b) i (0) = lim
= lim
= lim
= lim
=1
{<0
k<0
k<0
k<0
k
k
k
{
0
i ({) + i (k) + {2 k + {k2 3 i ({)
i ({ + k) 3 i ({)
i (k) + {2 k + {k2
= lim
= lim
(c) i ({) = lim
k<0
k<0
k<0
k
k
k
i (k)
+ {2 + {k = 1 + {2
= lim
k<0
k
0
14. We are given that |i ({)| $ {2 for all {. In particular, |i (0)| $ 0, but |d| D 0 for all d. The only conclusion is
2
{ i ({) 3 i (0) i ({) |i ({)|
i ({) 3 i (0)
{2
=
=
that i (0) = 0. Now $
=
= |{| i 3|{| $
$ |{|.
{30
{
|{|
|{|
|{|
{30
But lim (3|{|) = 0 = lim |{|, so by the Squeeze Theorem, lim
{<0
{<0
{<0
i ({) 3 i (0)
= 0. So by the de¿nition of a derivative,
{30
i is differentiable at 0 and, furthermore, i 0 (0) = 0.
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NOT FOR SALE
3
DIFFERENTIATION RULES
3.1 Derivatives of Polynomials and Exponential Functions
hk 3 1
= 1.
k<0
k
1. (a) h is the number such that lim
(b)
{
2=7{ 3 1
{
{
2=8{ 3 1
{
30=001
30=0001
0=001
0=0001
0=9928
0=9932
0=9937
0=9933
30=001
30=0001
0=001
0=0001
1=0291
1=0296
1=0301
1=0297
2. (a)
From the tables (to two decimal places),
lim
k<0
2=7k 3 1
2=8k 3 1
= 0=99 and lim
= 1=03.
k<0
k
k
Since 0=99 ? 1 ? 1=03, 2=7 ? h ? 2=8.
The function value at { = 0 is 1 and the slope at { = 0 is 1.
(b) i ({) = h{ is an exponential function and j({) = {h is a power function.
g {
g
(h ) = h{ and
({h ) = h{h31 .
g{
g{
(c) i ({) = h{ grows more rapidly than j({) = {h when { is large.
3. i ({) = 240 is a constant function, so its derivative is 0, that is, i 0 ({) = 0.
4. i ({) = h5 is a constant function, so its derivative is 0, that is, i 0 ({) = 0.
i i 0 (w) = 0 3
5. i (w) = 2 3 23 w
6. I ({) =
3 8
{
4
2
3
= 3 23
i I 0 ({) = 34 (8{7 ) = 6{7
7. i ({) = {3 3 4{ + 6
i i 0 ({) = 3{2 3 4(1) + 0 = 3{2 3 4
8. i (w) = 1=4w5 3 2=5w2 + 6=7
i i 0 (w) = 1=4(5w4 ) 3 2=5(2w) + 0 = 7w4 3 5w
9. j({) = {2 (1 3 2{) = {2 3 2{3
i j 0 ({) = 2{ 3 2(3{2 ) = 2{ 3 6{2
10. k({) = ({ 3 2)(2{ + 3) = 2{2 3 { 3 6
11. j(w) = 2w33@4
12. E(|) = f| 36
13. D(v) = 3
i
i k0 ({) = 2(2{) 3 1 3 0 = 4{ 3 1
j 0 (w) = 2(3 34 w37@4 ) = 3 32 w37@4
i E 0 (|) = f(36| 37 ) = 36f| 37
12
= 312v35
v5
14. | = {5@3 3 {2@3
i D0 (v) = 312(35v36 ) = 60v36
or 60@v6
i | 0 = 53 {2@3 3 23 {31@3
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170
¤
CHAPTER 3 DIFFERENTIATION RULES
i U0 (d) = 9(2d) + 6(1) + 0 = 18d + 6
15. U(d) = (3d + 1)2 = 9d2 + 6d + 1
16. k(w) =
I
4
w 3 4hw = w1@4 3 4hw
17. V(s) =
I
s 3 s = s1@2 3 s i V 0 (s) = 12 s31@2 3 1 or
18. | =
I
{ ({ 3 1) = {3@2 3 {1@2
or |0 =
1
I 31
2 s
i | 0 = 32 {1@2 3 12 {31@2 = 12 {31@2 (3{ 3 1) [factor out 12 {31@2 ]
3{ 3 1
I .
2 {
4
= 3h{ + 4{31@3
{
19. | = 3h{ + I
3
20. V(U) = 4U2
| 0 = 3(h{ ) + 4(3 13 ){34@3 = 3h{ 3 43 {34@3
i
i V 0 (U) = 4(2U) = 8U
i k0 (x) = D(3x2 ) + E(2x) + F(1) = 3Dx2 + 2Ex + F
21. k(x) = Dx3 + Ex2 + Fx
22. | =
k0 (w) = 14 w33@4 3 4(hw ) = 14 w33@4 3 4hw
i
I
I
{+{
{
{
=
+ 2 = {1@232 + {132 = {33@2 + {31
{2
{2
{
{2 + 4{ + 3
I
= {3@2 + 4{1@2 + 3{31@2
{
| 0 = 32 {1@2 + 4 12 {31@2 + 3 3 12 {33@2 =
23. | =
i | 0 = 3 32 {35@2 + (31{32 ) = 3 32 {35@2 3 {32
i
3
2
k
I l
note that {3@2 = {2@2 · {1@2 = { {
I
2
3
I
{+ I 3
{ 2{ {
3{2
4{
3
3{2 + 4{ 3 3
I +
I 3
I =
I
.
2{ {
2{ {
2{ {
2{ {
I
I
I
I
I I
I
I 24. j(x) = 2 x + 3x = 2 x + 3 x i j 0 (x) = 2 (1) + 3 12 x31@2 = 2 +
The last expression can be written as
25. m({) = {2=4 + h2=4
26. n(u) = hu + uh
I
I3
x
2
i m 0 ({) = 2=4{1=4 + 0 = 2=4{1=4
i n0 (u) = hu + huh31
27. We ¿rst expand using the Binomial Theorem (see Reference Page 1).
K({) = ({ + {31 )3 = {3 + 3{2 {31 + 3{({31 )2 + ({31 )3 = {3 + 3{ + 3{31 + {33
0
2
32
K ({) = 3{ + 3 + 3(31{
28. | = dhy +
29. x =
34
) + (33{
f
e
+ 2 = dhy + ey 31 + fy 32
y
y
I
I
5
w + 4 w5 = w1@5 + 4w5@2
2
) = 3{ + 3 3 3{
32
34
i
3 3{
i | 0 = dhy 3 ey 32 3 2fy 33 = dhy 3
i x0 = 15 w34@5 + 4 52 w3@2 = 15 w34@5 + 10w3@2
e
2f
3 3
y2
y
I I
5
or 1@ 5 w4 + 10 w3
2 2
I
I 2
I
1
1
1
I
I
{+ I
=
{
+
2
{
·
= { + 2{1@231@3 + 1@{2@3 = { + 2{1@6 + {32@3
+
3
3
3
{
{
{
1
2
3 s
y 0 = 1 + 2 16 {35@6 3 23 {35@3 = 1 + 13 {35@6 3 23 {35@3 or 1 + s
6
3
5
3 {
3 {5
30. y =
31. } =
D
+ Eh| = D| 310 + Eh|
| 10
i } 0 = 310D| 311 + Eh| = 3
32. | = h{+1 + 1 = h{ h1 + 1 = h · h{ + 1
10D
+ Eh|
| 11
i |0 = h · h{ = h{+1
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°
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i
SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS
33. | =
I
4
{ = {1@4
i | 0 = 14 {33@4 =
1
I
. At (1> 1), |0 =
4
4 {3
1
4
¤
171
and an equation of the tangent line is
| 3 1 = 14 ({ 3 1) or | = 14 { + 34 .
34. | = {4 + 2{2 3 {
i |0 = 4{3 + 4{ 3 1. At (1> 2), | 0 = 7 and an equation of the tangent line is
| 3 2 = 7({ 3 1) or | = 7{ 3 5.
35. | = {4 + 2h{
i | 0 = 4{3 + 2h{ .
At (0> 2), | 0 = 2 and an equation of the tangent line is | 3 2 = 2({ 3 0)
or | = 2{ + 2. The slope of the normal line is 3 12 (the negative reciprocal of 2) and an equation of the normal line is
| 3 2 = 3 12 ({ 3 0) or | = 3 12 { + 2.
i | 0 = 2{ 3 4{3 . At (1> 0), | 0 = 2 3 4 = 32 and an equation of the tangent line is | 3 0 = 32({ 3 1)
36. | = {2 3 {4
or | = 32{ + 2. The slope of the normal line is
|30=
1
({
2
3 1) or | =
37. | = 3{2 3 {3
1
{
2
3
1
2
(the negative reciprocal of 32) and an equation of the normal line is
1
.
2
i | 0 = 6{ 3 3{2 .
At (1> 2), |0 = 6 3 3 = 3, so an equation of the tangent line is
| 3 2 = 3({ 3 1) or | = 3{ 3 1.
38. | = { 3
I
{ i |0 = 1 3 12 {31@2 = 1 3
1
I .
2 {
At (1> 0), |0 = 12 , so an equation of the tangent line is
| 3 0 = 12 ({ 3 1) or | = 12 { 3 12 .
39. i ({) = {4 3 2{3 + {2
i i 0 ({) = 4{3 3 6{2 + 2{
Note that i 0 ({) = 0 when i has a horizontal tangent, i 0 is positive
when i is increasing, and i 0 is negative when i is decreasing.
40. i ({) = {5 3 2{3 + { 3 1
i i 0 ({) = 5{4 3 6{2 + 1
Note that i 0 ({) = 0 when i has a horizontal tangent, i 0 is positive
when i is increasing, and i 0 is negative when i is decreasing.
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NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
(b) From the graph in part (a), it appears that i 0 is zero at {1 E 31=25, {2 E 0=5,
41. (a)
and {3 E 3. The slopes are negative (so i 0 is negative) on (3"> {1 ) and
({2 > {3 ). The slopes are positive (so i 0 is positive) on ({1 > {2 ) and ({3 > ").
(c) i ({) = {4 3 3{3 3 6{2 + 7{ + 30 i
i 0 ({) = 4{3 3 9{2 3 12{ + 7
(b) From the graph in part (a), it appears that i 0 is zero at {1 E 0=2 and {2 E 2=8.
42. (a)
The slopes are positive (so i 0 is positive) on (3"> {1 ) and ({2 > "). The slopes
are negative (so i 0 is negative) on ({1 > {2 ).
(c) j({) = h{ 3 3{2
i j0 ({) = h{ 3 6{
43. i ({) = 10{10 + 5{5 3 {
44. J(u) =
I
I
3
u+ u
45. i ({) = 2{ 3 5{3@4
i i 0 ({) = 100{9 + 25{4 3 1 i i 00 ({) = 900{8 + 100{3
i J 0 (u) = 12 u31@2 + 13 u32@3
i i 0 ({) = 2 3
15 31@4
4 {
i J 00 (u) = 3 14 u33@2 3 29 u35@3
i i 00 ({) =
15 35@4
16 {
Note that i 0 is negative when i is decreasing and positive when i is
increasing. i 00 is always positive since i 0 is always increasing.
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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS
46. i ({) = h{ 3 {3
i i 0 ({) = h{ 3 3{2
¤
173
i i 00 ({) = h{ 3 6{
Note that i 0 ({) = 0 when i has a horizontal tangent and that i 00 ({) = 0
when i 0 has a horizontal tangent.
47. (a) v = w3 3 3w
i y(w) = v0 (w) = 3w2 3 3 i d(w) = y0 (w) = 6w
(b) d(2) = 6(2) = 12 m@s2
(c) y(w) = 3w2 3 3 = 0 when w2 = 1, that is, w = 1 [w D 0] and d(1) = 6 m@s2 .
48. (a) v = w4 3 2w3 + w2 3 w
i
(c)
y(w) = v0 (w) = 4w3 3 6w2 + 2w 3 1 i
d(w) = y 0 (w) = 12w2 3 12w + 2
(b) d(1) = 12(1)2 3 12(1) + 2 = 2 m@ s2
5=3
5=3
n
and S = 50 when Y = 0=106, so n = S Y = 50(0=106) = 5=3. Thus, S =
and Y =
.
Y
Y
S
49. (a) S =
(b) Y = 5=3S 31
5=3
gY
5=3
gY
= 5=3(31S 32 ) = 3 2 . When S = 50,
= 3 2 = 30=00212. The derivative is the
gS
S
gS
50
i
instantaneous rate of change of the volume with respect to the pressure at 25 C. Its units are m3@kPa.
50. (a) O = dS 2 + eS + f, where d r 30=275428, e r 19=74853, and f r 3273=55234.
(b)
gO
gO
gO
= 2dS + e. When S = 30,
r 3=2, and when S = 40,
r 32=3. The derivative is the instantaneous rate of
gS
gS
gS
change of tire life with respect to pressure. Its units are (thousands of miles)@(lb@in2 ). When
increasing, and when
gO
is positive, tire life is
gS
gO
? 0, tire life is decreasing.
gS
51. The curve | = 2{3 + 3{2 3 12{ + 1 has a horizontal tangent when | 0 = 6{2 + 6{ 3 12 = 0
C 6({2 + { 3 2) = 0 C
6({ + 2)({ 3 1) = 0 C { = 32 or { = 1. The points on the curve are (32> 21) and (1> 36).
52. i ({) = h{ 3 2{
i i 0 ({) = h{ 3 2. i 0 ({) = 0 i h{ = 2 i { = ln 2, so i has a horizontal tangent when
{ = ln 2=
53. | = 2h{ + 3{ + 5{3
i | 0 = 2h{ + 3 + 15{2 . Since 2h{ A 0 and 15{2 D 0, we must have |0 A 0 + 3 + 0 = 3, so no
tangent line can have slope 2.
I
{ = {3@2
i | 0 = 32 {1@2 . The slope of the line | = 1 + 3{ is 3, so the slope of any line parallel to it is also 3.
I
Thus, | 0 = 3 i 32 {1@2 = 3 i
{ = 2 i { = 4, which is the {-coordinate of the point on the curve at which the
I
slope is 3. The |-coordinate is | = 4 4 = 8, so an equation of the tangent line is | 3 8 = 3({ 3 4) or | = 3{ 3 4.
54. | = {
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CHAPTER 3 DIFFERENTIATION RULES
55. The slope of the line 12{ 3 | = 1 (or | = 12{ 3 1) is 12, so the slope of both lines tangent to the curve is 12.
| = 1 + {3
i | 0 = 3{2 . Thus, 3{2 = 12 i {2 = 4 i { = ±2, which are the {-coordinates at which the tangent
lines have slope 12. The points on the curve are (2> 9) and (32> 37), so the tangent line equations are | 3 9 = 12({ 3 2)
or | = 12{ 3 15 and | + 7 = 12({ + 2) or | = 12{ + 17.
56. The slope of | = 1 + 2h{ 3 3{ is given by p = | 0 = 2h{ 3 3.
The slope of 3{ 3 | = 5 C | = 3{ 3 5 is 3.
p = 3 i 2h{ 3 3 = 3 i h{ = 3 i { = ln 3.
This occurs at the point (ln 3> 7 3 3 ln 3) E (1=1> 3=7).
57. The slope of | = {2 3 5{ + 4 is given by p = | 0 = 2{ 3 5. The slope of { 3 3| = 5
C | = 13 { 3
5
3
is 13 ,
so the desired normal line must have slope 13 , and hence, the tangent line to the parabola must have slope 33. This occurs if
2{ 3 5 = 33 i 2{ = 2 i { = 1. When { = 1, | = 12 3 5(1) + 4 = 0, and an equation of the normal line is
| 3 0 = 13 ({ 3 1) or | = 13 { 3 13 .
58. | = i ({) = { 3 {2
i i 0 ({) = 1 3 2{.
So i 0 (1) = 31, and the slope of the normal line is the negative reciprocal of that
of the tangent line, that is, 31@(31) = 1. So the equation of the normal line at
(1> 0) is | 3 0 = 1({ 3 1) C | = { 3 1. Substituting this into the equation of
the parabola, we obtain { 3 1 = { 3 {2
C { = ±1. The solution { = 31 is
the one we require. Substituting { = 31 into the equation of the parabola to ¿nd
the |-coordinate, we have | = 32. So the point of intersection is (31> 32), as shown in the sketch.
Let d> d2 be a point on the parabola at which the tangent line passes
59.
through the point (0> 34). The tangent line has slope 2d and equation
| 3 (34) = 2d({ 3 0) C | = 2d{ 3 4. Since d> d2 also lies on the
line, d2 = 2d(d) 3 4, or d2 = 4. So d = ±2 and the points are (2> 4)
and (32> 4).
60. (a) If | = {2 + {, then | 0 = 2{ + 1. If the point at which a tangent meets the parabola is d> d2 + d , then the slope of the
tangent is 2d + 1. But since it passes through (2> 33), the slope must also be
Therefore, 2d + 1 =
{|
d2 + d + 3
=
.
{{
d32
d2 + d + 3
. Solving this equation for d we get d2 + d + 3 = 2d2 3 3d 3 2 C
d32
d2 3 4d 3 5 = (d 3 5)(d + 1) = 0 C d = 5 or 31. If d = 31, the point is (31> 0) and the slope is 31, so the
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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS
¤
175
equation is | 3 0 = (31)({ + 1) or | = 3{ 3 1. If d = 5, the point is (5> 30) and the slope is 11, so the equation is
| 3 30 = 11({ 3 5) or | = 11{ 3 25.
(b) As in part (a), but using the point (2> 7), we get the equation
2d + 1 =
d2 + d 3 7
d32
i 2d2 3 3d 3 2 = d2 + d 3 7 C d2 3 4d + 5 = 0.
The last equation has no real solution (discriminant = 316 ? 0), so there is no line
through the point (2> 7) that is tangent to the parabola. The diagram shows that the
point (2> 7) is “inside” the parabola, but tangent lines to the parabola do not pass
through points inside the parabola.
1
1
3
i ({ + k) 3 i({)
{ 3 ({ + k)
3k
31
1
{
+
k
{
= lim
= lim
= lim
= lim
=3 2
61. i ({) = lim
k<0
k<0
k<0 k{({ + k)
k<0 k{({ + k)
k<0 {({ + k)
k
k
{
0
62. (a) i ({) = {q
i i 0 ({) = q{q31
i i 00 ({) = q (q 3 1) {q32
i ···
i
i (q) ({) = q(q 3 1)(q 3 2) · · · 2 · 1{q3q = q!
(b) i ({) = {31
i i 0 ({) = (31){32
i i 00 ({) = (31)(32){33
i (q) ({) = (31)(32)(33) · · · (3q){3(q+1) = (31)q q!{3(q+1) or
i ···
i
(31)q q!
{q+1
63. Let S ({) = d{2 + e{ + f. Then S 0 ({) = 2d{ + e and S 00 ({) = 2d. S 00 (2) = 2
i 2d = 2 i d = 1.
0
S (2) = 3 i 2(1)(2) + e = 3 i 4 + e = 3 i e = 31.
S (2) = 5 i 1(2)2 + (31)(2) + f = 5 i 2 + f = 5 i f = 3. So S ({) = {2 3 { + 3.
64. | = D{2 + E{ + F
i | 0 = 2D{ + E
i | 00 = 2D. We substitute these expressions into the equation
|00 + | 0 3 2| = {2 to get
(2D) + (2D{ + E) 3 2(D{2 + E{ + F) = {2
2D + 2D{ + E 3 2D{2 3 2E{ 3 2F = {2
(32D){2 + (2D 3 2E){ + (2D + E 3 2F) = (1){2 + (0){ + (0)
The coef¿cients of {2 on each side must be equal, so 32D = 1 i D = 3 12 . Similarly, 2D 3 2E = 0 i
D = E = 3 12 and 2D + E 3 2F = 0 i 31 3
65. | = i({) = d{3 + e{2 + f{ + g
1
2
3 2F = 0 i F = 3 34 .
i i 0 ({) = 3d{2 + 2e{ + f. The point (32> 6) is on i, so i (32) = 6 i
38d + 4e 3 2f + g = 6 (1). The point (2> 0) is on i , so i (2) = 0 i 8d + 4e + 2f + g = 0 (2). Since there are
horizontal tangents at (32> 6) and (2> 0), i 0 (±2) = 0. i 0 (32) = 0 i 12d 3 4e + f = 0 (3) and i 0 (2) = 0 i
12d + 4e + f = 0 (4). Subtracting equation (3) from (4) gives 8e = 0 i e = 0. Adding (1) and (2) gives 8e + 2g = 6,
so g = 3 since e = 0. From (3) we have f = 312d, so (2) becomes 8d + 4(0) + 2(312d) + 3 = 0 i 3 = 16d i
3
3
3 3
= 3 94 and the desired cubic function is | = 16
. Now f = 312d = 312 16
{ 3 94 { + 3.
d = 16
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CHAPTER 3 DIFFERENTIATION RULES
66. | = d{2 + e{ + f
i |0 ({) = 2d{ + e. The parabola has slope 4 at { = 1 and slope 38 at { = 31, so | 0 (1) = 4 i
2d + e = 4 (1) and | 0 (31) = 38
i 32d + e = 38 (2). Adding (1) and (2) gives us 2e = 34 C e = 32. From
(1), 2d 3 2 = 4 C d = 3. Thus, the equation of the parabola is | = 3{2 3 2{ + f. Since it passes through the point
(2> 15), we have 15 = 3(2)2 3 2(2) + f i f = 7, so the equation is | = 3{2 3 2{ + 7.
67. i ({) =
+
{2 + 1
if { ? 1
{+1
if { D 1
Calculate the left- and right-hand derivatives as de¿ned in Exercise 2.8.56:
0
(1) = lim
i3
i(1 + k) 3 i(1)
[(1 + k)2 + 1] 3 (1 + 1)
k2 + 2k
= lim
= lim
= lim (k + 2) = 2 and
k
k
k
k<0
k<0
k<0
0
(1) = lim
i+
i (1 + k) 3 i (1)
[(1 + k) + 1] 3 (1 + 1)
k
= lim
= lim
= lim 1 = 1.
k
k
k<0+
k<0+ k
k<0+
k<0
k<0+
Since the left and right limits are different,
lim
k<0
i(1 + k) 3 i(1)
does not exist, that is, i 0 (1)
k
does not exist. Therefore, i is not differentiable at 1.
;
2{
A
?
68. j({) = 2{ 3 {2
A
=
23{
if { $ 0
if 0 ? { ? 2
if { D 2
Investigate the left- and right-hand derivatives at { = 0 and { = 2:
0
j3
(0) = lim
k<0
j(0 + k) 3 j(0)
2k 3 2(0)
= lim
= 2 and
k
k
k<0
0
(0) = lim
j+
j(0 + k) 3 j(0)
(2k 3 k2 ) 3 2(0)
= lim
= lim (2 3 k) = 2, so j is differentiable at { = 0.
+
k
k
k<0
k<0+
0
j3
(2) = lim
j(2 + k) 3 j(2)
2(2 + k) 3 (2 + k)2 3 (2 3 2)
32k 3 k2
= lim
= lim
= lim (32 3 k) = 32
k
k
k
k<0
k<0
k<0
k<0+
k<0
and
0
j+
(2) = lim
k<0+
j(2 + k) 3 j(2)
[2 3 (2 + k)] 3 (2 3 2)
3k
= lim
= lim
= lim (31) = 31,
k
k
k<0+
k<0+ k
k<0+
so j is not differentiable at { = 2. Thus, a formula for j 0 is
;
2
A
?
0
j ({) = 2 3 2{
A
=
31
if { $ 0
if 0 ? { ? 2
if { A 2
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS
69. (a) Note that {2 3 9 ? 0 for {2 ? 9
; 2
{ 39
A
A
?
i ({) =
3{2 + 9
A
A
= 2
{ 39
C |{| ? 3 C 33 ? { ? 3. So
;
2{
if { ? 33
A
?
0
32{ if 33 ? { ? 3
i i ({) =
if 33 ? { ? 3
A
=
2{
if { A 3
if { D 3
if { $ 33
To show that i 0 (3) does not exist we investigate lim
k<0
=
+
2{
if |{| A 3
32{
if |{| ? 3
¤
177
i (3 + k) 3 i (3)
by computing the left- and right-hand derivatives
k
de¿ned in Exercise 2.8.56.
i (3 + k) 3 i (3)
[3(3 + k)2 + 9] 3 0
0
= lim
= lim (36 3 k) = 36 and
(3) = lim
i3
k
k
k<0
k<0
k<0
2
(3 + k) 3 9 3 0
i (3 + k) 3 i (3)
6k + k2
0
i+
(3) = lim
= lim
= lim
= lim (6 + k) = 6.
+
+
+
k
k
k
k<0
k<0
k<0
k<0+
Since the left and right limits are different,
lim
k<0
(b)
i(3 + k) 3 i(3)
does not exist, that is, i 0 (3)
k
does not exist. Similarly, i 0 (33) does not exist.
Therefore, i is not differentiable at 3 or at 33.
70. If { D 1, then k({) = |{ 3 1| + |{ + 2| = { 3 1 + { + 2 = 2{ + 1.
If 32 ? { ? 1, then k({) = 3({ 3 1) + { + 2 = 3.
If { $ 32, then k({) = 3({ 3 1) 3 ({ + 2) = 32{ 3 1. Therefore,
;
;
32{ 3 1 if { $ 32
32
A
A
?
?
0
3
if 32 ? { ? 1
0
k({) =
i k ({) =
A
A
=
=
2{ + 1
if { D 1
2
To see that k0 (1) = lim
{<1
observe that lim
{<1
lim
{<1+
if { ? 32
if 32 ? { ? 1
if { A 1
k({) 3 k(1)
does not exist,
{31
k({) 3 k(1)
333
= lim
= 0 but
{31
{<1 3 3 1
k({) 3 k(1)
2{ 3 2
= lim
= 2. Similarly,
{31
{<1+ { 3 1
k0 (32) does not exist.
71. Substituting { = 1 and | = 1 into | = d{2 + e{ gives us d + e = 1 (1). The slope of the tangent line | = 3{ 3 2 is 3 and the
slope of the tangent to the parabola at ({> |) is |0 = 2d{ + e. At { = 1, |0 = 3 i 3 = 2d + e (2). Subtracting (1) from
(2) gives us 2 = d and it follows that e = 31. The parabola has equation | = 2{2 3 {.
72. | = {4 + d{3 + e{2 + f{ + g
have g = 1.
i |(0) = g.
Since the tangent line | = 2{ + 1 is equal to 1 at { = 0, we must
|0 = 4{3 + 3d{2 + 2e{ + f i |0 (0) = f. Since the slope of the tangent line | = 2{ + 1 at { = 0 is 2, we
must have f = 2. Now |(1) = 1 + d + e + f + g = d + e + 4 and the tangent line | = 2 3 3{ at { = 1 has |-coordinate 31,
so d + e + 4 = 31 or d + e = 35 (1). Also, | 0 (1) = 4 + 3d + 2e + f = 3d + 2e + 6 and the slope of the tangent line
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CHAPTER 3 DIFFERENTIATION RULES
| = 2 3 3{ at { = 1 is 33, so 3d + 2e + 6 = 33 or 3d + 2e = 39 (2). Adding 32 times (1) to (2) gives us d = 1 and
hence, e = 36. The curve has equation | = {4 + {3 3 6{2 + 2{ + 1.
73. | = i({) = d{2
i i 0 ({) = 2d{.
So the slope of the tangent to the parabola at { = 2 is p = 2d(2) = 4d. The slope
of the given line, 2{ + | = e C | = 32{ + e, is seen to be 32, so we must have 4d = 32 C d = 3 12 . So when
{ = 2, the point in question has |-coordinate 3 12 · 22 = 32. Now we simply require that the given line, whose equation is
2{ + | = e, pass through the point (2> 32): 2(2) + (32) = e C e = 2. So we must have d = 3 12 and e = 2.
I
{ is | 0 =
f
I and the slope of the tangent line | = 32 { + 6 is 32 . These must be equal at the
{
I
I
f
3
point of tangency d> f d , so I =
i f = 3 d. The |-coordinates must be equal at { = d, so
2
2 d
I I
I
I
3 d
d = 32 d + 6 i 3d = 32 d + 6 i 32 d = 6 i d = 4. Since f = 3 d, we have
f d = 32 d + 6 i
74. The slope of the curve | = f
f=3
2
I
4 = 6.
0
75. i is clearly differentiable for { ? 2 and for { A 2. For { ? 2, i 0 ({) = 2{, so i3
(2) = 4. For { A 2, i 0 ({) = p, so
0
0
0
(2) = p. For i to be differentiable at { = 2, we need 4 = i3
(2) = i+
(2) = p. So i ({) = 4{ + e. We must also have
i+
continuity at { = 2, so 4 = i (2) = lim i ({) = lim (4{ + e) = 8 + e. Hence, e = 34.
{<2+
76. (a) {| = f
i |=
{<2+
f
f
f
. Let S = d> . The slope of the tangent line at { = d is | 0 (d) = 3 2 . Its equation is
{
d
d
f
2f
f
f
2f
= 3 2 ({ 3 d) or | = 3 2 { + , so its |-intercept is . Setting | = 0 gives { = 2d, so the {-intercept is 2d.
d
d
d
d
d
f
2f
and (2d> 0) is d>
= S.
The midpoint of the line segment joining 0>
d
d
|3
(b) We know the {- and |-intercepts of the tangent line from part (a), so the area of the triangle bounded by the axes and the
tangent is 12 (base)(height) = 12 {| = 12 (2d)(2f@d) = 2f, a constant.
77. Solution 1:
i ({) 3 i (1)
{1000 3 1
= lim
.
{<1
{<1
{31
{31
Let i ({) = {1000 . Then, by the de¿nition of a derivative, i 0 (1) = lim
But this is just the limit we want to ¿nd, and we know (from the Power Rule) that i 0 ({) = 1000{999 , so
i 0 (1) = 1000(1)999 = 1000. So lim
{<1
{1000 3 1
= 1000.
{31
Solution 2: Note that ({1000 3 1) = ({ 3 1)({999 + {998 + {997 + · · · + {2 + { + 1). So
lim
{<1
{1000 3 1
({ 3 1)({999 + {998 + {997 + · · · + {2 + { + 1)
= lim
= lim ({999 + {998 + {997 + · · · + {2 + { + 1)
{<1
{<1
{31
{31
= 1 + 1 + 1 + · · · + 1 + 1 + 1 = 1000, as above.

~}
€
1000 ones
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179
78. In order for the two tangents to intersect on the |-axis, the points of tangency must be at
equal distances from the |-axis, since the parabola | = {2 is symmetric about the |-axis.
Say the points of tangency are d> d2 and 3d> d2 , for some d A 0. Then since the
derivative of | = {2 is g|@g{ = 2{, the left-hand tangent has slope 32d and equation
| 3 d2 = 32d({ + d), or | = 32d{ 3 d2 , and similarly the right-hand tangent line has
equation | 3 d2 = 2d({ 3 d), or | = 2d{ 3 d2 . So the two lines intersect at 0> 3d2 . Now if the lines are perpendicular,
then the product of their slopes is 31, so (32d)(2d) = 31 C d2 = 14 C d = 12 . So the lines intersect at 0> 3 14 .
79. | = {2
i | 0 = 2{, so the slope of a tangent line at the point (d> d2 ) is | 0 = 2d and the slope of a normal line is 31@(2d),
for d 6= 0. The slope of the normal line through the points (d> d2 ) and (0> f) is
d2 3 f = 3 12
d2 3 f
1
d2 3 f
, so
=3
d30
d
2d
i
i d2 = f 3 12 . The last equation has two solutions if f A 12 , one solution if f = 12 , and no solution if
f ? 12 . Since the |-axis is normal to | = {2 regardless of the value of f (this is the case for d = 0), we have three normal lines
if f A
1
2
and one normal line if f $ 12 .
80.
From the sketch, it appears that there may be a line that is tangent to both
curves. The slope of the line through the points S (d> d2 ) and
T(e> e2 3 2e + 2) is
e2 3 2e + 2 3 d2
. The slope of the tangent line at S
e3d
is 2d [| 0 = 2{] and at T is 2e 3 2 [| 0 = 2{ 3 2]. All three slopes are
equal, so 2d = 2e 3 2 C d = e 3 1.
Also, 2e 3 2 =
e2 3 2e + 2 3 d2
e3d
2e = 3 i e =
3
2
and d =
3
2
i 2e 3 2 =
e2 3 2e + 2 3 (e 3 1)2
e 3 (e 3 1)
i 2e 3 2 = e2 3 2e + 2 3 e2 + 2e 3 1 i
3 1 = 12 . Thus, an equation of the tangent line at S is | 3
| = { 3 14 .
1 2
2
= 2 12 { 3 12 or
APPLIED PROJECT Building a Better Roller Coaster
1. (a) i ({) = d{2 + e{ + f
i i 0 ({) = 2d{ + e.
The origin is at S :
i (0) = 0
0
i (0) = 0=8
The slope of the ascent is 0=8:
The slope of the drop is 31=6:
0
i
f=0
i
e = 0=8
i (100) = 31=6 i 200d + e = 31=6
(b) e = 0=8, so 200d + e = 31=6 i 200d + 0=8 = 31=6 i 200d = 32=4 i d = 3
2=4
= 30=012.
200
Thus, i ({) = 30=012{2 + 0=8{.
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CHAPTER 3 DIFFERENTIATION RULES
(c) Since O1 passes through the origin with slope 0=8, it has equation | = 0=8{.
The horizontal distance between S and T is 100, so the |-coordinate at T is
i (100) = 30=012(100)2 + 0=8(100) = 340. Since O2 passes through the
point (100> 340) and has slope 31=6, it has equation | + 40 = 31=6({ 3 100)
or | = 31=6{ + 120.
(d) The difference in elevation between S (0> 0) and T(100> 340) is 0 3 (340) = 40 feet.
2. (a)
Interval
Function
First Derivative
Second Derivative
O1 ({) = 0=8{
O01 ({) = 0=8
O001 ({) = 0
[0> 10)
j({) = n{3 + o{2 + p{ + q
j0 ({) = 3n{2 + 2o{ + p
j00 ({) = 6n{ + 2o
[10> 90]
t({) = d{2 + e{ + f
t 0 ({) = 2d{ + e
t 00 ({) = 2d
k({) = s{3 + t{2 + u{ + v
k0 ({) = 3s{2 + 2t{ + u
k00 ({) = 6s{ + 2t
O2 ({) = 31=6{ + 120
O02 ({)
O002 ({) = 0
(3"> 0)
(90> 100]
(100> ")
= 31=6
There are 4 values of { (0, 10, 90, and 100) for which we must make sure the function values are equal, the ¿rst derivative
values are equal, and the second derivative values are equal. The third column in the following table contains the value of
each side of the condition — these are found after solving the system in part (b).
At { =
0
Condition
j(0) = O1 (0)
0
j (0) =
00
j (0) =
10
Value
O01 (0)
O001 (0)
0
0
j (10) = t (10)
00
00
j (10) = t (10)
90
k(90) = t(90)
0
0
k (90) = t (90)
00
00
k (90) = t (90)
100
k(100) = O2 (100)
0
k (100) =
0
q=0
4
5
p = 0=8
0
2o = 0
68
9
2
3
2
3 75
j(10) = t(10)
O02 (100)
k00 (100) = O002 (100)
Resulting Equation
1000n + 100o + 10p + q = 100d + 10e + f
300n + 20o + p = 20d + e
60n + 2o = 2d
3 220
9
3 22
15
2
3 75
729,000s + 8100t + 90u + v = 8100d + 90e + f
340
1,000,000s + 10,000t + 100u + v = 340
3 85
0
24,300s + 180t + u = 180d + e
540s + 2t = 2d
30,000s + 200t + u = 31=6
600s + 2t = 0
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPLIED PROJECT BUILDING A BETTER ROLLER COASTER
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181
(b) We can arrange our work in a 12 × 12 matrix as follows.
d
e
f
n
o
p
q
s
t
u
v
constant
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0=8
0
0
0
0
2
0
0
0
0
0
0
0
3100
310
31
1000
100
10
1
0
0
0
0
0
320
31
0
300
20
1
0
0
0
0
0
0
32
0
0
60
2
0
0
0
0
0
0
0
38100
390
31
0
0
0
0
729,000
8100
90
1
0
3180
31
0
0
0
0
0
24,300
180
1
0
0
32
0
0
0
0
0
0
540
2
0
0
0
0
0
0
0
0
0
0
1,000,000
10,000
100
1
340
0
0
0
0
0
0
0
30,000
200
1
0
31=6
0
0
0
0
0
0
0
600
2
0
0
0
Solving the system gives us the formulas for t, j, and k.
<
1
d = 30=013 = 3 75
A
A
@
1 2
{ + 14
{ 3 49
t({) = 3 75
e = 0=93 = 14
15
15
A
A
>
f = 30=4 = 3 49
s = 0=0004 =
<
A
A
A
A
@
1
2250
2
t = 30=13 = 3 15
u = 11=73 =
176
15
v = 3324=4 = 3
(c) Graph of O1 , t, j, k, and O2 :
A
A
A
A
>
2920
k({) =
1 <
n = 30=0004 = 3 2250
A
A
A
A
@
o=0
p = 0=8 =
A
A
A
A
>
4
5
q=0
1
{3
2250
3
2 2
{
15
+
176
{
15
3
1
{3 + 45 {
j({) = 3 2250
2920
9
9
The graph of the ¿ve functions as a piecewise-de¿ned function:
This is the piecewise-de¿ned function assignment on a
A comparison of the graphs in part 1(c) and part 2(c):
TI-83/4 Plus calculator, where Y2 = O1 , Y6 = j, Y5 = t,
Y7 = k, and Y3 = O2 .
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CHAPTER 3 DIFFERENTIATION RULES
3.2 The Product and Quotient Rules
1. Product Rule: i({) = (1 + 2{2 )({ 3 {2 )
i
i 0 ({) = (1 + 2{2 )(1 3 2{) + ({ 3 {2 )(4{) = 1 3 2{ + 2{2 3 4{3 + 4{2 3 4{3 = 1 3 2{ + 6{2 3 8{3 .
Multiplying ¿rst: i ({) = (1 + 2{2 )({ 3 {2 ) = { 3 {2 + 2{3 3 2{4
2. Quotient Rule: I ({) =
I 0 ({) =
{4 3 5{3 +
{2
I
{
=
{4 3 5{3 + {1@2
{2
{2 (4{3 3 15{2 + 12 {31@2 ) 3 ({4 3 5{3 + {1@2 )(2{)
({2 )2
2{5 3 5{4 3 32 {3@2
= 2{ 3 5 3 32 {35@2
{4
I
{4 3 5{3 + {
Simplifying ¿rst: I ({) =
= {2 3 5{ + {33@2
{2
i i 0 ({) = 1 3 2{ + 6{2 3 8{3 (equivalent).
i
=
4{5 3 15{4 + 12 {3@2 3 2{5 + 10{4 3 2{3@2
{4
=
i I 0 ({) = 2{ 3 5 3 32 {35@2 (equivalent).
For this problem, simplifying ¿rst seems to be the better method.
3. By the Product Rule, i ({) = ({3 + 2{)h{
i
i 0 ({) = ({3 + 2{)(h{ )0 + h{ ({3 + 2{)0 = ({3 + 2{)h{ + h{ (3{2 + 2)
= h{ [({3 + 2{) + (3{2 + 2)] = h{ ({3 + 3{2 + 2{ + 2)
4. By the Product Rule, j({) =
I {
{ h = {1@2 h{
5. By the Quotient Rule, | =
{
h{
6. By the Quotient Rule, | =
h{
1 3 h{
PR
i |0 =
i j 0 ({) = {1@2 (h{ ) + h{
1 31@2
2{
= 12 {31@2 h{ (2{ + 1).
h{ (1) 3 {(h{ )
h{ (1 3 {)
13{
=
=
.
2
2
{
{
h{
(h )
(h )
i |0 =
(1 3 h{ )h{ 3 h{ (3h{ )
h{ 3 h2{ + h2{
h{
=
=
.
{
2
{
2
(1 3 h )
(1 3 h )
(1 3 h{ )2
QR
The notations i and i indicate the use of the Product and Quotient Rules, respectively.
7. j({) =
8. J({) =
1 + 2{
3 3 4{
i j0 ({) =
{2 3 2
2{ + 1
i J0 ({) =
QR
QR
(3 3 4{)(2) 3 (1 + 2{)(34)
6 3 8{ + 4 + 8{
10
=
=
(3 3 4{)2
(3 3 4{)2
(3 3 4{)2
(2{ + 1)(2{) 3 ({2 3 2)(2)
4{2 + 2{ 3 2{2 + 4
2{2 + 2{ + 4
=
=
2
2
(2{ + 1)
(2{ + 1)
(2{ + 1)2
I
I
PR
x )(x + x ) i
I
I
I
I
1
1
K 0 (x) = (x 3 x ) 1 + I
+ (x + x ) 1 3 I
= x + 12 x 3 x 3
2 x
2 x
9. K(x) = (x 3
1
2
+x3
1I
x
2
+
I
x3
1
2
= 2x 3 1.
An easier method is to simplify ¿rst and then differentiate as follows:
K(x) = (x 3
I
I
I
x )(x + x ) = x2 3 ( x )2 = x2 3 x i K 0 (x) = 2x 3 1
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NOT FOR SALE
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
10. M(y) = (y 3 3 2y)(y 34 + y 32 )
PR
i
M 0 (y) = (y 3 3 2y)(34y 35 3 2y 33 ) + (y 34 + y 32 )(3y 2 3 2)
= 34y 32 3 2y 0 + 8y 34 + 4y 32 + 3y 32 3 2y 34 + 3y 0 3 2y 32 = 1 + y 32 + 6y 34
11. I (|) =
1
3
PR
3
(| + 5|3 ) = (| 32 3 3| 34 )(| + 5| 3 ) i
|2
|4
I 0 (|) = (| 32 3 3| 34 )(1 + 15| 2 ) + (| + 5|3 )(32| 33 + 12| 35 )
= (| 32 + 15 3 3| 34 3 45| 32 ) + (32| 32 + 12| 34 3 10 + 60|32 )
= 5 + 14|32 + 9| 34 or 5 + 14@| 2 + 9@| 4
PR
12. i (}) = (1 3 h} )(} + h} )
i
i 0 (}) = (1 3 h} )(1 + h} ) + (} + h} )(3h} ) = 12 3 (h} )2 3 }h} 3 (h} )2 = 1 3 }h} 3 2h2}
13. | =
14. | =
{3
1 3 {2
QR
i |0 =
{+1
{3 + { 3 2
(1 3 {2 ) (3{2 ) 3 {3 (32{)
{2 (3 3 3{2 + 2{2 )
{2 (3 3 {2 )
=
=
(1 3 {2 )2
(1 3 {2 )2
(1 3 {2 )2
QR
i
|0 =
({3 + { 3 2)(1) 3 ({ + 1)(3{2 + 1)
{3 + { 3 2 3 3{3 3 3{2 3 { 3 1
32{3 3 3{2 3 3
=
=
3
2
3
2
({ + { 3 2)
({ + { 3 2)
({3 + { 3 2)2
or 3
2{3 + 3{2 + 3
({ 3 1)2 ({2 + { + 2)2
15. | =
|0 =
=
16. | =
|0 =
w2 + 2
w4 3 3w2 + 1
QR
i
(w4 3 3w2 + 1)(2w) 3 (w2 + 2)(4w3 3 6w)
2w[(w4 3 3w2 + 1) 3 (w2 + 2)(2w2 3 3)]
=
4
2
2
(w 3 3w + 1)
(w4 3 3w2 + 1)2
2w(w4 3 3w2 + 1 3 2w4 3 4w2 + 3w2 + 6)
2w(3w4 3 4w2 + 7)
=
(w4 3 3w2 + 1)2
(w4 3 3w2 + 1)2
w
w
= 2
(w 3 1)2
w 3 2w + 1
QR
i
(w2 3 2w + 1)(1) 3 w(2w 3 2)
(w 3 1)2 3 2w(w 3 1)
(w 3 1)[(w 3 1) 3 2w]
3w 3 1
=
=
=
[(w 3 1)2 ]2
(w 3 1)4
(w 3 1)4
(w 3 1)3
I
17. | = hs (s + s s ) = hs (s + s3@2 )
18. | =
1
v + nhv
19. | =
y 3 3 2y
y
QR
i |0 =
I
y
PR
I I
i | 0 = hs 1 + 32 s1@2 + (s + s3@2 )hs = hs 1 + 32 s + s + s s
(v + nhv )(0) 3 (1)(1 + nhv )
1 + nhv
=3
v
2
(v + nh )
(v + nhv )2
= y2 3 2
I
y = y 2 3 2y 1@2
i | 0 = 2y 3 2 12 y 31@2 = 2y 3 y 31@2 .
We can change the form of the answer as follows: 2y 3 y
31@2
I
1
2y y 3 1
2y 3@2 3 1
I
I
= 2y 3 I =
=
y
y
y
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184
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CHAPTER 3 DIFFERENTIATION RULES
20. } = z3@2 (z + fhz ) = z5@2 + fz3@2 hz
i } 0 = 52 z3@2 + f z3@2 · hz + hz · 32 z1@2 = 52 z3@2 + 12 fz1@2 hz (2z + 3)
1@2
I
1 31@2
(2
+
w
)(2)
3
2w
w
QR
2
4 + 2w1@2 3 w1@2
4 + w1@2
4+ w
I
I
I
I
i i 0 (w) =
=
=
or
(2 + w )2
(2 + w )2
(2 + w )2
(2 + w )2
21. i (w) =
2w
I
2+ w
22. j(w) =
I
w3 w
w
w1@2
= 1@3 3 1@3 = w2@3 3 w1@6
1@3
w
w
w
23. i ({) =
D
E + Fh{
24. i ({) =
1 3 {h{
{ + h{
QR
i i 0 ({) =
QR
i 0 ({) =
({ + h{ )(3{h{ )0 3 (1 3 {h{ )(1 + h{ )
({ + h{ )2
PR
i 0 ({) =
({ + h{ )[3({h{ + h{ · 1)] 3 (1 + h{ 3 {h{ 3 {h2{ )
({ + h{ )2
=
{
{ + f@{
26. i ({) =
d{ + e
f{ + g
27. i ({) = {4 h{
(E + Fh{ ) · 0 3 D(Fh{ )
DFh{
=
3
(E + Fh{ )2
(E + Fh{ )2
i
i
25. i ({) =
i j0 (w) = 23 w31@3 3 16 w35@6
i i 0 ({) =
i i 0 ({) =
3{2 h{ 3 {h{ 3 {h2{ 3 h2{ 3 1 3 h{ + {h{ + {h2{
3{2 h{ 3 h2{ 3 h{ 3 1
=
{
2
({ + h )
({ + h{ )2
({ + f@{)(1) 3 {(1 3 f@{2 )
{ + f@{ 3 { + f@{
2f@{
{2
2f{
=
=
·
= 2
2
2
2
2
2
f 2
{
({
+ f)2
({
+
f)
{ +f
{+
2
{
{
{
(f{ + g)(d) 3 (d{ + e)(f)
df{ + dg 3 df{ 3 ef
dg 3 ef
=
=
(f{ + g)2
(f{ + g)2
(f{ + g)2
i i 0 ({) = {4 h{ + h{ · 4{3 = {4 + 4{3 h{ or {3 h{ ({ + 4) i
i 00 ({) = ({4 + 4{3 )h{ + h{ (4{3 + 12{2 ) = ({4 + 4{3 + 4{3 + 12{2 )h{
= ({4 + 8{3 + 12{2 )h{ or {2 h{ ({ + 2)({ + 6)
l
k
i i 0 ({) = {5@2 h{ + h{ · 52 {3@2 = {5@2 + 52 {3@2 h{ or 12 {3@2 h{ (2{ + 5) i
l
k
{1@2 = {5@2 + 5{3@2 + 15
{1@2 h{ or 14 {1@2 h{ (4{2 + 20{ + 15)
i 00 ({) = {5@2 + 52 {3@2 h{ + h{ 52 {3@2 + 15
4
4
28. i ({) = {5@2 h{
29. i ({) =
{2
1 + 2{
i 00 ({) =
=
30. i ({) =
=
=
(1 + 2{)(2{) 3 {2 (2)
2{ + 4{2 3 2{2
2{2 + 2{
=
=
2
2
(1 + 2{)
(1 + 2{)
(1 + 2{)2
i
(1 + 2{)2 (4{ + 2) 3 (2{2 + 2{)(1 + 4{ + 4{2 )0
2(1 + 2{)2 (2{ + 1) 3 2{({ + 1)(4 + 8{)
=
2
2
[(1 + 2{) ]
(1 + 2{)4
2(1 + 2{)[(1 + 2{)2 3 4{({ + 1)]
2(1 + 4{ + 4{2 3 4{2 3 4{)
2
=
=
(1 + 2{)4
(1 + 2{)3
(1 + 2{)3
{2
i 00 ({) =
i i 0 ({) =
{
31
i i 0 ({) =
({2 3 1)(1) 3 {(2{)
{2 3 1 3 2{2
3{2 3 1
=
= 2
2
2
2
2
({ 3 1)
({ 3 1)
({ 3 1)2
i
({2 3 1)2 (32{) 3 (3{2 3 1)({4 3 2{2 + 1)0
({2 3 1)2 (32{) + ({2 + 1)(4{3 3 4{)
=
2
2
2
[({ 3 1) ]
({2 3 1)4
({2 3 1)2 (32{) + ({2 + 1)(4{)({2 3 1)
({2 3 1)[({2 3 1)(32{) + ({2 + 1)(4{)]
=
2
4
({ 3 1)
({2 3 1)4
32{3 + 2{ + 4{3 + 4{
2{3 + 6{
=
({2 3 1)3
({2 3 1)3
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
31. | =
|0 =
{2 3 1
+{+1
{2
185
i
({2 + { + 1)(2{) 3 ({2 3 1)(2{ + 1)
2{3 + 2{2 + 2{ 3 2{3 3 {2 + 2{ + 1
{2 + 4{ + 1
=
= 2
.
2
2
2
2
({ + { + 1)
({ + { + 1)
({ + { + 1)2
At (1> 0), |0 =
32. | =
¤
h{
{
6
2
= , and an equation of the tangent line is | 3 0 = 23 ({ 3 1), or | = 23 { 3 23 .
32
3
i |0 =
{ · h{ 3 h{ · 1
h{ ({ 3 1)
=
.
2
{
{2
At (1> h), | 0 = 0, and an equation of the tangent line is | 3 h = 0({ 3 1), or | = h.
33. | = 2{h{
i | 0 = 2({ · h{ + h{ · 1) = 2h{ ({ + 1).
At (0> 0), | 0 = 2h0 (0 + 1) = 2 · 1 · 1 = 2, and an equation of the tangent line is | 3 0 = 2({ 3 0), or | = 2{. The slope of
the normal line is 3 12 , so an equation of the normal line is | 3 0 = 3 12 ({ 3 0), or | = 3 12 {.
34. | =
2{
+1
{2
i |0 =
({2 + 1)(2) 3 2{(2{)
2 3 2{2
= 2
. At (1> 1), | 0 = 0, and an equation of the tangent line is
2
2
({ + 1)
({ + 1)2
| 3 1 = 0({ 3 1), or | = 1. The slope of the normal line is unde¿ned, so an equation of the normal line is { = 1.
35. (a) | = i ({) =
i 0 ({) =
1
1 + {2
i
(b)
(1 + {2 )(0) 3 1(2{)
32{
=
. So the slope of the
(1 + {2 )2
(1 + {2 )2
2
tangent line at the point 31> 12 is i 0 (31) = 2 =
2
equation is | 3
36. (a) | = i ({) =
i 0 ({) =
1
2
1
2
and its
= 12 ({ + 1) or | = 12 { + 1.
{
1 + {2
i
(b)
(1 + {2 )1 3 {(2{)
1 3 {2
=
. So the slope of the
2
2
(1 + { )
(1 + {2 )2
tangent line at the point (3> 0=3) is i 0 (3) =
38
100
and its equation is
| 3 0=3 = 30=08({ 3 3) or | = 30=08{ + 0=54.
37. (a) i ({) = ({3 3 {)h{
i i 0 ({) = ({3 3 {)h{ + h{ (3{2 3 1) = h{ ({3 + 3{2 3 { 3 1)
i 0 = 0 when i has a horizontal tangent line, i 0 is negative when i is
(b)
decreasing, and i 0 is positive when i is increasing.
38. (a) i ({) =
i 0 ({) =
h{
2{2 + { + 1
i
(2{2 + { + 1)h{ 3 h{ (4{ + 1)
h{ (2{2 + { + 1 3 4{ 3 1)
h{ (2{2 3 3{)
=
=
(2{2 + { + 1)2
(2{2 + { + 1)2
(2{2 + { + 1)2
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CHAPTER 3 DIFFERENTIATION RULES
i 0 = 0 when i has a horizontal tangent line, i 0 is negative when i is
(b)
decreasing, and i 0 is positive when i is increasing.
39. (a) i ({) =
{2 3 1
{2 + 1
i
i 0 ({) =
({2 + 1)(2{) 3 ({2 3 1)(2{)
(2{)[({2 + 1) 3 ({2 3 1)]
(2{)(2)
4{
=
= 2
= 2
2
2
({ + 1)
({2 + 1)2
({ + 1)2
({ + 1)2
i 00 ({) =
({2 + 1)2 (4) 3 4{({4 + 2{2 + 1)0
4({2 + 1)2 3 4{(4{3 + 4{)
=
[({2 + 1)2 ]2
({2 + 1)4
=
i
4({2 + 1)2 3 16{2 ({2 + 1)
4({2 + 1)[({2 + 1) 3 4{2 ]
4(1 3 3{2 )
=
=
2
4
2
4
({ + 1)
({ + 1)
({2 + 1)3
i 0 = 0 when i has a horizontal tangent and i 00 = 0 when i 0 has a
(b)
horizontal tangent. i 0 is negative when i is decreasing and positive when i
is increasing. i 00 is negative when i 0 is decreasing and positive when i 0 is
increasing. i 00 is negative when i is concave down and positive when i is
concave up.
40. (a) i ({) = ({2 3 1)h{
i i 0 ({) = ({2 3 1)h{ + h{ (2{) = h{ ({2 + 2{ 3 1) i
i 00 ({) = h{ (2{ + 2) + ({2 + 2{ 3 1)h{ = h{ ({2 + 4{ + 1)
We can see that our answers are plausible, since i has horizontal tangents
(b)
where i 0 ({) = 0, and i 0 has horizontal tangents where i 00 ({) = 0.
41. i ({) =
{2
1+{
i i 0 ({) =
i 00 ({) =
=
so i 00 (1) =
42. j({) =
{
h{
j 00 ({) =
(1 + {)(2{) 3 {2 (1)
2{ + 2{2 3 {2
{2 + 2{
=
= 2
2
2
(1 + {)
(1 + {)
{ + 2{ + 1
i
({2 + 2{ + 1)(2{ + 2) 3 ({2 + 2{)(2{ + 2)
(2{ + 2)({2 + 2{ + 1 3 {2 3 2{)
=
({2 + 2{ + 1)2
[({ + 1)2 ]2
2({ + 1)(1)
2
=
,
({ + 1)4
({ + 1)3
2
2
1
= = .
(1 + 1)3
8
4
i j0 ({) =
h{ (1 3 {)
h{ · 1 3 { · h{
13{
=
=
{
2
(h )
(h{ )2
h{
h{ · (31) 3 (1 3 {)h{
h{ [31 3 (1 3 {)]
{32
=
=
{
2
(h )
(h{ )2
h{
i
i
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
j 000 ({) =
h{ · 1 3 ({ 3 2)h{
h{ [1 3 ({ 3 2)]
33{
=
=
(h{ )2
(h{ )2
h{
j (4) ({) =
h{ · (31) 3 (3 3 {)h{
h{ [31 3 (3 3 {)]
{34
=
=
.
(h{ )2
(h{ )2
h{
The pattern suggests that j (q) ({) =
¤
187
i
({ 3 q)(31)q
. (We could use mathematical induction to prove this formula.)
h{
43. We are given that i(5) = 1, i 0 (5) = 6, j(5) = 33, and j0 (5) = 2.
(a) (i j)0 (5) = i (5)j 0 (5) + j(5)i 0 (5) = (1)(2) + (33)(6) = 2 3 18 = 316
0
j(5)i 0 (5) 3 i(5)j0 (5)
(33)(6) 3 (1)(2)
20
i
(5) =
=
=3
(b)
j
[j(5)]2
(33)2
9
(c)
0
i (5)j 0 (5) 3 j(5)i 0 (5)
(1)(2) 3 (33)(6)
j
(5) =
=
= 20
i
[i(5)]2
(1)2
44. We are given that i(2) = 33, j(2) = 4, i 0 (2) = 32, and j 0 (2) = 7.
(a) k({) = 5i ({) 3 4j({) i k0 ({) = 5i 0 ({) 3 4j 0 ({), so
k0 (2) = 5i 0 (2) 3 4j0 (2) = 5(32) 3 4(7) = 310 3 28 = 338.
(b) k({) = i ({)j({) i k0 ({) = i ({)j0 ({) + j({)i 0 ({), so
k0 (2) = i(2)j0 (2) + j(2)i 0 (2) = (33)(7) + (4)(32) = 321 3 8 = 329.
(c) k({) =
k0 (2) =
(d) k({) =
k0 (2) =
i ({)
j({)
j({)i 0 ({) 3 i({)j0 ({)
, so
[j({)]2
13
j(2)i 0 (2) 3 i (2)j0 (2)
4(32) 3 (33)(7)
38 + 21
=
.
=
=
42
16
16
[j(2)]2
j({)
1 + i({)
i k0 ({) =
[1 + i ({)] j0 ({) 3 j({)i 0 ({)
, so
[1 + i ({)]2
[1 + i(2)] j 0 (2) 3 j(2) i 0 (2)
[1 + (33)](7) 3 4(32)
314 + 8
36
3
=
=
=
=3 .
[1 + i ({)]2
[1 + (33)]2
(32)2
4
2
45. i ({) = h{ j({)
46.
i k0 ({) =
i i 0 ({) = h{ j 0 ({) + j({)h{ = h{ [j0 ({) + j({)]. i 0 (0) = h0 [j 0 (0) + j(0)] = 1(5 + 2) = 7
g k({)
{k0 ({) 3 k({) · 1
=
g{
{
{2
47. j({) = {i ({)
i
310
g k({)
2k0 (2) 3 k(2)
2(33) 3 (4)
=
= 32=5
=
=
g{
{ {=2
22
4
4
i j 0 ({) = {i 0 ({) + i({) · 1. Now j(3) = 3i (3) = 3 · 4 = 12 and
j 0 (3) = 3i 0 (3) + i(3) = 3(32) + 4 = 32. Thus, an equation of the tangent line to the graph of j at the point where { = 3
is | 3 12 = 32({ 3 3), or | = 32{ + 18.
48. i 0 ({) = {2 i({)
i i 00 ({) = {2 i 0 ({) + i ({) · 2{. Now i 0 (2) = 22 i (2) = 4(10) = 40, so
i 00 (2) = 22 (40) + 10(4) = 200.
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CHAPTER 3 DIFFERENTIATION RULES
49. (a) From the graphs of i and j, we obtain the following values: i (1) = 2 since the point (1> 2) is on the graph of i ;
j(1) = 1 since the point (1> 1) is on the graph of j; i 0 (1) = 2 since the slope of the line segment between (0> 0) and
(2> 4) is
034
430
= 2; j 0 (1) = 31 since the slope of the line segment between (32> 4) and (2> 0) is
= 31.
230
2 3 (32)
Now x({) = i ({)j({), so x0 (1) = i(1)j0 (1) + j(1) i 0 (1) = 2 · (31) + 1 · 2 = 0.
2 3 13 3 3 · 23
3 83
2
j(5)i 0 (5) 3 i (5)j0 (5)
=3
=
=
(b) y({) = i({)@j({), so y 0 (5) =
[j(5)]2
22
4
3
50. (a) S ({) = I ({) J({), so S 0 (2) = I (2) J0 (2) + J(2) I 0 (2) = 3 ·
(b) T({) = I ({)@J({), so T0 (7) =
51. (a) | = {j({)
i
1·
J(7) I 0 (7) 3 I (7) J0 (7)
=
[J(7)]2
+ 2 · 0 = 32 =
1
4
3 5 · 3 23
10
43
1
=
= +
12
4
3
12
| 0 = {j0 ({) + j({) · 1 = {j0 ({) + j({)
(b) | =
{
j({)
i |0 =
j({) · 1 3 {j0 ({)
j({) 3 {j 0 ({)
=
[j({)]2
[j({)]2
(c) | =
j({)
{
i |0 =
{j 0 ({) 3 j({) · 1
{j 0 ({) 3 j({)
=
2
({)
{2
52. (a) | = {2 i ({)
2
4
i | 0 = {2 i 0 ({) + i({)(2{)
(b) | =
i ({)
{2
i |0 =
{2 i 0 ({) 3 i ({)(2{)
{i 0 ({) 3 2i ({)
=
({2 )2
{3
(c) | =
{2
i ({)
i |0 =
i ({)(2{) 3 {2 i 0 ({)
[i ({)]2
(d) | =
1 + {i({)
I
{
i
I
1
{ [{i 0 ({) + i ({)] 3 [1 + {i ({)] I
2 {
0
| =
I 2
( {)
=
{3@2 i 0 ({) + {1@2 i ({) 3 12 {31@2 3 12 {1@2 i ({) 2{1@2
{i ({) + 2{2 i 0 ({) 3 1
· 1@2 =
{
2{
2{3@2
53. If | = i ({) =
|3
{
({ + 1)(1) 3 {(1)
1
=
. When { = d, the equation of the tangent line is
, then i 0 ({) =
{+1
({ + 1)2
({ + 1)2
1
1
d
d
=
=
({ 3 d). This line passes through (1> 2) when 2 3
(1 3 d) C
d+1
(d + 1)2
d+1
(d + 1)2
2(d + 1)2 3 d(d + 1) = 1 3 d
2d2 + 4d + 2 3 d2 3 d 3 1 + d = 0 C d2 + 4d + 1 = 0.
s
I
I
34 ± 42 3 4(1)(1)
34 ± 12
=
= 32 ± 3,
The quadratic formula gives the roots of this equation as d =
2(1)
2
C
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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
¤
so there are two such tangent lines. Since
I
I
I
I 32 ± 3
32 ± 3 31 ~ 3
I
I ·
I
i 32 ± 3 =
=
32 ± 3 + 1
31 ± 3 31 ~ 3
I
I
I
I
2±2 3~ 333
31 ± 3
1~ 3
=
=
=
,
133
32
2
I I
the lines touch the curve at D 32 + 3> 1 32 3 E (30=27> 30=37)
I
and E 32 3 3>
54. | =
{31
{+1
I 1+ 3
2
i |0 =
E (33=73> 1=37).
({ + 1)(1) 3 ({ 3 1)(1)
2
=
. If the tangent intersects
({ + 1)2
({ + 1)2
the curve when { = d, then its slope is 2@(d + 1)2 . But if the tangent is parallel to
{ 3 2| = 2, that is, | = 12 { 3 1, then its slope is 12 . Thus,
2
1
=
(d + 1)2
2
i
(d + 1)2 = 4 i d + 1 = ±2 i d = 1 or 33. When d = 1, | = 0 and the
equation of the tangent is | 3 0 = 12 ({ 3 1) or | = 12 { 3 12 . When d = 33, | = 2 and
the equation of the tangent is | 3 2 = 12 ({ + 3) or | = 12 { + 72 .
55. U =
i
j
i
U0 =
ji 0 3 i j 0
. For i({) = { 3 3{3 + 5{5 , i 0 ({) = 1 3 9{2 + 25{4 ,
j2
and for j({) = 1 + 3{3 + 6{6 + 9{9 , j0 ({) = 9{2 + 36{5 + 81{8 .
Thus, U0 (0) =
56. T =
i
j
i
j(0)i 0 (0) 3 i (0)j 0 (0)
1·130·0
1
=
= = 1.
[j(0)]2
12
1
T0 =
ji 0 3 i j0
. For i ({) = 1 + { + {2 + {h{ , i 0 ({) = 1 + 2{ + {h{ + h{ ,
j2
and for j({) = 1 3 { + {2 3 {h{ , j 0 ({) = 31 + 2{ 3 {h{ 3 h{ .
Thus, T0 (0) =
j(0)i 0 (0) 3 i (0)j 0 (0)
1 · 2 3 1 · (32)
4
=
= = 4.
[j(0)]2
12
1
57. If S (w) denotes the population at time w and D(w) the average annual income, then W (w) = S (w)D(w) is the total personal
income. The rate at which W (w) is rising is given by W 0 (w) = S (w)D0 (w) + D(w)S 0 (w) i
W 0 (1999) = S (1999)D0 (1999) + D(1999)S 0 (1999) = (961,400)($1400@yr) + ($30,593)(9200@yr)
= $1,345,960,000@yr + $281,455,600@yr = $1,627,415,600@yr
So the total personal income was rising by about $1.627 billion per year in 1999.
The term S (w)D0 (w) E $1.346 billion represents the portion of the rate of change of total income due to the existing
population’s increasing income. The term D(w)S 0 (w) E $281 million represents the portion of the rate of change of total
income due to increasing population.
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190
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
58. (a) i (20) = 10,000 means that when the price of the fabric is $20@yard, 10,000 yards will be sold.
i 0 (20) = 3350 means that as the price of the fabric increases past $20@yard, the amount of fabric which will be sold is
decreasing at a rate of 350 yards per (dollar per yard).
(b) U(s) = si (s) i U0 (s) = si 0 (s) + i (s) · 1 i U0 (20) = 20i 0 (20) + i (20) · 1 = 20(3350) + 10,000 = 3000.
This means that as the price of the fabric increases past $20@yard, the total revenue is increasing at $3000@($@yard). Note
that the Product Rule indicates that we will lose $7000@($@yard) due to selling less fabric, but this loss is more than made
up for by the additional revenue due to the increase in price.
59. (a) (i jk)0 = [(ij)k]0 = (i j)0 k + (i j)k0 = (i 0 j + ij 0 )k + (i j)k0 = i 0 jk + i j0 k + i jk0
(b) Putting i = j = k in part (a), we have
(c)
g
[i ({)]3 = (i i i)0 = i 0 i i + ii 0 i + i i i 0 = 3i i i 0 = 3[i({)]2 i 0 ({).
g{
g 3{
g { 3
(h ) =
(h ) = 3(h{ )2 h{ = 3h2{ h{ = 3h3{
g{
g{
60. (a) We use the Product Rule repeatedly: I = i j
i
I 0 = i 0 j + i j0
i
I 00 = (i 00 j + i 0 j 0 ) + (i 0 j 0 + i j 00 ) = i 00 j + 2i 0 j 0 + i j00 .
(b) I 000 = i 000 j + i 00 j0 + 2 (i 00 j 0 + i 0 j00 ) + i 0 j 00 + i j000 = i 000 j + 3i 00 j 0 + 3i 0 j00 + i j000
I (4) = i (4) j + i 000 j0 + 3 (i 000 j 0 + i 00 j 00 ) + 3 (i 00 j00 + i 0 j000 ) + i 0 j 000 + i j(4)
i
= i (4) j + 4i 000 j0 + 6i 00 j 00 + 4i 0 j 000 + i j (4)
(c) By analogy with the Binomial Theorem, we make the guess:
# $
# $
q (q32) 00
q (q3n) (n)
(q)
(q)
(q31) 0
I
= i j + qi
j +
i
j + ··· +
i
j + · · · + qi 0 j (q31) + i j (q) ,
2
n
# $
q
q!
q(q 3 1)(q 3 2) · · · (q 3 n + 1)
where
=
=
.
n
n! (q 3 n)!
n!
61. For i ({) = {2 h{ , i 0 ({) = {2 h{ + h{ (2{) = h{ ({2 + 2{). Similarly, we have
i 00 ({) = h{ ({2 + 4{ + 2)
i 000 ({) = h{ ({2 + 6{ + 6)
i (4) ({) = h{ ({2 + 8{ + 12)
i (5) ({) = h{ ({2 + 10{ + 20)
It appears that the coef¿cient of { in the quadratic term increases by 2 with each differentiation. The pattern for the
constant terms seems to be 0 = 1 · 0, 2 = 2 · 1, 6 = 3 · 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that
i (q) ({) = h{ [{2 + 2q{ + q(q 3 1)].
Proof: Let Vq be the statement that i (q) ({) = h{ [{2 + 2q{ + q(q 3 1)].
1. V1 is true because i 0 ({) = h{ ({2 + 2{).
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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
¤
2. Assume that Vn is true; that is, i (n) ({) = h{ [{2 + 2n{ + n(n 3 1)]. Then
g k (n) l
i (n+1) ({) =
i ({) = h{ (2{ + 2n) + [{2 + 2n{ + n(n 3 1)]h{
g{
= h{ [{2 + (2n + 2){ + (n2 + n)] = h{ [{2 + 2(n + 1){ + (n + 1)n]
This shows that Vn+1 is true.
3. Therefore, by mathematical induction, Vq is true for all q; that is, i (q) ({) = h{ [{2 + 2q{ + q(q 3 1)] for every
positive integer q.
g
62. (a)
g{
1
j({)
=
1
v + nhv
(b) | =
g
g
(1) 3 1 ·
[j({)]
g{
g{
2
[j({)]
j({) ·
i |0 = 3
g
g
({3q ) =
(c)
g{
g{
1
{q
[Quotient Rule] =
j({) · 0 3 1 · j0 ({)
0 3 j0 ({)
j 0 ({)
=
=3
2
2
[j({)]
[j({)]
[j({)]2
1 + nhv
(v + nhv )2
=3
({q )0
({q )2
[by the Reciprocal Rule] = 3
q{q31
= 3q{q3132q = 3q{3q31
{2q
3.3 Derivatives of Trigonometric Functions
1. i ({) = 3{2 3 2 cos {
2. i ({) =
i i 0 ({) = 6{ 3 2(3 sin {) = 6{ + 2 sin {
I
I
I
sin {
{ sin { i i 0 ({) = { cos { + sin { 12 {31@2 = { cos { + I
2 {
3. i ({) = sin { +
1
2
cot { i i 0 ({) = cos { 3
1
2
csc2 {
i | 0 = 2(sec { tan {) 3 (3 csc { cot {) = 2 sec { tan { + csc { cot {
4. | = 2 sec { 3 csc {
i |0 = sec (sec2 ) + tan (sec tan ) = sec (sec2 + tan2 ). Using the identity
5. | = sec tan 1 + tan2 = sec2 , we can write alternative forms of the answer as sec (1 + 2 tan2 ) or sec (2 sec2 3 1).
6. j() = h (tan 3 )
7. | = f cos w + w2 sin w
8. i (w) =
9. | =
cot w
hw
i
{
2 3 tan {
10. | = sin cos 11. i () =
sec 1 + sec i 0 () =
j0 () = h (sec2 3 1) + (tan 3 )h = h (sec2 3 1 + tan 3 )
i
| 0 = f(3 sin w) + w2 (cos w) + sin w (2w) = 3f sin w + w(w cos w + 2 sin w)
i
i 0 (w) =
i |0 =
hw (3 csc2 w) 3 (cot w)hw
hw (3 csc2 w 3 cot w)
csc2 w + cot w
=
=3
w
2
w
2
(h )
(h )
hw
(2 3 tan {)(1) 3 {(3 sec2 {)
2 3 tan { + { sec2 {
=
2
(2 3 tan {)
(2 3 tan {)2
i | 0 = sin (3 sin ) + cos (cos ) = cos2 3 sin2 [or cos 2]
i
(1 + sec )(sec tan ) 3 (sec )(sec tan )
(sec tan ) [(1 + sec ) 3 sec ]
sec tan =
=
(1 + sec )2
(1 + sec )2
(1 + sec )2
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192
¤
12. | =
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
cos {
1 3 sin {
i
(1 3 sin {)(3 sin {) 3 cos {(3 cos {)
3 sin { + sin2 { + cos2 {
=
(1 3 sin {)2
(1 3 sin {)2
3 sin { + 1
1
=
=
(1 3 sin {)2
1 3 sin {
|0 =
13. | =
|0 =
14. | =
w sin w
1+w
i
(1 + w)(w cos w + sin w) 3 w sin w(1)
(w2 + w) cos w + sin w
w cos w + sin w + w2 cos w + w sin w 3 w sin w
=
=
2
2
(1 + w)
(1 + w)
(1 + w)2
1 3 sec {
tan {
i
sec { 3 tan2 { 3 sec { + sec2 {
tan { (3 sec { tan {) 3 (1 3 sec {)(sec2 {)
sec { (1 3 sec {)
| =
=
=
2
2
tan
{
tan2 {
(tan {)
0
15. Using Exercise 3.2.59(a), i ({) = {h{ csc {
i
i 0 ({) = ({)0 h{ csc { + {(h{ )0 csc { + {h{ (csc {)0 = 1h{ csc { + {h{ csc { + {h{ (3 cot { csc {)
= h{ csc { (1 + { 3 { cot {)
16. Using Exercise 3.2.59(a), i ({) = {2 sin { tan {
i
i 0 ({) = ({2 )0 sin { tan { + {2 (sin {)0 tan { + {2 sin { (tan {)0 = 2{ sin { tan { + {2 cos { tan { + {2 sin { sec2 {
= 2{ sin { tan { + {2 sin { + {2 sin { sec2 { = { sin { (2 tan { + { + { sec2 {).
17.
g
g
(csc {) =
g{
g{
1
sin {
=
3 cos {
1
(sin {)(0) 3 1(cos {)
cos {
=
=3
·
= 3 csc { cot {
sin { sin {
sin2 {
sin2 {
18.
g
g
(sec {) =
g{
g{
1
cos {
=
(cos {)(0) 3 1(3 sin {)
sin {
1
sin {
=
=
·
= sec { tan {
cos2 {
cos2 {
cos { cos {
19.
sin2 { + cos2 {
1
g
g cos { (sin {)(3 sin {) 3 (cos {)(cos {)
=3
= 3 2 = 3 csc2 {
(cot {) =
=
2
g{
g{ sin {
sin {
sin2 {
sin {
20. i ({) = cos {
i
i({ + k) 3 i ({)
cos ({ + k) 3 cos {
cos { cos k 3 sin { sin k 3 cos {
= lim
= lim
k<0
k<0
k
k
k
cos k 3 1
sin k
sin k
cos k 3 1
3 sin {
= cos { lim
3 sin { lim
= lim cos {
k<0
k<0
k<0
k
k
k
k
i 0 ({) = lim
k<0
= (cos {)(0) 3 (sin {)(1) = 3 sin {
I
i |0 = sec { tan {, so | 0 ( 3 ) = sec 3 tan 3 = 2 3. An equation of the tangent line to the curve | = sec {
I
I
I at the point 3 > 2 is | 3 2 = 2 3 { 3 3 or | = 2 3 { + 2 3 23 3 .
21. | = sec {
22. | = h{ cos {
i | 0 = h{ (3 sin {) + (cos {)h{ = h{ (cos { 3 sin {)
i the slope of the tangent line at (0> 1) is
h0 (cos 0 3 sin 0) = 1(1 3 0) = 1 and an equation is | 3 1 = 1({ 3 0) or | = { + 1.
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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
¤
193
i | 0 = 3 sin { 3 cos {, so | 0 () = 3 sin 3 cos = 0 3 (31) = 1. An equation of the tangent
23. | = cos { 3 sin {
line to the curve | = cos { 3 sin { at the point (> 31) is | 3 (31) = 1({ 3 ) or | = { 3 3 1.
i | 0 = 1 + sec2 {, so | 0 () = 1 + (31)2 = 2. An equation of the tangent line to the curve
24. | = { + tan {
| = { + tan { at the point (> ) is | 3 = 2({ 3 ) or | = 2{ 3 .
i |0 = 2({ cos { + sin { · 1). At 2 > ,
| 0 = 2 2 cos 2 + sin 2 = 2(0 + 1) = 2, and an equation of the
tangent line is | 3 = 2 { 3 2 , or | = 2{.
25. (a) | = 2{ sin {
(b)
i | 0 = 3 3 6 sin {. At 3 > + 3 ,
I
I
| 0 = 3 3 6 sin 3 = 3 3 6 23 = 3 3 3 3, and an equation of the
I tangent line is | 3 ( + 3) = 3 3 3 3 { 3 3 , or
I I
| = 3 3 3 3 { + 3 + 3.
26. (a) | = 3{ + 6 cos {
27. (a) i ({) = sec { 3 {
(b)
i i 0 ({) = sec { tan { 3 1
(b)
Note that i 0 = 0 where i has a minimum. Also note that i 0 is negative
when i is decreasing and i 0 is positive when i is increasing.
28. (a) i ({) = h{ cos {
i i 0 ({) = h{ (3 sin {) + (cos {) h{ = h{ (cos { 3 sin {)
i
i 00 ({) = h{ (3 sin { 3 cos {) + (cos { 3 sin {) h{ = h{ (3 sin { 3 cos { + cos { 3 sin {) = 32h{ sin {
(b)
Note that i 0 = 0 where i has a minimum and i 00 = 0 where i 0 has a
minimum. Also note that i 0 is negative when i is decreasing and i 00 is
negative when i 0 is decreasing.
29. K() = sin i K 0 () = (cos ) + (sin ) · 1 = cos + sin i
K 00 () = (3 sin ) + (cos ) · 1 + cos = 3 sin + 2 cos 30. i (w) = csc w
i
so i 00 ( 6 ) = 2(22 +
i 0 (w) = 3 csc w cot w
I 2
3 ) = 2(4 + 3) = 14.
i
i 00 (w) = 3[csc w(3 csc2 w) + cot w(3 csc w cot w)] = csc w(csc2 w + cot2 w),
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NOT FOR SALE
¤
CHAPTER 3 DIFFERENTIATION RULES
31. (a) i ({) =
i 0 ({) =
tan { 3 1
sec {
i
sec {(sec2 {) 3 (tan { 3 1)(sec { tan {)
sec {(sec2 { 3 tan2 { + tan {)
1 + tan {
=
=
(sec {)2
sec 2 {
sec {
sin { 3 cos {
sin {
31
tan { 3 1
cos
{
cos {
=
= sin { 3 cos { i i 0 ({) = cos { 3 (3 sin {) = cos { + sin {
(b) i ({) =
=
1
1
sec {
cos {
cos {
(c) From part (a), i 0 ({) =
32. (a) j({) = i ({) sin {
1
tan {
1 + tan {
=
+
= cos { + sin {, which is the expression for i 0 ({) in part (b).
sec {
sec {
sec {
i j0 ({) = i ({) cos { + sin { · i 0 ({), so
j 0 ( 3 ) = i( 3 ) cos 3 + sin 3 · i 0 ( 3 ) = 4 ·
1
2
+
I
3
2
· (32) = 2 3
I
3
i ({) · (3 sin {) 3 cos { · i 0 ({)
, so
[i ({)]2
I I
I
0 4 3 23 3 12 (32)
i
(
)
·
(3
sin
)
3
cos
·
i
(
)
132 3
32 3 + 1
3
3
k0 ( 3 ) =
=
=
=
3 2 3
42
16
16
i
(b) k({) =
cos {
i ({)
i k0 ({) =
3
33. i ({) = { + 2 sin { has a horizontal tangent when i 0 ({) = 0
{=
2
3
+ 2q or
4
3
+ 2q, where q is an integer. Note that
1 + 2 cos { = 0 C cos { = 3 12
C
4
3
and
2
3
C
are ± 3 units from . This allows us to write the
solutions in the more compact equivalent form (2q + 1) ± 3 , q an integer.
34. i ({) = h{ cos { has a horizontal tangent when i 0 ({) = 0. i 0 ({) = h{ (3 sin {) + (cos {)h{ = h{ (cos { 3 sin {).
i 0 ({) = 0
C
35. (a) {(w) = 8 sin w
cos { 3 sin { = 0
C
cos { = sin {
C
tan { = 1
C
{=
4
+ q, q an integer.
i y(w) = {0 (w) = 8 cos w i d(w) = {00 (w) = 38 sin w
(b) The mass at time w =
2
3
I I
2
3
= 4 3, velocity y 2
=
8
sin
= 8 cos 2
has position { 2
=
8
= 8 3 12 = 34,
3
3
2
3
3
I I
= 38 sin 2
? 0, the particle is moving to the left.
and acceleration d 2
= 38 23 = 34 3. Since y 2
3
3
3
36. (a) v(w) = 2 cos w + 3 sin w
i y(w) = 32 sin w + 3 cos w i
(b)
d(w) = 32 cos w 3 3 sin w
(c) v = 0 i w2 E 2=55.
So the mass passes through the equilibrium
position for the ¿rst time when w E 2=55 s.
(d) y = 0 i w1 E 0=98, v(w1 ) E 3=61 cm.
So the mass travels
a maximum of about 3=6 cm (upward and downward) from its equilibrium position.
(e) The speed |y| is greatest when v = 0, that is, when w = w2 + q, q a positive integer.
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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
¤
195
From the diagram we can see that sin = {@10 C { = 10 sin . We want to ¿nd the rate
37.
of change of { with respect to , that is, g{@g. Taking the derivative of { = 10 sin , we get
g{@g = 10(cos ). So when =
38. (a) I =
(b)
Z
sin + cos i
g{
,
3 g
= 10 cos 3 = 10 12 = 5 ft@rad.
Z (sin 3 cos )
gI
( sin + cos )(0) 3 Z ( cos 3 sin )
=
=
g
( sin + cos )2
( sin + cos )2
gI
= 0 C Z (sin 3 cos ) = 0 C sin = cos g
From the graph of I =
(c)
C tan = C = tan31 0=6(50)
for 0 $ $ 1, we see that
0=6 sin + cos gI
= 0 i E 0=54. Checking this with part (b) and = 0=6, we
g
calculate = tan31 0=6 E 0=54. So the value from the graph is consistent
with the value in part (b).
39. lim
{<0
sin 3{
3 sin 3{
= lim
{<0
{
3{
sin 3{
= 3 lim
3{<0
3{
sin = 3 lim
<0
= 3(1)
[multiply numerator and denominator by 3]
[as { < 0, 3{ < 0]
[let = 3{]
[Equation 2]
=3
40. lim
sin 4{
= lim
sin 6{ {<0
41. lim
tan 6w
= lim
w<0
sin 2w
{<0
w<0
= 6 lim
w<0
sin 4{
{
·
{
sin 6{
{<0
4 sin 4{
6{
sin 4{ 1
6{
1
2
· lim
= 4 lim
· lim
= 4(1) · (1) =
{<0 6 sin 6{
{<0
4{
4{
6 {<0 sin 6{
6
3
= lim
w<0
6 sin 6w
1
2w
· lim
· lim
w<0 cos 6w w<0 2 sin 2w
6w
sin 6w
1
2w
1
1 1
· lim
· lim
= 6(1) · · (1) = 3
w<0 cos 6w
6w
2 w<0 sin 2w
1 2
cos 3 1
cos 3 1
lim
0
<0
=
= =0
sin sin 1
lim
<0
43. lim
sin 3{
= lim
5{3 3 4{ {<0
44. lim
sin 3{ sin 5{
= lim
{<0
{2
{<0
= lim
sin 6w
1
w
·
·
w
cos 6w sin 2w
cos 3 1
= lim
42. lim
<0
<0
sin {<0
sin 3{
3
· 2
3{
5{ 3 4
= lim
{<0
sin 3{
3
· lim
= 1·
{<0 5{2 3 4
3{
3
34
=3
3
4
3 sin 3{
5 sin 5{
3 sin 3{ 5 sin 5{
·
= lim
· lim
{<0
{<0
3{
5{
3{
5{
sin 3{
sin 5{
= 3 lim
· 5 lim
= 3(1) · 5(1) = 15
{<0
{<0
3{
5{
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196
NOT FOR SALE
¤
CHAPTER 3 DIFFERENTIATION RULES
45. Divide numerator and denominator by .
sin lim
= lim
<0 + tan <0
46. lim
{<0
47.
lim
{<1
49.
sin sin lim
1
1
<0
=
=
=
sin 1
1
sin 1+1·1
2
·
lim
1+
1 + lim
<0
cos <0 cos sin({2 )
sin({2 )
sin |
sin({2 )
= lim { ·
= lim { · lim
= 0 · lim
{<0
{<0
{<0
{
{·{
{2
|
|<0+
{<@4
48. lim
(sin also works.)
1 3 tan {
= lim
sin { 3 cos { {<@4
=0·1=0
sin {
13
· cos {
I
cos {
cos { 3 sin {
31
31
= lim
= lim
= I =3 2
(sin { 3 cos {) · cos { {<@4 (sin { 3 cos {) cos { {<@4 cos {
1@ 2
sin({ 3 1)
sin({ 3 1)
1
sin({ 3 1)
= lim
= lim
lim
=
{2 + { 3 2 {<1 ({ + 2)({ 3 1) {<1 { + 2 {<1 { 3 1
g
(sin {) = cos { i
g{
where | = {2
g2
(sin {) = 3 sin { i
g{2
1
3
·1=
1
3
g3
(sin {) = 3 cos { i
g{3
The derivatives of sin { occur in a cycle of four. Since 99 = 4(24) + 3, we have
g4
(sin {) = sin {.
g{4
g3
g99
(sin
{)
=
(sin {) = 3 cos {.
g{99
g{3
50. Let i({) = { sin { and k({) = sin {, so i ({) = {k({). Then i 0 ({) = k({) + {k0 ({),
i 00 ({) = k0 ({) + k0 ({) + {k00 ({) = 2k0 ({) + {k00 ({),
i 000 ({) = 2k00 ({) + k00 ({) + {k000 ({) = 3k00 ({) + {k000 ({)> · · · > i (q) ({) = qk(q31) ({) + {k(q) ({).
Since 34 = 4(8) + 2, we have k(34) ({) = k(2) ({) =
Thus,
g2
(sin {) = 3 sin { and k(35) ({) = 3 cos {.
g{2
g35
({ sin {) = 35k(34) ({) + {k(35) ({) = 335 sin { 3 { cos {.
g{35
51. | = D sin { + E cos {
i | 0 = D cos { 3 E sin { i | 00 = 3D sin { 3 E cos {. Substituting these
expressions for |, | 0 , and | 00 into the given differential equation |00 + | 0 3 2| = sin { gives us
(3D sin { 3 E cos {) + (D cos { 3 E sin {) 3 2(D sin { + E cos {) = sin { C
33D sin { 3 E sin { + D cos { 3 3E cos { = sin { C (33D 3 E) sin { + (D 3 3E) cos { = 1 sin {, so we must have
33D 3 E = 1 and D 3 3E = 0 (since 0 is the coef¿cient of cos { on the right side). Solving for D and E, we add the ¿rst
1
3
and D = 3 10
.
equation to three times the second to get E = 3 10
52. (a) Let =
1
1
1
sin . Then as { < ", < 0+ , and lim { sin = lim
sin = lim
= 1.
{<"
<0
{
{ <0+ (b) Since 31 $ sin (1@{) $ 1, we have (as illustrated in the ¿gure)
3 |{| $ { sin (1@{) $ |{|. We know that lim (|{|) = 0 and
{<0
lim (3 |{|) = 0; so by the Squeeze Theorem, lim { sin (1@{) = 0.
{<0
{<0
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NOT FOR SALE
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
¤
197
(c)
53. (a)
g
g sin {
tan { =
g{
g{ cos {
i sec2 { =
cos { cos { 3 sin { (3 sin {)
cos2 { + sin2 {
=
.
2
cos {
cos2 {
g 1
(cos {)(0) 3 1(3 sin {)
g
sec { =
i sec { tan { =
.
g{
g{ cos {
cos2 {
g
g 1 + cot {
(c)
(sin { + cos {) =
i
g{
g{ csc {
(b)
cos { 3 sin { =
=
So sec { tan { =
So sec2 { =
1
.
cos2 {
sin {
.
cos2 {
csc { (3 csc2 {) 3 (1 + cot {)(3 csc { cot {)
csc { [3 csc2 { + (1 + cot {) cot {]
=
2
csc {
csc2 {
3 csc2 { + cot2 { + cot {
31 + cot {
=
csc {
csc {
So cos { 3 sin { =
cot { 3 1
.
csc {
54. We get the following formulas for u and k in terms of :
sin
u
=
2
10
i
u = 10 sin
2
and cos
k
=
2
10
i
k = 10 cos
2
Now D() = 12 u2 and E() = 12 (2u)k = uk. So
lim
<0+
1
u2
D()
u
10 sin(@2)
= lim 2
= 12 lim
= 12 lim
E()
<0+ uk
<0+ k
<0+ 10 cos(@2)
= 12 lim tan(@2) = 0
<0+
55. By the de¿nition of radian measure, v = u, where u is the radius of the circle. By drawing the bisector of the angle , we can
see that sin
g@2
=
2
u
v
u
2 · (@2)
@2
i g = 2u sin . So lim
= lim
= lim
= lim
= 1.
<0 sin(@2)
2
<0+ g
<0+ 2u sin(@2)
<0+ 2 sin(@2)
[This is just the reciprocal of the limit lim
{<0
56. (a)
sin {
{
= 1 combined with the fact that as < 0,
2
< 0 also=]
{
has a jump discontinuity at { = 0.
It appears that i ({) = I
1 3 cos 2{
{
{
{
{
(b) Using the identity cos 2{ = 1 3 sin2 {, we have I
= s
= I
.
= I
1 3 cos 2{
2 |sin {|
1 3 (1 3 2 sin2 {)
2 sin2 {
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¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
Thus,
{
{
{
1
= lim I
= I lim
lim I
1 3 cos 2{ {<0 2 |sin {|
2 {<0 3(sin {)
I
1
2
1
1 1
= 3 I lim
= 3I · = 3
2
2 {<0 sin {@{
2 1
I
Evaluating lim i ({) is similar, but | sin {| = + sin {, so we get 12 2. These values appear to be reasonable values for
{<0
{<0+
the graph, so they con¿rm our answer to part (a).
Another method: Multiply numerator and denominator by
I
1 + cos 2{.
3.4 The Chain Rule
1. Let x = j({) = 1 + 4{ and | = i (x) =
I
g| gx
g|
4
3
=
= ( 13 x32@3 )(4) = s
x. Then
.
g{
gx g{
3 3 (1 + 4{)2
2. Let x = j({) = 2{3 + 5 and | = i(x) = x4 . Then
3. Let x = j({) = { and | = i (x) = tan x. Then
g|
g| gx
=
= (sec2 x)() = sec2 {.
g{
gx g{
4. Let x = j({) = cot { and | = i (x) = sin x. Then
5. Let x = j({) =
g| gx
g|
=
= (4x3 )(6{2 ) = 24{2 (2{3 + 5)3 .
g{
gx g{
g|
g| gx
=
= (cos x)(3 csc2 {) = 3 cos(cot {) csc2 {.
g{
gx g{
I
I
I
g|
1
h {
g| gx
{ and | = i (x) = hx . Then
=
= (hx ) 12 {31@2 = h { · I = I .
g{
gx g{
2 {
2 {
6. Let x = j({) = 2 3 h{ and | = i (x) =
7. I ({) = ({4 + 3{2 3 2)5
I
g|
h{
g| gx
x. Then
.
=
= ( 12 x31@2 )(3h{ ) = 3 I
g{
gx g{
2 2 3 h{
i I 0 ({) = 5({4 + 3{2 3 2)4 ·
or 10{({4 + 3{2 3 2)4 (2{2 + 3)
8. I ({) = (4{ 3 {2 )100
i I 0 ({) = 100(4{ 3 {2 )99 ·
or 200{99 ({ 3 2)({ 3 4)99
9. I ({) =
I
1 3 2{ = (1 3 2{)1@2
i
10. i ({) =
1
= (1 + sec {)32
(1 + sec {)2
11. i (}) =
1
= (} 2 + 1)31
}2 + 1
12. i (w) = sin(hw ) + hsin w
i
g 4
{ + 3{2 3 2 = 5({4 + 3{2 3 2)4 (4{3 + 6{)
g{
g
(4{ 3 {2 ) = 100(4{ 3 {2 )99 (4 3 2{)
g{
1
I 0 ({) = 12 (1 3 2{)31@2 (32) = 3 I
1 3 2{
i i 0 ({) = 32(1 + sec {)33 · sec { tan { =
i 0 (}) = 31(} 2 + 1)32 (2}) = 3
32 sec { tan {
(1 + sec {)3
2}
(} 2 + 1)2
i i 0 (w) = cos(hw ) · hw + hsin w · cos w = hw cos(hw ) + hsin w cos w
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.4 THE CHAIN RULE
13. | = cos(d3 + {3 )
14. | = d3 + cos3 {
15. | = {h3n{
i | 0 = 3 sin(d3 + {3 ) · 3{2
¤
[d3 is just a constant] = 33{2 sin(d3 + {3 )
i |0 = 3(cos {)2 (3 sin {) [d3 is just a constant] = 33 sin { cos2 {
i | 0 = { h3n{ (3n) + h3n{ · 1 = h3n{ (3n{ + 1) or (1 3 n{)h3n{
16. | = h32w cos 4w
i
| 0 = h32w (3 sin 4w · 4) + cos 4w[h32w (32)] = 32h32w (2 sin 4w + cos 4w)
17. i ({) = (2{ 3 3)4 ({2 + { + 1)5
i
i 0 ({) = (2{ 3 3)4 · 5({2 + { + 1)4 (2{ + 1) + ({2 + { + 1)5 · 4(2{ 3 3)3 · 2
= (2{ 3 3)3 ({2 + { + 1)4 [(2{ 3 3) · 5(2{ + 1) + ({2 + { + 1) · 8]
= (2{ 3 3)3 ({2 + { + 1)4 (20{2 3 20{ 3 15 + 8{2 + 8{ + 8) = (2{ 3 3)3 ({2 + { + 1)4 (28{2 3 12{ 3 7)
18. j({) = ({2 + 1)3 ({2 + 2)6
i
j 0 ({) = ({2 + 1)3 · 6({2 + 2)5 · 2{ + ({2 + 2)6 · 3({2 + 1)2 · 2{
= 6{({2 + 1)2 ({2 + 2)5 [2({2 + 1) + ({2 + 2)] = 6{({2 + 1)2 ({2 + 2)5 (3{2 + 4)
19. k(w) = (w + 1)2@3 (2w2 3 1)3
i
k0 (w) = (w + 1)2@3 · 3(2w2 3 1)2 · 4w + (2w2 3 1)3 · 23 (w + 1)31@3 = 23 (w + 1)31@3 (2w2 3 1)2 [18w(w + 1) + (2w2 3 1)]
= 23 (w + 1)31@3 (2w2 3 1)2 (20w2 + 18w 3 1)
20. I (w) = (3w 3 1)4 (2w + 1)33
i
I 0 (w) = (3w 3 1)4 (33)(2w + 1)34 (2) + (2w + 1)33 · 4(3w 3 1)3 (3)
= 6(3w 3 1)3 (2w + 1)34 [3(3w 3 1) + 2(2w + 1)] = 6(3w 3 1)3 (2w + 1)34 (w + 3)
21. | =
{2 + 1
{2 3 1
3
i
2
2
2
2
{ +1
{ +1
g {2 + 1
({2 3 1)(2{) 3 ({2 + 1)(2{)
=
3
·
·
|0 = 3 2
2
2
{ 31
g{ { 3 1
{ 31
({2 3 1)2
2
2
2
2
{ +1
{ +1
2{[{2 3 1 3 ({2 + 1)]
2{(32)
312{({2 + 1)2
=3 2
·
=
3
· 2
=
{ 31
({2 3 1)2
{2 3 1
({ 3 1)2
({2 3 1)4
u
v2 + 1
i
v2 + 4
31@2 2
1@2
(v + 4)(2v) 3 (v2 + 1)(2v)
2v[(v2 + 4) 3 (v2 + 1)]
1 v2 + 1
1 v2 + 4
=
i 0 (v) =
2
2
2
2
2 v +4
(v + 4)
2 v +1
(v2 + 4)2
22. i (v) =
=
23. | =
(v2 + 4)1@2 (2v)(3)
3v
= 2
2(v2 + 1)1@2 (v2 + 4)2
(v + 1)1@2 (v2 + 4)3@2
I
1 + 2h3{
i |0 =
1
1
3h3{
g
(2h3{ · 3) = I
(1 + 2h3{ )31@2
(1 + 2h3{ ) = I
3{
2
g{
2 1 + 2h
1 + 2h3{
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199
200
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
2
24. Using Formula 5 and the Chain Rule, | = 1013{
25. Using Formula 5 and the Chain Rule, | = 531@{
26. J(|) =
(| 3 1)4
(| 2 + 2|)5
J0 (|) =
=
=
2
i | 0 = 1013{ (ln 10) ·
2
g 1 3 {2 = 32{(ln 10)1013{ .
g{
i | 0 = 531@{ (ln 5) 31 · 3{32 = 531@{ (ln 5)@{2
i
(| 2 + 2|)5 · 4(| 3 1)3 · 1 3 (| 3 1)4 · 5(| 2 + 2|)4 (2| + 2)
2
(| 2 + 2|)5
2(| 2 + 2|)4 (| 3 1)3 2(| 2 + 2|) 3 5(| 3 1)(| + 1)
(| 2 + 2|)10
2(| 3 1)3 (2|2 + 4|) + (35| 2 + 5)
u
u2 + 1
27. | = I
(| 2 + 2|)6
=
2(| 3 1)3 33| 2 + 4| + 5
(| 2 + 2|)6
i
I
u2
I
u2 + 1 3 I
2
31@2
1
2
u + 1 (1) 3 u · 2 (u + 1)
(2u)
u2 + 1
=
=
|0 =
2
2
I
I
u2 + 1
u2 + 1
2
u + 1 3 u2
1
= I
or (u2 + 1)33@2
3 = 2
2
(u
+
1)3@2
u +1
I
I
u2 + 1 u2 + 1 3 u2
I
u2 + 1
2
I
u2 + 1
Another solution: Write | as a product and make use of the Product Rule. | = u(u2 + 1)31@2
i
| 0 = u · 3 12 (u2 + 1)33@2 (2u) + (u2 + 1)31@2 · 1 = (u2 + 1)33@2 [3u2 + (u2 + 1)1 ] = (u2 + 1)33@2 (1) = (u2 + 1)33@2 .
The step that students usually have trouble with is factoring out (u2 + 1)33@2 . But this is no different than factoring out {2
from {2 + {5 ; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, 3 32 is
smaller than 3 12 .
28. | =
|0 =
=
hx 3 h3x
hx + h3x
i
(hx + h3x )(hx 3 (3h3x )) 3 (hx 3 h3x )(hx + (3h3x ))
h2x + h0 + h0 + h32x 3 (h2x 3 h0 3 h0 + h32x )
=
(hx + h3x )2
(hx + h3x )2
4h0
4
=
(hx + h3x )2
(hx + h3x )2
29. By (9), I (w) = hw sin 2w
i I 0 (w) = hw sin 2w (w sin 2w)0 = hw sin 2w (w · 2 cos 2w + sin 2w · 1) = hw sin 2w (2w cos 2w + sin 2w)
6
y
i
y3 + 1
5 3
y
6y 5 (y 3 + 1 3 3y 3 )
6y 5 (1 3 2y 3 )
(y + 1)(1) 3 y(3y2 )
=
=
I 0 (y) = 6
y3 + 1
(y 3 + 1)2
(y 3 + 1)5 (y 3 + 1)2
(y 3 + 1)7
30. I (y) =
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.4 THE CHAIN RULE
i | 0 = cos(tan 2{) ·
31. | = sin(tan 2{)
g
g
(tan 2{) = cos(tan 2{) · sec2 (2{) ·
(2{) = 2 cos(tan 2{) sec2 (2{)
g{
g{
33. Using Formula 5 and the Chain Rule, | = 2 sin {
i | 0 = {2 h31@{
1 3 h2{
1 + h2{
i
g
(sin {) = 2 sin { (ln 2) · cos { · = 2 sin { ( ln 2) cos {
g{
| 0 = 2 sin { (ln 2) ·
35. | = cos
201
i | 0 = 2[sec(p)]1 sec(p) tan(p) · p = 2p sec2 (p) tan(p)
32. | = sec2 (p) = [sec(p)]2
34. | = {2 h31@{
¤
1
{2
+ h31@{ (2{) = h31@{ + 2{h31@{ = h31@{ (1 + 2{)
i
1 3 h2{
1 3 h2{
g 1 3 h2{
(1 + h2{ )(32h2{ ) 3 (1 3 h2{ )(2h2{ )
·
=
3
sin
·
| 0 = 3 sin
1 + h2{
g{ 1 + h2{
1 + h2{
(1 + h2{ )2
32h2{ (1 + h2{ ) + (1 3 h2{ )
32h2{ (2)
1 3 h2{
1 3 h2{
4h2{
1 3 h2{
= 3 sin
=
3
sin
=
·
sin
·
·
1 + h2{
(1 + h2{ )2
1 + h2{
(1 + h2{ )2
(1 + h2{ )2
1 + h2{
36. | =
I
1 + {h32{
i |0 =
1
2
31@2 h32{ (32{ + 1)
1 + {h32{
{ 32h32{ + h32{ = I
2 1 + {h32{
37. | = cot2 (sin ) = [cot(sin )]2
| 0 = 2[cot(sin )] ·
38. | = hn tan
I
{
g
[cot(sin )] = 2 cot(sin ) · [3 csc2 (sin ) · cos ] = 32 cos cot(sin ) csc2 (sin )
g
i | 0 = hn tan
39. i (w) = tan(hw ) + htan w
40. | = sin(sin(sin {))
2w
41. i (w) = sin2 hsin
i
I
{
·
n sec2 I{
I I
I
I g I
n tan { = hn tan { n sec2 { · 12 {31@2 =
hn tan {
g{
2 {
i i 0 (w) = sec2 (hw ) ·
i |0 = cos(sin(sin {))
k 2 l2
= sin hsin w
g w
g
(h ) + htan w ·
(tan w) = sec2 (hw ) · hw + htan w · sec2 w
gw
gw
g
(sin(sin {)) = cos(sin(sin {)) cos(sin {) cos {
g{
i
k 2 l g
2 2 g
2 2
i 0 (w) = 2 sin hsin w ·
sin hsin w = 2 sin hsin w · cos hsin w ·
hsin w
gw
gw
2 2 2 2 2
2
g
sin2 w = 2 sin hsin w cos hsin w hsin w · 2 sin w cos w
= 2 sin hsin w cos hsin w · hsin w ·
gw
2 2 2
= 4 sin hsin w cos hsin w hsin w sin w cos w
42. | =
t
s
s
I 31@2 I
I 31@2
1 + 12 { + {
1 + 12 {31@2
{ + { + { i |0 = 12 { + { + {
43. j({) = (2udu{ + q)s
i
j 0 ({) = s(2udu{ + q)s31 ·
g
(2udu{ + q) = s(2udu{ + q)s31 · 2udu{ (ln d) · u = 2u2 s(ln d)(2udu{ + q)s31 du{
g{
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NOT FOR SALE
¤
CHAPTER 3 DIFFERENTIATION RULES
{2
44. | = 2 3
45. | = cos
{2
i | 0 = 2 3 (ln 2)
{2
2
g {2 = 2 3 (ln 2)3{ (ln 3)(2{)
3
g{
s
sin(tan {) = cos(sin(tan {))1@2
i
g
g
(sin(tan {))1@2 = 3 sin(sin(tan {))1@2 · 12 (sin(tan {))31@2 ·
(sin(tan {))
g{
g{
s
s
3 sin sin(tan {)
3 sin sin(tan {)
g
s
s
=
· cos(tan {) ·
· cos(tan {) · sec2 ({) · tan { =
g{
2 sin(tan {)
2 sin(tan {)
s
3 cos(tan {) sec2 ({) sin sin(tan {)
s
=
2 sin(tan {)
| 0 = 3 sin(sin(tan {))1@2 ·
46. | = { + ({ + sin2 {)3
47. | = cos({2 )
i
4
3 i | 0 = 4 { + ({ + sin2 {)3 · 1 + 3({ + sin2 {)2 · (1 + 2 sin { cos {)
| 0 = 3 sin({2 ) · 2{ = 32{ sin({2 )
i
| 00 = 32{ cos({2 ) · 2{ + sin({2 ) · (32) = 34{2 cos({2 ) 3 2 sin({2 )
48. | = cos2 { = (cos {)2
i
| 0 = 2 cos {(3 sin {) = 32 cos { sin {
i
| 00 = (32 cos {) cos { + sin {(2 sin {) = 32 cos2 { + 2 sin2 {
Note: Many other forms of the answers exist. For example, | 0 = 3 sin 2{ and | 00 = 32 cos 2{.
49. | = h{ sin {
i | 0 = h{ · cos { + sin { · h{ = h{ ( cos { + sin {) i
|00 = h{ (3 2 sin { + cos {) + ( cos { + sin {) · h{
= h{ (3 2 sin { + cos { + cos { + 2 sin {) = h{ (2 sin { 3 2 sin { + 2 cos {)
= h{ (2 3 2 ) sin { + 2 cos {
50. | = hh
{
| 00 = hh
{
{
{
i | 0 = hh · (h{ )0 = hh · h{ i
{ 0
{
{
{
{
· (h{ )0 + h{ · hh
= hh · h{ + h{ · hh · h{ = hh · h{ (1 + h{ ) or hh +{ (1 + h{ )
51. | = (1 + 2{)10
i | 0 = 10(1 + 2{)9 · 2 = 20(1 + 2{)9 .
At (0> 1), |0 = 20(1 + 0)9 = 20, and an equation of the tangent line is | 3 1 = 20({ 3 0), or | = 20{ + 1.
52. | =
I
1 + {3 = (1 + {3 )1@2
i
3{2
3·4
| 0 = 12 (1 + {3 )31@2 · 3{2 = I
. At (2> 3), |0 = I = 2, and an equation of
2 1 + {3
2 9
the tangent line is | 3 3 = 2({ 3 2), or | = 2{ 3 1.
53. | = sin(sin {)
i | 0 = cos(sin {) · cos {. At (> 0), |0 = cos(sin ) · cos = cos(0) · (31) = 1(31) = 31, and an
equation of the tangent line is | 3 0 = 31({ 3 ), or | = 3{ + .
54. | = sin { + sin2 {
i | 0 = cos { + 2 sin { cos {.
At (0> 0), |0 = 1, and an equation of the tangent line is | 3 0 = 1({ 3 0), or | = {.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.4 THE CHAIN RULE
55. (a) | =
2
1 + h3{
At (0> 1), |0 =
i |0 =
(1 + h3{ )(0) 3 2(3h3{ )
2h3{
=
.
3{
2
(1 + h )
(1 + h3{ )2
¤
(b)
2h0
2(1)
2
1
=
= 2 = . So an equation of the
(1 + h0 )2
(1 + 1)2
2
2
tangent line is | 3 1 = 12 ({ 3 0) or | = 12 { + 1.
{
2 3 {2
56. (a) For { A 0, |{| = {, and | = i ({) = I
i
(b)
I
2 3 {2 (1) 3 { 12 (2 3 {2 )31@2 (32{) (2 3 {2 )1@2
·
i ({) =
I
2
(2 3 {2 )1@2
2 3 {2
0
=
(2 3 {2 ) + {2
2
=
(2 3 {2 )3@2
(2 3 {2 )3@2
So at (1> 1), the slope of the tangent line is i 0 (1) = 2 and its equation
is | 3 1 = 2({ 3 1) or | = 2{ 3 1.
I
57. (a) i ({) = { 2 3 {2 = {(2 3 {2 )1@2
i
2 3 2{2
i 0 ({) = { · 12 (2 3 {2 )31@2 (32{) + (2 3 {2 )1@2 · 1 = (2 3 {2 )31@2 3{2 + (2 3 {2 ) = I
2 3 {2
i 0 = 0 when i has a horizontal tangent line, i 0 is negative when i is
(b)
decreasing, and i 0 is positive when i is increasing.
58. (a)
From the graph of i , we see that there are 5 horizontal tangents, so there
must be 5 zeros on the graph of i 0 . From the symmetry of the graph of i ,
we must have the graph of i 0 as high at { = 0 as it is low at { = . The
intervals of increase and decrease as well as the signs of i 0 are indicated in
the ¿gure.
(b) i ({) = sin({ + sin 2{) i
i 0 ({) = cos({ + sin 2{) ·
g
({ + sin 2{) = cos({ + sin 2{)(1 + 2 cos 2{)
g{
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203
204
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
59. For the tangent line to be horizontal, i 0 ({) = 0. i({) = 2 sin { + sin2 {
i i 0 ({) = 2 cos { + 2 sin { cos { = 0 C
2 cos {(1 + sin {) = 0 C cos { = 0 or sin { = 31, so { = 2 + 2q or 3
+ 2q, where q is any integer. Now
2
= 31, so the points on the curve with a horizontal tangent are 2 + 2q> 3 and 3
+ 2q> 31 ,
i 2 = 3 and i 3
2
2
where q is any integer.
60. i ({) = sin 2{ 3 2 sin {
i i 0 ({) = 2 cos 2{ 3 2 cos { = 4 cos2 { 3 2 cos { 3 2, and 4 cos2 { 3 2 cos { 3 2 = 0 C
(cos { 3 1)(4 cos { + 2) = 0 C cos { = 1 or cos { = 3 12 . So { = 2q or (2q + 1) ± 3 , q any integer.
61. I ({) = i (j({))
62. k({) =
i I 0 ({) = i 0 (j({)) · j 0 ({), so I 0 (5) = i 0 (j(5)) · j 0 (5) = i 0 (32) · 6 = 4 · 6 = 24
s
4 + 3i ({) i k0 ({) =
1
2 (4
+ 3i ({))31@2 · 3i 0 ({), so
k0 (1) = 12 (4 + 3i (1))31@2 · 3i 0 (1) = 12 (4 + 3 · 7)31@2 · 3 · 4 =
63. (a) k({) = i (j({))
I6
25
=
6
5
i k0 ({) = i 0 (j({)) · j 0 ({), so k0 (1) = i 0 (j(1)) · j 0 (1) = i 0 (2) · 6 = 5 · 6 = 30.
(b) K({) = j(i ({)) i K 0 ({) = j 0 (i ({)) · i 0 ({), so K 0 (1) = j 0 (i (1)) · i 0 (1) = j 0 (3) · 4 = 9 · 4 = 36.
64. (a) I ({) = i (i ({))
i I 0 ({) = i 0 (i ({)) · i 0 ({), so I 0 (2) = i 0 (i (2)) · i 0 (2) = i 0 (1) · 5 = 4 · 5 = 20.
(b) J({) = j(j({)) i J0 ({) = j0 (j({)) · j0 ({), so J0 (3) = j 0 (j(3)) · j 0 (3) = j 0 (2) · 9 = 7 · 9 = 63.
65. (a) x({) = i (j({))
i x0 ({) = i 0 (j({))j 0 ({). So x0 (1) = i 0 (j(1))j 0 (1) = i 0 (3)j 0 (1). To ¿nd i 0 (3), note that i is
1
334
= 3 . To ¿nd j0 (1), note that j is linear from (0> 6) to (2> 0), so its slope
632
4
036
= 33. Thus, i 0 (3)j0 (1) = 3 14 (33) = 34 .
is
230
linear from (2> 4) to (6> 3), so its slope is
(b) y({) = j(i ({)) i y 0 ({) = j 0 (i ({))i 0 ({). So y 0 (1) = j 0 (i (1))i 0 (1) = j0 (2)i 0 (1), which does not exist since
j 0 (2) does not exist.
(c) z({) = j(j({)) i z0 ({) = j 0 (j({))j 0 ({). So z0 (1) = j0 (j(1))j 0 (1) = j0 (3)j 0 (1). To ¿nd j0 (3), note that j is
linear from (2> 0) to (5> 2), so its slope is
66. (a) k({) = i (i({))
2
230
= . Thus, j0 (3)j 0 (1) = 23 (33) = 32.
532
3
i k0 ({) = i 0 (i ({))i 0 ({). So k0 (2) = i 0 (i(2))i 0 (2) = i 0 (1)i 0 (2) E (31)(31) = 1.
(b) j({) = i ({2 ) i j 0 ({) = i 0 ({2 ) ·
g 2
{ = i 0 ({2 )(2{). So j 0 (2) = i 0 (2 2 )(2 · 2) = 4i 0 (4) E 4(2) = 8.
g{
67. The point (3> 2) is on the graph of i , so i (3) = 2. The tangent line at (3> 2) has slope
j({) =
s
i({) i j 0 ({) = 12 [i ({)]31@2 · i 0 ({) i
34
2
{|
=
=3 .
{{
6
3
I
1
j 0 (3) = 12 [i (3)]31@2 · i 0 (3) = 12 (2)31@2 (3 23 ) = 3 I or 3 16 2.
3 2
68. (a) I ({) = i ({ )
(b) J({) = [i ({)]
i I 0 ({) = i 0 ({ )
g
({ ) = i 0 ({ ){31
g{
i J0 ({) = [i ({)]31 i 0 ({)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.4 THE CHAIN RULE
69. (a) I ({) = i (h{ )
i I 0 ({) = i 0 (h{ )
i J0 ({) = hi ({)
(b) J({) = hi ({)
70. (a) j({) = hf{ + i ({)
¤
205
g {
(h ) = i 0 (h{ )h{
g{
g
i({) = hi ({) i 0 ({)
g{
i j0 ({) = hf{ · f + i 0 ({) i j0 (0) = h0 · f + i 0 (0) = f + 5.
j 0 ({) = fhf{ + i 0 ({) i j00 ({) = fhf{ · f + i 00 ({) i j 00 (0) = f2 h0 + i 00 (0) = f2 3 2.
(b) k({) = hn{ i ({) i k0 ({) = hn{ i 0 ({) + i ({) · nhn{
i k0 (0) = h0 i 0 (0) + i (0) · nh0 = 5 + 3n.
An equation of the tangent line to the graph of k at the point (0> k(0)) = (0> i (0)) = (0> 3) is
| 3 3 = (5 + 3n)({ 3 0) or | = (5 + 3n){ + 3.
71. u({) = i (j(k({)))
i u0 ({) = i 0 (j(k({))) · j 0 (k({)) · k0 ({), so
u0 (1) = i 0 (j(k(1))) · j 0 (k(1)) · k0 (1) = i 0 (j(2)) · j0 (2) · 4 = i 0 (3) · 5 · 4 = 6 · 5 · 4 = 120
72. i ({) = {j({2 )
i i 0 ({) = {j 0 ({2 ) 2{ + j({2 ) · 1 = 2{2 j 0 ({2 ) + j({2 ) i
i 00 ({) = 2{2 j 00 ({2 ) 2{ + j 0 ({2 ) 4{ + j 0 ({2 ) 2{ = 4{3 j00 ({2 ) + 4{j0 ({2 ) + 2{j 0 ({2 ) = 6{j0 ({2 ) + 4{3 j00 ({2 )
73. I ({) = i (3i (4i ({)))
i
I 0 ({) = i 0 (3i (4i ({))) ·
g
g
(3i (4i ({))) = i 0 (3i (4i ({))) · 3i 0 (4i({)) ·
(4i({))
g{
g{
= i 0 (3i (4i ({))) · 3i 0 (4i({)) · 4i 0 ({), so
I 0 (0) = i 0 (3i (4i(0))) · 3i 0 (4i(0)) · 4i 0 (0) = i 0 (3i (4 · 0)) · 3i 0 (4 · 0) · 4 · 2 = i 0 (3 · 0) · 3 · 2 · 4 · 2 = 2 · 3 · 2 · 4 · 2 = 96.
74. I ({) = i ({i ({i ({)))
i
I 0 ({) = i 0 ({i ({i({))) ·
g
g
({i ({i ({))) = i 0 ({i({i ({))) · { · i 0 ({i ({)) ·
({i({)) + i ({i ({)) · 1
g{
g{
= i 0 ({i ({i({))) · [{i 0 ({i ({)) · ({i 0 ({) + i ({) · 1) + i ({i ({))] , so
I 0 (1) = i 0 (i(i (1))) · [i 0 (i(1)) · (i 0 (1) + i(1)) + i (i(1))] = i 0 (i (2)) · [i 0 (2) · (4 + 2) + i(2)]
= i 0 (3) · [5 · 6 + 3] = 6 · 33 = 198.
75. | = h2{ (D cos 3{ + E sin 3{)
i
| 0 = h2{ (33D sin 3{ + 3E cos 3{) + (D cos 3{ + E sin 3{) · 2h2{
= h2{ (33D sin 3{ + 3E cos 3{ + 2D cos 3{ + 2E sin 3{)
= h2{ [(2D + 3E) cos 3{ + (2E 3 3D) sin 3{] i
|00 = h2{ [33(2D + 3E) sin 3{ + 3(2E 3 3D) cos 3{] + [(2D + 3E) cos 3{ + (2E 3 3D) sin 3{] · 2h2{
= h2{ {[33(2D + 3E) + 2(2E 3 3D)] sin 3{ + [3(2E 3 3D) + 2(2D + 3E)] cos 3{}
= h2{ [(312D 3 5E) sin 3{ + (35D + 12E) cos 3{]
Substitute the expressions for |, |0 , and | 00 in | 00 3 4| 0 + 13| to get
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¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
| 00 3 4| 0 + 13| = h2{ [(312D 3 5E) sin 3{ + (35D + 12E) cos 3{]
3 4h2{ [(2D + 3E) cos 3{ + (2E 3 3D) sin 3{] + 13h2{ (D cos 3{ + E sin 3{)
= h2{ [(312D 3 5E 3 8E + 12D + 13E) sin 3{ + (35D + 12E 3 8D 3 12E + 13D) cos 3{]
= h2{ [(0) sin 3{ + (0) cos 3{] = 0
Thus, the function | satis¿es the differential equation |00 3 4| 0 + 13| = 0.
76. | = hu{
| 0 = uhu{
i
u2 hu{ 3 4uhu{ + hu{ = 0
u2 3 4u + 1 = 0
i
| 00 = u2 hu{ . Substituting |, | 0 , and | 00 into | 00 3 4| 0 + | = 0 gives us
i
hu{ (u2 3 4u + 1) = 0. Since hu{ 6= 0, we must have
I
I
4 ± 16 3 4
= 2 ± 3.
u=
2
i
77. The use of G, G2 , = = =, Gq is just a derivative notation (see text page 157). In general, Gi (2{) = 2i 0 (2{),
G2 i (2{) = 4i 00 (2{), = = =, Gq i (2{) = 2q i (q) (2{). Since i ({) = cos { and 50 = 4(12) + 2, we have
i (50) ({) = i (2) ({) = 3 cos {, so G50 cos 2{ = 3250 cos 2{.
78. i ({) = {h3{ , i 0 ({) = h3{ 3 {h3{ = (1 3 {)h3{ , i 00 ({) = 3h3{ + (1 3 {)(3h3{ ) = ({ 3 2)h3{ .
Similarly,
i 000 ({) = (3 3 {)h3{ , i (4) ({) = ({ 3 4)h3{ , = = =, i (1000) ({) = ({ 3 1000)h3{ .
79. v(w) = 10 +
1
4
sin(10w) i the velocity after w seconds is y(w) = v0 (w) =
80. (a) v = D cos($w + )
1
4
cos(10w)(10) =
cos(10w) cm@s.
i velocity = v0 = 3$D sin($w + ).
(b) If D 6= 0 and $ 6= 0, then v0 = 0 C sin($w + ) = 0 C $w + = q
81. (a) E(w) = 4=0 + 0=35 sin
(b) At w = 1,
5
2
2w
5=4
i
gE
=
gw
C w=
q 3 , q an integer.
$
2w
2
0=7
2w
7
2w
0=35 cos
=
cos
=
cos
5=4
5=4
5=4
5=4
54
5=4
gE
7
2
=
cos
E 0=16.
gw
54
5=4
2
82. O(w) = 12 + 2=8 sin
365 (w
3 80)
2
2 .
i O0 (w) = 2=8 cos 365
(w 3 80) 365
On March 21, w = 80, and O0 (80) E 0=0482 hours per day. On May 21, w = 141, and O0 (141) E 0=02398, which is
approximately one-half of O0 (80).
83. v(w) = 2h31=5w sin 2w
i
y(w) = v0 (w) = 2[h31=5w (cos 2w)(2) + (sin 2w)h31=5w (31=5)] = 2h31=5w (2 cos 2w 3 1=5 sin 2w)
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.4 THE CHAIN RULE
84. (a) lim s(w) = lim
w<"
w<"
¤
207
1
1
= 1, since n A 0 i 3nw < 3" i h3nw < 0.
=
1 + dh3nw
1+d·0
(b) s(w) = (1 + dh3nw )31
i
gs
ndh3nw
= 3(1 + dh3nw )32 (3ndh3nw ) =
gw
(1 + dh3nw )2
(c)
From the graph of s(w) = (1 + 10h30=5w )31 , it seems that s(w) = 0=8
(indicating that 80% of the population has heard the rumor) when
w E 7=4 hours.
85. By the Chain Rule, d(w) =
gy
gy gv
gy
gy
=
=
y(w) = y(w) . The derivative gy@gw is the rate of change of the velocity
gw
gv gw
gv
gv
with respect to time (in other words, the acceleration) whereas the derivative gy@gv is the rate of change of the velocity with
respect to the displacement.
86. (a) The derivative gY @gu represents the rate of change of the volume with respect to the radius and the derivative gY @gw
represents the rate of change of the volume with respect to time.
(b) Since Y =
4 3 gY
gu
gY gu
u ,
=
= 4u2
.
3
gw
gu gw
gw
87. (a) Using a calculator or CAS, we obtain the model T = dew with d E 100=0124369 and e E 0=000045145933.
(b) Use T0 (w) = dew ln e (from Formula 5) with the values of d and e from part (a) to get T0 (0=04) E 3670=63 A.
The result of Example 2 in Section 2.1 was 3670 A.
88. (a) S = dew with d = 4=502714 × 10320 and e = 1=029953851,
where S is measured in thousands of people. The ¿t appears to be very good.
(b) For 1800: p1 =
5308 3 3929
7240 3 5308
= 137=9, p2 =
= 193=2.
1800 3 1790
1810 3 1800
So S 0 (1800) E (p1 + p2 )@2 = 165=55 thousand people@year.
For 1850: p1 =
23,192 3 17,063
31,443 3 23,192
= 612=9, p2 =
= 825=1.
1850 3 1840
1860 3 1850
So S 0 (1850) E (p1 + p2 )@2 = 719 thousand people@year.
(c) Using S 0 (w) = dew ln e (from Formula 7) with the values of d and e from part (a), we get S 0 (1800) E 156=85 and
S 0 (1850) E 686=07. These estimates are somewhat less than the ones in part (b).
(d) S (1870) E 41,946=56. The difference of 3=4 million people is most likely due to the Civil War (1861–1865).
89. (a) Derive gives j 0 (w) =
j 0 (w) = 9
45(w 3 2)8
without simplifying. With either Maple or Mathematica, we ¿rst get
(2w + 1)10
(w 3 2)8
(w 3 2)9
3 18
, and the simpli¿cation command results in the expression given by Derive.
9
(2w + 1)
(2w + 1)10
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208
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
(b) Derive gives | 0 = 2({3 3 { + 1)3 (2{ + 1)4 (17{3 + 6{2 3 9{ + 3) without simplifying. With either Maple or
Mathematica, we ¿rst get | 0 = 10(2{ + 1)4 ({3 3 { + 1)4 + 4(2{ + 1)5 ({3 3 { + 1)3 (3{2 3 1). If we use
Mathematica’s Factor or Simplify, or Maple’s factor, we get the above expression, but Maple’s simplify gives
the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.
90. (a) i ({) =
4
{ 3{+1
{4 + { + 1
give i 0 ({) = u
1@2
(3{ 3 1)
3{4 3 1
4
u
{4 3 { + 1
{4 + { + 1
whereas either Maple or Mathematica
. Derive gives i 0 ({) = 4
({ + { + 1)({4 3 { + 1)
4
{ 3{+1 4
({ + { + 1)2
{4 + { + 1
after simpli¿cation.
t
(b) i 0 ({) = 0 C 3{4 3 1 = 0 C { = ± 4 13 E ±0=7598.
(c) Yes. i 0 ({) = 0 where i has horizontal tangents. i 0 has two maxima and
one minimum where i has inÀection points.
91. (a) If i is even, then i ({) = i (3{). Using the Chain Rule to differentiate this equation, we get
i 0 ({) = i 0 (3{)
g
(3{) = 3i 0 (3{). Thus, i 0 (3{) = 3i 0 ({), so i 0 is odd.
g{
(b) If i is odd, then i({) = 3i (3{). Differentiating this equation, we get i 0 ({) = 3i 0 (3{)(31) = i 0 (3{), so i 0 is
even.
92.
i ({)
j({)
0
= i({) [j({)]31
=
0
= i 0 ({) [j({)]31 + (31) [j({)]32 j0 ({)i({)
i 0 ({)j({) 3 i ({)j0 ({)
i 0 ({)
i({)j 0 ({)
=
3
2
j({)
[j({)]
[j({)]2
This is an alternative derivation of the formula in the Quotient Rule. But part of the purpose of the Quotient Rule is to show
that if i and j are differentiable, so is i@j. The proof in Section 3.2 does that; this one doesn’t.
93. (a)
g
(sinq { cos q{) = q sinq31 { cos { cos q{ + sinq { (3q sin q{)
g{
= q sinq31 { (cos q{ cos { 3 sin q{ sin {)
[factor out q sinq31 {]
= q sinq31 { cos(q{ + {)
[Addition Formula for cosine]
q31
= q sin
(b)
[Product Rule]
{ cos[(q + 1){]
g
(cosq { cos q{) = q cosq31 { (3 sin {) cos q{ + cosq { (3q sin q{)
g{
[factor out {]
[Product Rule]
= 3q cosq31 { (cos q{ sin { + sin q{ cos {)
[factor out 3q cosq31 {]
= 3q cosq31 { sin(q{ + {)
[Addition Formula for sine]
q31
= 3q cos
{ sin[(q + 1){]
[factor out {]
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
SECTION 3.4 THE CHAIN RULE
94. “The rate of change of | 5 with respect to { is eighty times the rate of change of | with respect to {”
g|
g 5
| = 80
g{
g{
C
5| 4
g|
g|
= 80
g{
g{
C
C 5| 4 = 80 (Note that g|@g{ 6= 0 since the curve never has a
horizontal tangent) C |4 = 16 C | = 2 (since | A 0 for all {)
95. Since =
180
g
g rad, we have
(sin ) =
sin 180
=
g
g
96. (a) i ({) = |{| =
I
{2 = ({2 )1@2
180
cos 180
=
180
cos .
I
i i 0 ({) = 12 ({2 )31@2 (2{) = {@ {2 = {@ |{| for { 6= 0.
i is not differentiable at { = 0.
(b) i ({) = |sin {| =
I
sin2 { i
sin {
cos {
|sin {|
i 0 ({) = 12 (sin2 {)31@2 2 sin { cos { =
=
+
cos {
if sin { A 0
3 cos { if sin { ? 0
i is not differentiable when { = q, q an integer.
(c) j({) = sin |{| = sin
I
{2
i
+
cos {
if { A 0
{
{
=
cos { =
j ({) = cos |{| ·
|{|
|{|
3 cos { if { ? 0
0
j is not differentiable at 0.
g|
g| gx
=
, so
g{
gx g{
g g|
gx
g2 |
g g| gx
g| g gx
g g|
=
=
+
=
g{2
g{ g{
g{ gx g{
g{ gx
g{
gx g{ g{
2
g| g2 x
g2 | gx
g| g2 x
g g| gx gx
+
=
+
=
gx gx g{ g{
gx g{2
gx2 g{
gx g{2
97. The Chain Rule says that
g2 |
g2 |
98. From Exercise 97,
=
g{2
gx2
gx
g{
g g2 |
g
g3 |
=
=
3
g{
g{ g{2
g{
2
%
+
g2 |
gx2
g| g2 x
gx g{2
gx
g{
i
2 &
+
g g| g2 x
g{ gx g{2
%
gx
g{
2 &
g
=
g{
g2 |
gx2
gx
g{
g2 |
gx2
gx
g{
+3
gx g2 x g2 |
g| g3 x
+
g{ g{2 gx2
gx g{3
g
=
gx
=
g3 |
gx3
gx
g{
3
2
gx
g{
g
+
g{
2
[Product Rule]
2
g2 |
g|
g x
g g|
g g2 x
+
+
gx2
g{ gx
g{2
g{ g{2
gx
2 gx g2 x g2 |
g g| gx
g3 x g|
g x
+2
+
+ 3
2
2
2
g{ g{ gx
gx gx g{
g{
g{ gx
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°
licated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
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209
210
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
APPLIED PROJECT Where Should a Pilot Start Descent?
1. Condition (i) will hold if and only if all of the following four conditions hold:
() S (0) = 0
() S 0 (0) = 0 (for a smooth landing)
() S 0 (c) = 0 (since the plane is cruising horizontally when it begins its descent)
() S (c) = k.
First of all, condition implies that S (0) = g = 0, so S ({) = d{3 + e{2 + f{ i S 0 ({) = 3d{2 + 2e{ + f. But
S 0 (0) = f = 0 by condition . So S 0 (c) = 3dc2 + 2ec = c (3dc + 2e). Now by condition , 3dc + 2e = 0 i d = 3
Therefore, S ({) = 3
2e
.
3c
2e
2e 3
{ + e{2 . Setting S (c) = k for condition , we get S (c) = 3 c3 + ec2 = k i
3c
3c
2
3 ec2 + ec2 = k i
3
2k
2k
3k
. So | = S ({) = 3 3 {3 + 2 {2 .
c3
c
c
2 g |
g{
= 3y for all w, so { (w) = c 3 yw. Condition (iii) states that 2 $ n. By the Chain Rule,
2. By condition (ii),
gw
gw
we have
1 2
ec = k
3
i
e=
3k
c2
i d=3
g{
g| g{
2k 3k
6k{2 y
g{
6k{y
g|
=
= 3 3 3{2
+ 2 (2{)
=
3 2
gw
g{ gw
c
gw
c
gw
c3
c
(for { $ c) i
g2 |
12ky 2
6ky
g{ 6ky g{
6ky 2
3
=
3
=
(2{)
{
+
. In particular, when w = 0, { = c and so
gw2
c3
gw
c2 gw
c3
c2
g2 | g2 | 12ky 2
6ky 2
6ky 2
6ky 2
=
3
c
+
=
3
.
Thus,
=
$ n. (This condition also follows from taking { = 0.)
gw2 w=0
c3
c2
c2
gw2 w=0
c2
1 mi
, and y = 300 mi@h into the result of part (b):
5280 ft
u
1
(300)2
6 35,000 · 5280
35,000
$
860
i
c
D
300
6·
E 64=5 miles.
c2
5280 · 860
3. We substitute n = 860 mi@h2 , k = 35,000 ft ×
4. Substituting the values of k and c in Problem 3 into
S ({) = 3
2k 3 3k 2
{ + 2 { gives us S ({) = d{3 + e{2 ,
c3
c
where d E 34=937 × 1035 and e E 4=78 × 1033 .
3.5 Implicit Differentiation
g
g
9{
(9{2 3 | 2 ) =
(1) i 18{ 3 2| | 0 = 0 i 2| | 0 = 18{ i | 0 =
g{
g{
|
I
9{
(b) 9{2 3 | 2 = 1 i |2 = 9{2 3 1 i | = ± 9{2 3 1, so | 0 = ± 12 (9{2 3 1)31@2 (18{) = ± I
.
9{2 3 1
1. (a)
(c) From part (a), |0 =
9{
9{
, which agrees with part (b).
= I
|
± 9{2 3 1
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
¤
SECTION 3.5 IMPLICIT DIFFERENTIATION
2. (a)
211
g
g
4{ + | + 1
(2{2 + { + {|) =
(1) i 4{ + 1 + {| 0 + | · 1 = 0 i {| 0 = 34{ 3 | 3 1 i | 0 = 3
g{
g{
{
(b) 2{2 + { + {| = 1 i {| = 1 3 2{2 3 { i | =
1
1
3 2{ 3 1, so | 0 = 3 2 3 2
{
{
(c) From part (a),
|0 = 3
1
1
1
4{ + | + 1
= 34 3 | 3 = 34 3
{
{
{
{
1
1
3 2{ 3 1 3
{
{
= 34 3
1
1
1
1
+ 2 + 3 = 3 2 3 2, which
{2
{
{
{
agrees with part (b).
3. (a)
(b)
g
g{
1
1
+
{
|
=
1
1
+ =1 i
{
|
(c) | 0 = 3
4. (a)
g
1
1
1
1
(1) i 3 2 3 2 | 0 = 0 i 3 2 | 0 = 2
g{
{
|
|
{
1
1
{31
=13 =
|
{
{
7.
s
s
| sin { = 2 (5 3 cos {)2 = 2(5 3 cos {) sin { [since 5 3 cos { A 0].
g 3
g
{ + |3 =
(1) i 3{2 + 3| 2 · | 0 = 0 i 3| 2 | 0 = 33{2
g{
g{
s g I
1
g
1
2 {+ | =
(3) i 2 · {31@2 + | 31@2 · | 0 = 0 i
g{
g{
2
2
s
2 |
|0
1
0
s = 3I
i | =3 I
2 |
{
{
i |0 = 3
{2
|2
1
|0
I + s =0 i
{
2 |
g
g
({2 + {| 3 |2 ) =
(4) i 2{ + { · | 0 + | · 1 3 2| | 0 = 0 i
g{
g{
i ({ 3 2|) | 0 = 32{ 3 |
i |0 =
2{ + |
32{ 3 |
=
{ 3 2|
2| 3 {
g
g
(2{3 + {2 | 3 {| 3 ) =
(2) i 6{2 + {2 · | 0 + | · 2{ 3 ({ · 3| 2 | 0 + | 3 · 1) = 0 i
g{
g{
{2 | 0 3 3{| 2 | 0 = 36{2 3 2{| + | 3
9.
s
1
s · | 0 = sin { i |0 = 2 | sin {
2 |
s
s
|=5 i
| = 5 3 cos { i | = (5 3 cos {)2 , so |0 = 2(5 3 cos {)0 (sin {) = 2 sin {(5 3 cos {).
{| 0 3 2| | 0 = 32{ 3 |
8.
({ 3 1)(1) 3 ({)(1)
31
{
=
.
, so | 0 =
{31
({ 3 1)2
({ 3 1)2
s g g
cos { + | =
(5) i 3 sin { + 12 | 31@2 · | 0 = 0 i
g{
g{
(c) From part (a), |0 = 2
6.
|2
{2
|2
[{@({ 3 1)] 2
{2
1
=3
=3 2
=3
2
2
{
{
{ ({ 3 1)2
({ 3 1)2
(b) cos { +
5.
i |=
i |0 = 3
i ({2 3 3{| 2 ) | 0 = 36{2 3 2{| + | 3
i |0 =
36{2 3 2{| + | 3
{2 3 3{| 2
g 4
g 2
{ ({ + |) =
| (3{ 3 |) i {4 (1 + | 0 ) + ({ + |) · 4{3 = | 2 (3 3 | 0 ) + (3{ 3 |) · 2| | 0
g{
g{
{4 + {4 | 0 + 4{4 + 4{3 | = 3| 2 3 | 2 | 0 + 6{| | 0 3 2| 2 | 0
2
({4 + 3| 2 3 6{|) | 0 = 3| 2 3 5{4 3 4{3 |
i |0 =
i
i {4 | 0 + 3| 2 | 0 3 6{| |0 = 3| 2 3 5{4 3 4{3 |
4
3
3| 3 5{ 3 4{ |
{4 + 3| 2 3 6{|
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
licated, or posted to a publicly accessible website, in whole or in part.
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212
NOT FOR SALE
¤
CHAPTER 3 DIFFERENTIATION RULES
10.
g
g
({h| ) =
({ 3 |) i {h| | 0 + h| · 1 = 1 3 | 0
g{
g{
1 3 h|
|0 = |
{h + 1
11.
g
g 2
(| cos {) =
({ + | 2 ) i |(3 sin {) + cos { · | 0 = 2{ + 2| | 0
g{
g{
2{ + | sin {
| 0 (cos { 3 2|) = 2{ + | sin { i | 0 =
cos { 3 2|
12.
g
g
cos({|) =
(1 + sin |) i 3 sin({|)({| 0 + | · 1) = cos | · |0
g{
g{
|0 [3{ sin({|) 3 cos |] = | sin({|) i | 0 =
i {h| | 0 + | 0 = 1 3 h|
i 3{| 0 sin({|) 3 cos | · |0 = | sin({|) i
| sin({|)
| sin({|)
=3
3{ sin({|) 3 cos |
{ sin({|) + cos |
g
g
(4 cos { sin |) =
(1) i 4 [cos { · cos | · |0 + sin | · (3 sin {)] = 0 i
g{
g{
4 sin { sin |
= tan { tan |
| 0 (4 cos { cos |) = 4 sin { sin | i | 0 =
4 cos { cos |
14.
g |
g
(h sin {) =
({ + {|) i h| cos { + sin { · h| | 0 = 1 + {| 0 + | · 1 i
g{
g{
h| sin { · | 0 3 {| 0 = 1 + | 3 h| cos { i | 0 (h| sin { 3 {) = 1 + | 3 h| cos {
g {@|
g
g
(h ) =
({ 3 |) i h{@| ·
g{
g{
g{
h{@| ·
| · 1 3 { · |0
= 1 3 |0
|2
{h{@|
| 3 h{@|
|0 1 3
=
2
|
|
16.
17.
{
= 1 3 |0
|
h{@| ·
i
i
|0 =
1 + | 3 h| cos {
h| sin { 3 {
i
{h{@| 0
1
3
· | = 1 3 |0
|
|2
i
|0 3
{h{@| 0
h{@|
·| =13
2
|
|
i
| 3 h{@|
|(| 3 h{@| )
|
i |0 = 2
= 2
{@|
| 3 {h{@|
| 3 {h
|2
g s
g
(1 + {2 | 2 ) i
{+| =
g{
g{
1
2 ({
+ |)31@2 (1 + | 0 ) = {2 · 2| | 0 + | 2 · 2{ i
s
s
1
|0
s
+ s
= 2{2 | | 0 + 2{| 2 i 1 + |0 = 4{2 | { + | | 0 + 4{| 2 { + | i
2 {+|
2 {+|
s
s
s
s
| 0 3 4{2 | { + | | 0 = 4{| 2 { + | 3 1 i |0 1 3 4{2 | { + | = 4{| 2 { + | 3 1 i
s
4{| 2 { + | 3 1
0
s
| =
1 3 4{2 | { + |
g
g
tan31({2 |) =
({ + {| 2 ) i
g{
g{
1
({2 | 0 + | · 2{) = 1 + { · 2| | 0 + | 2 · 1 i
1 + ({2 |)2
{2
{2
2{|
2{|
0
0
2
0
| 3 2{| | = 1 + | 3
i |
3 2{| = 1 + |2 3
1 + {4 | 2
1 + {4 | 2
1 + {4 | 2
1 + {4 | 2
0
| =
1 + |2 3
2
2{|
1 + {4 | 2
{
3 2{|
1 + {4 | 2
or | 0 =
i
i cos { · |0 3 2| | 0 = 2{ + | sin { i
13.
15.
i | 0 ({h| + 1) = 1 3 h|
i
1 + {4 | 2 + | 2 + {4 | 4 3 2{|
{2 3 2{| 3 2{5 | 3
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.5 IMPLICIT DIFFERENTIATION
18.
213
g
g
({ sin | + | sin {) =
(1) i { cos | · | 0 + sin | · 1 + | cos { + sin { · | 0 = 0 i
g{
g{
{ cos | · |0 + sin { · | 0 = 3 sin | 3 | cos { i |0 ({ cos | + sin {) = 3 sin | 3 | cos { i | 0 =
19.
¤
3 sin | 3 | cos {
{ cos | + sin {
g
g |
(h cos {) =
[1 + sin({|)] i h| (3 sin {) + cos { · h| · | 0 = cos({|) · ({| 0 + | · 1) i
g{
g{
3h| sin { + h| cos { · | 0 = { cos({|) · | 0 + | cos({|) i h| cos { · | 0 3 { cos({|) · |0 = h| sin { + | cos({|) i
[h| cos { 3 { cos({|)] |0 = h| sin { + | cos({|) i |0 =
20. tan({ 3 |) =
|
1 + {2
h| sin { + | cos({|)
h| cos { 3 { cos({|)
i (1 + {2 ) sec2 ({ 3 |) · (1 3 | 0 ) + tan({ 3 |) · 2{ = | 0
i (1 + {2 ) tan({ 3 |) = |
(1 + {2 ) sec2 ({ 3 |) 3 (1 + {2 ) sec2 ({ 3 |) · |0 + 2{ tan({ 3 |) = | 0 i
(1 + {2 ) sec2 ({ 3 |) + 2{ tan({ 3 |) = 1 + (1 + {2 ) sec2 ({ 3 |) · | 0 i
|0 =
21.
(1 + {2 ) sec2 ({ 3 |) + 2{ tan({ 3 |)
1 + (1 + {2 ) sec2 ({ 3 |)
g
g
i ({) + {2 [i({)]3 =
(10) i i 0 ({) + {2 · 3[i ({)]2 · i 0 ({) + [i ({)]3 · 2{ = 0. If { = 1, we have
g{
g{
i 0 (1) + 12 · 3[i (1)]2 · i 0 (1) + [i (1)]3 · 2(1) = 0 i i 0 (1) + 1 · 3 · 22 · i 0 (1) + 23 · 2 = 0 i
i 0 (1) + 12i 0 (1) = 316 i 13i 0 (1) = 316 i i 0 (1) = 3 16
13 .
22.
g
g
[j({) + { sin j({)] =
({2 ) i j 0 ({) + { cos j({) · j 0 ({) + sin j({) · 1 = 2{. If { = 0, we have
g{
g{
j 0 (0) + 0 + sin j(0) = 2(0) i j 0 (0) + sin 0 = 0 i j0 (0) + 0 = 0 i j0 (0) = 0.
23.
g
g 4 2
({ | 3 {3 | + 2{| 3 ) =
(0) i {4 · 2| + | 2 · 4{3 {0 3 ({3 · 1 + | · 3{2 {0 ) + 2({ · 3| 2 + | 3 · {0 ) = 0 i
g|
g|
4{3 | 2 {0 3 3{2 | {0 + 2| 3 {0 = 32{4 | + {3 3 6{| 2
{0 =
24.
i (4{3 | 2 3 3{2 | + 2|3 ) {0 = 32{4 | + {3 3 6{| 2
i
32{4 | + {3 3 6{| 2
g{
= 3 2
g|
4{ | 3 3{2 | + 2| 3
g
g
(| sec {) =
({ tan |) i | · sec { tan { · {0 + sec { · 1 = { · sec2 | + tan | · {0
g|
g|
i
| sec { tan { · {0 3 tan | · {0 = { sec2 | 3 sec { i (| sec { tan { 3 tan |) {0 = { sec2 | 3 sec { i
{0 =
{ sec2 | 3 sec {
g{
=
g|
| sec { tan { 3 tan |
25. | sin 2{ = { cos 2|
i
| · cos 2{ · 2 + sin 2{ · |0 = {(3 sin 2| · 2| 0 ) + cos(2|) · 1
sin 2{ · | 0 + 2{ sin 2| · | 0 = 32| cos 2{ + cos 2|
|0 (sin 2{ + 2{ sin 2|) = 32| cos 2{ + cos 2|
|0 =
i
i
i
|0 =
32| cos 2{ + cos 2|
. When { =
sin 2{ + 2{ sin 2|
@2
1
(3@2)(31) + 0
=
= , so an equation of the tangent line is | 3
0+·1
2
4
2
and | =
,
4
we have
= 12 ({ 3 2 ), or | = 12 {.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
licated, or posted to a publicly accessible website, in whole or in part.
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214
NOT FOR SALE
¤
CHAPTER 3 DIFFERENTIATION RULES
26. sin({ + |) = 2{ 3 2|
cos({ + |) · (1 + |0 ) = 2 3 2| 0
i
| 0 [cos({ + |) + 2] = 2 3 cos({ + |)
i
|0 =
|0 =
cos({ + |) · | 0 + 2| 0 = 2 3 cos({ + |)
i
1
2 3 cos({ + |)
231
. When { = and | = , we have | 0 =
= , so
cos({ + |) + 2
1+2
3
an equation of the tangent line is | 3 = 13 ({ 3 ), or | = 13 { +
27. {2 + {| + | 2 = 3
i
2
.
3
i 2{ + { | 0 + | · 1 + 2||0 = 0 i { |0 + 2| | 0 = 32{ 3 |
i |0 ({ + 2|) = 32{ 3 |
i
32{ 3 |
32 3 1
33
. When { = 1 and | = 1, we have | 0 =
=
= 31, so an equation of the tangent line is
{ + 2|
1+2·1
3
| 3 1 = 31({ 3 1) or | = 3{ + 2.
i 2{ + 2({ |0 + | · 1) 3 2| | 0 + 1 = 0 i
28. {2 + 2{| 3 | 2 + { = 2
| 0 (2{ 3 2|) = 32{ 3 2| 3 1 i |0 =
|0 =
2{ |0 3 2| | 0 = 32{ 3 2| 3 1 i
32{ 3 2| 3 1
. When { = 1 and | = 2, we have
2{ 3 2|
37
7
32 3 4 3 1
=
= , so an equation of the tangent line is | 3 2 = 72 ({ 3 1) or | = 72 { 3 32 .
234
32
2
i 2{ + 2| |0 = 2(2{2 + 2| 2 3 {)(4{ + 4| | 0 3 1). When { = 0 and | = 12 , we have
29. {2 + | 2 = (2{2 + 2| 2 3 {)2
0 + | 0 = 2( 12 )(2| 0 3 1) i | 0 = 2| 0 3 1 i |0 = 1, so an equation of the tangent line is | 3
1
2
= 1({ 3 0)
or | = { + 12 .
30. {2@3 + | 2@3 = 4
i
2 31@3
{
3
+ 23 | 31@3 | 0 = 0 i
s
3
I
|
|0
1
0
I
s
I
+
. When { = 33 3
=
3
=
0
i
|
3
3
3
{
{
|
I 2@3
33 3
1
3
1
I
and | = 1, we have | = 3 I 1@3 = 3
= I = I , so an equation of the tangent line is
33 3
3 3
3
33 3
I
| 3 1 = I13 { + 3 3 or | = I13 { + 4.
0
31. 2({2 + | 2 )2 = 25({2 3 | 2 )
i 4({2 + | 2 )(2{ + 2| | 0 ) = 25(2{ 3 2| |0 ) i
4({ + | | 0 )({2 + | 2 ) = 25({ 3 | |0 )
|0 =
i
4| | 0 ({2 + | 2 ) + 25||0 = 25{ 3 4{({2 + | 2 ) i
25{ 3 4{({2 + | 2 )
. When { = 3 and | = 1, we have | 0 =
25| + 4|({2 + | 2 )
75 3 120
25 + 40
9
9
so an equation of the tangent line is | 3 1 = 3 13
({ 3 3) or | = 3 13
{+
32. | 2 (| 2 3 4) = {2 ({2 3 5)
i | 4 3 4| 2 = {4 3 5{2
9
= 3 45
= 3 13
,
65
40
.
13
i 4|3 | 0 3 8| | 0 = 4{3 3 10{.
When { = 0 and | = 32, we have 332| 0 + 16| 0 = 0 i 316| 0 = 0 i | 0 = 0, so an equation of the tangent line is
| + 2 = 0({ 3 0) or | = 32.
33. (a) | 2 = 5{4 3 {2
i 2| | 0 = 5(4{3 ) 3 2{ i | 0 =
So at the point (1> 2) we have | 0 =
10{3 3 {
.
|
(b)
9
10(1)3 3 1
= , and an equation
2
2
of the tangent line is | 3 2 = 92 ({ 3 1) or | = 92 { 3 52 .
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.5 IMPLICIT DIFFERENTIATION
i 2| |0 = 3{2 + 3(2{) i | 0 =
34. (a) | 2 = {3 + 3{2
|0 =
¤
215
3{2 + 6{
. So at the point (1> 32) we have
2|
3(1)2 + 6(1)
9
= 3 , and an equation of the tangent line is | + 2 = 3 94 ({ 3 1) or | = 3 94 { + 14 .
2(32)
4
(b) The curve has a horizontal tangent where | 0 = 0 C
(c)
3{2 + 6{ = 0 C 3{({ + 2) = 0 C { = 0 or { = 32.
But note that at { = 0, | = 0 also, so the derivative does not exist.
At { = 32, | 2 = (32)3 + 3(32)2 = 38 + 12 = 4, so | = ±2.
So the two points at which the curve has a horizontal tangent are
(32> 32) and (32> 2).
i 18{ + 2| |0 = 0 i 2| |0 = 318{ i | 0 = 39{@| i
| · 1 3 { · |0
9
| 3 {(39{@|)
| 2 + 9{2
| 00 = 39
= 39 · 3 [since x and y must satisfy the
= 39
= 39 ·
2
2
|
|
|3
|
35. 9{2 + | 2 = 9
original equation, 9{2 + | 2 = 9]. Thus, | 00 = 381@| 3 .
36.
s
|
1
|0
I + s = 0 i |0 = 3 I
i
2 { 2 |
{
%
#
&
$# s $
s
I
I
s
s
|
1
1
1
1
|
s |0 3 |
I
{
{ s
3I
3 | I
1+ I
2 |
2 {
|
{
{
{
00
=3
=
| =3
{
2{
2{
I
s
s
I
{+ |
1
I
I since { and | must satisfy the original equation, { + | = 1.
=
=
2{ {
2{ {
I
s
{+ | =1 i
37. {3 + | 3 = 1
| 00 = 3
i 3{2 + 3| 2 | 0 = 0 i | 0 = 3
{2
|2
i
| 2 (2{) 3 {2 · 2| | 0
2{| 2 3 2{2 |(3{2@| 2 )
2{| 4 + 2{4 |
2{|(| 3 + {3 )
2{
=3
=3
=3
=3 5,
2
2
4
6
(| )
|
|
|6
|
since { and | must satisfy the original equation, {3 + | 3 = 1.
i 4{3 + 4| 3 | 0 = 0 i 4|3 | 0 = 34{3 i | 0 = 3{3@| 3 i
3
| · 3{2 3 {3 · 3| 2 | 0
| 3 {(3{3@| 3 )
| 4 + {4
d4
33d4 {2
= 33{2 | 2 ·
= 33{2 ·
= 33{2 · 7 =
| 00 = 3
(| 3 )2
|6
|4 |3
|
|7
38. {4 + | 4 = d4
39. If { = 0 in {| + h| = h, then we get 0 + h| = h, so | = 1 and the point where { = 0 is (0> 1). Differentiating implicitly
with respect to { gives us {| 0 + | · 1 + h| | 0 = 0. Substituting 0 for { and 1 for | gives us
0 + 1 + h|0 = 0
i
h| 0 = 31
i
| 0 = 31@h. Differentiating {| 0 + | + h| | 0 = 0 implicitly with respect to { gives
us {|00 + | 0 · 1 + | 0 + h| | 00 + | 0 · h| | 0 = 0. Now substitute 0 for {, 1 for |, and 31@h for |0 .
1
1
1
1
1
1
2
0+ 3
+ 3
+ h|00 + 3
(h) 3
= 0 i 3 + h| 00 + = 0 i h| 00 =
h
h
h
h
h
h
h
40. If { = 1 in {2 + {| + | 3 = 1, then we get 1 + | + | 3 = 1
i
|3 + | = 0
i
0
|(| 2 + 1)
2
i
i
| 00 =
1
.
h2
| = 0, so the point
0
where { = 1 is (1> 0). Differentiating implicitly with respect to { gives us 2{ + {| + | · 1 + 3| · | = 0. Substituting 1 for
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216
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
{ and 0 for | gives us 2 + |0 + 0 + 0 = 0
i
| 0 = 32. Differentiating 2{ + {|0 + | + 3|2 | 0 = 0 implicitly with respect
to { gives us 2 + {|00 + | 0 · 1 + | 0 + 3(|2 | 00 + | 0 · 2|| 0 ) = 0. Now substitute 1 for {, 0 for |, and 32 for | 0 .
2 + | 00 + (32) + (32) + 3(0 + 0) = 0
i
|00 = 2. Differentiating 2 + {|00 + 2| 0 + 3| 2 | 00 + 6|(| 0 )2 = 0 implicitly
with respect to { gives us {|000 + | 00 · 1 + 2| 00 + 3(|2 | 000 + | 00 · 2|| 0 ) + 6[| · 2|0 | 00 + (| 0 )2 | 0 ] = 0. Now substitute 1 for {,
0 for |, 32 for |0 , and 2 for |00 . | 000 + 2 + 4 + 3(0 + 0) + 6[0 + (38)] = 0
i
| 000 = 32 3 4 + 48 = 42.
41. (a) There are eight points with horizontal tangents: four at { E 1=57735 and
four at { E 0=42265.
(b) | 0 =
3{2 3 6{ + 2
2(2| 3 3 3| 2 3 | + 1)
i |0 = 31 at (0> 1) and | 0 =
1
3
at (0> 2).
Equations of the tangent lines are | = 3{ + 1 and | = 13 { + 2.
(c) | 0 = 0 i 3{2 3 6{ + 2 = 0 i { = 1 ±
1
3
I
3
(d) By multiplying the right side of the equation by { 3 3, we obtain the ¿rst
graph. By modifying the equation in other ways, we can generate the other
graphs.
|(| 2 3 1)(| 3 2)
= {({ 3 1)({ 3 2)({ 3 3)
|(| 2 3 4)(| 3 2)
= {({ 3 1)({ 3 2)
{(| + 1)(|2 3 1)(| 3 2)
= |({ 3 1)({ 3 2)
|(| + 1)(| 2 3 1)(| 3 2)
= {({ 3 1)({ 3 2)
|(| 2 + 1)(| 3 2)
= {({2 3 1)({ 3 2)
(| + 1)(| 2 3 1)(| 3 2)
= ({ 3 1)({ 3 2)
|(| + 1)(| 2 3 2)
= {({ 3 1)({2 3 2)
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.5 IMPLICIT DIFFERENTIATION
42. (a)
(b)
¤
217
g
g 4
(2| 3 + | 2 3 | 5 ) =
({ 3 2{3 + {2 ) i
g{
g{
6| 2 | 0 + 2| | 0 3 5| 4 | 0 = 4{3 3 6{2 + 2{ i
|0 =
2{(2{2 3 3{ + 1)
2{(2{ 3 1)({ 3 1)
. From the graph and the
=
6| 2 + 2| 3 5| 4
|(6| + 2 3 5| 3 )
values for which | 0 = 0, we speculate that there are 9 points with horizontal
tangents: 3 at { = 0, 3 at { = 12 , and 3 at { = 1. The three horizontal
tangents along the top of the wagon are hard to ¿nd, but by limiting the
|-range of the graph (to [1=6> 1=7], for example) they are distinguishable.
43. From Exercise 31, a tangent to the lemniscate will be horizontal if | 0 = 0
{[25 3 4({2 + | 2 )] = 0
44.
we have
i an equation of the tangent line at ({0 > |0 ) is
{0 {
|0 |
{2
|2
+ 2 = 20 + 20 = 1.
2
d
e
d
e
| 3 |0 =
46.
(1). (Note that when { is 0, | is also 0, and there is no horizontal tangent
2
3e2 {0
|0
|0 |
|2
{0 {
{2
({ 3 {0 ). Multiplying both sides by 2 gives 2 3 20 = 3 2 + 20 . Since ({0 > |0 ) lies on the ellipse,
d2 |0
e
e
e
d
d
{2
|2
3
=1 i
d2
e2
we have
25
4
2
25
4
2{
2|| 0
e2 {
0
+
=
0
i
|
=
3
d2
e2
d2 |
|2
{2
+
=1 i
d2
e2
| 3 |0 =
45.
25
8
{2 + | 2 =
for { + | in the equation of the lemniscate, 2({2 + | 2 )2 = 25({2 3 | 2 ), we get
I
2
25
5 3
5
.
(2). Solving (1) and (2), we have {2 = 75
16 and | = 16 , so the four points are ± 4 > ± 4
at the origin.) Substituting
{2 3 | 2 =
i
i 25{ 3 4{({2 + | 2 ) = 0 i
2{
2|| 0
e2 {
0
3
=
0
i
|
=
d2
e2
d2 |
i an equation of the tangent line at ({0 > |0 ) is
e2 {0
|0
|0 |
|2
{0 { {2
({ 3 {0 ). Multiplying both sides by 2 gives 2 3 20 = 2 3 20 . Since ({0 > |0 ) lies on the hyperbola,
d2 |0
e
e
e
d
d
{0 {
|0 |
{2
|2
3 2 = 20 3 20 = 1.
2
d
e
d
e
s
|
|0
1
I + s = 0 i |0 = 3 I
i an equation of the tangent line at ({0 > |0 )
2 {
2 |
{
s
s
I s
|0
|0
is | 3 |0 = 3 I ({ 3 {0 ). Now { = 0 i | = |0 3 I (3{0 ) = |0 + {0 | 0 , so the |-intercept is
{0
{0
s
I
I s
|0
|0 {0
i
|0 + {0 | 0 . And | = 0 i 3|0 = 3 I ({ 3 {0 ) i { 3 {0 = s
{0
|0
I s
I s
{ = {0 + {0 | 0 , so the {-intercept is {0 + {0 | 0 . The sum of the intercepts is
I
s 2 I 2
I s I s I s
|0 + {0 | 0 + {0 + {0 | 0 = {0 + 2 {0 | 0 + |0 =
{0 + | 0
=
f
= f.
I
s
I
{+ | = f i
47. If the circle has radius u, its equation is {2 + | 2 = u2
at S ({0 > |0 ) is 3
{
i 2{ + 2|| 0 = 0 i |0 = 3 , so the slope of the tangent line
|
{0
31
|0
. The negative reciprocal of that slope is
=
, which is the slope of RS , so the tangent line at
|0
3{0 @|0
{0
S is perpendicular to the radius RS .
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218
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
i t| t31 | 0 = s{s31
48. | t = {s
i |0 =
49. | = (tan31 {)2
i | 0 = 2(tan31 {)1 ·
50. | = tan31 ({2 )
i
51. | = sin31 (2{ + 1)
|0 =
s{s31
s{s31 |
s{s31 {s@t
s
=
=
= {(s@t)31
t31
t
t|
t|
t{s
t
g
1
2 tan31 {
(tan31 {) = 2 tan31 { ·
=
2
g{
1+{
1 + {2
1
g 2
1
2{
({ ) =
·
· 2{ =
1 + ({2 )2 g{
1 + {4
1 + {4
i
2
1
1
g
1
·2= I
(2{ + 1) = s
·
= I
|0 = s
2
2 3{
2
2
g{
34{
3
4{
3{
1 3 (2{ + 1)
1 3 (4{ + 4{ + 1)
52. j({) =
I
I
{2 3 1 sec31 { i j 0 ({) = {2 3 1 ·
I 2
{ 3 1 + {2 sec31 {
I
or
{ {2 3 1
53. J({) =
31@2
{ sec31 {
1
1 2
1
I
{ 31
(2{) = + I
+ sec31 { ·
2
2
{
{ { 31
{2 3 1
I
I
31
1
{ arccos {
1 3 {2 arccos { i J0 ({) = 1 3 {2 · I
+ arccos { · (1 3 {2 )31@2 (32{) = 31 3 I
2
1 3 {2
1 3 {2
54. | = tan31 { 3
I
{2 + 1
i
I 2
1
{ +13{
{
1
I
I
I
1
3
=
I
2
{2 + 1
1 + {2 3 2{ {2 + 1 + {2 + 1
{2 + 1
1 + { 3 {2 + 1
I
I
I
{2 + 1 3 {
{2 + 1 3 {
{2 + 1 3 {
I
I
= I
= I
= 2 1 + {2 3 { {2 + 1
{2 + 1
2 {2 + 1 (1 + {2 ) 3 {({2 + 1)
2 (1 + {2 ) {2 + 1 3 {
|0 =
1
2(1 + {2 )
=
55. k(w) = cot31 (w) + cot31 (1@w)
k0 (w) = 3
i
1
1
1
1
g 1
w2
1
1
=
3
·
3
= 0.
3
·
3
+ 2
=3
1 + w2
1 + (1@w)2 gw w
1 + w2
w2 + 1
w2
1 + w2
w +1
Note that this makes sense because k(w) =
56. I () = arcsin
I
sin = arcsin(sin )1@2
3
for w A 0 and k(w) =
for w ? 0.
2
2
i
1
g
1
cos 1
1@2
I
= I
· (sin )31@2 · cos = I
I
2 · g (sin )
2
1 3 sin 2 1 3 sin sin 13
sin I 0 () = u
57. | = { sin31 { +
I
1 3 {2
i
1
1
{
{
+ (sin31 {)(1) + (1 3 {2 )31@2 (32{) = I
+ sin31 { 3 I
= sin31 {
|0 = { · I
2
1 3 {2
1 3 {2
1 3 {2
58. | = cos31 (sin31 w)
i
1
g
1
1
|0 = 3 s
·
·I
sin31 w = 3 s
31 2
31 2
gw
1 3 w2
1 3 (sin w)
1 3 (sin w)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.5 IMPLICIT DIFFERENTIATION
59. | = arccos
e + d cos {
d + e cos {
13
e + d cos {
d + e cos {
219
i
1
|0 = 3 v
¤
2
(d + e cos {)(3d sin {) 3 (e + d cos {)(3e sin {)
(d + e cos {)2
(d2 3 e2 ) sin {
1
= I
d2 + e2 cos2 { 3 e2 3 d2 cos2 { |d + e cos {|
I
(d2 3 e2 ) sin {
d2 3 e2 sin {
1
I
= I
=
2
2
2
|d
+
e
cos
{|
|d
+
e cos {| |sin {|
d 3 e 1 3 cos {
I
d2 3 e2
But 0 $ { $ , so |sin {| = sin {. Also d A e A 0 i e cos { D 3e A 3d, so d + e cos { A 0. Thus | =
.
d + e cos {
0
60. | = arctan
u
13{
= arctan
1+{
13{
1+{
1@2
i
1@2
31@2
1 13{
1
g 13{
1
(1 + {)(31) 3 (1 3 {)(1)
·
=
·
2 · g{ 1 + {
u
1
3
{
2
1
+
{
(1 + {)2
13{
1+
1+
1
+
{
1+{
1@2
1 1+{
1
32
1 + { 1 (1 + {)1@2
32
·
=
·
=
·
· ·
2
1+{
13{ 2 13{
(1 + {)
2
2 (1 3 {)1@2 (1 + {)2
+
1+{
1+{
|0 =
=
61. i ({) =
31
31
= I
2(1 3 {)1@2 (1 + {)1@2
2 1 3 {2
I
I
31@2
1
1
{ arcsin {
1 3 {2 arcsin { i i 0 ({) = 1 3 {2 · I
+ arcsin { ·
(32{) = 1 3 I
1 3 {2
2
2
13{
1 3 {2
Note that i 0 = 0 where the graph of i has a horizontal tangent. Also note
that i 0 is negative when i is decreasing and i 0 is positive when i is
increasing.
62. i ({) = arctan({2 3 {)
i i 0 ({) =
1
g 2
2{ 3 1
({ 3 {) =
·
1 + ({2 3 {)2 g{
1 + ({2 3 {)2
Note that i 0 = 0 where the graph of i has a horizontal tangent. Also note
that i 0 is negative when i is decreasing and i 0 is positive when i is
increasing.
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°
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220
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
63. Let | = cos31 {. Then cos | = { and 0 $ | $ i 3 sin |
g|
=1 i
g{
1
g|
1
1
. [Note that sin | D 0 for 0 $ | $ .]
= 3I
=3
= 3s
g{
sin |
1 3 {2
1 3 cos2 |
64. (a) Let | = sec31 {. Then sec | = { and | M 0>
2
> 3
. Differentiate with respect to {: sec | tan |
2
g|
g{
=1 i
s
1
1
1
g|
s
=
=
= I
. Note that tan2 | = sec2 | 3 1 i tan | = sec2 | 3 1
2
2
g{
sec | tan |
{ { 31
sec | sec | 3 1
since tan | A 0 when 0 ? | ?
2
or ? | ?
3
2 .
(b) | = sec31 { i sec | = { i sec | tan |
g|
=1 i
g{
g|
1
=
. Now tan2 | = sec2 | 3 1 = {2 3 1,
g{
sec | tan |
I
so tan | = ± {2 3 1. For | M 0> 2 , { D 1, so sec | = { = |{| and tan | D 0 i
I
g|
1
1
I
= I
=
. For | M 2 > , { $ 31, so |{| = 3{ and tan | = 3 {2 3 1 i
g{
{ {2 3 1
|{| {2 3 1
g|
1
1
1
1
I
I
=
=
= I
=
.
g{
sec | tan |
{ 3 {2 3 1
(3{) {2 3 1
|{| {2 3 1
65. {2 + | 2 = u2 is a circle with center R and d{ + e| = 0 is a line through R [assume d
and e are not both zero]. {2 + | 2 = u2
i 2{ + 2|| 0 = 0 i |0 = 3{@|, so the
slope of the tangent line at S0 ({0 > |0 ) is 3{0 @|0 . The slope of the line RS0 is |0 @{0 ,
which is the negative reciprocal of 3{0 @|0 . Hence, the curves are orthogonal, and the
families of curves are orthogonal trajectories of each other.
66. The circles {2 + | 2 = d{ and {2 + | 2 = e| intersect at the origin where the tangents are vertical and horizontal [assume d
and e are both nonzero]. If ({0 > |0 ) is the other point of intersection, then {20 + |02 = d{0 (1) and {20 + |02 = e|0 (2).
Now {2 + | 2 = d{ i 2{ + 2|| 0 = d i |0 =
2{ + 2|| 0 = e| 0
i
e 3 2|0
d 3 2{0
=3
2|0
2{0
|0 =
d 3 2{
and {2 + | 2 = e|
2|
i
2{
. Thus, the curves are orthogonal at ({0 > |0 ) C
e 3 2|
C 2d{0 3 4{20 = 4|02 3 2e|0
C d{0 + e|0 = 2({20 + |02 ),
which is true by (1) and (2).
67. | = f{2
i | 0 = 2f{ and {2 + 2| 2 = n [assume n A 0] i 2{ + 4|| 0 = 0 i
2||0 = 3{
i |0 = 3
{
{
1
=3
=3
, so the curves are orthogonal if
2(|)
2(f{2 )
2f{
f 6= 0. If f = 0, then the horizontal line | = f{2 = 0 intersects {2 + 2| 2 = n orthogonally
I
at ± n> 0 , since the ellipse {2 + 2| 2 = n has vertical tangents at those two points.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.5 IMPLICIT DIFFERENTIATION
¤
221
i | 0 = 3d{2 and {2 + 3| 2 = e [assume e A 0] i 2{ + 6|| 0 = 0 i
68. | = d{3
3||0 = 3{ i |0 = 3
{
{
1
, so the curves are orthogonal if
=3
=3
3(|)
3(d{3 )
3d{2
d 6= 0. If d = 0, then the horizontal line | = d{3 = 0 intesects {2 + 3| 2 = e orthogonally
I at ± e> 0 , since the ellipse {2 + 3| 2 = e has vertical tangents at those two points.
69. Since D2 ? d2 , we are assured that there are four points of intersection.
(1)
{2
|2
+ 2 =1 i
d2
e
(2)
|2
{2
3 2 =1 i
2
D
E
Now p1 p2 = 3
2{
2|| 0
+ 2 =0 i
d2
e
2{
2||0
3 2 =0 i
2
D
E
||0
{
=3 2
e2
d
|| 0
{
= 2
E2
D
i | 0 = p1 = 3
{E 2
.
|D2
i | 0 = p2 =
{e2 {E 2
e2 E 2 {2
{2
|2
{2
|2
·
= 3 2 2 · 2 (3). Subtracting equations, (1) 3 (2), gives us 2 + 2 3 2 + 2 = 0 i
2
2
|d |D
d D |
d
e
D
E
|2
|2
{2
{2
+
=
3
e2
E2
D2
d2
i
| 2 E 2 + | 2 e2
{2 d2 3 {2 D2
=
e2 E 2
D2 d2
i
| 2 (e2 + E 2 )
{2 (d2 3 D2 )
=
(4). Since
e2 E 2
d2 D2
d2 3 e2 = D2 + E 2 , we have d2 3 D2 = e2 + E 2 . Thus, equation (4) becomes
substituting for
{e2
.
|d2
|2
{2
=
e2 E 2
D2 d2
i
{2
D2 d2
=
, and
|2
e2 E 2
e2 E 2 d2 D2
{2
in equation (3) gives us p1 p2 = 3 2 2 · 2 2 = 31. Hence, the ellipse and hyperbola are orthogonal
2
|
d D e E
trajectories.
70. | = ({ + f)31
i | 0 = 3({ + f)32 and | = d({ + n)1@3
i | 0 = 13 d({ + n)32@3 , so the curves are othogonal if the
d
31
·
= 31 i d = 3({ + f)2 ({ + n)2@3 i
({ + f)2 3({ + n)2@3
2 I
1
1
| 2
d=3
[since | 2 = ({ + f)32 and | 2 = d2 ({ + n)2@3 ] i d = 3 2
i d3 = 3 i d = 3 3.
|
d
d
product of the slopes is 31, that is,
71. (a)
q2 d
S + 2 (Y 3 qe) = qUW
Y
i S Y 3 S qe +
q2 d
q3 de
= qUW
3
Y
Y2
g
g
(S Y 3 S qe + q2 dY 31 3 q3 deY 32 ) =
(qUW )
gS
gS
i
i
S Y 0 + Y · 1 3 qe 3 q2 dY 32 · Y 0 + 2q3 deY 33 · Y 0 = 0 i Y 0 (S 3 q2 dY 32 + 2q3 deY 33 ) = qe 3 Y
Y0 =
S3
Y 3 (qe 3 Y )
qe 3 Y
gY
=
or
3
33
3
+ 2q deY
gS
S Y 3 q2 dY + 2q3 de
q2 dY 32
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
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i
222
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
(b) Using the last expression for gY @gS from part (a), we get
(10 L)3 [(1 mole)(0=04267 L@mole) 3 10 L]
gY
= 5
6
gS
(2=5 atm)(10 L)3 3 (1 mole)2 (3=592 L2 - atm@ mole2 )(10 L)
8
7
+ 2(1 mole)3 (3=592 L2 - atm@ mole2 )(0=04267 L@ mole)
=
39957=33 L4
E 34=04 L@ atm=
2464=386541 L3 - atm
72. (a) {2 + {| + | 2 + 1 = 0
i
2{ + {| 0 + | · 1 + 2|| 0 + 0 = 0
i
| 0 ({ + 2|) = 32{ 3 |
i
|0 =
32{ 3 |
{ + 2|
(b) Plotting the curve in part (a) gives us an empty graph, that is, there are no points that satisfy the equation. If there were any
points that satis¿ed the equation, then { and | would have opposite signs; otherwise, all the terms are positive and their
sum can not equal 0.
{2 + {| + | 2 + 1 = 0
i
{2 + 2{| + | 2 3 {| + 1 = 0
i
({ + |)2 = {| 3 1. The left
side of the last equation is nonnegative, but the right side is at most 31, so that proves there are no points that satisfy the
equation.
Another solution: {2 + {| + | 2 + 1 = 12 {2 + {| + 12 | 2 + 12 {2 + 12 | 2 + 1 = 12 ({2 + 2{| + |2 ) + 12 ({2 + | 2 ) + 1
= 12 ({ + |)2 + 12 ({2 + | 2 ) + 1 D 1
Another solution: Regarding {2 + {| + | 2 + 1 = 0 as a quadratic in {, the discriminant is | 2 3 4(|2 + 1) = 33| 2 3 4.
This is negative, so there are no real solutions.
(c) The expression for | 0 in part (a) is meaningless; that is, since the equation in part (a) has no solution, it does not implicitly
de¿ne a function | of {, and therefore it is meaningless to consider | 0 .
73. To ¿nd the points at which the ellipse {2 3 {| + | 2 = 3 crosses the {-axis, let | = 0 and solve for {.
I
I | = 0 i {2 3 {(0) + 02 = 3 C { = ± 3. So the graph of the ellipse crosses the {-axis at the points ± 3> 0 .
Using implicit differentiation to ¿nd | 0 , we get 2{ 3 {| 0 3 | + 2||0 = 0 i | 0 (2| 3 {) = | 3 2{ C |0 =
So |0 at
| 3 2{
.
2| 3 {
I
I
I I 032 3
0+2 3
I = 2 and | 0 at 3 3> 0 is
I = 2. Thus, the tangent lines at these points are parallel.
3> 0 is
2(0) 3 3
2(0) + 3
74. (a) We use implicit differentiation to ¿nd | 0 =
of the tangent line at (31> 1) is p =
normal line is 3
| 3 2{
as in Exercise 73. The slope
2| 3 {
(b)
3
1 3 2(31)
= = 1, so the slope of the
2(1) 3 (31)
3
1
= 31, and its equation is | 3 1 = 31({ + 1) C
p
| = 3{. Substituting 3{ for | in the equation of the ellipse, we get
{2 3 {(3{) + (3{)2 = 3 i 3{2 = 3 C { = ±1. So the normal line
must intersect the ellipse again at { = 1, and since the equation of the line is
| = 3{, the other point of intersection must be (1> 31).
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NOT FOR SALE
SECTION 3.5 IMPLICIT DIFFERENTIATION
i {2 · 2|| 0 + | 2 · 2{ + { · | 0 + | · 1 = 0 C |0 (2{2 | + {) = 32{| 2 3 |
75. {2 | 2 + {| = 2
|0 = 3
¤
223
C
2{|2 + |
2{| 2 + |
. So 3 2
= 31 C 2{|2 + | = 2{2 | + { C |(2{| + 1) = {(2{| + 1) C
2
2{ | + {
2{ | + {
|(2{| + 1) 3 {(2{| + 1) = 0 C (2{| + 1)(| 3 {) = 0 C {| = 3 12 or | = {. But {| = 3 12
{2 | 2 + {| =
1
4
3
1
2
i
6= 2, so we must have { = |. Then {2 | 2 + {| = 2 i {4 + {2 = 2 C {4 + {2 3 2 = 0 C
({2 + 2)({2 3 1) = 0. So {2 = 32, which is impossible, or {2 = 1 C { = ±1. Since { = |, the points on the curve
where the tangent line has a slope of 31 are (31> 31) and (1> 1).
i 2{ + 8||0 = 0 i | 0 = 3
76. {2 + 4| 2 = 36
through (12> 3). The tangent line is then | 3 3 = 3
{
. Let (d> e) be a point on {2 + 4| 2 = 36 whose tangent line passes
4|
d
d
({ 3 12), so e 3 3 = 3 (d 3 12). Multiplying both sides by 4e
4e
4e
gives 4e2 3 12e = 3d2 + 12d, so 4e2 + d2 = 12(d + e). But 4e2 + d2 = 36, so 36 = 12(d + e) i d + e = 3 i
e = 3 3 d. Substituting 3 3 d for e into d2 + 4e2 = 36 gives d2 + 4(3 3 d)2 = 36 C
d2 + 36 3 24d + 4d2 = 36 C
5d2 3 24d = 0 C d(5d 3 24) = 0, so d = 0 or d = 24
. If d = 0, e = 3 3 0 = 3, and if d =
5
24
So the two points on the ellipse are (0> 3) and 5 > 3 95 . Using
24
,
5
e=33
24
5
= 3 95 .
d
({ 3 12) with (d> e) = (0> 3) gives us the tangent line
4e
9
| 3 3 = 0 or | = 3. With (d> e) = 24
, we have
5 > 35
|33=3
24@5
({ 3 12) C | 3 3 = 23 ({ 3 12) C | = 23 { 3 5.
| 3 3 = 3 4(39@5)
A graph of the ellipse and the tangent lines con¿rms our results.
77. (a) If | = i 31 ({), then i (|) = {. Differentiating implicitly with respect to { and remembering that | is a function of {,
we get i 0 (|)
g|
1
g|
= 1, so
= 0
g{
g{
i (|)
i
31 0
i
({) =
0
(b) i (4) = 5 i i 31 (5) = 4. By part (a), i 31 (5) =
1
.
i 0 (i 31 ({))
1
1
= 0
= 1 23 = 32 .
i 0 (i 31 (5))
i (4)
78. (a) Assume d ? e. Since h{ is an increasing function, hd ? he , and hence, d + hd ? e + he ; that is, i (d) ? i(e)=
So i ({) = { + h{ is an increasing function and therefore one-to-one.
(b) i 31 (1) = d C i (d) = 1, so we need to ¿nd d such that i (d) = 1. By inspection, we see that i (0) = 0 + h0 = 1, so
d = 0, and hence, i 31 (1) = 0.
(c) (i 31 )0 (1) =
1
1
= 0
[by part (b)]. Now i ({) = { + h{
i 0 (i 31 (1))
i (0)
i i 0 ({) = 1 + h{ , so i 0 (0) = 1 + h0 = 2.
Thus, (i 31 )0 (1) = 12 .
79. (a) | = M({) and {| 00 + | 0 + {| = 0
i {M 00 ({) + M 0 ({) + {M({) = 0. If { = 0, we have 0 + M 0 (0) + 0 = 0,
so M 0 (0) = 0.
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¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
(b) Differentiating {| 00 + | 0 + {| = 0 implicitly, we get {| 000 + | 00 · 1 + | 00 + {| 0 + | · 1 = 0 i
{|000 + 2| 00 + {| 0 + | = 0, so {M 000 ({) + 2M 00 ({) + {M 0 ({) + M({) = 0. If { = 0, we have
0 + 2M 00 (0) + 0 + 1 [M(0) = 1 is given] = 0 i 2M 00 (0) = 31 i M 00 (0) = 3 12 .
80. {2 + 4| 2 = 5
i 2{ + 4(2||0 ) = 0 i |0 = 3
{
. Now let k be the height of the lamp, and let (d> e) be the point of
4|
tangency of the line passing through the points (3> k) and (35> 0). This line has slope (k 3 0)@[3 3 (35)] = 18 k. But the
slope of the tangent line through the point (d> e) can be expressed as | 0 = 3
passes through (35> 0) and (d> e)], so 3
e
d
=
4e
d+5
e30
e
d
, or as
=
[since the line
4e
d 3 (35)
d+5
C 4e2 = 3d2 3 5d C d2 + 4e2 = 35d. But d2 + 4e2 = 5
[since (d> e) is on the ellipse], so 5 = 35d C d = 31. Then 4e2 = 3d2 3 5d = 31 3 5(31) = 4 i e = 1, since the
point is on the top half of the ellipse. So
{-axis.
k
e
1
1
=
=
=
8
d+5
31 + 5
4
i k = 2. So the lamp is located 2 units above the
LABORATORY PROJECT Families of Implicit Curves
1. (a) There appear to be nine points of intersection. The “inner four” near the origin are about (±0=2> 30=9) and (±0=3> 31=1).
The “outer ¿ve” are about (2=0> 38=9), (32=8> 38=8), (37=5> 37=7), (37=8> 34=7), and (38=0> 1=5).
(b) We see from the graphs with f = 5 and f = 10, and for other values of f, that the curves change shape but the nine points
of intersection are the same.
2. (a) If f = 0, the graph is the unit circle. As f increases, the graph looks more diamondlike and then more crosslike (see the
graph for f D 0).
For 31 ? f ? 0 (see the graph), there are four hyperboliclike branches as well as an ellipticlike curve bounded by
|{| $ 1 and ||| $ 1 for values of f close to 0. As f gets closer to 31, the branches and the curve become more rectangular,
approaching the lines |{| = 1 and ||| = 1.
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NOT FOR SALE
SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS
¤
225
For f = 31, we get the lines { = ±1 and | = ±1. As f decreases, we get four test-tubelike curves (see the graph)
that are bounded by |{| = 1 and ||| = 1, and get thinner as |f| gets larger.
f0
1 ? f ? 0
f 1
(b) The curve for f = 31 is described in part (a). When f = 31, we get {2 + | 2 3 {2 | 2 = 1 C
0 = {2 | 2 3 {2 3 | 2 + 1 C 0 = ({2 3 1)(| 2 3 1) C { = ±1 or | = ±1, which algebraically proves that the
graph consists of the stated lines.
(c)
g 2
g
({ + | 2 + f{2 | 2 ) =
(1) i 2{ + 2| | 0 + f({2 · 2| | 0 + | 2 · 2{) = 0 i
g{
g{
2| | 0 + 2f{2 | | 0 = 32{ 3 2f{| 2
For f = 31, | 0 = 3
i 2|(1 + f{2 )| 0 = 32{(1 + f|2 ) i | 0 = 3
{(1 + f| 2 )
.
|(1 + f{2 )
{(1 3 | 2 )
{(1 + |)(1 3 |)
=3
, so | 0 = 0 when | = ±1 or { = 0 (which leads to | = ±1)
|(1 3 {2 )
|(1 + {)(1 3 {)
and | 0 is unde¿ned when { = ±1 or | = 0 (which leads to { = ±1). Since the graph consists of the lines { = ±1 and
| = ±1, the slope at any point on the graph is unde¿ned or 0, which is consistent with the expression found for |0 .
3.6 Derivatives of Logarithmic Functions
1. The differentiation formula for logarithmic functions,
2. i ({) = { ln { 3 {
3. i ({) = sin(ln {)
i
i 0 ({) = { ·
1
+ (ln {) · 1 3 1 = 1 + ln { 3 1 = ln {
{
i i 0 ({) = cos(ln {) ·
cos(ln {)
g
1
ln { = cos(ln {) · =
g{
{
{
4. i ({) = ln(sin2 {) = ln(sin {)2 = 2 ln |sin {|
5. i ({) = ln
1
{
i i 0 ({) =
Another solution: i({) = ln
6. | =
1
= (ln {)31
ln {
1 g
1@{ g{
g
1
(logd {) =
, is simplest when d = h because ln h = 1.
g{
{ ln d
i i 0 ({) = 2 ·
1
· cos { = 2 cot {
sin {
1
1
1
={ 3 2 =3 .
{
{
{
1
1
= ln 1 3 ln { = 3 ln { i i 0 ({) = 3 .
{
{
i | 0 = 31(ln {)32 ·
31
1
=
{
{(ln {)2
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¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
7. i ({) = log10 ({3 + 1)
i i 0 ({) =
i i 0 ({) =
8. i ({) = log5 ({h{ )
({3
g 3
1
3{2
({ + 1) = 3
+ 1) ln 10 g{
({ + 1) ln 10
g
{+1
1
1
h{ ({ + 1)
({h{ ) = {
({h{ + h{ · 1) =
=
ln 5 g{
{h ln 5
{h{ ln 5
{ ln 5
{h{
Another solution: We can change the form of the function by ¿rst using logarithm properties.
i ({) = log5 ({h{ ) = log5 { + log5 h{
9. i ({) = sin { ln(5{)
10. i (x) =
x
1 + ln x
i i 0 ({) =
i i 0 ({) = sin { ·
i i 0 (x) =
1
1
1
1
1+{
+ {
· h{ =
+
or
{ ln 5
h ln 5
{ ln 5 ln 5
{ ln 5
1 g
sin { · 5
sin {
·
(5{) + ln(5{) · cos { =
+ cos { ln(5{) =
+ cos { ln(5{)
5{ g{
5{
{
(1 + ln x)(1) 3 x(1@x)
1 + ln x 3 1
ln x
=
=
(1 + ln x)2
(1 + ln x)2
(1 + ln x)2
I
{2 3 1 = ln { + ln({2 3 1)1@2 = ln { +
11. j({) = ln {
j 0 ({) =
1
2
ln({2 3 1) i
1
1
1
1
{
{2 3 1 + { · {
2{2 3 1
+ · 2
· 2{ = + 2
=
=
2
{ 2 { 31
{
{ 31
{({ 3 1)
{({2 3 1)
I
12. k({) = ln { + {2 3 1
i k0 ({) =
1
I
{ + {2 3 1
I
1
{2 3 1 + {
{
1
I
· I
= I
1+ I
=
2
2
2
2
{ 31
{+ { 31
{ 31
{ 31
(2| + 1)5
= ln(2| + 1)5 3 ln(| 2 + 1)1@2 = 5 ln(2| + 1) 3 12 ln(|2 + 1) i
|2 + 1
1
1
10
|
8| 2 3 | + 10
1
J0 (|) = 5 ·
·23 · 2
· 2| =
3 2
or
2| + 1
2 | +1
2| + 1 | + 1
(2| + 1)(|2 + 1)
13. J(|) = ln s
14. j(u) = u2 ln(2u + 1)
15. I (v) = ln ln v
i j0 (u) = u2 ·
1 1
1
1 g
ln v =
· =
ln v gv
ln v v
v ln v
i I 0 (v) =
16. | = ln 1 + w 3 w3 i |0 =
17. | = tan [ln(d{ + e)]
18. | = ln |cos(ln {)|
2u2
1
· 2 + ln(2u + 1) · 2u =
+ 2u ln(2u + 1)
2u + 1
2u + 1
1
1 3 3w2
g
3
)
=
(1
+
w
3
w
1 + w 3 w3 gw
1 + w 3 w3
i | 0 = sec2 [ln(d{ + e)] ·
i |0 =
1
d
· d = sec2 [ln(d{ + e)]
d{ + e
d{ + e
1
tan(ln {)
1
(3 sin(ln {)) = 3
cos(ln {)
{
{
19. | = ln(h3{ + {h3{ ) = ln(h3{ (1 + {)) = ln(h3{ ) + ln(1 + {) = 3{ + ln(1 + {)
|0 = 31 +
i
31 3 { + 1
{
1
=
=3
1+{
1+{
1+{
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS
20. K(}) = ln
u
1
2
ln(d2 3 } 2 ) 3
21. | = 2{ log10
I
{ = 2{ log10 {1@2 = 2{ ·
1
2
log10 { = { log10 { i | 0 = { ·
| 0 = 3 log2 h +
1
1
+ log10 { · 1 =
+ log10 {
{ ln 10
ln 10
i
g
1
3 sin {
(cos {) = 3 log2 h +
= 3 log2 h 3
tan {
cos { (ln 2) g{
cos { (ln 2)
ln 2
ln h
1
=
= log2 h, so the answer could be written as 3 log2 h 3 log2 h tan { = (3 log2 h) (1 + tan {).
ln 2
ln 2
23. | = {2 ln(2{)
| 00 = 1 + 2{ ·
ln {
{2
| 00 =
i | 0 = {2 ·
1
· 2 + ln(2{) · (2{) = { + 2{ ln(2{) i
2{
1
· 2 + ln(2{) · 2 = 1 + 2 + 2 ln(2{) = 3 + 2 ln(2{)
2{
i |0 =
{2 (1@{) 3 (ln {)(2{)
{(1 3 2 ln {)
1 3 2 ln {
=
=
{4
{3
({2 )2
i
{3 (32@{) 3 (1 3 2 ln {)(3{2 )
{2 (32 3 3 + 6 ln {)
6 ln { 3 5
=
=
3
2
({ )
{6
{4
25. | = ln { +
I
i
1 + {2
k
l
I
g 1
1
I
I
1 + 12 (1 + {2 )31@2 (2{)
{ + 1 + {2 =
2
2
{ + 1 + { g{
{+ 1+{
I
1 + {2 + {
{
1
1
1
I
I
1+ I
=
· I
= I
=
{ + 1 + {2
1 + {2
{ + 1 + {2
1 + {2
1 + {2
|0 =
ln(d2 + } 2 ) i
ln h
1
1
=
= log10 h, so the answer could be written as
+ log10 { = log10 h + log10 { = log10 h{.
ln 10
ln 10
ln 10
22. | = log2 (h3{ cos {) = log2 h3{ + log2 cos { = 3{ log2 h + log2 cos {
24. | =
227
2d2 }
} 3 + }d2 3 } 3 + }d2
= 4
2
2
2
2
(} 3 d )(} + d )
} 3 d4
=
Note:
1
2
¤
1
1
1
1
}
}
}(} 2 + d2 ) 3 }(} 2 3 d2 )
· 2
· (32}) 3 · 2
· (2}) = 2
3 2
=
2
2
2
2
2 d 3}
2 d +}
} 3d
} +d
(} 2 3 d2 )(} 2 + d2 )
K 0 (}) =
Note:
1@2
2
2
d2 3 } 2
d 3 }2
d 3 }2
1
=
= ln 2
= ln 2
d2 + } 2
d + }2
2
d + }2
| 00 = 3 12 (1 + {2 )33@2 (2{) =
26. | = ln(sec { + tan {)
27. i ({) =
0
i ({) =
=
{
1 3 ln({ 3 1)
i
3{
(1 + {2 )3@2
i |0 =
sec { tan { + sec2 {
= sec { i | 00 = sec { tan {
sec { + tan {
i
31
({ 3 1)[1 3 ln({ 3 1)] + {
{ 3 1 3 ({ 3 1) ln({ 3 1) + {
{31 =
{31
=
2
2
[1
3
ln({
3
1)]
({ 3 1)[1 3 ln({ 3 1)]2
[1 3 ln({ 3 1)]
[1 3 ln({ 3 1)] · 1 3 { ·
2{ 3 1 3 ({ 3 1) ln({ 3 1)
({ 3 1)[1 3 ln({ 3 1)]2
[continued]
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¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
Dom(i) = {{ | { 3 1 A 0 and 1 3 ln({ 3 1) 6= 0} = {{ | { A 1 and
= { | { A 1 and { 3 1 6= h
28. i ({) =
I
2 + ln { = (2 + ln {)1@2
1
ln({ 3 1) 6= 1}
= {{ | { A 1 and { 6= 1 + h} = (1> 1 + h) (1 + h> ")
i i 0 ({) =
1
1
1
(2 + ln {)31@2 · = I
2
{
2{ 2 + ln {
Dom(i) = {{ | 2 + ln { D 0} = {{ | ln { D 32} = {{ | { D h32 } = [h32 > ").
29. i ({) = ln({2 3 2{)
i i 0 ({) =
2({ 3 1)
1
(2{ 3 2) =
.
{2 3 2{
{({ 3 2)
Dom(i ) = {{ | {({ 3 2) A 0} = (3"> 0) (2> ").
30. i ({) = ln ln ln {
i i 0 ({) =
1
1
1
·
· .
ln ln { ln { {
Dom(i ) = {{ | ln ln { A 0} = {{ | ln { A 1} = {{ | { A h} = (h> ").
31. i ({) =
ln {
{2
so i 0 (1) =
{2 (1@{) 3 (ln {)(2{)
{ 3 2{ ln {
{(1 3 2 ln {)
1 3 2 ln {
=
=
=
,
({2 )2
{4
{4
{3
i i 0 ({) =
1 3 2 ln 1
132·0
= 1.
=
13
1
32. i ({) = ln(1 + h2{ )
i i 0 ({) =
33. | = ln({2 3 3{ + 1)
|0 =
i
1
2h2{
2h0
2(1)
(2h2{ ) =
, so i 0 (0) =
=
= 1.
2{
2{
1+h
1+h
1 + h0
1+1
1
· (2{ 3 3)
{2 3 3{ + 1
i
| 0 (3) =
1
1
· 3 = 3, so an equation of a tangent line at
(3> 0) is | 3 0 = 3({ 3 3), or | = 3{ 3 9.
34. | = {2 ln {
i | 0 = {2 ·
1
+ (ln {)(2{) i |0 (1) = 1 + 0 = 1 , so an equation of a tangent line at (1> 0) is
{
| 3 0 = 1({ 3 1), or | = { 3 1.
35. i ({) = sin { + ln {
i i 0 ({) = cos { + 1@{.
This is reasonable, because the graph shows that i increases when i 0 is
positive, and i 0 ({) = 0 when i has a horizontal tangent.
36. | =
ln {
{
| 0 (1) =
i |0 =
{(1@{) 3 ln {
1 3 ln {
=
.
{2
{2
130
131
= 1 and | 0 (h) =
= 0 i equations of tangent
12
h2
lines are | 3 0 = 1({ 3 1) or | = { 3 1 and | 3 1@h = 0({ 3 h)
or | = 1@h.
37. i ({) = f{ + ln(cos {)
i 0 ( 4 ) = 6
i
i
i 0 ({) = f +
f 3 tan 4 = 6
i
1
· (3 sin {) = f 3 tan {.
cos {
f31 =6
i
f = 7.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS
38. i ({) = logd (3{2 3 2)
i 0 (1) = 3
i 0 ({) =
i
1
·6=3
ln d
i
i
¤
1
· 6{.
(3{2 3 2) ln d
2 = ln d
i
d = h2 .
39. | = ({2 + 2)2 ({4 + 4)4
i ln | = ln[({2 + 2)2 ({4 + 4)4 ] i ln | = 2 ln({2 + 2) + 4 ln({4 + 4) i
1 0
1
16{3
1
4{
| =2· 2
· 2{ + 4 · 4
· 4{3 i | 0 = | 2
+ 4
i
|
{ +2
{ +4
{ +2 { +4
4{
16{3
|0 = ({2 + 2)2 ({4 + 4)4
+
{2 + 2
{4 + 4
40. | =
h3{ cos2 {
{2 + { + 1
i ln | = ln
h3{ cos2 {
{2 + { + 1
i
ln | = ln h3{ + ln | cos { |2 3 ln({2 + { + 1) = 3{ + 2 ln | cos { | 3 ln({2 + { + 1) i
1 0
1
1
2{ + 1
| = 31 + 2 ·
(3 sin {) 3 2
(2{ + 1) i |0 = | 31 3 2 tan { 3 2
i
|
cos {
{ +{+1
{ +{+1
2{ + 1
h3{ cos2 {
|0 = 3 2
1 + 2 tan { + 2
{ +{+1
{ +{+1
41. | =
u
{31
{4 + 1
1@2
{31
i ln | = ln 4
{ +1
1 1
1 0
1 1
| =
3
· 4{3
|
2{31
2 {4 + 1
42. | =
I {2 3{
{h
({ + 1)2@3
1
1
ln({ 3 1) 3 ln({4 + 1) i
2
2
u
1
{31
2{3
1
2{3
0
0
i | =|
3 4
i | =
3
2({ 3 1)
{ +1
{4 + 1 2{ 3 2
{4 + 1
i ln | =
l
k
2
i
i ln | = ln {1@2 h{ 3{ ({ + 1)2@3
1 0
1 1
2
1
| = · + 2{ 3 1 + ·
i
|
2 {
3 {+1
I
2
1
2
2
1
+ 2{ 3 1 +
i |0 = { h{ 3{ ({ + 1)2@3
+ 2{ 3 1 +
|0 = |
2{
3{ + 3
2{
3{ + 3
1
2
ln | =
43. | = {{
ln { + ({2 3 {) +
2
3
i ln | = ln {{
ln({ + 1) i
i ln | = { ln { i | 0 @| = {(1@{) + (ln {) · 1 i | 0 = |(1 + ln {) i
| 0 = {{ (1 + ln {)
1 0
1
| = cos { · + ln { · (3 sin {) i
|
{
cos {
cos {
3 ln { sin {
i |0 = {cos {
3 ln { sin {
|0 = |
{
{
44. | = {cos {
i ln | = ln {cos {
45. | = { sin {
i ln | = ln { sin {
|0 = |
sin {
+ ln { cos {
{
i
i ln | = cos { ln { i
|0
1
= (sin {) · + (ln {)(cos {) i
|
{
sin {
| 0 = { sin {
+ ln { cos {
{
i
ln | = sin { ln { i
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229
230
¤
46. | =
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
I {
{
|0 = |
1
2
i ln | = ln
+
1
2
I {
{
i ln | = { ln {1@2
i ln | = 12 { ln { i
I
ln { i | 0 = 12 { { (1 + ln {)
47. | = (cos {){
i ln | = ln(cos {){
i ln | = { ln cos { i
1
1
1 0
1
| = { · + ln { ·
|
2
{
2
i
1 0
1
| ={·
· (3 sin {) + ln cos { · 1 i
|
cos {
{ sin {
i | 0 = (cos {){ (ln cos { 3 { tan {)
| 0 = | ln cos { 3
cos {
1
1 0
1
| = ln { ·
· cos { + ln sin { ·
|
sin {
{
ln sin {
cos {
ln sin {
+
i | 0 = (sin {)ln { ln { cot { +
| 0 = | ln { ·
sin {
{
{
48. | = (sin {)ln {
i ln | = ln(sin {)ln {
i ln | = ln { · ln sin { i
i
1
ln tan { i
{
1
ln tan {
1
1
sec2 {
1 0
i |0 = |
i
| = ·
· sec2 { + ln tan { · 3 2
3
|
{ tan {
{
{ tan {
{2
sec2 {
ln tan {
ln tan {
0
1@{ 1
=
(tan
{)
·
| 0 = (tan {)1@{
or
|
3
csc
{
sec
{
3
{ tan {
{2
{
{
49. | = (tan {)1@{
i ln | = ln(tan {)1@{
50. | = (ln {)cos {
i ln | = cos { ln(ln {) i
| 0 = (ln {)cos {
cos {
{ ln {
51. | = ln({2 + | 2 )
i ln | =
|0
1
1
= cos { ·
· + (ln ln {)(3 sin {) i
|
ln { {
3 sin { ln ln {
i |0 =
1
2{ + 2|| 0
g
({2 + | 2 ) i | 0 = 2
{2 + | 2 g{
{ + |2
{2 | 0 + | 2 | 0 3 2|| 0 = 2{ i ({2 + | 2 3 2|)| 0 = 2{ i |0 =
52. {| = | {
|0 =
i | ln { = { ln |
i |·
i {2 | 0 + | 2 | 0 = 2{ + 2||0
i
2{
{2 + | 2 3 2|
1
1
+ (ln {) · | 0 = { · · | 0 + ln |
{
|
i |0 ln { 3
{ 0
|
| = ln | 3
|
{
i
ln | 3 |@{
ln { 3 {@|
53. i ({) = ln({ 3 1)
i i 0 ({) =
i (4) ({) = 32 · 3({ 3 1)34
1
= ({ 3 1)31
({ 3 1)
i ···
i i 00 ({) = 3({ 3 1)32
i i 000 ({) = 2({ 3 1)33
i i (q) ({) = (31)q31 · 2 · 3 · 4 · · · · · (q 3 1)({ 3 1)3q = (31)q31
54. | = {8 ln {, so G9 | = G8 | 0 = G8(8{7 ln { + {7 ). But the eighth derivative of {7 is 0, so we now have
G8(8{7 ln {) = G7(8 · 7{6 ln { + 8{6 ) = G7(8 · 7{6 ln {) = G6(8 · 7 · 6{5 ln {) = · · · = G(8! {0 ln {) = 8!@{=
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°
© Cengage Learning. All Rights Reserved.
i
(q 3 1)!
({ 3 1)q
NOT FOR SALE
SECTION 3.7
55. If i ({) = ln (1 + {), then i 0 ({) =
Thus, lim
{<0
RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES
¤
231
1
, so i 0 (0) = 1.
1+{
ln(1 + {)
i ({)
i({) 3 i(0)
= lim
= lim
= i 0 (0) = 1.
{<0
{<0
{
{
{30
56. Let p = q@{. Then q = {p, and as q < ", p < ".
q
p{ p {
{
1
1
= lim 1 +
= lim 1 +
= h{ by Equation 6.
1+
q<"
p<"
p<"
q
p
p
Therefore, lim
3.7 Rates of Change in the Natural and Social Sciences
1. (a) v = i(w) = w3 3 12w2 + 36w (in feet)
i y(w) = i 0 (w) = 3w2 3 24w + 36 (in ft@s)
(b) y(3) = 27 3 72 + 36 = 39 ft@s
(c) The particle is at rest when y(w) = 0. 3w2 3 24w + 36 = 0 C 3(w 3 2)(w 3 6) = 0 C w = 2 s or 6 s.
(d) The particle is moving in the positive direction when y(w) A 0. 3(w 3 2)(w 3 6) A 0 C 0 $ w ? 2 or w A 6.
(e) Since the particle is moving in the positive direction and in the
(f)
negative direction, we need to calculate the distance traveled in the
intervals [0> 2], [2> 6], and [6> 8] separately.
|i (2) 3 i (0)| = |32 3 0| = 32.
|i (6) 3 i (2)| = |0 3 32| = 32.
|i (8) 3 i (6)| = |32 3 0| = 32.
The total distance is 32 + 32 + 32 = 96 ft.
(g) y(w) = 3w2 3 24w + 36 i
(h)
d(w) = y 0 (w) = 6w 3 24.
d(3) = 6(3) 3 24 = 36 (ft@s)@s or ft@s2 .
(i) The particle is speeding up when y and d have the same sign. This occurs when 2 ? w ? 4 [y and d are both negative] and
when w A 6 [y and d are both positive]. It is slowing down when y and d have opposite signs; that is, when 0 $ w ? 2 and
when 4 ? w ? 6.
2. (a) v = i(w) = 0=01w4 3 0=04w3 (in feet)
i y(w) = i 0 (w) = 0=04w3 3 0=12w2 (in ft@s)
(b) y(3) = 0=04(3)3 3 0=12(3)2 = 0 ft@s
(c) The particle is at rest when y(w) = 0. 0=04w3 3 0=12w2 = 0 C 0=04w2 (w 3 3) = 0 C w = 0 s or 3 s.
(d) The particle is moving in the positive direction when y(w) A 0. 0=04w2 (w 3 3) A 0 C w A 3.
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NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
(e) See Exercise 1(e).
(f)
|i (3) 3 i (0)| = |30=27 3 0| = 0=27.
|i (8) 3 i (3)| = |20=48 3 (30=27)| = 20=75.
The total distance is 0=27 + 20=75 = 21=02 ft.
(g) y(w) = 0=04w3 3 0=12w2
i d(w) = y 0 (w) = 0=12w2 3 0=24w. d(3) = 0=12(3)2 3 0=24(3) = 0=36 (ft@s)@s or ft@s2 .
(h) Here we show the graph of v, y, and d
for 0 $ w $ 4 and 4 $ w $ 8.
(i) The particle is speeding up when y and d have the same sign. This occurs when 0 ? w ? 2 [y and d are both negative]
and when w A 3 [y and d are both positive]. It is slowing down when y and d have opposite signs; that is,
when 2 ? w ? 3.
3. (a) v = i(w) = cos(w@4)
i y(w) = i 0 (w) = 3 sin(w@4) · (@4)
= 3 4 ·
(b) y(3) = 3 4 sin 3
4
I
2
2
I
= 3 8 2 ft@s [E 30=56]
= 0 i sin w
=0 i
(c) The particle is at rest when y(w) = 0. 3 4 sin w
4
4
w
4
= q i w = 0, 4, 8 s.
A 0 i sin w
? 0 i 4 ? w ? 8.
(d) The particle is moving in the positive direction when y(w) A 0. 3 4 sin w
4
4
(e) From part (c), y(w) = 0 for w = 0> 4> 8. As in Exercise 1, we’ll
(f)
¿nd the distance traveled in the intervals [0> 4] and [4> 8].
|i (4) 3 i (0)| = |31 3 1| = 2
|i (8) 3 i (4)| = |1 3 (31)| = 2.
The total distance is 2 + 2 = 4 ft.
(g) y(w) = 3
w
sin
4
4
i
(h)
w 2
w
cos
· = 3 cos .
4
4 4
16
4
I I
2
2 2
2
3
2
=3
3
=
(ft@s)@s or ft@s2 .
d(3) = 3 cos
16
4
16
2
32
d(w) = y 0 (w) = 3
(i) The particle is speeding up when y and d have the same sign. This occurs when 0 ? w ? 2 or 8 ? w ? 10 [y and d are
both negative] and when 4 ? w ? 6 [y and d are both positive]. It is slowing down when y and d have opposite signs;
that is, when 2 ? w ? 4 and when 6 ? w ? 8.
4. (a) v = i(w) = wh3w@2
i y(w) = i 0 (w) = wh3w@2 3 12 + h3w@2 · 1 = h3w@2 3 12 w + 1
(b) y(3) = h33@2 3 32 + 1 = 3 12 h33@2 ft@s [E 30=11]
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°
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SECTION 3.7
RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES
¤
233
(c) The particle is at rest when y(w) = 0. h3w@2 3 12 w + 1 = 0 C w = 2 s.
(d) The particle is moving in the positive direction when y(w) A 0. h3w@2 3 12 w + 1 A 0 i 3 12 w + 1 A 0 i
3 12 w A 31 i 0 ? w ? 2.
(f)
(e) See Exercise 1(a).
|i (2) 3 i (0)| = 2h31 3 0 = 2h31 .
|i (8) 3 i (2)| = 8h34 3 2h31 = 2h31 3 8h34 .
The total distance is 2h31 + 2h31 3 8h34 = 4h31 3 8h34 ft. [E 1=32]
(g) y(w) = h3w@2 3 12 w + 1 i
(h)
d(w) = y 0 (w) = h3w@2 3 12 + 3 12 w + 1 h3w@2 3 12
= 3 12 h3w@2 1 + 3 12 w + 1
= 3 12 h3w@2 3 12 w + 2 = 14 h3w@2 (w 3 4)
d(3) = 3 14 h33@2 ft@s2
(i) The particle is speeding up when y and d have the same sign. This occurs when 2 ? w ? 4 [y and d are both negative].
It is slowing down when y and d have opposite signs; that is, when 0 ? w ? 2 and when w A 4.
5. (a) From the ¿gure, the velocity y is positive on the interval (0> 2) and negative on the interval (2> 3). The acceleration d is
positive (negative) when the slope of the tangent line is positive (negative), so the acceleration is positive on the interval
(0> 1), and negative on the interval (1> 3). The particle is speeding up when y and d have the same sign, that is, on the
interval (0> 1) when y A 0 and d A 0, and on the interval (2> 3) when y ? 0 and d ? 0. The particle is slowing down
when y and d have opposite signs, that is, on the interval (1> 2) when y A 0 and d ? 0.
(b) y A 0 on (0> 3) and y ? 0 on (3> 4). d A 0 on (1> 2) and d ? 0 on (0> 1) and (2> 4). The particle is speeding up on (1> 2)
[y A 0, d A 0] and on (3> 4) [y ? 0, d ? 0]. The particle is slowing down on (0> 1) and (2> 3) [y A 0, d ? 0].
6. (a) The velocity y is positive when v is increasing, that is, on the intervals (0> 1) and (3> 4); and it is negative when v is
decreasing, that is, on the interval (1> 3). The acceleration d is positive when the graph of v is concave upward (y is
increasing), that is, on the interval (2> 4); and it is negative when the graph of v is concave downward (y is decreasing), that
is, on the interval (0> 2). The particle is speeding up on the interval (1> 2) [y ? 0, d ? 0] and on (3> 4) [y A 0, d A 0].
The particle is slowing down on the interval (0> 1) [y A 0, d ? 0] and on (2> 3) [y ? 0, d A 0].
(b) The velocity y is positive on (3> 4) and negative on (0> 3). The acceleration d is positive on (0> 1) and (2> 4) and negative
on (1> 2). The particle is speeding up on the interval (1> 2) [y ? 0, d ? 0] and on (3> 4) [y A 0, d A 0]. The particle is
slowing down on the interval (0> 1) [y ? 0, d A 0] and on (2> 3) [y ? 0, d A 0].
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NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
7. (a) k(w) = 2 + 24=5w 3 4=9w2
i y(w) = k0 (w) = 24=5 3 9=8w. The velocity after 2 s is y(2) = 24=5 3 9=8(2) = 4=9 m@s
and after 4 s is y(4) = 24=5 3 9=8(4) = 314=7 m@s.
(b) The projectile reaches its maximum height when the velocity is zero. y(w) = 0 C 24=5 3 9=8w = 0 C
w=
24=5
= 2=5 s.
9=8
(c) The maximum height occurs when w = 2=5. k(2=5) = 2 + 24=5(2=5) 3 4=9(2=5)2 = 32=625 m or 32 58 m .
(d) The projectile hits the ground when k = 0 C 2 + 24=5w 3 4=9w2 = 0 C
s
324=5 ± 24=52 3 4(34=9)(2)
i w = wi E 5=08 s [since w D 0]=
w=
2(34=9)
(e) The projectile hits the ground when w = wi . Its velocity is y(wi ) = 24=5 3 9=8wi E 325=3 m@s [downward].
8. (a) At maximum height the velocity of the ball is 0 ft@s. y(w) = v0 (w) = 80 3 32w = 0
2
So the maximum height is v 52 = 80 52 3 16 52 = 200 3 100 = 100 ft.
C 32w = 80 C w = 52 .
(b) v(w) = 80w 3 16w2 = 96 C 16w2 3 80w + 96 = 0 C 16(w2 3 5w + 6) = 0 C 16(w 3 3)(w 3 2) = 0.
So the ball has a height of 96 ft on the way up at w = 2 and on the way down at w = 3. At these times the velocities are
y(2) = 80 3 32(2) = 16 ft@s and y(3) = 80 3 32(3) = 316 ft@s, respectively.
9. (a) k(w) = 15w 3 1=86w2
i
y(w) = k0 (w) = 15 3 3=72w. The velocity after 2 s is y(2) = 15 3 3=72(2) = 7=56 m@s.
(b) 25 = k C 1=86w2 3 15w + 25 = 0 C w =
15 ±
s
152 3 4(1=86)(25)
2(1=86)
C w = w1 E 2=35 or w = w2 E 5=71.
The velocities are y(w1 ) = 15 3 3=72w1 E 6=24 m@s [upward] and y(w2 ) = 15 3 3=72w2 E 36=24 m@s [downward].
10. (a) v(w) = w4 3 4w3 3 20w2 + 20w
i y(w) = v0 (w) = 4w3 3 12w2 3 40w + 20. y = 20 C
4w3 3 12w2 3 40w + 20 = 20 C 4w3 3 12w2 3 40w = 0 C 4w(w2 3 3w 3 10) = 0 C
4w(w 3 5)(w + 2) = 0 C w = 0 s or 5 s [for w D 0].
(b) d(w) = y 0 (w) = 12w2 3 24w 3 40. d = 0 C 12w2 3 24w 3 40 = 0 C 4(3w2 3 6w 3 10) = 0 C
s
6 ± 62 3 4(3)(310)
1I
=1±
39 E 3=08 s [for w D 0]. At this time, the acceleration changes from negative to
w=
2(3)
3
positive and the velocity attains its minimum value.
11. (a) D({) = {2
i D0 ({) = 2{. D0 (15) = 30 mm2 @mm is the rate at which
the area is increasing with respect to the side length as { reaches 15 mm.
(b) The perimeter is S ({) = 4{, so D0 ({) = 2{ = 12 (4{) = 12 S ({). The
¿gure suggests that if {{ is small, then the change in the area of the square
is approximately half of its perimeter (2 of the 4 sides) times {{. From the
¿gure, {D = 2{ ({{) + ({{)2 . If {{ is small, then {D E 2{ ({{) and
so {D@{{ E 2{.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.7
12. (a) Y ({) = {3
i
RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES
¤
gY
gY = 3(3)2 = 27 mm3 @mm is the
= 3{2 .
g{
g{ {=3
rate at which the volume is increasing as { increases past 3 mm.
(b) The surface area is V({) = 6{2 , so Y 0 ({) = 3{2 = 12 (6{2 ) = 12 V({).
The ¿gure suggests that if {{ is small, then the change in the volume of the
cube is approximately half of its surface area (the area of 3 of the 6 faces)
times {{. From the ¿gure, {Y = 3{2 ({{) + 3{({{)2 + ({{)3 .
If {{ is small, then {Y E 3{2 ({{) and so {Y @{{ E 3{2 .
13. (a) Using D(u) = u2 , we ¿nd that the average rate of change is:
(i)
(iii)
D(3) 3 D(2)
9 3 4
=
= 5
332
1
(ii)
D(2=5) 3 D(2)
6=25 3 4
=
= 4=5
2=5 3 2
0=5
D(2=1) 3 D(2)
4=41 3 4
=
= 4=1
2=1 3 2
0=1
i D0 (u) = 2u, so D0 (2) = 4.
(b) D(u) = u2
(c) The circumference is F(u) = 2u = D0 (u). The ¿gure suggests that if {u is small,
then the change in the area of the circle (a ring around the outside) is approximately
equal to its circumference times {u. Straightening out this ring gives us a shape that
is approximately rectangular with length 2u and width {u, so {D E 2u({u).
Algebraically, {D = D(u + {u) 3 D(u) = (u + {u)2 3 u2 = 2u({u) + ({u)2 .
So we see that if {u is small, then {D E 2u({u) and therefore, {D@{u E 2u.
14. After w seconds the radius is u = 60w, so the area is D(w) = (60w)2 = 3600w2
(a) D0 (1) = 7200 cm2 @s
(b) D0 (3) = 21> 600 cm2 @s
i D0 (w) = 7200w i
(c) D0 (5) = 36,000 cm2 @s
As time goes by, the area grows at an increasing rate. In fact, the rate of change is linear with respect to time.
15. V(u) = 4u2
0
i V 0 (u) = 8u
2
i
(a) V (1) = 8 ft @ft
(b) V 0 (2) = 16 ft2 @ft
(c) V 0 (3) = 24 ft2 @ft
As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the
radius.
16. (a) Using Y (u) =
3
4
3 u ,
we ¿nd that the average rate of change is:
4
(512)
3
3 43 (125)
= 172 m3 @m
3
4
(216)
3
3 43 (125)
= 121=3 m3 @m
1
(i)
Y (8) 3 Y (5)
=
835
(ii)
Y (6) 3 Y (5)
=
635
(iii)
Y (5=1) 3 Y (5)
=
5=1 3 5
4
(5=1)3
3
3 43 (5)3
= 102=013 m3 @m
0=1
(b) Y 0 (u) = 4u2 , so Y 0 (5) = 100 m3 @m.
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236
¤
NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
i Y 0 (u) = 4u2 = V(u). By analogy with Exercise 13(c), we can say that the change in the volume
(c) Y (u) = 43 u3
of the spherical shell, {Y , is approximately equal to its thickness, {u, times the surface area of the inner sphere. Thus,
{Y E 4u2 ({u) and so {Y @{u E 4u2 .
17. The mass is i ({) = 3{2 , so the linear density at { is ({) = i 0 ({) = 6{.
(a) (1) = 6 kg@m
(b) (2) = 12 kg@m
(c) (3) = 18 kg@m
Since is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end.
i Y 0 (w) = 5000 · 2 1 3
5
= 3218=75 gal@min
(a) Y 0 (5) = 3250 1 3 40
= 3125 gal@min
(c) Y 0 (20) = 3250 1 3 20
40
18. Y(w) = 5000 1 3
2
1
w
40
1 3 40 = 3250 1 3
1
w
40
1
w
40
(b) Y 0 (10) = 3250 1 3
(d) Y 0 (40) = 3250 1 3
10
40
40
40
= 3187=5 gal@min
= 0 gal@min
The water is Àowing out the fastest at the beginning — when w = 0, Y 0 (w) = 3250 gal@min. The water is Àowing out the
slowest at the end — when w = 40, Y 0 (w) = 0. As the tank empties, the water Àows out more slowly.
19. The quantity of charge is T(w) = w3 3 2w2 + 6w + 2, so the current is T0 (w) = 3w2 3 4w + 6.
(a) T0 (0=5) = 3(0=5)2 3 4(0=5) + 6 = 4=75 A
(b) T0 (1) = 3(1)2 3 4(1) + 6 = 5 A
The current is lowest when T0 has a minimum. T00 (w) = 6w 3 4 ? 0 when w ? 23 . So the current decreases when w ?
increases when w A 23 . Thus, the current is lowest at w =
20. (a) I =
JpP
= (JpP)u32
u2
i
2
3
2
3
and
s.
gI
2JpP
, which is the rate of change of the force with
= 32(JpP)u33 = 3
gu
u3
respect to the distance between the bodies. The minus sign indicates that as the distance u between the bodies increases,
the magnitude of the force I exerted by the body of mass p on the body of mass P is decreasing.
(b) Given I 0 (20,000) = 32, ¿nd I 0 (10,000). 32 = 3
I 0 (10> 000) = 3
21. With p = p0 1 3
2JpP
20,0003
i JpP = 20,0003 .
2(20,0003 )
= 32 · 23 = 316 N@km
10,0003
y2
f2
31@2
,
%
31@2
33@2 & g
1
2y g
y2
g
g
y2
I=
· d + y · p0 3
3 2
(py) = p (y) + y
(p) = p0 1 3 2
13 2
(y)
gw
gw
gw
f
2
f
f
gw
33@2 y2
y2
y2
p0 d
·d 13 2 + 2 =
= p0 1 3 2
f
f
f
(1 3 y 2 @f2 )3@2
Note that we factored out (1 3 y 2 @f2 )33@2 since 33@2 was the lesser exponent. Also note that
22. (a) G(w) = 7 + 5 cos[0=503(w 3 6=75)]
g
(y) = d.
gw
i G0 (w) = 35 sin[0=503(w 3 6=75)](0=503) = 32=515 sin[0=503(w 3 6=75)].
At 3:00 AM, w = 3, and G0 (3) = 32=515 sin[0=503(33=75)] E 2=39 m@h (rising).
(b) At 6:00 AM , w = 6, and G0 (6) = 32=515 sin[0=503(30=75)] E 0=93 m@h (rising).
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RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES
¤
237
(c) At 9:00 AM, w = 9, and G0 (9) = 32=515 sin[0=503(2=25)] E 32=28 m@h (falling).
(d) At noon, w = 12, and G0 (12) = 32=515 sin[0=503(5=25)] E 31=21 m@h (falling).
23. (a) To ¿nd the rate of change of volume with respect to pressure, we ¿rst solve for Y in terms of S .
SY = F
i Y =
F
S
i
gY
F
= 3 2.
gS
S
(b) From the formula for gY @gS in part (a), we see that as S increases, the absolute value of gY @gS decreases.
Thus, the volume is decreasing more rapidly at the beginning.
(c) = 3
24. (a) [C] =
1 gY
1
=3
Y gS
Y
d2 nw
dnw + 1
F
F
1
F
3 2 =
=
=
S
(S Y )S
FS
S
i rate of reaction =
g[C]
(dnw + 1)(d2 n) 3 (d2 nw)(dn)
d2 n(dnw + 1 3 dnw)
d2 n
=
=
=
2
2
gw
(dnw + 1)
(dnw + 1)
(dnw + 1)2
d2 nw
d2 nw + d 3 d2 nw
d
=
=
.
dnw + 1
dnw + 1
dnw + 1
2
g{
d
d2 n
g[C]
[from part (a)] =
.
=
=
So n(d 3 {)2 = n
dnw + 1
(dnw + 1)2
gw
gw
(b) If { = [C], then d 3 { = d 3
(c) As w < ", [C] =
(d) As w < ",
d2 nw
(d2 nw)@w
d2 n
d2 n
=
=
<
= d moles@L.
dnw + 1
(dnw + 1)@w
dn + (1@w)
dn
d2 n
g[C]
=
< 0.
gw
(dnw + 1)2
(e) As w increases, nearly all of the reactants A and B are converted into product C. In practical terms, the reaction virtually
stops.
25. In Example 6, the population function was q = 2w q0 . Since we are tripling instead of doubling and the initial population is
400, the population function is q(w) = 400 · 3w . The rate of growth is q0 (w) = 400 · 3w · ln 3, so the rate of growth after
2=5 hours is q0 (2=5) = 400 · 32=5 · ln 3 E 6850 bacteria@hour=
26. q = i(w) =
d
1 + eh30=7w
i (0) = 20 i 20 =
12
e
=
14
1+e
i q0 = 3
d
1+e
d · eh30=7w (30=7)
(1 + eh30=7w )2
[Reciprocal Rule]. When w = 0, q = 20 and q0 = 12.
i d = 20(1 + e). i 0 (0) = 12 i 12 =
0=7de
(1 + e)2
i 12 =
0=7(20)(1 + e)e
(1 + e)2
i
i 6(1 + e) = 7e i 6 + 6e = 7e i e = 6 and d = 20(1 + 6) = 140. For the long run, we let w
increase without bound.
lim i(w) = lim
w<"
w<"
140
140
= 140, indicating that the yeast population stabilizes
=
1 + 6h30=7w
1+6·0
at 140 cells.
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CHAPTER 3
27. (a) 1920: p1 =
DIFFERENTIATION RULES
110
210
1860 3 1750
2070 3 1860
=
= 11, p2 =
=
= 21,
1920 3 1910
10
1930 3 1920
10
(p1 + p2 )/ 2 = (11 + 21)@2 = 16 million@year
1980: p1 =
4450 3 3710
5280 3 4450
740
830
=
= 74, p2 =
=
= 83,
1980 3 1970
10
1990 3 1980
10
(p1 + p2 )/ 2 = (74 + 83)@2 = 78=5 million@year
(b) S (w) = dw3 + ew2 + fw + g (in millions of people), where d E 0=0012937063, e E 37=061421911, f E 12,822=97902,
and g E 37,743,770=396.
(c) S (w) = dw3 + ew2 + fw + g i S 0 (w) = 3dw2 + 2ew + f (in millions of people per year)
(d) S 0 (1920) = 3(0=0012937063)(1920)2 + 2(37=061421911)(1920) + 12,822=97902
E 14.48 million@year [smaller than the answer in part (a), but close to it]
0
S (1980) E 75.29 million@year (smaller, but close)
(e) S 0 (1985) E 81.62 million@year, so the rate of growth in 1985 was about 81=62 million@year.
28. (a) D(w) = dw4 + ew3 + fw2 + gw + h, where d E 33=076923 × 1036 ,
(b)
e E 0=0243620824, f E 372=33147086, g E 95442=50365, and
h E 347,224,986=6.
(c) D(w) = dw4 + ew3 + fw2 + gw + h i
D0 (w) = 4dw3 + 3ew2 + 2fw + g.
(d) Part (b) gives D0 (1990) E 0=11 years of age per year.
29. (a) Using y =
y(u) =
(b) y(u) =
S
(U2 3 u2 ) with U = 0=01, o = 3, S = 3000, and = 0=027, we have y as a function of u:
4o
3000
(0=012 3 u2 ). y(0) = 0=925 cm@s, y(0=005) = 0=694 cm@s, y(0=01) = 0.
4(0=027)3
S
S
Su
(U2 3 u2 ) i y 0 (u) =
(32u) = 3
. When o = 3, S = 3000, and = 0=027, we have
4o
4o
2o
y0 (u) = 3
3000u
. y0 (0) = 0, y 0 (0=005) = 392=592 (cm@s)@cm, and y 0 (0=01) = 3185=185 (cm@s)@cm.
2(0=027)3
(c) The velocity is greatest where u = 0 (at the center) and the velocity is changing most where u = U = 0=01 cm
(at the edge).
30. (a) (i) i =
1
2O
u
W
=
u 1 W
O31
2 i
gi
=3
gO
u
u 1
W
1 W
O32 = 3 2
2
2O
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1
2O
u
RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES
¤
239
gi
1
1
1
1
I
W 1@2 i
=
W 31@2 =
I
I
2O gW
2 2O 4O W I
u
I I gi
1
W
W
W
W
1
(iii) i =
=
31@2 i
=3
33@2 = 3
2O
2O
g
2 2O
4O3@2
(ii) i =
W
=
(b) Note: Illustrating tangent lines on the generic ¿gures may help to explain the results.
(i)
gi
? 0 and O is decreasing i i is increasing i higher note
gO
(ii)
gi
A 0 and W is increasing i i is increasing i higher note
gW
(iii)
gi
? 0 and is increasing i i is decreasing i lower note
g
31. (a) F({) = 1200 + 12{ 3 0=1{2 + 0=0005{3
i F 0 ({) = 12 3 0=2{ + 0=0015{2 $@yard, which is the marginal cost
function.
(b) F 0 (200) = 12 3 0=2(200) + 0=0015(200)2 = $32@yard, and this is the rate at which costs are increasing with respect to
the production level when { = 200. F 0 (200) predicts the cost of producing the 201st yard=
(c) The cost of manufacturing the 201st yard of fabric is F(201) 3 F(200) = 3632=2005 3 3600 E $32=20, which is
approximately F 0 (200).
32. (a) F({) = 339 + 25{ 3 0=09{2 + 0=0004{3
i F 0 ({) = 25 3 0=18{ + 0=0012{2 . F 0 (100) = $19@item, and this is
the rate at which costs are increasing with respect to the production level when { = 100.
(b) The cost of producing the 101st item is F(101) 3 F(100) = 2358=0304 3 2339 = $19=03, which is approximately
F 0 (100).
33. (a) D({) =
s({)
{
i D0 ({) =
{s0 ({) 3 s({) · 1
{s0 ({) 3 s({)
=
.
2
{
{2
D0 ({) A 0 i D({) is increasing; that is, the average productivity increases as the size of the workforce increases.
(b) s0 ({) is greater than the average productivity i s0 ({) A D({) i s0 ({) A
{s0 ({) 3 s({) A 0 i
34. (a) V =
s({)
{
i {s0 ({) A s({) i
{s0 ({) 3 s({)
A 0 i D0 ({) A 0.
{2
(1 + 4{0=4 )(9=6{30=6 ) 3 (40 + 24{0=4 )(1=6{30=6 )
gU
=
g{
(1 + 4{0=4 )2
=
9=6{30=6 + 38=4{30=2 3 64{30=6 3 38=4{30=2
54=4{30=6
=3
0=4
2
(1 + 4{ )
(1 + 4{0=4 )2
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DIFFERENTIATION RULES
At low levels of brightness, U is quite large [U(0) = 40] and is quickly
(b)
decreasing, that is, V is negative with large absolute value. This is to be
expected: at low levels of brightness, the eye is more sensitive to slight
changes than it is at higher levels of brightness.
35. S Y = qUW
i W =
SY
1
SY
=
=
(S Y ). Using the Product Rule, we have
qU
(10)(0=0821)
0=821
1
gW
1
=
[S (w)Y 0 (w) + Y (w)S 0 (w)] =
[(8)(30=15) + (10)(0=10)] E 30=2436 K@min.
gw
0=821
0=821
36. (a) If gS@gw = 0, the population is stable (it is constant).
gS
S
(b)
S
= 0 i S = u0 1 3
gw
Sf
i
S
=13
u0
Sf
i
S
=13
Sf
u0
i S = Sf 1 3
.
u0
If Sf = 10,000, u0 = 5% = 0=05, and = 4% = 0=04, then S = 10,000 1 3 45 = 2000.
(c) If = 0=05, then S = 10,000 1 3 55 = 0. There is no stable population.
37. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is,
gF
gZ
= 0 and
= 0.
gw
gw
(b) “The caribou go extinct” means that the population is zero, or mathematically, F = 0.
(c) We have the equations
gZ
gF
= dF 3 eFZ and
= 3fZ + gFZ . Let gF@gw = gZ@gw = 0, d = 0=05, e = 0=001,
gw
gw
f = 0=05, and g = 0=0001 to obtain 0=05F 3 0=001FZ = 0 (1) and 30=05Z + 0=0001FZ = 0 (2). Adding 10 times
(2) to (1) eliminates the FZ -terms and gives us 0=05F 3 0=5Z = 0 i F = 10Z . Substituting F = 10Z into (1)
results in 0=05(10Z ) 3 0=001(10Z )Z = 0 C 0=5Z 3 0=01Z 2 = 0 C 50Z 3 Z 2 = 0 C
Z (50 3 Z ) = 0 C Z = 0 or 50. Since F = 10Z , F = 0 or 500. Thus, the population pairs (F> Z ) that lead to
stable populations are (0> 0) and (500> 50). So it is possible for the two species to live in harmony.
3.8 Exponential Growth and Decay
1. The relative growth rate is
gS
1 gS
= 0=7944, so
= 0=7944S and, by Theorem 2, S (w) = S (0)h0=7944w = 2h0=7944w .
S gw
gw
Thus, S (6) = 2h0=7944(6) E 234=99 or about 235 members.
2. (a) By Theorem 2, S (w) = S (0)hnw = 60hnw . In 20 minutes ( 13 hour), there are 120 cells, so S
hn@3 = 2 i n@3 = ln 2 i n = 3 ln 2 = ln 23 = ln 8.
1
3
= 60hn@3 = 120 i
(b) S (w) = 60h(ln 8)w = 60 · 8w
(c) S (8) = 60 · 88 = 60 · 224 = 1,006,632,960
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EXPONENTIAL GROWTH AND DECAY
¤
241
i S 0(8) = nS (8) = (ln 8)S (8) E 2=093 billion cells@h
(d) gS@gw = nS
(e) S (w) = 20,000 i 60 · 8w = 20,000 i 8w = 1000@3 i w ln 8 = ln(1000@3) i
w=
ln(1000@3)
E 2=79 h
ln 8
3. (a) By Theorem 2, S (w) = S (0)hnw = 100hnw . Now S (1) = 100hn(1) = 420
i hn =
420
100
i n = ln 4=2.
So S (w) = 100h(ln 4=2)w = 100(4=2)w .
(b) S (3) = 100(4=2)3 = 7408=8 E 7409 bacteria
i S 0 (3) = n · S (3) = (ln 4=2) 100(4=2)3 [from part (a)] E 10,632 bacteria@h
(c) gS@gw = nS
(d) S (w) = 100(4=2)w = 10,000 i (4=2)w = 100 i w = (ln 100)@(ln 4=2) E 3=2 hours
4. (a) |(w) = |(0)hnw
i |(2) = |(0)h2n = 400 and |(6) = |(0)h6n = 25,600. Dividing these equations, we get
h6n @h2n = 25,600@400 i h4n = 64 i 4n = ln 26 = 6 ln 2 i n =
(b) 400 = |(0)h2n
3
2
ln 2 E 1=0397, about 104% per hour.
3
i |(0) = 400@h3 ln 2 = 400@ hln 2 = 400@23 = 50.
i |(0) = 400@h2n
(c) |(w) = |(0)hnw = 50h(3@2)(ln 2)w = 50(hln 2 )(3@2)w
i |(w) = 50(2)1=5w
(d) |(4=5) = 50(2)1=5(4=5) = 50(2)6=75 E 5382 bacteria
(e)
g|
= n| =
gw
3
ln 2 (50(2)6=75 ) E 5596 bacteria@h
2
(f ) |(w) = 50,000 i 50,000 = 50(2)1=5w
w=
i 1000 = (2)1.5w
i ln 1000 = 1.5w ln 2 i
ln 1000
E 6.64 h
1.5 ln 2
5. (a) Let the population (in millions) in the year w be S (w). Since the initial time is the year 1750, we substitute w 3 1750 for w in
Theorem 2, so the exponential model gives S (w) = S (1750)hn(w31750) . Then S (1800) = 980 = 790hn(180031750)
980
790
= hn(50)
i ln 980
= 50n
790
i n=
1
50
i
ln 980
E 0=0043104. So with this model, we have
790
S (1900) = 790hn(190031750) E 1508 million, and S (1950) = 790hn(195031750) E 1871 million. Both of these
estimates are much too low.
(b) In this case, the exponential model gives S (w) = S (1850)hn(w31850)
ln
1650
1260
= n(50) i n =
1
50
ln
1650
1260
i S (1900) = 1650 = 1260hn(190031850)
i
E 0=005393. So with this model, we estimate
S (1950) = 1260hn(195031850) E 2161 million. This is still too low, but closer than the estimate of S (1950) in part (a).
(c) The exponential model gives S (w) = S (1900)hn(w31900)
ln 2560
= n(50) i n =
1650
1
50
i S (1950) = 2560 = 1650hn(195031900)
i
ln 2560
E 0=008785. With this model, we estimate
1650
S (2000) = 1650hn(200031900) E 3972 million. This is much too low. The discrepancy is explained by the fact that the
world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate
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DIFFERENTIATION RULES
(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the ¿rst
part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will
remain constant.
6. (a) Let S (w) be the population (in millions) in the year w. Since the initial time is the year 1951, we substitute w 3 1951 for w in
Theorem 2, and ¿nd that the exponential model gives S (w) = S (1951)hn(w31951)
S (1961) = 92 = 76hn(196131951)
i n=
1
10
i
ln 439
E 0=0196. With this model, we estimate
361
S (2001) = 361hn(200131951) E 960 million. This estimate is slightly lower than the given value, 1029 million.
(b) Substituting w 3 1961 for w in Theorem 2, we ¿nd that the exponential model gives S (w) = S (1961)hn(w31961)
S (1981) = 653 = 439hn(198131961)
i n=
1
20
i
ln 653
E 0=0199. With this model, we estimate
439
S (2001) = 439hn(200131961) E 971 million, which is better than the estimate in part (a). The further estimates are
S (2010) = 439h49n E 1161 million and S (2020) = 439h49n E 1416 million.
Both models are reasonable; in fact, their graphs are nearly
(c)
indistinguishable.
7. (a) If | = [N2 O5 ] then by Theorem 2,
(b) |(w) = Fh30=0005w = 0=9F
g|
= 30=0005|
gw
i |(w) = |(0)h30=0005w = Fh30=0005w .
i h30=0005w = 0=9 i 30=0005w = ln 0=9 i w = 32000 ln 0=9 E 211 s
8. (a) The mass remaining after w days is |(w) = |(0) hnw = 50hnw . Since the half-life is 28 days, |(28) = 50h28n = 25
h28n =
1
2
i 28n = ln 12
i n = 3(ln 2)@28, so |(w) = 50h3(ln 2)w@28 = 50 · 23w@28 .
(b) |(40) = 50 · 2340@28 E 18=6 mg
(c) |(w) = 2 i 2 = 50 · 23w@28
1
(3w@28) ln 2 = ln 25
(d)
= 23w@28 i
1
@ ln 2 E 130 days
i w = 328 ln 25
i
2
50
9. (a) If |(w) is the mass (in mg) remaining after w years, then |(w) = |(0)hnw = 100hnw .
|(30) = 100h30n = 12 (100) i h30n =
1
2
i n = 3(ln 2)@30 i |(w) = 100h3(ln 2)w@30 = 100 · 23w@30
(b) |(100) = 100 · 23100@30 E 9=92 mg
1
(c) 100h3(ln 2)w@30 = 1 i 3(ln 2)w@30 = ln 100
i w = 330 lnln0=01
E 199=3 years
2
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EXPONENTIAL GROWTH AND DECAY
¤
243
10. (a) If |(w) is the mass after w days and |(0) = D, then |(w) = Dhnw .
|(1) = Dhn = 0=945D i hn = 0=945 i n = ln 0=945.
Then Dh(ln 0=945)w = 12 D C ln h(ln 0=945)w = ln 12
(b) Dh(ln 0=945)w = 0=20D C (ln 0=945)w = ln 15
C (ln 0=945)w = ln 12
2
C w = 3 ln ln
E 12=25 years.
0=945
5
C w = 3 ln ln
E 28=45 years
0=945
11. Let |(w) be the level of radioactivity. Thus, |(w) = |(0)h3nw and n is determined by using the half-life:
|(5730) = 12 |(0) i |(0)h3n(5730) = 12 |(0) i h35730n =
1
2
i 35730n = ln 12
If 74% of the 14 C remains, then we know that |(w) = 0=74|(0) i 0=74 = h3w(ln 2)@5730
w=3
i n=3
ln 12
ln 2
=
.
5730
5730
i ln 0=74 = 3
w ln 2
5730
i
5730(ln 0=74)
E 2489 E 2500 years.
ln 2
12. From the information given, we know that
5 = Fh2(0)
g|
= 2|
g{
i | = Fh2{ by Theorem 2. To calculate F we use the point (0> 5):
i F = 5. Thus, the equation of the curve is | = 5h2{ .
13. (a) Using Newton’s Law of Cooling,
gW
gW
= n(W 3 Wv ), we have
= n(W 3 75). Now let | = W 3 75, so
gw
gw
|(0) = W (0) 3 75 = 185 3 75 = 110, so | is a solution of the initial-value problem g|@gw = n| with |(0) = 110 and by
Theorem 2 we have |(w) = |(0)hnw = 110hnw .
|(30) = 110h30n = 150 3 75 i h30n =
45
75
110
=
15
22
i n=
1
30
1
15
w ln( 22 )
30
ln 15
and
22 , so |(w) = 110h
15
|(45) = 110h 30 ln( 22 ) E 62 F. Thus, W (45) E 62 + 75 = 137 F.
1
15
1
15
(b) W (w) = 100 i |(w) = 25. |(w) = 110h 30 w ln( 22 ) = 25 i h 30 w ln( 22 ) =
w=
25
110
i
1
w ln 15
30
22
25
= ln 110
i
25
30 ln 110
E 116 min.
ln 15
22
14. Let W (w) be the temperature of the body w hours after 1:30 PM . Then W (0) = 32=5 and W (1) = 30=3. Using Newton’s Law of
Cooling,
gW
gW
= n(W 3 Wv ), we have
= n(W 3 20). Now let | = W 3 20, so |(0) = W (0) 3 20 = 32=5 3 20 = 12=5,
gw
gw
so | is a solution to the initial value problem g|@gw = n| with |(0) = 12=5 and by Theorem 2 we have
|(w) = |(0)hnw = 12=5hnw .
|(1) = 30=3 3 20 i 10=3 = 12=5hn(1)
i hn =
|(w) = 37 3 20 i 12.5hnw = 17 i hnw =
17
12.5
10=3
12.5
i n = ln 10=3
. The murder occurred when
12.5
17 17
@ ln 10=3
i nw = ln 12.5
i w = ln 12.5
E 31=588 h
12.5
E 395 minutes. Thus, the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.
15.
gW
g|
= n(W 3 20). Letting | = W 3 20, we get
= n|, so |(w) = |(0)hnw . |(0) = W (0) 3 20 = 5 3 20 = 315, so
gw
gw
|(25) = |(0)h25n = 315h25n , and |(25) = W (25) 3 20 = 10 3 20 = 310, so 315h25n = 310 i h25n = 23 . Thus,
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¤
CHAPTER 3
DIFFERENTIATION RULES
25n = ln 23 and n =
hnw =
2 w@25
1
25
ln 23 , so |(w) = |(0)hnw = 315h(1@25) ln(2@3)w . More simply, h25n =
i |(w) = 315 ·
3
2 w@25
3
(a) W (50) = 20 + |(50) = 20 3 15 ·
(w@25) ln
16.
2
3
i hn =
.
2 50@25
3
(b) 15 = W (w) = 20 + |(w) = 20 3 15 ·
2
3
= 20 3 15 ·
2 w@25
3
i 15 ·
2 2
3
= 20 3
2 w@25
3
20
3
2 1@25
3
= 13=3̄ C
=5 i
= ln 13
i w = 25 ln 13 ln 23 E 67=74 min.
2 w@25
3
=
1
3
i
gW
g|
= n(W 3 20). Let | = W 3 20. Then
= n|, so |(w) = |(0)hnw = |(0) = W (0) 3 20 = 95 3 20 = 75,
gw
gw
so |(w) = 75hnw . When W (w) = 70,
gW
g|
= 31 C@min. Equivalently,
= 31 when |(w) = 50. Thus,
gw
gw
g|
= n|(w) = 50n and 50 = |(w) = 75hnw . The ¿rst relation implies n = 31@50, so the second relation says
gw
i w = 350 ln 23 E 20=27 min.
50 = 75h3w@50 . Thus, h3w@50 = 23 i 3w@50 = ln 23
31 =
17. (a) Let S (k) be the pressure at altitude k. Then gS@gk = nS
i S (k) = S (0)hnk = 101=3hnk .
i n=
S (1000) = 101=3h1000n = 87=14 i 1000n = ln 87=14
101=3
1
87=14
1
1000
87=14
S (k) = 101=3 h 1000 k ln( 101=3 ) , so S (3000) = 101=3h3 ln( 101=3 ) E 64=5 kPa.
6187
87=14 101=3
i
87=14
(b) S (6187) = 101=3 h 1000 ln( 101=3 ) E 39=9 kPa
u qw
18. (a) Using D = D0 1 +
with D0 = 1000, u = 0=08, and w = 3, we have:
q
1·3
(i) Annually: q = 1;
D = 1000 1 + 0=08
= $1259=71
1
0=08 4·3
(ii) Quarterly: q = 4;
D = 1000 1 + 4
= $1268=24
12·3
= $1270=24
(iii) Monthly: q = 12;
D = 1000 1 + 0=08
12
52·3
= $1271=01
(iv) Weekly: q = 52
D = 1000 1 + 0=08
52
365·3
(v) Daily: q = 365;
D = 1000 1 + 0=08
= $1271=22
365
365·24·3
0=08
(vi) Hourly: q = 365 · 24; D = 1000 1 + 365
= $1271=25
· 24
(vii) Continuously:
ln
(b)
D = 1000h(0=08)3 = $1271=25
D0=10 (3) = $1349=86,
D0=08 (3) = $1271=25, and
D0=06 (3) = $1197=22.
u qw
with D0 = 3000, u = 0=05, and w = 5, we have:
q
1·5
= $3828=84
Annually: q = 1;
D = 3000 1 + 0=05
1
0=05 2·5
= $3840=25
Semiannually: q = 2;
D = 3000 1 + 2
12·5
Monthly: q = 12;
D = 3000 1 + 0=05
= $3850=08
12
0=05 52·5
Weekly: q = 52;
D = 3000 1 + 52
= $3851=61
365·5
Daily: q = 365;
D = 3000 1 + 0=05
= $3852=01
365
19. (a) Using D = D0 1 +
(i)
(ii)
(iii)
(iv)
(v)
(vi) Continuously:
D = 3000h(0=05)5 = $3852=08
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i
NOT FOR SALE
SECTION 3.9
RELATED RATES
¤
245
(b) gD@gw = 0=05D and D(0) = 3000.
20. (a) D0 h0=06w = 2D0
C h0=06w = 2 C 0=06w = ln 2 C w =
50
3
ln 2 E 11=55, so the investment will
double in about 11=55 years.
(b) The annual interest rate in D = D0 (1 + u)w is u. From part (a), we have D = D0 h0=06w . These amounts must be equal,
so (1 + u)w = h0=06w
i 1 + u = h0=06
i u = h0=06 3 1 E 0=0618 = 6=18%, which is the equivalent annual
interest rate.
3.9 Related Rates
1. Y = {3
i
2. (a) D = u2
g{
gY g{
gY
=
= 3{2
gw
g{ gw
gw
i
gD gu
gu
gD
=
= 2u
gw
gu gw
gw
(b)
gD
gu
= 2u
= 2(30 m)(1 m@s) = 60 m2@s
gw
gw
3. Let v denote the side of a square. The square’s area D is given by D = v2 . Differentiating with respect to w gives us
gD
gv
gv
gD
= 2v . When D = 16, v = 4. Substitution 4 for v and 6 for
gives us
= 2(4)(6) = 48 cm2@s.
gw
gw
gw
gw
4. D = cz
i
gz
gc
gD
=c·
+z·
= 20(3) + 10(8) = 140 cm2@s.
gw
gw
gw
5. Y = u2 k = (5)2 k = 25k
6. Y =
3
4
3 u
7. (a) | =
(b) | =
i
i
gu
gY
= 43 · 3u2
gw
gw
I
g{
=3 i
2{ + 1 and
gw
I
2{ + 1
i
gY
gk
= 25
gw
gw
i
i 3 = 25
gk
gw
i
gk
3
=
m@min.
gw
25
2
gY
= 4 12 · 80 (4) = 25,600 mm3@s.
gw
g|
g| g{
1
3
g|
3
=
= (2{ + 1)31@2 · 2 · 3 = I
= I = 1.
. When { = 4,
gw
g{ gw
2
gw
2{ + 1
9
| 2 = 2{ + 1 i 2{ = |2 3 1 i { = 12 | 2 3
1
2
and
g|
=5 i
gw
I
g{
g{ g|
g{
=
= | · 5 = 5|. When { = 12, | = 25 = 5, so
= 5(5) = 25.
gw
g| gw
gw
8. (a)
g
g
g{
g|
g{
g|
(4{2 + 9| 2 ) =
(36) i 8{
+ 18|
= 0 i 4{
+ 9|
=0 i
gw
gw
gw
gw
gw
gw
I
2I
1
g{
1I
g{
g{
+9
=0 i 8
= 32 5 i
=3
5
5
4(2)
gw
3
3
gw
gw
4
(b) 4{
9.
I g|
g{
g|
2I
g|
+ 9|
= 0 i 4(32)(3) + 9
=0 i 6 5
= 24 i
5
gw
gw
3
gw
gw
g|
4
= I
gw
5
g 2
g{
g|
g}
g{
g|
g}
g{
g|
g
({ + | 2 + } 2 ) =
(9) i 2{
+ 2|
+ 2}
=0 i {
+|
+}
= 0. If
= 5,
= 4 and
gw
gw
gw
gw
gw
gw
gw
gw
gw
gw
({> |> }) = (2> 2> 1), then 2(5) + 2(4) + 1
g}
=0 i
gw
g}
= 318.
gw
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246
10.
¤
NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
g
g
g|
g{
g|
g{
({|) =
(8) i {
+|
= 0. If
= 33 cm@s and ({> |) = (4> 2), then 4(33) + 2
=0 i
gw
gw
gw
gw
gw
gw
g{
= 6. Thus, the {-coordinate is increasing at a rate of 6 cm@s.
gw
11. (a) Given: a plane Àying horizontally at an altitude of 1 mi and a speed of 500 mi@h passes directly over a radar station.
If we let w be time (in hours) and { be the horizontal distance traveled by the plane (in mi), then we are given
that g{@gw = 500 mi@h.
(b) Unknown: the rate at which the distance from the plane to the station is increasing
(c)
when it is 2 mi from the station. If we let | be the distance from the plane to the station,
then we want to ¿nd g|@gw when | = 2 mi.
(d) By the Pythagorean Theorem, | 2 = {2 + 1 i 2| (g|@gw) = 2{ (g{@gw).
(e)
I
I
I
3
{ g{
{
g|
g|
=
= (500). Since | 2 = {2 + 1, when | = 2, { = 3, so
=
(500) = 250 3 E 433 mi@h.
gw
| gw
|
gw
2
12. (a) Given: the rate of decrease of the surface area is 1 cm2@min. If we let w be
(b)
2
time (in minutes) and V be the surface area (in cm ), then we are given that
gV@gw = 31 cm2@s.
(c) Unknown: the rate of decrease of the diameter when the diameter is 10 cm.
If we let { be the diameter, then we want to ¿nd g{@gw when { = 10 cm.
(d) If the radius is u and the diameter { = 2u, then u = 12 { and
V = 4u2 = 4
(e) 31 =
1 2
{ = {2
2
g{
gV
= 2{
gw
gw
i
i
gV g{
g{
gV
=
= 2{
.
gw
g{ gw
gw
g{
1
g{
1
1
=3
. When { = 10,
=3
. So the rate of decrease is
cm@min.
gw
2{
gw
20
20
13. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15-ft-tall pole at a rate of 5 ft@s. If we let w be time (in s)
and { be the distance from the pole to the man (in ft), then we are given that g{@gw = 5 ft@s.
(b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft
(c)
from the pole. If we let | be the distance from the man to the tip of his
shadow (in ft), then we want to ¿nd
(d) By similar triangles,
{+|
15
=
6
|
g
({ + |) when { = 40 ft.
gw
i 15| = 6{ + 6|
(e) The tip of the shadow moves at a rate of
i 9| = 6{ i | = 23 {.
g
g
2
5 g{
({ + |) =
{+ { =
= 53 (5) =
gw
gw
3
3 gw
25
3
ft@s.
14. (a) Given: at noon, ship A is 150 km west of ship B; ship A is sailing east at 35 km@h, and ship B is sailing north at 25 km@h.
If we let w be time (in hours), { be the distance traveled by ship A (in km), and | be the distance traveled by ship B (in km),
then we are given that g{@gw = 35 km@h and g|@gw = 25 km@h.
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°
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NOT FOR SALE
SECTION 3.9
(b) Unknown: the rate at which the distance between the ships is changing at
RELATED RATES
¤
(c)
4:00 PM. If we let } be the distance between the ships, then we want to ¿nd
g}@gw when w = 4 h.
g{
g|
g}
= 2(150 3 {) 3
+ 2|
gw
gw
gw
s
I
(e) At 4:00 PM, { = 4(35) = 140 and | = 4(25) = 100 i } = (150 3 140)2 + 1002 = 10,100.
215
g}
1
g{
g|
310(35) + 100(25)
I
= I
So
=
({ 3 150)
+|
=
E 21=4 km@h.
gw
}
gw
gw
10,100
101
(d) } 2 = (150 3 {)2 + | 2
15.
i 2}
We are given that
g|
g{
= 60 mi@h and
= 25 mi@h. } 2 = {2 + | 2
gw
gw
i
g}
1
g{
g|
=
{
+|
.
gw
}
gw
gw
I
After 2 hours, { = 2 (60) = 120 and | = 2 (25) = 50 i } = 1202 + 502 = 130,
1
g{
g|
120(60) + 50(25)
g}
=
{
+|
=
= 65 mi@h.
so
gw
}
gw
gw
130
2}
16.
g}
g{
g|
= 2{
+ 2|
gw
gw
gw
We are given that
i }
g}
g{
g|
={
+|
gw
gw
gw
i
|
2
g{
= 1=6 m@s. By similar triangles,
=
gw
12
{
i |=
24
{
i
g|
24 g{
24
24(1=6)
g|
=3 2
= 3 2 (1=6). When { = 8,
=3
= 30=6 m@s, so the shadow
gw
{ gw
{
gw
64
is decreasing at a rate of 0=6 m@s.
g{
g|
= 4 ft@s and
= 5 ft@s. } 2 = ({ + |)2 + 5002 i
gw
gw
g}
g{ g|
2}
= 2({ + |)
+
. 15 minutes after the woman starts, we have
gw
gw
gw
17.
We are given that
{ = (4 ft@s)(20 min)(60 s@min) = 4800 ft and | = 5 · 15 · 60 = 4500 i
s
I
} = (4800 + 4500)2 + 5002 = 86,740,000, so
837
g}
{ + | g{ g|
4800 + 4500
(4 + 5) = I
=
+
= I
E 8=99 ft@s=
gw
}
gw
gw
86,740,000
8674
18. We are given that
g{
= 24 ft@s.
gw
g|
g{
= 2(90 3 {) 3
. When { = 45,
gw
gw
I
I
90 3 {
g{
45
g|
24
I (324) = 3 I ,
=
3
=
| = 452 + 902 = 45 5, so
gw
|
gw
45 5
5
| 2 = (90 3 {)2 + 902
(a)
i 2|
so the distance from second base is decreasing at a rate of
24
I
5
E 10=7 ft@s.
(b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer — and we do.
} 2 = {2 + 902
i 2}
I
g{
45
g}
g}
24
I (24) = I E 10=7 ft@s.
= 2{
. When { = 45, } = 45 5, so
=
gw
gw
gw
45 5
5
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°
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247
248
¤
19. D =
NOT FOR SALE
CHAPTER 3
1
ek,
2
DIFFERENTIATION RULES
where e is the base and k is the altitude. We are given that
gD
gk
= 1 cm@min and
= 2 cm2 @min. Using the
gw
gw
1
gk
ge
gD
=
e
+k
. When k = 10 and D = 100, we have 100 = 12 e(10) i
Product Rule, we have
gw
2
gw
gw
1
ge
ge
ge
4 3 20
e = 20, so 2 =
20 · 1 + 10
i 4 = 20 + 10
i
=
= 31=6 cm@min.
2
gw
gw
gw
10
1
e
2
= 10 i
20.
g{
g{
g|
g|
= 31 m@s, ¿nd
when { = 8 m. | 2 = {2 + 1 i 2|
= 2{
i
gw
gw
gw
gw
I
I
65
g{
| g|
|
g{
=
= 3 . When { = 8, | = 65, so
=3
. Thus, the boat approaches
gw
{ gw
{
gw
8
I
65
the dock at
E 1=01 m@s.
8
21.
g{
g|
= 35 km@h and
= 25 km@h. } 2 = ({ + |)2 + 1002 i
gw
gw
g}
g{ g|
2}
= 2({ + |)
+
. At 4:00 PM, { = 4(35) = 140 and | = 4(25) = 100 i
gw
gw
gw
s
I
} = (140 + 100)2 + 1002 = 67,600 = 260, so
{ + | g{
g|
140 + 100
720
g}
=
+
=
(35 + 25) =
E 55=4 km@h.
gw
}
gw
gw
260
13
Given
We are given that
s
{2 + | 2 , so } 2 = {2 + [2 sin({@2)]2 i
g{
g}
g{
i }
={
+ 2 sin { cos {
. When
gw
gw
2
2
gw
22. The distance } of the particle to the origin is given by } =
g{
g}
g{
= 2{
+ 4 · 2 sin { cos { ·
gw
gw
2
2
2 gw
v u
2
1
1
10
1I
1I
I
1 I g}
>1 , } =
=
=
cos · 10 i
+ 12 =
10, so
10
10 + 2 sin
({> |) =
3
3
9
3
3
gw
3
6
6
I
1 g}
1
1
1I
3 3
g}
= + 2
=1+
cm@s.
3
i
3 gw
3
2
2
gw
2
2}
23. If F = the rate at which water is pumped in, then
gY
= F 3 10,000, where
gw
Y = 13 u2 k is the volume at time w. By similar triangles,
Y = 13 1 2
k k=
3
3
k
27
i
k
u
=
2
6
i u=
1
k i
3
gY
gk
= k2
. When k = 200 cm,
gw
9
gw
gk
800,000
= 20 cm@min, so F 3 10,000 = (200)2 (20) i F = 10,000 +
E 289,253 cm3@min.
gw
9
9
24. By similar triangles,
e
3
= , so e = 3k. The trough has volume
1
k
Y = 12 ek(10) = 5(3k)k = 15k2
When k = 12 ,
i 12 =
gk
gY
= 30k
gw
gw
i
gk
2
=
.
gw
5k
gk
4
2
= ft@min.
=
gw
5
5 · 12
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.9
RELATED RATES
The ¿gure is labeled in meters. The area D of a trapezoid is
25.
1
(base1
2
+ base2 )(height), and the volume Y of the 10-meter-long trough is 10D.
Thus, the volume of the trapezoid with height k is Y = (10) 12 [0=3 + (0=3 + 2d)]k.
By similar triangles,
Now
d
0=25
1
=
= , so 2d = k i Y = 5(0=6 + k)k = 3k + 5k2 .
k
0=5
2
gY
gY gk
=
gw
gk gw
i
0=2 = (3 + 10k)
gk
gw
i
gk
0=2
=
. When k = 0=3,
gw
3 + 10k
0=2
0=2
1
10
gk
=
=
m@min =
m@min or
cm@min.
gw
3 + 10(0=3)
6
30
3
26. The ¿gure is drawn without the top 3 feet.
Y = 12 (e + 12)k(20) = 10(e + 12)k and, from similar triangles,
{
6
|
16
8
8k
11k
= and =
= , so e = { + 12 + | = k + 12 +
= 12 +
.
k
6
k
6
3
3
3
110k2
gY
220
gk
11k
k = 240k +
and so 0=8 =
= 240 +
k
.
Thus, Y = 10 24 +
3
3
gw
3
gw
When k = 5,
0=8
3
gk
=
=
E 0=00132 ft@min.
gw
240 + 5(220@3)
2275
27. We are given that
2
k
gY
1
1
k3
= 30 ft3 @min. Y = u2 k = k=
gw
3
3
2
12
k2 gk
4 gw
gY
gY gk
=
gw
gk gw
i 30 =
When k = 10 ft,
120
6
gk
= 2 =
E 0=38 ft@min.
gw
10 5
28. We are given g{@gw = 8 ft@s. cot =
g{
g
= 3100 csc2 gw
gw
i
i
{
100
gk
120
=
.
gw
k2
i { = 100 cot i
g
sin2 100
1
=3
· 8. When | = 200, sin =
=
gw
100
200
2
(1@2)2
g
1
=3
·8=3
rad@s. The angle is decreasing at a rate of
gw
100
50
29. D =
1
2 ek,
i
but e = 5 m and sin =
k
4
1
50
i
rad@s.
i k = 4 sin , so D = 12 (5)(4 sin ) = 10 sin .
gD
gD g
g
= 0=06 rad@s, so
=
= (10 cos )(0=06) = 0=6 cos .
gw
gw
g gw
gD
= 0=6 cos
= (0=6) 12 = 0=3 m2@s.
When = ,
3 gw
3
We are given
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249
250
¤
NOT FOR SALE
CHAPTER 3
30. cos =
{
10
DIFFERENTIATION RULES
i 3 sin g
1 g{
g{
=
. From Example 2,
= 1 and
gw
10 gw
gw
when { = 6, | = 8, so sin =
Thus, 3
1
8 g
=
(1) i
10 gw
10
8
.
10
g
1
= 3 rad@s.
gw
8
31. From the ¿gure and given information, we have {2 + | 2 = O2 ,
g|
= 30=15 m@ s, and
gw
g{
= 0=2 m@ s when { = 3 m. Differentiating implicitly with respect to w, we get
gw
{2 + | 2 = O2
i 2{
g{
g|
g|
g{
+ 2|
=0 i |
= 3{ . Substituting the given
gw
gw
gw
gw
information gives us |(30=15) = 33(0=2) i | = 4 m. Thus, 32 + 42 = O2
i
2
O = 25 i O = 5 m.
32. The volume of a hemisphere is 23 u3 , so the volume of a hemispherical basin of radius 30 cm is 23 (30)3 = 18,000 cm3 .
i 9000 = 30k2 3 13 k3
i
If the basin is half full, then Y = uk2 3 13 k3
1 3
3k
3 30k2 + 9000 = 0 i
k = K E 19=58 [from a graph or numerical root¿nder; the other two solutions are less than 0 and greater than 30].
gk
gk
gY
gk
L
cm3
i
= 60k
3 k2
i
2
1000
= (60k 3 k2 )
i
Y = 30k2 3 13 k3
gw
gw
gw
min
L
gw
2000
gk
=
E 0=804 cm@min.
gw
(60K 3 K 2 )
33. Differentiating both sides of S Y = F with respect to w and using the Product Rule gives us S
gS
gY
+Y
=0 i
gw
gw
gY
Y gS
gS
gY
600
=3
. When Y = 600, S = 150 and
= 20, so we have
=3
(20) = 380. Thus, the volume is
gw
S gw
gw
gw
150
decreasing at a rate of 80 cm3@min.
34. S Y 1=4 = F
i S · 1=4Y 0=4
When Y = 400, S = 80 and
rate of
250
7
gY
gS
+ Y 1=4
=0 i
gw
gw
gS
gY
Y 1=4
Y gS
=3
=3
.
gw
S · 1=4Y 0=4 gw
1=4S gw
gS
gY
400
250
= 310, so we have
=3
(310) =
. Thus, the volume is increasing at a
gw
gw
1=4(80)
7
E 36 cm3@min.
1
1
180
9
400
1
1
1
1
1
1
=
+
=
=
, so U =
. Differentiating
=
+
=
+
U
U1
U2
80 100
8000
400
9
U
U1
U2
1 gU
1 gU1
1 gU1
1 gU2
gU
1 gU2
2
with respect to w, we have 3 2
=3 2
3 2
i
=U
+ 2
. When U1 = 80 and
U gw
U1 gw
U2 gw
gw
U12 gw
U2 gw
4002 1
gU
1
107
= 2
E 0=132 l@s.
(0=3) +
(0=2) =
U2 = 100,
gw
9
802
1002
810
35. With U1 = 80 and U2 = 100,
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.9
36. We want to ¿nd
RELATED RATES
¤
gE
when O = 18 using E = 0=007Z 2@3 and Z = 0=12O2=53 .
gw
gE
gE gZ gO 20 3 15
=
= 0=007 · 23 Z 31@3 (0=12 · 2=53 · O1=53 )
gw
gZ gO gw
10,000,000
k
l
5
= 0=007 · 23 (0=12 · 18 2=53 )31@3 0=12 · 2=53 · 181=53
E 1=045 × 1038 g@yr
107
37. We are given g@gw = 2@min =
90
rad@min. By the Law of Cosines,
{2 = 122 + 152 3 2(12)(15) cos = 369 3 360 cos 2{
g{
g
= 360 sin gw
gw
i
i
g{
180 sin g
=
. When = 60 ,
gw
{
gw
I
I
I
I
I
7
180 sin 60 3
g{
I
=
= I =
E 0=396 m@min.
{ = 369 3 360 cos 60 = 189 = 3 21, so
gw
21
3 21 90
3 21
g{
g|
= 32 ft@s and need to ¿nd
when { = 35.
gw
gw
s
I
Using the Pythagorean Theorem twice, we have {2 + 122 + | 2 + 122 = 39,
38. Using T for the origin, we are given
the total length of the rope. Differentiating with respect to w, we get
s
{ | 2 + 122 g{
g{
g|
{
|
g|
I
+s
= 0, so
=3 I
.
gw
{2 + 122 gw
| {2 + 122 gw
| 2 + 122 gw
s
s
s
Now when { = 35, 39 = (35)2 + 122 + | 2 + 122 = 13 + | 2 + 122
|=
C
s
| 2 + 122 = 26, and
I
I
(35)(26)
g|
10
= 3I
262 3 122 = 532. So when { = 35,
(32) = 3 I
E 30=87 ft@s.
gw
532 (13)
133
So cart E is moving towards T at about 0=87 ft@s.
39. (a) By the Pythagorean Theorem, 40002 + | 2 = c2 . Differentiating with respect to w,
gc
g|
g|
= 2c . We know that
= 600 ft@s, so when | = 3000 ft,
gw
gw
gw
I
I
c = 40002 + 30002 = 25,000,000 = 5000 ft
we obtain 2|
and
gc
| g|
3000
1800
=
=
(600) =
= 360 ft@s.
gw
c gw
5000
5
|
4000
(b) Here tan =
| = 3000 ft,
i
g
g | (tan ) =
gw
gw 4000
i
sec2 i
g
cos2 g|
=
. When
gw
4000 gw
(4@5)2
4000
4000
4
g
g|
= 600 ft@s, c = 5000 and cos =
=
= , so
=
(600) = 0=096 rad@s.
gw
c
5000
5
gw
4000
40. We are given that
g
= 4(2) = 8 rad@min. { = 3 tan gw
i
2
g{
g
= 3 sec2 . When { = 1, tan = 13 , so sec2 = 1 + 13 =
gw
gw
and
g
1 g|
=
gw
4000 gw
g{
(8) =
= 3 10
9
gw
80
3 10
9
E 83=8 km@min.
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CHAPTER 3
DIFFERENTIATION RULES
1 g{
2 1 g{
{
g
3
i 3 csc2 =
i 3 csc
=
5
gw
5 gw
3
6
5 gw
2
g{
5 2
10
I
=
km@min [E 130 mi@h]
=
gw
6
9
3
41. cot =
42. We are given that
i
g
2 rad
=
= rad@min= By the Pythagorean Theorem, when
gw
2 min
k = 6, { = 8, so sin =
6
10
and cos =
8
.
10
From the ¿gure, sin =
g
8
gk
= 10 cos = 10
= 8 m@min.
k = 10 sin , so
gw
gw
10
k
10
i
g{
= 300 km@h. By the Law of Cosines,
gw
| 2 = {2 + 12 3 2(1)({) cos 120 = {2 + 1 3 2{ 3 12 = {2 + { + 1, so
43. We are given that
2|
g|
g{ g{
= 2{
+
gw
gw
gw
|=
i
g|
2{ + 1 g{
=
. After 1 minute, { =
gw
2| gw
I
I
52 + 5 + 1 = 31 km i
300
60
= 5 km i
2(5) + 1
g|
1650
I
(300) = I
E 296 km@h.
=
gw
2 31
31
g{
g|
= 3 mi@h and
= 2 mi@h. By the Law of Cosines,
gw
gw
I
} 2 = {2 + | 2 3 2{| cos 45 = {2 + | 2 3 2 {| i
44. We are given that
2}
g| I
g{
g}
g{
g| I
= 2{
+ 2|
3 2{
3 2|
. After 15 minutes =
gw
gw
gw
gw
gw
1
4
h ,
s
I
13 3 6 2
and
we have { = and | = =
i } = 4 + 4
4
I
I 3
I 1 I
1
3
2
2
13 3 6 2 s
g}
= s
= 13 3 6 2 E 2=125 mi@h.
I 2 4 3+2 2 23 2 4 23 2 2 3 = s
I
gw
2
13 3 6 2
13 3 6 2
3
4
2
4
2
1
2
3 2
2 2
I i }=
3 2 34 24
45. Let the distance between the runner and the friend be c. Then by the Law of Cosines,
c 2 = 2002 + 1002 3 2 · 200 · 100 · cos = 50,000 3 40,000 cos (B). Differentiating
implicitly with respect to w, we obtain 2c
g
gc
= 340,000(3 sin ) . Now if G is the
gw
gw
distance run when the angle is radians, then by the formula for the length of an arc
on a circle, v = u, we have G = 100, so =
1
G
100
i
g
1 gG
7
=
=
. To substitute into the expression for
gw
100 gw
100
gc
, we must know sin at the time when c = 200, which we ¿nd from (B): 2002 = 50,000 3 40,000 cos gw
C
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS
1
4
cos =
gc@gw =
7
t
2
i sin = 1 3 14 =
I
15
4
I
15
.
4
Substituting, we get 2(200)
¤
I gc
7
i
= 40,000 415 100
gw
E 6=78 m@s. Whether the distance between them is increasing or decreasing depends on the direction in which
the runner is running.
46. The hour hand of a clock goes around once every 12 hours or, in radians per hour,
2
12
=
6
rad@h. The minute hand goes around once an hour, or at the rate of 2 rad@h.
So the angle between them (measuring clockwise from the minute hand to the hour
hand) is changing at the rate of g@gw =
6
3 2 = 3 11
rad@h. Now, to relate to c,
6
we use the Law of Cosines: c2 = 42 + 82 3 2 · 4 · 8 · cos = 80 3 64 cos (B).
gc
g
= 364(3 sin ) . At 1:00, the angle between the two hands is
gw
gw
s
I
s
80 3 64 cos 6 = 80 3 32 3.
one-twelfth of the circle, that is, 2
12 = 6 radians. We use (B) to ¿nd c at 1:00: c =
64 12 3 11
gc
11
gc
88
6
Substituting, we get 2c
= 64 sin
3
i
= s
I =3 s
I E 318=6.
gw
6
6
gw
2 80 3 32 3
3 80 3 32 3
Differentiating implicitly with respect to w, we get 2c
So at 1:00, the distance between the tips of the hands is decreasing at a rate of 18=6 mm@h E 0=005 mm@s.
3.10 Linear Approximations and Differentials
1. i ({) = {4 + 3{2
i i 0 ({) = 4{3 + 6{, so i (31) = 4 and i 0 (31) = 310.
Thus, O({) = i (31) + i 0 (31)({ 3 (31)) = 4 + (310)({ + 1) = 310{ 3 6.
2. i ({) = sin {
O({) = i
3. i ({) =
6
i i 0 ({) = cos {, so i
+ i 0 6 { 3 6 =
1
2
+
1
2
6
=
1
2
and i 0
6
=
1
2
I I
3 { 3 6 = 12 3 { +
I
3. Thus,
1
2
3
1
12
I
3 .
I
I
{ i i 0 ({) = 12 {31@2 = 1@(2 { ), so i (4) = 2 and i 0 (4) = 14 . Thus,
O({) = i (4) + i 0 (4)({ 3 4) = 2 + 14 ({ 3 4) = 2 + 14 { 3 1 = 14 { + 1.
4. i ({) = {3@4
i i 0 ({) = 34 {31@4 , so i(16) = 8 and i 0 (16) = 38 .
Thus, O({) = i (16) + i 0 (16)({ 3 16) = 8 + 38 ({ 3 16) = 38 { + 2.
5. i ({) =
253
I
1 3 { i i 0 ({) =
2
31
I
, so i (0) = 1 and i 0 (0) = 3 12 .
13{
Therefore,
I
1 3 { = i ({) E i(0) + i 0 (0)({ 3 0) = 1 + 3 12 ({ 3 0) = 1 3 12 {.
I
I
So 0=9 = 1 3 0=1 E 1 3 12 (0=1) = 0=95
I
I
and 0=99 = 1 3 0=01 E 1 3 12 (0=01) = 0=995.
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¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
I
3
1 + { = (1 + {)1@3 i j 0 ({) = 13 (1 + {)32@3 , so j(0) = 1 and
I
j 0 (0) = 13 . Therefore, 3 1 + { = j({) E j(0) + j 0 (0)({ 3 0) = 1 + 13 {.
s
I
So 3 0=95 = 3 1 + (30=05) E 1 + 13 (30=05) = 0=983,
I
I
and 3 1=1 = 3 1 + 0=1 E 1 + 13 (0=1) = 1=03.
6. j({) =
i i 0 ({) =
7. i ({) = ln(1 + {)
1
, so i (0) = 0 and i 0 (0) = 1.
1+{
Thus, i({) E i (0) + i 0 (0)({ 3 0) = 0 + 1({) = {. We need
ln(1 + {) 3 0=1 ? { ? ln(1 + {) + 0=1, which is true when
30=383 ? { ? 0=516.
8. i ({) = (1 + {)33
i i 0 ({) = 33(1 + {)34 , so i (0) = 1 and
i 0 (0) = 33. Thus, i ({) E i (0) + i 0 (0)({ 3 0) = 1 3 3{. We need
(1 + {)33 3 0=1 ? 1 3 3{ ? (1 + {)33 + 0=1, which is true when
30=116 ? { ? 0=144.
9. i ({) =
I
4
1 + 2{ i i 0 ({) = 14 (1 + 2{)33@4 (2) = 12 (1 + 2{)33@4 , so
i (0) = 1 and i 0 (0) = 12 . Thus, i ({) E i (0) + i 0 (0)({ 3 0) = 1 + 12 {.
I
I
We need 4 1 + 2{ 3 0=1 ? 1 + 12 { ? 4 1 + 2{ + 0=1, which is true when
30=368 ? { ? 0=677.
10. i ({) = h{ cos {
i i 0 ({) = h{ (3 sin {) + (cos {)h{ = h{ (cos { 3 sin {),
so i (0) = 1 and i 0 (0) = 1. Thus, i ({) E i (0) + i 0 (0)({ 3 0) = 1 + {.
We need h{ cos { 3 0=1 ? 1 + { ? h{ cos { + 0=1, which is true when
30=762 ? { ? 0=607.
11. (a) The differential g| is de¿ned in terms of g{ by the equation g| = i 0 ({) g{. For | = i ({) = {2 sin 2{,
i 0 ({) = {2 cos 2{ · 2 + sin 2{ · 2{ = 2{({ cos 2{ + sin 2{), so g| = 2{({ cos 2{ + sin 2{) g{.
(b) For | = i (w) = ln
12. (a) For | = i (v) =
I
1 + w2 =
1
2
ln(1 + w2 ), i 0 (w) =
1
w
w
1
· 2w =
, so g| =
gw.
·
2 1 + w2
1 + w2
1 + w2
v
(1 + 2v)(1) 3 v(2)
1
1
, i 0 (v) =
=
, so g| =
gv.
1 + 2v
(1 + 2v)2
(1 + 2v)2
(1 + 2v)2
(b) For | = i (x) = h3x cos x, i 0 (x) = h3x (3 sin x) + cos x(3h3x ) = 3h3x (sin x + cos x), so
g| = 3h3x (sin x + cos x) gx.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS
13. (a) For | = i (w) = tan
I 1
I 0
w, i (w) = sec2 w · w31@2
2
I
I
sec2 w
sec2 w
I , so g| =
I gw.
=
2 w
2 w
1 3 y2
,
1 + y2
(1 + y 2 )(32y) 3 (1 3 y2 )(2y)
32y[(1 + y2 ) + (1 3 y 2 )]
32y(2)
34y
=
=
=
,
i 0 (y) =
2
2
(1 + y )
(1 + y 2 )2
(1 + y 2 )2
(1 + y 2 )2
34y
gy.
so g| =
(1 + y 2 )2
(b) For | = i (y) =
14. (a) For | = i (w) = htan w , i 0 (w) = htan w · sec2 (w) · , so g| = sec2 (w)htan w gw.
(b) For | = i (}) =
15. (a) | = h{@10
I
1
1
I
1 + ln }, i 0 (}) = 12 (1 + ln })31@2 · , so g| =
g}.
}
2} 1 + ln }
i g| = h{@10 ·
1
10
(b) { = 0 and g{ = 0=1 i g| =
16. (a) | = cos {
(b) { =
17. (a) | =
1
3
g{ =
1 {@10
h
g{
10
1 0@10
h
(0=1)
10
= 0=01.
i g| = 3sin { · g{ = 3 sin { g{
and g{ = 30=02 i g| = 3 sin 3 (30=02) = I
3 + {2
i g| =
I
I
3@2 (0=02) = 0=01 3 E 0=054.
1
{
g{
(3 + {2 )31@2 (2{) g{ = I
2
3 + {2
1
1
(b) { = 1 and g{ = 30=1 i g| = I
(30=1) = (30=1) = 30=05.
2
3 + 12
18. (a) | =
{+1
{31
i g| =
({ 3 1)(1) 3 ({ + 1)(1)
32
g{ =
g{
({ 3 1)2
({ 3 1)2
(b) { = 2 and g{ = 0=05 i g| =
32
(0=05) = 32(0=05) = 30=1.
(2 3 1)2
19. | = i ({) = 2{ 3 {2 , { = 2, {{ = 30=4
i
{| = i (1=6) 3 i (2) = 0=64 3 0 = 0=64
g| = (2 3 2{) g{ = (2 3 4)(30=4) = 0=8
I
{, { = 1, {{ = 1 i
I
I
I
{| = i (2) 3 i(1) = 2 3 1 = 2 3 1 E 0=414
20. | = i ({) =
g| =
2
1
I g{ = 12 (1) = 0=5
{
21. | = i ({) = 2@{, { = 4, {{ = 1
{| = i (5) 3 i(4) =
g| = 3
2
5
3
2
4
i
= 30=1
2
2
g{ = 3 2 (1) = 30=125
{2
4
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255
256
NOT FOR SALE
¤
CHAPTER 3 DIFFERENTIATION RULES
22. | = i ({) = h{ , { = 0, {{ = 0=5
i
I
{| = i (0=5) 3 i (0) = h 3 1 [E 0=65]
g| = h{ g{ = h0 (0=5) = 0=5
23. To estimate (1=999)4 , we’ll ¿nd the linearization of i ({) = {4 at d = 2. Since i 0 ({) = 4{3 , i (2) = 16, and
i 0 (2) = 32, we have O({) = 16 + 32({ 3 2). Thus, {4 E 16 + 32({ 3 2) when { is near 2, so
(1=999)4 E 16 + 32(1=999 3 2) = 16 3 0=032 = 15=968.
24. To estimate h30=015 , we’ll ¿nd the linearization of i ({) = h{ at d = 0. Since i 0 ({) = h{ , i(0) = 1, and i 0 (0) = 1, we have
O({) = 1 + 1({ 3 0) = { + 1. Thus, h{ E { + 1 when { is near 0, so h30=015 E 30=015 + 1 = 0=985.
I
3
{ i g| = 13 {32@3 g{. When { = 1000 and g{ = 1, g| = 13 (1000)32@3 (1) =
25. | = i ({) =
I
3
1001 = i (1001) E i (1000) + g| = 10 +
26. | = i ({) = 1@{
1
4=002
so
= 10=003 E 10=003.
1
1
i g| = 31@{2 g{. When { = 4 and g{ = 0=002, g| = 3 16
(0=002) = 3 8000
, so
E i (4) + g| =
1
4
3
1
8000
=
1999
8000
= 0=249875.
i g| = sec2 { g{. When { = 45 and g{ = 31 ,
27. | = i ({) = tan {
g| = sec2 45 (3@180) =
28. | = i ({) =
1
300
1
,
300
I 2
2 (3@180) = 3@90, so tan 44 = i (44 ) E i (45 ) + g| = 1 3 @90 E 0=965.
I
{ i g| =
1
1
I g{. When { = 100 and g{ = 30=2, g| = I
(30=2) = 30=01, so
2
100
2 {
I
99=8 = i (99=8) E i (100) + g| = 10 3 0=01 = 9=99.
29. | = i ({) = sec {
i i 0 ({) = sec { tan {, so i (0) = 1 and i 0 (0) = 1 · 0 = 0. The linear approximation of i at 0 is
i (0) + i 0 (0)({ 3 0) = 1 + 0({) = 1. Since 0=08 is close to 0, approximating sec 0=08 with 1 is reasonable.
30. If | = {6 , | 0 = 6{5 and the tangent line approximation at (1> 1) has slope 6. If the change in { is 0=01, the change in | on the
tangent line is 0=06, and approximating (1=01)6 with 1=06 is reasonable.
31. | = i ({) = ln {
i i 0 ({) = 1@{, so i (1) = 0 and i 0 (1) = 1. The linear approximation of i at 1 is
i (1) + i 0 (1)({ 3 1) = 0 + 1({ 3 1) = { 3 1. Now i (1=05) = ln 1=05 E 1=05 3 1 = 0=05, so the approximation
is reasonable.
32. (a) i ({) = ({ 3 1)2
i i 0 ({) = 2({ 3 1), so i (0) = 1 and i 0 (0) = 32.
Thus, i ({) E Oi ({) = i (0) + i 0 (0)({ 3 0) = 1 3 2{.
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NOT FOR SALE
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS
j({) = h32{
¤
257
i j0 ({) = 32h32{ , so j(0) = 1 and j 0 (0) = 32.
Thus, j({) E Oj ({) = j(0) + j0 (0)({ 3 0) = 1 3 2{.
k({) = 1 + ln(1 3 2{) i k0 ({) =
32
, so k(0) = 1 and k0 (0) = 32.
1 3 2{
Thus, k({) E Ok ({) = k(0) + k0 (0)({ 3 0) = 1 3 2{.
Notice that Oi = Oj = Ok . This happens because i , j, and k have the same function values and the same derivative
values at d = 0.
(b) The linear approximation appears to be the best for the function i since it is
closer to i for a larger domain than it is to j and k. The approximation
looks worst for k since k moves away from O faster than i and j do.
33. (a) If { is the edge length, then Y = {3
i gY = 3{2 g{. When { = 30 and g{ = 0=1, gY = 3(30)2 (0=1) = 270, so the
maximum possible error in computing the volume of the cube is about 270 cm3 . The relative error is calculated by dividing
the change in Y , {Y , by Y . We approximate {Y with gY .
{Y
gY
3{2 g{
0=1
g{
Relative error =
E
=
=
3
= 0=01.
=
3
Y
Y
{3
{
30
Percentage error = relative error × 100% = 0=01 × 100% = 1%.
(b) V = 6{2
i gV = 12{ g{. When { = 30 and g{ = 0=1, gV = 12(30)(0=1) = 36, so the maximum possible error in
computing the surface area of the cube is about 36 cm2 .
{V
gV
12{ g{
0=1
g{
Relative error =
E
=
=
2
= 0=006.
=
2
V
V
6{2
{
30
Percentage error = relative error × 100% = 0=006 × 100% = 0=6%.
34. (a) D = u2
i gD = 2u gu. When u = 24 and gu = 0=2, gD = 2(24)(0=2) = 9=6, so the maximum possible error
in the calculated area of the disk is about 9=6 E 30 cm2 .
(b) Relative error =
{D
gD
2u gu
2(0=2)
0=2
1
2 gu
E
=
=
=
=
= 0=016.
=
D
D
u2
u
24
12
60
Percentage error = relative error ×100% = 0=016 × 100% = 1=6%.
35. (a) For a sphere of radius u, the circumference is F = 2u and the surface area is V = 4u2 , so
u=
F
2
2
F
F2
i V = 4
=
2
so the maximum error is about
i gV =
2
2
84
F gF. When F = 84 and gF = 0=5, gV = (84)(0=5) =
,
84@
1
gV
84
E 27 cm2 . Relative error E
= 2
=
E 0=012 = 1=2%
V
84 @
84
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258
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
(b) Y =
3
4 3
4
F3
F
=
u = 3
3
2
62
gY =
i gY =
1 2
F gF. When F = 84 and gF = 0=5,
22
1764
1764
1
(84)2 (0=5) =
, so the maximum error is about 2 E 179 cm3 .
22
2
The relative error is approximately
36. For a hemispherical dome, Y =
gY = 2(25)2 (0=0005) =
37. (a) Y = u2 k
5
,
8
1764@2
gY
1
=
=
E 0=018 = 1=8%.
Y
(84)3@(62 )
56
3
2
3 u
i gY = 2u2 gu. When u = 12 (50) = 25 m and gu = 0=05 cm = 0=0005 m,
so the amount of paint needed is about
5
8
E 2 m3 .
i {Y E gY = 2uk gu = 2uk {u
(b) The error is
{Y 3 gY = [(u + {u)2 k 3 u2 k] 3 2uk {u = u2 k + 2uk {u + ({u)2 k 3 u2 k 3 2uk {u = ({u)2 k.
38. (a) sin =
20
{
i { = 20 csc i
g{ = 20(3 csc cot ) g = 320 csc 30 cot 30 (±1 )
I
I 2 3
= 320(2) 3 ±
=±
180
9
I
So the maximum error is about ± 29 3 E ±1=21 cm.
(b) The relative error is
39. Y = UL
i L=
Y
U
I
I
±2 3
3
g{
{{
E
= 9
=±
E ±0=03, so the percentage error is approximately ±3%.
{
{
20(2)
180
i gL = 3
Y
gL
3(Y @U2 ) gU
gU
{L
E
=
=3
.
gU. The relative error in calculating L is
2
U
L
L
Y @U
U
Hence, the relative error in calculating L is approximately the same (in magnitude) as the relative error in U.
40. I = nU4
i gI = 4nU3 gU i
gI
gU
4nU3 gU
=
4
=
. Thus, the relative change in I is about 4 times the
I
nU4
U
relative change in U. So a 5% increase in the radius corresponds to a 20% increase in blood Àow.
41. (a) gf =
gf
g{ = 0 g{ = 0
g{
(b) g(fx) =
g
gx
(fx) g{ = f
g{ = f gx
g{
g{
gx
g
gy
gx
gy
(c) g(x + y) =
(x + y) g{ =
+
g{ =
g{ +
g{ = gx + gy
g{
g{
g{
g{
g{
gy
gx
gy
gx
g
(xy) g{ = x
+y
g{ = x
g{ + y
g{ = x gy + y gx
(d) g(xy) =
g{
g{
g{
g{
g{
gx
gy
gy
gx
x
y
3x
g{ 3 x
g{
y
y gx 3 x gy
g x
g{
g{
g{
g{
(e) g
g{ =
=
=
g{ =
y
g{ y
y2
y2
y2
g
(f ) g ({q ) =
({q ) g{ = q{q31 g{
g{
42. (a) i ({) = sin {
i i 0 ({) = cos {, so i (0) = 0 and i 0 (0) = 1. Thus, i({) E i(0) + i 0 (0)({ 3 0) = 0 + 1({ 3 0) = {.
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NOT FOR SALE
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS
¤
259
(b)
We want to know the values of { for which | = { approximates | = sin { with less than a 2% difference; that is, the
values of { for which
{ 3 sin { sin { ? 0=02
C
30=02 ?
{ 3 sin {
? 0=02
sin {
+
30=02 sin { ? { 3 sin { ? 0=02 sin { if sin { A 0
30=02 sin { A { 3 sin { A 0=02 sin { if sin { ? 0
C
C
+
0=98 sin { ? { ? 1=02 sin { if sin { A 0
1=02 sin { ? { ? 0=98 sin { if sin { ? 0
In the ¿rst ¿gure, we see that the graphs are very close to each other near { = 0. Changing the viewing rectangle
and using an intersect feature (see the second ¿gure) we ¿nd that | = { intersects | = 1=02 sin { at { E 0=344.
By symmetry, they also intersect at { E 30=344 (see the third ¿gure). Converting 0=344 radians to degrees, we get
E 19=7 E 20 , which veri¿es the statement.
0=344 180
43. (a) The graph shows that i 0 (1) = 2, so O({) = i (1) + i 0 (1)({ 3 1) = 5 + 2({ 3 1) = 2{ + 3.
i (0=9) E O(0=9) = 4=8 and i (1=1) E O(1=1) = 5=2.
(b) From the graph, we see that i 0 ({) is positive and decreasing. This means that the slopes of the tangent lines are positive,
but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too
large.
44. (a) j 0({) =
I
I
{2 + 5 i j0 (2) = 9 = 3. j(1=95) E j(2) + j 0(2)(1=95 3 2) = 34 + 3(30=05) = 34=15.
j(2=05) E j(2) + j0(2)(2=05 3 2) = 34 + 3(0=05) = 33=85.
(b) The formula j0 ({) =
I
{2 + 5 shows that j 0 ({) is positive and increasing. This means that the slopes of the tangent lines
are positive and the tangents are getting steeper. So the tangent lines lie below the graph of j. Hence, the estimates in
part (a) are too small.
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¤
CHAPTER 3 DIFFERENTIATION RULES
LABORATORY PROJECT Taylor Polynomials
1. We ¿rst write the functions described in conditions (i), (ii), and (iii):
S ({) = D + E{ + F{2
i ({) = cos {
S 0 ({) = E + 2F{
i 0 ({) = 3 sin {
S 00 ({) = 2F
i 00 ({) = 3 cos {
So, taking d = 0, our three conditions become
S (0) = i(0):
0
0
S (0) = i (0):
S 00 (0) = i 00 (0):
D = cos 0 = 1
E = 3 sin 0 = 0
2F = 3 cos 0 = 31 i F = 3 12
The desired quadratic function is S ({) = 1 3 12 {2 , so the quadratic approximation is cos { E 1 3 12 {2 .
The ¿gure shows a graph of the cosine function together with its linear
approximation O({) = 1 and quadratic approximation S ({) = 1 3 12 {2
near 0. You can see that the quadratic approximation is much better than the
linear one.
2. Accuracy to within 0=1 means that cos { 3 1 3 12 {2 ? 0=1
C 30=1 ? cos { 3 1 3 12 {2 ? 0=1
C
0=1 A 1 3 12 {2 3 cos { A 30.1 C cos { + 0=1 A 1 3 12 {2 A cos { 3 0=1 C cos { 3 0=1 ? 1 3 12 {2 ? cos { + 0=1.
From the ¿gure we see that this is true between D and E. Zooming in or
using an intersect feature, we ¿nd that the {-coordinates of E and D are
about ±1=26. Thus, the approximation cos { E 1 3 12 {2 is accurate to
within 0=1 when 31=26 ? { ? 1=26.
3. If S ({) = D + E({ 3 d) + F({ 3 d)2 , then S 0 ({) = E + 2F({ 3 d) and S 00 ({) = 2F. Applying the conditions (i), (ii),
and (iii), we get
S (d) = i (d):
0
0
D = i (d)
S (d) = i (d):
E = i 0 (d)
S 00 (d) = i 00 (d):
2F = i 00 (d)
i F = 12 i 00 (d)
Thus, S ({) = D + E({ 3 d) + F({ 3 d)2 can be written in the form S ({) = i (d) + i 0 (d)({ 3 d) + 12 i 00 (d)({ 3 d)2 .
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¤
LABORATORY PROJECT TAYLOR POLYNOMIALS
4. From Example 1 in Section 3.10, we have i (1) = 2, i 0 (1) =
i 0 ({) = 12 ({ + 3)31@2 . So i 00 ({) = 3 14 ({ + 3)33@2
1
4,
261
and
1
i i 00 (1) = 3 32
.
From Problem 3, the quadratic approximation S ({) is
I
1
{ + 3 E i (1) + i 0 (1)({ 3 1) + 12 i 00 (1)({ 3 1)2 = 2 + 14 ({ 3 1) 3 64
({ 3 1)2 .
The ¿gure shows the function i ({) =
approximation O({) = 14 { +
7
4
I
{ + 3 together with its linear
and its quadratic approximation S ({). You can see that S ({) is a better approximation than
O({) and this is borne out by the numerical values in the following chart.
I
3=98
I
4=05
I
4=2
from O({)
actual value
from S ({)
1=9950
1=99499373 = = =
1=99499375
2=0125
2=01246118 = = =
2=01246094
2=0500
2=04939015 = = =
2=04937500
5. Wq ({) = f0 + f1 ({ 3 d) + f2 ({ 3 d)2 + f3 ({ 3 d)3 + · · · + fq ({ 3 d)q . If we put { = d in this equation,
then all terms after the ¿rst are 0 and we get Wq (d) = f0 . Now we differentiate Wq ({) and obtain
Wq0 ({) = f1 + 2f2 ({ 3 d) + 3f3 ({ 3 d)2 + 4f4 ({ 3 d)3 + · · · + qfq ({ 3 d)q31 . Substituting { = d gives Wq0 (d) = f1 .
Differentiating again, we have Wq00 ({) = 2f2 + 2 · 3f3 ({ 3 d) + 3 · 4f4 ({ 3 d2 ) + · · · + (q 3 1)qfq ({ 3 d)q32 and so
Wq00 (d) = 2f2 . Continuing in this manner, we get Wq000 ({) = 2 · 3f3 + 2 · 3 · 4f4 ({ 3 d) + · · · + (q 3 2)(q 3 1)qfq ({ 3 d)q33
and Wq000 (d) = 2 · 3f3 . By now we see the pattern. If we continue to differentiate and substitute { = d, we obtain
(4)
(n)
Wq (d) = 2 · 3 · 4f4 and in general, for any integer n between 1 and q, Wq (d) = 2 · 3 · 4 · 5 · · · · · nfn = n! fn
i
(n)
fn =
Wq (d)
i (n) (d)
. Because we want Wq and i to have the same derivatives at d, we require that fn =
for
n!
n!
n = 1> 2> = = = > q.
6. Wq ({) = i (d) + i 0 (d)({ 3 d) +
i 00 (d)
i (q) (d)
({ 3 d)2 + · · · +
({ 3 d)q . To compute the coef¿cients in this equation we
2!
q!
need to calculate the derivatives of i at 0:
i ({) = cos {
0
i ({) = 3 sin {
i 00 ({) = 3 cos {
i(0) = cos 0 = 1
i 0 (0) = 3 sin 0 = 0
i 00 (0) = 31
i 000 ({) = sin {
i 000 (0) = 0
i (4) ({) = cos {
i (4) (0) = 1
We see that the derivatives repeat in a cycle of length 4, so i (5) (0) = 0, i (6) (0) = 31, i (7) (0) = 0, and i (8) (0) = 1.
[continued]
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NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
From the original expression for Wq ({), with q = 8 and d = 0, we have
W8 ({) = i (0) + i 0 (0)({ 3 0) +
=1+0·{+
i 00 (0)
i 000 (0)
i (8) (0)
({ 3 0)2 +
({ 3 0)3 + · · · +
({ 3 0)8
2!
3!
8!
{4
{6
{8
31 2
1
31 6
1
{2
{ + 0 · {3 + {4 + 0 · {5 +
{ + 0 · {7 + {8 = 1 3
+
3
+
2!
4!
6!
8!
2!
4!
6!
8!
and the desired approximation is cos { E 1 3
{2
{4
{6
{8
+
3
+
. The Taylor polynomials W2 , W4 , and W6 consist of the
2!
4!
6!
8!
initial terms of W8 up through degree 2, 4, and 6, respectively. Therefore, W2 ({) = 1 3
W6 ({) = 1 3
{2
{2
{4
, W4 ({) = 1 3
+
, and
2!
2!
4!
{4
{6
{2
+
3
. We graph W2 , W4 , W6 , W8 , and i :
2!
4!
6!
Notice that W2 ({) is a good approximation to cos {
near 0, W4 ({) is a good approximation on a larger
interval, W6 ({) is a better approximation, and
W8 ({) is better still. Each successive Taylor
polynomial is a good approximation on a larger
interval than the previous one.
3.11 Hyperbolic Functions
1. (a) sinh 0 =
2. (a) tanh 0 =
1
(h0
2
3 h0 ) = 0
(b) cosh 0 = 12 (h0 + h0 ) = 12 (1 + 1) = 1
(h0 3 h30 )@2
=0
(h0 + h30 )@2
3. (a) sinh(ln 2) =
(b) tanh 1 =
h1 3 h31
h2 3 1
E 0=76159
= 2
1
31
h +h
h +1
23
hln 2 3 (hln 2 )31
2 3 231
hln 2 3 h3ln 2
=
=
=
2
2
2
2
1
2
=
3
4
(b) sinh 2 = 12 (h2 3 h32 ) E 3=62686
+ h33 ) E 10=06766
4. (a) cosh 3 =
1
(h3
2
5. (a) sech 0 =
1
1
= =1
cosh 0
1
6. (a) sinh 1 =
1
(h1
2
(b) cosh(ln 3) =
3+
hln 3 + h3 ln 3
=
2
2
1
3
=
5
3
(b) cosh31 1 = 0 because cosh 0 = 1.
3 h31 ) E 1=17520
I I
(b) Using Equation 3, we have sinh31 1 = ln 1 + 12 + 1 = ln 1 + 2 E 0=88137.
7. sinh(3{) = 12 [h3{ 3 h3(3{) ] =
8. cosh(3{) =
1 3{
2 [h
1
(h3{
2
3 h{ ) = 3 12 (h3{ 3 h{ ) = 3 sinh {
+ h3(3{) ] = 12 (h3{ + h{ ) = 12 (h{ + h3{ ) = cosh {
9. cosh { + sinh { = 12 (h{ + h3{ ) + 12 (h{ 3 h3{ ) =
1
(2h{ )
2
= h{
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.11
10. cosh { 3 sinh { = 12 (h{ + h3{ ) 3 12 (h{ 3 h3{ ) =
1
11. sinh { cosh | + cosh { sinh | =
2
3{
1
)
2 (2h
HYPERBOLIC FUNCTIONS
¤
= h3{
(h{ 3 h3{ ) 12 (h| + h3| ) + 12 (h{ + h3{ ) 12 (h| 3 h3| )
= 14 [(h{+| + h{3| 3 h3{+| 3 h3{3| ) + (h{+| 3 h{3| + h3{+| 3 h3{3| )]
= 14 (2h{+| 3 2h3{3| ) = 12 [h{+| 3 h3({+|) ] = sinh({ + |)
1
{
2 (h
12. cosh { cosh | + sinh { sinh | =
+ h3{ ) 12 (h| + h3| ) + 12 (h{ 3 h3{ ) 12 (h| 3 h3| )
{+|
(h
+ h{3| + h3{+| + h3{3| ) + (h{+| 3 h{3| 3 h3{+| + h3{3| )
k
l
= 14 (2h{+| + 2h3{3| ) = 12 h{+| + h3({+|) = cosh({ + |)
1
4
=
13. Divide both sides of the identity cosh2 { 3 sinh2 { = 1 by sinh2 {:
1
cosh2 { sinh2 {
3
=
sinh2 {
sinh2 {
sinh2 {
C coth2 { 3 1 = csch2 {.
sinh { cosh |
cosh { sinh |
+
sinh { cosh | + cosh { sinh |
sinh({ + |)
cosh { cosh |
cosh { cosh |
=
=
14. tanh({ + |) =
cosh { cosh |
sinh { sinh |
cosh({ + |)
cosh { cosh | + sinh { sinh |
+
cosh { cosh |
cosh { cosh |
tanh { + tanh |
=
1 + tanh { tanh |
15. Putting | = { in the result from Exercise 11, we have
sinh 2{ = sinh({ + {) = sinh { cosh { + cosh { sinh { = 2 sinh { cosh {.
16. Putting | = { in the result from Exercise 12, we have
cosh 2{ = cosh({ + {) = cosh { cosh { + sinh { sinh { = cosh2 { + sinh2 {.
17. tanh(ln {) =
18.
(hln { 3 h3 ln { )@2
{ 3 (hln { )31
({2 3 1)@{
{2 3 1
sinh(ln {)
{ 3 {31
{ 3 1@{
= ln {
=
=
=
=
=
cosh(ln {)
(h
+ h3 ln { )@2
{ + (hln { )31
{ + {31
{ + 1@{
({2 + 1)@{
{2 + 1
1 + tanh {
1 + (sinh {)@ cosh {
cosh { + sinh {
=
=
=
1 3 tanh {
1 3 (sinh {) @ cosh {
cosh { 3 sinh {
Or: Using the results of Exercises 9 and 10,
{
1
2 (h
1
(h{
2
+ h3{ ) + 12 (h{ 3 h3{ )
h{
=
= h2{
1
h3{
+ h3{ ) 3 2 (h{ 3 h3{ )
h{
cosh { + sinh {
= 3{ = h2{
cosh { 3 sinh {
h
19. By Exercise 9, (cosh { + sinh {)q = (h{ )q = hq{ = cosh q{ + sinh q{.
1
1
13
=
=
.
tanh {
12@13
12
2
25
= 169
i sech { =
sech2 { = 1 3 tanh2 { = 1 3 12
13
20. coth { =
1
tanh {
i coth { =
5
13
[sech, like cosh, is positive].
cosh { =
1
sech {
i cosh { =
tanh { =
sinh {
cosh {
i sinh { = tanh { cosh { i sinh { =
csch { =
1
sinh {
i csch { =
1
13
=
.
5@13
5
12 13
12
·
=
.
13 5
5
1
5
=
.
12@5
12
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264
¤
NOT FOR SALE
CHAPTER 3
21. sech { =
1
cosh {
DIFFERENTIATION RULES
1
3
= .
5@3
5
i sech { =
cosh2 { 3 sinh2 { = 1 i sinh2 { = cosh2 { 3 1 =
1
3
= .
4@3
4
csch { =
1
sinh {
tanh { =
sinh {
cosh {
i tanh { =
4@3
4
= .
5@3
5
coth { =
1
tanh {
i coth { =
1
5
= .
4@5
4
i csch { =
5 2
3
31 =
16
9
i sinh { =
4
3
[because { A 0].
22. (a)
h{ 3 h3{ h3{
1 3 h32{
130
=1
· 3{ = lim
=
{
3{
{<" h + h
{<" 1 + h32{
h
1+0
23. (a) lim tanh { = lim
{<"
h{ 3 h3{ h{
h2{ 3 1
031
=
= 31
· { = lim 2{
{
3{
{<3" h + h
{<3" h
h
+1
0+1
(b) lim tanh { = lim
{<3"
(c) lim sinh { = lim
{<"
{<"
h{ 3 h3{
="
2
(d) lim sinh { = lim
{<3"
{<3"
(e) lim sech { = lim
{<"
{<"
h{ 3 h3{
= 3"
2
2
=0
h{ + h3{
1+0
h{ + h3{ h3{
1 + h32{
· 3{ = lim
=
= 1 [Or: Use part (a)]
{
3{
{<" h 3 h
{<" 1 3 h32{
h
130
(f ) lim coth { = lim
{<"
(g) lim coth { = lim
cosh {
= ", since sinh { < 0 through positive values and cosh { < 1.
sinh {
(h) lim coth { = lim
cosh {
= 3", since sinh { < 0 through negative values and cosh { < 1.
sinh {
{<0+
{<0
{<0+
{<0
(i) lim csch { = lim
{<3"
{<3"
2
=0
h{ 3 h3{
g 1 {
g
3{
(cosh {) =
) = 12 (h{ 3 h3{ ) = sinh {
2 (h + h
g{
g{
g
sinh {
cosh { cosh { 3 sinh { sinh {
cosh2 { 3 sinh2 {
1
g
(tanh {) =
=
=
=
= sech2 {
(b)
2
g{
g{ cosh {
cosh {
cosh2 {
cosh2 {
1
g
g
1
cosh {
cosh {
=3
(c)
(csch {) =
=3
·
= 3 csch { coth {
2
g{
g{ sinh {
sinh
{
sinh {
sinh {
24. (a)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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NOT FOR SALE
SECTION 3.11
(d)
g
g
(sech {) =
g{
g{
(e)
g
g
(coth {) =
g{
g{
1
cosh {
cosh {
sinh {
=3
=
HYPERBOLIC FUNCTIONS
¤
265
1
sinh {
sinh {
=3
·
= 3 sech { tanh {
cosh { cosh {
cosh2 {
sinh { sinh { 3 cosh { cosh {
sinh2 { 3 cosh2 {
1
=
=3
= 3 csch2 {
2
sinh {
sinh2 {
sinh2 {
25. Let | = sinh31 {. Then sinh | = { and, by Example 1(a), cosh2 | 3 sinh2 | = 1
i [with cosh | A 0]
s
I
I
I
cosh | = 1 + sinh2 | = 1 + {2 . So by Exercise 9, h| = sinh | + cosh | = { + 1 + {2 i | = ln { + 1 + {2 .
s
I
cosh2 | 3 1 = {2 3 1. So, by Exercise 9,
I
I
h| = cosh | + sinh | = { + {2 3 1 i | = ln { + {2 3 1 .
Another method: Write { = cosh | = 12 h| + h3| and solve a quadratic, as in Example 3.
26. Let | = cosh31 {. Then cosh | = { and | D 0, so sinh | =
27. (a) Let | = tanh31 {. Then { = tanh | =
1 + { = h2| 3 {h2|
(h| 3 h3| )@2 h|
sinh |
h2| 3 1
= |
· | = 2|
3|
cosh |
(h + h )@2 h
h +1
i 1 + { = h2| (1 3 {) i h2| =
1+{
13{
i {h2| + { = h2| 3 1 i
1+{
i 2| = ln
i |=
13{
1
2
1+{
.
ln
13{
(b) Let | = tanh31 {. Then { = tanh |, so from Exercise 18 we have
1+{
1+{
1 + tanh |
1+{
=
i 2| = ln
i | = 12 ln
.
h2| =
1 3 tanh |
13{
13{
13{
28. (a) (i) | = csch31 {
C csch | = { ({ 6= 0)
(ii) We sketch the graph of csch31 by reÀecting the graph of csch (see Exercise 22)
about the line | = {.
2
i {h| 3 {h3| = 2 i
h| 3 h3|
I
1 ± {2 + 1
. But h| A 0, so for { A 0,
{(h| )2 3 2h| 3 { = 0 i h| =
{
I
I
I
1 + {2 + 1
1 3 {2 + 1
1
{2 + 1
h| =
and for { ? 0, h| =
. Thus, csch31 { = ln
+
.
{
{
{
|{|
(iii) Let | = csch31 {. Then { = csch | =
(b) (i) | = sech31 { C sech | = { and | A 0=
(ii) We sketch the graph of sech31 by reÀecting the graph of sech (see Exercise 22)
about the line | = {.
2
i {h| + {h3| = 2 i
h| + h3|
I
1 ± 1 3 {2
{ (h| )2 3 2h| + { = 0 C h| =
. But | A 0 i h| A 1.
{
I
I
I
1 3 1 3 {2
This rules out the minus sign because
A 1 C 1 3 1 3 {2 A { C 1 3 { A 1 3 {2
{
(iii) Let | = sech31 {, so { = sech | =
1 3 2{ + {2 A 1 3 {2 C {2 A { C { A 1, but { = sech | $ 1.
I
I
1 + 1 3 {2
1 + 1 3 {2
|
31
Thus, h =
i sech { = ln
.
{
{
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C
266
¤
NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
(c) (i) | = coth31 { C coth | = {
(ii) We sketch the graph of coth31 by reÀecting the graph of coth (see Exercise 22)
about the line | = {.
(iii) Let | = coth31 {. Then { = coth | =
{h| 3 {h3| = h| + h3|
2| = ln
{+1
{31
h| + h3|
h| 3 h3|
i
i ({ 3 1)h| = ({ + 1)h3|
i coth31 { =
i h2| =
1 {+1
ln
2 {31
29. (a) Let | = cosh31 {. Then cosh | = { and | D 0
g|
1
1
1
= I
=
= s
2
2 31
g{
sinh |
{
cosh | 3 1
i sinh |
{+1
{31
i
g|
=1 i
g{
[since sinh | D 0 for | D 0]. Or: Use Formula 4.
(b) Let | = tanh31 {. Then tanh | = { i sech2 |
g|
=1 i
g{
g|
1
1
1
=
.
=
=
2
2
g{
1
3
{2
sech |
1 3 tanh |
Or: Use Formula 5.
g|
=1 i
g{
g|
1
=3
. By Exercise 13,
g{
csch | coth |
s
I
I
coth | = ± csch2 | + 1 = ± {2 + 1. If { A 0, then coth | A 0, so coth | = {2 + 1. If { ? 0, then coth | ? 0,
(c) Let | = csch31 {. Then csch | = { i 3 csch | coth |
I
1
1
g|
=3
=3 I
.
so coth | = 3 {2 + 1. In either case we have
g{
csch | coth |
|{| {2 + 1
(d) Let | = sech31 {. Then sech | = { i 3 sech | tanh |
g|
=1 i
g{
g|
1
1
1
s
=3
=3
. [Note that | A 0 and so tanh | A 0.]
=3 I
2
g{
sech | tanh |
{
1
3 {2
sech | 1 3 sech |
(e) Let | = coth31 {. Then coth | = { i 3 csch2 |
g|
=1 i
g{
g|
1
1
1
=3
=
=
2
2
g{
1
3
{2
csch |
1 3 coth |
by Exercise 13.
30. i ({) = tanh(1 + h2{ )
i i 0 ({) = sech2 (1 + h2{ )
31. i ({) = { sinh { 3 cosh {
32. j({) = cosh(ln {)
g
(1 + h2{ ) = 2h2{ sech2 (1 + h2{ )
g{
i i 0 ({) = { (sinh {)0 + sinh { · 1 3 sinh { = { cosh {
i j0 ({) = sinh(ln {) · (ln {)0 =
1
{
sinh(ln {)
Or: j({) = cosh(ln {) = 12 (hln { + h3 ln { ) = 12 ({ + {31 ) i j 0 ({) = 12 (1 3 {32 ) =
33. k({) = ln(cosh {)
34. | = { coth(1 + {2 )
35. | = hcosh 3{
i k0 ({) =
1
2
3 1@(2{2 )
1
sinh {
(cosh {)0 =
= tanh {
cosh {
cosh {
i | 0 = {[3 csch2 (1 + {2 ) · 2{] + coth(1 + {2 ) · 1 = 32{2 csch2 (1 + {2 ) + coth(1 + {2 )
i | 0 = hcosh 3{ · sinh 3{ · 3 = 3hcosh 3{ sinh 3{
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 3.11
36. i (w) = csch w (1 3 ln csch w)
i 0 (w) = csch w 3
HYPERBOLIC FUNCTIONS
i
1
(3 csch w coth w) + (1 3 ln csch w)(3 csch w coth w)
csch w
= csch w coth w 3 (1 3 ln csch w) csch w coth w = csch w coth w [1 3 (1 3 ln csch w)] = csch w coth w ln csch w
37. i (w) = sech2 (hw ) = [sech(hw )]2
i
i 0 (w) = 2[sech(hw )] [sech(hw )]0 = 2 sech(hw ) 3 sech(hw ) tanh(hw ) · hw = 32hw sech2 (hw ) tanh(hw )
i | 0 = cosh(cosh {) · sinh {
38. | = sinh(cosh {)
39. J({) =
J0 ({) =
=
1 3 cosh {
1 + cosh {
i
(1 + cosh {)(3 sinh {) 3 (1 3 cosh {)(sinh {)
3 sinh { 3 sinh { cosh { 3 sinh { + sinh { cosh {
=
(1 + cosh {)2
(1 + cosh {)2
32 sinh {
(1 + cosh {)2
40. | = sinh31 (tan {)
I
41. | = cosh31 {
g
sec2 {
1
| sec2 { |
(tan {) = I
=| sec { |
i |0 = s
=
2
2
| sec { |
1 + (tan {) g{
sec {
g I
1
1
1
1
I = s
( {) = I
i |0 = t
I 2
g{
2 {
{
3
1
2
{({
3 1)
( {) 3 1
I
1 3 {2 = { tanh31 { + 12 ln(1 3 {2 ) i
1
{
1
+
(32{) = tanh31 {
| 0 = tanh31 { +
1 3 {2
2 1 3 {2
42. | = { tanh31 { + ln
43. | = { sinh31 ({@3) 3
| 0 = sinh31
{
45. | = coth31 (sec {)
i |0 = 3
i
h3{
g 3{
1
1
1
s
I
(h ) = 3
(3h3{ ) = I
h3{ 1 3 h32{
1 3 h32{
1 3 (h3{ )2 g{
g
sec { tan {
sec { tan {
sec { tan {
1
(sec {) =
=
=
1 3 (sec {)2 g{
1 3 sec2 {
1 3 (tan2 { + 1)
3 tan2 {
=3
46.
i
{
{
1@3
{
2{
{
+ {s
+I
3 I
= sinh31
3I
= sinh31
3
3
3
2 9 + {2
9 + {2
9 + {2
1 + ({@3)2
44. | = sech 31 (h3{ )
|0 =
I
9 + {2
sec {
1@ cos {
1
=3
=3
= 3 csc {
tan {
sin {@ cos {
sin {
1 + tanh {
1 + (sinh {)@ cosh {
cosh { + sinh {
h{
=
=
= 3{ [by Exercises 9 and 10] = h2{ , so
1 3 tanh {
1 3 (sinh {)@ cosh {
cosh { 3 sinh {
h
u
u
I
g {@2
g 4 1 + tanh {
1
4
4 1 + tanh {
= h2{ = h{@2 . Thus,
=
(h ) = h{@2 .
1 3 tanh {
g{ 1 3 tanh {
g{
2
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°
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¤
267
268
47.
¤
NOT FOR SALE
CHAPTER 3
DIFFERENTIATION RULES
g
g
1@ cosh2 {
1
sech2 {
=
arctan(tanh {) =
(tanh {) =
2
2
g{
1 + (tanh {) g{
1 + tanh {
1 + (sinh2 {)@ cosh2 {
=
1
1
[by Exercise 16] = sech 2{
=
cosh 2{
cosh2 { + sinh2 {
48. (a) Let d = 0=03291765. A graph of the central curve,
| = i ({) = 211=49 3 20=96 cosh d{, is shown.
(b) i (0) = 211=49 3 20=96 cosh 0 = 211=49 3 20=96(1) = 190=53 m=
(c) | = 100 i 100 = 211=49 3 20=96 cosh d{ i
20=96 cosh d{ = 111=49 i cosh d{ =
d{ = ± cosh31
111=49
20=96
111=49
20=96
i
111=49
1
i { = ± cosh31
E ±71=56 m. The points are approximately (±71=56> 100).
d
20=96
(d) i ({) = 211=49 3 20=96 cosh d{ i i 0 ({) = 320=96 sinh d{ · d.
111=49
1
1
31 111=49
31 111=49
i 0 ± cosh31
= 320=96d sinh d ± cosh
= 320=96d sinh ± cosh
E ~3=6.
d
20=96
d
20=96
20=96
So the slope at (71=56> 100) is about 33=6 and the slope at (371=56> 100) is about 3=6.
2g
2g
gets large, and from Figure 3 or Exercise 23(a), tanh
O
O
v
u
u
2g
jO
jO
jO
tanh
E
(1) =
.
approaches 1. Thus, y =
2
O
2
2
49. As the depth g of the water gets large, the fraction
50.
For | = d cosh({@d) with d A 0, we have the |-intercept equal to d.
As d increases, the graph Àattens.
51. (a) | = 20 cosh({@20) 3 15
i | 0 = 20 sinh({@20) ·
1
20
= sinh({@20). Since the right pole is positioned at { = 7,
7
we have | 0 (7) = sinh 20
E 0=3572.
7
, so
(b) If is the angle between the tangent line and the {-axis, then tan = slope of the line = sinh 20
7
E 0=343 rad E 19=66 . Thus, the angle between the line and the pole is = 90 3 E 70=34 .
= tan31 sinh 20
52. We differentiate the function twice, then substitute into the differential equation: | =
j{ j
g|
W
j{
=
sinh
= sinh
g{
j
W
W
W
W
j{
cosh
j
W
i
j{ j
g2 |
j
j{
=
cosh
. We evaluate the two sides
= cosh
2
g{
W
W
W
W
v
2
u
g2 |
j{
j{
j
j
j{
j
g|
j
separately: LHS =
cosh
and
RHS
=
=
cosh
,
=
1
+
=
1 + sinh2
g{2
W
W
W
g{
W
W
W
W
i
by the identity proved in Example 1(a).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 3.11
HYPERBOLIC FUNCTIONS
¤
269
j{ W
cosh
. The shape is symmetric about the
53. (a) From Exercise 52, the shape of the cable is given by | = i ({) =
j
W
W
|-axis, so the lowest point is (0> i (0)) = 0>
and the poles are at { = ±100. We want to ¿nd W when the lowest
j
kg-m
W
= 60 i W = 60j = (60 m)(2 kg@m)(9=8 m@s2 ) = 1176
, or 1176 N (newtons).
j
s2
W
j · 100
100
The height of each pole is i(100) =
cosh
= 60 cosh
E 164=50 m.
j
W
60
point is 60 m, so
(b) If the tension is doubled from W to 2W , then the low point is doubled since
W
= 60 i
j
2W
= 120. The height of the
j
2W
j · 100
100
poles is now i (100) =
cosh
= 120 cosh
E 164=13 m, just a slight decrease.
j
2W
120
54. lim
{<"
sinh {
h{ 3 h3{
1 3 h32{
130
1
=
=
= lim
= lim
{
{
{<"
{<"
h
2h
2
2
2
55. (a) | = D sinh p{ + E cosh p{
i |0 = pD cosh p{ + pE sinh p{ i
|00 = p2 D sinh p{ + p2 E cosh p{ = p2 (D sinh p{ + E cosh p{) = p2 |
(b) From part (a), a solution of | 00 = 9| is |({) = D sinh 3{ + E cosh 3{. So 34 = |(0) = D sinh 0 + E cosh 0 = E, so
E = 34. Now | 0 ({) = 3D cosh 3{ 3 12 sinh 3{ i 6 = | 0 (0) = 3D i D = 2, so | = 2 sinh 3{ 3 4 cosh 3{.
l 1
1
1 k ln(sec +tan )
h
sec + tan +
+ h3 ln(sec +tan ) =
2
2
sec + tan 1
sec 3 tan 1
sec 3 tan =
sec + tan +
=
sec + tan +
2
(sec + tan )(sec 3 tan )
2
sec2 3 tan2 56. cosh { = cosh[ln(sec + tan )] =
= 12 (sec + tan + sec 3 tan ) = sec 57. The tangent to | = cosh { has slope 1 when | 0 = sinh { = 1
Since sinh { = 1 and | = cosh { =
I i { = sinh31 1 = ln 1 + 2 , by Equation 3.
s
I
I I 1 + sinh2 {, we have cosh { = 2. The point is ln 1 + 2 , 2 .
58. iq ({) = tanh(q sin {), where q is a positive integer. Note that iq ({ + 2) = iq ({); that is, iq is periodic with period 2.
Also, from Figure 3, 31 ? tanh { ? 1, so we can choose a viewing rectangle of [0> 2] × [31> 1]. From the graph, we see
that iq ({) becomes more rectangular looking as q increases. As q becomes
large, the graph of iq approaches the graph of | = 1 on the intervals
(2n> (2n + 1)) and | = 31 on the intervals ((2n 3 1)> 2n).
59. If dh{ + eh3{ = cosh({ + )
dh{ + eh3{
[or sinh({ + )], then
= 2 h{+ ± h3{3 = 2 h{ h ± h3{ h3 = 2 h h{ ± 2 h3 h3{ . Comparing coef¿cients of h{
and h3{ , we have d =
h
2
(1) and e = ± 2 h3 (2). We need to ¿nd and . Dividing equation (1) by equation (2)
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¤
CHAPTER 3 DIFFERENTIATION RULES
gives us
h =
= ±h2
d
e
i =
i (B) 2 = ln ± de
2d
2d
and h = ± , so
=±
2e
2e
(B) If
d
e
1
2
ln ± de . Solving equations (1) and (2) for h gives us
i 2 = ±4de i = 2
A 0, we use the + sign and obtain a cosh function, whereas if
I
±de.
d
e
? 0, we use the 3 sign and obtain a sinh
function.
In summary, if d and e have the same sign, we have dh{ + eh3{ = 2
I
opposite sign, then dh{ + eh3{ = 2 3de sinh { + 12 ln 3 de .
I
de cosh { +
1
2
ln de , whereas, if d and e have the
3 Review
1. (a) The Power Rule: If q is any real number, then
g
({q ) = q{q31 . The derivative of a variable base raised to a constant
g{
power is the power times the base raised to the power minus one.
(b) The Constant Multiple Rule: If f is a constant and i is a differentiable function, then
g
g
[fi ({)] = f
i ({).
g{
g{
The derivative of a constant times a function is the constant times the derivative of the function.
(c) The Sum Rule: If i and j are both differentiable, then
g
g
g
[i ({) + j({)] =
i ({) +
j({). The derivative of a sum
g{
g{
g{
of functions is the sum of the derivatives.
(d) The Difference Rule: If i and j are both differentiable, then
g
g
g
[i ({) 3 j({)] =
i ({) 3
j({). The derivative of a
g{
g{
g{
difference of functions is the difference of the derivatives.
(e) The Product Rule: If i and j are both differentiable, then
g
g
g
[i ({) j({)] = i ({)
j({) + j({)
i ({). The
g{
g{
g{
derivative of a product of two functions is the ¿rst function times the derivative of the second function plus the second
function times the derivative of the ¿rst function.
j({) g i ({) 3 i ({) g j({)
g i ({)
g{
g{
(f ) The Quotient Rule: If i and j are both differentiable, then
=
.
g{ j({)
[ j({)]2
The derivative of a quotient of functions is the denominator times the derivative of the numerator minus the numerator
times the derivative of the denominator, all divided by the square of the denominator.
(g) The Chain Rule: If i and j are both differentiable and I = i j is the composite function de¿ned by I ({) = i (j({)),
then I is differentiable and I 0 is given by the product I 0 ({) = i 0 (j({)) j 0 ({). The derivative of a composite function is
the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
2. (a) | = {q
i | 0 = q{q31
(b) | = h{
(c) | = d{
i | 0 = d{ ln d
(d) | = ln { i | 0 = 1@{
(e) | = logd { i |0 = 1@({ ln d)
i | 0 = h{
(f ) | = sin { i | 0 = cos {
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
CHAPTER 3 REVIEW
(g) | = cos { i | 0 = 3 sin {
(h) | = tan { i | 0 = sec2 {
(i) | = csc { i | 0 = 3 csc { cot {
(j) | = sec { i |0 = sec { tan {
¤
271
I
(l) | = sin31 { i | 0 = 1@ 1 3 {2
(k) | = cot { i | 0 = 3 csc2 {
I
(m) | = cos31 { i | 0 = 31@ 1 3 {2
(n) | = tan31 { i | 0 = 1@(1 + {2 )
(o) | = sinh { i |0 = cosh {
(p) | = cosh { i | 0 = sinh {
I
(r) | = sinh31 { i |0 = 1@ 1 + {2
(q) | = tanh { i | 0 = sech2 {
I
(s) | = cosh31 { i |0 = 1@ {2 3 1
(t) | = tanh31 { i |0 = 1@(1 3 {2 )
hk 3 1
= 1.
k<0
k
3. (a) h is the number such that lim
(b) h = lim (1 + {)1@{
{<0
(c) The differentiation formula for | = d{
[| 0 = d{ ln d] is simplest when d = h because ln h = 1.
(d) The differentiation formula for | = logd { [| 0 = 1@({ ln d)] is simplest when d = h because ln h = 1.
4. (a) Implicit differentiation consists of differentiating both sides of an equation involving { and | with respect to {, and then
solving the resulting equation for | 0 .
(b) Logarithmic differentiation consists of taking natural logarithms of both sides of an equation | = i ({), simplifying,
differentiating implicitly with respect to {, and then solving the resulting equation for | 0 .
5. See the examples in Section 3.7 as well as the text following Example 8.
6. (a)
g|
= n|, where | is a quantity and n A 0 is a constant.
gw
(b) The equation in part (a) is an appropriate model for population growth, assuming that there is enough room and nutrition to
support the growth.
(c) If |(0) = |0 , then the solution is |(w) = |0 hnw .
7. (a) The linearization O of i at { = d is O({) = i (d) + i 0 (d)({ 3 d).
(b) If | = i ({), then the differential g| is given by g| = i 0 ({) g{.
(c) See Figure 5 in Section 3.10.
1. True.
This is the Sum Rule.
2. False.
See the warning before the Product Rule.
3. True.
This is the Chain Rule.
4. True.
i 0 ({)
g s
g
1
i ({) =
[i ({)]1@2 = [i ({)]31@2 i 0 ({) = s
g{
g{
2
2 i ({)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
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licated, or posted to a publicly accessible website, in whole or in part.
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272
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CHAPTER 3 DIFFERENTIATION RULES
5. False.
I
I
I
g
i 0( { )
i 0 ({)
I , which is not I .
i ( { ) = i 0 ( { ) · 12 {31@2 =
g{
2 {
2 {
6. False.
h2 is a constant, so |0 = 0.
7. False.
g
(10{ ) = 10{ ln 10, which is not equal to {10{31 .
g{
8. False.
ln 10 is a constant, so its derivative,
9. True.
g
g
(tan2 {) = 2 tan { sec2 {, and
(sec2 {) = 2 sec { (sec { tan {) = 2 tan { sec2 {.
g{
g{
g
(ln 10), is 0, not
g{
1
.
10
g
g
g
(sec2 {) =
(1 + tan2 {) =
(tan2 {).
g{
g{
g{
i ({) = {2 + { = {2 + { for { D 0 or { $ 31 and {2 + { = 3({2 + {) for 31 ? { ? 0.
Or:
10. False.
So i 0 ({) = 2{ + 1 for { A 0 or { ? 31 and i 0 ({) = 3(2{ + 1) for 31 ? { ? 0. But |2{ + 1| = 2{ + 1
for { D 3 12 and |2{ + 1| = 32{ 3 1 for { ? 3 12 .
If s({) = dq {q + dq31 {q31 + · · · + d1 { + d0 , then s0 ({) = qdq {q31 + (q 3 1)dq31 {q32 + · · · + d1 , which is
11. True.
a polynomial.
12. True.
i ({) = ({6 3 {4 )5 is a polynomial of degree 30, so its 31st derivative, i (31) ({), is 0.
13. True.
If u({) =
s({)
t({)s0 ({) 3 s({)t 0 ({)
, which is a quotient of polynomials, that is, a rational
, then u0 ({) =
t({)
[t({)]2
function.
A tangent line to the parabola | = {2 has slope g|@g{ = 2{, so at (32> 4) the slope of the tangent is 2(32) = 34
14. False.
and an equation of the tangent line is | 3 4 = 34({ + 2). [The given equation, | 3 4 = 2{({ + 2), is not even
linear!]
j({) = {5
15. True.
lim
{<2
i | 0 = 4({2 + {3 )3 (2{ + 3{2 ) = 4({2 )3 (1 + {)3 {(2 + 3{) = 4{7 ({ + 1)3 (3{ + 2)
1
= {31@2 3 {33@5
{3
2. | = I 3 I
5
1
3
3
1
1 38@5
I
{
i | 0 = 3 {33@2 + {38@5 or
3 I or
(35{1@10 + 6)
5
2
5
10
2{ {
5{ {3
3. | =
{2 3 { + 2
I
= {3@2 3 {1@2 + 2{31@2
{
4. | =
tan {
1 + cos {
5. | = {2 sin {
i j 0 (2) = 5(2)4 = 80, and by the de¿nition of the derivative,
j({) 3 j(2)
= j 0 (2) = 80.
{32
1. | = ({2 + {3 )4
1
{
i j0 ({) = 5{4
i |0 =
i |0 =
3 1@2 1 31@2
3I
1
1
3 {
3 {33@2 =
{3 I 3 I
{
2
2
2
2 {
{3
(1 + cos {) sec2 { 3 tan {(3 sin {)
(1 + cos {) sec2 { + tan { sin {
=
(1 + cos {)2
(1 + cos {)2
i | 0 = {2 (cos {) + (sin {)(2{) = {({ cos { + 2 sin {)
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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CHAPTER 3 REVIEW
1
{
6. | = { cos31 { i | 0 = { 3 I
+ (cos31 {)(1) = cos31 { 3 I
1 3 {2
1 3 {2
w4 3 1
w4 + 1
7. | =
8.
i |0 =
(w4 + 1)4w3 3 (w4 3 1)4w3
4w3 [(w4 + 1) 3 (w4 3 1)]
8w3
=
=
(w4 + 1)2
(w4 + 1)2
(w4 + 1)2
g
g
({h| ) =
(| sin {) i {h| | 0 + h| · 1 = | cos { + sin { · |0
g{
g{
({h| 3 sin {)| 0 = | cos { 3 h|
i |0 =
9. | = ln({ ln {)
i |0 =
0
p{
| =h
i
| cos { 3 h|
{h| 3 sin {
1
1
({ ln {)0 =
{ ln {
{ ln {
1
1 + ln {
{ · + ln { · 1 =
{
{ ln {
Another method: | = ln({ ln {) = ln { + ln ln { i | 0 =
10. | = hp{ cos q{
i {h| | 0 3 sin { · | 0 = | cos { 3 h|
1
1
ln { + 1
1
+
· =
{ ln { {
{ ln {
i
(cos q{) + cos q{ (hp{ )0 = hp{ (3 sin q{ · q) + cos q{ (hp{ · p) = hp{ (p cos q{ 3 q sin q{)
0
I
I
{ cos { i
l
I I I 0
I I 0 I k
I { = { 3 sin { 12 {31@2 + cos { 12 {31@2
| 0 = { cos { + cos {
11. | =
=
1 31@2
2{
I
I
I
I
I
I cos { 3 { sin {
I
3 { sin { + cos { =
2 {
1
4 arcsin 2{
i | 0 = 2(arcsin 2{) · (arcsin 2{)0 = 2 arcsin 2{ · s
·2= I
1 3 4{2
1 3 (2{)2
12. | = (arcsin 2{)2
h1@{
13. | = 2
{
0
{2 (h1@{ )0 3 h1@{ {2
{2 (h1@{ )(31@{2 ) 3 h1@{ (2{)
3h1@{ (1 + 2{)
i | =
=
=
2
2
4
({ )
{
{4
14. | = ln sec {
15.
0
i |0 =
1
1 g
(sec {) =
(sec { tan {) = tan {
sec { g{
sec {
g
g
(| + { cos |) =
({2 |) i | 0 + {(3 sin | · | 0 ) + cos | · 1 = {2 | 0 + | · 2{ i
g{
g{
| 0 3 { sin | · |0 3 {2 | 0 = 2{| 3 cos |
i (1 3 { sin | 3 {2 )| 0 = 2{| 3 cos |
i |0 =
2{| 3 cos |
1 3 { sin | 3 {2
4
x31
i
x2 + x + 1
3
3 2
(x + x + 1)(1) 3 (x 3 1)(2x + 1)
x31
g
x31
x31
0
| =4
=4 2
x2 + x + 1 gx x2 + x + 1
x +x+1
(x2 + x + 1)2
16. | =
=
17. | =
4(x 3 1)3 x2 + x + 1 3 2x2 + x + 1
4(x 3 1)3 (3x2 + 2x + 2)
=
2
3
2
2
(x + x + 1)
(x + x + 1)
(x2 + x + 1)5
I
g
1
1
(arctan {) = I
arctan { i | 0 = (arctan {)31@2
2
g{
2 arctan { (1 + {2 )
18. | = cot(csc {)
i | 0 = 3 csc2 (csc {)
g
(csc {) = 3 csc2 (csc {) · (3 csc { cot {) = csc2 (csc {) csc { cot {
g{
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273
274
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¤
CHAPTER 3 DIFFERENTIATION RULES
w
i
1 + w2
1 3 w2
w
g
w
(1 + w2 )(1) 3 w(2w)
w
w
0
2
2
2
=
sec
= sec
·
| = sec
1 + w2 gw 1 + w2
1 + w2
(1 + w2 )2
(1 + w2 )2
1 + w2
19. | = tan
20. | = h{ sec {
21. | = 3{ ln {
g
({ sec {) = h{ sec { ({ sec { tan { + sec { · 1) = sec { h{ sec { ({ tan { + 1)
g{
g
1
({ ln {) = 3{ ln { (ln 3) { · + ln { · 1 = 3{ ln { (ln 3)(1 + ln {)
i | 0 = 3{ ln { (ln 3)
g{
{
i | 0 = h{ sec {
i | 0 = 2{ sec(1 + {2 ) tan(1 + {2 )
22. | = sec(1 + {2 )
23. | = (1 3 {31 )31
0
i
31 32
| = 31(1 3 {
24. | = { +
)
I 31@3
{
25. sin({|) = {2 3 |
[3(31{32 )] = 3(1 3 1@{)32 {32 = 3(({ 3 1)@{)32 {32 = 3({ 3 1)32
I 34@3
1+
i | 0 = 3 13 { + {
i cos({|)({|0 + | · 1) = 2{ 3 | 0
| 0 [{ cos({|) + 1] = 2{ 3 | cos({|) i |0 =
26. | =
1
I
2 {
i { cos({|)| 0 + | 0 = 2{ 3 | cos({|) i
2{ 3 | cos({|)
{ cos({|) + 1
I
I 31@2 I 1 s I
cos {
I
cos {
sin { i | 0 = 12 sin {
= s
I
2 {
4 { sin {
27. | = log5 (1 + 2{)
28. | = (cos {){
i |0 =
g
2
1
(1 + 2{) =
(1 + 2{) ln 5 g{
(1 + 2{) ln 5
i ln | = ln(cos {){ = { ln cos { i
1
|0
={·
· (3 sin {) + ln cos { · 1 i
|
cos {
| 0 = (cos {){ (ln cos { 3 { tan {)
29. | = ln sin { 3
30. | =
1
2
sin2 { i | 0 =
({2 + 1)4
(2{ + 1)3 (3{ 3 1)5
ln | = ln
1
· cos { 3
sin {
1
2
· 2 sin { · cos { = cot { 3 sin { cos {
i
({2 + 1)4
= ln({2 + 1)4 3 ln[(2{ + 1)3 (3{ 3 1)5 ] = 4 ln({2 + 1) 3 [ln(2{ + 1)3 + ln(3{ 3 1)5 ]
(2{ + 1)3 (3{ 3 1)5
= 4 ln({2 + 1) 3 3 ln(2{ + 1) 3 5 ln(3{ 3 1) i
|0
1
1
1
({2 + 1)4
=4· 2
· 2{ 3 3 ·
·235·
· 3 i |0 =
|
{ +1
2{ + 1
3{ 3 1
(2{ + 1)3 (3{ 3 1)5
[The answer could be simpli¿ed to | 0 = 3
31. | = { tan31 (4{)
i |0 = { ·
8{
6
15
3
3
.
{2 + 1
2{ + 1
3{ 3 1
({2 + 56{ + 9)({2 + 1)3
, but this is unnecessary.]
(2{ + 1)4 (3{ 3 1)6
1
4{
· 4 + tan31 (4{) · 1 =
+ tan31 (4{)
1 + (4{)2
1 + 16{2
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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NOT FOR SALE
CHAPTER 3 REVIEW
32. | = hcos { + cos(h { )
i
1
5 sec 5{ (tan 5{ + sec 5{)
(sec 5{ tan 5{ · 5 + sec2 5{ · 5) =
= 5 sec 5{
sec 5{ + tan 5{
sec 5{ + tan 5{
i | 0 = 10 tan · ln 10 · sec2 · = (ln 10)10 tan sec2 34. | = 10 tan i | 0 = 3 csc2 (3{2 + 5)(6{) = 36{ csc2 (3{2 + 5)
35. | = cot(3{2 + 5)
36. | =
275
i | 0 = hcos { (3 sin {) + [3 sin(h { ) · h { ] = 3 sin { hcos { 3 h { sin(h { )
33. | = ln |sec 5{ + tan 5{|
|0 =
¤
s
w ln(w4 ) i
1
1
1
ln(w4 ) + 4
1
4 31@2 g
4
4
3
s
· 1 · ln(w ) + w · 4 · 4w = s
· [ln(w4 ) + 4] = s
[w ln(w )] =
| = [w ln(w )]
2
gw
w
2 w ln(w4 )
2 w ln(w4 )
2 w ln(w4 )
0
I
I
Or: Since y is only de¿ned for w A 0, we can write | = w · 4 ln w = 2 w ln w. Then
1
ln w + 1
1
· 1 · ln w + w ·
. This agrees with our ¿rst answer since
= I
|0 = 2 · I
w
2 w ln w
w ln w
ln(w4 ) + 4
4 ln w + 4
4(ln w + 1)
ln w + 1
s
I
= I
=
= I
.
2 w ln(w4 )
2 w · 4 ln w
2 · 2 w ln w
w ln w
37. | = sin tan
I
I
I
I
i | 0 = cos tan 1 + {3 sec2 1 + {3 3{2 2 1 + {3
1 + {3
38. | = arctan arcsin
I {
i |0 =
39. | = tan2 (sin ) = [tan(sin )]2
40. {h| = | 3 1
1
1
1
I 2 · I1 3 { · I
2 {
1 + arcsin {
i | 0 = 2[tan(sin )] · sec2 (sin ) · cos i {h| | 0 + h| = | 0
i h| = | 0 3 {h| | 0
i | 0 = h| @(1 3 {h| )
I
{ + 1 (2 3 {)5
1
35
7
|0
=
+
3
i ln | = 12 ln({ + 1) + 5 ln(2 3 {) 3 7 ln({ + 3) i
7
({ + 3)
|
2({ + 1) 2 3 { { + 3
I
(2 3 {)4 (3{2 3 55{ 3 52)
{ + 1 (2 3 {)5
1
5
7
0
I
| =
.
3
3
or | 0 =
7
({ + 3)
2({ + 1)
23{ {+3
2 { + 1 ({ + 3)8
41. | =
42. | =
({ + )4
{4 + 4
43. | = { sinh({2 )
i |0 =
({4 + 4 )(4)({ + )3 3 ({ + )4(4{3 )
4({ + )3 (4 3 {3 )
=
4 2
4
({ + )
({4 + 4 )2
i |0 = { cosh({2 ) · 2{ + sinh({2 ) · 1 = 2{2 cosh({2 ) + sinh({2 )
44. | = (sin p{)@{
i | 0 = (p{ cos p{ 3 sin p{)@{2
45. | = ln(cosh 3{)
i | 0 = (1@ cosh 3{)(sinh 3{)(3) = 3 tanh 3{
2
{ 3 4
2({ + 1)({ + 4)
2
= ln {2 3 4 3 ln |2{ + 5| i | 0 = 2{ 3
or
2{ + 5
{2 3 4
2{ + 5
({ + 2)({ 3 2)(2{ + 5)
46. | = ln 47. | = cosh31 (sinh {)
1
cosh {
i |0 = s
· cosh { = s
2
(sinh {) 3 1
sinh2 { 3 1
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276
¤
NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
I
48. | = { tanh31 {
I
tan 3{
49. | = cos h
I
i |0 = tanh31 { + {
I
I
1
1
{
31
I
=
tanh
{
+
I 2
2(1 3 {)
2
{
13
{
i
I
I
I
0
I
| 0 = 3 sin h tan 3{ · h tan 3{ = 3 sin h tan 3{ h tan 3{ · 12 (tan 3{)31@2 · sec2 (3{) · 3
I
I
33 sin h tan 3{ h tan 3{ sec2 (3{)
I
=
2 tan 3{
50. | = sin2 cos
k l2
I
I
sin { = sin cos sin {
i
k lk l0
0
I
I
I
I
I
|0 = 2 sin cos sin {
sin cos sin {
= 2 sin cos sin { cos cos sin { cos sin {
I
0
I
I
I
= 2 sin cos sin { cos cos sin { 3 sin sin {
sin {
I
I
I
= 32 sin cos sin { cos cos sin { sin sin { · 12 (sin {)31@2 (sin {)0
I
I
I
3 sin cos sin { cos cos sin { sin sin {
I
· cos { · =
sin {
I
I
I
3 sin cos sin { cos cos sin { sin sin { cos {
I
=
sin {
51. i (w) =
I
4w + 1 i i 0 (w) =
1
(4w
2
+ 1)31@2 · 4 = 2(4w + 1)31@2
i
4
i 00 (w) = 2(3 12 )(4w + 1)33@2 · 4 = 34@(4w + 1)3@2 , so i 00 (2) = 34@93@2 = 3 27
.
i j 0 () = cos + sin · 1 i j 00 () = (3 sin ) + cos · 1 + cos = 2 cos 3 sin ,
I
I
so j 00 (@6) = 2 cos(@6) 3 (@6) sin(@6) = 2 3@2 3 (@6)(1@2) = 3 3 @12.
52. j() = sin 53. {6 + | 6 = 1
i 6{5 + 6| 5 | 0 = 0 i | 0 = 3{5@| 5
i
5{4 | 4 | 3 {(3{5@| 5 )
5{4 (| 6 + {6 )@| 5
| 5 (5{4 ) 3 {5 (5| 4 | 0 )
5{4
00
| =3
=
3
=
3
=
3
(| 5 )2
| 10
|6
| 11
54. i ({) = (2 3 {)31
i i 0 ({) = (2 3 {)32
i i 00 ({) = 2(2 3 {)33
i i 000 ({) = 2 · 3(2 3 {)34
i (4) ({) = 2 · 3 · 4(2 3 {)35 . In general, i (q) ({) = 2 · 3 · 4 · · · · · q(2 3 {)3(q+1) =
55. We ¿rst show it is true for q = 1: i ({) = {h{
i
q!
.
(2 3 {)(q+1)
i i 0 ({) = {h{ + h{ = ({ + 1)h{ . We now assume it is true
for q = n: i (n) ({) = ({ + n)h{ . With this assumption, we must show it is true for q = n + 1:
g k (n) l
g
i ({) =
[({ + n)h{ ] = ({ + n)h{ + h{ = [({ + n) + 1]h{ = [{ + (n + 1)]h{ .
i (n+1) ({) =
g{
g{
Therefore, i (q) ({) = ({ + q)h{ by mathematical induction.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
CHAPTER 3 REVIEW
56. lim
w<0
277
w3
w3 cos3 2w
cos3 2w
1
1
1
= lim
= lim =
= lim cos3 2w ·
3 =
3
3
3
w<0
w<0
w<0
tan 2w
8 · 13
8
sin 2w
sin 2w
sin 2w
8
8 lim
(2w)3
w<0
2w
i | 0 = 4 · 2 sin { cos {. At 6 > 1 , | 0 = 8 ·
I I
I
is | 3 1 = 2 3 { 3 6 , or | = 2 3 { + 1 3 3@3.
57. | = 4 sin2 {
58. | =
¤
{2 3 1
{2 + 1
i |0 =
1
2
·
I
3
2
=2
I
3, so an equation of the tangent line
({2 + 1)(2{) 3 ({2 3 1)(2{)
4{
= 2
.
({ + 1)2
({2 + 1)2
At (0> 31), | 0 = 0, so an equation of the tangent line is | + 1 = 0({ 3 0), or | = 31.
59. | =
I
2 cos {
1 + 4 sin { i |0 = 12 (1 + 4 sin {)31@2 · 4 cos { = I
.
1 + 4 sin {
2
At (0> 1), |0 = I = 2, so an equation of the tangent line is | 3 1 = 2({ 3 0), or | = 2{ + 1.
1
The slope of the normal line is 3 12 , so an equation of the normal line is | 3 1 = 3 12 ({ 3 0), or | = 3 12 { + 1.
60. {2 + 4{| + | 2 = 13
i 2{ + 4({|0 + | · 1) + 2|| 0 = 0 i { + 2{|0 + 2| + ||0 = 0 i
2{| 0 + || 0 = 3{ 3 2|
At (2> 1), | 0 =
i | 0 (2{ + |) = 3{ 3 2|
i |0 =
3{ 3 2|
.
2{ + |
4
32 3 2
= 3 , so an equation of the tangent line is | 3 1 = 3 45 ({ 3 2), or | = 3 45 { +
4+1
5
13
5 .
The slope of the normal line is 54 , so an equation of the normal line is | 3 1 = 54 ({ 3 2), or | = 54 { 3 32 .
61. | = (2 + {)h3{
i | 0 = (2 + {)(3h3{ ) + h3{ · 1 = h3{ [3(2 + {) + 1] = h3{ (3{ 3 1).
At (0> 2), | 0 = 1(31) = 31, so an equation of the tangent line is | 3 2 = 31({ 3 0), or | = 3{ + 2.
The slope of the normal line is 1, so an equation of the normal line is | 3 2 = 1({ 3 0), or | = { + 2.
62. i ({) = {hsin {
i i 0 ({) = {[hsin { (cos {)] + hsin { (1) = hsin { ({ cos { + 1). As a check on our work, we notice from the
graphs that i 0 ({) A 0 when i is increasing. Also, we see in the larger viewing rectangle a certain similarity in the graphs of i
and i 0 : the sizes of the oscillations of i and i 0 are linked.
I
53{ i
I
I
I
3{
2(5 3 {)
1
3{
2 53{
0
31@2
+ 53{· I
= I
+ I
(31) + 5 3 { = I
i ({) = { (5 3 {)
2
2 53{
2 53{
2 53{
2 53{
63. (a) i ({) = {
=
10 3 3{
3{ + 10 3 2{
I
= I
2 53{
2 53{
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NOT FOR SALE
CHAPTER 3 DIFFERENTIATION RULES
(b) At (1> 2): i 0 (1) = 74 .
(c)
So an equation of the tangent line is | 3 2 =
7
4 ({
3 1) or | =
7
4{
+
1
4.
At (4> 4): i 0 (4) = 3 22 = 31.
So an equation of the tangent line is | 3 4 = 31({ 3 4) or | = 3{ + 8.
The graphs look reasonable, since i 0 is positive where i has tangents with
(d)
positive slope, and i 0 is negative where i has tangents with negative slope.
64. (a) i ({) = 4{ 3 tan {
i i 0 ({) = 4 3 sec2 { i i 00 ({) = 32 sec { (sec { tan {) = 32 sec2 { tan {.
We can see that our answers are reasonable, since the graph of i 0 is 0 where
(b)
i has a horizontal tangent, and the graph of i 0 is positive where i has
tangents with positive slope and negative where i has tangents with
negative slope. The same correspondence holds between the graphs of i 0
and i 00 .
i | 0 = cos { 3 sin { = 0 C cos { = sin { and 0 $ { $ 2
I I are 4 > 2 and 5
>3 2 .
4
65. | = sin { + cos {
C {=
4
or
5
4 ,
so the points
i 2{ + 4||0 = 0 i |0 = 3{@(2|) = 1 C { = 32|. Since the points lie on the ellipse,
we have (32|)2 + 2| 2 = 1 i 6| 2 = 1 i | = ± I16 . The points are 3 I26 > I16 and I26 > 3 I16 .
66. {2 + 2| 2 = 1
67. i ({) = ({ 3 d)({ 3 e)({ 3 f)
So
i i 0 ({) = ({ 3 e)({ 3 f) + ({ 3 d)({ 3 f) + ({ 3 d)({ 3 e).
i 0 ({)
({ 3 e)({ 3 f) + ({ 3 d)({ 3 f) + ({ 3 d)({ 3 e)
1
1
1
=
=
+
+
.
i({)
({ 3 d)({ 3 e)({ 3 f)
{3d
{3e
{3f
Or: i({) = ({ 3 d)({ 3 e)({ 3 f) i ln |i ({)| = ln |{ 3 d| + ln |{ 3 e| + ln |{ 3 f| i
1
1
1
i 0 ({)
=
+
+
i ({)
{3d
{3e
{3f
68. (a) cos 2{ = cos2 { 3 sin2 {
i 32 sin 2{ = 32 cos { sin { 3 2 sin { cos { C sin 2{ = 2 sin { cos {
(b) sin({ + d) = sin { cos d + cos { sin d i cos({ + d) = cos { cos d 3 sin { sin d.
69. (a) k({) = i ({) j({)
i k0 ({) = i ({) j 0 ({) + j({) i 0 ({) i
k0 (2) = i(2) j 0 (2) + j(2) i 0 (2) = (3)(4) + (5)(32) = 12 3 10 = 2
(b) I ({) = i (j({)) i I 0 ({) = i 0 (j({)) j 0 ({) i I 0 (2) = i 0 (j(2)) j0 (2) = i 0 (5)(4) = 11 · 4 = 44
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CHAPTER 3 REVIEW
¤
279
i S 0 ({) = i ({) j0 ({) + j({) i 0 ({) i
033
+
(4)
= (1)(2) + (4)(31) = 2 3 4 = 32
S 0 (2) = i (2) j 0 (2) + j(2) i 0 (2) = (1) 630
330
330
70. (a) S ({) = i ({) j({)
(b) T({) =
i({)
j({)
T0 (2) =
i T0 ({) =
j({) i 0 ({) 3 i ({) j0 ({)
[j({)]2
i
j(2) i 0 (2) 3 i (2) j0 (2)
(4)(31) 3 (1)(2)
36
3
=
=
=3
[j(2)]2
42
16
8
(c) F({) = i (j({)) i F 0 ({) = i 0 (j({))j 0 ({) i
(2) = (3)(2) = 6
F 0 (2) = i 0 (j(2))j 0 (2) = i 0 (4)j 0 (2) = 630
533
i i 0 ({) = {2 j 0 ({) + j({)(2{) = {[{j0 ({) + 2j({)]
71. i ({) = {2 j({)
i i 0 ({) = j0 ({2 )(2{) = 2{j0 ({2 )
72. i ({) = j({2 )
2
73. i ({) = [ j({)]
i i 0 ({) = 2[ j({)] · j 0 ({) = 2j({) j0 ({)
74. i ({) = j(j({))
i i 0 ({) = j0 (j({)) j 0 ({)
75. i ({) = j(h{ )
i i 0 ({) = j0 (h{ ) h{
76. i ({) = hj({)
i i 0 ({) = hj({) j 0 ({)
77. i ({) = ln |j({)|
78. i ({) = j(ln {)
i i 0 ({) =
i i 0 ({) = j 0 (ln {) ·
i ({) j({)
i ({) + j({)
79. k({) =
k0 ({) =
1 0
j 0 ({)
j ({) =
j({)
j({)
j 0 (ln {)
1
=
{
{
i
[i ({) + j({)] [i ({) j 0 ({) + j({) i 0 ({)] 3 i ({) j({) [i 0 ({) + j 0 ({)]
[i ({) + j({)]2
=
[i ({)]2 j 0 ({) + i({) j({) i 0 ({) + i ({) j({) j0 ({) + [ j({)]2 i 0 ({) 3 i ({) j({) i 0 ({) 3 i ({) j({) j 0 ({)
[i({) + j({)]2
=
i 0 ({) [ j({)]2 + j 0 ({) [i ({)]2
[i ({) + j({)]2
80. k({) =
v
i ({)
j({)
i k0 ({) =
i 0 ({) j({) 3 i ({) j 0 ({)
i 0 ({) j({) 3 i ({) j0 ({)
s
s
=
2
2 i ({)@j({) [j({)]
2[j({)]3@2 i ({)
81. Using the Chain Rule repeatedly, k({) = i (j(sin 4{))
k0 ({) = i 0 (j(sin 4{)) ·
i
g
g
(j(sin 4{)) = i 0 (j(sin 4{)) · j 0 (sin 4{) ·
(sin 4{) = i 0 (j(sin 4{))j 0 (sin 4{)(cos 4{)(4).
g{
g{
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280
¤
CHAPTER 3 DIFFERENTIATION RULES
82. (a)
(b) The average rate of change is larger on [2> 3].
(c) The instantaneous rate of change (the slope of the tangent) is larger at { = 2.
(d) i ({) = { 3 2 sin { i i 0 ({) = 1 3 2 cos {,
so i 0 (2) = 1 3 2 cos 2 E 1=8323 and i 0 (5) = 1 3 2 cos 5 E 0=4327.
So i 0 (2) A i 0 (5), as predicted in part (c).
2
i | 0 = 2[ln({ + 4)]1 ·
83. | = [ln({ + 4)]
{ + 4 = h0
i
ln({ + 4)
1
·1=2
and |0 = 0 C ln({ + 4) = 0 C
{+4
{+4
{ + 4 = 1 C { = 33, so the tangent is horizontal at the point (33> 0).
when | 0 = h{ = 14 i { = ln 14 = 3 ln 4.
and the point of tangency is 3 ln 4> 14 . Thus, an equation of the tangent line
84. (a) The line { 3 4| = 1 has slope 14 . A tangent to | = h{ has slope
Since | = h{ , the |-coordinate is
is | 3
1
4
1
4
1
4
= 14 ({ + ln 4) or | = 14 { + 14 (ln 4 + 1).
g { = hd . Thus, an equation of the tangent line is
(b) The slope of the tangent at the point (d> h ) is
h
g{ { = d
d
| 3 hd = hd ({ 3 d). We substitute { = 0, | = 0 into this equation, since we want the line to pass through the origin:
0 3 hd = hd (0 3 d) C 3hd = hd (3d) C d = 1. So an equation of the tangent line at the point (d> hd ) = (1> h)
is | 3 h = h({ 3 1) or | = h{.
85. | = i({) = d{2 + e{ + f
i i 0 ({) = 2d{ + e. We know that i 0 (31) = 6 and i 0 (5) = 32, so 32d + e = 6 and
10d + e = 32. Subtracting the ¿rst equation from the second gives 12d = 38 i d = 3 23 . Substituting 3 23 for d in the
¿rst equation gives e =
14
.
3
Now i (1) = 4 i 4 = d + e + f, so f = 4 +
2
3
3
14
3
= 0 and hence, i ({) = 3 23 {2 +
14
{.
3
86. (a) lim F(w) = lim [N(h3dw 3 h3ew )] = N lim (h3dw 3 h3ew ) = N(0 3 0) = 0 because 3dw < 3" and 3ew < 3"
w<"
w<"
w<"
as w < ".
(b) F(w) = N(h3dw 3 h3ew ) i F 0 (w) = N(h3dw (3d) 3 h3ew (3e)) = N(3dh3dw + eh3ew )
(c) F 0 (w) = 0 C eh3ew = dh3dw
87. v(w) = Dh3fw cos($w + )
C
e
= h(3d+e)w
d
C ln
ln(e@d)
e
= (e 3 d)w C w =
d
e3d
i
y(w) = v0 (w) = D{h3fw [3$ sin($w + )] + cos($w + )(3fh3fw )} = 3Dh3fw [$ sin($w + ) + f cos($w + )] i
d(w) = y 0 (w) = 3D{h3fw [$2 cos($w + ) 3 f$ sin($w + )] + [$ sin($w + ) + f cos($w + )](3fh3fw )}
= 3Dh3fw [$2 cos($w + ) 3 f$ sin($w + ) 3 f$ sin($w + ) 3 f2 cos($w + )]
= 3Dh3fw [($2 3 f2 ) cos($w + ) 3 2f$ sin($w + )] = Dh3fw [(f2 3 $2 ) cos($w + ) + 2f$ sin($w + )]
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NOT FOR SALE
CHAPTER 3 REVIEW
I
I
i y(w) = {0 = 1@ 2 e2 + f2 w2 2f2 w = f2 w@ e2 + f2 w2
I
I
f2 e2 + f2 w2 3 f2 w f2 w@ e2 + f2 w2
e2 f2
d(w) = y 0 (w) =
=
2
2
2
e +f w
(e2 + f2 w2 )3@2
88. (a) { =
I
e2 + f2 w2
¤
281
i
(b) y(w) A 0 for w A 0, so the particle always moves in the positive direction.
89. (a) | = w3 3 12w + 3
i y(w) = | 0 = 3w2 3 12 i d(w) = y 0 (w) = 6w
(b) y(w) = 3(w2 3 4) A 0 when w A 2, so it moves upward when w A 2 and downward when 0 $ w ? 2.
(c) Distance upward = |(3) 3 |(2) = 36 3 (313) = 7,
Distance downward = |(0) 3 |(2) = 3 3 (313) = 16. Total distance = 7 + 16 = 23.
(e) The particle is speeding up when y and d have the same sign, that is,
(d)
when w A 2. The particle is slowing down when y and d have opposite
signs; that is, when 0 ? w ? 2.
90. (a) Y = 13 u2 k
i gY @gk = 13 u2
[u constant]
1
u2 k
3
2
uk
3
[k constant]
(b) Y =
i gY @gu =
91. The linear density is the rate of change of mass p with respect to length {.
I p = { 1 + { = { + {3@2
i = gp@g{ = 1 +
92. (a) F({) = 920 + 2{ 3 0=02{2 + 0=00007{3
3
2
I
I
{, so the linear density when { = 4 is 1 + 32 4 = 4 kg@m.
i F 0 ({) = 2 3 0=04{ + 0=00021{2
(b) F 0 (100) = 2 3 4 + 2=1 = $0=10@unit. This value represents the rate at which costs are increasing as the hundredth unit is
produced, and is the approximate cost of producing the 101st unit.
(c) The cost of producing the 101st item is F(101) 3 F(100) = 990=10107 3 990 = $0=10107, slightly larger than F 0 (100).
93. (a) |(w) = |(0)hnw = 200hnw
i |(0=5) = 200h0=5n = 360 i h0=5n = 1=8 i 0=5n = ln 1=8 i
n = 2 ln 1=8 = ln(1=8)2 = ln 3=24
i
|(w) = 200h(ln 3=24)w = 200(3=24)w
(b) |(4) = 200(3=24)4 E 22,040 bacteria
(c) |0 (w) = 200(3=24)w · ln 3=24, so | 0 (4) = 200(3=24)4 · ln 3=24 E 25,910 bacteria per hour
(d) 200(3=24)w = 10,000 i
(3=24)w = 50 i w ln 3=24 = ln 50 i w = ln 50@ ln 3=24 E 3=33 hours
94. (a) If |(w) is the mass remaining after w years, then |(w) = |(0)hnw = 100hnw . |(5=24) = 100h5=24n =
h5=24n =
1
2
1
2
· 100 i
1
i 5=24n = 3 ln 2 i n = 3 5=24
ln 2 i |(w) = 100h3(ln 2)w@5=24 = 100 · 23w@5=24 . Thus,
|(20) = 100 · 2320@5=24 E 7=1 mg.
(b) 100 · 23w@5=24 = 1 i 23w@5=24 =
1
100
i 3
w
1
ln 2 = ln
5=24
100
i w = 5=24
ln 100
E 34=8 years
ln 2
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CHAPTER 3 DIFFERENTIATION RULES
95. (a) F 0 (w) = 3nF(w)
i F(w) = F(0)h3nw by Theorem 3.8.2. But F(0) = F0 , so F(w) = F0 h3nw .
(b) F(30) = 12 F0 since the concentration is reduced by half. Thus, 12 F0 = F0 h330n
1
ln 12 =
n = 3 30
1
30
such that F(w) =
1
F .
10 0
i ln 12 = 330n
i
ln 2. Since 10% of the original concentration remains if 90% is eliminated, we want the value of w
Therefore,
96. (a) If | = x 3 20, x(0) = 80
1
F
10 0
= F0 h3w(ln 2)@30
i ln 0=1 = 3w(ln 2)@30 i w = 3 ln302 ln 0=1 E 100 h.
i |(0) = 80 3 20 = 60, and the initial-value problem is g|@gw = n| with |(0) = 60.
So the solution is |(w) = 60hnw . Now |(0=5) = 60hn(0=5) = 60 3 20 i h0=5n =
so |(w) = 60h(ln 4@9)w = 60( 49 )w . Thus, |(1) = 60( 49 )1 =
80
3
40
60
=
2
3
i n = 2 ln 23 = ln 49 ,
= 26 23 C and x(1) = 46 23 C.
w
w
4
1
4
= 20 i
=
(b) x(w) = 40 i |(w) = 20. |(w) = 60
9
9
3
i w ln
4
1
= ln
9
3
i w=
ln 13
E 1=35 h
ln 49
or 81=3 min.
97. If { = edge length, then Y = {3
i gY @gw = 3{2 g{@gw = 10 i g{@gw = 10@(3{2 ) and V = 6{2
gV@gw = (12{) g{@gw = 12{[10@(3{2 )] = 40@{. When { = 30, gV@gw =
98. Given gY @gw = 2, ¿nd gk@gw when k = 5. Y =
triangles,
2=
3
u
=
k
10
3
i Y =
9 2 gk
gY
=
k
gw
100
gw
i
3k
10
2
k=
1
u2 k
3
40
30
=
4
3
i
cm2 @min.
and, from similar
3 3
k , so
100
gk
200
200
8
=
cm@s
=
2 =
gw
9k2
9
9 (5)
when k = 5.
99. Given gk@gw = 5 and g{@gw = 15, ¿nd g}@gw. } 2 = {2 + k2
2}
g}
g{
gk
= 2{
+ 2k
gw
gw
gw
so
g}
1
=
[15(45) + 5(60)] = 13 ft@s.
gw
75
i
g}
1
= (15{ + 5k). When w = 3,
gw
}
I
k = 45 + 3(5) = 60 and { = 15(3) = 45 i } = 452 + 602 = 75,
i
100. We are given g}@gw = 30 ft@s. By similar triangles,
4
|
= I
}
241
i
g|
4
4 g}
120
}, so
E 7=7 ft@s.
|= I
= I
= I
gw
241
241 gw
241
101. We are given g@gw = 30=25 rad@h. tan = 400@{
{ = 400 cot i
i
g{
g
= 3400 csc2 . When =
gw
gw
,
6
g{
= 3400(2)2 (30=25) = 400 ft@h.
gw
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NOT FOR SALE
CHAPTER 3 REVIEW
102. (a) i ({) =
I
25 3 {2
i i 0 ({) =
32{
I
= 3{(25 3 {2 )31@2 .
2 25 3 {2
(b)
So the linear approximation to i({) near 3
is i ({) E i (3) + i 0 (3)({ 3 3) = 4 3 34 ({ 3 3).
I
(c) For the required accuracy, we want 25 3 {2 3 0=1 ? 4 3 34 ({ 3 3) and
I
4 3 34 ({ 3 3) ? 25 3 {2 + 0=1. From the graph, it appears that these both
hold for 2=24 ? { ? 3=66.
I
3
1 + 3{ = (1 + 3{)1@3
i i 0 ({) = (1 + 3{)32@3 , so the linearization of i at d = 0 is
I
O({) = i (0) + i 0 (0)({ 3 0) = 11@3 + 132@3 { = 1 + {. Thus, 3 1 + 3{ E 1 + { i
s
I
3
1=03 = 3 1 + 3(0=01) E 1 + (0=01) = 1=01.
103. (a) i ({) =
I
(b) The linear approximation is 3 1 + 3{ E 1 + {, so for the required accuracy
I
I
we want 3 1 + 3{ 3 0=1 ? 1 + { ? 3 1 + 3{ + 0=1. From the graph,
it appears that this is true when 30=235 ? { ? 0=401.
104. | = {3 3 2{2 + 1
i g| = (3{2 3 4{) g{. When { = 2 and g{ = 0=2, g| = 3(2)2 3 4(2) (0=2) = 0=8.
1 2 { = 1 + 8 {2 i gD = 2 + 4 { g{. When { = 60
2
, so the maximum error is
and g{ = 0=1, gD = 2 + 4 60(0=1) = 12 + 3
2
105. D = {2 + 12 approximately 12 +
106. lim
{<1
3
2
E 16=7 cm2 .
{17 3 1
g 17
= 17(1)16 = 17
=
{
{31
g{
{=1
I
4
1 33@4 1
1
16 + k 3 2
g I
4
107. lim
{
= {
= I
=
3 =
4
k<0
k
g{
4
32
4 16
{ = 16
{ = 16
108.
lim
<@3
I
cos 3 0=5
3
g
=
cos = 3 sin = 3
3 @3
g
3
2
= @3
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283
284
¤
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CHAPTER 3 DIFFERENTIATION RULES
I
I
I
I
I
I
1 + tan { 3 1 + sin {
1 + tan { + 1 + sin {
1 + tan { 3 1 + sin {
I
I
109. lim
=
lim
{<0
{<0
{3
{3
1 + tan { + 1 + sin {
= lim
{<0
= lim
{<0
{3
{3
cos {
(1 + tan {) 3 (1 + sin {)
sin { (1@ cos { 3 1)
= lim I
·
I
I
I
{<0 {3
cos {
1 + tan { + 1 + sin {
1 + tan { + 1 + sin {
1 + cos {
sin { (1 3 cos {)
I
I
·
1 + tan { + 1 + sin { cos { 1 + cos {
sin { · sin2 {
I
I
{<0 {3
1 + tan { + 1 + sin { cos { (1 + cos {)
3
sin {
1
I
lim I
= lim
{<0
{<0
{
1 + tan { + 1 + sin { cos { (1 + cos {)
= lim
1
1
I = 13 · I
=
4
1 + 1 · 1 · (1 + 1)
110. Differentiating the ¿rst given equation implicitly with respect to { and using the Chain Rule, we obtain i (j({)) = {
i 0 (j({)) j0 ({) = 1 i j 0 ({) =
gives j 0 ({) =
111.
i
1
. Using the second given equation to expand the denominator of this expression
i 0 (j({))
1
1
. But the ¿rst given equation states that i (j({)) = {, so j 0 ({) =
.
1 + [i(j({))]2
1 + {2
g
[i (2{)] = {2
g{
i i 0 (2{) · 2 = {2
i i 0 (2{) = 12 {2 . Let w = 2{. Then i 0 (w) =
112. Let (e> f) be on the curve, that is, e2@3 + f2@3 = d2@3 . Now {2@3 + | 2@3 = d2@3
i
1
2
1 2
w = 18 w2 , so i 0 ({) = 18 {2 .
2
2 31@3
{
3
+ 23 | 31@3
g|
= 0, so
g{
| 1@3
| 1@3
g|
, so at (e> f) the slope of the tangent line is 3(f@e)1@3 and an equation of the tangent line is
= 3 1@3 = 3
g{
{
{
| 3 f = 3(f@e)1@3 ({ 3 e) or | = 3(f@e)1@3 { + (f + e2@3 f1@3 ). Setting | = 0, we ¿nd that the {-intercept is
e1@3 f2@3 + e = e1@3 (f2@3 + e2@3 ) = e1@3 d2@3 and setting { = 0 we ¿nd that the |-intercept is
f + e2@3 f1@3 = f1@3 (f2@3 + e2@3 ) = f1@3 d2@3 . So the length of the tangent line between these two points is
s
s
I
(e1@3 d2@3 )2 + (f1@3 d2@3 )2 = e2@3 d4@3 + f2@3 d4@3 = (e2@3 + f2@3 )d4@3
I
I
= d2@3 d4@3 = d2 = d = constant
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
PROBLEMS PLUS
1. Let d be the {-coordinate of T. Since the derivative of | = 1 3 {2 is | 0 = 32{, the slope at T is 32d. But since the triangle
I
I
I
I
3@1, so the slope at T is 3 3. Therefore, we must have that 32d = 3 3 i d = 23 .
I
I
I 2 I
3
= 23 > 14 and by symmetry, S has coordinates 3 23 > 14 .
Thus, the point T has coordinates 2 > 1 3 23
is equilateral, DR@RF =
2. | = {3 3 3{ + 4
i |0 = 3{2 3 3, and | = 3({2 3 {) i | 0 = 6{ 3 3.
The slopes of the tangents of the two curves are equal when 3{2 3 3 = 6{ 3 3;
that is, when { = 0 or 2. At { = 0, both tangents have slope 33, but the curves do
not intersect. At { = 2, both tangents have slope 9 and the curves intersect at
(2> 6). So there is a common tangent line at (2> 6), | = 9{ 3 12.
We must show that u (in the ¿gure) is halfway between s and t, that is,
3.
u = (s + t)@2. For the parabola | = d{2 + e{ + f, the slope of the tangent line is
given by | 0 = 2d{ + e. An equation of the tangent line at { = s is
| 3 (ds2 + es + f) = (2ds + e)({ 3 s). Solving for | gives us
| = (2ds + e){ 3 2ds2 3 es + (ds2 + es + f)
| = (2ds + e){ + f 3 ds2
or
(1)
Similarly, an equation of the tangent line at { = t is
| = (2dt + e){ + f 3 dt 2
(2)
We can eliminate | and solve for { by subtracting equation (1) from equation (2).
[(2dt + e) 3 (2ds + e)]{ 3 dt 2 + ds2 = 0
(2dt 3 2ds){ = dt 2 3 ds2
2d(t 3 s){ = d(t 2 3 s2 )
{=
d(t + s)(t 3 s)
s+t
=
2d(t 3 s)
2
Thus, the {-coordinate of the point of intersection of the two tangent lines, namely u, is (s + t)@2.
4. We could differentiate and then simplify or we can simplify and then differentiate. The latter seems to be the simpler method.
sin2 {
sin {
cos2 {
sin2 {
cos2 {
sin3 {
cos3 {
cos {
·
+
=
+
=
+
·
cos
{
sin { cos {
1 + cot {
1 + tan {
sin {
sin { + cos { cos { + sin {
1+
1+
sin {
cos {
=
sin3 { + cos3 {
sin { + cos {
[factor sum of cubes]
=
(sin { + cos {)(sin2 { 3 sin { cos { + cos2 {)
sin { + cos {
= sin2 { 3 sin { cos { + cos2 { = 1 3 sin { cos { = 1 3 12 (2 sin { cos {) = 1 3
Thus,
g
g{
sin2 {
cos2 {
+
1 + cot { 1 + tan {
=
g 13
g{
1
2
sin 2{ = 3 12 cos 2{ · 2 = 3 cos 2{.
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1
2
sin 2{
285
286
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CHAPTER 3 PROBLEMS PLUS
5. Using i 0 (d) = lim
{<d
i ({) 3 i (d)
sec w 3 sec {
, we recognize the given expression, i ({) = lim
, as
w<{
{3d
w3{
j 0 ({) with j({) = sec {. Now i 0 ( 4 ) = j00 ( 4 ), so we will ¿nd j00 ({).
j 0 ({) = sec { tan {
j 00 ( 4 ) =
i
j 00 ({) = sec { sec2 { + tan { sec { tan { = sec {(sec2 { + tan2 {), so
I I 2
I
I
2( 2 + 12 ) = 2(2 + 1) = 3 2.
I
3
I
i ({) 3 i (0)
d{ + e 3 2
5
, we see that for the given equation, lim
=
, we have i ({) = 3 d{ + e,
6. Using i (0) = lim
{<0
{<0
{30
{
12
I
3
5
. Now i (0) = 2 C
e = 2 C e = 8. Also i 0 ({) = 13 (d{ + e)32@3 · d, so
i (0) = 2, and i 0 (0) = 12
0
i 0 (0) =
5
12
C
1
(8)32@3
3
·d=
5
12
C
1 1
( )d
3 4
=
5
12
C
d = 5.
7. Let | = tan31 {. Then tan | = {, so from the triangle we see that
{
= Using this fact we have that
sin(tan31 {) = sin | = I
1 + {2
sinh {
sinh {
sin(tan31 (sinh {)) = s
=
= tanh {.
cosh {
1 + sinh2 {
Hence, sin31 (tanh {) = sin31 (sin(tan31 (sinh {))) = tan31 (sinh {).
8. We ¿nd the equation of the parabola by substituting the point (3100> 100), at which the car is situated, into the general
equation | = d{2 : 100 = d(3100)2
i d=
({0 > |0 ). We can show that | 0 = d(2{) =
1
100 .
1
100 (2{)
=
Now we ¿nd the equation of a tangent to the parabola at the point
1
50 {,
so an equation of the tangent is | 3 |0 =
Since the point ({0 > |0 ) is on the parabola, we must have |0 =
|=
2
1
100 {0
+
1
50 {0 ({
1
2
3 {0 ).
so our equation of the tangent can be simpli¿ed to
3 {0 ). We want the statue to be located on the tangent line, so we substitute its coordinates (100> 50)
into this equation: 50 =
{0 =
2
1
100 {0 ,
1
50 {0 ({
2
1
100 {0
+
1
50 {0 (100
3 {0 ) i {20 3 200{0 + 5000 = 0 i
k
l
s
I
200 ± 2002 3 4(5000)
i {0 = 100 ± 50 2. But {0 ? 100, so the car’s headlights illuminate the statue
I I
when it is located at the point 100 3 50 2> 150 3 100 2 E (29=3> 8=6), that is, about 29=3 m east and 8=6 m north of
the origin.
9. We use mathematical induction. Let Vq be the statement that
gq
(sin4 { + cos4 {) = 4q31 cos(4{ + q@2).
g{q
V1 is true because
g
(sin4 { + cos4 {) = 4 sin3 { cos { 3 4 cos3 { sin { = 4 sin { cos { sin2 { 3 cos2 { {
g{
= 34 sin { cos { cos 2{ = 32 sin 2{ cos 2 = 3 sin 4{ = sin(34{)
= cos 2 3 (34{) = cos 2 + 4{ = 4q31 cos 4{ + q 2 when q = 1
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[continued]
NOT FOR SALE
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CHAPTER 3 PROBLEMS PLUS
287
gn 4
sin { + cos4 { = 4n31 cos 4{ + n 2 . Then
n
g{
n
g
g n31
g
4
4
(sin4 { + cos4 {) =
(sin
{
+
cos
{)
=
cos 4{ + n 2
4
n
g{ g{
g{
Now assume Vn is true, that is,
gn+1
g{n+1
which shows that Vn+1 is true.
Therefore,
g 4{ + n 2 = 34n sin 4{ + n 2
= 34n31 sin 4{ + n 2 ·
g{
n
n
= 4 sin 34{ 3 n 2 = 4 cos 2 3 34{ 3 n 2 = 4n cos 4{ + (n + 1) 2
gq
(sin4 { + cos4 {) = 4q31 cos 4{ + q 2 for every positive integer q, by mathematical induction.
g{q
Another proof: First write
sin4 { + cos4 { = (sin2 { + cos2 {)2 3 2 sin2 { cos2 { = 1 3
Then we have
gq
gq
(sin4 { + cos4 {) =
q
g{
g{q
1
2
sin2 2{ = 1 3 14 (1 3 cos 4{) =
3
4
+
1
4
cos 4{
3
1
1
+ cos 4{ = · 4q cos 4{ + q
= 4q31 cos 4{ + q .
4
4
4
2
2
10. If we divide 1 3 { into {q by long division, we ¿nd that i ({) =
{q
1
= 3{q31 3 {q32 3 · · · 3 { 3 1 +
.
13{
13{
This can also be seen by multiplying the last expression by 1 3 { and canceling terms on the right-hand side. So we let
(q)
1
1
3 j (q) ({).
j({) = 1 + { + {2 + · · · + {q31 , so that i ({) =
3 j({) i i (q) ({) =
13{
13{
(q)
1
. Now
But j is a polynomial of degree (q 3 1), so its qth derivative is 0, and therefore i (q) ({) =
13{
g
g2
(1 3 {)31 = (31)(1 3 {)32 (31) = (1 3 {)32 , 2 (1 3 {)31 = (32)(1 3 {)33 (31) = 2(1 3 {)33 ,
g{
g{
4
g3
31
34
34 g
(1
3
{)
=
(33)
·
2(1
3
{)
(31)
=
3
·
2(1
3
{)
,
(1 3 {)31 = 4 · 3 · 2(1 3 {)35 , and so on.
g{3
g{4
(q)
1
q!
=
.
So after q differentiations, we will have i (q) ({) =
13{
(1 3 {)q+1
11. We must ¿nd a value {0 such that the normal lines to the parabola | = {2 at { = ±{0 intersect at a point one unit from the
points ±{0 > {20 . The normals to | = {2 at { = ±{0 have slopes 3
1
and pass through ±{0 > {20 respectively, so the
±2{0
1
1
1
({ 3 {0 ) and | 3 {20 =
({ + {0 ). The common |-intercept is {20 + .
2{0
2{0
2
We want to ¿nd the value of {0 for which the distance from 0> {20 + 12 to {0 > {20 equals 1. The square of the distance is
normals have the equations | 3 {20 = 3
2
= {20 +
({0 3 0)2 + {20 3 {20 + 12
the center of the circle is at 0> 54 .
1
4
= 1 C {0 = ±
I
3
2 .
For these values of {0 , the |-intercept is {20 +
1
2
= 54 , so
Another solution: Let the center of the circle be (0> d). Then the equation of the circle is {2 + (| 3 d)2 = 1.
Solving with the equation of the parabola, | = {2 , we get {2 + ({2 3 d)2 = 1 C {2 + {4 3 2d{2 + d2 = 1 C
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CHAPTER 3 PROBLEMS PLUS
{4 + (1 3 2d){2 + d2 3 1 = 0. The parabola and the circle will be tangent to each other when this quadratic equation in {2
has equal roots; that is, when the discriminant is 0. Thus, (1 3 2d)2 3 4(d2 3 1) = 0 C
1 3 4d + 4d2 3 4d2 + 4 = 0 C 4d = 5, so d = 54 . The center of the circle is 0> 54 .
%
I
I &
I i ({) 3 i (d)
i ({) 3 i (d) I
i ({) 3 i (d)
{+ d
I
I
I ·I
I
= lim
·
12. lim I
{+ d
= lim
{<d
{<d
{<d
{3d
{3 d
{3 d
{+ d
= lim
{<d
I
I
I I I
i ({) 3 i (d)
· lim
{ + d = i 0 (d) ·
d + d = 2 d i 0 (d)
{<d
{3d
13. We can assume without loss of generality that = 0 at time w = 0, so that = 12w rad. [The angular velocity of the wheel
is 360 rpm = 360 · (2 rad)@(60 s) = 12 rad@s.] Then the position of D as a function of time is
40 sin sin 1
|
=
=
= sin 12w.
1=2 m
120
3
3
1
g
= · 12 · cos 12w = 4 cos . When = , we have
(a) Differentiating the expression for sin , we get cos ·
gw
3
3
v
I
I
I 2 u
4
cos
4 3
3
3
11
1
g
2
3
= I
=
sin = sin =
E 6=56 rad@s.
, so cos = 1 3
and
=
= s
3
6
6
12
gw
cos 11
11@12
D = (40 cos > 40 sin ) = (40 cos 12w> 40 sin 12w), so sin =
(b) By the Law of Cosines, |DS |2 = |RD|2 + |RS |2 3 2 |RD| |RS | cos i
1202 = 402 + |RS |2 3 2 · 40 |RS | cos i |RS |2 3 (80 cos ) |RS | 3 12,800 = 0 i
I
I
I
|RS | = 12 80 cos ± 6400 cos2 + 51,200 = 40 cos ± 40 cos2 + 8 = 40 cos + 8 + cos2 cm
[since |RS | A 0]. As a check, note that |RS | = 160 cm when = 0 and |RS | = 80
I
2 cm when =
.
2
I
(c) By part (b), the {-coordinate of S is given by { = 40 cos + 8 + cos2 , so
g{
g{ g
2 cos sin cos =
= 40 3 sin 3 I
· 12 = 3480 sin 1 + I
cm@s.
gw
g gw
2 8 + cos2 8 + cos2 In particular, g{@gw = 0 cm@s when = 0 and g{@gw = 3480 cm@s when =
2.
14. The equation of W1 is | 3 {21 = 2{1 ({ 3 {1 ) = 2{1 { 3 2{21 or | = 2{1 { 3 {21 .
The equation of W2 is | = 2{2 { 3 {22 . Solving for the point of intersection, we
get 2{({1 3 {2 ) = {21 3 {22 i { = 12 ({1 + {2 ). Therefore, the coordinates
of S are 12 ({1 + {2 )> {1 {2 . So if the point of contact of W is d> d2 , then
T1 is 12 (d + {1 )> d{1 and T2 is 12 (d + {2 )> d{2 . Therefore,
|S T1 |2 = 14 (d 3 {2 )2 + {21 (d 3 {2 )2 = (d 3 {2 )2 14 + {21 and
|S S1 |2 = 14 ({1 3 {2 )2 + {21 ({1 3 {2 )2 = ({1 3 {2 )2 14 + {21 .
So
|S T1 |2
(d 3 {2 )2
|S T2 |2
({1 3 d)2
|S T1 |
{1 3 d
|S T2 |
d 3 {2
. Finally,
+
= 1.
+
=
2 =
2 , and similarly
2 =
|S S1 |
|S S2 |
{1 3 {2
{1 3 {2
|S S1 |
({1 3 {2 )
|S S2 |
({1 3 {2 )2
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CHAPTER 3 PROBLEMS PLUS
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289
q
15. Consider the statement that
g
(hd{ sin e{) = uq hd{ sin(e{ + q). For q = 1,
g{q
g d{
(h sin e{) = dhd{ sin e{ + ehd{ cos e{, and
g{
d{
uh
d{
d{
sin(e{ + ) = uh [sin e{ cos + cos e{ sin ] = uh
since tan =
e
d
i sin =
d
e
sin e{ + cos e{ = dhd{ sin e{ + ehd{ cos e{
u
u
e
d
and cos = . So the statement is true for q = 1.
u
u
Assume it is true for q = n. Then
gn+1 d{
g n d{
u h sin(e{ + n) = un dhd{ sin(e{ + n) + un hd{ e cos(e{ + n)
(h sin e{) =
n+1
g{
g{
= un hd{ [d sin(e{ + n) + e cos(e{ + n)]
But
sin[e{ + (n + 1)] = sin[(e{ + n) + ] = sin(e{ + n) cos + sin cos(e{ + n) =
d
u
sin(e{ + n) +
e
u
cos(e{ + n).
Hence, d sin(e{ + n) + e cos(e{ + n) = u sin[e{ + (n + 1)]. So
gn+1 d{
(h sin e{) = un hd{ [d sin(e{+ n)+ e cos(e{+n)] = un hd{ [u sin(e{ +(n + 1))] = un+1 hd{ [sin(e{+ (n + 1))].
g{n+1
Therefore, the statement is true for all q by mathematical induction.
16. We recognize this limit as the de¿nition of the derivative of the function i ({) = hsin { at { = , since it is of the form
lim
{<
i ({) 3 i ()
. Therefore, the limit is equal to i 0 () = (cos )hsin = 31 · h0 = 31.
{3
17. It seems from the ¿gure that as S approaches the point (0> 2) from the right, {W < " and |W < 2+ . As S approaches the
point (3> 0) from the left, it appears that {W < 3+ and |W < ". So we guess that {W M (3> ") and |W M (2> "). It is
more dif¿cult to estimate the range of values for {Q and |Q . We might perhaps guess that {Q M (0> 3),
and |Q M (3"> 0) or (32> 0).
In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to ¿nd the equation of the
tangent line:
{2
|2
+
=1 i
9
4
tangent line is | 3 |0 = 3
2{
2| 0
4{
+
| = 0, so |0 = 3
. So at the point ({0 > |0 ) on the ellipse, an equation of the
9
4
9|
{2
|2
4 {0
{0 { |0 |
+
= 0 + 0 = 1,
({ 3 {0 ) or 4{0 { + 9|0 | = 4{20 + 9|02 . This can be written as
9 |0
9
4
9
4
because ({0 > |0 ) lies on the ellipse. So an equation of the tangent line is
Therefore, the {-intercept {W for the tangent line is given by
by
{0 { |0 |
+
= 1.
9
4
{0 {W
9
, and the |-intercept |W is given
= 1 C {W =
9
{0
4
|0 |W
= 1 C |W = .
4
|0
So as {0 takes on all values in (0> 3), {W takes on all values in (3> "), and as |0 takes on all values in (0> 2), |W takes on
all values in (2> ").
At the point ({0 > |0 ) on the ellipse, the slope of the normal line is 3
1
9 |0
, and its
=
| 0 ({0 > |0 )
4 {0
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CHAPTER 3 PROBLEMS PLUS
9 |0
9 |0
({ 3 {0 ). So the {-intercept {Q for the normal line is given by 0 3 |0 =
({Q 3 {0 ) i
4 {0
4 {0
equation is | 3 |0 =
4{0
5{0
9 |0
9|0
5|0
+ {0 =
, and the |-intercept |Q is given by |Q 3 |0 =
+ |0 = 3
.
(0 3 {0 ) i |Q = 3
9
9
4 {0
4
4
So as {0 takes on all values in (0> 3), {Q takes on all values in 0> 53 , and as |0 takes on all values in (0> 2), |Q takes on
all values in 3 52 > 0 .
{Q = 3
18. lim
{<0
sin(3 + {)2 3 sin 9
= i 0 (3) where i({) = sin {2 . Now i 0 ({) = (cos {2 )(2{), so i 0 (3) = 6 cos 9.
{
19. (a) If the two lines O1 and O2 have slopes p1 and p2 and angles of
inclination !1 and !2 , then p1 = tan !1 and p2 = tan !2 . The triangle
in the ¿gure shows that !1 + + (180 3 !2 ) = 180 and so
= !2 3 !1 . Therefore, using the identity for tan({ 3 |), we have
tan = tan(!2 3 !1 ) =
tan !2 3 tan !1
p2 3 p1
and so tan =
.
1 + tan !2 tan !1
1 + p1 p2
(b) (i) The parabolas intersect when {2 = ({ 3 2)2
i { = 1. If | = {2 , then | 0 = 2{, so the slope of the tangent
to | = {2 at (1> 1) is p1 = 2(1) = 2. If | = ({ 3 2)2 , then | 0 = 2({ 3 2), so the slope of the tangent to
| = ({ 3 2)2 at (1> 1) is p2 = 2(1 3 2) = 32. Therefore, tan =
so = tan31
4
3
4
p2 3 p1
32 3 2
= and
=
1 + p1 p2
1 + 2(32)
3
E 53 [or 127 ].
(ii) {2 3 | 2 = 3 and {2 3 4{ + | 2 + 3 = 0 intersect when {2 3 4{ + ({2 3 3) + 3 = 0 C 2{({ 3 2) = 0 i
{ = 0 or 2, but 0 is extraneous. If { = 2, then | = ±1. If {2 3 | 2 = 3 then 2{ 3 2|| 0 = 0 i | 0 = {@| and
{2 3 4{ + | 2 + 3 = 0 i 2{ 3 4 + 2|| 0 = 0 i |0 =
p2 = 0, so tan =
so tan =
20. | 2 = 4s{
p2 =
032
1+2·0
23{
. At (2> 1) the slopes are p1 = 2 and
|
= 32 i E 117 . At (2> 31) the slopes are p1 = 32 and p2 = 0>
0 3 (32)
= 2 i E 63 [or 117 ].
1 + (32)(0)
i 2||0 = 4s i | 0 = 2s@|
i slope of tangent at S ({1 > |1 ) is p1 = 2s@|1 . The slope of I S is
|1
, so by the formula from Problem 19(a),
{1 3 s
2s
|1
3
| 2 3 2s({1 3 s)
| ({ 3 s)
{1 3 s
|
1 · 1 1
= 1
tan =
|1 ({1 3 s)
|1 ({1 3 s) + 2s|1
2s
|1
1+
|1
{1 3 s
=
4s{1 3 2s{1 + 2s2
2s(s + {1 )
2s
=
=
{1 |1 3 s|1 + 2s|1
|1 (s + {1 )
|1
= slope of tangent at S = tan Since 0 $ , $
2,
this proves that = .
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CHAPTER 3 PROBLEMS PLUS
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291
21. Since _URT = _RTS = , the triangle TRU is isosceles, so
|TU| = |UR| = {. By the Law of Cosines, {2 = {2 + u2 3 2u{ cos . Hence,
2u{ cos = u2 , so { =
u
u2
=
. Note that as | < 0+ , < 0+ (since
2u cos 2 cos sin = |@u), and hence { <
u
u
= . Thus, as S is taken closer and closer
2 cos 0
2
to the {-axis, the point U approaches the midpoint of the radius DR.
i ({) 3 i (0)
i({) 3 i(0)
lim
i ({)
i({) 3 0
i ({) 3 i (0)
i 0 (0)
{<0
{
3
0
{30
=
= 0
= lim
= lim
= lim
22. lim
{<0 j({)
{<0 j({) 3 0
{<0 j({) 3 j(0)
{<0 j({) 3 j(0)
j({) 3 j(0)
j (0)
lim
{<0
{30
{30
23. lim
{<0
sin(d + 2{) 3 2 sin(d + {) + sin d
{2
sin d cos 2{ + cos d sin 2{ 3 2 sin d cos { 3 2 cos d sin { + sin d
= lim
{<0
{2
sin d (cos 2{ 3 2 cos { + 1) + cos d (sin 2{ 3 2 sin {)
= lim
{<0
{2
sin d (2 cos2 { 3 1 3 2 cos { + 1) + cos d (2 sin { cos { 3 2 sin {)
{<0
{2
sin d (2 cos {)(cos { 3 1) + cos d (2 sin {)(cos { 3 1)
= lim
{<0
{2
2(cos { 3 1)[sin d cos { + cos d sin {](cos { + 1)
= lim
{<0
{2 (cos { + 1)
2
32 sin2 { [sin(d + {)]
sin(d + 0)
sin(d + {)
sin {
·
= lim
=
32
lim
= 32(1)2
= 3 sin d
{<0
{<0
{2 (cos { + 1)
{
cos { + 1
cos 0 + 1
= lim
24. (a) i ({) = {({ 3 2)({ 3 6) = {3 3 8{2 + 12{
i
i 0 ({) = 3{2 3 16{ + 12. The average of the ¿rst pair of zeros is
(0 + 2)@2 = 1. At { = 1, the slope of the tangent line is i 0 (1) = 31, so an
equation of the tangent line has the form | = 31{ + e. Since i (1) = 5, we
have 5 = 31 + e i e = 6 and the tangent has equation | = 3{ + 6.
2+6
0+6
= 3, | = 39{ + 18; at { =
= 4, | = 34{. From the graph, we see that each tangent line
Similarly, at { =
2
2
drawn at the average of two zeros intersects the graph of i at the third zero.
(b) A CAS gives i 0 ({) = ({ 3 e)({ 3 f) + ({ 3 d)({ 3 f) + ({ 3 d)({ 3 e) or
i 0 ({) = 3{2 3 2(d + e + f){ + de + df + ef. Using the Simplify command, we get
(d 3 e)2
d+e
(d 3 e)2
d+e
0 d+e
i
=3
and i
=3
(d + e 3 2f), so an equation of the tangent line at { =
2
4
2
8
2
(d 3 e)2
(d 3 e)2
d+e
is | = 3
{3
3
(d + e 3 2f). To ¿nd the {-intercept, let | = 0 and use the Solve
4
2
8
command. The result is { = f=
[continued]
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CHAPTER 3 PROBLEMS PLUS
Using Derive, we can begin by authoring the expression ({ 3 d)({ 3 e)({ 3 f). Now load the utility ¿le
DifferentiationApplications. Next we author tangent (#1> {> (d + e)@2)—this is the command to ¿nd an
equation of the tangent line of the function in #1 whose independent variable is { at the {-value (d + e)@2. We then
simplify that expression and obtain the equation | = #4. The form in expression #4 makes it easy to see that the
{-intercept is the third zero, namely f. In a similar fashion we see that e is the {-intercept for the tangent line at (d + f)@2
and d is the {-intercept for the tangent line at (e + f)@2.
Let i ({) = h2{ and j({) = n
25.
I
{ [n A 0]. From the graphs of i and j,
we see that i will intersect j exactly once when i and j share a tangent
line. Thus, we must have i = j and i 0 = j 0 at { = d.
i (d) = j(d) i h2d = n
and
So we must have n
n = 2h1@2 = 2
I
d=
n
I
4 d
I
h E 3=297.
i
I 2
n
d
=
4n
i 0 (d) = j 0 (d) i 2h2d =
2
I
d (B)
n
I
d
i d = 14 . From (B), h2(1@4) = n
i h2d =
4
n
I .
d
s
1@4 i
26. We see that at { = 0, i ({) = d{ = 1 + { = 1, so if | = d{ is to lie above | = 1 + {,
the two curves must just touch at (0> 1), that is, we must have i 0 (0) = 1. [To see this
analytically, note that d{ D 1 + { i d{ 3 1 D { i
d{ 3 1
D 1 for { A 0, so
{
i 0 (0) = lim
d{ 3 1
D 1. Similarly, for { ? 0, d{ 3 1 D { i
{
i 0 (0) = lim
d{ 3 1
$ 1.
{
{<0+
{<0
d{ 3 1
$ 1, so
{
Since 1 $ i 0 (0) $ 1, we must have i 0 (0) = 1.] But i 0 ({) = d{ ln d i i 0 (0) = ln d, so we have ln d = 1 C d = h.
Another method: The inequality certainly holds for { $ 31, so consider { A 31, { 6= 0. Then d{ D 1 + { i
d D (1 + {)1@{ for { A 0 i d D lim (1 + {)1@{ = h, by Equation 3.6.5. Also, d{ D 1 + { i d $ (1 + {)1@{
{<0+
for { ? 0 i d $ lim (1 + {)1@{ = h. So since h $ d $ h, we must have d = h.
{<0
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CHAPTER 3 PROBLEMS PLUS
27. | = I
¤
293
I
{
2
sin {
I
3I
arctan
. Let n = d + d2 3 1. Then
2
2
31
d 31
d + d 3 1 + cos {
d2
cos {(n + cos {) + sin2 {
2
1
1
3I
·
·
|0 = I
2
(n + cos {)2
d2 3 1
d2 3 1 1 + sin {@(n + cos {)2
2
n cos { + cos2 { + sin2 {
2
n cos { + 1
1
1
3I
·
3I
·
= I
= I
2
2
(n + cos {)2 + sin2 {
d 31
d 31
d2 3 1
d2 3 1 n2 + 2n cos { + 1
=
n2 3 1
n2 + 2n cos { + 1 3 2n cos { 3 2
I
= I
2
2
2
d 3 1 (n + 2n cos { + 1)
d 3 1 (n2 + 2n cos { + 1)
But n2 = 2d2 + 2d
I
I
d2 3 1 3 1 = 2d d + d2 3 1 3 1 = 2dn 3 1, so n2 + 1 = 2dn, and n2 3 1 = 2(dn 3 1).
I
I
2(dn 3 1)
dn 3 1
= I
. But dn 3 1 = d2 + d d2 3 1 3 1 = n d2 3 1,
So | 0 = I
2
2
d 3 1 (2dn + 2n cos {)
d 3 1n (d + cos {)
so | 0 = 1@(d + cos {).
28. Suppose that | = p{ + f is a tangent line to the ellipse. Then it intersects the ellipse at only one point, so the discriminant
of the equation
{2
(p{ + f)2
+
= 1 C (e2 + d2 p2 ){2 + 2pfd2 { + d2 f2 3 d2 e2 = 0 must be 0; that is,
d2
e2
0 = (2pfd2 )2 3 4(e2 + d2 p2 )(d2 f2 3 d2 e2 ) = 4d4 f2 p2 3 4d2 e2 f2 + 4d2 e4 3 4d4 f2 p2 + 4d4 e2 p2
= 4d2 e2 (d2 p2 + e2 3 f2 )
Therefore, d2 p2 + e2 3 f2 = 0.
Now if a point (> ) lies on the line | = p{ + f, then f = 3 p, so from above,
0 = d2 p2 + e2 3 ( 3 p)2 = (d2 3 2 )p2 + 2p + e2 3 2
C p2 +
2
e2 3 2
p
+
= 0.
d2 3 2
d2 3 2
(a) Suppose that the two tangent lines from the point (> ) to the ellipse
1
1
. Then p and
are roots of the equation
p
p
2
e2 3 2
1
=0 C
}
+
=
0.
This
implies
that
(}
3
p)
}
3
}2 + 2
d 3 2
d2 3 2
p
1
1
}2 3 p +
}+p
= 0, so equating the constant terms in the two
p
p
1
e2 3 2
= 1, and hence e2 3 2 = d2 3 2 . So (> ) lies on the
quadratic equations, we get 2
=
p
d 3 2
p
have slopes p and
hyperbola {2 3 | 2 = d2 3 e2 .
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CHAPTER 3 PROBLEMS PLUS
(b) If the two tangent lines from the point (> ) to the ellipse have slopes p
1
1
, then p and 3 are roots of the quadratic equation, and so
p
p
1
= 0, and equating the constant terms as in part (a), we get
(} 3 p) } +
p
and 3
e2 3 2
= 31, and hence e2 3 2 = 2 3 d2 . So the point (> ) lies on the
d2 3 2
circle {2 + | 2 = d2 + e2 .
29. | = {4 3 2{2 3 {
i | 0 = 4{3 3 4{ 3 1. The equation of the tangent line at { = d is
| 3 (d4 3 2d2 3 d) = (4d3 3 4d 3 1)({ 3 d) or | = (4d3 3 4d 3 1){ + (33d4 + 2d2 ) and similarly for { = e. So if at
{ = d and { = e we have the same tangent line, then 4d3 3 4d 3 1 = 4e3 3 4e 3 1 and 33d4 + 2d2 = 33e4 + 2e2 . The ¿rst
equation gives d3 3 e3 = d 3 e i (d 3 e)(d2 + de + e2 ) = (d 3 e). Assuming d 6= e, we have 1 = d2 + de + e2 .
The second equation gives 3(d4 3 e4 ) = 2(d2 3 e2 ) i 3(d2 3 e2 )(d2 + e2 ) = 2(d2 3 e2 ) which is true if d = 3e.
Substituting into 1 = d2 + de + e2 gives 1 = d2 3 d2 + d2
i d = ±1 so that d = 1 and e = 31 or vice versa. Thus,
the points (1> 32) and (31> 0) have a common tangent line.
As long as there are only two such points, we are done. So we show that these are in fact the only two such points.
Suppose that d2 3 e2 6= 0. Then 3(d2 3 e2 )(d2 + e2 ) = 2(d2 3 e2 ) gives 3(d2 + e2 ) = 2 or d2 + e2 = 23 .
Thus, de = (d2 + de + e2 ) 3 (d2 + e2 ) = 1 3
1
1
2
1
2
= , so e =
. Hence, d2 + 2 = , so 9d4 + 1 = 6d2
3
3
3d
9d
3
0 = 9d4 3 6d2 + 1 = (3d2 3 1)2 . So 3d2 3 1 = 0 i d2 =
1
3
i e2 =
i
1
1
= = d2 , contradicting our assumption
9d2
3
that d2 6= e2 .
30. Suppose that the normal lines at the three points d1 > d21 , d2 > d22 , and d3 > d23 intersect at a common point. Now if one of
the dl is 0 (suppose d1 = 0) then by symmetry d2 = 3d3 , so d1 + d2 + d3 = 0. So we can assume that none of the dl is 0.
1
The slope of the tangent line at dl > d2l is 2dl , so the slope of the normal line is 3
and its equation is
2dl
1
({ 3 dl ). We solve for the {-coordinate of the intersection of the normal lines from d1 > d21 and d2 > d22 :
2dl
1
1
1
1
= d22 3 d21 i
| = d21 3
({ 3 d1 ) = d22 3
({ 3 d2 ) i {
3
2d1
2d2
2d2
2d1
d1 3 d2
{
= (3d1 3 d2 )(d1 + d2 ) C { = 32d1 d2 (d1 + d2 ) (1). Similarly, solving for the {-coordinate of the
2d1 d2
intersections of the normal lines from d1 > d21 and d3 > d23 gives { = 32d1 d3 (d1 + d3 ) (2).
| 3 d2l = 3
Equating (1) and (2) gives d2 (d1 + d2 ) = d3 (d1 + d3 ) C d1 (d2 3 d3 ) = d23 3 d22 = 3(d2 + d3 )(d2 3 d3 ) C
d1 = 3(d2 + d3 ) C d1 + d2 + d3 = 0.
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CHAPTER 3 PROBLEMS PLUS
¤
295
31. Because of the periodic nature of the lattice points, it suf¿ces to consider the points in the 5 × 2 grid shown. We can see that
the minimum value of u occurs when there is a line with slope
2
5
which touches the circle centered at (3> 1) and the circles
centered at (0> 0) and (5> 2).
To ¿nd S , the point at which the line is tangent to the circle at (0> 0), we simultaneously solve {2 + | 2 = u2 and
| = 3 52 { i {2 +
25
4
{2 = u2
i {2 =
4
29
u2
I2
29
i {=
u, | = 3 I529 u. To ¿nd T, we either use symmetry or
solve ({ 3 3)2 + (| 3 1)2 = u2 and | 3 1 = 3 52 ({ 3 3). As above, we get { = 3 3
the line S T is
5
2
5,
1+
so pS T =
I
I
29 + 50u = 6 29 3 8u
I5
29
33
u 3 3 I529 u
I2
29
u3
C 58u =
I2
29
u
=
1+
33
I
29 C u =
I10
29
I4
29
I
29
58 .
u, | = 1 +
I
29 + 10u
2
= I
=
5
u
3 29 3 4u
u
I5
29
u. Now the slope of
i
So the minimum value of u for which any line with slope
intersects circles with radius u centered at the lattice points on the plane is u =
32.
I2
29
I
29
58
E 0=093.
Assume the axes of the cone and the cylinder are parallel. Let K denote the initial
height of the water. When the cone has been dropping for w seconds, the water level has
risen { centimeters, so the tip of the cone is { + 1w centimeters below the water line.
We want to ¿nd g{@gw when { + w = k (when the cone is completely submerged).
Using similar triangles,
volume of water and cone at time w
2
=
u
u1
=
{+w
k
original volume of water
2
+
U (K + {)
=
U K
+
U2 K + U2 {
=
U2 K
+
3k2 U2 {
=
u2 ({ + w)3
Differentiating implicitly with respect to w gives us 3k2 U2
u2 ({ + w)2
g{
= 2 2
gw
k U 3 u2 ({ + w)2
U2
i u1 =
i
u
({ + w).
k
volume of submerged part of cone
1
u12 ({
3
2
1
3
+ w)
u
({ + w)3
k2
g{
g{
gw
= u2 3({ + w)2
+ 3({ + w)2
gw
gw
gw
i
g{ u2 k2
u2
= 2 2
= 2
. Thus, the water level is rising at a rate of
2
2
gw { + w = k
k U 3u k
U 3 u2
u2
cm@s at the instant the cone is completely submerged.
3 u2
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5
296
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CHAPTER 3 PROBLEMS PLUS
k
5k
u
=
i u=
. The volume of the cone is
5
16
16
2
5k
25 3
gY
25 2 gk
k=
Y = 13 u2 k = 13 k , so
=
k
. Now the rate of
16
768
gw
256
gw
33.
By similar triangles,
change of the volume is also equal to the difference of what is being added
(2 cm3 @min) and what is oozing out (nuo, where uo is the area of the cone and n
is a proportionality constant). Thus,
gY
= 2 3 nuo.
gw
gk
5(10)
25
o
10
gY
and substituting k = 10,
= 30=3, u =
=
, and I
C
=
gw
gw
16
8
16
281
I
25
25 5 I
125n 281
5I
750
2
281, we get
281 C
o=
(10) (30=3) = 2 3 n
·
=2+
. Solving for n gives us
8
256
8 8
64
256
Equating the two expressions for
n=
256 + 375
I
. To maintain a certain height, the rate of oozing, nuo, must equal the rate of the liquid being poured in;
250 281
that is,
gY
= 0. Thus, the rate at which we should pour the liquid into the container is
gw
I
256 + 375
256 + 375
25 5 281
I
·
=
E 11=204 cm3@min
nuo =
··
8
8
128
250 281
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4
APPLICATIONS OF DIFFERENTIATION
4.1 Maximum and Minimum Values
1. A function i has an absolute minimum at { = f if i(f) is the smallest function value on the entire domain of i, whereas
i has a local minimum at f if i(f) is the smallest function value when { is near f.
2. (a) The Extreme Value Theorem
(b) See the Closed Interval Method.
3. Absolute maximum at v, absolute minimum at u, local maximum at f, local minima at e and u, neither a maximum nor a
minimum at d and g.
4. Absolute maximum at u; absolute minimum at d; local maxima at e and u; local minimum at g; neither a maximum nor a
minimum at f and v.
5. Absolute maximum value is i(4) = 5; there is no absolute minimum value; local maximum values are i (4) = 5 and
i (6) = 4; local minimum values are i (2) = 2 and i(1) = i (5) = 3.
6. There is no absolute maximum value; absolute minimum value is j(4) = 1; local maximum values are j(3) = 4 and
j(6) = 3; local minimum values are j(2) = 2 and j(4) = 1.
7. Absolute minimum at 2, absolute maximum at 3,
local minimum at 4
8. Absolute minimum at 1, absolute maximum at 5,
local maximum at 2, local minimum at 4
9. Absolute maximum at 5, absolute minimum at 2,
local maximum at 3, local minima at 2 and 4
10. i has no local maximum or minimum, but 2 and 4 are
critical numbers
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298
¤
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
11. (a)
(b)
12. (a) Note that a local maximum cannot
(c)
(b)
occur at an endpoint.
Note: By the Extreme Value Theorem, i must not be continuous.
13. (a) Note: By the Extreme Value Theorem,
(b)
i must not be continuous; because if it
were, it would attain an absolute
minimum.
14. (a)
15. i ({) =
(b)
1
(3{
2
3 1), { $ 3. Absolute maximum
16. i ({) = 2 3 13 {, { D 32. Absolute maximum
i (3) = 4; no local maximum. No absolute or local
i (32) = 83 ; no local maximum. No absolute or local
minimum.
minimum.
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SECTION 4.1
17. i ({) = 1@{, { D 1. Absolute maximum i (1) = 1;
no local maximum. No absolute or local minimum.
19. i ({) = sin {, 0 $ { ? @2. No absolute or local
No absolute or local minimum.
20. i ({) = sin {, 0 ? { $ @2. Absolute maximum
i
minimum.
minimum.
i
= 1; no local maximum. Absolute minimum
2
i 3 2 = 31; no local minimum.
23. i ({) = ln {, 0 ? { $ 2. Absolute maximum
i (2) = ln 2 r 0=69; no local maximum. No absolute
¤
299
18. i ({) = 1@{, 1 ? { ? 3. No absolute or local maximum.
maximum. Absolute minimum i(0) = 0; no local
21. i ({) = sin {, 3@2 $ { $ @2. Absolute maximum
MAXIMUM AND MINIMUM VALUES
2
= 1; no local maximum. No absolute or local
22. i (w) = cos w, 3 3
$w$
2
3
.
2
Absolute and local
maximum i(0) = 1; absolute and local minima
i (±> 31).
24. i ({) =| { |. No absolute or local maximum. Absolute
and local minimum i (0) = 0.
or local minimum.
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CHAPTER 4
25. i ({) = 1 3
APPLICATIONS OF DIFFERENTIATION
I
{. Absolute maximum i (0) = 1;
26. i ({) = h{ . No absolute or local maximum or
no local maximum. No absolute or local minimum.
27. i ({) =
+
13{
if 0 $ { ? 2
2{ 3 4
if 2 $ { $ 3
minimum value.
28. i ({) =
+
4 3 {2
if 3 2 $ { ? 0
2{ 3 1
if 0 $ { $ 2
Absolute maximum i(3) = 2; no local maximum.
Absolute and local minimum i (0) = 31.
No absolute or local minimum.
No absolute or local maximum.
i i 0 ({) =
29. i ({) = 4 + 13 { 3 12 {2
30. i ({) = {3 + 6{2 3 15{
1
3
3 {. i 0 ({) = 0 i { = 13 . This is the only critical number.
i i 0 ({) = 3{2 + 12{ 3 15 = 3({2 + 4{ 3 5) = 3({ + 5)({ 3 1).
i 0 ({) = 0 i { = 35, 1. These are the only critical numbers.
i i 0 ({) = 6{2 3 6{ 3 36 = 6({2 3 { 3 6) = 6({ + 2)({ 3 3).
31. i ({) = 2{3 3 3{2 3 36{
i 0 ({) = 0 C { = 32, 3. These are the only critical numbers.
i i 0 ({) = 6{2 + 2{ + 2 = 2(3{2 + { + 1). Using the quadratic formula, i 0 ({) = 0 C
I
31 ± 311
{=
. Since the discrimininant, 311, is negative, there are no real soutions, and hence, there are no critical
6
32. i ({) = 2{3 + {2 + 2{
numbers.
33. j(w) = w4 + w3 + w2 + 1
i j 0 (w) = 4w3 + 3w2 + 2w = w(4w2 + 3w + 2). Using the quadratic formula, we see that
4w2 + 3w + 2 = 0 has no real solution (its discriminant is negative), so j 0 (w) = 0 only if w = 0. Hence, the only critical number
is 0.
34. j(w) = |3w 3 4| =
j 0 (w) =
+
+
3w 3 4
if 3w 3 4 D 0
3(3w 3 4)
if 3w 3 4 ? 0
3
if w A
33
if w ?
4
3
4
3
=
+
3w 3 4
if w D
4 3 3w
if w ?
and j 0 (w) does not exist at w = 43 , so w =
4
3
4
3
4
3
is a critical number.
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SECTION 4.1
|31
|2 3 | + 1
35. j(|) =
MAXIMUM AND MINIMUM VALUES
¤
301
i
(| 2 3 | + 1)(1) 3 (| 3 1)(2| 3 1)
| 2 3 | + 1 3 (2|2 3 3| + 1)
3| 2 + 2|
|(2 3 |)
=
= 2
= 2
.
2
2
2
2
(| 3 | + 1)
(| 3 | + 1)
(| 3 | + 1)2
(| 3 | + 1)2
j 0 (|) =
j 0 (|) = 0 i | = 0, 2. The expression | 2 3 | + 1 is never equal to 0, so j 0 (|) exists for all real numbers.
The critical numbers are 0 and 2.
36. k(s) =
s31
s2 + 4
i k0 (s) =
k0 (s) = 0 i s =
37. k(w) = w
3@4
32 ±
I
I
I
4 + 16
= 1 ± 5. The critical numbers are 1 ± 5. [k0 (s) exists for all real numbers.]
32
0
1@4
3 2w
i k (w) =
I
w=2
k0 (w) = 0 i 3
(s2 + 4)(1) 3 (s 3 1)(2s)
s2 + 4 3 2s2 + 2s
3s2 + 2s + 4
=
=
.
(s2 + 4)2
(s2 + 4)2
(s2 + 4)2
38. j({) = {1@3 3 {32@3
i
3 31@4
w
4
I
w=
3
2
3
2 33@4
w
4
=
1 33@4
w
(3w1@2
4
3 2) =
3
I
w32
I
.
4
4 w3
i w = 49 . k0 (w) does not exist at w = 0, so the critical numbers are 0 and 49 .
i j0 ({) = 13 {32@3 + 23 {35@3 = 13 {35@3 ({ + 2) =
{+2
.
3{5@3
j 0 (32) = 0 and j 0 (0) does not exist, but 0 is not in the domain of j, so the only critical number is 32.
39. I ({) = {4@5 ({ 3 4)2
i
I 0 ({) = {4@5 · 2({ 3 4) + ({ 3 4)2 · 45 {31@5 = 15 {31@5 ({ 3 4)[5 · { · 2 + ({ 3 4) · 4]
({ 3 4)(14{ 3 16)
2({ 3 4)(7{ 3 8)
=
5{1@5
5{1@5
=
I 0 ({) = 0 i { = 4, 87 . I 0 (0) does not exist. Thus, the three critical numbers are 0, 87 , and 4.
i j 0 () = 4 3 sec2 . j 0 () = 0 i sec2 = 4 i sec = ±2 i cos = ± 12
40. j() = 4 3 tan =
3
+ 2q,
5
3
+ 2q,
2
3
+ 2q, and
4
3
i
+ 2q are critical numbers.
Note: The values of that make j 0 () unde¿ned are not in the domain of j.
41. i () = 2 cos + sin2 i i 0 () = 32 sin + 2 sin cos . i 0 () = 0 i 2 sin (cos 3 1) = 0 i sin = 0
or cos = 1 i = q [q an integer] or = 2q. The solutions = q include the solutions = 2q, so the critical
numbers are = q.
I
1
1
i k0 (w) = 3 3 I
. k0 (w) = 0 i 3 = I
i
1 3 w2 =
2
2
13w
13w
I
i w2 = 89 i w = ± 23 2 E ±0=94, both in the domain of k, which is [31> 1].
42. k(w) = 3w 3 arcsin w
1 3 w2 =
1
9
43. i ({) = {2 h33{
1
3
i i 0 ({) = {2 (33h33{ ) + h33{ (2{) = {h33{ (33{ + 2). i 0 ({) = 0 i { = 0,
i
2
3
[h33{ is never equal to 0]. i 0 ({) always exists, so the critical numbers are 0 and 23 .
44. i ({) = {32 ln {
i i 0 ({) = {32 (1@{) + (ln {)(32{33 ) = {33 3 2{33 ln { = {33 (1 3 2 ln {) =
i 0 ({) = 0 i 1 3 2 ln { = 0 i ln { =
I
of i , so the only critical number is h.
1
2
1 3 2 ln {
.
{3
i { = h1@2 r 1=65. i 0 (0) does not exist, but 0 is not in the domain
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302
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
45. The graph of i 0 ({) = 5h30=1|{| sin { 3 1 has 10 zeros and exists
everywhere, so i has 10 critical numbers.
46. A graph of i 0 ({) =
100 cos2 {
3 1 is shown. There are 7 zeros
10 + {2
between 0 and 10, and 7 more zeros since i 0 is an even function.
i 0 exists everywhere, so i has 14 critical numbers.
47. i ({) = 12 + 4{ 3 {2 , [0> 5].
i 0 ({) = 4 3 2{ = 0 C { = 2. i (0) = 12, i (2) = 16, and i (5) = 7.
So i (2) = 16 is the absolute maximum value and i (5) = 7 is the absolute minimum value.
48. i ({) = 5 + 54{ 3 2{3 , [0> 4].
i 0 ({) = 54 3 6{2 = 6(9 3 {2 ) = 6(3 + {)(3 3 {) = 0 C { = 33, 3. i(0) = 5,
i (3) = 113, and i (4) = 93. So i (3) = 113 is the absolute maximum value and i (0) = 5 is the absolute minimum value.
49. i ({) = 2{3 3 3{2 3 12{ + 1, [32> 3].
i 0 ({) = 6{2 3 6{ 3 12 = 6({2 3 { 3 2) = 6({ 3 2)({ + 1) = 0 C
{ = 2> 31. i (32) = 33, i (31) = 8, i (2) = 319, and i (3) = 38. So i(31) = 8 is the absolute maximum value and
i (2) = 319 is the absolute minimum value.
50. {3 3 6{2 + 5, [33> 5].
i 0 ({) = 3{2 3 12{ = 3{({ 3 4) = 0 C { = 0, 4. i(33) = 376, i (0) = 5, i(4) = 327,
and i (5) = 320. So i (0) = 5 is the absolute maximum value and i (33) = 376 is the absolute minimum value.
51. i ({) = 3{4 3 4{3 3 12{2 + 1, [32> 3].
i 0 ({) = 12{3 3 12{2 3 24{ = 12{({2 3 { 3 2) = 12{({ + 1)({ 3 2) = 0 C
{ = 31, 0, 2. i (32) = 33, i (31) = 34, i (0) = 1, i (2) = 331, and i (3) = 28. So i(32) = 33 is the absolute maximum
value and i(2) = 331 is the absolute minimum value.
52. i ({) = ({2 3 1)3 , [31> 2].
i 0 ({) = 3({2 3 1)2 (2{) = 6{({ + 1)2 ({ 3 1)2 = 0 C { = 31> 0> 1. i (±1) = 0,
i (0) = 31, and i (2) = 27. So i (2) = 27 is the absolute maximum value and i(0) = 31 is the absolute minimum value.
53. i ({) = { +
1
1
{2 3 1
({ + 1)({ 3 1)
, [0=2> 4]. i 0 ({) = 1 3 2 =
=
= 0 C { = ±1, but { = 31 is not in the given
{
{
{2
{2
interval, [0=2> 4]. i 0 ({) does not exist when { = 0, but 0 is not in the given interval, so 1 is the only critical nuumber.
i (0=2) = 5=2, i (1) = 2, and i (4) = 4=25. So i (0=2) = 5=2 is the absolute maximum value and i (1) = 2 is the absolute
minimum value.
54. i ({) =
i 0 ({) =
{
, [0> 3].
{2 3 { + 1
({2 3 { + 1) 3 {(2{ 3 1)
{2 3 { + 1 3 2{2 + {
1 3 {2
(1 + {)(1 3 {)
=
= 2
=
= 0 C { = ±1,
2
2
2
2
({ 3 { + 1)
({ 3 { + 1)
({ 3 { + 1)2
({2 3 { + 1)2
but { = 31 is not in the given interval, [0> 3]. i(0) = 0, i (1) = 1, and i(3) = 37 . So i(1) = 1 is the absolute maximum
value and i(0) = 0 is the absolute minimum value.
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SECTION 4.1
55. i (w) = w
MAXIMUM AND MINIMUM VALUES
¤
303
I
4 3 w2 , [31> 2].
I
3w2
3w2 + (4 3 w2 )
4 3 2w2
I
+ 4 3 w2 =
= I
.
i 0 (w) = w · 12 (4 3 w2 )31@2 (32w) + (4 3 w2 )1@2 · 1 = I
2
2
43w
43w
4 3 w2
I
I
i 0 (w) = 0 i 4 3 2w2 = 0 i w2 = 2 i w = ± 2, but w = 3 2 is not in the given interval, [31> 2].
I
I i 0 (w) does not exist if 4 3 w2 = 0 i w = ±2, but 32 is not in the given interval. i (31) = 3 3, i 2 = 2, and
I
I i (2) = 0. So i 2 = 2 is the absolute maximum value and i (31) = 3 3 is the absolute minimum value.
56. i (w) =
I
3
w (8 3 w), [0> 8]. i (w) = 8w1@3 3 w4@3
4(2 3 w)
I
.
3
3 w2
i i 0 (w) = 83 w32@3 3 43 w1@3 = 43 w32@3 (2 3 w) =
I
i 0 (w) = 0 i w = 2. i 0 (w) does not exist if w = 0. i (0) = 0, i (2) = 6 3 2 E 7=56, and i (8) = 0.
I
So i (2) = 6 3 2 is the absolute maximum value and i (0) = i(8) = 0 is the absolute minimum value.
57. i (w) = 2 cos w + sin 2w, [0, @2].
i 0 (w) = 32 sin w + cos 2w · 2 = 32 sin w + 2(1 3 2 sin2 w) = 32(2 sin2 w + sin w 3 1) = 32(2 sin w 3 1)(sin w + 1).
I
I
I
i 0 (w) = 0 i sin w = 12 or sin w = 31 i w = 6 . i (0) = 2, i ( 6 ) = 3 + 12 3 = 32 3 r 2=60, and i ( 2 ) = 0.
I
So i ( 6 ) = 32 3 is the absolute maximum value and i ( 2 ) = 0 is the absolute minimum value.
58. i (w) = w + cot(w@2), [@4> 7@4].
i 0 (w) = 1 3 csc2 (w@2) · 12 .
I
i 0 (w) = 0 i 12 csc2 (w@2) = 1 i csc2 (w@2) = 2 i csc(w@2) = ± 2 i 12 w = 4 or 12 w = 3
4
I
$ w $ 7
i 8 $ 12 w $ 7
and csc(w@2) 6= 3 2 in the last interval i w = 2 or w = 3
.
4
4
8
2
3
3
= 3
i 4 = 4 + cot 8 r 3=20, i 2 = 2 + cot 4 = 2 + 1 r 2=57, i 3
2
2 + cot 2 = 2 3 1 r 3=71, and
7
i 7
r 3=08. So i 3
3 1 is the absolute maximum value and i 2 = 2 + 1 is the absolute
= 4 + cot 7
= 3
4
8
2
2
minimum value.
2 @8
59. i ({) = {h3{
, [31> 4]. i 0 ({) = { · h3{
2 @8
· (3 {4 ) + h3{
2 @8
2 @8
· 1 = h3{
2
2 @8
(3 {4 + 1). Since h3{
is never 0,
i 0 ({) = 0 i 3{2 @4 + 1 = 0 i 1 = {2 @4 i {2 = 4 i { = ±2, but 32 is not in the given interval, [31> 4].
I
i (31) = 3h31@8 E 30=88, i (2) = 2h31@2 E 1=21, and i(4) = 4h32 E 0=54. So i (2) = 2h31@2 = 2@ h is the absolute
I
maximum value and i (31) = 3h31@8 = 31@ 8 h is the absolute minimum value.
60. i ({) = { 3 ln {, [ 12 > 2].
i
1
2
=
1
2
i 0 ({) = 1 3
1
{
=
{31
.
{
i 0 ({) = 0 i { = 1. [Note that 0 is not in the domain of i .]
3 ln 12 E 1=19, i (1) = 1, and i(2) = 2 3 ln 2 E 1=31. So i(2) = 2 3 ln 2 is the absolute maximum value and
i (1) = 1 is the absolute minimum value.
1
· (2{ + 1) = 0 C { = 3 12 . Since {2 + { + 1 A 0 for all {, the
{2 + { + 1
domain of i and i 0 is R. i (31) = ln 1 = 0, i 3 12 = ln 34 r 30=29, and i(1) = ln 3 r 1=10. So i (1) = ln 3 r 1=10 is
the absolute maximum value and i 3 12 = ln 34 r 30=29 is the absolute minimum value.
61. i ({) = ln({2 + { + 1), [31> 1]. i 0 ({) =
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304
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
62. i ({) = { 3 2 tan31 {, [0> 4].
i 0 ({) = 1 3 2 ·
{ = ±1. i(0) = 0, i(1) = 1 3
maximum value and i (1) = 1 3
2
2
1
2
=0 C 1=
1 + {2
1 + {2
C 1 + {2 = 2 C {2 = 1 C
E 30=57, and i(4) = 4 3 2 tan31 4 E 1= 35. So i (4) = 4 3 2 tan31 4 is the absolute
is the absolute minimum value.
63. i ({) = {d (1 3 {)e , 0 $ { $ 1, d A 0, e A 0.
i 0 ({) = {d · e(1 3 {)e31 (31) + (1 3 {)e · d{d31 = {d31 (1 3 {)e31 [{ · e(31) + (1 3 {) · d]
= {d31 (1 3 {)e31 (d 3 d{ 3 e{)
At the endpoints, we have i(0) = i (1) = 0 [the minimum value of i ]. In the interval (0> 1), i 0 ({) = 0 C { =
d
=
d+e
d e
e
d
d
d+e3d
dd
dd
ee
dd ee
d
=
=
=
·
=
.
13
d+e
d+e
d+e
(d + e)d
d+e
(d + e)d (d + e)e
(d + e)d+e
d
dd ee
So i
is the absolute maximum value.
=
d+e
(d + e)d+e
i
We see that i 0 ({) = 0 at about { = 0=0 and 2=0, and that i 0 ({) does not
64.
exist at about { = 30=7, 1=0, and 2=7, so the critical numbers of
i ({) = {3 3 3{2 + 2 are about 30=7, 0=0, 1=0, 2=0, and 2=7.
65. (a)
From the graph, it appears that the absolute maximum value is about
i (30=77) = 2=19, and the absolute minimum value is about i (0=77) = 1=81.
t
(b) i ({) = {5 3 {3 + 2 i i 0 ({) = 5{4 3 3{2 = {2 (5{2 3 3). So i 0 ({) = 0 i { = 0, ± 35 .
t
t t 5 t 3
2 t 3
3
3
i 3 35 = 3 35 3 3 35 + 2 = 3 35
+
+2
5
5
5
t
t
3
9
6
3
= 35 3 25
+ 2 = 25
+ 2 (maximum)
5
5
t
t 3
6
3
= 3 25
+ 2 (minimum).
and similarly, i
5
5
66. (a)
From the graph, it appears that the absolute maximum value
is about i (1) = 2=85, and the absolute minimum value is about
i(0=23) = 1=89.
(b) i ({) = h{ + h32{
i i 0 ({) = h{ 3 2h32{ = h32{ (h3{ 3 2). So i 0 ({) = 0 C h3{ = 2 C 3{ = ln 2 C
{ = 13 ln 2 [E 0=23]. i 13 ln 2 = (hln 2 )1@3 + (hln 2 )32@3 = 21@3 + 232@3 [E 1=89], the minimum value.
i (1) = h1 + h32 [E 2=85], the maximum.
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SECTION 4.1
67. (a)
MAXIMUM AND MINIMUM VALUES
¤
305
From the graph, it appears that the absolute maximum value is about
i (0=75) = 0=32, and the absolute minimum value is i (0) = i(1) = 0;
that is, at both endpoints.
(b) i ({) = {
68. (a)
I
{ 3 {2
i i 0 ({) = { ·
I
1 3 2{
({ 3 2{2 ) + (2{ 3 2{2 )
3{ 3 4{2
I
I
+ { 3 {2 =
= I
.
2 { 3 {2
2 { 3 {2
2 { 3 {2
So i 0 ({) = 0 i 3{ 3 4{2 = 0 i {(3 3 4{) = 0 i { = 0 or 34 .
t
t
I
2
3
i (0) = i (1) = 0 (minimum), and i 34 = 34 34 3 34 = 34 16
= 3163 (maximum).
From the graph, it appears that the absolute maximum value is about
i (32) = 31=17, and the absolute minimum value is about
i (30=52) = 32=26.
(b) i ({) = { 3 2 cos { i i 0 ({) = 1 + 2 sin {. So i 0 ({) = 0 i sin { = 3 12
i { = 3 6 on [32> 0].
I I
i (32) = 32 3 2 cos(32) (maximum) and i 3 6 = 3 6 3 2 cos 3 6 = 3 6 3 2 23 = 3 6 3 3 (minimum).
69. The density is de¿ned as =
[since
1000
mass
=
(in g@cm3 ). But a critical point of will also be a critical point of Y
volume
Y (W )
gY
g
= 31000Y 32
and Y is never 0], and Y is easier to differentiate than .
gW
gW
Y (W ) = 999=87 3 0=06426W + 0=0085043W 2 3 0=0000679W 3
i Y 0 (W ) = 30=06426 + 0=0170086W 3 0=0002037W 2 .
Setting this equal to 0 and using the quadratic formula to ¿nd W , we get
W =
30=0170086 ±
I
0=01700862 3 4 · 0=0002037 · 0=06426
E 3=9665 C or 79=5318 C. Since we are only interested
2(30=0002037)
in the region 0 C $ W $ 30 C, we check the density at the endpoints and at 3=9665 C: (0) E
(30) E
1000
E 1=00013;
999=87
1000
1000
E 0=99625; (3=9665) E
E 1=000255. So water has its maximum density at
1003=7628
999=7447
about 3=9665 C.
70. I =
So
Z
sin + cos i
gI
( sin + cos )(0) 3 Z ( cos 3 sin )
3Z ( cos 3 sin )
=
=
.
2
g
( sin + cos )
( sin + cos )2
gI
= 0 i cos 3 sin = 0
g
i =
sin = tan . Substituting tan for in I gives us
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NOT FOR SALE
¤
CHAPTER 4
I =
APPLICATIONS OF DIFFERENTIATION
(tan )Z
Z sin Z tan Z tan cos =
=
= Z sin .
=
2
2
2
(tan ) sin + cos 1
sin + cos sin + cos cos If tan = , then sin = s
(see the ¿gure), so I = s
Z.
2
+1
2 + 1
We compare this with the value of I at the endpoints: I (0) = Z and I 2 = Z .
Now because s
$ 1 and s
$ , we have that s
Z is less than or equal to each of I (0) and I 2 .
2 + 1
2 + 1
2 + 1
Hence, s
Z is the absolute minimum value of I (), and it occurs when tan = .
2 + 1
71. Let d = 30=000 032 37, e = 0=000 903 7, f = 30=008 956, g = 0=03629, h = 30=04458, and i = 0=4074.
Then V(w) = dw5 + ew4 + fw3 + gw2 + hw + i and V 0 (w) = 5dw4 + 4ew3 + 3fw2 + 2gw + h.
We now apply the Closed Interval Method to the continuous function V on the interval 0 $ w $ 10. Since V 0 exists for all w,
the only critical numbers of V occur when V 0 (w) = 0. We use a root¿nder on a CAS (or a graphing device) to ¿nd that
V 0 (w) = 0 when w1 r 0=855, w2 r 4=618, w3 r 7=292, and w4 r 9=570. The values of V at these critical numbers are
V(w1 ) r 0=39, V(w2 ) r 0=43645, V(w3 ) r 0=427, and V(w4 ) r 0=43641. The values of V at the endpoints of the interval are
V(0) r 0=41 and V(10) r 0=435. Comparing the six numbers, we see that sugar was most expensive at w2 r 4=618
(corresponding roughly to March 1998) and cheapest at w1 r 0=855 (June 1994).
72. (a) The equation of the graph in the ¿gure is
y(w) = 0=00146w3 3 0=11553w2 + 24=98169w 3 21=26872.
(b) d(w) = y 0 (w) = 0=00438w2 3 0=23106w + 24=98169 i
d0 (w) = 0=00876w 3 0=23106.
d0 (w) = 0 i w1 =
0=23106
0=00876
E 26=4. d(0) E 24=98, d(w1 ) E 21=93,
and d(125) E 64=54.
The maximum acceleration is about 64=5 ft@s2 and the minimum acceleration is about 21=93 ft@s2 .
i y 0 (u) = 2nu0 u 3 3nu2 . y 0 (u) = 0 i nu(2u0 3 3u) = 0 i
u = 0 or 23 u0 (but 0 is not in the interval). Evaluating y at 12 u0 , 23 u0 , and u0 , we get y 12 u0 = 18 nu03 , y 23 u0 =
73. (a) y(u) = n(u0 3 u)u2 = nu0 u2 3 nu3
and y(u0 ) = 0. Since
4
27
A 18 , y attains its maximum value at u = 23 u0 . This supports the statement in the text.
(b) From part (a), the maximum value of y is
74. j({) = 2 + ({ 3 5)3
4
nu03 ,
27
4
nu03 .
27
i j 0 ({) = 3({ 3 5)2
(c)
i j 0 (5) = 0, so 5 is a critical number. But j(5) = 2 and j takes on
values A 2 and values ? 2 in any open interval containing 5, so j does not have a local maximum or minimum at 5.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
APPLIED PROJECT
75. i ({) = {101 + {51 + { + 1
THE CALCULUS OF RAINBOWS
¤
307
i i 0 ({) = 101{100 + 51{50 + 1 D 1 for all {, so i 0 ({) = 0 has no solution. Thus, i ({)
has no critical number, so i ({) can have no local maximum or minimum.
76. Suppose that i has a minimum value at f, so i ({) D i (f) for all { near f. Then j({) = 3i({) $ 3i(f) = j(f) for all {
near f, so j({) has a maximum value at f.
77. If i has a local minimum at f, then j({) = 3i ({) has a local maximum at f, so j 0 (f) = 0 by the case of Fermat’s Theorem
proved in the text. Thus, i 0 (f) = 3j 0 (f) = 0.
78. (a) i ({) = d{3 + e{2 + f{ + g, d 6= 0. So i 0 ({) = 3d{2 + 2e{ + f is a quadratic and hence has either 2, 1, or 0 real roots,
so i ({) has either 2, 1 or 0 critical numbers.
Case (i) [2 critical numbers]:
Case (ii) [1 critical number]:
3
3
i ({) = { 3 3{ i
i ({) = {
Case (iii) [no critical number]:
i ({) = {3 + 3{ i
i
i 0 ({) = 3{2 3 3, so { = 31, 1
i 0 ({) = 3{2 , so { = 0
i 0 ({) = 3{2 + 3,
are critical numbers.
is the only critical number.
so there is no critical number.
(b) Since there are at most two critical numbers, it can have at most two local extreme values and by (i) this can occur. By (iii)
it can have no local extreme value. However, if there is only one critical number, then there is no local extreme value.
APPLIED PROJECT
The Calculus of Rainbows
1. From Snell’s Law, we have sin = n sin E
4
3
sin C E arcsin 34 sin . We substitute this into
G() = + 2 3 4 = + 2 3 4 arcsin 34 sin , and then differentiate to ¿nd the minimum:
k
2 l31@2 3
3 cos 3 cos . This is 0 when t
=2 C
cos = 2 3 t
G0 () = 2 3 4 1 3 34 sin 4
9
9
1 3 16
sin2 1 3 16
sin2 t
2
2
2
2
9
9
9
9
7
7
1 3 cos2 C 27
C cos = 27
C
4 cos = 1 3 16 sin C
4 cos = 1 3 16
16 cos = 16
t
7
= arccos 27
E 59=4 , and so the local minimum is G(59=4 ) E 2=4 radians E 138 .
To see that this is an absolute minimum, we check the endpoints, which in this case are = 0 and =
G(0) = radians = 180 , and G 2 E 166 .
Another method: We ¿rst calculate
G0 () = 2 3 4
g
=0 C
g
substitute sin =
4
3
g
: sin =
g
4
3
sin C cos =
4
3
cos g
g
C
2:
g
3 cos =
, so since
g
4 cos g
1
= , the minimum occurs when 3 cos = 2 cos . Now we square both sides and
g
2
sin , leading to the same result.
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308
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
2. If we repeat Problem 1 with n in place of 43 , we get G() = + 2 3 4 arcsin
4 cos 2 cos =
G () = 2 3 v
2 , which is 0 when
n
sin n 13
n
0
v
13
4 cos2 = n2 3 sin2 C 3 cos2 = n2 3 1 C = arccos
u
sin n
2
1
sin i
n
C
2 cos n
2
=13
sin n
2
C
n2 3 1
. So for n E 1=3318 (red light) the minimum
3
occurs at 1 E 1=038 radians, and so the rainbow angle is about 3 G(1 ) E 42=3 . For n E 1=3435 (violet light) the
minimum occurs at 2 E 1=026 radians, and so the rainbow angle is about 3 G(2 ) E 40=6 .
Another method: As in Problem 1, we can instead ¿nd G0 () in terms of
g
g
cos , and then substitute
=
.
g
g
n cos 3. At each reÀection or refraction, the light is bent in a counterclockwise direction: the bend at D is 3 , the bend at E is
3 2, the bend at F is again 3 2, and the bend at G is 3 . So the total bend is
sin and differentiate, to get
G() = 2( 3 ) + 2( 3 2) = 2 3 6 + 2, as required. We substitute = arcsin
n
v
2
6 cos 3 cos sin 0
,
which
is
0
when
1
3
C 9 cos2 = n2 3 sin2 C
G () = 2 3 v
=
2
n
n
sin n 13
n
t
8 cos2 = n2 3 1 C cos = 18 (n2 3 1). If n = 43 , then the minimum occurs at
u
(4@3)2 3 1
1 = arccos
E 1=254 radians. Thus, the minimum
8
counterclockwise rotation is G(1 ) E 231 , which is equivalent to a
clockwise rotation of 360 3 231 = 129 (see the ¿gure). So the rainbow
angle for the secondary rainbow is about 180 3 129 = 51 , as required.
In general, the rainbow angle for the secondary rainbow is
3 [2 3 G()] = G() 3 .
4. In the primary rainbow, the rainbow angle gets smaller as n gets larger, as we found in Problem 2, so the colors appear from
top to bottom in order of increasing n. But in the secondary rainbow, the rainbow angle gets larger as n gets larger. To see this,
we ¿nd the minimum deviations for red light and for violet light in the secondary rainbow. For n E 1=3318 (red light) the
u
1=33182 3 1
E 1=255 radians, and so the rainbow angle is G(1 ) 3 E 50=6 . For
minimum occurs at 1 E arccos
8
u
1=34352 3 1
n E 1=3435 (violet light) the minimum occurs at 2 E arccos
E 1=248 radians, and so the rainbow angle is
8
G(2 ) 3 E 53=6 . Consequently, the rainbow angle is larger for colors with higher indices of refraction, and the colors
appear from bottom to top in order of increasing n, the reverse of their order in the primary rainbow.
Note that our calculations above also explain why the secondary rainbow is more spread out than the primary rainbow: in
the primary rainbow, the difference between rainbow angles for red and violet light is about 1=7 , whereas in the secondary
rainbow it is about 3 .
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NOT FOR SALE
SECTION 4.2
THE MEAN VALUE THEOREM
¤
309
4.2 The Mean Value Theorem
1. i ({) = 5 3 12{ + 3{2 , [1, 3].
Since i is a polynomial, it is continuous and differentiable on R, so it is continuous on [1> 3]
and differentiable on (1> 3). Also i (1) = 34 = i (3). i 0 (f) = 0 C 312 + 6f = 0 C f = 2, which is in the open
interval (1> 3), so f = 2 satis¿es the conclusion of Rolle’s Theorem.
2. i ({) = {3 3 {2 3 6{ + 2, [0> 3].
Since i is a polynomial, it is continuous and differentiable on R, so it is continuous on
[0> 3] and differentiable on (0> 3). Also, i (0) = 2 = i (3). i 0 (f) = 0 C 3f2 3 2f 3 6 = 0 C
I
I
I
2 ± 4 + 72
= 13 ± 13 19 [r 1=79, r 31=12], so f = 13 + 13 19 satis¿es the conclusion of Rolle’s Theorem.
f=
6
3. i ({) =
I
{ 3 13 {, [0> 9]. i , being the difference of a root function and a polynomial, is continuous and differentiable
on [0> "), so it is continuous on [0> 9] and differentiable on (0> 9). Also, i (0) = 0 = i (9). i 0 (f) = 0 C
2
I
I
3
1
1
I 3 =0 C 2 f=3 C
f=
3
2
f
i f=
9
9
, which is in the open interval (0> 9), so f = satis¿es the
4
4
conclusion of Rolle’s Theorem.
4. i ({) = cos 2{, [@8> 7@8].
i , being the composite of the cosine function and the polynomial 2{, is continuous and
I
.
differentiable on R, so it is continuous on [@8> 7@8] and differentiable on (@8> 7@8). Also, i 8 = 12 2 = i 7
8
i 0 (f) = 0 C 32 sin 2f = 0 C sin 2f = 0 C 2f = q C f =
interval (@8> 7@8), so f =
5. i ({) = 1 3 {2@3 .
2
2 q.
If q = 1, then f =
2,
which is in the open
satis¿es the conclusion of Rolle’s Theorem.
i (31) = 1 3 (31)2@3 = 1 3 1 = 0 = i (1). i 0 ({) = 3 23 {31@3 , so i 0 (f) = 0 has no solution. This
does not contradict Rolle’s Theorem, since i 0 (0) does not exist, and so i is not differentiable on (31> 1).
i (0) = tan 0 = 0 = tan = i (). i 0 ({) = sec2 { D 1, so i 0 (f) = 0 has no solution. This does not
contradict Rolle’s Theorem, since i 0 2 does not exist, and so i is not differentiable on (0> ). (Also, i ({) is not continuous
6. i ({) = tan {.
on [0> ].)
7. i 0 (f) =
i 0 (f) =
8. i 0 (f) =
i 0 (f) =
i (8) 3 i (0)
330
3
=
= . It appears that
830
8
8
3
8
when f E 0=3, 3, and 6=3.
i (7) 3 i (1)
1=6 3 0=6
1
=
= . It appears that
731
6
6
1
6
when f E 3=2 and 6=1.
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NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
9. i ({) = 2{2 3 3{ + 1, [0> 2].
i is continuous on [0> 2] and differentiable on (0> 2) since polynomials are continuous and
differentiable on R. i 0 (f) =
i (e) 3 i (d)
e3d
i (2) 3 i (0)
331
=
= 1 C 4f = 4 C f = 1, which
230
2
C 4f 3 3 =
is in (0> 2)=
10. i ({) = {3 3 3{ + 2, [32> 2].
differentiable on R. i 0 (f) =
f2 =
4
3
i is continuous on [32> 2] and differentiable on (32> 2) since polynomials are continuous and
i (e) 3 i (d)
e3d
430
i (2) 3 i(32)
=
= 1 C 3f2 = 4 C
2 3 (32)
4
C 3f2 3 3 =
2
C f = ± I , which are both in (32> 2).
3
11. i ({) = ln {, [1> 4].
i (e) 3 i (d)
e3d
i 0 (f) =
12. i ({) =
i is continuous and differentiable on (0> "), so i is continuous on [1> 4] and differentiable on (1> 4).
C
1
i (4) 3 i (1)
ln 4 3 0
ln 4
=
=
=
f
431
3
3
C f=
3
E 2=16, which is in (1> 4).
ln 4
1
, [1> 3]. i is continuous and differentiable on (3"> 0) (0> "), so i is continous on [1> 3] and differentiable
{
i (e) 3 i (d)
e3d
on (1> 3). i 0 (f) =
C 3
1
i (3) 3 i (1)
=
=
f2
331
1
3
31
1
=3
2
3
C
I
I
f2 = 3 C f = ± 3, but only 3
is in (1> 3).
13. i ({) =
I
i (4) 3 i (0)
{, [0> 4]. i 0 (f) =
430
1
1
I =
2
2 f
C
C
1
230
I =
4
2 f
C
I
f = 1 C f = 1. The secant line and the tangent line
are parallel.
14. i ({) = h3{ , [0> 2].
h3f =
1 3 h32
2
f = 3 ln
i 0 (f) =
i (2) 3 i (0)
230
C 3f = ln
1 3 h32
2
C 3h3f =
h32 3 1
2
C
C
1 3 h32
E 0=8386. The secant line and the tangent line are
2
parallel.
15. i ({) = ({ 3 3)32
3
36
=
4
(f 3 3)3
i i 0 ({) = 32 ({ 3 3)33 . i (4) 3 i (1) = i 0 (f)(4 3 1) i
1
1
32
3
=
·3 i
12
(32)2
(f 3 3)3
i (f 3 3)3 = 38 i f 3 3 = 32 i f = 1, which is not in the open interval (1> 4). This does not
contradict the Mean Value Theorem since i is not continuous at { = 3.
16. i ({) = 2 3 |2{ 3 1| =
+
2 3 (2{ 3 1)
if 2{ 3 1 D 0
2 3 [3(2{ 3 1)]
if 2{ 3 1 ? 0
i (3) 3 i (0) = i 0 (f)(3 3 0)
i 33 3 1 = i 0 (f) · 3
=
+
3 3 2{
if { D
1 + 2{
if { ?
1
2
1
2
0
i i ({) =
+
32 if { A
2 if { ?
1
2
1
2
i i 0 (f) = 3 43 [not ± 2]. This does not contradict the Mean
Value Theorem since i is not differentiable at { = 12 .
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NOT FOR SALE
SECTION 4.2
THE MEAN VALUE THEOREM
¤
311
17. Let i ({) = 2{ + cos {. Then i(3) = 32 3 1 ? 0 and i(0) = 1 A 0. Since i is the sum of the polynomial 2{ and the
trignometric function cos {, i is continuous and differentiable for all {. By the Intermediate Value Theorem, there is a number
f in (3> 0) such that i (f) = 0. Thus, the given equation has at least one real root. If the equation has distinct real roots d and
e with d ? e, then i (d) = i (e) = 0. Since i is continuous on [d> e] and differentiable on (d> e), Rolle’s Theorem implies that
there is a number u in (d> e) such that i 0 (u) = 0. But i 0 (u) = 2 3 sin u A 0 since sin u $ 1. This contradiction shows that the
given equation can’t have two distinct real roots, so it has exactly one root.
18. Let i ({) = {3 + h{ . Then i (31) = 31 + 1@h ? 0 and i(0) = 1 A 0. Since i is the sum of a polynomial and the natural
exponential function, i is continous and differentiable for all {. By the Intermediate Value Theorem, there is a number f in
(31> 0) such that i (f) = 0. Thus, the given equation has at least one real root. If the equation has distinct real roots d and e
with d ? e, then i (d) = i (e) = 0. Since i is continuous on [d> e] and differentiable on (d> e), Rolle’s Theorem implies that
there is a number u in (d> e) such that i 0 (u) = 0. But i 0 (u) = 3u2 + hu A 0. This contradiction shows that the given equation
can’t have two distinct real roots, so it has exactly one root.
19. Let i ({) = {3 3 15{ + f for { in [32> 2]. If i has two real roots d and e in [32> 2], with d ? e, then i (d) = i (e) = 0. Since
the polynomial i is continuous on [d> e] and differentiable on (d> e), Rolle’s Theorem implies that there is a number u in (d> e)
such that i 0 (u) = 0. Now i 0 (u) = 3u2 3 15. Since u is in (d> e), which is contained in [32> 2], we have |u| ? 2, so u2 ? 4.
It follows that 3u2 3 15 ? 3 · 4 3 15 = 33 ? 0. This contradicts i 0 (u) = 0, so the given equation can’t have two real roots
in [32> 2]. Hence, it has at most one real root in [32> 2].
20. i ({) = {4 + 4{ + f. Suppose that i ({) = 0 has three distinct real roots d, e, g where d ? e ? g. Then
i (d) = i(e) = i(g) = 0. By Rolle’s Theorem there are numbers f1 and f2 with d ? f1 ? e and e ? f2 ? g
and 0 = i 0 (f1 ) = i 0 (f2 ), so i 0 ({) = 0 must have at least two real solutions. However
0 = i 0 ({) = 4{3 + 4 = 4({3 + 1) = 4({ + 1)(({2 3 { + 1) has as its only real solution { = 31. Thus, i ({) can have at
most two real roots.
21. (a) Suppose that a cubic polynomial S ({) has roots d1 ? d2 ? d3 ? d4 , so S (d1 ) = S (d2 ) = S (d3 ) = S (d4 ).
By Rolle’s Theorem there are numbers f1 , f2 , f3 with d1 ? f1 ? d2 , d2 ? f2 ? d3 and d3 ? f3 ? d4 and
S 0 (f1 ) = S 0 (f2 ) = S 0 (f3 ) = 0. Thus, the second-degree polynomial S 0 ({) has three distinct real roots, which is
impossible.
(b) We prove by induction that a polynomial of degree q has at most q real roots. This is certainly true for q = 1. Suppose
that the result is true for all polynomials of degree q and let S ({) be a polynomial of degree q + 1. Suppose that S ({) has
more than q + 1 real roots, say d1 ? d2 ? d3 ? · · · ? dq+1 ? dq+2 . Then S (d1 ) = S (d2 ) = · · · = S (dq+2 ) = 0.
By Rolle’s Theorem there are real numbers f1 > = = = > fq+1 with d1 ? f1 ? d2 > = = = > dq+1 ? fq+1 ? dq+2 and
S 0 (f1 ) = · · · = S 0 (fq+1 ) = 0. Thus, the qth degree polynomial S 0 ({) has at least q + 1 roots. This contradiction shows
that S ({) has at most q + 1 real roots.
22. (a) Suppose that i (d) = i(e) = 0 where d ? e. By Rolle’s Theorem applied to i on [d> e] there is a number f such that
d ? f ? e and i 0 (f) = 0.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(b) Suppose that i(d) = i (e) = i (f) = 0 where d ? e ? f. By Rolle’s Theorem applied to i ({) on [d> e] and [e> f] there are
numbers d ? g ? e and e ? h ? f with i 0 (g) = 0 and i 0 (h) = 0. By Rolle’s Theorem applied to i 0 ({) on [g> h] there is a
number j with g ? j ? h such that i 00 (j) = 0.
(c) Suppose that i is q times differentiable on R and has q + 1 distinct real roots. Then i (q) has at least one real root.
23. By the Mean Value Theorem, i (4) 3 i (1) = i 0 (f)(4 3 1) for some f M (1> 4). But for every f M (1> 4) we have
i 0 (f) D 2. Putting i 0 (f) D 2 into the above equation and substituting i (1) = 10, we get
i (4) = i (1) + i 0 (f)(4 3 1) = 10 + 3i 0 (f) D 10 + 3 · 2 = 16. So the smallest possible value of i (4) is 16.
24. If 3 $ i 0 ({) $ 5 for all {, then by the Mean Value Theorem, i (8) 3 i (2) = i 0 (f) · (8 3 2) for some f in [2> 8].
(i is differentiable for all {, so, in particular, i is differentiable on (2> 8) and continuous on [2> 8]. Thus, the hypotheses of the
Mean Value Theorem are satis¿ed.) Since i (8) 3 i (2) = 6i 0 (f) and 3 $ i 0 (f) $ 5, it follows that
6 · 3 $ 6i 0 (f) $ 6 · 5 i 18 $ i (8) 3 i(2) $ 30=
25. Suppose that such a function i exists. By the Mean Value Theorem there is a number 0 ? f ? 2 with
i 0 (f) =
i (2) 3 i (0)
5
= . But this is impossible since i 0 ({) $ 2 ?
230
2
5
2
for all {, so no such function can exist.
26. Let k = i 3 j. Then since i and j are continuous on [d> e] and differentiable on (d> e), so is k, and thus k satis¿es the
assumptions of the Mean Value Theorem. Therefore, there is a number f with d ? f ? e such that
k(e) = k(e) 3 k(d) = k0 (f)(e 3 d). Since k0 (f) ? 0, k0 (f)(e 3 d) ? 0, so i (e) 3 j(e) = k(e) ? 0 and hence i(e) ? j(e).
27. We use Exercise 26 with i ({) =
I
1 + {, j({) = 1 + 12 {, and d = 0. Notice that i (0) = 1 = j(0) and
I
1
1
I
? = j 0 ({) for { A 0. So by Exercise 26, i (e) ? j(e) i
1 + e ? 1 + 12 e for e A 0.
2
1+{
I
I
Another method: Apply the Mean Value Theorem directly to either i ({) = 1 + 12 { 3 1 + { or j({) = 1 + { on [0> e].
i 0 ({) =
2
28. i satis¿es the conditions for the Mean Value Theorem, so we use this theorem on the interval [3e> e]:
i (e) 3 i(3e)
= i 0 (f)
e 3 (3e)
for some f M (3e> e). But since i is odd, i (3e) = 3i (e). Substituting this into the above equation, we get
i (e) + i (e)
= i 0 (f) i
2e
i (e)
= i 0 (f).
e
29. Let i ({) = sin { and let e ? d. Then i ({) is continuous on [e> d] and differentiable on (e> d). By the Mean Value Theorem,
there is a number f M (e> d) with sin d 3 sin e = i (d) 3 i (e) = i 0 (f)(d 3 e) = (cos f)(d 3 e). Thus,
|sin d 3 sin e| $ |cos f| |e 3 d| $ |d 3 e|. If d ? e, then |sin d 3 sin e| = |sin e 3 sin d| $ |e 3 d| = |d 3 e|. If d = e, both
sides of the inequality are 0.
30. Suppose that i 0 ({) = f. Let j({) = f{, so j 0 ({) = f. Then, by Corollary 7, i ({) = j({) + g, where g is a constant, so
i ({) = f{ + g.
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SECTION 4.2
THE MEAN VALUE THEOREM
¤
313
31. For { A 0, i ({) = j({), so i 0 ({) = j0 ({). For { ? 0, i 0 ({) = (1@{)0 = 31@{2 and j0 ({) = (1 + 1@{)0 = 31@{2 , so
again i 0 ({) = j 0 ({). However, the domain of j({) is not an interval [it is (3"> 0) (0> ")] so we cannot conclude that
i 3 j is constant (in fact it is not).
2
4{
2
4{
I
3s
= I
3
=0
1 3 {2
1 3 {2
2{ 1 3 {2
1 3 (1 3 2{2 )2
32. Let i ({) = 2 sin31 { 3 cos31 (1 3 2{2 ). Then i 0 ({) = I
[since { D 0]. Thus, i 0 ({) = 0 for all { M (0> 1). Thus, i ({) = F on (0> 1). To ¿nd F, let { = 0=5. Thus,
2 sin31 (0=5) 3 cos31 (0=5) = 2 6 3 3 = 0 = F. We conclude that i({) = 0 for { in (0> 1). By continuity of i , i ({) = 0
on [0> 1]. Therefore, we see that i ({) = 2 sin31 { 3 cos31 (1 3 2{2 ) = 0 i 2 sin31 { = cos31 (1 3 2{2 ).
33. Let i ({) = arcsin
{31
{+1
3 2 arctan
I
{ + 2 . Note that the domain of i is [0> "). Thus,
({ + 1) 3 ({ 3 1)
1
1
2
1
1
· I = I
3
3I
= 0.
2
2
1
+
{
2
{
{
({
+
1)
{
({
+ 1)
({
+
1)
{31
13
{+1
i 0 ({) = v
Then i ({) = F on (0> ") by Theorem 5. By continuity of i , i ({) = F on [0> "). To ¿nd F, we let { = 0 i
arcsin(31) 3 2 arctan(0) + 2 = F i 3 2 3 0 +
I
{31
= 2 arctan { 3 2 .
arcsin
{+1
2
= 0 = F. Thus, i ({) = 0 i
34. Let y(w) be the velocity of the car w hours after 2:00 PM. Then
Theorem, there is a number f such that 0 ? f ?
1
6
50 3 30
y(1@6) 3 y(0)
=
= 120. By the Mean Value
1@6 3 0
1@6
with y 0 (f) = 120. Since y 0 (w) is the acceleration at time w, the acceleration
f hours after 2:00 PM is exactly 120 mi@h2 .
35. Let j(w) and k(w) be the position functions of the two runners and let i(w) = j(w) 3 k(w). By hypothesis,
i (0) = j(0) 3 k(0) = 0 and i (e) = j(e) 3 k(e) = 0, where e is the ¿nishing time. Then by the Mean Value Theorem,
there is a time f, with 0 ? f ? e, such that i 0 (f) =
i (e) 3 i (0)
. But i(e) = i(0) = 0, so i 0 (f) = 0. Since
e30
i 0 (f) = j 0 (f) 3 k0 (f) = 0, we have j0 (f) = k0 (f). So at time f, both runners have the same speed j 0 (f) = k0 (f).
36. Assume that i is differentiable (and hence continuous) on R and that i 0 ({) 6= 1 for all {. Suppose i has more than one ¿xed
point. Then there are numbers d and e such that d ? e, i (d) = d, and i (e) = e. Applying the Mean Value Theorem to the
function i on [d> e], we ¿nd that there is a number f in (d> e) such that i 0 (f) =
i (e) 3 i(d)
e3d
. But then i 0 (f) =
= 1,
e3d
e3d
contradicting our assumption that i 0 ({) 6= 1 for every real number {. This shows that our supposition was wrong, that is, that
i cannot have more than one ¿xed point.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
4.3 How Derivatives Affect the Shape of a Graph
1. (a) i is increasing on (1> 3) and (4> 6).
(b) i is decreasing on (0> 1) and (3> 4).
(c) i is concave upward on (0> 2).
(d) i is concave downward on (2> 4) and (4> 6).
(e) The point of inÀection is (2> 3).
2. (a) i is increasing on (0> 1) and (3> 7).
(b) i is decreasing on (1> 3).
(c) i is concave upward on (2> 4) and (5> 7).
(d) i is concave downward on (0> 2) and (4> 5).
(e) The points of inÀection are (2> 2), (4> 3), and (5> 4).
3. (a) Use the Increasing/Decreasing (I/D) Test.
(b) Use the Concavity Test.
(c) At any value of { where the concavity changes, we have an inÀection point at ({> i({)).
4. (a) See the First Derivative Test.
(b) See the Second Derivative Test and the note that precedes Example 7.
5. (a) Since i 0 ({) A 0 on (1> 5), i is increasing on this interval. Since i 0 ({) ? 0 on (0> 1) and (5> 6), i is decreasing on these
intervals.
(b) Since i 0 ({) = 0 at { = 1 and i 0 changes from negative to positive there, i changes from decreasing to increasing and has
a local minimum at { = 1. Since i 0 ({) = 0 at { = 5 and i 0 changes from positive to negative there, i changes from
increasing to decreasing and has a local maximum at { = 5.
6. (a) i 0 ({) A 0 and i is increasing on (0> 1) and (3> 5). i 0 ({) ? 0 and i is decreasing on (1> 3) and (5> 6).
(b) Since i 0 ({) = 0 at { = 1 and { = 5 and i 0 changes from positive to negative at both values, i changes from increasing to
decreasing and has local maxima at { = 1 and { = 5. Since i 0 ({) = 0 at { = 3 and i 0 changes from negative to positive
there, i changes from decreasing to increasing and has a local minimum at { = 3.
7. (a) There is an IP at { = 3 because the graph of i changes from CD to CU there. There is an IP at { = 5 because the graph
of i changes from CU to CD there.
(b) There is an IP at { = 2 and at { = 6 because i 0 ({) has a maximum value there, and so i 00 ({) changes from positive to
negative there. There is an IP at { = 4 because i 0 ({) has a minimum value there and so i 00 ({) changes from negative to
positive there.
(c) There is an inÀection point at { = 1 because i 00 ({) changes from negative to positive there, and so the graph of i changes
from concave downward to concave upward. There is an inÀection point at { = 7 because i 00 ({) changes from positive to
negative there, and so the graph of i changes from concave upward to concave downward.
8. (a) i is increasing on the intervals where i 0 ({) A 0, namely, (2> 4) and (6> 9).
(b) i has a local maximum where it changes from increasing to decreasing, that is, where i 0 changes from positive to negative
(at { = 4). Similarly, where i 0 changes from negative to positive, i has a local minimum (at { = 2 and at { = 6).
(c) When i 0 is increasing, its derivative i 00 is positive and hence, i is concave upward. This happens on (1> 3), (5> 7), and
(8> 9). Similarly, i is concave downward when i 0 is decreasing—that is, on (0> 1), (3> 5), and (7> 8).
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SECTION 4.3
¤
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
315
(d) i has inÀection points at { = 1, 3, 5, 7, and 8, since the direction of concavity changes at each of these values.
9. (a) i ({) = 2{3 + 3{2 3 36{
i i 0 ({) = 6{2 + 6{ 3 36 = 6({2 + { 3 6) = 6({ + 3)({ 3 2).
We don’t need to include the “6” in the chart to determine the sign of i 0 ({).
Interval
{+3
{32
i 0 ({)
i
{ ? 33
3
3
+
increasing on (3"> 33)
{A2
+
+
+
33 ? { ? 2
+
3
3
decreasing on (33> 2)
increasing on (2> ")
(b) i changes from increasing to decreasing at { = 33 and from decreasing to increasing at { = 2. Thus, i (33) = 81 is a
local maximum value and i (2) = 344 is a local minimum value.
(c) i 0 ({) = 6{2 + 6{ 3 36 i i 00 ({) = 12{ + 6. i 00 ({) = 0 at { = 3 12 , i 00 ({) A 0 C { A 3 12 , and
i 00 ({) ? 0 C { ? 3 12 . Thus, i is concave upward on 3 12 > " and concave downward on 3"> 3 12 . There is an
.
inÀection point at 3 12 > i 3 12 = 3 12 > 37
2
10. (a) i ({) = 4{3 + 3{2 3 6{ + 1
i i 0 ({) = 12{2 + 6{ 3 6 = 6(2{2 + { 3 1) = 6(2{ 3 1)({ + 1). Thus,
i 0 ({) A 0 C { ? 31 or { A
i is decreasing on 31> 12 .
1
2
and i 0 ({) ? 0 C 31 ? { ? 12 . So i is increasing on (3"> 31) and
1
2>"
and
(b) i changes from increasing to decreasing at { = 31 and from decreasing to increasing at { = 12 . Thus, i (31) = 6 is a
local maximum value and i 12 = 3 34 is a local minimum value.
(c) i 00 ({) = 24{ + 6 = 6(4{ + 1). i 00 ({) A 0 C { A 3 14 and i 00 ({) ? 0 C { ? 3 14 . Thus, i is concave upward
.
on 3 14 > " and concave downward on 3"> 3 14 . There is an inÀection point at 3 14 > i 3 14 = 3 14 > 21
8
11. (a) i ({) = {4 3 2{2 + 3
i i 0 ({) = 4{3 3 4{ = 4{ {2 3 1 = 4{({ + 1)({ 3 1).
Interval
{+1
{
{31
i 0 ({)
i
{ ? 31
3
3
3
3
decreasing on (3"> 31)
0?{?1
+
+
+
3
decreasing on (0> 1)
{A1
3
31 ? { ? 0
+
3
+
3
+
+
+
increasing on (31> 0)
increasing on (1> ")
(b) i changes from increasing to decreasing at { = 0 and from decreasing to increasing at { = 31 and { = 1. Thus,
i (0) = 3 is a local maximum value and i (±1) = 2 are local minimum values.
I I I
I
(c) i 00 ({) = 12{2 3 4 = 12 {2 3 13 = 12 { + 1@ 3 { 3 1@ 3 . i 00 ({) A 0 C { ? 31@ 3 or { A 1@ 3 and
I
I
I
I
3@3> " and concave
i 00 ({) ? 0 C 31@ 3 ? { ? 1@ 3. Thus, i is concave upward on 3"> 3 3@3 and
I
I
I
.
downward on 3 3@3> 3@3 . There are inÀection points at ± 3@3> 22
9
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316
¤
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CHAPTER 4
12. (a) i ({) =
{2
APPLICATIONS OF DIFFERENTIATION
{
+1
i i 0 ({) =
({2 + 1)(1) 3 {(2{)
1 3 {2
({ + 1)({ 3 1)
= 2
=3
. Thus, i 0 ({) A 0 if
2
2
({ + 1)
({ + 1)2
({2 + 1)2
({ + 1)({ 3 1) ? 0 C 31 ? { ? 1, and i 0 ({) ? 0 if { ? 31 or { A 1. So i is increasing on (31> 1) and i is
decreasing on (3"> 31) and (1> ").
(b) i changes from decreasing to increasing at { = 31 and from increasing to decreasing at { = 1. Thus, i (31) = 3 12 is a
local minimum value and i (1) =
1
2
is a local maximum value.
({2 + 1)2 (32{) 3 (1 3 {2 )[2({2 + 1)(2{)]
({2 + 1)(32{)[({2 + 1) + 2(1 3 {2 )]
2{({2 3 3)
=
=
.
2
2
2
2
4
[({ + 1) ]
({ + 1)
({2 + 1)3
I
I
I
I
i 00 ({) A 0 C 3 3 ? { ? 0 or { A 3, and i 00 ({) ? 0 C { ? 3 3 or 0 ? { ? 3. Thus, i is concave
I I I
I upward on 3 3> 0 and
3> " and concave downward on 3"> 3 3 and 0> 3 . There are inÀection points at
I
I I
I
3> 3@4 .
3 3> 3 3@4 , (0> 0), and
(c) i 00 ({) =
13. (a) i ({) = sin { + cos {, 0 $ { $ 2.
i 0 ({) = cos { 3 sin { = 0 i cos { = sin { i 1 =
sin {
cos {
i
Thus, i 0 ({) A 0 C cos { 3 sin { A 0 C cos { A sin { C 0 ? { ? 4 or
5
? { ? 2 and i 0 ({) ? 0 C cos { ? sin { C 4 ? { ? 5
. So i is increasing on 0> 4 and 5
> 2 and i
4
4
4
.
is decreasing on 4 > 5
4
tan { = 1 i { =
4
or
5
.
4
(b) i changes from increasing to decreasing at { = 4 and from decreasing to increasing at { =
I
local maximum value and i 5
= 3 2 is a local minimum value.
4
(c) i 00 ({) = 3 sin { 3 cos { = 0 i 3 sin { = cos { i tan { = 31 i { =
3
4
or
I
2 is a
5
.
4
Thus, i
7
.
4
Divide the interval
4
=
(0> 2) into subintervals with these numbers as endpoints and complete a second derivative chart.
Interval
3 0> 4
3 7 > 4
4
7
> 2
4
There are inÀection points at
3
4
i 00 ({) = 3 sin { 3 cos {
i 00 2 = 31 ? 0
i 00 () = 1 A 0
1
i 00 11
= 23
6
1
2
I
3?0
Concavity
downward
upward
downward
> 0 and 7
>0 .
4
14. (a) i ({) = cos2 { 3 2 sin {, 0 $ { $ 2.
i 0 ({) = 32 cos { sin { 3 2 cos { = 32 cos { (1 + sin {). Note that
1 + sin { D 0 [since sin { D 31], with equality C sin { = 31 C { =
3
2
[since 0 $ { $ 2] i
and i 0 ({) ? 0 C cos { A 0 C 0 ? { ? 2
cos { = 0. Thus, i 0 ({) A 0 C cos { ? 0 C 2 ? { ? 3
2
or 3
? { ? 2. Thus, i is increasing on 2 > 3
> 2 .
and i is decreasing on 0> 2 and 3
2
2
2
(b) i changes from decreasing to increasing at { = 2 and from increasing to decreasing at { = 3
. Thus, i 2 = 32 is a
2
local minimum value and i 3
= 2 is a local maximum value.
2
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SECTION 4.3
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
317
(c) i 00 ({) = 2 sin { (1 + sin {) 3 2 cos2 { = 2 sin { + 2 sin2 { 3 2(1 3 sin2 {)
= 4 sin2 { + 2 sin { 3 2 = 2(2 sin { 3 1)(sin { + 1)
so i 00 ({) A 0 C sin { A
and i 00 ({) ? 0 C sin { ? 12 and sin { 6= 31 C
? { ? 3
or 3
? { ? 2= Thus, i is concave upward on 6 > 5
and concave downward on 0> 6 ,
0 ? { ? 6 or 5
6
2
2
6
5 3 > 2 , and 3
> 2 . There are inÀection points at 6 > 3 14 and 5
> 3 14 .
6
2
6
15. (a) i ({) = h2{ + h3{
1
2
C
6
?{?
5
,
6
i i 0 ({) = 2h2{ 3 h3{ . i 0 ({) A 0 C 2h2{ A h3{
C h3{ A
1
2
C 3{ A ln 12 C
{ A 13 (ln 1 3 ln 2) C { A 3 13 ln 2 [r 30=23] and i 0 ({) ? 0 if { ? 3 13 ln 2. So i is increasing on 3 13 ln 2> "
and i is decreasing on 3"> 3 13 ln 2 .
(b) i changes from decreasing to increasing at { = 3 13 ln 2. Thus,
s
I
I
I
I
s
I
3
3
3
3
i 3 13 ln 2 = i ln 3 1@2 = h2 ln 1@2 + h3 ln 1@2 = hln 1@4 + hln 2 = 3 1@4 + 3 2 = 232@3 + 21@3 [r 1=89]
is a local minimum value.
(c) i 00 ({) = 4h2{ + h3{ A 0 [the sum of two positive terms]. Thus, i is concave upward on (3"> ") and there is no
point of inÀection.
16. (a) i ({) = {2 ln {
i i 0 ({) = {2 (1@{) + (ln {)(2{) = { + 2{ ln { = {(1 + 2 ln {). The domain of i is (0> "), so
the sign of i 0 is determined solely by the factor 1 + 2 ln {. i 0 ({) A 0 C ln { A 3 12
C { A h31@2 [r 0=61]
and i 0 ({) ? 0 C 0 ? { ? h31@2 . So i is increasing on (h31@2 > ") and i is decreasing on (0> h31@2 ).
(b) i changes from decreasing to increasing at { = h31@2 . Thus, i(h31@2 ) = (h31@2 )2 ln(h31@2 ) = h31 (31@2) = 31@(2h)
[r 30=18] is a local minimum value.
(c) i 0 ({) = {(1 + 2 ln {) i i 00 ({) = {(2@{) + (1 + 2 ln {) · 1 = 2 + 1 + 2 ln { = 3 + 2 ln {. i 00 ({) A 0 C
3 + 2 ln { A 0 C ln { A 33@2 C { A h33@2 [r 0=22]. Thus, i is concave upward on (h33@2 > ") and i is
concave downward on (0> h33@2 ). i(h33@2 ) = (h33@2 )2 ln h33@2 = h33 (33@2) = 33@(2h3 ) [r 30=07]. There is a
point of inÀection at h33@2 > i (h33@2 ) = h33@2 > 33@(2h3 ) .
17. (a) i ({) = {2 3 { 3 ln {
i i 0 ({) = 2{ 3 1 3
2{2 3 { 3 1
(2{ + 1)({ 3 1)
1
=
=
. Thus, i 0 ({) A 0 if { A 1
{
{
{
[note that { A 0] and i 0 ({) ? 0 if 0 ? { ? 1. So i is increasing on (1> ") and i is decreasing on (0> 1).
(b) i changes from decreasing to increasing at { = 1. Thus, i(1) = 0 is a local minimum value.
(c) i 00 ({) = 2 + 1@{2 A 0 for all {, so i is concave upward on (0> "). There is no inÀection point.
18. (a) i ({) = {4 h3{
i i 0 ({) = {4 (3h3{ ) + h3{ (4{3 ) = {3 h3{ (3{ + 4). Thus, i 0 ({) A 0 if 0 ? { ? 4 and i 0 ({) ? 0
if { ? 0 or { A 4. So i is increasing on (0> 4) and decreasing on (3"> 0) and (4> ").
(b) i changes from decreasing to increasing at { = 0 and from increasing to decreasing at { = 4. Thus, i (0) = 0 is a local
minimum value and i (4) = 256@h4 is a local maximum value.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(c) i 0 ({) = h3{ (3{4 + 4{3 ) i
i 00 ({) = h3{ (34{3 + 12{2 ) + (3{4 + 4{3 )(3h3{ ) = h3{ [(34{3 + 12{2 ) 3 (3{4 + 4{3 )]
= h3{ ({4 3 8{3 + 12{2 ) = {2 h3{ ({2 3 8{ + 12) = {2 h3{ ({ 3 2)({ 3 6)
i 00 ({) A 0 C { ? 2 [excluding 0] or { A 6 and i 00 ({) ? 0 C 2 ? { ? 6. Thus, i is concave upward on
(3"> 2) and (6> ") and i is concave downward on (2> 6). There are inÀection points at (2> 16h32 ) and (6> 1296h36 ).
19. i ({) = 1 + 3{2 3 2{3
i i 0 ({) = 6{ 3 6{2 = 6{(1 3 {).
First Derivative Test: i 0 ({) A 0 i 0 ? { ? 1 and i 0 ({) ? 0 i { ? 0 or { A 1. Since i 0 changes from negative
to positive at { = 0, i (0) = 1 is a local minimum value; and since i 0 changes from positive to negative at { = 1, i(1) = 2 is
a local maximum value.
Second Derivative Test: i 00 ({) = 6 3 12{. i 0 ({) = 0 C { = 0> 1. i 00 (0) = 6 A 0 i i (0) = 1 is a local
minimum value. i 00 (1) = 36 ? 0 i i (1) = 2 is a local maximum value.
Preference: For this function, the two tests are equally easy.
20. i ({) =
{2
{31
i i 0 ({) =
({ 3 1)(2{) 3 {2 (1)
{2 3 2{
{({ 3 2)
=
=
.
({ 3 1)2
({ 3 1)2
({ 3 1)2
First Derivative Test: i 0 ({) A 0 i { ? 0 or { A 2 and i 0 ({) ? 0 i 0 ? { ? 1 or 1 ? { ? 2. Since i 0 changes
from positive to negative at { = 0, i (0) = 0 is a local maximum value; and since i 0 changes from negative to positive at
{ = 2, i (2) = 4 is a local minimum value.
Second Derivative Test:
i 00 ({) =
({ 3 1)2 (2{ 3 2) 3 ({2 3 2{)2({ 3 1)
2({ 3 1)[({ 3 1)2 3 ({2 3 2{)]
2
=
=
.
2
2
[({ 3 1) ]
({ 3 1)4
({ 3 1)3
i 0 ({) = 0 C { = 0> 2. i 00 (0) = 32 ? 0 i i (0) = 0 is a local maximum value. i 00 (2) = 2 A 0 i i (2) = 4 is a
local minimum value.
Preference: Since calculating the second derivative is fairly dif¿cult, the First Derivative Test is easier to use for this
function.
I
I
1
{ 3 4 { i i 0 ({) = {31@2 3
2
I
First Derivative Test: 2 4 { 3 1 A 0 i
21. i ({) =
I
1 33@4
1
24 {31
I
= {33@4 (2{1@4 3 1) =
{
4
4
4
4 {3
{A
Since i 0 changes from negative to positive at { =
1
16 ,
1
,
16
so i 0 ({) A 0 i { A
1
i ( 16
)=
1
4
3
1
2
1
16
and i 0 ({) ? 0 i 0 ? { ?
1
16 .
= 3 14 is a local minimum value.
1
3
1
3
.
Second Derivative Test: i 00 ({) = 3 {33@2 + {37@4 = 3 I + I
4
4
16
4 {3
16 {7
1
1
1
= 316 + 24 = 8 A 0 i i 16
= 3 14 is a local minimum value.
. i 00 16
i 0 ({) = 0 C { = 16
Preference: The First Derivative Test may be slightly easier to apply in this case.
22. (a) i ({) = {4 ({ 3 1)3
i i 0 ({) = {4 · 3({ 3 1)2 + ({ 3 1)3 · 4{3 = {3 ({ 3 1)2 [3{ + 4({ 3 1)] = {3 ({ 3 1)2 (7{ 3 4)
The critical numbers are 0, 1, and 47 .
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SECTION 4.3
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
319
(b) i 00 ({) = 3{2 ({ 3 1)2 (7{ 3 4) + {3 · 2({ 3 1)(7{ 3 4) + {3 ({ 3 1)2 · 7
= {2 ({ 3 1) [3({ 3 1)(7{ 3 4) + 2{(7{ 3 4) + 7{({ 3 1)]
Now i 00 (0) = i 00 (1) = 0, so the Second Derivative Test gives no information for { = 0 or { = 1.
i 00
4
7
=
4 2 4
7
7
2 3 1 0 + 0 + 7 47 47 3 1 = 47 3 37 (4) 3 37 A 0, so there is a local minimum at { = 47 .
(c) i 0 is positive on (3"> 0), negative on 0> 47 , positive on 47 > 1 , and positive on (1> "). So i has a local maximum at
{ = 0, a local minimum at { = 47 , and no local maximum or minimum at { = 1.
23. (a) By the Second Derivative Test, if i 0 (2) = 0 and i 00 (2) = 35 ? 0, i has a local maximum at { = 2.
(b) If i 0 (6) = 0, we know that i has a horizontal tangent at { = 6. Knowing that i 00 (6) = 0 does not provide any additional
information since the Second Derivative Test fails. For example, the ¿rst and second derivatives of | = ({ 3 6)4 ,
| = 3({ 3 6)4 , and | = ({ 3 6)3 all equal zero for { = 6, but the ¿rst has a local minimum at { = 6, the second has a
local maximum at { = 6, and the third has an inÀection point at { = 6.
24. Vertical asymptote { = 0
i 0 ({) A 0 if { ? 32 i i is increasing on (3"> 32).
i 0 ({) ? 0 if { A 32 ({ 6= 0) i i is decreasing on (32> 0) and (0> ").
i 00 ({) ? 0 if { ? 0 i i is concave downward on (3"> 0).
i 00 ({) A 0 if { A 0 i i is concave upward on (0> ").
25. i 0 (0) = i 0 (2) = i 0 (4) = 0
i horizontal tangents at { = 0, 2, 4.
0
i ({) A 0 if { ? 0 or 2 ? { ? 4 i i is increasing on (3"> 0) and (2> 4).
i 0 ({) ? 0 if 0 ? { ? 2 or { A 4 i i is decreasing on (0> 2) and (4> ").
i 00 ({) A 0 if 1 ? { ? 3 i i is concave upward on (1> 3).
i 00 ({) ? 0 if { ? 1 or { A 3 i i is concave downward on (3"> 1)
and (3> "). There are inÀection points when { = 1 and 3.
26. i 0 (1) = i 0 (31) = 0
i horizontal tangents at { = ±1.
0
i ({) ? 0 if |{| ? 1 i i is decreasing on (31> 1).
i 0 ({) A 0 if 1 ? |{| ? 2 i i is increasing on (32> 31) and (1> 2).
i 0 ({) = 31 if |{| A 2 i the graph of i has constant slope 31 on (3"> 32)
and (2> ").
i 00 ({) ? 0 if 32 ? { ? 0 i i is concave downward on (32> 0). The point (0> 1) is an inÀection point.
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°
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¤
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
27. i 0 ({) A 0 if |{| ? 2
i i is increasing on (32> 2).
i 0 ({) ? 0 if |{| A 2 i i is decreasing on (3"> 32)
and (2> ").
i 0 (32) = 0 i horizontal tangent at
{ = 32. lim |i 0 ({)| = " i there is a vertical
{<2
asymptote or vertical tangent (cusp) at { = 2. i 00 ({) A 0 if { 6= 2 i i is concave upward on (3"> 2) and (2> ").
28. i 0 ({) A 0 if |{| ? 2
i i is increasing on (32> 2). i 0 ({) ? 0 if |{| A 2 i
i is decreasing on (3"> 32) and (2> "). i 0 (2) = 0, so i has a horizontal tangent
(and local maximum) at { = 2. lim i ({) = 1 i | = 1 is a horizontal asymptote.
{<"
i (3{) = 3i ({) i i is an odd function (its graph is symmetric about the origin).
Finally, i 00 ({) ? 0 if 0 ? { ? 3 and i 00 ({) A 0 if { A 3, so i is CD on (0> 3) and
CU on (3> ").
29. The function must be always decreasing (since the ¿rst derivative is always
negative) and concave downward (since the second derivative is always
negative).
30. (a) i (3) = 2
i the point (3> 2) is on the graph of i . i 0 (3) =
1
2
i the slope of
the tangent line at (3> 2) is 12 . i 0 ({) A 0 for all { i i is increasing on R.
i 00 ({) ? 0 for all { i i is concave downward on R. A possible graph for i
is shown.
(b) The tangent line at (3> 2) has equation | 3 2 = 12 ({ 3 3), or | = 12 { + 12 , and {-intercept 31. Since i is concave
downward on R, i is below the {-axis at { = 31, and hence changes sign at least once. Since i is increasing on R,
it changes sign at most once. Thus, it changes sign exactly once and there is one solution of the equation i({) = 0.
(c) i 00 ? 0 i i 0 is decreasing. Since i 0 (3) = 12 , i 0 (2) must be greater than 12 , so no, it is not possible that i 0 (2) = 13 .
31. (a) i is increasing where i 0 is positive, that is, on (0> 2), (4> 6), and (8> "); and decreasing where i 0 is negative, that is, on
(2> 4) and (6> 8).
(b) i has local maxima where i 0 changes from positive to negative, at { = 2 and at { = 6, and local minima where i 0 changes
from negative to positive, at { = 4 and at { = 8.
(c) i is concave upward (CU) where i 0 is increasing, that is, on (3> 6) and (6> "), and concave downward (CD) where i 0 is
decreasing, that is, on (0> 3).
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.3
(d) There is a point of inÀection where i changes from
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
321
(e)
being CD to being CU, that is, at { = 3.
32. (a) i is increasing where i 0 is positive, on (1> 6) and (8> "), and decreasing where i 0 is negative, on (0> 1) and (6> 8).
(b) i has a local maximum where i 0 changes from positive to negative, at { = 6, and local minima where i 0 changes from
negative to positive, at { = 1 and at { = 8.
(c) i is concave upward where i 0 is increasing, that is, on (0> 2), (3> 5), and (7> "), and concave downward where i 0 is
decreasing, that is, on (2> 3) and (5> 7).
(d) There are points of inÀection where i changes its
(e)
direction of concavity, at { = 2, { = 3, { = 5 and
{ = 7.
33. (a) i ({) = {3 3 12{ + 2
i i 0 ({) = 3{2 3 12 = 3({2 3 4) = 3({ + 2)({ 3 2). i 0 ({) A 0 C { ? 32 or { A 2
and i 0 ({) ? 0 C 32 ? { ? 2. So i is increasing on (3"> 32) and (2> ") and i is decreasing on (32> 2).
(b) i changes from increasing to decreasing at { = 32, so i (32) = 18 is a local maximum value. i changes from decreasing
to increasing at { = 2, so i (2) = 314 is a local minimum value.
(c) i 00 ({) = 6{. i 00 ({) = 0 C { = 0. i 00 ({) A 0 on (0> ") and
(d)
i 00 ({) ? 0 on (3"> 0). So i is concave upward on (0> ") and i is
concave downward on (3"> 0). There is an inÀection point at (0> 2).
34. (a) i ({) = 36{ + 3{2 3 2{3
i i 0 ({) = 36 + 6{ 3 6{2 = 36({2 3 { 3 6) = 36({ + 2)({ 3 3). i 0 ({) A 0 C
32 ? { ? 3 and i 0 ({) ? 0 C { ? 32 or { A 3. So i is increasing on (32> 3) and i is decreasing on (3"> 32)
and (3> ").
(b) i changes from increasing to decreasing at { = 3, so i (3) = 81 is a local maximum value. i changes from decreasing to
increasing at { = 32, so i (32) = 344 is a local minimum value.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(c) i 00 ({) = 6 3 12{. i 00 ({) = 0 C { = 12 . i 00 ({) A 0 on 3"> 12
and i 00 ({) ? 0 on 12 > " . So i is CU on 3"> 12 and i is CD on
1
.
> " . There is an inÀection point at 12 > 37
2
2
35. (a) i ({) = 2 + 2{2 3 {4
(d)
i i 0 ({) = 4{ 3 4{3 = 4{(1 3 {2 ) = 4{(1 + {)(1 3 {). i 0 ({) A 0 C { ? 31 or
0 ? { ? 1 and i 0 ({) ? 0 C 31 ? { ? 0 or { A 1. So i is increasing on (3"> 31) and (0> 1) and i is decreasing
on (31> 0) and (1> ").
(b) i changes from increasing to decreasing at { = 31 and { = 1, so i (31) = 3 and i (1) = 3 are local maximum values.
i changes from decreasing to increasing at { = 0, so i (0) = 2 is a local minimum value.
(c) i 00 ({) = 4 3 12{2 = 4(1 3 3{2 ). i 00 ({) = 0 C 1 3 3{2 = 0 C
I
I
I {2 = 13 C { = ±1@ 3. i 00 ({) A 0 on 31@ 3> 1@ 3 and i 00 ({) ? 0
I I
on 3"> 31@ 3 and 1@ 3> " . So i is concave upward on
I I I
31@ 3> 1@ 3 and i is concave downward on 3"> 31@ 3 and
I I
1@ 3> " . i ±1@ 3 = 2 + 23 3 19 = 23
. There are points of inÀection
9
I 23 at ±1@ 3> 9 .
36. (a) j({) = 200 + 8{3 + {4
(d)
i j0 ({) = 24{2 + 4{3 = 4{2 (6 + {) = 0 when { = 36 and when { = 0.
j 0 ({) A 0 C { A 36 [{ 6= 0] and j 0 ({) ? 0 C { ? 36, so j is decreasing on (3"> 36) and j is increasing
on (36> "), with a horizontal tangent at { = 0.
(b) j(36) = 3232 is a local minimum value.
(d)
There is no local maximum value.
(c) j 00 ({) = 48{ + 12{2 = 12{(4 + {) = 0 when { = 34 and when { = 0.
j 00 ({) A 0 C { ? 34 or { A 0 and j 00 ({) ? 0 C 34 ? { ? 0, so j is
CU on (3"> 34) and (0> "), and j is CD on (34> 0). There are inÀection
points at (34> 356) and (0> 200).
37. (a) k({) = ({ + 1)5 3 5{ 3 2
i k0 ({) = 5({ + 1)4 3 5. k0 ({) = 0 C 5({ + 1)4 = 5 C ({ + 1)4 = 1 i
({ + 1)2 = 1 i { + 1 = 1 or { + 1 = 31 i { = 0 or { = 32. k0 ({) A 0 C { ? 32 or { A 0 and
k0 ({) ? 0 C 32 ? { ? 0. So k is increasing on (3"> 32) and (0> ") and k is decreasing on (32> 0).
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NOT FOR SALE
SECTION 4.3
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
(b) k(32) = 7 is a local maximum value and k(0) = 31 is a local minimum value.
¤
323
(d)
(c) k00 ({) = 20({ + 1)3 = 0 C { = 31. k00 ({) A 0 C { A 31 and
k00 ({) ? 0 C { ? 31, so k is CU on (31> ") and k is CD on (3"> 31).
There is a point of inÀection at (31> k(31)) = (31> 3).
i k0 ({) = 15{2 3 15{4 = 15{2 (1 3 {2 ) = 15{2 (1 + {)(1 3 {). k0 ({) A 0 C
38. (a) k({) = 5{3 3 3{5
31 ? { ? 0 and 0 ? { ? 1 [note that k0 (0) = 0] and k0 ({) ? 0 C { ? 31 or { A 1. So k is increasing on (31> 1)
and k is decreasing on (3"> 31) and (1> ").
(b) k changes from decreasing to increasing at { = 31, so k(31) = 32 is a local minimum value. k changes from increasing
to decreasing at { = 1, so k(1) = 2 is a local maximum value.
(c) k00 ({) = 30{ 3 60{3 = 30{(1 3 2{2 ). k00 ({) = 0 C { = 0 or
(d)
I
I
1 3 2{2 = 0 C { = 0 or { = ±1@ 2. k00 ({) A 0 on 3"> 31@ 2 and
I I I
0> 1@ 2 , and k00 ({) ? 0 on 31@ 2> 0 and 1@ 2> " . So k is CU on
I I I I
3"> 31@ 2 and 0> 1@ 2 , and k is CD on 31@ 2> 0 and 1@ 2> " .
I
I I
I There are inÀection points at 31@ 2> 37@ 4 2 , (0> 0), and 1@ 2> 7@ 4 2 .
I
39. (a) I ({) = { 6 3 {
i
I 0 ({) = { · 12 (6 3 {)31@2 (31) + (6 3 {)1@2 (1) = 12 (6 3 {)31@2 [3{ + 2(6 3 {)] =
33{ + 12
I
.
2 63{
I 0 ({) A 0 C 33{ + 12 A 0 C { ? 4 and I 0 ({) ? 0 C 4 ? { ? 6. So I is increasing on (3"> 4) and I is
decreasing on (4> 6).
I
(b) I changes from increasing to decreasing at { = 4, so I (4) = 4 2 is a local maximum value. There is no local minimum
value.
(c) I 0 ({) = 3 32 ({ 3 4)(6 3 {)31@2 i
l
k
I 00 ({) = 3 32 ({ 3 4) 3 12 (6 3 {)33@2 (31) + (6 3 {)31@2 (1)
(d)
3 1
3({ 3 8)
= 3 · (6 3 {)33@2 [({ 3 4) + 2(6 3 {)] =
2 2
4(6 3 {)3@2
I 00 ({) ? 0 on (3"> 6), so I is CD on (3"> 6). There is no inÀection point.
40. (a) J({) = 5{2@3 3 2{5@3
i J0 ({) =
10 31@3
{
3
3
10 2@3
{
3
=
10 31@3
{
(1
3
3 {) =
10(1 3 {)
.
3{1@3
J0 ({) A 0 C 0 ? { ? 1 and J0 ({) ? 0 C { ? 0 or { A 1. So J is increasing on (0> 1) and J is decreasing on
(3"> 0) and (1> ").
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324
¤
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(b) J changes from decreasing to increasing at { = 0, so J(0) = 0 is a local minimum value. J changes from increasing to
decreasing at { = 1, so J(1) = 3 is a local maximum value. Note that the First Derivative Test applies at { = 0 even
though J0 is not de¿ned at { = 0, since J is continuous at 0.
(c) J00 ({) = 3 10
{34@3 3
9
20 31@3
{
9
= 3 10
{34@3 (1 + 2{). J00 ({) A 0 C
9
(d)
{ ? 3 12 and J00 ({) ? 0 C 3 12 ? { ? 0 or { A 0. So J is CU on
3"> 3 12 and J is CD on 3 12 > 0 and (0> "). The only change in concavity
I occurs at { = 3 12 , so there is an inÀection point at 3 12 > 6@ 3 4 .
i F 0 ({) = 43 {1@3 + 43 {32@3 = 43 {32@3 ({ + 1) =
41. (a) F({) = {1@3 ({ + 4) = {4@3 + 4{1@3
4({ + 1)
I
. F 0 ({) A 0 if
3
3 {2
31 ? { ? 0 or { A 0 and F 0 ({) ? 0 for { ? 31, so F is increasing on (31> ") and F is decreasing on (3"> 31).
(b) F(31) = 33 is a local minimum value.
(d)
(c) F 00 ({) = 49 {32@3 3 89 {35@3 = 49 {35@3 ({ 3 2) =
4({ 3 2)
I
.
3
9 {5
F 00 ({) ? 0 for 0 ? { ? 2 and F 00 ({) A 0 for { ? 0 and { A 2, so F is
concave downward on (0> 2) and concave upward on (3"> 0) and (2> ").
I There are inÀection points at (0> 0) and 2> 6 3 2 E (2> 7=56).
42. (a) i ({) = ln({4 + 27)
i i 0 ({) =
4{3
. i 0 ({) A 0 if { A 0 and i 0 ({) ? 0 if { ? 0, so i is increasing on (0> ")
+ 27
{4
and i is decreasing on (3"> 0).
(b) i (0) = ln 27 E 3=3 is a local minimum value.
(c) i 00 ({) =
=
4{2 3({4 + 27) 3 4{4
({4 + 27)(12{2 ) 3 4{3 (4{3 )
=
({4 + 27)2
({4 + 27)2
(d)
4{2 (81 3 {4 )
34{2 ({2 + 9)({ + 3)({ 3 3)
=
4
2
({ + 27)
({4 + 27)2
i 00 ({) A 0 if 33 ? { ? 0 and 0 ? { ? 3, and i 00 ({) ? 0 if { ? 33 or
{ A 3. Thus, i is concave upward on (33> 0) and (0> 3) [hence on (33> 3)]
and i is concave downward on (3"> 33) and (3> "). There are inÀection
points at (±3> ln 108) E (±3> 4=68).
43. (a) i () = 2 cos + cos2 , 0 $ $ 2
i i 0 () = 32 sin + 2 cos (3 sin ) = 32 sin (1 + cos ).
i 0 () = 0 C = 0> > and 2. i 0 () A 0 C ? ? 2 and i 0 () ? 0 C 0 ? ? . So i is increasing
on (> 2) and i is decreasing on (0> ).
(b) i () = 31 is a local minimum value.
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SECTION 4.3
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
325
(c) i 0 () = 32 sin (1 + cos ) i
i 00 () = 32 sin (3 sin ) + (1 + cos )(32 cos ) = 2 sin2 3 2 cos 3 2 cos2 = 2(1 3 cos2 ) 3 2 cos 3 2 cos2 = 34 cos2 3 2 cos + 2
= 32(2 cos2 + cos 3 1) = 32(2 cos 3 1)(cos + 1)
(d)
and
Since 32(cos + 1) ? 0 [for 6= ], i 00 () A 0 i 2 cos 3 1 ? 0 i cos ? 12 i 3 ? ? 5
3
? ? 2. So i is CU on 3 > 5
i 00 () ? 0 i cos A 12 i 0 ? ? 3 or 5
and i is CD on 0> 3 and
3
3
5 5 5 5 5
= 3 > 4 and 5
= 3>4 .
3 > 2 . There are points of inÀection at 3 > i 3
3 >i 3
44. (a) V({) = { 3 sin {, 0 $ { $ 4
i V 0 ({) = 1 3 cos {. V 0 ({) = 0 C cos { = 1 C { = 0, 2, and 4.
V 0 ({) A 0 C cos { ? 1, which is true for all { except integer multiples of 2, so V is increasing on (0> 4)
since V 0 (2) = 0.
(b) There is no local maximum or minimum.
00
(d)
00
00
(c) V ({) = sin {. V ({) A 0 if 0 ? { ? or 2 ? { ? 3, and V ({) ? 0 if
? { ? 2 or 3 ? { ? 4. So V is CU on (0> ) and (2> 3), and V is CD
on (> 2) and (3> 4). There are inÀection points at (> ), (2> 2), and
(3> 3).
1
1
has domain (3"> 0) (0> ").
3
{ {2
2
{ +{31
1
1
1
1
(a) lim
1 + 3 2 = 1, so | = 1 is a HA. lim 1 + 3 2 = lim
= 3" since
{<±"
{
{
{ {
{2
{<0+
{<0+
45. i ({) = 1 +
({2 + { 3 1) < 31 and {2 < 0 as { < 0+ [a similar argument can be made for { < 03 ], so { = 0 is a VA.
(b) i 0 ({) = 3
1
2
1
+ 3 = 3 3 ({ 3 2). i 0 ({) = 0 C { = 2. i 0 ({) A 0 C 0 ? { ? 2 and i 0 ({) ? 0 C { ? 0
{2
{
{
or { A 2. So i is increasing on (0> 2) and i is decreasing on (3"> 0) and (2> ").
(c) i changes from increasing to decreasing at { = 2, so i (2) =
5
4
is a local
(e)
maximum value. There is no local minimum value.
(d) i 00 ({) =
2
6
2
3 4 = 4 ({ 3 3). i 00 ({) = 0 C { = 3. i 00 ({) A 0 C
{3
{
{
{ A 3 and i 00 ({) ? 0 C { ? 0 or 0 ? { ? 3. So i is CU on (3> ") and i
.
is CD on (3"> 0) and (0> 3). There is an inÀection point at 3> 11
9
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326
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CHAPTER 4
46. i ({) =
APPLICATIONS OF DIFFERENTIATION
{2 3 4
has domain R.
{2 + 4
(a) lim
{<±"
{2 3 4
1 3 4@{2
1
=
lim
= = 1, so | = 1 is a HA. There is no vertical asymptote.
{2 + 4 {<±" 1 + 4@{2
1
(b) i 0 ({) =
({2 + 4)(2{) 3 ({2 3 4)(2{)
2{[({2 + 4) 3 ({2 3 4)]
16{
=
= 2
. i 0 ({) A 0 C { A 0 and
2
2
({ + 4)
({2 + 4)2
({ + 4)2
i 0 ({) ? 0 C { ? 0. So i is increasing on (0> ") and i is decreasing on (3"> 0).
(c) i changes from decreasing to increasing at { = 0, so i (0) = 31 is a local minimum value.
({2 + 4)2 (16) 3 16{ · 2({2 + 4)(2{)
16({2 + 4)[({2 + 4) 3 4{2 ]
16(4 3 3{2 )
=
=
.
2
2
2
2
4
[({ + 4) ]
({ + 4)
({2 + 4)3
I
I
I
i 00 ({) = 0 C { = ±2@ 3. i 00 ({) A 0 C 32@ 3 ? { ? 2@ 3
I
I
and i 00 ({) ? 0 C { ? 32@ 3 or { A 2@ 3. So i is CU on
(e)
I
I I I
32@ 3> 2@ 3 and i is CD on 3"> 32@ 3 and 2@ 3> " .
I
There are inÀection points at ±2@ 3> 3 12 .
(d) i 00 ({) =
47. (a)
lim
{<3"
I
{2 + 1 3 { = " and
I
I
I
{2 + 1 + {
1
{2 + 1 3 { = lim
{2 + 1 3 { I
= lim I
= 0, so | = 0 is a HA.
{<"
{<"
{2 + 1 + { {<" {2 + 1 + {
lim
(b) i ({) =
I
{
{
3 1. Since I
? 1 for all {, i 0 ({) ? 0, so i is decreasing on R.
{2 + 1 3 { i i 0 ({) = I
{2 + 1
{2 + 1
(c) No minimum or maximum
(d) i 00 ({) =
=
({2 + 1)1@2 (1) 3 { · 12 ({2 + 1)31@2 (2{)
I
2
{2 + 1
({2 + 1)1@2 3
{2
({2
{2
+ 1)1@2
=
+1
(e)
1
({2 + 1) 3 {2
= 2
A 0,
({2 + 1)3@2
({ + 1)3@2
so i is CU on R. No IP
48. i ({) =
h{
has domain {{ | 1 3 h{ 6= 0} = {{ | h{ 6= 1} = {{ | { 6= 0}.
1 3 h{
h{
h{ @h{
1
1
=
= 31, so | = 31 is a HA.
= lim
= lim
{
{<" 1 3 h
{<" (1 3 h{ )@h{
{<" 1@h{ 3 1
031
(a) lim
lim
{<3"
h{
0
= 0, so | = 0 is a HA.
=
1 3 h{
130
{
(b) i 0 ({) =
{
{
{
{
{
lim
{<0+
h{
h{
=
3"
and
lim
= ", so { = 0 is a VA.
1 3 h{
{<0 1 3 h{
(1 3 h )h 3 h (3h )
h [(1 3 h ) + h{ ]
h{
=
=
. i 0 ({) A 0 for { 6= 0, so i is increasing on
{
2
{
2
(1 3 h )
(1 3 h )
(1 3 h{ )2
(3"> 0) and (0> ").
(c) There is no local maximum or minimum.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.3
(d) i 00 ({) =
=
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
(1 3 h{ )2 h{ 3 h{ · 2(1 3 h{ )(3h{ )
[(1 3 h{ )2 ]2
¤
327
(e)
(1 3 h{ )h{ [(1 3 h{ ) + 2h{ ]
h{ (h{ + 1)
=
{
4
(1 3 h )
(1 3 h{ )3
i 00 ({) A 0 C (1 3 h{ )3 A 0 C h{ ? 1 C { ? 0 and
i 00 ({) ? 0 C { A 0. So i is CU on (3"> 0) and i is CD on (0> ").
There is no inÀection point.
49. (a)
2
lim h3{ = lim
{<±"
(b) i ({) = h3{
{<±"
2
1
= 0, so | = 0 is a HA. There is no VA.
h{2
2
i i 0 ({) = h3{ (32{). i 0 ({) = 0 C { = 0. i 0 ({) A 0 C { ? 0 and i 0 ({) ? 0 C
{ A 0. So i is increasing on (3"> 0) and i is decreasing on (0> ").
(c) i changes from increasing to decreasing at { = 0, so i(0) = 1 is a local maximum value. There is no local minimum
value.
2
2
2
(d) i 00 ({) = h3{ (32) + (32{)h3{ (32{) = 32h3{ (1 3 2{2 ).
I
i 00 ({) = 0 C {2 = 12 C { = ±1@ 2. i 00 ({) A 0 C
I
I
I
I
{ ? 31@ 2 or { A 1@ 2 and i 00 ({) ? 0 C 31@ 2 ? { ? 1@ 2. So
I
I I I
i is CU on 3"> 31@ 2 and 1@ 2> " , and i is CD on 31@ 2> 1@ 2 .
I
There are inÀection points at ±1@ 2> h31@2 .
50. i ({) = { 3 16 {2 3
(e)
2
3
ln { has domain (0> ").
(a) lim { 3 16 {2 3 23 ln { = " since ln { < 3" as { < 0+ , so { = 0 is a VA. There is no HA.
{<0+
(b) i 0 ({) = 1 3
2
3{ 3 {2 3 2
3({2 3 3{ + 2)
3({ 3 1)({ 3 2)
1
{3
=
=
=
. i 0 ({) A 0 C
3
3{
3{
3{
3{
({ 3 1)({ 3 2) ? 0 C 1 ? { ? 2 and i 0 ({) ? 0 C 0 ? { ? 1 or { A 2. So i is increasing on (1> 2) and
i is decreasing on (0> 1) and (2> ").
(c) i changes from decreasing to increasing at { = 1, so i (1) =
decreasing at { = 2, so i (2) =
4
3
3
2
3
5
6
is a local minimum value. i changes from increasing to
ln 2 r 0=87 is a local maximum value.
I
1
2 3 {2
2
(d) i 00 ({) = 3 + 2 =
. i 00 ({) A 0 C 0 ? { ? 2 and
3
3{
3{2
I
I i 00 ({) ? 0 C { A 2. So i is CU on 0> 2 and i is CD on
I I
I
2> " . There is an inÀection point at
2> 2 3 13 3 13 ln 2 .
(e)
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328
¤
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
51. i ({) = ln(1 3 ln {) is de¿ned when { A 0 (so that ln { is de¿ned) and 1 3 ln { A 0 [so that ln(1 3 ln {) is de¿ned].
The second condition is equivalent to 1 A ln { C { ? h> so i has domain (0> h).
(a) As { < 0+ > ln { < 3"> so 1 3 ln { < " and i ({) < "= As { < h3 > ln { < 13 > so 1 3 ln { < 0+ and
i ({) < 3". Thus, { = 0 and { = h are vertical asymptotes. There is no horizontal asymptote.
1
1
1
(b) i 0 ({) =
3
=3
? 0 on (0> h) = Thus, i is decreasing on its domain, (0> h) =
1 3 ln {
{
{(1 3 ln {)
(c) i 0 ({) 6= 0 on (0> h) > so i has no local maximum or minimum value.
(d) i 00 ({) = 3
=3
(e)
3 [{(1 3 ln {)]0
{(31@{) + (1 3 ln {)
=
{2 (1 3 ln {)2
[{(1 3 ln {)]2
ln {
{2 (1 3 ln {)2
so i 00 ({) A 0 C ln { ? 0 C 0 ? { ? 1= Thus, i is CU on (0> 1)
and CD on (1> h) = There is an inÀection point at (1> 0) =
52. (a) lim arctan { =
{<"
2,
so lim harctan { = h@2 [r 4=81], so | = h@2 is a HA.
{<"
lim harctan { = h3@2 [r 0=21], so | = h3@2 is a HA. No VA.
{<3"
(b) i ({) = harctan {
i i 0 ({) = harctan { .
1
A 0 for all {. Thus, i is increasing on R.
1 + {2
(c) There is no local maximum or minimum.
(d) i 00 ({) = harctan {
=
32{
1
1
+
· harctan { ·
(1 + {2 )2
1 + {2
1 + {2
(e)
harctan {
(32{ + 1)
(1 + {2 )2
i 00 ({) A 0 C 32{ + 1 A 0 C { ? 12 and i 00 ({) ? 0 C
{ A 12 , so i is CU on 3"> 12 and i is CD on 12 > " . There is an
inÀection point at 12 > i 12 = 12 > harctan(1@2) E 12 > 1=59 .
53. The nonnegative factors ({ + 1)2 and ({ 3 6)4 do not affect the sign of i 0 ({) = ({ + 1)2 ({ 3 3)5 ({ 3 6)4 .
So i 0 ({) A 0 i ({ 3 3)5 A 0 i { 3 3 A 0 i { A 3. Thus, i is increasing on the interval (3> ").
54. | = i ({) = {3 3 3d2 { + 2d3 , d A 0. The |-intercept is i (0) = 2d3 .
| 0 = 3{2 3 3d2 = 3({2 3 d2 ) = 3({ + d)({ 3 d).
The critical numbers are 3d and d. i 0 ? 0 on (3d> d), so i is decreasing on (3d> d) and i is increasing on (3"> 3d) and
(d> "). i(3d) = 4d3 is a local maximum value and i (d) = 0 is a local minimum value. Since i (d) = 0, d is an {-intercept,
and { 3 d is a factor of i . Synthetically dividing | = {3 3 3d2 { + 2d3 by { 3 d gives us the following result:
| = {3 3 3d2 { + 2d3 = ({ 3 d)({2 + d{ 3 2d2 ) = ({ 3 d)({ 3 d)({ + 2d) = ({ 3 d)2 ({ + 2d), which tells us
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NOT FOR SALE
SECTION 4.3
that the only {-intercepts are 32d and d. | 0 = 3{2 3 3d2
HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
329
i | 00 = 6{, so |00 A 0
on (0> ") and | 00 ? 0 on (3"> 0). This tells us that i is CU on (0> ") and CD on
(3"> 0). There is an inÀection point at (0> 2d3 ). The graph illustrates these features.
What the curves in the family have in common is that they are all CD on (3"> 0),
CU on (0> "), and have the same basic shape. But as d increases, the four key points
shown in the ¿gure move further away from the origin.
55. (a)
From the graph, we get an estimate of i(1) E 1=41 as a local maximum
value, and no local minimum value.
13{
.
({2 + 1)3@2
I
i 0 ({) = 0 C { = 1. i (1) = I22 = 2 is the exact value.
{+1
i ({) = I
{2 + 1
i i 0 ({) =
(b) From the graph in part (a), i increases most rapidly somewhere between { = 3 12 and { = 3 14 . To ¿nd the exact value,
we need to ¿nd the maximum value of i 0 , which we can do by ¿nding the critical numbers of i 0 .
I
I
3 + 17
2{2 3 3{ 3 1
3 ± 17
.
{
=
corresponds to the minimum value of i 0 .
=
0
C
{
=
i 00 ({) =
4
4
({2 + 1)5@2
The maximum value of i 0 occurs at { =
56. (a)
33
I
4
17
E 30=28.
Tracing the graph gives us estimates of i(0) = 0 for a local minimum value
and i (2) = 0=54 for a local maximum value.
i({) = {2 h3{
i i 0 ({) = {h3{ (2 3 {)= i 0 ({) = 0 C { = 0 or 2.
i(0) = 0 and i (2) = 4h32 are the exact values.
(b) From the graph in part (a), i increases most rapidly around { = 34 . To ¿nd the exact value, we need to ¿nd the maximum
value of i 0 , which we can do by ¿nding the critical numbers of i 0 . i 00 ({) = h3{ {2 3 4{ + 2 = 0 i
I
I
{ = 2 ± 2. { = 2 + 2 corresponds to the minimum value of i 0 . The maximum value of i 0 is at
I I 2
I 2 3 2> 2 3 2 h32+ 2 E (0=59> 0=19).
57. i ({) = cos { +
(a)
1
2
cos 2{ i i 0 ({) = 3 sin { 3 sin 2{ i i 00 ({) = 3 cos { 3 2 cos 2{
From the graph of i , it seems that i is CD on (0> 1), CU on (1> 2=5), CD on
(2=5> 3=7), CU on (3=7> 5=3), and CD on (5=3> 2). The points of inÀection
appear to be at (1> 0=4), (2=5> 30=6), (3=7> 30=6), and (5=3> 0=4).
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¤
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
From the graph of i 00 (and zooming in near the zeros), it seems that i is CD
(b)
on (0> 0=94), CU on (0=94> 2=57), CD on (2=57> 3=71), CU on (3=71> 5=35),
and CD on (5=35> 2). Re¿ned estimates of the inÀection points are
(0=94> 0=44), (2=57> 30=63), (3=71> 30=63), and (5=35> 0=44).
58. i ({) = {3 ({ 3 2)4
i
i 0 ({) = {3 · 4({ 3 2)3 + ({ 3 2)4 · 3{2 = {2 ({ 3 2)3 [4{ + 3({ 3 2)] = {2 ({ 3 2)3 (7{ 3 6) i
i 00 ({) = (2{)({ 3 2)3 (7{ 3 6) + {2 · 3({ 3 2)2 (7{ 3 6) + {2 ({ 3 2)3 (7)
= {({ 3 2)2 [2({ 3 2)(7{ 3 6) + 3{(7{ 3 6) + 7{({ 3 2)]
= {({ 3 2)2 [42{2 3 72{ + 24] = 6{({ 3 2)2 (7{2 3 12{ + 4)
From the graph of i , it seems that i is CD on (3"> 0), CU on (0> 0=5), CD
(a)
on (0=5> 1=3), and CU on (1=3> "). The points of inÀection appear to be at
(0> 0), (0=5> 0=5), and (1=3> 0=6).
From the graph of i 00 (and zooming in near the zeros), it seems that i is CD
(b)
on (3"> 0), CU on (0> 0=45), CD on (0=45> 1=26), and CU on (1=26> ").
Re¿ned estimates of the inÀection points are (0> 0), (0=45> 0=53),
and (1=26> 0=60).
{4 + {3 + 1
. In Maple, we de¿ne i and then use the command
{2 + { + 1
59. i ({) = I
plot(diff(diff(f,x),x),x=-2..2);. In Mathematica, we de¿ne i
and then use Plot[Dt[Dt[f,x],x],{x,-2,2}]. We see that i 00 A 0 for
{ ? 30=6 and { A 0=0 [E 0=03] and i 00 ? 0 for 30=6 ? { ? 0=0. So i is CU
on (3"> 30=6) and (0=0> ") and CD on (30=6> 0=0).
60. i ({) =
{2 tan31 {
. It appears that i 00 is positive (and thus i is concave
1 + {3
upward) on (3"> 31), (0> 0=7), and (2=5> "); and i 00 is negative (and thus i is
concave downward) on (31> 0) and (0=7> 2=5).
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¤
331
61. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about
w = 8 hours, and decreases toward 0 as the population begins to level off.
(b) The rate of increase has its maximum value at w = 8 hours.
(c) The population function is concave upward on (0> 8) and concave downward on (8> 18).
(d) At w = 8, the population is about 350, so the inÀection point is about (8> 350).
62. (a) I’m very unhappy. It’s uncomfortably hot and i 0 (3) = 2 indicates that the
temperature is increasing, and i 00 (3) = 4 indicates that the rate of increase
is increasing. (The temperature is rapidly getting warmer.)
(b) I’m still unhappy, but not as unhappy as in part (a). It’s uncomfortably hot
and i 0 (3) = 2 indicates that the temperature is increasing, but i 00 (3) = 34
indicates that the rate of increase is decreasing. (The temperature is slowly
getting warmer.)
(c) I’m somewhat happy. It’s uncomfortably hot and i 0 (3) = 32 indicates that
the temperature is decreasing, but i 00 (3) = 4 indicates that the rate of
change is increasing. (The rate of change is negative but it’s becoming less
negative. The temperature is slowly getting cooler.)
(d) I’m very happy. It’s uncomfortably hot and i 0 (3) = 32 indicates that the
temperature is decreasing, and i 00 (3) = 34 indicates that the rate of change
is decreasing, that is, becoming more negative. (The temperature is rapidly
getting cooler.)
63. Most students learn more in the third hour of studying than in the eighth hour, so N(3) 3 N(2) is larger than N(8) 3 N(7).
In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so N 0 (w)
decreases and the graph of N is concave downward.
64. At ¿rst the depth increases slowly because the base of the mug is wide.
But as the mug narrows, the coffee rises more quickly. Thus, the depth
g increases at an increasing rate and its graph is concave upward. The
rate of increase of g has a maximum where the mug is narrowest; that is,
when the mug is half full. It is there that the inÀection point (IP) occurs.
Then the rate of increase of g starts to decrease as the mug widens and
the graph becomes concave down.
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APPLICATIONS OF DIFFERENTIATION
65. V(w) = Dws h3nw with D = 0=01, s = 4, and n = 0=07. We will ¿nd the
zeros of i 00 for i(w) = ws h3nw .
i 0 (w) = ws (3nh3nw ) + h3nw (sws31 ) = h3nw (3nws + sws31 )
i 00 (w) = h3nw (3nsws31 + s(s 3 1)ws32 ) + (3nws + sws31 )(3nh3nw )
= ws32 h3nw [3nsw + s(s 3 1) + n2 w2 3 nsw]
= ws32 h3nw (n2 w2 3 2nsw + s2 3 s)
Using the given values of s and n gives us i 00 (w) = w2 h30=07w (0=0049w2 3 0=56w + 12). So V 00 (w) = 0=01i 00 (w) and its zeros
are w = 0 and the solutions of 0=0049w2 3 0=56w + 12 = 0, which are w1 =
200
7
600
7
r 28=57 and w2 =
r 85=71.
At w1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at w2 minutes, the rate of
decrease is the greatest.
66. (a) As |{| < ", w = 3{2 @(22 ) < 3", and hw < 0. The HA is | = 0. Since w takes on its maximum value at { = 0, so
2
does hw . Showing this result using derivatives, we have i ({) = h3{ @(2
2)
2
2
i i 0 ({) = h3{ @(2 ) (3{@2 ).
i 0 ({) = 0 C { = 0. Because i 0 changes from positive to negative at { = 0, i (0) = 1 is a local maximum. For
l 31
2
2
2
2
2
2
1 k
inÀection points, we ¿nd i 00 ({) = 3 2 h3{ @(2 ) · 1 + {h3{ @(2 ) (3{@2 ) = 2 h3{ @(2 ) (1 3 {2@2 ).
i 00 ({) = 0 C {2 = 2
C { = ±. i 00 ({) ? 0 C {2 ? 2
C 3 ? { ? .
So i is CD on (3> ) and CU on (3"> 3) and (> "). IP at (±> h31@2 ).
(b) Since we have IP at { = ±, the inÀection points move away from the |-axis as increases.
From the graph, we see that as increases, the graph tends to spread out and
(c)
there is more area between the curve and the {-axis.
67. i ({) = d{3 + e{2 + f{ + g
i i 0 ({) = 3d{2 + 2e{ + f.
We are given that i(1) = 0 and i (32) = 3, so i (1) = d + e + f + g = 0 and
i (32) = 38d + 4e 3 2f + g = 3. Also i 0 (1) = 3d + 2e + f = 0 and
i 0 (32) = 12d 3 4e + f = 0 by Fermat’s Theorem. Solving these four equations, we get
d = 29 , e = 13 , f = 3 43 , g = 79 , so the function is i ({) = 19 2{3 + 3{2 3 12{ + 7 .
68. i ({) = d{he{
2
k
l
2
2
2
i i 0 ({) = d {he{ · 2e{ + he{ · 1 = dhe{ (2e{2 + 1). For i (2) = 1 to be a maximum value, we
must have i 0 (2) = 0. i(2) = 1 i 1 = 2dh4e and i 0 (2) = 0 i 0 = (8e + 1)dh4e . So 8e + 1 = 0 [d 6= 0] i
I
e = 3 18 and now 1 = 2dh31@2 i d = h/2.
I
I
i i 0 ({) = 3{2 + 2d{ + e. i has the local minimum value 3 29 3 at { = 1@ 3, so
I
I
I
I
i 0 ( I13 ) = 0 i 1 + I23 d + e = 0 (1) and i ( I13 ) = 3 29 3 i 19 3 + 13 d + 13 3e = 3 29 3 (2).
69. (a) i ({) = {3 + d{2 + e{
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333
Rewrite the system of equations as
2
3
I
3d +
1
d
3
e = 31
I
I
1
3e = 3 13 3
3
+
I
and then multiplying (4) by 32 3 gives us the system
I
2
3d + e = 31
3
I
3 23 3d 3 2e = 2
(3)
(4)
Adding the equations gives us 3e = 1 i e = 31. Substituting 31 for e into (3) gives us
I
I
2
3d 3 1 = 31 i 23 3d = 0 i d = 0. Thus, i ({) = {3 3 {.
3
(b) To ¿nd the smallest slope, we want to ¿nd the minimum of the slope function, i 0 , so we’ll ¿nd the critical
numbers of i 0 . i ({) = {3 3 { i i 0 ({) = 3{2 3 1 i i 00 ({) = 6{. i 00 ({) = 0 C { = 0.
At { = 0, | = 0, i 0 ({) = 31, and i 00 changes from negative to positive. Thus, we have a minimum for i 0 and
| 3 0 = 31({ 3 0), or | = 3{, is the tangent line that has the smallest slope.
70. The original equation can be written as ({2 + e)| + d{ = 0. Call this (1). Since (2> 2=5) is on this curve, we have
(4 + e) 52 + 2d = 0, or 20 + 5e + 4d = 0. Let’s rewrite that as 4d + 5e = 320 and call it (A). Differentiating (1) gives
(after regrouping) ({2 + e)| 0 = 3(2{| + d). Call this (2). Differentiating again gives ({2 + e)| 00 + (2{)| 0 = 32{| 0 3 2|,
or ({2 + e)| 00 + 4{| 0 + 2| = 0. Call this (3). At (2> 2=5), equations (2) and (3) say that (4 + e)| 0 = 3(10 + d) and
(4 + e)| 00 + 8| 0 + 5 = 0. If (2> 2=5) is an inÀection point, then |00 = 0 there, so the second condition becomes 8| 0 + 5 = 0,
or |0 = 3 58 . Substituting in the ¿rst condition, we get 3(4 + e) 58 = 3(10 + d), or 20 + 5e = 80 + 8d, which simpli¿es to
38d + 5e = 60. Call this (B). Subtracting (B) from (A) yields 12d = 380, so d = 3 20
. Substituting that value in (A) gives
3
60
3 80
3 + 5e = 320 = 3 3 , so 5e =
20
3
and e = 43 . Thus far we’ve shown that IF the curve has an inÀection point at (2> 2=5),
4
then d = 3 20
3 and e = 3 .
4
To prove the converse, suppose that d = 3 20
3 and e = 3 . Then by (1), (2), and (3), our curve satis¿es
2 4
{ + 3 | = 20
{
(4)
3
2 4 0
20
(5)
{ + 3 | = 32{| + 3
2 4 00
and
(6)
{ + 3 | + 4{| 0 + 2| = 0.
2
+ 2 20
{ = 0, or
Multiply (6) by {2 + 43 and substitute from (4) and (5) to obtain {2 + 43 | 00 + 4{ 32{| + 20
3
3
2 4 2 00
{ + 3 | 3 8{2 | + 40{ = 0. Now multiply by ({2 + e) again and substitute from the ¿rst equation to obtain
3
2 4 3 00
2 4
3
00
= 0, or {2 + 43 | 00 3 40
{ + 3 | 3 8{2 20
3 { + 40{ { + 3
3 ({ 3 4{) = 0. The coef¿cient of | is positive, so
the sign of | 00 is the same as the sign of
3
40
3 ({
3 4{), which is a positive multiple of {({ + 2)({ 3 2). It is clear from this
00
expression that | changes sign at { = 0, { = 32, and { = 2, so the curve changes its direction of concavity at those values
4
of {. By (4), the corresponding |-values are 0, 32=5, and 2=5, respectively. Thus when d = 3 20
3 and e = 3 , the curve has
inÀection points, not only at (2> 2=5), but also at (0> 0) and (32> 32=5).
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CHAPTER 4
71. | =
1+{
1 + {2
| 00 =
=
APPLICATIONS OF DIFFERENTIATION
i |0 =
(1 + {2 )(1) 3 (1 + {)(2{)
1 3 2{ 3 {2
=
2
2
(1 + { )
(1 + {2 )2
i
(1 + {2 )2 (32 3 2{) 3 (1 3 2{ 3 {2 ) · 2(1 + {2 )(2{)
2(1 + {2 )[(1 + {2 )(31 3 {) 3 (1 3 2{ 3 {2 )(2{)]
=
2
2
2
[(1 + { ) ]
(1 + {2 )4
2(31 3 { 3 {2 3 {3 3 2{ + 4{2 + 2{3 )
2({3 + 3{2 3 3{ 3 1)
2({ 3 1)({2 + 4{ + 1)
=
=
2
3
2
3
(1 + { )
(1 + { )
(1 + {2 )3
I I
I
I
So | 00 = 0 i { = 1, 32 ± 3. Let d = 32 3 3, e = 32 + 3, and f = 1= We can show that i (d) = 14 1 3 3 ,
I i (e) = 14 1 + 3 , and i (f) = 1. To show that these three points of inÀection lie on one straight line, we’ll show that the
slopes pdf and pef are equal.
I I
3
1 3 14 1 3 3
+ 14 3
i (f) 3 i (d)
1
I = 4
I =
=
f3d
4
1 3 32 3 3
3+ 3
I
I
3
1 3 14 1 + 3
3 14 3
i (f) 3 i (e)
1
I = 4
I =
=
=
f3e
4
1 3 32 + 3
33 3
pdf =
pef
72. | = i ({) = h3{ sin {
i
| 0 = h3{ cos { + sin {(3h3{ ) = h3{ (cos { 3 sin {)
i
| 00 = h3{ (3 sin { 3 cos {) + (cos { 3 sin {)(3h3{ ) = h3{ (3 sin { 3 cos { 3 cos { + sin {) = h3{ (32 cos {).
So | 00 = 0 i cos { = 0 i { = 2 + q. At these values of {, i has points of inÀection and since
sin 2 + q = ±1, we get | = ±h3{ , so i intersects the other curves at its inÀection points.
Let j({) = h3{ and k({) = 3h3{ , so that j 0 ({) = 3h3{ and k0 ({) = h3{ . Now
i 0 2 + q = h3(@2+q) cos 2 + q 3 sin 2 + q = 3h3(@2+q) sin 2 + q . If q is odd, then
i 0 2 + q = h3(@2+q) = k0 2 + q . If q is even, then i 0 2 + q = 3h3(@2+q) = j0 2 + q .
Thus, at { =
73. | = { sin {
2
+ q, i has the same slope as either j or k> and hence, j and k touch i at its inÀection points.
i | 0 = { cos { + sin { i |00 = 3{ sin { + 2 cos {. | 00 = 0 i 2 cos { = { sin { [which is |] i
(2 cos {)2 = ({ sin {)2
cos2 {(4 + {2 ) = {2
i 4 cos2 { = {2 sin2 { i 4 cos2 { = {2 (1 3 cos2 {) i 4 cos2 { + {2 cos2 { = {2
i 4 cos2 {({2 + 4) = 4{2
i
i | 2 ({2 + 4) = 4{2 since | = 2 cos { when |00 = 0.
74. (a) We will make use of the converse of the Concavity Test (along with the stated assumptions); that is, if i is concave upward
on L, then i 00 A 0 on L. If i and j are CU on L, then i 00 A 0 and j 00 A 0 on L, so (i + j)00 = i 00 + j 00 A 0 on L
i
i + j is CU on L.
(b) Since i is positive and CU on L, i A 0 and i 00 A 0 on L. So j ({) = [i ({)]2
00
0 0
00
0 2
i j 0 = 2i i 0
i
00
j = 2i i + 2i i = 2 (i ) + 2i i A 0 i j is CU on L.
75. (a) Since i and j are positive, increasing, and CU on L with i 00 and j 00 never equal to 0, we have i A 0, i 0 D 0, i 00 A 0,
j A 0, j0 D 0, j 00 A 0 on L. Then (i j)0 = i 0 j + i j 0
i (ij)00 = i 00 j + 2i 0 j 0 + i j 00 D i 00 j + i j00 A 0 on L
i
i j is CU on L.
(b) In part (a), if i and j are both decreasing instead of increasing, then i 0 $ 0 and j 0 $ 0 on L, so we still have 2i 0 j 0 D 0
on L. Thus, (i j)00 = i 00 j + 2i 0 j0 + ij 00 D i 00 j + i j 00 A 0 on L
i i j is CU on L as in part (a).
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HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
335
(c) Suppose i is increasing and j is decreasing [with i and j positive and CU]. Then i 0 D 0 and j0 $ 0 on L, so 2i 0 j0 $ 0
on L and the argument in parts (a) and (b) fails.
Example 1.
L = (0> "), i ({) = {3 , j({) = 1@{. Then (i j)({) = {2 , so (i j)0 ({) = 2{ and
(ij)00 ({) = 2 A 0 on L. Thus, i j is CU on L.
Example 2.
Example 3.
I
I
I
L = (0> "), i ({) = 4{ {, j({) = 1@{. Then (i j)({) = 4 {, so (ij)0 ({) = 2@ { and
I
(ij)00 ({) = 31@ {3 ? 0 on L. Thus, i j is CD on L.
L = (0> "), i ({) = {2 , j({) = 1@{. Thus, (i j)({) = {, so i j is linear on L.
76. Since i and j are CU on (3"> "), i 00 A 0 and j 00 A 0 on (3"> ").
k({) = i (j({)) i
k0 ({) = i 0 (j({)) j0 ({) i k00 ({) = i 00 (j({)) j 0 ({) j0 ({) + i 0 (j({)) j 00 ({) = i 00 (j({))[j0 ({)]2 + i 0 (j({)) j00 ({) A 0
if i 0 A 0. So k is CU if i is increasing.
77. i ({) = tan { 3 {
i i 0 ({) = sec2 { 3 1 A 0 for 0 ? { ?
on 0> 2 . Thus, i ({) A i (0) = 0 for 0 ? { ?
2
2
since sec2 { A 1 for 0 ? { ?
2.
So i is increasing
i tan { 3 { A 0 i tan { A { for 0 ? { ?
.
2
78. (a) Let i ({) = h{ 3 1 3 {. Now i (0) = h0 3 1 = 0, and for { D 0, we have i 0 ({) = h{ 3 1 D 0. Now, since i (0) = 0
and i is increasing on [0> "), i({) D 0 for { D 0 i h{ 3 1 3 { D 0 i h{ D 1 + {.
(b) Let i({) = h{ 3 1 3 { 3 12 {2 . Thus, i 0 ({) = h{ 3 1 3 {, which is positive for { D 0 by part (a). Thus, i({) is
increasing on (0> "), so on that interval, 0 = i (0) $ i ({) = h{ 3 1 3 { 3 12 {2
(c) By part (a), the result holds for q = 1. Suppose that h{ D 1 + { +
Let i ({) = h{ 3 1 3 { 3
i h{ D 1 + { + 12 {2 .
{n
{2
+ ··· +
for { D 0.
2!
n!
{n
{n+1
{2
{n
3···3
3
. Then i 0 ({) = h{ 3 1 3 { 3 · · · 3
D 0 by assumption. Hence,
2!
n!
(n + 1)!
n!
i ({) is increasing on (0> "). So 0 $ { implies that 0 = i (0) $ i ({) = h{ 3 1 3 { 3 · · · 3
h{ D 1 + { + · · · +
{n+1
{n
3
, and hence
n!
(n + 1)!
{n
{2
{n+1
{q
+
for { D 0. Therefore, for { D 0, h{ D 1 + { +
+ ··· +
for every positive
n!
(n + 1)!
2!
q!
integer q, by mathematical induction.
79. Let the cubic function be i({) = d{3 + e{2 + f{ + g
i i 0 ({) = 3d{2 + 2e{ + f i i 00 ({) = 6d{ + 2e.
So i is CU when 6d{ + 2e A 0 C { A 3e@(3d), CD when { ? 3e@(3d), and so the only point of inÀection occurs
when { = 3e@(3d). If the graph has three {-intercepts {1 , {2 and {3 , then the expression for i ({) must factor as
i ({) = d({ 3 {1 )({ 3 {2 )({ 3 {3 ). Multiplying these factors together gives us
i ({) = d[{3 3 ({1 + {2 + {3 ){2 + ({1 {2 + {1 {3 + {2 {3 ){ 3 {1 {2 {3 ]
Equating the coef¿cients of the {2 -terms for the two forms of i gives us e = 3d({1 + {2 + {3 ). Hence, the {-coordinate of
the point of inÀection is 3
80. S ({) = {4 + f{3 + {2
e
3d({1 + {2 + {3 )
{1 + {2 + {3
=3
=
.
3d
3d
3
i S 0 ({) = 4{3 + 3f{2 + 2{ i S 00 ({) = 12{2 + 6f{ + 2. The graph of S 00 ({) is a
parabola. If S 00 ({) has two roots, then it changes sign twice and so has two inÀection points. This happens when the
discriminant of S 00 ({) is positive, that is, (6f)2 3 4 · 12 · 2 A 0 C 36f2 3 96 A 0 C |f| A
2
I
6
3
E 1=63= If
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
I
36f2 3 96 = 0 C f = ± 2 3 6 , S 00 ({) is 0 at one point, but there is still no inÀection point since S 00 ({) never changes
sign, and if 36f2 3 96 ? 0 C |f| ?
2
I
3
6
, then S 00 ({) never changes sign, and so there is no inÀection point.
f=6
f=
I
2 6
3
f=3
f = 1=8
f=0
f = 32
For large positive f, the graph of i has two inÀection points and a large dip to the left of the |-axis. As f decreases, the graph
of i becomes Àatter for { ? 0, and eventually the dip rises above the {-axis, and then disappears entirely, along with the
inÀection points. As f continues to decrease, the dip and the inÀection points reappear, to the right of the origin.
81. By hypothesis j = i 0 is differentiable on an open interval containing f. Since (f> i (f)) is a point of inÀection, the concavity
changes at { = f, so i 00 ({) changes signs at { = f. Hence, by the First Derivative Test, i 0 has a local extremum at { = f.
Thus, by Fermat’s Theorem i 00 (f) = 0.
82. i ({) = {4
i i 0 ({) = 4{3
i i 00 ({) = 12{2
i i 00 (0) = 0. For { ? 0, i 00 ({) A 0, so i is CU on (3"> 0);
for { A 0, i 00 ({) A 0, so i is also CU on (0> "). Since i does not change concavity at 0, (0> 0) is not an inÀection point.
83. Using the fact that |{| =
I
I
{2 , we have that j({) = { |{| = { {2
i j0 ({) =
I
I
I
{2 + {2 = 2 {2 = 2 |{| i
31@2
2{
? 0 for { ? 0 and j00 ({) A 0 for { A 0, so (0> 0) is an inÀection point. But j 00 (0) does not
=
j 00 ({) = 2{ {2
|{|
exist.
84. There must exist some interval containing f on which i 000 is positive, since i 000 (f) is positive and i 000 is continuous. On this
interval, i 00 is increasing (since i 000 is positive), so i 00 = (i 0 )0 changes from negative to positive at f. So by the First
Derivative Test, i 0 has a local minimum at { = f and thus cannot change sign there, so i has no maximum or minimum at f.
But since i 00 changes from negative to positive at f, i has a point of inÀection at f (it changes from concave down to
concave up).
85. Suppose that i is differentiable on an interval L and i 0 ({) A 0 for all { in L except { = f. To show that i is increasing on L,
let {1 , {2 be two numbers in L with {1 ? {2 .
Case 1 {1 ? {2 ? f. Let M be the interval {{ M L | { ? f}. By applying the Increasing/Decreasing Test to i
on M, we see that i is increasing on M, so i ({1 ) ? i ({2 ).
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HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH
¤
337
Case 2 f ? {1 ? {2 . Apply the Increasing/Decreasing Test to i on N = {{ M L | { A f}.
Case 3 {1 ? {2 = f. Apply the proof of the Increasing/Decreasing Test, using the Mean Value Theorem (MVT)
on the interval [{1 > {2 ] and noting that the MVT does not require i to be differentiable at the endpoints
of [{1 > {2 ].
Case 4 f = {1 ? {2 . Same proof as in Case 3.
Case 5 {1 ? f ? {2 . By Cases 3 and 4, i is increasing on [{1 > f] and on [f> {2 ], so i ({1 ) ? i (f) ? i({2 ).
In all cases, we have shown that i ({1 ) ? i({2 ). Since {1 , {2 were any numbers in L with {1 ? {2 , we have shown that i is
increasing on L.
86. i ({) = f{ +
1
{2 + 3
i i 0 ({) = f 3
2{
2{
. i 0 ({) A 0 C f A 2
({2 + 3)2
({ + 3)2
[call this j({)].
Now i 0 is positive (and hence i increasing) if f A j, so we’ll ¿nd the maximum value of j.
j 0 ({) =
({2 + 3)2 · 2 3 2{ · 2({2 + 3) · 2{
2({2 + 3)[({2 + 3) 3 4{2 ]
2(3 3 3{2 )
6(1 + {)(1 3 {)
=
=
=
.
[({2 + 3)2 ]2
({2 + 3)4
({2 + 3)3
({2 + 3)3
j 0 ({) = 0 C { = ±1. j0 ({) A 0 on (0> 1) and j 0 ({) ? 0 on (1> "), so j is increasing on (0> 1) and decreasing on
(1> "), and hence j has a maximum value on (0> ") of j(1) =
2
16
= 18 . Also since j({) $ 0 if { $ 0, the maximum value
of j on (3"> ") is 18 . Thus, when f A 18 , i is increasing. When f = 18 , i 0 ({) A 0 on (3"> 1) and (1> "), and hence i is
increasing on these intervals. Since i is continuous, we may conclude that i is also increasing on (3"> ") if f = 18 .
Therefore, i is increasing on (3"> ") if f D 18 .
87. (a) i ({) = {4 sin
1
{
i i 0 ({) = {4 cos
1
{
1
1
1
1
3 2 + sin (4{3 ) = 4{3 sin 3 {2 cos .
{
{
{
{
1
= 2{4 + i({) i j 0 ({) = 8{3 + i 0 ({).
j({) = {4 2 + sin
{
1
k({) = {4 32 + sin
= 32{4 + i ({) i k0 ({) = 38{3 + i 0 ({).
{
1
{4 sin 3 0
i ({) 3 i (0)
1
{
=
lim
= lim {3 sin . Since
It is given that i (0) = 0, so i (0) = lim
{<0
{<0
{<0
{30
{
{
1
3 {3 $ {3 sin $ {3 and lim {3 = 0, we see that i 0 (0) = 0 by the Squeeze Theorem. Also,
{<0
{
0
j 0 (0) = 8(0)3 + i 0 (0) = 0 and k0 (0) = 38(0)3 + i 0 (0) = 0, so 0 is a critical number of i , j, and k.
For {2q =
1
1
1
= sin 2q = 0 and cos
= cos 2q = 1, so i 0 ({2q ) = 3{22q ? 0.
[q a nonzero integer], sin
2q
{2q
{2q
For {2q+1 =
1
1
1
= sin(2q + 1) = 0 and cos
= cos(2q + 1) = 31, so
, sin
(2q + 1)
{2q+1
{2q+1
i 0 ({2q+1 ) = {22q+1 A 0. Thus, i 0 changes sign in¿nitely often on both sides of 0.
Next, j 0 ({2q ) = 8{32q + i 0 ({2q ) = 8{32q 3 {22q = {22q (8{2q 3 1) ? 0 for {2q ? 18 , but
j 0 ({2q+1 ) = 8{32q+1 + {22q+1 = {22q+1 (8{2q+1 + 1) A 0 for {2q+1 A 3 18 , so j 0 changes sign in¿nitely often on both
sides of 0.
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¤
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
Last, k0 ({2q ) = 38{32q + i 0 ({2q ) = 38{32q 3 {22q = 3{22q (8{2q + 1) ? 0 for {2q A 3 18 and
k0 ({2q+1 ) = 38{32q+1 + {22q+1 = {22q+1 (38{2q+1 + 1) A 0 for {2q+1 ? 18 , so k0 changes sign in¿nitely often on both
sides of 0.
(b) i (0) = 0 and since sin
1
1
and hence {4 sin is both positive and negative ini¿nitely often on both sides of 0, and
{
{
arbitrarily close to 0, i has neither a local maximum nor a local minimum at 0.
1
1
Since 2 + sin D 1, j({) = {4 2 + sin
A 0 for { 6= 0, so j(0) = 0 is a local minimum.
{
{
1
1
4
Since 32 + sin $ 31, k({) = { 32 + sin
? 0 for { 6= 0, so k(0) = 0 is a local maximum.
{
{
4.4 Indeterminate Forms and l'Hospital's Rule
H
Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign: =
1. (a) lim
{<d
i ({)
0
is an indeterminate form of type .
j({)
0
(b) lim
i ({)
= 0 because the numerator approaches 0 while the denominator becomes large.
s({)
(c) lim
k({)
= 0 because the numerator approaches a ¿nite number while the denominator becomes large.
s({)
{<d
{<d
(d) If lim s({) = " and i ({) < 0 through positive values, then lim
{<d
{<d
and i({) = {2 .] If i ({) < 0 through negative values, then lim
{<d
s({)
= ". [For example, take d = 0, s({) = 1@{2 ,
i ({)
s({)
= 3". [For example, take d = 0, s({) = 1@{2 ,
i({)
and i ({) = 3{2 .] If i ({) < 0 through both positive and negative values, then the limit might not exist. [For example,
take d = 0, s({) = 1@{2 , and i ({) = {.]
(e) lim
{<d
s({)
"
is an indeterminate form of type
.
t({)
"
2. (a) lim [i ({)s({)] is an indeterminate form of type 0 · ".
{<d
(b) When { is near d, s({) is large and k({) is near 1, so k({)s({) is large. Thus, lim [k({)s({)] = ".
{<d
(c) When { is near d, s({) and t({) are both large, so s({)t({) is large. Thus, lim [s({)t({)] = ".
{<d
3. (a) When { is near d, i({) is near 0 and s({) is large, so i ({) 3 s({) is large negative. Thus, lim [i ({) 3 s({)] = 3".
{<d
(b) lim [ s({) 3 t({)] is an indeterminate form of type " 3 ".
{<d
(c) When { is near d, s({) and t({) are both large, so s({) + t({) is large. Thus, lim [ s({) + t({)] = ".
{<d
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SECTION 4.4
4. (a) lim [i ({)]
j({)
{<d
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
339
is an indeterminate form of type 00 .
(b) If | = [i ({)]s({) , then ln | = s({) ln i ({). When { is near d, s({) < " and ln i ({) < 3", so ln | < 3".
Therefore, lim [i ({)]s({) = lim | = lim hln | = 0, provided i s is de¿ned.
{<d
{<d
{<d
(c) lim [k({)]s({) is an indeterminate form of type 1" .
{<d
(d) lim [ s({)]i ({) is an indeterminate form of type "0 .
{<d
(e) If | = [ s({)]t({) , then ln | = t({) ln s({). When { is near d, t({) < " and ln s({) < ", so ln | < ". Therefore,
lim [ s({)]t({) = lim | = lim hln | = ".
{<d
{<d
{<d
s
(f ) lim t({) s({) = lim [ s({)]1@t({) is an indeterminate form of type "0 .
{<d
{<d
5. From the graphs of i and j, we see that lim i ({) = 0 and lim j({) = 0, so l’Hospital’s Rule applies.
{<2
lim
{<2
0
lim i 0 ({)
{<2
0
i ({)
i ({)
i (2)
1=8
9
{<2
= lim 0
=
= 0
= 4 =
{<2 j ({)
j({)
lim j 0 ({)
j (2)
4
5
{<2
6. From the graphs of i and j, we see that lim i ({) = 0 and lim j({) = 0, so l’Hospital’s Rule applies.
{<2
lim
{<2
0
lim i 0 ({)
{<2
0
i ({)
i ({)
i (2)
1=5
3
{<2
= lim 0
=
= 0
=
=3
{<2 j ({)
j({)
lim j 0 ({)
j (2)
31
2
{<2
7. This limit has the form 00 . We can simply factor and simplify to evaluate the limit.
lim
{<1
{2 3 1
({ + 1)({ 3 1)
{+1
1+1
= lim
= lim
=
=2
{<1
{2 3 { {<1
{({ 3 1)
{
1
8. This limit has the form 00 . We can simply factor and simplify to evaluate the limit.
lim
{<2
{2 + { 3 6
({ + 3)({ 3 2)
= lim
= lim ({ + 3) = 2 + 3 = 5
{<2
{<2
{32
{32
9. This limit has the form 00 .
1
{3 3 2{2 + 1 H
3{2 3 4{
= lim
=3
3
{<1
{ 31
3{2
3
lim
{<1
Note: Alternatively, we could factor and simplify.
10. This limit has the form 00 .
6{2 + 5{ 3 4 H
12{ + 5
6+5
11
= lim
=
=
{<1@2 4{2 + 16{ 3 9
{<1@2 8{ + 16
4 + 16
20
lim
Note: Alternatively, we could factor and simplify.
11. This limit has the form 00 .
lim
{<(@2)+
cos { H
3 sin {
= lim
= lim tan { = 3".
1 3 sin { {<(@2)+ 3 cos { {<(@2)+
12. This limit has the form 00 .
{<0
lim
4
sin 4{ H
4 cos 4{
4(1)
= lim
=
=
tan 5{ {<0 5 sec2 (5{)
5(1)2
5
13. This limit has the form 00 .
lim
h2w 3 1 H
2h2w
2(1)
= lim
=
=2
w<0 cos w
sin w
1
w<0
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APPLICATIONS OF DIFFERENTIATION
{2
2{
2
2
H
= =2
= lim = lim
1 3 cos { {<0 sin { {<0 sin {
1
{
14. This limit has the form 00 .
{<0
15. This limit has the form 00 .
<@2
16.
lim
<@2
lim
lim
1 3 sin H
3 cos H
sin 1
=
= lim
= lim
<@2 32 sin 2
<@2 34 cos 2
1 + cos 2
4
1 3 sin 0
= = 0. L’Hospital’s Rule does not apply.
csc 1
ln { H
1@{
2
lim I = lim 1 31@2 = lim I = 0
{<" {
{<"
{
{
2
17. This limit has the form
"
.
"
18. This limit has the form
"
. lim
" {<"
{<"
{ + {2 H
1 + 2{ H
2
1
=3 .
= lim
= lim
{<" 34{
{<" 34
1 3 2{2
2
1
+1
{ + {2
1
0+1
{
=3 .
=
lim
=
2
{<" 1 3 2{
{<" 1
032
2
32
{2
A better method is to divide the numerator and the denominator by {2 : lim
19. lim [(ln {)@{] = 3" since ln { < 3" as { < 0+ and dividing by small values of { just increases the magnitude of the
{<0+
quotient (ln {)@{. L’Hospital’s Rule does not apply.
20. This limit has the form
1
I
1
ln { H
ln {
2
2{ = lim 1 = 0
=
lim
=
lim
=
{<"
{<"
{<"
{<" 4{2
{2
{2
2{
"
.
"
lim
lim
8
8
w8 3 1 H
8w7
8
=
lim
= lim w3 = (1) =
w5 3 1 w<1 5w4
5 w<1
5
5
lim
8w 3 5w H
8w ln 8 3 5w ln 5
8
= ln 8 3 ln 5 = ln
= lim
w<0
w
1
5
21. This limit has the form 00 .
w<1
22. This limit has the form 00 .
w<0
23. This limit has the form 00 .
lim
{<0
24. This limit has the form
"
.
"
x@10
lim
x<"
I
I
1
(1 + 2{)31@2 · 2 3 12 (1 3 4{)31@2 (34)
1 + 2{ 3 1 3 4{ H
= lim 2
{<0
{
1
2
1
1
2
+I
= I +I =3
= lim I
{<0
1 + 2{
1 3 4{
1
1
h
·
hx@10 H
= lim
x<"
x3
3x2
1
10
25. This limit has the form 00 . lim
{<0
26. This limit has the form 00 .
H
=
1
lim
30 x<"
hx@10 ·
2x
1
10
H
=
1
lim
600 x<"
hx@10 ·
1
1
10
=
1
lim hx@10
6000 x<"
="
h{ 3 1 3 { H
h{ 3 1 H
h{
1
=
= lim
= lim
2
{<0
{<0 2
{
2{
2
lim
{<0
sinh { 3 { H
cosh { 3 1 H
sinh { H
cosh {
1
= lim
= lim
= lim
=
{<0
{<0
{<0
{3
3{2
6{
6
6
tanh { H
sech 2 {
sech2 0
1
=
= =1
= lim
2
{<0 tan {
{<0 sec {
sec2 0
1
27. This limit has the form 00 . lim
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INDETERMINATE FORMS AND L’HOSPITAL’S RULE
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341
28. This limit has the form 00 .
lim
{<0
{ 3 sin { H
1 3 cos { H
3(3 sin {)
1
= 3 lim
= lim
= lim
{ 3 tan { {<0 1 3 sec2 { {<0 32 sec { (sec { tan {)
2 {<0
sin {
cos { sin {
sec2 {
= 3 12 lim cos3 { = 3 12 (1)3 = 3 12
{<0
sin {
{ .
tan {
{<0
13
{
Another method is to write the limit as lim
29. This limit has the form 00 .
30. This limit has the form
"
.
"
31. This limit has the form 00 .
32. This limit has the form
lim
{<0
33. lim
{<0
lim
{<0
I
1
sin31 { H
1@ 1 3 {2
1
= lim
= =1
= lim I
{<0
{<0
{
1
1
1 3 {2
lim
{<"
lim
(ln {)2 H
2(ln {)(1@{)
ln { H
1@{
= 2 lim
= 2(0) = 0
= lim
= 2 lim
{<"
{<" {
{<" 1
{
1
{3{ H
{3{ ln 3 + 3{
3{ ({ ln 3 + 1)
{ ln 3 + 1
1
= lim
= lim
=
= lim
{
{<0
{<0
3 1 {<0
3 ln 3
3{ ln 3
ln 3
ln 3
{<0 3{
0
.
0
cos p{ 3 cos q{ H
3p sin p{ + q sin q{ H
3p2 cos p{ + q2 cos q{
= 12 q2 3 p2
=
lim
=
lim
{<0
{<0
{2
2{
2
{ + sin {
0+0
0
=
= = 0. L’Hospital’s Rule does not apply.
{ + cos {
0+1
1
34. This limit has the form 00 . lim
{
H
{<0 tan31 (4{)
35. This limit has the form 00 .
36. lim
13
{<0+
lim
{<1
= lim
{<0
1
1
·4
1 + (4{)2
= lim
{<0
1 + 16{2
1
=
4
4
1 3 { + ln { H
31 + 1@{ H
31@{2
31
1
=
=3 2
= lim
= lim
{<1 3 sin {
{<1 3 2 cos {
1 + cos {
3 2 (31)
{{ 3 1
. From Example 9, lim {{ = 1, so lim ({{ 3 1) = 0. As { < 0+ , ln { < 3", so
ln { + { 3 1
{<0+
{<0+
ln { + { 3 1 < 3" as { < 0+ . Thus, lim
{<0+
37. This limit has the form 00 .
lim
{<1
{{ 3 1
= 0.
ln { + { 3 1
{d 3 d{ + d 3 1 H
d{d31 3 d H
d(d 3 1){d32
d(d 3 1)
=
= lim
= lim
2
{<1 2({ 3 1)
{<1
({ 3 1)
2
2
h{ 3 h3{ 3 2{ H
h{ + h3{ 3 2 H
h{ 3 h3{ H
h{ + h3{
1+1
=
=2
= lim
= lim
= lim
{<0
{<0
{<0
{<0
{ 3 sin {
1 3 cos {
sin {
cos {
1
38. This limit has the form 00 . lim
39. This limit has the form 00 . lim
{<0
cos { 3 1 + 12 {2 H
3 sin { + { H
3 cos { + 1 H
sin { H
cos {
1
=
= lim
= lim
= lim
= lim
{<0
{<0
{<0 24{
{<0 24
{4
4{3
12{2
24
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¤
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
40. This limit has the form
"
".
cos { ln({ 3 d)
ln({ 3 d) H
= lim cos { lim
lim
= cos d lim
{
d
ln(h{ 3 hd )
{<d+
{<d+
{<d+ ln(h 3 h )
{<d+
= cos d lim
{<d+
1
{3d
1
· h{
h{ 3 hd
1
h{ 3 hd H
h{
1
1
= cos d · d · hd = cos d
· lim
= cos d · d lim
{
h {<d+ { 3 d
h {<d+ 1
h
41. This limit has the form " · 0. We’ll change it to the form 00 .
sin(@{) H
cos(@{)(3@{2 )
= lim
= lim cos(@{) = (1) = {<"
{<"
{<"
1@{
31@{2
lim { sin(@{) = lim
{<"
42. This limit has the form " · 0. We’ll change it to the form
lim
{<"
"
.
"
I
1 31@2
{
{ H
1
2
=
lim
= lim I {@2 = 0
{<" h{@2
{<" 1 h{@2
{<"
{h
2
I 3{@2
{h
= lim
43. This limit has the form " · 0. We’ll change it to the form 00 .
lim cot 2{ sin 6{ = lim
{<0
{<0
sin 6{ H
6 cos 6{
6(1)
=
= lim
=3
tan 2{ {<0 2 sec2 2{
2(1)2
44. This limit has the form 0 · (3").
lim sin { ln { = lim
{<0+
{<0+
ln { H
1@{
= 3 lim
= lim
csc { {<0+ 3 csc { cot {
{<0+
sin {
sin {
· tan { = 3 lim
lim tan {
{
{
{<0+
{<0+
= 31 · 0 = 0
2
{3 H
3{2
3{ H
3
=
lim
2
2 = lim
2 = lim
2 = 0
{<" h{
{<" 2{h{
{<" 2h{
{<" 4{h{
45. This limit has the form " · 0. lim {3 h3{ = lim
{<"
46. This limit has the form " · 0.
tan(1@{) H
sec2 (1@{)(31@{2 )
= lim
= lim sec2 (1@{) = 12 = 1
{<"
{<"
{<"
1@{
31@{2
lim { tan(1@{) = lim
{<"
47. This limit has the form 0 · (3").
2
ln {
1@{
1
H
= lim
=3
=
cot({@2) {<1+ (3@2) csc2 ({@2)
(3@2)(1)2
lim ln { tan({@2) = lim
{<1+
{<1+
48. This limit has the form 0 · ".
lim
{<(@2)
cos { sec 5{ =
lim
{<(@2)
cos { H
3 sin {
31
1
=
=
=
lim
cos 5{ {<(@2) 35 sin 5{
35
5
49. This limit has the form " 3 ".
lim
{<1
{
1
3
{31
ln {
= lim
{<1
H
= lim
{ ln { 3 ({ 3 1) H
{(1@{) + ln { 3 1
ln {
= lim
= lim
{<1 ({ 3 1)(1@{) + ln {
{<1 1 3 (1@{) + ln {
({ 3 1) ln {
1@{
{
{2
1
1
· 2 = lim
=
=
{<1 1 + {
+ 1@{ {
1+1
2
{<1 1@{2
50. This limit has the form " 3 ". lim (csc { 3 cot {) = lim
{<0
{<0
1
cos {
3
sin {
sin {
= lim
{<0
1 3 cos { H
sin {
=0
= lim
{<0 cos {
sin {
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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NOT FOR SALE
SECTION 4.4
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
51. This limit has the form " 3 ".
lim
{<0+
1
1
3 {
{
h 31
= lim
{<0+
1
h{ 3 1 3 { H
h{ 3 1
h{
1
H
=
lim
=
lim
=
=
{(h{ 3 1)
0+1+1
2
{<0+ {h{ + h{ 3 1
{<0+ {h{ + h{ + h{
52. This limit has the form " 3 ".
{ cos { 3 sin { H
{(3 sin {) + cos { 3 cos {
1
cos {
1
= lim
cot { 3
= lim
3
= lim
{<0
{<0
{<0
{<0
{
sin {
{
{ sin {
{ cos { + sin {
lim
= 3 lim
{<0
{ sin {
{ cos { + sin {
0+0
H
=3
=0
= 3 lim
{<0 {(3 sin {) + cos { + cos {
{ cos { + sin {
0+1+1
53. The limit has the form " 3 " and we will change the form to a product by factoring out {.
ln { H
1@{
ln {
lim ({ 3 ln {) = lim { 1 3
= " since lim
= 0.
= lim
{<"
{<"
{<" {
{<" 1
{
54. This limit has the form " 3 ".
lim [ln({7 3 1) 3 ln({5 3 1)] = lim ln
{<1+
{<1+
I
55. | = {
{
i ln | =
{<0+
{<0+
I
{
lim {
{<0+
I
{ ln {, so
I
{ ln { = lim
lim ln | = lim
{<0+
I
ln { H
1@{
= lim
{=0 i
1 33@2 = 32 lim+
31@2
+
{
{<0 3 {
{<0
2
= lim hln | = h0 = 1.
{<0+
56. | = (tan 2{){
i ln | = { · ln tan 2{, so
lim ln | = lim { · ln tan 2{ = lim
{<0+
{7 3 1 H
7{6
{7 3 1
7
= ln lim 5
= ln lim
= ln
5
+
+
{ 31
5
{<1 { 3 1
{<1 5{4
{<0+
= lim
{<0+
{<0+
ln tan 2{ H
(1@ tan 2{)(2 sec2 2{)
32{2 cos 2{
= lim
=
lim
2
1@{
31@{2
{<0+
{<0+ sin 2{ cos 2{
2{
3{
· lim
=1·0=0
sin 2{ {<0+ cos 2{
i
lim (tan 2{){ = lim hln | = h0 = 1.
{<0+
{<0+
57. | = (1 3 2{)1@{
i ln | =
ln(1 3 2{) H
32@(1 3 2{)
1
ln(1 3 2{), so lim ln | = lim
= 32 i
= lim
{<0
{<0
{<0
{
{
1
lim (1 3 2{)1@{ = lim hln | = h32 .
{<0
{<0
d
i ln | = e{ ln 1 +
, so
{
1
d
e
3 2
1 + d@{
{
e ln(1 + d@{) H
de
= de i
= lim
= lim
lim ln | = lim
{<"
{<"
{<"
{<" 1 + d@{
1@{
31@{2
58. | =
d e{
1+
{
d e{
lim 1 +
= lim hln | = hde .
{<"
{<"
{
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°
licated, or posted to a publicly accessible website, in whole or in part.
par
© Cengage Learning. All Rights Reserved.
¤
343
344
¤
CHAPTER 4
59. | = {1@(13{)
APPLICATIONS OF DIFFERENTIATION
1
1
ln { H
1@{
= lim
ln {, so lim ln | = lim
ln { = lim
= 31 i
+
+
+
+
13{
1
3
{
1
3
{
31
{<1
{<1
{<1
{<1
i ln | =
lim {1@(13{) = lim hln | = h31 =
{<1+
{<1+
60. | = {(ln 2)@(1 + ln {)
lim ln | = lim
{<"
{<"
61. | = {1@{
i ln | =
1
.
h
ln 2
ln { i
1 + ln {
(ln 2)(ln {) H
(ln 2)(1@{)
= lim
= lim ln 2 = ln 2, so lim {(ln 2)@(1 + ln {) = lim hln | = hln 2 = 2.
{<"
{<"
{<"
{<"
1 + ln {
1@{
i ln | = (1@{) ln { i
lim ln | = lim
{<"
{<"
ln { H
1@{
=0 i
= lim
{<" 1
{
lim {1@{ = lim hln | = h0 = 1
{<"
{<"
62. | = (h{ + {)1@{
i ln | =
ln(h{ + {) H
h{ + 1 H
h{
h{
H
= lim {
= lim {
= lim { = 1 i
{<" h + {
{<" h + 1
{<" h
{
so lim ln | = lim
{<"
{<"
1@{
{
lim (h + {)
{<"
1
ln(h{ + {),
{
= lim hln | = h1 = h.
{<"
4
ln(4{ + 1) H
4{
+1 = 4 i
i ln | = cot { ln(4{ + 1), so lim ln | = lim
= lim
2
tan {
{<0+
{<0+
{<0+ sec {
cot {
63. | = (4{ + 1)
lim (4{ + 1)cot { = lim hln | = h4 .
{<0+
{<0+
64. | = (2 3 {)tan({@2)
{ ln(2 3 {) i
i ln | = tan
2
{ 1
(31)
k { l
sin2
2
ln (2 3 {) H
2 3{ 2
{ = lim
ln(2 3 {) = lim
lim
lim ln | = lim tan
{
= {<1
{<1
{<1
{<1
{<1
2
23{
cot
3csc2
·
2
2
2
=
2
2 12
·
=
1
65. | = (cos {)1@{
2
i
i ln | =
lim (2 3 {)tan({@2) = lim hln | = h(2@)
{<1
{<1
1
ln cos { i
{2
ln cos { H
3 tan { H
3 sec2 {
1
=3
= lim
= lim
2
{
2{
2
2
{<0+
{<0+
{<0+
{<0+
I
2
lim (cos {)1@{ = lim hln | = h31@2 = 1@ h
lim ln | = lim
{<0+
66. | =
i
{<0+
2{ 3 3
2{ + 5
2{+1
i
2{ 3 3
ln | = (2{ + 1) ln
i
2{ + 5
ln(2{ 3 3) 3 ln(2{ + 5) H
2@(2{ 3 3) 3 2@(2{ + 5)
38(2{ + 1)2
= lim
= lim
2
{<"
{<"
1@(2{ + 1)
32@(2{ + 1)
(2{ 3 3)(2{ + 5)
2{+1
2
38(2 + 1@{)
2{ 3 3
= lim
= h38
= 38 i lim
{<" (2 3 3@{)(2 + 5@{)
{<"
2{ + 5
lim ln | = lim
{<"
{<"
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.4
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
345
From the graph, if { = 500, | E 7=36. The limit has the form 1" .
{
2
2
Now | = 1 +
i ln | = { ln 1 +
i
{
{
1
2
3 2
1 + 2@{
{
ln(1 + 2@{) H
= lim
lim ln | = lim
{<"
{<"
{<"
1@{
31@{2
67.
= 2 lim
{<"
1
= 2(1) = 2
1 + 2@{
i
{
2
= lim hln | = h2 [E 7=39]
1+
{<"
{<"
{
lim
68.
From the graph, as { < 0, | E 0=55. The limit has the form 00 .
lim
{<0
ln 54
5{ 3 4{ H
5{ ln 5 3 4{ ln 4
ln 5 3 ln 4
= lim {
[E 0=55]
=
=
{
{
{
{<0 3 ln 3 3 2 ln 2
3 32
ln 3 3 ln 2
ln 32
i ({)
i 0 ({)
= lim 0
= 0=25=
{<0 j({)
{<0 j ({)
69.
From the graph, it appears that lim
We calculate lim
{<0
i ({)
h{ 3 1 H
h{
1
= lim 3
= .
= lim
2
{<0 { + 4{
{<0 3{ + 4
j({)
4
i ({)
i 0 ({)
= lim 0
= 4. We calculate
{<0 j({)
{<0 j ({)
70.
From the graph, it appears that lim
lim
{<0
i ({)
2{ sin { H
2({ cos { + sin {)
= lim
= lim
j({) {<0 sec { 3 1 {<0
sec { tan {
H
= lim
{<0
71. lim
{<"
h{ H
h{
h{
h{
H
H
H
="
= lim
= lim
= · · · = lim
q
q31
q32
{<" q{
{<" q(q 3 1){
{<" q!
{
72. This limit has the form
"
. lim
" {<"
1@{
ln { H
1
= lim
= lim
= 0 since s A 0.
{<" s{s31
{<" s{s
{s
{
1
H
= lim
= lim
{<"
{2 + 1 {<" 12 ({2 + 1)31@2 (2{)
73. lim I
{<"
2(3{ sin { + cos { + cos {)
4
= =4
sec {(sec2 {) + tan {(sec { tan {)
1
I
{2 + 1
. Repeated applications of l’Hospital’s Rule result in the
{
original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator
{
{@{
1
1
= lim s
= =1
= lim s
by {: lim I
{<"
{<"
1
{2 + 1 {<" {2 @{2 + 1@{2
1 + 1@{2
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°
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346
74.
NOT FOR SALE
¤
CHAPTER 4
lim
{<(@2)
APPLICATIONS OF DIFFERENTIATION
sec { H
sec { tan {
tan {
=
lim
. Repeated applications of l’Hospital’s Rule result in the
=
lim
tan { {<(@2) sec2 {
{<(@2) sec {
original limit or the limit of the reciprocal of the function. Another method is to simplify ¿rst:
lim
{<(@2)
sec {
1@cos {
1
1
=
lim
=
lim
= =1
tan { {<(@2) sin {@cos { {<(@2) sin {
1
i i 0 ({) = h{ 3 f = 0 C h{ = f C { = ln f, f A 0. i 00 ({) = h{ A 0, so i is CU on
{
h{ H
h{
h
3f
= O1 . Now lim
= ", so O1 = ", regardless
= lim
(3"> "). lim (h{ 3 f{) = lim {
{<"
{<"
{<" {
{<" 1
{
75. i ({) = h{ 3 f{
of the value of f. For O = lim (h{ 3 f{), h{ < 0, so O is determined
{<3"
by 3f{. If f A 0, 3f{ < ", and O = ". If f ? 0, 3f{ < 3", and
O = 3". Thus, i has an absolute minimum for f A 0. As f increases, the
minimum points (ln f> f 3 f ln f), get farther away from the origin.
pj
pj
pj 1 3 h3fw@p =
lim 1 3 h3fw@p =
(1 3 0) [because 3fw@p < 3" as w < "]
w<" f
f w<"
f
pj
, which is the speed the object approaches as time goes on, the so-called limiting velocity.
=
f
76. (a) lim y = lim
w<"
(b) lim y = lim
f<0+
f<0+
pj
1 3 h3fw@p
(1 3 h3fw@p ) = pj lim
f
f
f<0+
H
= pj lim
f<0+
[form is 00 ]
(3h3fw@p ) · (3w@p)
pjw
=
lim h3fw@p = jw(1) = jw
1
p f<0+
The velocity of a falling object in a vacuum is directly proportional to the amount of time it falls.
u qw
u qw
u
1+
, so
, which is of the form 1" . | = 1 +
i ln | = qw ln 1 +
q<"
q
q
q
3u@q2
ln(1 + u@q) H
u
u
= w lim
= w lim
= wu
= w lim
lim ln | = lim qw ln 1 +
q<"
q<"
q<"
q<" (1 + u@q)(31@q2 )
q<" 1 + l@q
q
1@q
u qw
< D0 huw .
lim | = huw . Thus, as q < ", D = D0 1 +
q<"
q
77. First we will ¿nd lim
u
$
I
ln cosh df
[let d = j@(pw)]
f
f<0+
f<0+
# I $
I
1
d
I (sinh df )
I
I
I
cosh
df
2 f
tanh df
p d
H
I
= p lim
=
lim
1
2 f<0+
f<0+
f
I kI I l
I
2
sech df
d@ 2 f
I
d
j
pd
pj
pd
H p
I
lim
lim sech2 df =
(1)2 =
=
=
=
2 f<0+
2 f<0+
2
2pw
2w
1@ (2 f )
78. lim v(w) = lim
f<0+
#
p
ln cosh
f
jf
pw
= p lim
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°
© Cengage Learning. All Rights Reserved.
i
NOT FOR SALE
SECTION 4.4
79.
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
hH + h3H
1
3
hH 3 h3H
H
H hH + h3H 3 1 hH 3 h3H
HhH + Hh3H 3 hH + h3H
= lim
= lim
H
3H
+
+
(h 3 h ) H
HhH 3 Hh3H
H<0
H<0
HhH + hH · 1 + H 3h3H + h3H · 1 3 hH + 3h3H
H
= lim
HhH + hH · 1 3 [H(3h3H ) + h3H · 1]
H<0+
lim S (H) = lim
H<0+
H<0+
= lim
H<0+
=
HhH 3 Hh3H
= lim
HhH + hH + Hh3H 3 h3H
H<0+
hH 3 h3H
0
, where O = lim
2+O
H
H<0+
Thus, lim S (H) =
H<0+
¤
347
0
= 0.
2+2
hH 3 h3H
hH
h3H
hH +
+ h3H 3
H
H
form is 00
H
= lim
H<0+
form is 00
[divide by H]
hH + h3H
1+1
=
=2
1
1
% &
2
u
1
u
1
u 2
2
80. (a) lim y = lim 3f
ln
ln
= 3fu lim
= 3fu2 · 2 · ln 1 = 3f · 0 = 0
U
U
U
U
u
U<u +
U<u +
U<u +
As the insulation of a metal cable becomes thinner, the velocity of an electrical impulse in the cable approaches zero.
k
u l
u
u 2
f
[form is 0 · "]
ln
3f
= 3 2 lim u2 ln
U
U
U u<0+
U
u<0+
u
U 1
2
·
ln
f
f
f
u
H
U
[form is "@"] = 3 2 lim u U = 3 2 lim 3
= 3 2 lim
=0
1
U u<0+
U u<0+ 32
U u<0+
2
2
3
u
u
(b) lim y = lim
u<0+
As the radius of the metal cable approaches zero, the velocity of an electrical impulse in the cable approaches zero.
81. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:
lim
{<d
I
I
3
4 31@2
1
(2d3 3 4{3 ) 3 d 13 (dd{)32@3 d2
2d3 { 3 {4 3 d 3 dd{ H
2 (2d { 3 { )
I
=
lim
4
{<d
3 14 (d{3 )33@4 (3d{2 )
d 3 d{3
=
=
1
(2d3 d
2
3 d4 )31@2 (2d3 3 4d3 ) 3 13 d3 (d2 d)32@3
3 14 (dd3 )33@4 (3dd2 )
3d 3 13 d
(d4 )31@2 (3d3 ) 3 13 d3 (d3 )32@3
=
= 43 43 d =
3 3 4 33@4
3 4 d (d )
3 34
16
9 d
82. Let the radius of the circle be u. We see that D() is the area of the whole ¿gure (a sector of the circle with radius 1), minus
the area of 4RS U. But the area of the sector of the circle is 12 u2 (see Reference Page 1), and the area of the triangle is
1
u |S T|
2
= 12 u(u sin ) = 12 u2 sin . So we have D() = 12 u2 3 12 u2 sin = 12 u2 ( 3 sin ). Now by elementary
trigonometry, E() =
1
2
|TU| |S T| = 12 (u 3 |RT|) |S T| = 12 (u 3 u cos )(u sin ) = 12 u2 (1 3 cos ) sin .
[continued]
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°
licated, or posted to a publicly accessible website, in whole or in part.
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348
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
So the limit we want is
lim
<0+
D()
= lim
E() <0+
1 2
u ( 3 sin )
H
2
= lim
1 2
<0+
u
(1
3
cos
)
sin
2
1 3 cos (1 3 cos ) cos + sin (sin )
= lim
1 3 cos sin H
= lim
cos 3 cos2 + sin2 <0+ 3 sin 3 2 cos (3 sin ) + 2 sin (cos )
= lim
sin 1
1
1
= lim
=
=
3 sin + 4 sin cos 31 + 4 cos 0
3
<0+ 31 + 4 cos <0+
<0+
1+{
1
{ 3 {2 ln
= lim { 3 {2 ln
+ 1 . Let w = 1@{, so as { < ", w < 0+ .
{<"
{<"
{
{
83. The limit, O = lim
O = lim
w<0+
1
13
w 3 ln(w + 1) H
1
1
1
1
w
+
1 = lim w@(w + 1) = lim
= lim
3 2 ln(w + 1) = lim
=
w
w
w2
2w
2w
2
w<0+
w<0+
w<0+
w<0+ 2 (w + 1)
Note: Starting the solution by factoring out { or {2 leads to a more complicated solution.
84. | = [i ({)]j({)
i ln | = j({) ln i ({). Since i is a positive function, ln i ({) is de¿ned. Now
lim ln | = lim j({) ln i({) = 3" since lim j({) = " and lim i ({) = 0 i
{<d
{<d
{<d
{<d
lim ln i ({) = 3". Thus, if w = ln |,
{<d
lim | = lim hw = 0. Note that the limit, lim j({) ln i ({), is not of the form " · 0.
{<d
w<3"
{<d
85. Since i (2) = 0, the given limit has the form 00 .
lim
{<0
i (2 + 3{) + i (2 + 5{) H
i 0 (2 + 3{) · 3 + i 0 (2 + 5{) · 5
= i 0 (2) · 3 + i 0 (2) · 5 = 8i 0 (2) = 8 · 7 = 56
= lim
{<0
{
1
86. O = lim
{<0
sin 2{
e
+d+ 2
{3
{
= lim
{<0
sin 2{ + d{3 + e{ H
2 cos 2{ + 3d{2 + e
= lim
. As { < 0, 3{2 < 0, and
3
{<0
{
3{2
(2 cos 2{ + 3d{2 + e) < e + 2, so the last limit exists only if e + 2 = 0, that is, e = 32. Thus,
lim
{<0
2 cos 2{ + 3d{2 3 2 H
34 sin 2{ + 6d{ H
38 cos 2{ + 6d
6d 3 8
= lim
= lim
=
, which is equal to 0 if and only
{<0
{<0
3{2
6{
6
6
if d = 43 . Hence, O = 0 if and only if e = 32 and d = 43 .
87. Since lim [i ({ + k) 3 i({ 3 k)] = i ({) 3 i ({) = 0 (i is differentiable and hence continuous) and lim 2k = 0, we use
k<0
k<0
l’Hospital’s Rule:
i ({ + k) 3 i({ 3 k) H
i 0 ({ + k)(1) 3 i 0 ({ 3 k)(31)
i 0 ({) + i 0 ({)
2i 0 ({)
=
=
= i 0 ({)
= lim
k<0
k<0
2k
2
2
2
lim
i ({ + k) 3 i ({ 3 k)
is the slope of the secant line between
2k
({ 3 k> i ({ 3 k)) and ({ + k> i ({ + k)). As k < 0, this line gets closer
to the tangent line and its slope approaches i 0 ({).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.4
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
349
88. Since lim [i({ + k) 3 2i ({) + i ({ 3 k)] = i ({) 3 2i({) + i({) = 0 [i is differentiable and hence continuous]
k<0
and lim k2 = 0, we can apply l’Hospital’s Rule:
k<0
lim
k<0
i ({ + k) 3 2i ({) + i ({ 3 k) H
i 0 ({ + k) 3 i 0 ({ 3 k)
= lim
= i 00 ({)
k<0
k2
2k
At the last step, we have applied the result of Exercise 87 to i 0 ({).
89. (a) We show that lim
{<0
1
i ({)
= 0 for every integer q D 0. Let | = 2 . Then
{q
{
2
i ({)
h31@{
|q H
q| q31 H
q!
H
= lim
= · · · = lim | = 0 i
lim 2q = lim
q = lim
2
|
{<0 {
{<0 ({ )
|<" h
|<"
|<" h
h|
lim
{<0
i ({)
i ({)
i ({)
i ({) 3 i (0)
i ({)
= lim {q 2q = lim {q lim 2q = 0. Thus, i 0 (0) = lim
= lim
= 0.
{<0
{<0
{<0 {
{<0
{<0
{q
{
{30
{
(b) Using the Chain Rule and the Quotient Rule we see that i (q) ({) exists for { 6= 0. In fact, we prove by induction that for
each q D 0, there is a polynomial sq and a non-negative integer nq with i (q) ({) = sq ({)i ({)@{nq for { 6= 0. This is
true for q = 0; suppose it is true for the qth derivative. Then i 0 ({) = i ({)(2@{3 ), so
i (q+1) ({) = {nq [s0q ({) i({) + sq ({) i 0 ({)] 3 nq {nq 31 sq ({) i ({) {32nq
= {nq s0q ({) + sq ({)(2@{3 ) 3 nq {nq 31 sq ({) i ({){32nq
= {nq +3 s0q ({) + 2sq ({) 3 nq {nq +2 sq ({) i ({){3(2nq +3)
which has the desired form.
Now we show by induction that i (q) (0) = 0 for all q. By part (a), i 0 (0) = 0. Suppose that i (q) (0) = 0. Then
i (q) ({) 3 i (q) (0)
i (q) ({)
sq ({) i ({)@{nq
sq ({) i ({)
= lim
= lim
= lim
{<0
{<0
{<0
{<0
{30
{
{
{nq +1
i (q+1) (0) = lim
= lim sq ({) lim
{<0
{<0
i ({)
= sq (0) · 0 = 0
{nq +1
90. (a) For i to be continuous, we need lim i({) = i (0) = 1. We note that for { 6= 0, ln i({) = ln |{|{ = { ln |{|.
{<0
So lim ln i ({) = lim { ln |{| = lim
{<0
{<0
{<0
ln |{| H
1@{
= lim
= 0. Therefore, lim i ({) = lim hln i ({) = h0 = 1.
{<0 31@{2
{<0
{<0
1@{
So i is continuous at 0.
(b) From the graphs, it appears that i is differentiable at 0.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(c) To ¿nd i 0 , we use logarithmic differentiation: ln i ({) = { ln |{| i
i 0 ({)
1
={
+ ln |{| i
i({)
{
i 0 ({) = i({)(1 + ln |{|) = |{|{ (1 + ln |{|), { 6= 0. Now i 0 ({) < 3" as { < 0 [since |{|{ < 1 and
(1 + ln |{|) < 3"], so the curve has a vertical tangent at (0> 1) and is therefore not differentiable there.
The fact cannot be seen in the graphs in part (b) because ln |{| < 3" very slowly as { < 0.
4.5 Summary of Curve Sketching
1. | = i ({) = {3 3 12{2 + 36{ = {({2 3 12{ + 36) = {({ 3 6)2
A. i is a polynomial, so G = R.
B. {-intercepts are 0 and 6, |-intercept = i (0) = 0 C. No symmetry D. No asymptote
E. i 0 ({) = 3{2 3 24{ + 36 = 3({2 3 8{ + 12) = 3({ 3 2)({ 3 6) ? 0 C
H.
2 ? { ? 6, so i is decreasing on (2> 6) and increasing on (3"> 2) and (6> ").
F. Local maximum value i (2) = 32, local minimum value i (6) = 0
G. i 00 ({) = 6{ 3 24 = 6({ 3 4) A 0 C { A 4, so i is CU on (4> ") and
CD on (3"> 4). IP at (4> 16)
2. | = i ({) = 2 + 3{2 3 {3
A. G = R B. |-intercept = i (0) = 2 C. No
0
H.
2
symmetry D. No asymptote E. i ({) = 6{ 3 3{ = 3{(2 3 {) A 0 C
0 ? { ? 2, so i is increasing on (0> 2) and decreasing on (3"> 0) and (2> ").
F. Local maximum value i (2) = 6, local minimum value i(0) = 2
G. i 00 ({) = 6 3 6{ = 6(1 3 {) A 0 C { ? 1, so i is CU on (3"> 1) and
CD on (1> "). IP at (1> 4)
3. | = i ({) = {4 3 4{ = {({3 3 4)
A. G = R B. {-intercepts are 0 and
I
3
4,
H.
|-intercept = i (0) = 0 C. No symmetry D. No asymptote
E. i 0 ({) = 4{3 3 4 = 4({3 3 1) = 4({ 3 1)({2 + { + 1) A 0 C { A 1, so
i is increasing on (1> ") and decreasing on (3"> 1). F. Local minimum value
i (1) = 33, no local maximum G. i 00 ({) = 12{2 A 0 for all {, so i is CU on
(3"> "). No IP
4. | = i ({) = {4 3 8{2 + 8
A. G = R B. |-intercept i (0) = 8; {-intercepts: i ({) = 0 i [by the quadratic formula]
s
I
{ = ± 4 ± 2 2 r ±2=61> ±1=08 C. i (3{) = i ({), so i is even and symmetric about the |-axis D. No asymptote
E. i 0 ({) = 4{3 3 16{ = 4{({2 3 4) = 4{({ + 2)({ 3 2) A 0 C 32 ? { ? 0 or { A 2, so i is increasing on (32> 0)
and (2> "), and i is decreasing on (3"> 32) and (0> 2).
H.
F. Local maximum value i (0) = 8, local minimum values i (±2) = 38
I
G. i 00 ({) = 12{2 3 16 = 4(3{2 3 4) A 0 i |{| A 2@ 3 [E 1=15], so i is
I
I I I
CU on 3"> 32@ 3 and 2@ 3> " , and i is CD on 32@ 3> 2@ 3 .
I
IP at ±2@ 3> 3 89
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SECTION 4.5 SUMMARY OF CURVE SKETCHING
5. | = i ({) = {({ 3 4)3
¤
351
A. G = R B. {-intercepts are 0 and 4, |-intercept i (0) = 0 C. No symmetry
H.
D. No asymptote
E. i 0 ({) = { · 3({ 3 4)2 + ({ 3 4)3 · 1 = ({ 3 4)2 [3{ + ({ 3 4)]
= ({ 3 4)2 (4{ 3 4) = 4({ 3 1)({ 3 4)2 A 0
C
{ A 1, so i is increasing on (1> ") and decreasing on (3"> 1).
F. Local minimum value i(1) = 327, no local maximum value
G. i 00 ({) = 4[({ 3 1) · 2({ 3 4) + ({ 3 4)2 · 1] = 4({ 3 4)[2({ 3 1) + ({ 3 4)]
= 4({ 3 4)(3{ 3 6) = 12({ 3 4)({ 3 2) ? 0
C
2 ? { ? 4, so i is CD on (2> 4) and CU on (3"> 2) and (4> "). IPs at (2> 316) and (4> 0)
6. | = i ({) = {5 3 5{ = {({4 3 5)
I
A. G = R B. {-intercepts ± 4 5 and 0, |-intercept = i(0) = 0
C. i (3{) = 3i ({), so i is odd; the curve is symmetric about the origin. D. No asymptote
E. i 0 ({) = 5{4 3 5 = 5({4 3 1) = 5({2 3 1)({2 + 1)
2
= 5({ + 1)({ 3 1)({ + 1) A 0
H.
C
{ ? 31 or { A 1, so i is increasing on (3"> 31) and (1> "), and i is decreasing
on (31> 1). F. Local maximum value i (31) = 4, local minimum value
i (1) = 34 G. i 00 ({) = 20{3 A 0 C { A 0, so i is CU on (0> ") and CD
on (3"> 0). IP at (0> 0)
7. | = i ({) =
1 5
{
5
3 83 {3 + 16{ = { 15 {4 3 83 {2 + 16 A. G = R B. {-intercept 0, |-intercept = i(0) = 0
C. i (3{) = 3i ({), so i is odd; the curve is symmetric about the origin. D. No asymptote
E. i 0 ({) = {4 3 8{2 + 16 = ({2 3 4)2 = ({ + 2)2 ({ 3 2)2 A 0 for all {
H.
except ±2, so i is increasing on R. F. There is no local maximum or
minimum value.
G. i 00 ({) = 4{3 3 16{ = 4{({2 3 4) = 4{({ + 2)({ 3 2) A 0 C
32 ? { ? 0 or { A 2, so i is CU on (32> 0) and (2> "), and i is CD on
, (0> 0), and 2> 256
(3"> 32) and (0> 2). IP at 32> 3 256
15
15
8. | = i ({) = (4 3 {2 )5
A. G = R B. |-intercept: i (0) = 45 = 1024; {-intercepts: ±2 C. i (3{) = i ({) i
i is even; the curve is symmetric about the y-axis. D. No asymptote E. i 0 ({) = 5(4 3 {2 )4 (32{) = 310{(4 3 {2 )4 ,
so for { 6= ±2 we have i 0 ({) A 0 C { ? 0 and i 0 ({) ? 0 C { A 0. Thus, i is increasing on (3"> 0) and
decreasing on (0> "). F. Local maximum value i (0) = 1024
G. i 00 ({) = 310{ · 4(4 3 {2 )3 (32{) + (4 3 {2 )4 (310)
H.
= 310(4 3 {2 )3 [38{2 + 4 3 {2 ] = 310(4 3 {2 )3 (4 3 9{2 )
so i 00 ({) = 0 C { = ±2> ± 23 . i 00 ({) A 0 C 32 ? { ? 3 23 and
? { ? 2 and i 00 ({) ? 0 C { ? 32, 3 23 ? { ? 23 , and { A 2, so i is
CU on (3"> 2), 3 23 > 23 , and (2> "), and CD on 32> 3 23 and 23 > 2 .
5 E (±0=67> 568=25)
IP at (±2> 0) and ± 23 > 32
9
2
3
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
9. | = i ({) = {@({ 3 1)
A. G = {{ | { 6= 1} = (3"> 1) (1> ") B. {-intercept = 0, |-intercept = i (0) = 0
{
{
{
= 1, so | = 1 is a HA. lim
= 3", lim
= ", so { = 1 is a VA.
C. No symmetry D. lim
{<±" { 3 1
{<1 { 3 1
{<1+ { 3 1
E. i 0 ({) =
({ 3 1) 3 {
31
=
? 0 for { 6= 1, so i is
({ 3 1)2
({ 3 1)2
H.
decreasing on (3"> 1) and (1> ") = F. No extreme values
G. i 00 ({) =
2
A 0 C { A 1, so i is CU on (1> ") and
({ 3 1)3
CD on (3"> 1). No IP
10. | = i ({) =
({ + 2)({ 3 2)
{+2
2
{2 3 4
=
=
= 1 + for { 6= 2. There is a hole in the graph at (2> 2).
{2 3 2{
{({ 3 2)
{
{
A. G = {{ | { 6= 0> 2} = (3"> 0) (0> 2) (2> ") B. |-intercept: none; {-intercept: 32 C. No symmetry
D.
lim
{<±"
lim
{<0+
{+2
{+2
= 1, so | = 1 is a HA. lim
= 3",
{
{
{<0
H.
{+2
= ", so { = 0 is a VA. E. i 0 ({) = 32@{2 ? 0 [{ 6= 0> 2]
{
so i is decreasing on (3"> 0), (0> 2), and (2> "). F. No extrema
G. i 00 ({) = 4@{3 A 0 C { A 0, so i is CU on (0> 2) and (2> ") and
CD on (3"> 0). No IP
11. | = i ({) =
{(1 3 {)
{ 3 {2
{
=
=
for { 6= 1. There is a hole in the graph at (1> 1).
2 3 3{ + {2
(1 3 {)(2 3 {)
23{
A. G = {{ | { 6= 1> 2} = (3"> 1) (1> 2) (2> ") B. {-intercept = 0, |-intercept = i (0) = 0 C. No symmetry
{
{
{
= 31, so | = 31 is a HA. lim
= ", lim
= 3", so { = 2 is a VA.
D. lim
{<±" 2 3 {
{<2 2 3 {
{<2+ 2 3 {
E. i 0 ({) =
(2 3 {)(1) 3 {(31)
2
=
A 0 [{ 6= 1> 2], so i is
(2 3 {)2
(2 3 {)2
H.
increasing on (3"> 1), (1> 2), and (2> "). F. No extrema
G. i 0 ({) = 2(2 3 {)32
i
i 00 ({) = 34(2 3 {)33 (31) =
4
A 0 C { ? 2, so i is CU on
(2 3 {)3
(3"> 1) and (1> 2), and i is CD on (2> "). No IP
12. | = i({) = {@({2 3 9)
A. G = {{ | { 6= ±3} = (3"> 33) (33> 3) (3> ") B. {-intercept = 0,
|-intercept = i(0) = 0= C. i (3{) = 3i ({), so i is odd; the curve is symmetric about the origin.
{
{
{
{
= 0, so | = 0 is a HA. lim 2
= ", lim 2
= 3", lim
= ",
D. lim
2
{<±" {2 3 9
{<3+ { 3 9
{<3 { 3 9
{<33+ { 3 9
lim
{<33
{
({2 3 9) 3 {(2{)
{2 + 9
= 3", so { = 3 and { = 33 are VA. E. i 0 ({) =
=3 2
? 0 [{ 6= ±3]
{2 3 9
({2 3 9)2
({ 3 9)2
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SECTION 4.5 SUMMARY OF CURVE SKETCHING
so i is decreasing on (3"> 33), (33> 3), and (3> ").
¤
353
H.
F. No extreme values
G. i 00 ({) = 3
=
2{({2 3 9)2 3 ({2 + 9) · 2({2 3 9)(2{)
({2 3 9)4
2{({2 + 27)
A 0 when 33 ? { ? 0 or { A 3,
({2 3 9)3
so i is CU on (33> 0) and (3> "); CD on (3"> 33) and (0> 3). IP at (0> 0)
13. | = i({) = 1@({2 3 9)
A. G = {{ | { 6= ±3} = (3"> 33) (33> 3) (3> ") B. |-intercept = i (0) = 3 19 , no
{-intercept C. i (3{) = i ({) i i is even; the curve is symmetric about the |-axis. D.
is a HA. lim
{<3
1
= 0, so | = 0
{2 3 9
1
1
1
1
= 3", lim 2
= ", lim
= ", lim
= 3", so { = 3 and { = 33
{2 3 9
{<3+ { 3 9
{<33 {2 3 9
{<33+ {2 3 9
are VA. E. i 0 ({) = 3
2{
A 0 C { ? 0 ({ 6= 33) so i is increasing on (3"> 33) and (33> 0) and
({2 3 9)2
decreasing on (0> 3) and (3> "). F. Local maximum value i (0) = 3 19 .
G. | 00 =
lim
{<±"
H.
32({2 3 9)2 + (2{)2({2 3 9)(2{)
6({2 + 3)
=
A0 C
({2 3 9)4
({2 3 9)3
{2 A 9 C { A 3 or { ? 33, so i is CU on (3"> 33) and (3> ") and
CD on (33> 3). No IP
14. | = i ({) = {2 @({2 + 9)
A. G = R B. |-intercept: i (0) = 0; {-intercept: i ({) = 0 C { = 0
C. i(3{) = i ({), so i is even and symmetric about the |-axis. D. lim {2 @({2 + 9) = 1, so | = 1 is a HA; no VA
{<±"
2
E. i 0 ({) =
2
({ + 9)(2{) 3 { (2{)
18{
= 2
A 0 C { A 0, so i is increasing on (0> ")
({2 + 9)2
({ + 9)2
and decreasing on (3"> 0).
2
F. Local minimum value i(0) = 0; no local maximum
2
({ + 9) (18) 3 18{ · 2({2 + 9) · 2{
18({2 + 9)[({2 + 9) 3 4{2 ]
18(9 3 3{2 )
=
=
2
2
2
2
4
[({ + 9) ]
({ + 9)
({2 + 9)3
I I I
I
354 { + 3 { 3 3
=
A0 C 3 3?{? 3
H.
2
3
({ + 9)
G. i 00 ({) =
I I I I
so i is CU on 3 3> 3 and CD on 3"> 3 3 and
3> " .
I
There are two inÀection points: ± 3> 14 .
15. | = i ({) = {@({2 + 9)
A. G = R B. |-intercept: i (0) = 0; {-intercept: i({) = 0 C { = 0
C. i (3{) = 3i ({), so i is odd and the curve is symmetric about the origin. D.
HA; no VA E. i 0 ({) =
lim [{@({2 + 9)] = 0, so | = 0 is a
{<±"
({2 + 9)(1) 3 {(2{)
9 3 {2
(3 + {)(3 3 {)
= 2
=
A 0 C 33 ? { ? 3, so i is increasing
2
2
({ + 9)
({ + 9)2
({2 + 9)2
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
on (33> 3) and decreasing on (3"> 33) and (3> "). F. Local minimum value i (33) = 3 16 , local maximum
value i (3) =
1
6
(2{)({2 + 9) 3({2 + 9) 3 2(9 3 {2 )
({2 + 9)2 (32{) 3 (9 3 {2 ) · 2({2 + 9)(2{)
=
[({2 + 9)2 ]2
({2 + 9)4
I
I
2{({2 3 27)
=
= 0 C { = 0, ± 27 = ±3 3
H.
({2 + 9)3
I
I
I i 00 ({) A 0 C 33 3 ? { ? 0 or { A 3 3, so i is CU on 33 3> 0
I I I
and 3 3> " , and CD on 3"> 33 3 and 0> 3 3 . There are three
I I
1
inÀection points: (0> 0) and ±3 3> ± 12
3 .
G. i 00 ({) =
16. | = i ({) = 1 +
1
1
{2 + { + 1
+ 2 =
{
{
{2
A. G = (3"> 0) (0> ") B. |-intercept: none [{ 6= 0];
{-intercepts: i ({) = 0 C {2 + { + 1 = 0, there is no real solution, and hence, no {-intercept C. No symmetry
1
1
2
3{ 3 2
1
D. lim
.
1 + + 2 = 1, so | = 1 is a HA. lim i ({) = ", so { = 0 is a VA. E. i 0 ({) = 3 2 3 3 =
{<±"
{<0
{
{
{
{
{3
i 0 ({) A 0 C 32 ? { ? 0 and i 0 ({) ? 0 C { ? 32 or { A 0, so i is increasing on (32> 0) and decreasing
on (3"> 32) and (0> "). F. Local minimum value i (32) = 34 ; no local
maximum
G. i 00 ({) =
H.
2
6
2{ + 6 00
+ 4 =
. i ({) ? 0 C { ? 33 and
{3
{
{4
i 00 ({) A 0 C 33 ? { ? 0 and { A 0, so i is CD on (3"> 33) and CU on
(33> 0) and (0> "). IP at 33> 79
17. | = i ({) =
{31
{2
C. No symmetry D.
E. i 0 ({) =
A. G = {{ | { 6= 0} = (3"> 0) (0> ") B. No |-intercept; {-intercept: i ({) = 0 C { = 1
lim
{<±"
{31
{31
= 0, so | = 0 is a HA. lim
= 3", so { = 0 is a VA.
{<0 {2
{2
{2 · 1 3 ({ 3 1) · 2{
3({ 3 2)
3{2 + 2{
=
=
, so i 0 ({) A 0 C 0 ? { ? 2 and i 0 ({) ? 0 C
({2 )2
{4
{3
{ ? 0 or { A 2. Thus, i is increasing on (0> 2) and decreasing on (3"> 0)
and (2> "). F. No local minimum, local maximum value i(2) =
G. i 00 ({) =
H.
1
.
4
{3 · (31) 3 [3({ 3 2)] · 3{2
2{3 3 6{2
2({ 3 3)
=
=
.
3
2
({ )
{6
{4
i 00 ({) is negative on (3"> 0) and (0> 3) and positive on (3> "), so i is CD
on (3"> 0) and (0> 3) and CU on (3> "). IP at 3> 29
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SECTION 4.5 SUMMARY OF CURVE SKETCHING
{
{3 3 1
18. | = i ({) =
C. No symmetry D.
¤
355
A. G = (3"> 1) (1> ") B. |-intercept: i (0) = 0; {-intercept: i ({) = 0 C { = 0
lim
{<±"
{
= 0, so | = 0 is a HA. lim i ({) = 3" and lim i ({) = ", so { = 1 is a VA.
{3 3 1
{<1
{<1+
s
s
({3 3 1)(1) 3 {(3{2 )
32{3 3 1 0
=
. i ({) = 0 i { = 3 3 1@2. i 0 ({) A 0 C { ? 3 3 1@2 and
({3 3 1)2
({3 3 1)2
s
s
s
i 0 ({) ? 0 C 3 3 1@2 ? { ? 1 and { A 1, so i is increasing on 3"> 3 3 1@2 and decreasing on 3 3 1@2> 1
s
s
and (1> "). F. Local maximum value i 3 3 1@2 = 23 3 1@2; no local minimum
E. i 0 ({) =
({3 3 1)2 (36{2 ) 3 (32{3 3 1)2({3 3 1)(3{2 )
[({3 3 1)2 ]2
G. i 00 ({) =
H.
36{2 ({3 3 1)[({3 3 1) 3 (2{3 + 1)]
6{2 ({3 + 2)
=
=
3
4
({ 3 1)
({3 3 1)3
I
I
i 00 ({) A 0 C { ? 3 3 2 and { A 1, i 00 ({) ? 0 C 3 3 2 ? { ? 0 and
I I 0 ? { ? 1, so i is CU on 3"> 3 3 2 and (1> ") and CD on 3 3 2> 1 .
I I
IP at 3 3 2> 13 3 2
=
({2 + 3) 3 3
3
{2
=
=13 2
+3
{2 + 3
{ +3
19. | = i ({) =
{2
A. G = R B. |-intercept: i(0) = 0;
{-intercepts: i ({) = 0 C { = 0 C. i (3{) = i ({), so i is even; the graph is symmetric about the |-axis.
D.
32{
6{
{2
= 2
.
= 1, so | = 1 is a HA. No VA. E. Using the Reciprocal Rule, i 0 ({) = 33 · 2
+3
({ + 3)2
({ + 3)2
lim
{<±" {2
i 0 ({) A 0 C { A 0 and i 0 ({) ? 0 C { ? 0, so i is decreasing on (3"> 0) and increasing on (0> ").
F. Local minimum value i(0) = 0, no local maximum.
G. i 00 ({) =
=
({2 + 3)2 · 6 3 6{ · 2({2 + 3) · 2{
[({2 + 3)2 ]2
H.
6({2 + 3)[({2 + 3) 3 4{2 ]
6(3 3 3{2 )
318({ + 1)({ 3 1)
=
=
2
4
({ + 3)
({2 + 3)3
({2 + 3)3
i 00 ({) is negative on (3"> 31) and (1> ") and positive on (31> 1),
so i is CD on (3"> 31) and (1> ") and CU on (31> 1). IP at ±1> 14
20. | = i ({) =
{3
8
= {2 + 2{ + 4 +
[by long division] A. G = (3"> 2) (2> ") B. {-intercept = 0,
{32
{32
|-intercept = i (0) = 0 C. No symmetry D. lim
{<2
{3
{3
= 3" and lim
= ", so { = 2 is a VA.
{32
{<2+ { 3 2
There are no horizontal or slant asymptotes. Note: Since lim
{<±"
asymptotically as { < ±".
E. i 0 ({) =
8
= 0, the parabola | = {2 + 2{ + 4 is approached
{32
({ 3 2)(3{2 ) 3 {3 (1)
{2 [3({ 3 2) 3 {]
{2 (2{ 3 6)
2{2 ({ 3 3)
=
=
=
A 0 C { A 3 and
({ 3 2)2
({ 3 2)2
({ 3 2)2
({ 3 2)2
i 0 ({) ? 0 C { ? 0 or 0 ? { ? 2 or 2 ? { ? 3, so i is increasing on (3> ") and i is decreasing on (3"> 2) and (2> 3).
F. Local minimum value i(3) = 27, no local maximum value G. i 0 ({) = 2
{3 3 3{2
({ 3 2)2
i
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CHAPTER 4
i 00 ({) = 2
=2
APPLICATIONS OF DIFFERENTIATION
({ 3 2)2 (3{2 3 6{) 3 ({3 3 3{2 )2({ 3 2)
[({ 3 2)2 ]2
H.
({ 3 2){[({ 3 2)(3{ 3 6) 3 ({2 3 3{)2]
({ 3 2)4
=
2{(3{2 3 12{ + 12 3 2{2 + 6{)
({ 3 2)3
=
2{({2 3 6{ + 12)
A0
({ 3 2)3
C
{ ? 0 or { A 2, so i is CU on (3"> 0) and (2> "), and i is CD on (0> 2). IP at (0> 0)
I
21. | = i ({) = ({ 3 3) { = {3@2 3 3{1@2
A. G = [0> ") B. {-intercepts: 0> 3; |-intercept = i (0) = 0 C. No
symmetry D. No asymptote E. i 0 ({) = 32 {1@2 3 32 {31@2 = 32 {31@2 ({ 3 1) =
3({ 3 1)
I
A0
2 {
so i is increasing on (1> ") and decreasing on (0> 1).
C
{ A 1,
H.
F. Local minimum value i(1) = 32, no local maximum value
G. i 00 ({) = 34 {31@2 + 34 {33@2 = 34 {33@2 ({ + 1) =
3({ + 1)
A 0 for { A 0,
4{3@2
so i is CU on (0> "). No IP
22. | = i ({) = 2
I
{ 3 { A. G = [0> ") B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 i
I
{ = { i 4{ = {2
i 4{ 3 {2 = 0 i {(4 3 {) = 0 i { = 0, 4 C. No symmetry D. No asymptote
I 1
1 E. i 0 ({) = I 3 1 = I 1 3 { . This is positive for { ? 1 and negative for { A 1, decreasing on (1> ").
{
{
2
so i is increasing on (0> 1) and decreasing on (1> ").
H.
F. Local maximum value i (1) = 1, no local minimum.
G. i 00 ({) = ({31@2 3 1)0 = 3 12 {33@2 =
31
? 0 for { A 0,
2{3@2
so i is CD on (0> "). No IP
23. | = i ({) =
s
I
{2 + { 3 2 = ({ + 2)({ 3 1) A. G = {{ | ({ + 2)({ 3 1) D 0} = (3"> 32] [1> ")
B. |-intercept: none; {-intercepts: 32 and 1 C. No symmetry D. No asymptote
E. i 0 ({) = 12 ({2 + { 3 2)31@2 (2{ + 1) =
2
2{ + 1
I
, i 0 ({) = 0 if { = 3 12 , but 3 12 is not in the domain.
{2 + { 3 2
i 0 ({) A 0 i { A 3 12 and i 0 ({) ? 0 i { ? 3 12 , so (considering the domain) i is increasing on (1> ") and
i is decreasing on (3"> 32). F. No local extrema
2({2 + { 3 2)1@2 (2) 3 (2{ + 1) · 2 · 12 ({2 + { 3 2)31@2 (2{ + 1)
I
2
2 {2 + { 3 2
({2 + { 3 2)31@2 4({2 + { 3 2) 3 (4{2 + 4{ + 1)
=
4({2 + { 3 2)
G. i 00 ({) =
=
H.
39
?0
4({2 + { 3 2)3@2
so i is CD on (3"> 32) and (1> "). No IP
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.5 SUMMARY OF CURVE SKETCHING
¤
357
s
I
{2 + { 3 { = {({ + 1) 3 { A. G = (3"> 31] [0> ") B. |-intercept: i (0) = 0;
I
{-intercepts: i ({) = 0 i
{2 + { = { i {2 + { = {2 i { = 0 C. No symmetry
I
I
{2 + { + {
{2 + { 3 {2
{2 + { 3 { I
= lim I
D. lim i ({) = lim
2
{<"
{<"
{<"
{ +{+{
{2 + { + {
24. | = i ({) =
{@{
1
1
= lim s
= , so | =
= lim I
{<"
2
{2 + { + { @{ {<" 1 + 1@{ + 1
E. i 0 ({) = 12 ({2 + {)31@2 (2{ + 1) 3 1 =
{+
1
2
A
2
1
2
is a HA. No VA
I
2{ + 1
I
3 1 A 0 C 2{ + 1 A 2 {2 + { C
{2 + {
t
2
{ + 12 3 14 . Keep in mind that the domain excludes the interval (31> 0). When { +
the last inequality is true since the value of the radical is less than { + 12 . When { +
1
2
1
2
is positive (for { D 0),
is negative (for { $ 31), the last
inequality is false since the value of the radical is positive. So i is increasing on (0> ") and decreasing on (3"> 31).
H.
F. No local extrema
2
G. i 00 ({) =
=
1@2
2({ + {)
(2) 3 (2{ + 1) · 2 ·
I
2 {2 + {
1
({2
2
2
31@2
+ {)
(2{ + 1)
31
({2 + {)31@2 [4({2 + {) 3 (2{ + 1)2 ]
=
=
4({2 + {)
4({2 + {)3@2
i 00 ({) ? 0 when it is de¿ned, so i is CD on (3"> 31) and (0> "). No IP
I
25. | = i ({) = {@ {2 + 1
A. G = R B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 i { = 0
C. i (3{) = 3i ({), so i is odd; the graph is symmetric about the origin.
{
{@{
{@{
1
1
I = lim s
= lim I
= lim I
=1
= I
D. lim i ({) = lim I
{<"
{<"
{<"
1+0
{2 + 1 {<" {2 + 1@{ {<" {2 + 1@ {2
1 + 1@{2
and
{
{@{
{@{
1
I = lim
s
= lim I
= lim I
lim i({) = lim I
{<3"
{<3" 3
{2 + 1 {<3" {2 + 1@{ {<3" {2 + 1@ 3 {2
1 + 1@{2
{<3"
=
No VA.
E. i 0 ({) =
1
I
= 31 so | = ±1 are HA.
3 1+0
I
{2 + 1 3 { ·
2{
I
{2 + 1 3 {2
1
{2 + 1
=
=
A 0 for all {, so i is increasing on R.
3@2
2
2
[({2 + 1)1@2 ]2
({ + 1)
({ + 1)3@2
2
H.
F. No extreme values
33{
, so i 00 ({) A 0 for { ? 0
G. i 00 ({) = 3 32 ({2 + 1)35@2 · 2{ = 2
({ + 1)5@2
and i 00 ({) ? 0 for { A 0. Thus, i is CU on (3"> 0) and CD on (0> ").
IP at (0> 0)
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°
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358
NOT FOR SALE
¤
CHAPTER 4
26. | = i ({) = {
APPLICATIONS OF DIFFERENTIATION
I I A. G = 3 2> 2
B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 i
I
2 3 {2
I
{ = 0, ± 2. C. i (3{) = 3i ({), so i is odd; the graph is symmetric about the origin. D. No asymptote
I
I
3{
3{2 + 2 3 {2
2(1 + {)(1 3 {) 0
I
I
E. i 0 ({) = { · I
+ 2 3 {2 =
=
. i ({) is negative for 3 2 ? { ? 31
2
2
2
23{
23{
23{
I
I
I and 1 ? { ? 2, and positive for 31 ? { ? 1, so i is decreasing on 3 2> 31 and 1> 2 and increasing on (31> 1).
F. Local minimum value i(31) = 31, local maximum value i(1) = 1.
I
3{
2 3 {2 (34{) 3 (2 3 2{2 ) I
2 3 {2
00
G. i ({) =
2
[(2 3 {2 )1@2 ]
H.
(2 3 {2 )(34{) + (2 3 2{2 ){
2{({2 3 3)
2{3 3 6{
=
=
2
3@2
2
3@2
(2 3 { )
(2 3 { )
(2 3 {2 )3@2
I
I I Since {2 3 3 ? 0 for { in 3 2> 2 , i 00 ({) A 0 for 3 2 ? { ? 0 and
I
I I i 00 ({) ? 0 for 0 ? { ? 2. Thus, i is CU on 3 2> 0 and CD on 0> 2 .
=
The only IP is (0> 0).
I
1 3 {2 @{ A. G = {{ | |{| $ 1, { 6= 0} = [31> 0) (0> 1] B. {-intercepts ±1, no |-intercept
I
I
1 3 {2
1 3 {2
= ", lim
= 3",
C. i (3{) = 3i ({), so the curve is symmetric about (0> 0) = D. lim
{
{
{<0+
{<0
I
2I
3{ @ 1 3 {2 3 1 3 {2
1
=3 I
? 0, so i is decreasing
so { = 0 is a VA. E. i 0 ({) =
{2
{2 1 3 {2
27. | = i({) =
on (31> 0) and (0> 1). F. No extreme values
G. i 00 ({) =
2 3 3{2
{3 (1
{2 )3@2
H.
t
t
A 0 C 31 ? { ? 3 23 or 0 ? { ? 23 , so
3
t t
t t
2
>1 .
i is CU on 31> 3 23 and 0> 23 and CD on 3 23 > 0 and
3
t
IP at ± 23 > ± I12
I
28. | = i ({) = {@ {2 3 1
A. G = (3"> 31) (1> ") B. No intercepts C. i (3{) = 3i ({), so i is odd;
{
{
the graph is symmetric about the origin. D. lim I
= 1 and lim I
= 31, so | = ±1 are HA.
{<"
{<3"
{2 3 1
{2 3 1
lim i ({) = +" and
{<1+
lim i ({) = 3", so { = ±1 are VA.
{<31
I
{
{2 3 1 3 { · I
{2 3 1 3 {2
31
{2 3 1
= 2
= 2
? 0, so i is decreasing on (3"> 31) and (1> ").
E. i ({) =
2
1@2
2
[({ 3 1) ]
({ 3 1)3@2
({ 3 1)3@2
0
F. No extreme values
G. i 00 ({) = (31) 3 32 ({2 3 1)35@2 · 2{ =
H.
3{
.
({2 3 1)5@2
i 00 ({) ? 0 on (3"> 31) and i 00 ({) A 0 on (1> "), so i is CD on (3"> 31)
and CU on (1> "). No IP
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NOT FOR SALE
SECTION 4.5 SUMMARY OF CURVE SKETCHING
29. | = i ({) = { 3 3{1@3
A. G = R B. |-intercept: i (0) = 0; {-intercepts: i ({) = 0 i { = 3{1@3
{3 = 27{ i {3 3 27{ = 0 i {({2 3 27) = 0 i { = 0, ±3
¤
i
I
3 C. i (3{) = 3i ({), so i is odd;
the graph is symmetric about the origin. D. No asymptote E. i 0 ({) = 1 3 {32@3 = 1 3
1
{2@3
=
{2@3 3 1
.
{2@3
i 0 ({) A 0 when |{| A 1 and i 0 ({) ? 0 when 0 ? |{| ? 1, so i is increasing on (3"> 31) and (1> "), and
decreasing on (31> 0) and (0> 1) [hence decreasing on (31> 1) since i is
H.
continuous on (31> 1)]. F. Local maximum value i (31) = 2, local minimum
value i (1) = 32 G. i 00 ({) = 23 {35@3 ? 0 when { ? 0 and i 00 ({) A 0
when { A 0, so i is CD on (3"> 0) and CU on (0> "). IP at (0> 0)
30. | = i ({) = {5@3 3 5{2@3 = {2@3 ({ 3 5)
D.
A. G = R B. {-intercepts 0, 5; |-intercept 0 C. No symmetry
lim {2@3 ({ 3 5) = ±", so there is no asymptote E. i 0 ({) = 53 {2@3 3
{<±"
10 31@3
{
3
{ ? 0 or { A 2, so i is increasing on (3"> 0), (2> ") and
= 53 {31@3 ({ 3 2) A 0 C
H.
decreasing on (0> 2).
F. Local maximum value i (0) = 0, local minimum value i(2) = 33
G. i 00 ({) =
10 31@3
9 {
+
10 34@3
9 {
=
10 34@3
({
9 {
I
3
4
+ 1) A 0 C { A 31, so
i is CU on (31> 0) and (0> "), CD on (3"> 31). IP at (31> 36)
31. | = i ({) =
I
3
{2 3 1 A. G = R B. |-intercept: i (0) = 31; {-intercepts: i ({) = 0 C {2 3 1 = 0 C
{ = ±1 C. i (3{) = i ({), so the curve is symmetric about the |-axis= D. No asymptote
E. i 0 ({) = 13 ({2 3 1)32@3 (2{) =
3
2{
s
. i 0 ({) A 0 C { A 0 and i 0 ({) ? 0 C { ? 0, so i is
3
({2 3 1)2
increasing on (0> ") and decreasing on (3"> 0). F. Local minimum value i (0) = 31
G. i 00 ({) =
=
2
2@3
2
31@3
(2{)
2 ({ 3 1) (1) 3 { · 23 ({ 3 1)
·
3
[({2 3 1)2@3 ]2
H.
2 ({2 3 1)31@3 [3({2 3 1) 3 4{2 ]
2({2 + 3)
·
=3
2
4@3
9
({ 3 1)
9({2 3 1)5@3
i 00 ({) A 0 C 31 ? { ? 1 and i 00 ({) ? 0 C { ? 31 or { A 1, so
i is CU on (31> 1) and i is CD on (3"> 31) and (1> "). IP at (±1> 0)
32. | = i ({) =
I
3
{3 + 1 A. G = R B. |-intercept: i (0) = 1; {-intercept: i ({) = 0 C {3 + 1 = 0 i { = 31
{2
. i 0 ({) A 0 if { ? 31,
C. No symmetry D. No asymptote E. i 0 ({) = 13 ({3 + 1)32@3 (3{2 ) = s
3
({3 + 1)2
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359
360
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
31 ? { ? 0, and { A 0, so i is increasing on R. F. No local extrema
G. i 00 ({) =
=
({3 + 1)2@3 (2{) 3 {2 · 23 ({3 + 1)31@3 (3{2 )
[({3 + 1)2@3 ]2
H.
.
{({3 + 1)31@3 [2({3 + 1) 3 2{3 ]
2{
= 3
({3 + 1)4@3
({ + 1)5@3
i 00 ({) A 0 C { ? 31 or { A 0 and i 00 ({) ? 0 C 31 ? { ? 0, so i is
CU on (3"> 31) and (0> ") and CD on (31> 0). IP at (31> 0) and (0> 1)
33. | = i ({) = sin3 {
A. G = R B. {-intercepts: i ({) = 0 C { = q, q an integer; |-intercept = i (0) = 0
C. i (3{) = 3i ({), so i is odd and the curve is symmetric about the origin. Also, i ({ + 2) = i({), so i is periodic
with period 2, and we determine E–G for 0 $ { $ . Since i is odd, we can reÀect the graph of i on [0> ] about the
origin to obtain the graph of i on [3> ], and then since i has period 2, we can extend the graph of i for all real numbers.
D. No asymptote E. i 0 ({) = 3 sin2 { cos { A 0 C cos { A 0 and sin { 6= 0 C 0 ? { ? 2 , so i is increasing on
0> 2 and i is decreasing on 2 > . F. Local maximum value i 2 = 1 local minimum value i 3 2 = 31
G. i 00 ({) = 3 sin2 { (3 sin {) + 3 cos { (2 sin { cos {) = 3 sin { (2 cos2 { 3 sin2 {)
= 3 sin { [2(1 3 sin2 {) 3 sin2 {] = 3 sin {(2 3 3 sin2 {) A 0 C
t
t k
t l
sin { A 0 and sin2 { ? 23 C 0 ? { ? and 0 ? sin { ? 23 C 0 ? { ? sin31 23 let = sin31 23 or
3 ? { ? , so i is CU on (0> ) and ( 3 > ), and i is CD on (> 3 ). There are inÀection points at { = 0, , ,
and { = 3 .
H.
34. | = i ({) = { + cos {
A. G = R B. |-intercept: i (0) = 1; the {-intercept is about 30=74 and can be found using
Newton’s method C. No symmetry D. No asymptote E. i 0 ({) = 1 3 sin { A 0 except for { =
so i is increasing on R. F. No local extrema
2
+ 2q,
H.
G. i 00 ({) = 3 cos {. i 00 ({) A 0 i 3 cos { A 0 i cos { ? 0 i
+ 2q and i 00 ({) ? 0 i
{ is in 2 + 2q> 3
2
+ 2q and CD on
{ is in 3 2 + 2q> 2 + 2q , so i is CU on 2 + 2q> 3
2
3 2 + 2q> 2 + 2q . IP at 2 + q> i 2 + q = 2 + q> 2 + q
[on the line | = {]
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
¤
SECTION 4.5 SUMMARY OF CURVE SKETCHING
35. | = i ({) = { tan {, 3 2 ? { ?
symmetric about the |-axis. D.
2
B. Intercepts are 0 C. i(3{) = i ({), so the curve is
A. G = 3 2 > 2
lim
{<(@2)
{ tan { = " and
lim
{<3(@2)+
{ tan { = ", so { =
E. i 0 ({) = tan { + { sec2 { A 0 C 0 ? { ? 2 , so i increases on 0> 2
and decreases on 3 2 > 0 . F. Absolute and local minimum value i (0) = 0.
G. | 00 = 2 sec2 { + 2{ tan { sec2 { A 0 for 3 2 ? { ?
CU on 3 2 > 2 . No IP
36. | = i ({) = 2{ 3 tan {, 3 2 ? { ?
lim
{<(3@2)+
(2{ 3 tan {) = " and
2,
2
and { = 3 2 are VA.
H.
so i is
B. |-intercept: i (0) = 0; {-intercepts: i (0) = 0 C
A. G = 3 2 > 2
2
2{ = tan { C { = 0 or { E ±1=17
D.
361
C. i(3{) = 3i({), so i is odd; the graph is symmetric about the origin.
lim
{<(@2)
(2{ 3 tan {) = 3", so { = ± 2 are VA. No HA.
I
I
E. i 0 ({) = 2 3 sec2 { ? 0 C |sec {| A 2 and i 0 ({) A 0 C |sec {| ? 2, so i is decreasing on 3 2 > 3 4 ,
increasing on 3 4 > 4 , and decreasing again on 4 > 2
F. Local maximum value i 4 = 2 3 1,
local minimum value i 3 4 = 3 2 + 1
H.
G. i 00 ({) = 32 sec { · sec { tan { = 32 tan { sec2 { = 32 tan {(tan2 { + 1)
so i 00 ({) A 0 C tan { ? 0 C 3 2 ? { ? 0, and i 00 ({) ? 0 C
tan { A 0 C 0 ? { ? 2 . Thus, i is CU on 3 2 > 0 and CD on 0> 2 .
IP at (0> 0)
37. | = i ({) =
1
{
2
3 sin {, 0 ? { ? 3
A. G = (0> 3) B. No |-intercept. The {-intercept, approximately 1=9, can be
found using Newton’s Method. C. No symmetry D. No asymptote E. i 0 ({) = 12 3 cos { A 0 C cos { ?
5 5
7
7
and 7
and 5
.
3 ? { ? 3 or 3 ? { ? 3, so i is increasing on 3 > 3
3 > 3 and decreasing on 0> 3
3 > 3
I
H.
F. Local minimum value i 3 = 6 3 23 , local maximum value
i
5 3
=
5
6
+
I
3
2 ,
local minimum value i
7 3
=
7
6
3
1
2
I
3
2
G. i 00 ({) = sin { A 0 C 0 ? { ? or 2 ? { ? 3, so i is CU on
(0> ) and (2> 3) and CD on (> 2). IPs at > 2 and (2> )
38. | = i ({) = sec { + tan {, 0 ? { ? @2
B. |-intercept: none (0 not in domain); {-intercept: none,
A. G = 0> 2
since sec { and tan { are both positive on the domain
C. No symmetry D.
lim i ({) = ", so { =
{<(@2)
0
2
E. i ({) = sec { tan { + sec { = sec { (tan { + sec {) A 0 on 0> 2 , so i is increasing on 0> 2 .
is a VA.
H.
F. No local extrema
00
2
2
G. i ({) = sec { (sec { + sec { tan {) + (tan { + sec {) sec { tan {
= sec { (sec { + tan {) sec { + sec { (sec { + tan {) tan {
= sec { (sec { + tan {)(sec { + tan {) = sec { (sec { + tan {)2 A 0
on 0> 2 , so i is CU on 0> 2 . No IP
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°
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362
¤
NOT FOR SALE
CHAPTER 4
39. | = i ({) =
APPLICATIONS OF DIFFERENTIATION
sin {
1 + cos {
5
9
7
6
when
cos { 6= 1
=
1 3 cos {
1 3 cos {
sin { (1 3 cos {)
sin {
:
=
·
=
= csc { 3 cot {8
1 + cos { 1 3 cos {
sin {
sin2 {
A. The domain of i is the set of all real numbers except odd integer multiples of ; that is, all reals except (2q + 1), where
q is an integer. B. |-intercept: i (0) = 0; {-intercepts: { = 2q, q an integer. C. i (3{) = 3i ({), so i is an odd
function; the graph is symmetric about the origin and has period 2. D. When q is an odd integer,
lim
{<(q)
i ({) = " and
E. i 0 ({) =
lim
{<(q)+
i ({) = 3", so { = q is a VA for each odd integer q. No HA.
(1 + cos {) · cos { 3 sin {(3 sin {)
1 + cos {
1
. i 0 ({) A 0 for all { except odd multiples of
=
=
(1 + cos {)2
(1 + cos {)2
1 + cos {
, so i is increasing on ((2n 3 1)> (2n + 1)) for each integer n. F. No extreme values
G. i 00 ({) =
sin {
A 0 i sin { A 0 i
(1 + cos {)2
H.
{ M (2n> (2n + 1)) and i 00 ({) ? 0 on ((2n 3 1)> 2n) for each
integer n. i is CU on (2n> (2n + 1)) and CD on ((2n 3 1)> 2n)
for each integer n. i has IPs at (2n> 0) for each integer n.
40. | = i ({) =
sin {
2 + cos {
A. G = R B. |-intercept: i (0) = 0; {-intercepts: i({) = 0 C sin { = 0 C { = q
C. i (3{) = 3i({), so the curve is symmetric about the origin. i is periodic with period 2, so we determine E–G for
0 $ { $ 2. D. No asymptote
(2 + cos {) cos { 3 sin {(3 sin {)
2 cos { + cos2 { + sin2 {
2 cos { + 1
=
=
.
2
(2 + cos {)
(2 + cos {)2
(2 + cos {)2
4
i 0 ({) A 0 C 2 cos { + 1 A 0 C cos { A 3 12 C { is in 0> 2
or 3 > 2 , so i is increasing
3
> 2 , and i is decreasing on 2
> 4
on 0> 2
and 4
.
3
3
3
3
I
I
I
I
4 2 3@2
3
3
3 3@2
=
and local minimum value i 3 =
=3
F. Local maximum value i 3 =
2 3 (1@2)
3
2 3 (1@2)
3
E. i 0 ({) =
G. i 00 ({) =
=
(2 + cos {)2 (32 sin {) 3 (2 cos { + 1)2(2 + cos {)(3 sin {)
[(2 + cos {)2 ]2
32 sin { (2 + cos {)[(2 + cos {) 3 (2 cos { + 1)]
32 sin { (1 3 cos {)
=
(2 + cos {)4
(2 + cos {)3
i 00 ({) A 0 C 32 sin { A 0
C sin { ? 0 C { is in (> 2) [i is CU] and i 00 ({) ? 0 C
{ is in (0> ) [i is CD]. The inÀection points are (0> 0), (> 0), and (2> 0).
H.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.5 SUMMARY OF CURVE SKETCHING
41. | = i ({) = arctan(h{ )
C. No symmetry D.
E. i 0 ({) =
G. i 00 ({) =
=
A. G = R B. |-intercept = i (0) = arctan 1 =
{
{
lim arctan(h ) = 0 and lim arctan(h ) =
{<3"
{<"
2,
4.
¤
363
i({) A 0 so there are no {-intercepts.
so | = 0 and | =
2
are HAs. No VA
g {
1
h{
h =
A 0, so i is increasing on (3"> "). F. No local extrema
{
2
1 + (h ) g{
1 + h2{
(1 + h2{ )h{ 3 h{ (2h2{ )
h{ [(1 + h2{ ) 3 2h2{ ]
=
(1 + h2{ )2
(1 + h2{ )2
h{ (1 3 h2{ )
A0
(1 + h2{ )2
H.
C
1 3 h2{ A 0 C h2{ ? 1 C 2{ ? 0 C { ? 0, so i is CU on
(3"> 0) and CD on (0> "). IP at 0> 4
42. | = i ({) = (1 3 {)h{
A. G = R B. {-intercept 1, |-intercept = i (0) = 1 C. No symmetry
H
13{ 31
form "
D. lim
= lim
= 0, so | = 0 is a HA. No VA
"
{<3" h3{
{<3" 3h3{
E. i 0 ({) = (1 3 {)h{ + h{ (31) = h{ [(1 3 {) + (31)] = 3{h{ A 0 C { ? 0, so i is increasing on (3"> 0)
and decreasing on (0> ").
H.
F. Local maximum value i (0) = 1, no local minimum value
G. i 00 ({) = 3{h{ + h{ (31) = h{ (3{ 3 1) = 3({ + 1)h{ A 0 C
{ ? 31, so i is CU on (3"> 31) and CD on (31> "). IP at (31> 2@h)
43. | = 1@(1 + h3{ )
A. G = R B. No {-intercept; |-intercept = i (0) = 12 = C. No symmetry
D. lim 1@(1 + h3{ ) =
{<"
1
1+0
= 1 and lim 1@(1 + h3{ ) = 0 since lim h3{ = ", so i has horizontal asymptotes
0
{<3"
{<3"
3{ 32
| = 0 and | = 1. E. i ({) = 3(1 + h
F. No extreme values G. i 00 ({) =
)
(3h
3{ 2
(1 + h
3{
3{
)=h
3{
) (3h
3{ 2
@(1 + h
) . This is positive for all {, so i is increasing on R.
3{
) 3 h (2)(1 + h3{ )(3h3{ )
h3{ (h3{ 3 1)
=
(1 + h3{ )4
(1 + h3{ )3
The second factor in the numerator is negative for { A 0 and positive for { ? 0,
H.
and the other factors are always positive, so i is CU on (3", 0) and CD
on (0> "). IP at 0, 12
44. | = i ({) = h3{ sin {, 0 $ { $ 2
A. G = R B. |-intercept: i (0) = 0; {-intercepts: i({) = 0 C sin { = 0 C
{ = 0, , and 2. C. No symmetry D. No asymptote E. i 0 ({) = h3{ cos { + sin { (3h3{ ) = h3{ (cos { 3 sin {).
i 0 ({) = 0 C cos { = sin { C { = 4 , 5
. i 0 ({) A 0 if { is in 0> 4 or 5
> 2 [ i is increasing] and
4
4
[ i is decreasing]. F. Local maximum value i 4 and local minimum value i 5
i 0 ({) ? 0 if { is in 4 > 5
4
4
G. i 00 ({) = h3{ (3 sin { 3 cos {) + (cos { 3 sin {)(3h3{ ) = h3{ (32 cos {). i 00 ({) A 0 C 32 cos { A 0 C
cos { ? 0 i { is in 2 > 3
[i is CU] and i 00 ({) ? 0 C
H.
2
3
cos { A 0 i { is in 0> 2 or 2 > 2 [i is CD].
IP at 2 + q> i 2 + q
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NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
45. | = i({) = { 3 ln {
A. G = (0> ") B. |-intercept: none (0 is not in the domain); {-intercept: i({) = 0 C
{ = ln {, which has no solution, so there is no {-intercept. C. No symmetry D. lim ({ 3 ln {) = "> so { = 0
{<0+
is a VA. E. i 0 ({) = 1 3 1@{ A 0 i 1 A 1@{ i { A 1 and
H.
0
i ({) ? 0 i 0 ? { ? 1, so i is increasing on (1> ") and i is decreasing
on (0> 1). F. Local minimum value i(1) = 1; no local maximum value
G. i 00 ({) =
1
A 0 for all {, so i is CU on (0> "). No IP
{2
46. | = i ({) = h2{ 3 h{
A. G = R B. |-intercept: i(0) = 0; {-intercepts: i ({) = 0 i h2{ = h{
{ = 0= C. No symmetry D.
so i 0 ({) A 0 C h{ A
i h{ = 1 i
lim h2{ 3 h{ = 0, so | = 0 is a HA. No VA. E. i 0 ({) = 2h2{ 3 h{ = h{ (2h{ 3 1),
{<3"
C { A ln 12 = 3 ln 2 and i 0 ({) ? 0 C
h{ ? 12 C { ? ln 12 , so i is decreasing on 3"> ln 12
and increasing on ln 12 > " .
1
2
2
F. Local minimum value i ln 12 = h2 ln(1@2) 3 hln(1@2) = 12 3
1
2
H.
= 3 14
G. i 00 ({) = 4h2{ 3 h{ = h{ (4h{ 3 1), so i 00 ({) A 0 C h{ A 14 C
{ A ln 14 and i 00 ({) ? 0 C { ? ln 14 . Thus, i is CD on 3"> ln 14 and
2
3
CU on ln 14 > " . IP at ln 14 > 14 3 14 = ln 14 > 3 16
47. | = i ({) = (1 + h{ )32 =
1
(1 + h{ )2
A. G = R B. |-intercept: i (0) = 14 . {-intercepts: none [since i ({) A 0]
C. No symmetry D. lim i ({) = 0 and lim i ({) = 1, so | = 0 and | = 1 are HA; no VA
{<"
{<3"
{
E. i 0 ({) = 32(1 + h{ )33 h{ =
32h
? 0, so i is decreasing on R F. No local extrema
(1 + h{ )3
G. i 00 ({) = (1 + h{ )33 (32h{ ) + (32h{ )(33)(1 + h{ )34 h{
= 32h{ (1 + h{ )34 [(1 + h{ ) 3 3h{ ] =
H.
32h{ (1 3 2h{ )
=
(1 + h{ )4
i 00 ({) A 0 C 1 3 2h{ ? 0 C h{ A 12 C { A ln 12 and
i 00 ({) ? 0 C { ? ln 12 , so i is CU on ln 12 > " and CD on 3"> ln 12 .
IP at ln 12 > 49
48. | = i ({) = h{ @{2
lim
{<0
A. G = (3"> 0) (0> ") B. No intercept C. No symmetry D.
lim
{<3"
h{
= 0, so | = 0 is HA.
{2
h{
{2 h{ 3 h{ (2{)
{h{ ({ 3 2)
h{ ({ 3 2)
0
=
",
so
{
=
0
is
a
VA.
E.
i
({)
=
=
=
A 0 C { ? 0 or { A 2,
{2
({2 )2
{4
{3
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SECTION 4.5 SUMMARY OF CURVE SKETCHING
so i is increasing on (3"> 0) and (2> "), and i is decreasing on (0> 2).
¤
365
H.
2
F. Local minimum value i(2) = h @4 r 1=85, no local maximum value
G. i 00 ({) =
=
{3 [h{ (1) + ({ 3 2)h{ ] 3 h{ ({ 3 2)(3{2 )
({3 )2
{2 h{ [{({ 3 1) 3 3({ 3 2)]
h{ ({2 3 4{ + 6)
=
A0
{6
{4
for all { in the domain of i ; that is, i is CU on (3"> 0) and (0> "). No IP
49. | = i ({) = ln(sin {)
A. G = {{ in R | sin { A 0} =
"
V
q=3"
(2q> (2q + 1) ) = · · · (34> 33) (32> 3) (0> ) (2> 3) · · ·
B. No |-intercept; {-intercepts: i ({) = 0 C ln(sin {) = 0 C sin { = h0 = 1 C
integer q.
C. i is periodic with period 2. D.
lim
{<(2q)+
i({) = 3" and
lim
{<[(2q+1)]
{ = 2q +
2
for each
i ({) = 3", so the lines
cos {
= cot {, so i 0 ({) A 0 when 2q ? { ? 2q + 2 for each
sin {
? { ? (2q + 1). Thus, i is increasing on 2q> 2q + 2 and
{ = q are VAs for all integers q. E. i 0 ({) =
integer q, and i 0 ({) ? 0 when 2q + 2
decreasing on 2q + 2 > (2q + 1) for each integer q.
F. Local maximum values i 2q + 2 = 0, no local minimum.
H.
G. i 00 ({) = 3 csc2 { ? 0, so i is CD on (2q> (2q + 1)) for
each integer q= No IP
50. | = i ({) = ln({2 3 3{ + 2) = ln [({ 3 1)({ 3 2)]
A. G = { in R | {2 3 3{ + 2 A 0 = (3"> 1) (2> ").
B. |-intercept: i (0) = ln 2; {-intercepts: i ({) = 0 C {2 3 3{ + 2 = h0
{=
I
3± 5
2
C {2 3 3{ + 1 = 0 C
i { E 0=38, 2=62 C. No symmetry D. lim i ({) = lim i ({) = 3", so { = 1 and { = 2 are VAs.
{<1
No HA E. i 0 ({) =
{<2+
2({ 3 3@2)
2{ 3 3
=
, so i 0 ({) ? 0 for { ? 1 and i 0 ({) A 0 for { A 2. Thus, i is
{2 3 3{ + 2
({ 3 1)({ 3 2)
decreasing on (3"> 1) and increasing on (2> "). F. No extreme values
G. i 00 ({) =
=
({2 3 3{ + 2) · 2 3 (2{ 3 3)2
({2 3 3{ + 2)2
H.
2{2 3 6{ + 4 3 4{2 + 12{ 3 9
32{2 + 6{ 3 5
= 2
2
2
({ 3 3{ + 2)
({ 3 3{ + 2)2
The numerator is negative for all { and the denominator is positive, so
i 00 ({) ? 0 for all { in the domain of i . Thus, i is CD on (3"> 1) and (2> ") =
No IP
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
51. | = i ({) = {h31@{
A. G = (3"> 0) (0> ") B. No intercept C. No symmetry
h31@{ H
h31@{ (1@{2 )
= lim
= 3 lim h31@{ = 3", so { = 0 is a VA. Also, lim {h31@{ = 0, so the graph
1@{
31@{2
{<0
{<0
{<0+
1
{+1
31@{
31@{ 1
(1)
=
h
approaches the origin as { < 0+ . E. i 0 ({) = {h31@{
+
h
+
1
= 1@{ A 0 C
{2
{
{h
D. lim
{<0
{ ? 31 or { A 0, so i is increasing on (3"> 31) and (0> "), and i is decreasing on (31> 0).
F. Local maximum value i (31) = 3h, no local minimum value
1
G. i 0 ({) = h31@{
+1
i
{
1
1
1
00
31@{
31@{
3 2 +
+1 h
i ({) = h
{
{
{2
1
1
1
= 2 h31@{ 31 +
+1
= 3 1@{ A 0 C
{
{
{ h
H.
{ A 0, so i is CU on (0> ") and CD on (3"> 0). No IP
52. | = i ({) =
ln {
{2
A. G = (0> ") B. |-intercept: none; {-intercept: i ({) = 0 C ln { = 0 C { = 1
C. No symmetry D. lim i({) = 3", so { = 0 is a VA; lim
{<"
{<0+
1@{
ln { H
= lim
= 0, so | = 0 is a HA.
{<" 2{
{2
{2 (1@{) 3 (ln {)(2{)
{(1 3 2 ln {)
1 3 2 ln { 0
=
=
. i ({) A 0 C 1 3 2 ln { A 0 C ln { ?
({2 )2
{4
{3
I I
0 ? { ? h1@2 and i 0 ({) ? 0 i { A h1@2 , so i is increasing on 0> h and decreasing on
h> " .
E. i 0 ({) =
F. Local maximum value i (h1@2 ) =
G. i 00 ({) =
=
1
1@2
=
h
2h
1
2
i
H.
{3 (32@{) 3 (1 3 2 ln {)(3{2 )
({3 )2
{2 [32 3 3(1 3 2 ln {)]
35 + 6 ln {
=
{6
{4
i 00 ({) A 0 C 35 + 6 ln { A 0 C ln { A
5
6
i { A h5@6 [ i is CU]
and i 00 ({) ? 0 C 0 ? { ? h5@6 [ i is CD]. IP at (h5@6 > 5@(6h5@3 ))
53. | = i ({) = h3{ + h32{
0
3{
E. i ({) = 3h
h5{ A
2
3
32{
3 2h
A. G = R B. |-intercept = i (0) = 2; no {-intercept C. No symmetry D. No asymptote
, so i 0 ({) A 0 C 3h3{ A 2h32{ [multiply by h2{ ] C
C 5{ A ln 23
C {A
{ ? 15 ln 23 . i is decreasing on 3">
F. Local minimum value i
1
5
H.
ln 23 E 30=081. Similarly, i 0 ({) ? 0 C
1
2
and increasing on 15 ln 23 > " .
5 ln 3
1
5
3@5 2 32@5
ln 23 = 23
+ 3
E 1=96; no local maximum.
G. i 00 ({) = 9h3{ + 4h32{ , so i 00 ({) A 0 for all {, and i is CU on (3"> "). No IP
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NOT FOR SALE
SECTION 4.5 SUMMARY OF CURVE SKETCHING
54. | = i ({) = tan31
{31
{+1
A. G = {{ | { 6= 31} B. {-intercept = 1, |-intercept = i(0) = tan31 (31) = 3 4
E. i 0 ({) =
lim tan31
{31
{+1
1 3 1@{
{<±"
{<±"
1 + 1@{
{31
{31
= 3 and lim tan31
= .
Also lim tan31
{+1
2
{+1
2
{<31+
{<31
C. No symmetry D.
¤
= lim tan31
= tan31 1 =
,
4
so | =
4
is a HA.
({ + 1) 3 ({ 3 1)
1
2
1
=
= 2
A 0,
1 + [({ 3 1)@({ + 1)]2
({ + 1)2
({ + 1)2 + ({ 3 1)2
{ +1
so i is increasing on (3"> 31) and (31> ") = F. No extreme values
2
G. i 00 ({) = 32{@ {2 + 1 A 0 C { ? 0, so i is CU on (3"> 31)
and (31> 0), and CD on (0> "). IP at 0> 3 4
H.
p0
. The m-intercept is i (0) = p0 . There are no y-intercepts. lim i (y) = ", so y = f is a VA.
y<f
1 3 y 2 @f2
p0 y
p0 y
p0 fy
i 0 (y) = 3 12 p0 (1 3 y 2 @f2 )33@2 (32y@f2 ) = 2
= 2 2
= 2
A 0, so i is
f (1 3 y 2 @f2 )3@2
(f 3 y 2 )3@2
f (f 3 y 2 )3@2
f3
increasing on (0> f). There are no local extreme values.
55. p = i(y) = s
i 00 (y) =
=
(f2 3 y 2 )3@2 (p0 f) 3 p0 fy · 32 (f2 3 y 2 )1@2 (32y)
[(f2 3 y 2 )3@2 ]2
p0 f(f2 3 y 2 )1@2 [(f2 3 y 2 ) + 3y 2 ]
p0 f(f2 + 2y 2 )
=
A 0,
2
2
3
(f 3 y )
(f2 3 y 2 )5@2
so i is CU on (0> f). There are no inÀection points.
56. Let d = p20 f4 and e = k2 f2 , so the equation can be written as H = i () =
s
d + e@2 =
u
d2 + e
=
2
s
d2 + e
= ", so = 0 is a VA.
<0+
s
s
s
I
I
d2 + e@
d + e@2
d2 + e
lim
= lim
= lim
= d, so H = d = p0 f2 is a HA.
<"
<"
<"
@
1
s
d2 + e
.
lim
i 0 () =
· 12 (d2 + e)31@2 (2d) 3 (d2 + e)1@2 (1)
(d2 + e)31@2 [d2 3 (d2 + e)]
3e
s
? 0,
=
=
2
2
2
d2 + e
so i is decreasing on (0> "). Using the Reciprocal Rule,
i 00 () = e ·
=
2 · 12 (d2 + e)31@2 (2d) + (d2 + e)1@2 (2)
2
s
2 d2 + e
e(d2 + e)31@2 [d2 + 2(d2 + e)]
e(3d2 + 2e)
= 3
A 0,
2
s
(d2 + e)3@2
2 d2 + e
so i is CU on (0> "). There are no extrema or inÀection points. The graph
shows that as decreases, the energy increases and as increases, the energy
decreases. For large wavelengths, the energy is very close to the energy at rest.
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368
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CHAPTER 4
57. (a) s(w) =
1
2
APPLICATIONS OF DIFFERENTIATION
1
1
=
2
1 + dh3nw
i
ln h3nw = ln d31
C 1 + dh3nw = 2 C dh3nw = 1
C 3nw = 3 ln d C w =
(b) The rate of spread is given by s0 (w) =
C
h3nw =
1
d
C
ln d
, which is when half the population will have heard the rumor.
n
dnh3nw
. To ¿nd the greatest rate of spread, we’ll apply the First Derivative
(1 + dh3nw )2
Test to s0 (w) [not s(w)].
[s0 (w)]0 = s00 (w) =
=
(1 + dh3nw )2 (3dn2 h3nw ) 3 dnh3nw · 2(1 + dh3nw )(3dnh3nw )
[(1 + dh3nw )2 ]2
(1 + dh3nw )(3dnh3nw )[n(1 + dh3nw ) 3 2dnh3nw ]
3dnh3nw (n)(1 3 dh3nw )
dn2 h3nw (dh3nw 3 1)
=
=
3nw
4
3nw
3
(1 + dh )
(1 + dh )
(1 + dh3nw )3
ln d
ln d
, so s0 (w) is increasing for w ?
and s0 (w) is
n
n
ln d
ln d
decreasing for w A
. Thus, s0 (w), the rate of spread of the rumor, is greatest at the same time,
, as when half the
n
n
population [by part (a)] has heard it.
s00 (w) A 0 C dh3nw A 1 C 3nw A ln d31
(c) s(0) =
C w?
1
and lim s(w) = 1. The graph is shown
w<"
1+d
with d = 4 and n = 12 .
58. F(w) = N(h3dw 3 h3ew ), where N A 0 and e A d A 0.
F(0) = N(1 3 1) = 0 is the only intercept.
F = 0 is a HA. F 0 (w) = N(3dh3dw + eh3ew ) A 0 C eh3ew A dh3dw
(d 3 e)w A ln
d
e
C wA
C hdw h3ew A
d
e
lim F(w) = 0, so
w<"
C h(d3e)w A
d
e
C
ln(d@e)
ln(e@d)
or
[call this value G]. F is increasing for w ? G and decreasing for w A G, so
d3e
e3d
F(G) is a local maximum [and absolute] value. F 00 (w) = N(d2 h3dw 3 e2 h3ew ) A 0 C d2 h3dw A e2 h3ew C
2
2
e2
e
e
2 ln(e@d)
hew h3dw A 2 C h(e3d)w A
C (e 3 d)w A ln
C wA
= 2G, so F is CU on (2G> ") and
d
d
d
e3d
CD on (0> 2G). The inÀection point is (2G> F(2G)). For the graph shown,
N = 1, d = 1, e = 2, G = ln 2, F(G) = 14 , and F(2G) =
3
16 .
The graph tells
us that when the drug is injected into the bloodstream, its concentration rises
rapidly to a maximum at time G, then falls, reaching its maximum rate of
decrease at time 2G, then continues to decrease more and more slowly,
approaching 0 as w < ".
59. | = 3
=
Z 4
Z O 3 Z O2 2
Z 2 2
{ +
{ 3
{ =3
{ { 3 2O{ + O2
24HL
12HL
24HL
24HL
3Z 2
{ ({ 3 O)2 = f{2 ({ 3 O)2
24HL
where f = 3
Z
is a negative constant and 0 $ { $ O. We sketch
24HL
i ({) = f{2 ({ 3 O)2 for f = 31. i (0) = i (O) = 0.
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NOT FOR SALE
SECTION 4.5 SUMMARY OF CURVE SKETCHING
¤
i 0 ({) = f{2 [2({ 3 O)] + ({ 3 O)2 (2f{) = 2f{({ 3 O)[{ + ({ 3 O)] = 2f{({ 3 O)(2{ 3 O). So for 0 ? { ? O,
i 0 ({) A 0 C {({ 3 O)(2{ 3 O) ? 0 [since f ? 0] C O@2 ? { ? O and i 0 ({) ? 0 C 0 ? { ? O@2.
Thus, i is increasing on (O@2> O) and decreasing on (0> O@2), and there is a local and absolute
minimum at the point (O@2> i (O@2)) = O@2> fO4 @16 . i 0 ({) = 2f[{({ 3 O)(2{ 3 O)] i
i 00 ({) = 2f[1({ 3 O)(2{ 3 O) + {(1)(2{ 3 O) + {({ 3 O)(2)] = 2f(6{2 3 6O{ + O2 ) = 0 C
I
I
6O ± 12O2
{=
= 12 O ± 63 O, and these are the {-coordinates of the two inÀection points.
12
60. I ({) = 3
I 0 ({) =
n
n
+
, where n A 0 and 0 ? { ? 2. For 0 ? { ? 2, { 3 2 ? 0, so
{2
({ 3 2)2
2n
2n
3
A 0 and I is increasing. lim I ({) = 3" and lim
{3
{<0+
{<2
({ 3 2)3
I ({) = ", so { = 0 and { = 2 are vertical asymptotes. Notice that when the middle
particle is at { = 1, the net force acting on it is 0. When { A 1, the net force is positive,
meaning that it acts to the right. And if the particle approaches { = 2, the force on it
rapidly becomes very large. When { ? 1, the net force is negative, so it acts to the left.
If the particle approaches 0, the force becomes very large to the left.
61. | =
{2 + 1
. Long division gives us:
{+1
{31
2
{+1 {
+1
{2 + {
3{+1
3{31
2
Thus, | = i ({) =
{2 + 1
2
2
={31+
and i ({) 3 ({ 3 1) =
=
{+1
{+1
{+1
2
{
1+
1
{
[for { 6= 0] < 0 as { < ±".
So the line | = { 3 1 is a slant asymptote (SA).
62. | =
2{3 + {2 + { + 3
. Long division gives us:
{2 + 2{
2
3
2
2{ 3 3
{ + 2{ 2{ + { + { + 3
2{3 + 4{2
3 3{2 + {
3 3{2 3 6{
7{ + 3
7
3
+ 2
2{3 + {2 + { + 3
7{ + 3
7{ + 3
{
{
Thus, | = i ({) =
= 2{ 3 3 + 2
and i ({) 3 (2{ 3 3) = 2
=
2
{2 + 2{
{ + 2{
{ + 2{
1+
{
[for { 6= 0] < 0 as { < ±". So the line | = 2{ 3 3 is a SA.
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370
¤
63. | =
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
4{3 3 2{2 + 5
. Long division gives us:
2{2 + { 3 3
2
3
2{ 3 2
2
2{ + { 3 3 4{ 3 2{
+5
4{3 + 2{2 3 6{
3 4{2 + 6{ + 5
3 4{2 3 2{ + 6
8{ 3 1
4{3 3 2{2 + 5
8{ 3 1
8{ 3 1
Thus, | = i ({) =
= 2{ 3 2 + 2
and i ({) 3 (2{ 3 2) = 2
=
2{2 + { 3 3
2{ + { 3 3
2{ + { 3 3
8
1
3 2
{
{
3
1
2+ 3 2
{
{
[for { 6= 0] < 0 as { < ±". So the line | = 2{ 3 2 is a SA.
64. | =
5{4 + {2 + {
. Long division gives us:
{3 3 {2 + 2
5{ + 5
3
2
4
2
{ 3 { + 2 5{
+ { +
5{4 3 5{3
{
+ 10{
5{3 + {2 3 9{
5{3 3 5{2
+ 10
Thus, | = i ({) =
6{2 3 9{ 3 10
5{4 + {2 + {
=
5{
+
5
+
and
{3 3 {2 + 2
{3 3 {2 + 2
6
9
10
3 2 3 3
6{2 3 9{ 3 10
{
{
{
=
i ({) 3 (5{ + 5) =
2
1
{3 3 {2 + 2
13 + 3
{
{
65. | = i ({) =
6{2 3 9{ 3 10
1
{2
={+1+
{31
{31
[for { 6= 0] < 0 as { < ±". So the line | = 5{ + 5 is a SA.
A. G = (3"> 1) (1> ") B. {-intercept: i ({) = 0 C { = 0;
|-intercept: i(0) = 0 C. No symmetry D. lim i ({) = 3" and lim i({) = ", so { = 1 is a VA.
{<1
lim [i ({) 3 ({ + 1)] = lim
{<±"
{<±"
E. i 0 ({) = 1 3
{<1+
1
= 0, so the line | = { + 1 is a SA.
{31
1
({ 3 1)2 3 1
{2 3 2{
{({ 3 2)
=
=
=
A 0 for
2
2
({ 3 1)
({ 3 1)
({ 3 1)2
({ 3 1)2
H.
{ ? 0 or { A 2, so i is increasing on (3"> 0) and (2> "), and i is decreasing
on (0> 1) and (1> 2). F. Local maximum value i (0) = 0, local minimum value
i (2) = 4 G. i 00 ({) =
2
A 0 for { A 1, so i is CU on (1> ") and i
({ 3 1)3
is CD on (3"> 1). No IP
3
1 + 5{ 3 2{2
= 32{ + 1 +
A. G = (3"> 2) (2> ") B. {-intercepts: i ({) = 0 C
{32
{32
I
35 ± 33
1 + 5{ 3 2{2 = 0 i { =
i { r 30=19, 2=69; |-intercept: i(0) = 3 12 C. No symmetry
34
66. | = i ({) =
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SECTION 4.5 SUMMARY OF CURVE SKETCHING
D. lim i ({) = 3" and lim i ({) = ", so { = 2 is a VA.
{<2
{<2+
lim [i({) 3 (32{ + 1)] = lim
{<±"
{<±"
¤
371
3
= 0, so
{32
| = 32{ + 1 is a SA.
E. i 0 ({) = 32 3
=
3
32({2 3 4{ + 4) 3 3
=
2
({ 3 2)
({ 3 2)2
H.
32{2 + 8{ 3 11
?0
({ 3 2)2
for { 6= 2, so i is decreasing on (3"> 2) and (2> "). F. No local extrema
G. i 00 ({) =
6
A 0 for { A 2, so i is CU on (2> ") and CD on (3"> 2).
({ 3 2)3
No IP
67. | = i ({) =
{3 + 4
4
={+ 2
{2
{
I
A. G = (3"> 0) (0> ") B. {-intercept: i ({) = 0 C { = 3 3 4; no |-intercept
C. No symmetry D. lim i ({) = ", so { = 0 is a VA.
{<0
E. i 0 ({) = 1 3
lim [i ({) 3 {] = lim
{<±"
{<±"
8
{3 3 8
=
A 0 for { ? 0 or { A 2, so i is increasing on
3
{
{3
4
= 0, so | = { is a SA.
{2
H.
(3"> 0) and (2> "), and i is decreasing on (0> 2). F. Local minimum value
i (2) = 3, no local maximum value G. i 00 ({) =
24
A 0 for { 6= 0, so i is CU
{4
on (3"> 0) and (0> "). No IP
68. | = i ({) =
{3
3{ + 2
={32+
({ + 1)2
({ + 1)2
C. No symmetry D.
lim i ({) = 3" and lim i ({) = 3", so { = 31 is a VA.
{<31
lim [i ({) 3 ({ 3 2)] = lim
{<±"
{<±"
E. i 0 ({) =
A. G = (3"> 31) (31> ") B. {-intercept: 0; |-intercept: i(0) = 0
{<31+
3{ + 2
= 0, so | = { 3 2 is a SA.
({ + 1)2
({ + 1)2 (3{2 ) 3 {3 · 2({ + 1)
{2 ({ + 1)[3({ + 1) 3 2{]
{2 ({ + 3)
=
=
A 0 C { ? 33 or
2
2
4
[({ + 1) ]
({ + 1)
({ + 1)3
{ A 31 [{ 6= 0], so i is increasing on (3"> 33) and (31> "), and i is decreasing on (33> 31).
F. Local maximum value i (33) = 3 27
, no local minimum
4
G. i 00 ({) =
({ + 1)3 (3{2 + 6{) 3 ({3 + 3{2 ) · 3({ + 1)2
[({ + 1)3 ]2
=
3{({ + 1)2 [({ + 1)({ + 2) 3 ({2 + 3{)]
({ + 1)6
=
3{({2 + 3{ + 2 3 {2 3 3{)
6{
=
A0
({ + 1)4
({ + 1)4
H.
C
{ A 0, so i is CU on (0> ") and i is CD on (3"> 31) and (31> 0). IP at (0> 0)
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372
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
69. | = i ({) = 1 + 12 { + h3{
D. No VA or HA.
1
2
3{
Ah
A. G = R B. |-intercept = i (0) = 2, no {-intercept [see part F] C. No symmetry
lim i ({) 3 1 + 12 { = lim h3{ = 0, so | = 1 + 12 { is a SA. E. i 0 ({) = 12 3 h3{ A 0 C
{<"
C 3{ ? ln
{<"
31
1
2
C { A 3 ln 2
C { A ln 2, so i is increasing on (ln 2> ") and decreasing
on (3"> ln 2). F. Local and absolute minimum value
i (ln 2) = 1 +
=1+
1
2
1
2
3 ln 2
ln 2 + h
ln 2 +
1
2
=
3
2
=1+
+
1
2
1
2
H.
ln 2 31
ln 2 + (h
)
ln 2 r 1=85,
no local maximum value G. i 00 ({) = h3{ A 0 for all {, so i is CU
on (3"> "). No IP
70. | = i ({) = 1 3 { + h1+{@3
A. G = R B. |-intercept = i (0) = 1 + h, no {-intercept [see part F]
C. No symmetry D. No VA or HA
lim [i ({) 3 (1 3 {)] = lim h1+{@3 = 0, so | = 1 3 { is a SA.
{<3"
E. i 0 ({) = 31 + 13 h1+{@3 A 0 C
1 1+{@3
3h
{<3"
A 1 C h1+{@3 A 3 C 1 +
{ A 3(ln 3 3 1) r 0=3, so i isincreasing on (3 ln 3 3 3> ") and decreasing
{
A ln 3 C
3
{
A ln 3 3 1 C
3
H.
on (3"> 3 ln 3 3 3). F. Local and absolute minimum value
i (3 ln 3 3 3) = 1 3 (3 ln 3 3 3) + h1+ln 331 = 4 3 3 ln 3 + 3 = 7 3 3 ln 3 r 3=7,
no local maximum value G. i 00 ({) = 19 h1+{@3 A 0 for all {, so i is CU
on (3"> "). No IP
71. | = i ({) = { 3 tan31 {, i 0 ({) = 1 3
1
1 + {2 3 1
{2
=
=
,
2
2
1+{
1+{
1 + {2
(1 + {2 )(2{) 3 {2 (2{)
2{(1 + {2 3 {2 )
2{
=
=
.
2
2
(1 + { )
(1 + {2 )2
(1 + {2 )2
lim i ({) 3 { 3 2 = lim 2 3 tan31 { = 2 3 2 = 0, so | = { 3
i 00 ({) =
{<"
{<"
Also, lim
{<3"
so | = { +
2
is a SA.
i({) 3 { + 2 = lim 3 2 3 tan31 { = 3 2 3 3 2 = 0,
{<3"
2
is also a SA. i 0 ({) D 0 for all {, with equality C { = 0, so i is
increasing on R. i 00 ({) has the same sign as {, so i is CD on (3"> 0) and CU on
(0> "). i (3{) = 3i ({), so i is an odd function; its graph is symmetric about the
origin. i has no local extreme values. Its only IP is at (0> 0).
72. | = i ({) =
s
I
{2 + 4{ = {({ + 4). {({ + 4) D 0 C { $ 34 or { D 0, so G = (3"> 34] [0> ").
y-intercept: i (0) = 0; x-intercepts: i ({) = 0 i { = 34, 0.
I
I
I
{2 + 4{ ~ ({ + 2)
{2 + 4{ ± ({ + 2)
({2 + 4{) 3 ({2 + 4{ + 4)
2
I
=
{ + 4{ ~ ({ + 2) =
·I
1
{2 + 4{ ± ({ + 2)
{2 + 4{ ± ({ + 2)
34
= I
{2 + 4{ ± ({ + 2)
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SECTION 4.5 SUMMARY OF CURVE SKETCHING
¤
373
so lim [i({) ~ ({ + 2)] = 0. Thus, the graph of i approaches the slant asymptote | = { + 2 as { < " and it approaches
{<±"
{+2
, so i 0 ({) ? 0 for { ? 34 and i 0 ({) A 0 for { A 0;
the slant asymptote | = 3({ + 2) as { < 3". i 0 ({) = I
{2 + 4{
that is, i is decreasing on (3"> 34) and increasing on (0> "). There are no local
extreme values. i 0 ({) = ({ + 2)({2 + 4{)31@2 i
i 00 ({) = ({ + 2) · 3 12 ({2 + 4{)33@2 · (2{ + 4) + ({2 + 4{)31@2
= ({2 + 4{)33@2 3({ + 2)2 + ({2 + 4{) = 34({2 + 4{)33@2 ? 0 on G
so i is CD on (3"> 34) and (0> "). No IP
73.
{2
|2
eI 2
3
=1 i | =±
{ 3 d2 . Now
d2
e2
d
I
I
{2 3 d2 + {
3d2
eI 2
e
e
e
lim
{ 3 d2 3 { = · lim
{2 3 d2 3 { I
= · lim I
= 0,
{<" d
d
d {<"
d {<" {2 3 d2 + {
{2 3 d2 + {
e
{ is a slant asymptote. Similarly,
d
e
e
3d2
eI 2
e
= 0, so | = 3 { is a slant asymptote.
lim 3
{ 3 d2 3 3 {
= 3 · lim I
{<"
d
d
d {<" {2 3 d2 + {
d
which shows that | =
74. i ({) 3 {2 =
1
1
{3 + 1
{3 + 1 3 {3
3 {2 =
= , and lim
= 0= Therefore, lim [i ({) 3 {2 ] = 0,
{<±" {
{<±"
{
{
{
and so the graph of i is asymptotic to that of | = {2 . For purposes of differentiation, we will use i ({) = {2 + 1@{=
A. G = {{ | { 6= 0} B. No |-intercept; to ¿nd the {-intercept, we set | = 0 C { = 31=
{3 + 1
{3 + 1
= " and lim
= 3",
{
{
{<0+
{<0
so { = 0 is a vertical asymptote. Also, the graph is asymptotic to the parabola
C. No symmetry D. lim
1
| = {2 , as shown above. E. i 0 ({) = 2{ 3 1@{2 A 0 C { A I
3 , so i
2
1
1
.
is increasing on I
> " and decreasing on (3"> 0) and 0> I
3
3
2
2
I
1
333
F. Local minimum value i I
, no local maximum
=
3
2
2
H.
y
21
0
G. i 00 ({) = 2 + 2@{3 A 0 C { ? 31 or { A 0, so i is CU on
(3"> 31) and (0> "), and CD on (31> 0). IP at (31> 0)
75.
{4 + 1
1
{4
i({) 3 {3 = lim
3
= lim
= 0, so the graph of i is asymptotic to that of | = {3 =
{<±"
{<±"
{<±" {
{
{
1
= 3" and
A. G = {{ | { 6= 0} B. No intercept C. i is symmetric about the origin. D. lim {3 +
{
{<0
lim
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374
¤
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
1
lim {3 +
= ", so { = 0 is a vertical asymptote, and as shown above, the graph of i is asymptotic to that of | = {3 .
{
{<0+
1
1
1
I
I
and
E. i 0 ({) = 3{2 3 1@{2 A 0 C {4 A 13 C |{| A I
,
so
i
is
increasing
on
3">
3
>
"
and
4
4
4
3
3
3
1
1
I
H.
>
0
and
0>
. F. Local maximum value
decreasing on 3 I
4
4
3
3
1
1
35@4
I
,
local
minimum
value
i
i 3I
=
34
·
3
= 4 · 335@4
4
4
3
3
G. i 00 ({) = 6{ + 2@{3 A 0 C { A 0, so i is CU on (0> ") and CD
on (3"> 0). No IP
76.
lim [i ({) 3 cos {] = lim 1@{2 = 0, so the graph of i is asymptotic to
{<±"
{<±"
that of cos {. The intercepts can only be found approximately.
1
i ({) = i (3{), so i is even. lim cos { + 2 = ", so { = 0 is a
{<0
{
vertical asymptote. We don’t need to calculate the derivatives, since we know
the asymptotic behavior of the curve.
4.6 Graphing with Calculus and Calculators
1. i ({) = 4{4 3 32{3 + 89{2 3 95{ + 29
i i 0 ({) = 16{3 3 96{2 + 178{ 3 95 i i 00 ({) = 48{2 3 192{ + 178.
i ({) = 0 C { E 0=5, 1=60; i 0 ({) = 0 C { E 0=92, 2=5, 2=58 and i 00 ({) = 0
C
{ E 1=46, 2=54.
From the graphs of i 0 , we estimate that i 0 ? 0 and that i is decreasing on (3"> 0=92) and (2=5> 2=58), and that i 0 A 0 and i
is increasing on (0=92> 2=5) and (2=58> ") with local minimum values i (0=92) E 35=12 and i (2=58) E 3=998 and local
maximum value i (2=5) = 4. The graphs of i 0 make it clear that i has a maximum and a minimum near { = 2=5, shown more
clearly in the fourth graph.
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SECTION 4.6
GRAPHING WITH CALCULUS AND CALCULATORS
¤
375
From the graph of i 00 , we estimate that i 00 A 0 and that i is CU on
(3"> 1=46) and (2=54> "), and that i 00 ? 0 and i is CD on (1=46> 2=54).
There are inÀection points at about (1=46> 31=40) and (2=54> 3=999).
i i 0 ({) = 6{5 3 75{4 + 300{3 3 375{2 3 1 i
2. i ({) = {6 3 15{5 + 75{4 3 125{3 3 {
i 00 ({) = 30{4 3 300{3 + 900{2 3 750{.
i ({) = 0 C { = 0 or { E 5=33; i 0 ({) = 0 C { E 2=50, 4=95, or 5=05;
i 00 ({) = 0 C { = 0, 5 or { E 1=38, 3=62.
From the graphs of i 0 , we estimate that i is decreasing on (3"> 2=50), increasing on (2=50> 4=95), decreasing on
(4=95> 5=05), and increasing on (5=05> "), with local minimum values i (2=50) E 3246=6 and i(5=05) E 35=03, and local
maximum value i (4=95) E 34=965 (notice the second graph of i ). From the graph of i 00 , we estimate that i is CU on
(3"> 0), CD on (0> 1=38), CU on (1=38> 3=62), CD on (3=62> 5), and CU on (5> "). There are inÀection points at (0> 0) and
(5> 35), and at about (1=38> 3126=38) and (3=62> 3128=62).
3. i ({) = {6 3 10{5 3 400{4 + 2500{3
00
4
3
i i 0 ({) = 6{5 3 50{4 3 1600{3 + 7500{2
i
2
i ({) = 30{ 3 200{ 3 4800{ + 1500{
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
From the graph of i 0 , we estimate that i is decreasing on (3"> 315), increasing on (315> 4=40), decreasing
on (4=40> 18=93), and increasing on (18=93> "), with local minimum values of i (315) E 39,700,000 and
i (18.93) E 312,700,000 and local maximum value i (4=40) E 53,800. From the graph of i 00 , we estimate that i is CU on
(3"> 311=34), CD on (311=34> 0), CU on (0> 2=92), CD on (2=92> 15=08), and CU on (15=08> "). There is an inÀection
point at (0> 0) and at about (311=34> 36,250,000), (2=92> 31,800), and (15=08> 38,150,000).
4. i ({) =
{2 3 1
40{3 + { + 1
i i 0 ({) =
340{4 + 121{2 + 2{ + 1
(40{3 + { + 1)2
i i 00 ({) =
80{(40{5 3 243{3 3 7{2 3 3{ + 3)
(40{3 + { + 1)3
From the ¿rst graph of i , we see that there is a VA at { E 30=26. From the graph of i 0 , we estimate that i is decreasing on
(3"> 31=73)> increasing on (31=73> 30=26), increasing on (30=26> 1=75), and decreasing on (1=75> "), with local
minimum value i (31=73) E 30=01 and local maximum value i (1=75) E 0=01.
From the graphs of i 00 , we estimate that i is CD on (3"> 32=45), CU on (32=45> 30=26), CD on (30=26> 0), CU on
(0> 0=21), CD on (0=21> 2=48)> and CU on (2=48> "). There is an inÀection point at (0> 31) and at about (32=45> 30=01),
(0=21> 30=62), and (2=48> 0=00).
5. i ({) =
{3
{
+ {2 + 1
i i 0 ({) = 3
2{3 + {2 3 1
({3 + {2 + 1)2
i i 00 ({) =
2{(3{4 + 3{3 + {2 3 6{ 3 3)
({3 + {2 + 1)3
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SECTION 4.6
GRAPHING WITH CALCULUS AND CALCULATORS
¤
377
From the graph of i , we see that there is a VA at { r 31=47. From the graph of i 0 , we estimate that i is increasing on
(3"> 31=47), increasing on (31=47> 0=66), and decreasing on (0=66> "), with local maximum value i(0=66) r 0=38.
From the graph of i 00 , we estimate that i is CU on (3"> 31=47), CD on (31=47> 30=49), CU on (30=49> 0), CD on
(0> 1=10), and CU on (1=10> "). There is an inÀection point at (0> 0) and at about (30=49> 30=44) and (1=10> 0=31).
6. i ({) = 6 sin { 3 {2 , 35 $ { $ 3
i i 0 ({) = 6 cos { 3 2{ i i 00 ({) = 36 sin { 3 2
From the graph of i 0 , which has two negative zeros, we estimate that i is increasing on (35> 32=94), decreasing on
(32=94> 32=66), increasing on (32=66> 1=17), and decreasing on (1=17> 3), with local maximum values i (32=94) r 39=84
and i (1=17) r 4=16, and local minimum value i (32=66) r 39=85.
From the graph of i 00 , we estimate that i is CD on (35> 32=80), CU on (32=80> 30=34), and CD on (30=34> 3). There are
inÀection points at about (32=80> 39=85) and (30=34> 32=12).
7. i ({) = 6 sin { + cot {, 3 $ { $ i i 0 ({) = 6 cos { 3 csc2 { i i 00 ({) = 36 sin { + 2 csc2 { cot {
From the graph of i , we see that there are VAs at { = 0 and { = ±. i is an odd function, so its graph is symmetric about
the origin. From the graph of i 0 , we estimate that i is decreasing on (3> 31=40), increasing on (31=40> 30=44), decreasing
on (30=44> 0), decreasing on (0> 0=44), increasing on (0=44> 1=40), and decreasing on (1=40> ), with local minimum values
i (31=40) r 36=09 and i(0=44) r 4=68, and local maximum values i (30=44) r 34=68 and i (1=40) r 6=09.
From the graph of i 00 , we estimate that i is CU on (3> 30=77), CD on (30=77> 0), CU on (0> 0=77), and CD on
(0=77> ). There are IPs at about (30=77> 35=22) and (0=77> 5=22).
8. i ({) = h{ 3 0=186{4
i i 0 ({) = h{ 3 0=744{3
i i 00 ({) = h{ 3 2=232{2
From the graph of i 0 , which has two positive zeros, we estimate that i is increasing on (3"> 2=973), decreasing on
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(2=973> 3=027), and increasing on (3=027> "), with local maximum value i (2=973) r 5=01958 and local minimum value
i (3=027) r 5=01949.
From the graph of i 00 , we estimate that i is CD on (3"> 30=52), CU on (30=52> 1=25), CD on (1=25> 3=00), and CU
on (3=00> "). There are inÀection points at about (30=52> 0=58), (1=25> 3=04) and (3=00> 5=01954).
9. i ({) = 1 +
i 00 ({) =
8
1
1
+ 2 + 3
{
{
{
i i 0 ({) = 3
1
16
3
1
3 3 3 4 = 3 4 ({2 + 16{ + 3) i
{2
{
{
{
2
48
12
2
+ 4 + 5 = 5 ({2 + 24{ + 6).
{3
{
{
{
From the graphs, it appears that i increases on (315=8> 30=2) and decreases on (3"> 315=8), (30=2> 0), and (0> "); that i
has a local minimum value of i(315=8) E 0=97 and a local maximum value of i(30=2) E 72; that i is CD on (3"> 324)
and (30=25> 0) and is CU on (324> 30=25) and (0> "); and that i has IPs at (324> 0=97) and (30=25> 60).
I
I
316 ± 256 3 12
0
To ¿nd the exact values, note that i = 0 i { =
= 38 ± 61 [E 30=19 and 315=81].
2
I
I I i 0 is positive (i is increasing) on 38 3 61> 38 + 61 and i 0 is negative (i is decreasing) on 3"> 38 3 61 ,
I
I
I
324 ± 576 3 24
38 + 61> 0 , and (0> "). i 00 = 0 i { =
= 312 ± 138 [E 30=25 and 323=75]. i 00 is
2
I
I
I
positive (i is CU) on 312 3 138> 312 + 138 and (0> ") and i 00 is negative (i is CD) on 3"> 312 3 138
I
and 312 + 138> 0 .
10. i ({) =
1
f
3 4
{8
{
i 0 ({) = 3
i 00 ({) =
[f = 2 × 108 ] i
8
4f
4
+ 5 = 3 9 (2 3 f{4 ) i
{9
{
{
72
20f
4
3 6 = 10 (18 3 5f{4 ).
{10
{
{
From the graph, it appears that i increases on (30=01> 0) and (0=01> ") and decreases on (3"> 30=01) and (0> 0=01);
that i has a local minimum value of i (±0=01) = 31016 ; and that i is CU on (30=012> 0) and (0> 0=012) and i is CD
on (3"> 30=012) and (0=012> ").
To ¿nd the exact values, note that i 0 = 0 i {4 =
2
f
i {±
t
4
2
f
1
= ± 100
[f = 2 × 108 ]. i 0 is positive
(i is increasing) on (30=01> 0) and (0=01> ") and i 0 is negative (i is decreasing) on (3"> 30=01) and (0> 0=01).
u
1 I
18
1 I
18
4
4
00
4
i {=±4
=±
1=8 [E ±0=0116]. i 00 is positive (i is CU) on 3 100
1=8> 0
i =0 i { =
5f
5f
100
I 1 I
1 I
4
4
1 4
1=8 and i 00 is negative (i is CD) on 3"> 3 100
1=8 and 100
1=8> " .
and 0> 100
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379
11. (a) i ({) = {2 ln {. The domain of i is (0> ").
(b) lim {2 ln { = lim
{<0+
{<0+
2
ln { H
1@{
{
3
= 0.
=
lim
=
lim
3
1@{2
2
{<0+ 32@{
{<0+
There is a hole at (0> 0).
(c) It appears that there is an IP at about (0=2> 30=06) and a local minimum at (0=6> 30=18). i ({) = {2 ln { i
i 0 ({) = {2 (1@{) + (ln {)(2{) = {(2 ln { + 1) A 0 C ln { A 3 12 C { A h31@2 , so i is increasing on
I
I I 1@ h> " , decreasing on 0> 1@ h . By the FDT, i 1@ h = 31@(2h) is a local minimum value. This point is
approximately (0=6065> 30=1839), which agrees with our estimate.
i 00 ({) = {(2@{) + (2 ln { + 1) = 2 ln { + 3 A 0 C ln { A 3 32
C { A h33@2 , so i is CU on (h33@2 > ")
and CD on (0> h33@2 ). IP is (h33@2 > 33@(2h3 )) E (0=2231> 30=0747).
12. (a) i ({) = {h1@{ . The domain of i is (3"> 0) (0> ").
(b) lim {h1@{ = lim
{<0+
{<0+
h1@{ 31@{2
h1@{ H
= lim
= lim h1@{ = ",
1@{
31@{2
{<0+
{<0+
so { = 0 is a VA.
Also lim {h1@{ = 0 since 1@{ < 3" i h1@{ < 0.
{<0
(c) It appears that there is a local minimum at (1> 2=7). There are no IP and i is CD on (3"> 0) and CU on (0> ").
1
1
1
A0 C
? 1 C { ? 0 or { A 1,
i ({) = {h1@{ i i 0 ({) = {h1@{ 3 2 + h1@{ = h1@{ 1 3
{
{
{
so i is increasing on (3"> 0) and (1> "), and decreasing on (0> 1). By the FDT, i (1) = h is a local minimum value,
which agrees with our estimate.
i 00 ({) = h1@{ (1@{2 ) + (1 3 1@{)h1@{ (31@{2 ) = (h1@{ @{2 )(1 3 1 + 1@{) = h1@{ @{3 A 0
C
{ A 0, so i is
CU on (0, ") and CD on (3"> 0). No IP
13.
i ({) =
({ + 4)({ 3 3)2
has VA at { = 0 and at { = 1 since lim i ({) = 3",
{<0
{4 ({ 3 1)
lim i ({) = 3" and lim i ({) = ".
{<1
{<1+
{ + 4 ({ 3 3)2 ·
(1 + 4@{)(1 3 3@{)2
dividing numerator
{2
i ({) = { 4
=
<0
3
and
denominator
by
{
{({ 3 1)
{
·
({
3
1)
{3
as { < ±", so i is asymptotic to the {-axis.
Since i is unde¿ned at { = 0, it has no |-intercept. i({) = 0 i ({ + 4)({ 3 3)2 = 0 i { = 34 or { = 3, so i has
{-intercepts 34 and 3. Note, however, that the graph of i is only tangent to the {-axis and does not cross it at { = 3, since i is
positive as { < 33 and as { < 3+ .
[continued]
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From these graphs, it appears that i has three maximum values and one minimum value. The maximum values are
approximately i (35=6) = 0=0182, i (0=82) = 3281=5 and i(5=2) = 0=0145 and we know (since the graph is tangent to the
{-axis at { = 3) that the minimum value is i(3) = 0.
14.
i ({) =
(2{ + 3)2 ({ 3 2)5
has VAs at { = 0 and { = 5 since lim i ({) = ",
{3 ({ 3 5)2
{<0
lim i ({) = 3", and lim i({) = ". No HA since lim i ({) = ".
{<0+
{<5
{<±"
Since i is unde¿ned at { = 0, it has no |-intercept.
i ({) = 0 C (2{ + 3)2 ({ 3 2)5 = 0 C { = 3 32 or { = 2, so i
has {-intercepts at 3 32 and 2. Note, however, that the graph of i is only tangent to
3
the {-axis and does not cross it at { = 3 32 , since i is positive as { < 3 32
and
3
3 +
as { < 3 2 . There is a local minimum value of i 3 2 = 0.
The only “mystery” feature is the local minimum to the right of the VA
{ = 5. From the graph, we see that i (7=98) E 609 is a local minimum
value.
15. i ({) =
{2 ({ + 1)3
({ 3 2)2 ({ 3 4)4
i i 0 ({) = 3
{({ + 1)2 ({3 + 18{2 3 44{ 3 16)
({ 3 2)3 ({ 3 4)5
[from CAS].
From the graphs of i 0 , it seems that the critical points which indicate extrema occur at { E 320, 30=3, and 2=5, as estimated
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381
in Example 3. (There is another critical point at { = 31, but the sign of i 0 does not change there.) We differentiate again,
obtaining i 00 ({) = 2
({ + 1)({6 + 36{5 + 6{4 3 628{3 + 684{2 + 672{ + 64)
.
({ 3 2)4 ({ 3 4)6
From the graphs of i 00 , it appears that i is CU on (335=3> 35=0), (31> 30=5), (30=1> 2), (2> 4) and (4> ") and CD
on (3"> 335=3), (35=0> 31) and (30=5> 30=1). We check back on the graphs of i to ¿nd the |-coordinates of the
inÀection points, and ¿nd that these points are approximately (335=3> 30=015), (35=0> 30=005), (31> 0), (30=5> 0=00001),
and (30=1> 0=0000066).
16. From a CAS,
and
i 0 ({) =
2({ 3 2)4 (2{ + 3)(2{3 3 14{2 3 10{ 3 45)
{4 ({ 3 5)3
i 00 ({) =
2({ 3 2)3 (4{6 3 56{5 + 216{4 + 460{3 + 805{2 + 1710{ + 5400)
{5 ({ 3 5)4
From Exercise 14 and i 0 ({) above, we know that the zeros of i 0 are 31=5, 2,
and 7=98. From the graph of i 0 , we conclude that i is decreasing on (3"> 31=5),
increasing on (31=5> 0) and (0> 5), decreasing on (5> 7=98), and increasing
on (7=98> ").
From i 00 ({), we know that { = 2 is a zero, and the graph of i 00 shows us that
{ = 2 is the only zero of i 00 . Thus, i is CU on (3"> 0), CD on (0> 2), CU on
(2> 5), and CU on (5> ").
17. i ({) =
{3 + 5{2 + 1
3{({5 + 10{4 + 6{3 + 4{2 3 3{ 3 22)
0
({)
=
and
.
From
a
CAS,
i
{4 + {3 3 {2 + 2
({4 + {3 3 {2 + 2)2
i 00 ({) =
2({9 + 15{8 + 18{7 + 21{6 3 9{5 3 135{4 3 76{3 + 21{2 + 6{ + 22)
({4 + {3 3 {2 + 2)3
The ¿rst graph of i shows that | = 0 is a HA. As { < ", i ({) < 0 through positive values. As { < 3", it is not clear if
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APPLICATIONS OF DIFFERENTIATION
i ({) < 0 through positive or negative values. The second graph of i shows that i has an {-intercept near 35, and will have a
local minimum and inÀection point to the left of 35.
From the two graphs of i 0 , we see that i 0 has four zeros. We conclude that i is decreasing on (3"> 39=41), increasing on
(39=41> 31=29), decreasing on (31=29> 0), increasing on (0> 1=05), and decreasing on (1=05> "). We have local minimum
values i(39=41) r 30=056 and i (0) = 0=5, and local maximum values i(31=29) r 7=49 and i (1=05) r 2=35.
From the two graphs of i 00 , we see that i 00 has ¿ve zeros. We conclude that i is CD on (3"> 313=81), CU on
(313=81> 31=55), CD on (31=55> 31=03), CU on (31=03> 0=60), CD on (0=60> 1=48), and CU on (1=48> "). There are ¿ve
inÀection points: (313=81> 30=05), (31=55> 5=64), (31=03> 5=39), (0=60> 1=52), and (1=48> 1=93).
18. | = i ({) =
{2@3
10{4 + { 3 2
2(65{8 3 14{5 3 80{4 + 2{2 3 8{ 3 1)
. From a CAS, | 0 = 3 1@3 4
and | 00 =
4
2
1+{+{
3{ ({ + { + 1)
9{4@3 ({4 + { + 1)3
i 0 ({) does not exist at { = 0 and i 0 ({) = 0 C { E 30=72 and 0=61, so i is increasing on (3"> 30=72), decreasing on
(30=72> 0), increasing on (0> 0=61), and decreasing on (0=61> "). There is a local maximum value of i (30=72) E 1=46 and a
local minimum value of i (0=61) E 0=41. i 00 ({) does not exist at { = 0 and i 00 ({) = 0 C { E 30=97, 30=46, 30=12,
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383
and 1=11, so i is CU on (3"> 30=97), CD on (30=97> 30=46), CU on (30=46> 30=12), CD on (30=12> 0), CD on (0> 1=11),
and CU on (1=11> "). There are inÀection points at (30=97> 1=08), (30=46> 1=01), (30=12> 0=28), and (1=11> 0=29).
19. | = i ({) =
I
{ + 5 sin {, { $ 20.
5 cos { + 1
10 cos { + 25 sin2 { + 10{ sin { + 26
I
.
and | 00 = 3
4({ + 5 sin {)3@2
2 { + 5 sin {
s
We’ll start with a graph of j({) = { + 5 sin {. Note that i({) = j({) is only de¿ned if j({) D 0. j({) = 0 C { = 0
From a CAS, |0 =
or { E 34=91, 34=10, 4=10, and 4=91. Thus, the domain of i is [34=91> 34=10] [0> 4=10] [4=91> 20].
From the expression for | 0 , we see that | 0 = 0 C 5 cos { + 1 = 0 i {1 = cos31 3 15 E 1=77 and
{2 = 2 3 {1 E 34=51 (not in the domain of i ). The leftmost zero of i 0 is {1 3 2 E 34=51. Moving to the right, the
zeros of i 0 are {1 , {1 + 2, {2 + 2, {1 + 4, and {2 + 4. Thus, i is increasing on (34=91> 34=51), decreasing on
(34=51> 34=10), increasing on (0> 1=77), decreasing on (1=77> 4=10), increasing on (4=91> 8=06), decreasing on (8=06> 10=79),
increasing on (10=79> 14=34), decreasing on (14=34> 17=08), and increasing on (17=08> 20). The local maximum values are
i (34=51) E 0=62, i (1=77) E 2=58, i (8=06) E 3=60, and i(14=34) E 4=39. The local minimum values are i (10=79) E 2=43
and i (17=08) E 3=49.
i is CD on (34=91> 34=10), (0> 4=10), (4=91> 9=60), CU on (9=60> 12=25),
CD on (12=25> 15=81), CU on (15=81> 18=65), and CD on (18=65> 20). There are
inÀection points at (9=60> 2=95), (12=25> 3=27), (15=81> 3=91), and (18=65> 4=20).
20. | = i ({) = ({2 3 1)harctan { .
From a CAS, |0 = 3harctan {
3
2
2
00
arctan { 2{ + { + 6{ 3 1
3
2{
3
1
and
|
=
h
+
2
.
{2 + 1
({2 + 1)2
From the graphs of i and i 0 , we conclude that i is decreasing on (3"> 0=38) and increasing on (0=38> "). There is a local
minimum value of i (0=38) = 31=23. From the graph of i 00 , we conclude that i is CU on (3"> 31), CD on (31> 30=20),
and CU on (30=20> "). There are inÀection points at (31> 0) and (30=20> 30=79).
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CHAPTER 4
21. | = i ({) =
APPLICATIONS OF DIFFERENTIATION
1 3 h1@{
2h1@{
32h1@{ (1 3 h1@{ + 2{ + 2{h1@{ )
. From a CAS, | 0 = 2
and | 00 =
.
1@{
1@{
2
1+h
{ (1 + h )
{4 (1 + h1@{ )3
i is an odd function de¿ned on (3"> 0) (0> "). Its graph has no x- or y-intercepts. Since lim i ({) = 0, the x-axis
{<±"
0
is a HA. i ({) A 0 for { 6= 0, so i is increasing on (3"> 0) and (0> "). It has no local extreme values.
i 00 ({) = 0 for { E ±0=417, so i is CU on (3"> 30=417), CD on (30=417> 0), CU on (0> 0=417), and CD on (0=417> ").
i has IPs at (30=417> 0=834) and (0=417> 30=834).
22. | = i ({) =
| 00 = 3
1
htan {
. From a CAS, | 0 = 3 2
and
tan
{
1+h
cos { (1 + htan { )2
htan { [htan { (2 sin { cos { 3 1) + 2 sin { cos { + 1]
. i is a periodic function with period that has positive values
cos4 { (1 + htan { )3
, ± 5
, and so on).
throughout its domain, which consists of all real numbers except odd multiples of 2 (that is, ± 2 , ± 3
2
2
i has |-intercept 12 , but no {-intercepts. We graph i , i 0 , and i 00 on one period, 3 2 > 2 .
Since i 0 ({) ? 0 for all { in the domain of i , i is decreasing on the intervals between odd multiples of 2 . i 00 ({) = 0 for
{ = 0 + q and for { E ±1=124 + q, so i is CD on 3 2 > 31=124 , CU on (31=124> 0), CD on (0> 1=124), and CU on
1=124> 2 . Since i is periodic, this behavior repeats on every interval of length . i has IPs at (31=124 + q> 0=890),
q> 12 , and (1=124 + q> 0=110).
23. i ({) =
1 3 cos({4 )
D 0. i is an even function, so its graph is symmetric with respect to the |-axis. The ¿rst graph shows
{8
that i levels off at | =
1
2
for |{| ? 0=7. It also shows that i then drops to the {-axis. Your graphing utility may show some
severe oscillations near the origin, but there are none. See the discussion in Section 2.2 after Example 2, as well as “Lies My
Calculator and Computer Told Me” on the website.
The second graph indicates that as |{| increases, i has progressively smaller humps.
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SECTION 4.6
24. i ({) = h{ + ln |{ 3 4|.
GRAPHING WITH CALCULUS AND CALCULATORS
¤
385
The ¿rst graph shows the big picture of i but conceals hidden behavior.
The second graph shows that for large negative values of {, i looks like j({) = ln |{|. It also shows a minimum value and
a point of inÀection.
The third graph hints at the vertical asymptote that we know exists at { = 4 because lim (h{ + ln |{ 3 4|) = 3".
{<4
A graphing calculator is unable to show much of the dip of the curve toward the vertical asymptote because of limited
resolution. A computer can show more if we restrict ourselves to a narrow interval around { = 4. See the solution to
Exercise 42 in Section 2.2 for a hand-drawn graph of this function.
25. (a) i ({) = {1@{
(b) Recall that de = he ln d . lim {1@{ = lim h(1@{) ln { . As { < 0+ ,
{<0+
{<0+
ln {
< 3", so {1@{ = h(1@{) ln { < 0. This
{
indicates that there is a hole at (0> 0). As { < ", we have the indeterminate form "0 . lim {1@{ = lim h(1@{) ln { ,
{<"
but lim
{<"
{<"
ln { H
1@{
= 0, so lim {1@{ = h0 = 1. This indicates that | = 1 is a HA.
= lim
{<" 1
{<"
{
(c) Estimated maximum: (2=72> 1=45). No estimated minimum. We use logarithmic differentiation to ¿nd any critical
1 3 ln {
|0
1 1
1
1
= · + (ln {) 3 2
i |0 = {1@{
=0 i
numbers. | = {1@{ i ln | = lnx i
{
|
{ {
{
{2
ln { = 1 i { = h. For 0 ? { ? h, | 0 A 0 and for { A h, |0 ? 0, so i (h) = h1@h is a local maximum value. This
point is approximately (2=7183> 1=4447), which agrees with our estimate.
(d)
From the graph, we see that i 00 ({) = 0 at { E 0=58 and { E 4=37. Since i 00
changes sign at these values, they are {-coordinates of inÀection points.
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APPLICATIONS OF DIFFERENTIATION
26. (a) i ({) = (sin {)sin { is continuous where sin { A 0, that is, on intervals
of the form (2q> (2q + 1)), so we have graphed i on (0> ).
(b) | = (sin {)sin {
i ln | = sin { ln sin {, so
lim ln | = lim sin { ln sin { = lim
{<0+
{<0+
{<0+
= lim (3 sin {) = 0
{<0+
i
ln sin { H
cot {
= lim
csc {
{<0+ 3 csc { cot {
lim | = h0 = 1=
{<0+
(c) It appears that we have a local maximum at (1=57> 1) and local minima at (0=38> 0=69) and (2=76> 0=69).
| = (sin {)sin {
i ln | = sin { ln sin { i
cos { |0
= (sin {)
+ (ln sin {) cos { = cos { (1 + ln sin {) i
|
sin {
| 0 = (sin {)sin { (cos {)(1 + ln sin {). | 0 = 0 i cos { = 0 or ln sin { = 31 i {2 =
On (0> ), sin { = h31
2
or sin { = h31 .
i {1 = sin31 (h31 ) and {3 = 3 sin31 (h31 ). Approximating these points gives us
({1 > i ({1 )) E (0=3767> 0=6922), ({2 > i({2 )) E (1=5708> 1), and ({3 > i ({3 )) E (2=7649> 0=6922). The approximations
con¿rm our estimates.
(d)
From the graph, we see that i 00 ({) = 0 at { E 0=94 and { E 2=20.
Since i 00 changes sign at these values, they are x-coordinates of inÀection
points.
27.
From the graph of i ({) = sin({ + sin 3{) in the viewing rectangle [0> ] by [31=2> 1=2], it looks like i has two maxima
and two minima. If we calculate and graph i 0 ({) = [cos({ + sin 3{)] (1 + 3 cos 3{) on [0> 2], we see that the graph of i 0
appears to be almost tangent to the {-axis at about { = 0=7. The graph of
i 00 = 3 [sin({ + sin 3{)] (1 + 3 cos 3{)2 + cos({ + sin 3{)(39 sin 3{)
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GRAPHING WITH CALCULUS AND CALCULATORS
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387
is even more interesting near this {-value: it seems to just touch the {-axis.
If we zoom in on this place on the graph of i 00 , we see that i 00 actually does cross the axis twice near { = 0=65,
indicating a change in concavity for a very short interval. If we look at the graph of i 0 on the same interval, we see that it
changes sign three times near { = 0=65, indicating that what we had thought was a broad extremum at about { = 0=7 actually
consists of three extrema (two maxima and a minimum). These maximum values are roughly i (0=59) = 1 and i (0=68) = 1,
and the minimum value is roughly i (0=64) = 0=99996. There are also a maximum value of about i (1=96) = 1 and minimum
values of about i (1=46) = 0=49 and i (2=73) = 30=51. The points of inÀection on (0> ) are about (0=61> 0=99998),
(0=66> 0=99998), (1=17> 0=72), (1=75> 0=77), and (2=28> 0=34). On (> 2), they are about (4=01> 30=34), (4=54> 30=77),
(5=11> 30=72), (5=62> 30=99998), and (5=67> 30=99998). There are also IP at (0> 0) and (> 0). Note that the function is odd
and periodic with period 2, and it is also rotationally symmetric about all points of the form ((2q + 1)> 0), q an integer.
28. i ({) = {3 + f{ = {({2 + f)
f = 36
i i 0 ({) = 3{2 + f i i 00 ({) = 6{
f=0
f=6
I
{-intercepts: When f D 0, 0 is the only {-intercept. When f ? 0, the {-intercepts are 0 and ± 3f.
|-intercept = i (0) = 0. i is odd, so the graph is symmetric with respect to the origin. i 00 ({) ? 0 for { ? 0 and
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NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
i 00 ({) A 0 for { A 0, so i is CD on (3"> 0) and CU on (0> ") = The origin is the only inÀection point.
If f A 0, then i 0 ({) A 0 for all {, so i is increasing and has no local maximum or minimum.
If f = 0, then i 0 ({) D 0 with equality at { = 0, so again i is increasing and has no local maximum or minimum.
s
s
s
If f ? 0, then i 0 ({) = 3[{2 3 (3f@3)] = 3 { + 3f@3 { 3 3f@3 , so i 0 ({) A 0 on 3"> 3 3f@3
and
s
s
s
3f@3> " ; i 0 ({) ? 0 on 3 3f@3> 3f@3 . It follows that
s
s
i 3 3f@3 = 3 23 f 3f@3 is a local maximum value and
i
s
s
3f@3 = 23 f 3f@3 is a local minimum value. As f decreases
(toward more negative values), the local maximum and minimum move
further apart.
There is no absolute maximum or minimum value. The only transitional
value of f corresponding to a change in character of the graph is f = 0.
29. i ({) =
I
I
{(2{2 + f)
{4 + f{2 = |{| {2 + f i i 0 ({) = I
{4 + f{2
i i 00 ({) =
{4 (2{2 + 3f)
({4 + f{2 )3@2
I
{-intercepts: When f D 0, 0 is the only {-intercept. When f ? 0, the {-intercepts are ± 3f.
|-intercept = i (0) = 0 when f D 0. When f ? 0, there is no |-intercept.
i is an even function, so its graph is symmetric with respect to the |-axis.
i 0 ({) =
2{2 + f
2{2 + f
{(2{2 + f)
I
= 3I
for { ? 0 and I
for { A 0, so i has a corner or “point” at { = 0 that gets
|{| {2 + f
{2 + f
{2 + f
sharper as f increases. There is an absolute minimum at 0 for f D 0. There are no other maximums nor minimums.
s
s
{4 (2{2 + 3f)
, so i 00 ({) = 0 i { = ± 33f@2 [only for f ? 0]. i 00 changes sign at ± 33f@2, so i is
|{|3 ({2 + f)3@2
s
s
I
s
s
I 33f@2> " , and i is CD on 3 33f@2> 3 3f and
3f> 33f@2 . There are
CU on 3"> 3 33f@2 and
i 00 ({) =
s
s
IPs at ± 33f@2> 3f2 @4 . The more negative f becomes, the farther the IPs move from the origin. The only transitional
value is f = 0.
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SECTION 4.6
30. With f = 0 in | = i ({) = {
GRAPHING WITH CALCULUS AND CALCULATORS
¤
389
I
f2 3 {2 , the graph of i is just the point (0> 0). Since (3f)2 = f2 , we only
consider f A 0. Since i (3{) = 3i ({), the graph is symmetric about the origin. The domain of i is found by
solving f2 3 {2 D 0 C {2 $ f2
C |{| $ f, which gives us [3f> f].
f2 3 2{2
i 0 ({) = { · 12 (f2 3 {2 )31@2 (32{) + (f2 3 {2 )1@2 (1) = (f2 3 {2 )31@2 [3{2 + (f2 3 {2 )] = I
.
f2 3 {2
I
i 0 ({) A 0 C f2 3 2{2 A 0 C {2 ? f2 @2 C |{| ? f@ 2, so i is increasing on
I
I I I 3f@ 2> f@ 2 and decreasing on 3f> 3f@ 2 and f@ 2> f . There is a local minimum value of
I s
I I I I f2 3 f2 @2 = 3f@ 2 f@ 2 = 3f2 @2 and a local maximum value of i f@ 2 = f2 @2.
i 3f@ 2 = 3f@ 2
31@2
(f2 3 {2 )1@2 (34{) 3 (f2 3 2{2 ) 12 f2 3 {2
(32{)
2
2
1@2
2
[(f 3 { ) ]
2
2 31@2
2
2{ 2{2 3 3f2
[(f 3 {2 )(34) + (f2 3 2{2 )]
{(f 3 { )
=
=
,
(f2 3 {2 )1
(f2 3 {2 )3@2
t
so i 00 ({) = 0 C { = 0 or { = ± 32 f, but only 0 is in the domain of i .
i 00 ({) =
i 00 ({) ? 0 for 0 ? { ? f and i 00 ({) A 0 for 3f ? { ? 0, so i is CD on (0> f)
and CU on (3f> 0). There is an IP at (0> 0). So as |f| gets larger, the maximum and
minimum values increase in magnitude. The value of f does not affect the
concavity of i .
31. i ({) = h{ + fh3{ . i = 0
i fh3{ = 3h{
i 0 ({) = h{ 3 fh3{ . i 0 = 0 i fh3{ = h{
i f = 3h2{
i f = h2{
i 2{ = ln(3f) i { =
i 2{ = ln f i { =
1
2
1
2
ln(3f)=
ln f.
i 00 ({) = h{ + fh3{ = i ({).
The only transitional value of f is 0. As f increases from 3" to 0,
1
2
ln(3f) is both the the {-intercept and inÀection point,
and this decreases from " to 3". Also i 0 A 0, so i is increasing. When f = 0, i ({) = i 0 ({) = i 00 ({) = h{ , i is positive,
increasing, and concave upward. As f increases from 0 to ", the absolute minimum occurs at { =
1
2
ln f, which increases
from 3" to ". Also, i = i 00 A 0, so i is positive and concave upward. The
value of the |-intercept is i (0) = 1 + f> and this increases as f increases from
3" to ".
Note: The minimum point
|=2
1
2
ln f> 2
I f can be parameterized by { =
1
2
ln f,
I
f, and after eliminating the parameter f, we see that the minimum point
lies on the graph of | = 2h{ .
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
32. We see that if f $ 0, i ({) = ln({2 + f) is only de¿ned for {2 A 3f
lim
I
{< 3f+
i ({) =
lim
I
{<3 3f
i |{| A
I
3f, and
i({) = 3", since ln | < 3" as | < 0. Thus, for f ? 0, there are vertical asymptotes at
I
{ = ± f, and as f decreases (that is, |f| increases), the asymptotes get further apart. For f = 0, lim i ({) = 3", so there is
{<0
a vertical asymptote at { = 0. If f A 0, there are no asymptotes. To ¿nd the extrema and inÀection points, we differentiate:
i ({) = ln({2 + f) i i 0 ({) =
1
(2{), so by the First Derivative Test there is a local and absolute minimum at
{2 + f
{ = 0. Differentiating again, we get i 00 ({) =
{2
2(f 3 {2 )
1
(2) + 2{ 3({2 + f)32 (2{) = 2
.
+f
({ + f)2
Now if f $ 0, i 00 is always negative, so i is concave down on both of the intervals
on which it is de¿ned. If f A 0, then i 00 changes sign when f = {2 C
I
I
{ = ± f. So for f A 0 there are inÀection points at { = ± f, and as f increases,
the inÀection points get further apart.
33. Note that f = 0 is a transitional value at which the graph consists of the {-axis. Also, we can see that if we substitute 3f for f,
the function i ({) =
f{
will be reÀected in the {-axis, so we investigate only positive values of f (except f = 31, as a
1 + f2 {2
demonstration of this reÀective property). Also, i is an odd function.
for all f. We calculate i 0 ({) =
lim i ({) = 0, so | = 0 is a horizontal asymptote
{<±"
(1 + f2 {2 )f 3 f{(2f2 {)
f(f2 {2 3 1) 0
=3
. i ({) = 0 C f2 {2 3 1 = 0 C
2
2
2
(1 + f { )
(1 + f2 {2 )2
{ = ±1@f. So there is an absolute maximum value of i (1@f) =
1
2
and an absolute minimum value of i (31@f) = 3 12 .
These extrema have the same value regardless of f, but the maximum points move closer to the |-axis as f increases.
i 00 ({) =
=
(32f3 {)(1 + f2 {2 )2 3 (3f3 {2 + f)[2(1 + f2 {2 )(2f2 {)]
(1 + f2 {2 )4
(32f3 {)(1 + f2 {2 ) + (f3 {2 3 f)(4f2 {)
2f3 {(f2 {2 3 3)
=
2
2
3
(1 + f { )
(1 + f2 {2 )3
I
i 00 ({) = 0 C { = 0 or ± 3@f, so there are inÀection points at (0> 0) and
I
I
at ± 3@f> ± 3@4 . Again, the |-coordinate of the inÀection points does not depend on f, but as f increases, both inÀection
points approach the |-axis.
34. i ({) = {2 + fh3{
i i 0 ({) = 2{ 3 fh3{
i i 00 ({) = 2 + fh3{ . When f = 0, we get i({) = {2 .
When f A 0, i 00 ({) A 2, so i is CU on R. Also i 0 ({) = 0 i 2{ = fh3{ . We can see that the solution to this
equation is the {-coordinate of the point of intersection of the graphs of | = 2{ and | = fh3{ , which moves away from the
origin as f increases.
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SECTION 4.6
When f ? 0, i 00 ({) A 0 C fh3{ A 32 C h3{ ? 3
GRAPHING WITH CALCULUS AND CALCULATORS
¤
391
2
C 3{ ? ln 3
C
f
2
f
f
2
= ln 3
[call this value d]. So i has an IP at (d> i (d)), is CU on (d> "), and CD on (3"> d). Solving
{ A 3 ln 3
f
2
i 0 ({) = 0 with f ? 0, we can see that the graph of | = 2{ intersects the graph of | = fh3{ once, twice, or not at all.
The tangent line for | = fh3{ at { = e is | 3 fh3e = 3fh3e ({ 3 e), or
| = 3fh3e { + fh3e (e + 1). If this line coincides with the line | = 2{ + 0, then
fh3e (e + 1) = 0 i e = 31, and 3fh3(31) = 2 i f = 3
Thus, for 3
2
r 30=74.
h
2
? f ? 0, i has a local maximum and a local minimum, and for
h
2
f $ 3 , i has no local extrema. The transitional values of f are 0 and 32@h.
h
35. i ({) = f{ + sin {
i i 0 ({) = f + cos { i i 00 ({) = 3 sin {
i (3{) = 3i ({), so i is an odd function and its graph is symmetric with respect to the origin.
i ({) = 0 C sin { = 3f{, so 0 is always an {-intercept.
i 0 ({) = 0 C cos { = 3f, so there is no critical number when |f| A 1. If |f| $ 1, then there are in¿nitely
many critical numbers. If {1 is the unique solution of cos { = 3f in the interval [0> ], then the critical numbers are 2q ± {1 ,
where q ranges over the integers. (Special cases: When f = 31, {1 = 0; when f = 0, { =
2;
and when f = 1, {1 = .)
i 00 ({) ? 0 C sin { A 0, so i is CD on intervals of the form (2q> (2q + 1)). i is CU on intervals of the form
((2q 3 1)> 2q). The inÀection points of i are the points (q> qf), where q is an integer.
If f D 1, then i 0 ({) D 0 for all {, so i is increasing and has no extremum. If f $ 31, then i 0 ({) $ 0 for all {, so i is
decreasing and has no extremum. If |f| ? 1, then i 0 ({) A 0 C cos { A 3f C { is in an interval of the form
(2q 3 {1 > 2q + {1 ) for some integer q. These are the intervals on which i is increasing. Similarly, we
¿nd that i is decreasing on the intervals of the form (2q + {1 > 2(q + 1) 3 {1 ). Thus, i has local maxima at the points
I
2q + {1 , where i has the values f(2q + {1 ) + sin {1 = f(2q + {1 ) + 1 3 f2 , and i has local minima at the points
I
2q 3 {1 , where we have i (2q 3 {1 ) = f(2q 3 {1 ) 3 sin {1 = f(2q 3 {1 ) 3 1 3 f2 .
The transitional values of f are 31 and 1. The inÀection points move vertically, but not horizontally, when f changes.
When |f| D 1, there is no extremum. For |f| ? 1, the maxima are spaced
2 apart horizontally, as are the minima. The horizontal spacing between
maxima and adjacent minima is regular (and equals ) when f = 0, but
the horizontal space between a local maximum and the nearest local
minimum shrinks as |f| approaches 1.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
36. For i (w) = F(h3dw 3 h3ew ), F affects only vertical stretching, so we let F = 1. From the ¿rst ¿gure, we notice that the
graphs all pass through the origin, approach the w-axis as w increases, and approach 3" as w < 3". Next we let d = 2 and
produce the second ¿gure.
Here, as e increases, the slope of the tangent at the origin increases and the local maximum value increases.
i (w) = h32w 3 h3ew
i i 0 (w) = eh3ew 3 2h32w . i 0 (0) = e 3 2, which increases as e increases.
i 0 (w) = 0 i eh3ew = 2h32w
i
e
= h(e32)w
2
i ln
e
ln e 3 ln 2
= (e 3 2)w i w = w1 =
, which decreases as
2
e32
e increases (the maximum is getting closer to the |-axis). i (w1 ) =
(e 3 2)22@(e32)
. We can show that this value increases as e
e1+2@(e32)
increases by considering it to be a function of e and graphing its derivative with respect to e, which is always positive.
37. If f ? 0, then lim i ({) = lim {h3f{ = lim
{<3"
{<3"
{<3"
{ H
1
= lim
= 0, and lim i({) = ".
{<3" fhf{
{<"
hf{
H
If f A 0, then lim i ({) = 3", and lim i ({) = lim
{<3"
{<"
{<"
1
= 0.
fhf{
If f = 0, then i ({) = {, so lim i ({) = ±", respectively.
{<±"
So we see that f = 0 is a transitional value. We now exclude the case f = 0, since we know how the function behaves
in that case. To ¿nd the maxima and minima of i , we differentiate: i ({) = {h3f{
i
i 0 ({) = {(3fh3f{ ) + h3f{ = (1 3 f{)h3f{ . This is 0 when 1 3 f{ = 0 C { = 1@f. If f ? 0 then this
represents a minimum value of i (1@f) = 1@(fh), since i 0 ({) changes from negative to positive at { = 1@f;
and if f A 0, it represents a maximum value. As |f| increases, the maximum or
minimum point gets closer to the origin. To ¿nd the inÀection points, we
differentiate again: i 0 ({) = h3f{ (1 3 f{)
i
i 00 ({) = h3f{ (3f) + (1 3 f{)(3fh3f{ ) = (f{ 3 2)fh3f{ . This changes sign
when f{ 3 2 = 0
C
{ = 2@f. So as |f| increases, the points of inÀection get
closer to the origin.
38. For f = 0, there is no inÀection point; the curve is CU everywhere. If f increases, the curve simply becomes steeper, and there
are still no inÀection points. If f starts at 0 and decreases, a slight upward bulge appears near { = 0, so that there are two
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SECTION 4.6
GRAPHING WITH CALCULUS AND CALCULATORS
¤
393
inÀection points for any f ? 0. This can be seen algebraically by calculating the second derivative:
i ({) = {4 + f{2 + { i i 0 ({) = 4{3 + 2f{ + 1 i i 00 ({) = 12{2 + 2f. Thus, i 00 ({) A 0 when f A 0. For f ? 0,
t
there are inÀection points when { = ± 3 16 f. For f = 0, the graph has one critical number, at the absolute minimum
somewhere around { = 30=6. As f increases, the number of critical points does not change. If f instead decreases from 0, we
see that the graph eventually sprouts another local minimum, to the right of the origin, somewhere between { = 1 and { = 2.
Consequently, there is also a maximum near { = 0.
After a bit of experimentation, we ¿nd that at f = 31=5, there appear to be two critical numbers: the absolute minimum at
about { = 31, and a horizontal tangent with no extremum at about { = 0=5. For any f smaller than this there will be
3 critical points, as shown in the graphs with f = 33 and with f = 35.
To prove this algebraically, we calculate i 0 ({) = 4{3 + 2f{ + 1. Now if
we substitute our value of f = 31=5, the formula for i 0 ({) becomes
4{3 3 3{ + 1 = ({ + 1)(2{ 3 1)2 . This has a double root at { = 12 , indicating
that the function has two critical points: { = 31 and { = 12 , just as we had
guessed from the graph.
39. (a) i ({) = f{4 3 2{2 + 1. For f = 0, i({) = 32{2 + 1, a parabola whose vertex, (0> 1), is the absolute maximum. For
f A 0, i ({) = f{4 3 2{2 + 1 opens upward with two minimum points. As f < 0, the minimum points spread apart and
move downward; they are below the {-axis for 0 ? f ? 1 and above for f A 1. For f ? 0, the graph opens downward, and
has an absolute maximum at { = 0 and no local minimum.
(b) i 0 ({) = 4f{3 3 4{ = 4f{({2 3 1@f) [f 6= 0]. If f $ 0, 0 is the only critical number.
i 00 ({) = 12f{2 3 4, so i 00 (0) = 34 and there is a local maximum at
(0> i (0)) = (0> 1), which lies on | = 1 3 {2 . If f A 0, the critical
I
numbers are 0 and ±1@ f. As before, there is a local maximum at
(0> i (0)) = (0> 1), which lies on | = 1 3 {2 .
I i 00 ±1@ f = 12 3 4 = 8 A 0, so there is a local minimum at
I I
{ = ±1@ f. Here i ±1@ f = f(1@f2 ) 3 2@f + 1 = 31@f + 1.
I
I 2
But ±1@ f> 31@f + 1 lies on | = 1 3 {2 since 1 3 ±1@ f
= 1 3 1@f.
40. (a) i ({) = 2{3 + f{2 + 2{
i i 0 ({) = 6{2 + 2f{ + 2 = 2(3{2 + f{ + 1). i 0 ({) = 0 C { =
So i has critical points C f2 3 12 D 0 C |f| D 2
3f ±
I
f2 3 12
.
6
I
I
3. For f = ±2 3, i 0 ({) D 0 on (3"> "), so i 0 does not
change signs at 3f@6, and there is no extremum. If f2 3 12 A 0, then i 0 changes from positive to negative at
I
I
3f 3 f2 3 12
3f + f2 3 12
{=
and from negative to positive at { =
. So i has a local maximum at
6
6
I
I
3f + f2 3 12
3f 3 f2 3 12
and a local minimum at { =
.
{=
6
6
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(b) Let {0 be a critical number for i ({). Then i 0 ({0 ) = 0 i
31 3 3{20
. Now
{0
31 3 3{20
+ 2{0
i ({0 ) = 2{30 + f{20 + 2{0 = 2{30 + {20
{0
3{20 + f{0 + 1 = 0 C f =
= 2{30 3 {0 3 3{30 + 2{0 = {0 3 {30
So the point is ({0 > |0 ) = {0 > {0 3 {30 ; that is, the point lies
on the curve | = { 3 {3 .
4.7 Optimization Problems
We needn’t consider pairs where the ¿rst number is larger
1. (a)
First Number
Second Number
Product
1
22
22
2
21
42
3
20
60
4
19
76
5
18
90
6
17
102
7
16
112
8
15
120
9
14
126
10
13
130
11
12
132
than the second, since we can just interchange the numbers
in such cases. The answer appears to be 11 and 12, but we
have considered only integers in the table.
(b) Call the two numbers { and |. Then { + | = 23, so | = 23 3 {. Call the product S . Then
S = {| = {(23 3 {) = 23{ 3 {2 , so we wish to maximize the function S ({) = 23{ 3 {2 . Since S 0 ({) = 23 3 2{,
we see that S 0 ({) = 0 C { =
23
2
= 11=5. Thus, the maximum value of S is S (11=5) = (11=5)2 = 132=25 and it
occurs when { = | = 11=5.
Or: Note that S 00 ({) = 32 ? 0 for all {, so S is everywhere concave downward and the local maximum at { = 11=5
must be an absolute maximum.
2. The two numbers are { + 100 and {. Minimize i ({) = ({ + 100){ = {2 + 100{. i 0 ({) = 2{ + 100 = 0
i { = 350.
Since i 00 ({) = 2 A 0, there is an absolute minimum at { = 350. The two numbers are 50 and 350.
3. The two numbers are { and
100
100
{2 3 100
100
. The critical
, where { A 0. Minimize i ({) = { +
. i 0 ({) = 1 3 2 =
{
{
{
{2
number is { = 10. Since i 0 ({) ? 0 for 0 ? { ? 10 and i 0 ({) A 0 for { A 10, there is an absolute minimum at { = 10.
The numbers are 10 and 10.
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SECTION 4.7
OPTIMIZATION PROBLEMS
¤
395
4. Call the two numbers { and |. Then { + | = 16, so | = 16 3 {. Call the sum of their squares V. Then
V = {2 + | 2 = {2 + (16 3 {)2
i V 0 = 2{ + 2(16 3 {)(31) = 2{ 3 32 + 2{ = 4{ 3 32. V 0 = 0 i { = 8.
Since V 0 ({) ? 0 for 0 ? { ? 8 and V 0 ({) A 0 for { A 8, there is an absolute minimum at { = 8= Thus, | = 16 3 8 = 8
and V = 82 + 82 = 128.
5. Let the vertical distance be given by y({) = ({ + 2) 3 {2 , 31 $ { $ 2.
y 0 ({) = 1 3 2{ = 0 C { = 12 . y(31) = 0, y
1
2
= 94 , and y(2) = 0, so
there is an absolute maximum at { = 12 . The maximum distance is
y 12 = 12 + 2 3 14 = 94 .
6. Let the vertical distance be given by
y({) = ({2 + 1) 3 ({ 3 {2 ) = 2{2 3 { + 1. y 0 ({) = 4{ 3 1 = 0 C
{ = 14 . y 0 ({) ? 0 for { ?
and y 0 ({) A 0 for { A 14 , so there is an absolute
minimum at { = 14 . The minimum distance is y 14 = 18 3 14 + 1 = 78 .
1
4
7. If the rectangle has dimensions { and |, then its perimeter is 2{ + 2| = 100 m, so | = 50 3 {. Thus, the area is
D = {| = {(50 3 {). We wish to maximize the function D({) = {(50 3 {) = 50{ 3 {2 , where 0 ? { ? 50. Since
D0 ({) = 50 3 2{ = 32({ 3 25), D0 ({) A 0 for 0 ? { ? 25 and D0 ({) ? 0 for 25 ? { ? 50. Thus, D has an absolute
maximum at { = 25, and D(25) = 252 = 625 m2 . The dimensions of the rectangle that maximize its area are { = | = 25 m.
(The rectangle is a square.)
8. If the rectangle has dimensions { and |, then its area is {| = 1000 m2 , so | = 1000@{. The perimeter
S = 2{ + 2| = 2{ + 2000@{. We wish to minimize the function S ({) = 2{ + 2000@{ for { A 0.
I
S 0 ({) = 2 3 2000@{2 = (2@{2 )({2 3 1000), so the only critical number in the domain of S is { = 1000.
I
I
S 00 ({) = 4000@{3 A 0, so S is concave upward throughout its domain and S 1000 = 4 1000 is an absolute minimum
I
I
value. The dimensions of the rectangle with minimal perimeter are { = | = 1000 = 10 10 m. (The rectangle is a square.)
9. We need to maximize \ for Q D 0.
\ 0 (Q) =
\ (Q) =
nQ
1 + Q2
i
(1 + Q 2 )n 3 nQ(2Q)
n(1 3 Q 2 )
n(1 + Q)(1 3 Q)
=
=
. \ 0 (Q) A 0 for 0 ? Q ? 1 and \ 0 (Q) ? 0
(1 + Q 2 )2
(1 + Q 2 )2
(1 + Q 2 )2
for Q A 1. Thus, \ has an absolute maximum of \ (1) = 12 n at Q = 1.
10. We need to maximize S for L D 0.
S 0 (L) =
S (L) =
100L
L2 + L + 4
i
(L 2 + L + 4)(100) 3 100L(2L + 1)
100(L 2 + L + 4 3 2L 2 3 L)
3100(L 2 3 4)
3100(L + 2)(L 3 2)
=
= 2
=
.
2
2
2
2
(L + L + 4)
(L + L + 4)
(L + L + 4)2
(L 2 + L + 4)2
S 0 (L) A 0 for 0 ? L ? 2 and S 0 (L) ? 0 for L A 2. Thus, S has an absolute maximum of S (2) = 20 at L = 2.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
11. (a)
The areas of the three ¿gures are 12,500, 12,500, and 9000 ft2 . There appears to be a maximum area of at least 12,500 ft2 .
(b) Let { denote the length of each of two sides and three dividers.
Let | denote the length of the other two sides.
(c) Area D = length × width = | · {
(d) Length of fencing = 750 i 5{ + 2| = 750
(e) 5{ + 2| = 750 i | = 375 3 52 { i D({) = 375 3 52 { { = 375{ 3 52 {2
(f ) D0 ({) = 375 3 5{ = 0 i { = 75. Since D00 ({) = 35 ? 0 there is an absolute maximum when { = 75. Then
375 = 14,062=5 ft2 . These values of { and | are between the values in the ¿rst
| = 375
2 = 187=5. The largest area is 75
2
and second ¿gures in part (a). Our original estimate was low.
12. (a)
The volumes of the resulting boxes are 1, 1=6875, and 2 ft3 . There appears to be a maximum volume of at least 2 ft3 .
(b) Let { denote the length of the side of the square being cut out. Let |
denote the length of the base.
(c) Volume Y = length × width × height i Y = | · | · { = {|2
(d) Length of cardboard = 3 i { + | + { = 3 i | + 2{ = 3
(e) | + 2{ = 3 i | = 3 3 2{ i Y ({) = {(3 3 2{)2
(f ) Y ({) = {(3 3 2{)2
i
Y 0 ({) = { · 2(3 3 2{)(32) + (3 3 2{)2 · 1 = (3 3 2{)[34{ + (3 3 2{)] = (3 3 2{)(36{ + 3),
so the critical numbers are { = 32 and { = 12 . Now 0 $ { $ 32 and Y (0) = Y 32 = 0, so the maximum is
Y 12 = 12 (2)2 = 2 ft3 , which is the value found from our third ¿gure in part (a).
13.
{| = 1=5 × 106 , so | = 1=5 × 106@{. Minimize the amount of fencing, which is
3{ + 2| = 3{ + 2(1=5 × 106@{) = 3{ + 3 × 106@{ = I ({).
I 0 ({) = 3 3 3 × 106@{2 = 3({2 3 106 )@{2 . The critical number is { = 103 and
I 0 ({) ? 0 for 0 ? { ? 103 and I 0 ({) A 0 if { A 103 , so the absolute minimum
occurs when { = 103 and | = 1=5 × 103 .
The ¿eld should be 1000 feet by 1500 feet with the middle fence parallel to the short side of the ¿eld.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.7
14. Let e be the length of the base of the box and k the height. The volume is 32,000 = e2 k
¤
OPTIMIZATION PROBLEMS
i
397
k = 32,000@e2 .
The surface area of the open box is V = e2 + 4ke = e2 + 4(32,000@e2 )e = e2 + 4(32,000)@e.
I
So V 0 (e) = 2e 3 4(32,000)@e2 = 2 e3 3 64,000 @e2 = 0 C e = 3 64,000 = 40. This gives an absolute minimum
since V 0 (e) ? 0 if 0 ? e ? 40 and V 0 (e) A 0 if e A 40. The box should be 40 × 40 × 20.
15. Let e be the length of the base of the box and k the height. The surface area is 1200 = e2 + 4ke
i k = (1200 3 e2 )@(4e).
The volume is Y = e2 k = e2 (1200 3 e2 )@4e = 300e 3 e3@4 i Y 0 (e) = 300 3 34 e2 .
I
Y 0 (e) = 0 i 300 = 34 e2 i e2 = 400 i e = 400 = 20. Since Y 0 (e) A 0 for 0 ? e ? 20 and Y 0 (e) ? 0 for
e A 20, there is an absolute maximum when e = 20 by the First Derivative Test for Absolute Extreme Values (see page 328).
If e = 20, then k = (1200 3 202 )@(4 · 20) = 10, so the largest possible volume is e2 k = (20)2 (10) = 4000 cm3 .
Y = ozk i 10 = (2z)(z)k = 2z2 k, so k = 5@z2 .
16.
The cost is 10(2z2 ) + 6[2(2zk) + 2(kz)] = 20z2 + 36zk, so
F(z) = 20z2 + 36z 5@z2 = 20z2 + 180@z.
F 0 (z) = 40z 3 180@z2 = 40 z3 3 92 z2
when z =
F
17.
t 3
9
2
t
3
9
2
since F 0 (z) ? 0 for 0 ? z ?
t 2
180
E $163=54.
= 20 3 92 + s
3
9@2
i z=
t
3
9
2
t
3
9
2
is the critical number. There is an absolute minimum for F
and F 0 (z) A 0 for z A
t
3
9
.
2
10 = (2z)(z)k = 2z2 k, so k = 5@z2 . The cost is
F(z) = 10(2z2 ) + 6[2(2zk) + 2kz] + 6(2z2 )
= 32z2 + 36zk = 32z2 + 180@z
F 0 (z) = 64z 3 180@z2 = 4(16z3 3 45)@z2
F 0 (z) A 0 for z A
t
3
45
.
16
i z=
The minimum cost is F
t
3
45
16
t
3
45
16
is the critical number. F 0 (z) ? 0 for 0 ? z ?
I
= 32(2=8125)2@3 + 180 3 2=8125 E $191=28.
t
3
45
16
and
18. (a) Let the rectangle have sides { and | and area D, so D = {| or | = D@{. The problem is to minimize the
perimeter = 2{ + 2| = 2{ + 2D@{ = S ({). Now S 0 ({) = 2 3 2D@{2 = 2 {2 3 D @{2 . So the critical number is
I
I
I
I
{ = D. Since S 0 ({) ? 0 for 0 ? { ? D and S 0 ({) A 0 for { A D, there is an absolute minimum at { = D.
I
I
I
The sides of the rectangle are D and D@ D = D, so the rectangle is a square.
(b) Let s be the perimeter and { and | the lengths of the sides, so s = 2{ + 2| i 2| = s 3 2{ i | = 12 s 3 {.
The area is D({) = { 12 s 3 { = 12 s{ 3 {2 . Now D0 ({) = 0 i 12 s 3 2{ = 0 i 2{ = 12 s i { = 14 s. Since
D00 ({) = 32 ? 0, there is an absolute maximum for D when { = 14 s by the Second Derivative Test. The sides of the
rectangle are 14 s and 12 s 3 14 s = 14 s, so the rectangle is a square.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
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398
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
19. The distance g from the origin (0> 0) to a point ({> 2{ + 3) on the line is given by g =
s
({ 3 0)2 + (2{ + 3 3 0)2 and the
square of the distance is V = g2 = {2 + (2{ + 3)2 . V 0 = 2{ + 2(2{ + 3)2 = 10{ + 12 and V 0 = 0 C { = 3 65 . Now
V 00 = 10 A 0, so we know that V has a minimum at { = 3 65 . Thus, the |-value is 2 3 65 + 3 = 35 and the point is 3 65 > 35 .
20. The distance g from the point (3> 0) to a point ({>
I
{ ) on the curve is given by g =
t
I
2
({ 3 3)2 + ( { 3 0) and the square
of the distance is V = g2 = ({ 3 3)2 + {. V 0 = 2({ 3 3) + 1 = 2{ 3 5 and V 0 = 0 C { = 52 . Now V 00 = 2 A 0, so we
t
t know that V has a minimum at { = 52 . Thus, the |-value is 52 and the point is 52 > 52 .
From the ¿gure, we see that there are two points that are farthest away from
21.
D(1> 0). The distance g from D to an arbitrary point S ({> |) on the ellipse is
s
g = ({ 3 1)2 + (| 3 0)2 and the square of the distance is
V = g 2 = {2 3 2{ + 1 + | 2 = {2 3 2{ + 1 + (4 3 4{2 ) = 33{2 3 2{ + 5.
V 0 = 36{ 3 2 and V 0 = 0 i { = 3 13 . Now V 00 = 36 ? 0, so we know
that V has a maximum at { = 3 13 . Since 31 $ { $ 1, V(31) = 4,
t
16
,
and
V(1)
=
0,
we
see
that
the
maximum
distance
is
V 3 13 = 16
3
3 . The corresponding |-values are
t
t
I
I 2
| = ± 4 3 4 3 13 = ± 32
= ± 43 2 E ±1=89. The points are 3 13 > ± 43 2 .
9
22. The distance g from the point (4> 2) to a point ({> sin {) on the curve is given by g =
s
({ 3 4)2 + (sin { 3 2)2 and the
square of the distance is V = g2 = ({ 3 4)2 + (sin { 3 2)2 . V 0 = 2({ 3 4) + 2(sin { 3 2) cos {. Using a calculator, it is
clear that V has a minimum between 0 and 5, and from a graph of V 0 , we ¿nd that V 0 = 0 i { E 2=65, so the point is
about (2=65> 0=47).
The area of the rectangle is (2{)(2|) = 4{|. Also u2 = {2 + | 2 so
I
I
| = u2 3 {2 , so the area is D({) = 4{ u2 3 {2 . Now
I
{2
u2 3 2{2
u2 3 {2 3 I
. The critical number is
D0 ({) = 4
=4I
2
2
u 3{
u2 3 {2
23.
{=
|=
u
u2 3
and 2| =
1
I
u
2
I
2 u.
2
=
t
1 2
u
2
=
I1 u
2
I1 u.
2
Clearly this gives a maximum.
= {, which tells us that the rectangle is a square. The dimensions are 2{ =
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
I
2u
SECTION 4.7
24.
OPTIMIZATION PROBLEMS
The area of the rectangle is (2{)(2|) = 4{|. Now
¤
399
{2
|2
+ 2 = 1 gives
2
d
e
eI 2
e I
d 3 {2 , so we maximize D({) = 4 { d2 3 {2 .
d
d
l
4e k 1 2
{ · 2 (d 3 {2 )31@2 (32{) + (d2 3 {2 )1@2 · 1
D0 ({) =
d
4e 2
4e
=
(d 3 {231@2 [3{2 + d2 3 {2 ] = I
[d2 3 2{2 ]
d
d d2 3 {2
|=
So the critical number is { =
I1
2
d, and this clearly gives a maximum. Then | =
1
I
2
e, so the maximum area
is 4 I12 d I12 e = 2de.
25.
The height k of the equilateral triangle with sides of length O is
since k2 + (O@2)2 = O2
k=
I
3
O.
2
I
3{ =
i k2 = O2 3 14 O2 = 34 O2
I
Using similar triangles,
I
3
2 O
3|
3
O
2
3|
=
I
3
O
2
=
{
O@2
I
i | = 23 O 3 3 { i | =
I
I
3
2
O,
i
I
3 i
I
3
2 (O
3 2{).
I
I
I
The area of the inscribed rectangle is D({) = (2{)| = 3 {(O 3 2{) = 3 O{ 3 2 3 {2 , where 0 $ { $ O@2. Now
I
I
I I 0 = D0 ({) = 3 O 3 4 3 { i { = 3 O 4 3 = O@4. Since D(0) = D(O@2) = 0, the maximum occurs when
{ = O@4, and | =
I
3
O
2
26.
3
I
3
O
4
=
I
3
O,
4
so the dimensions are O@2 and
I
3
O.
4
The area D of a trapezoid is given by D = 12 k(E + e). From the diagram,
k = |, E = 2, and e = 2{, so D = 12 |(2 + 2{) = |(1 + {). Since it’s easier to
substitute for | 2 , we’ll let W = D2 = | 2 (1 + {)2 = (1 3 {2 )(1 + {)2 . Now
W 0 = (1 3 {2 )2(1 + {) + (1 + {)2 (32{) = 32(1 + {)[3(1 3 {2 ) + (1 + {){]
= 32(1 + {)(2{2 + { 3 1) = 32(1 + {)(2{ 3 1)({ + 1)
W 0 = 0 C { = 31 or { = 12 . W 0 A 0 if { ? 12 and W 0 ? 0 if { A 12 , so we get a maximum at { = 12 [{ = 31 gives us
t
I
I I
2
D = 0]. Thus, | = 1 3 12 = 23 and the maximum area is D = |(1 + {) = 23 1 + 12 = 3 4 3 .
27.
The area of the triangle is
D({) = 12 (2w)(u + {) = w(u + {) =
I
u2 3 {2 (u + {). Then
I
32{
32{
+ u2 3 {2 + { I
0 = D0 ({) = u I
2
2
2 u 3{
2 u2 3 {2
I
{2 + u{
= 3I
+ u2 3 {2 i
2
2
u 3{
I
{2 + u{
I
= u2 3 {2
2
2
u 3{
i {2 + u{ = u2 3 {2
i 0 = 2{2 + u{ 3 u2 = (2{ 3 u)({ + u) i
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°
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400
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
{ = 12 u or { = 3u. Now D(u) = 0 = D(3u) i the maximum occurs where { = 12 u, so the triangle has
t
t
I
2
height u + 12 u = 32 u and base 2 u2 3 12 u = 2 34 u2 = 3 u.
28.
The rectangle has area {|. By similar triangles
33|
3
=
{
4
i
34| + 12 = 3{ or | = 3 34 { + 3. So the area is
D({) = { 3 34 { + 3 = 3 34 {2 + 3{ where 0 $ { $ 4. Now
29.
0 = D0 ({) = 3 32 { + 3 i { = 2 and | = 32 . Since D(0) = D(4) = 0,
the maximum area is D(2) = 2 32 = 3 cm2 .
The cylinder has volume Y = |2 (2{). Also {2 + | 2 = u2
i | 2 = u2 3 {2 , so
Y ({) = (u2 3 {2 )(2{) = 2(u2 { 3 {3 ), where 0 $ { $ u.
I
Y 0 ({) = 2 u2 3 3{2 = 0 i { = u@ 3. Now Y (0) = Y (u) = 0, so there is a
I
I I I maximum when { = u@ 3 and Y u@ 3 = (u2 3 u2 @3) 2u@ 3 = 4u3 @ 3 3 .
30.
By similar triangles, |@{ = k@u, so | = k{@u. The volume of the cylinder is
{2 (k 3 |) = k{2 3 (k@u){3 = Y ({). Now
Y 0 ({) = 2k{ 3 (3k@u){2 = k{(2 3 3{@u).
So Y 0 ({) = 0 i { = 0 or { = 23 u. The maximum clearly occurs when
{ = 23 u and then the volume is
k{2 3 (k@u){3 = k{2 (1 3 {@u) = 31.
The cylinder has surface area
2 2 2
=
3u k 1 3 3
2
4
27 u k.
2(area of the base) + (lateral surface area) = 2(radius)2 + 2(radius)(height)
= 2|2 + 2|(2{)
I
| = u2 3 {2 , so the surface area is
I
V({) = 2(u2 3 {2 ) + 4{ u2 3 {2 > 0 $ { $ u
I
= 2u2 3 2{2 + 4 { u2 3 {2
k
l
Thus,
V 0 ({) = 0 3 4{ + 4 { · 12 (u2 3 {2 )31@2 (32{) + (u2 3 {2 )1@2 · 1
I
I
{2
3{ u2 3 {2 3 {2 + u2 3 {2
I
= 4 3{ 3 I
+ u2 3 {2 = 4 ·
u2 3 {2
u2 3 {2
I
2
I
V 0 ({) = 0 i { u2 3 {2 = u2 3 2{2 (B) i { u2 3 {2 = (u2 3 2{2 )2 i
Now {2 + | 2 = u2
{2 (u2 3 {2 ) = u4 3 4u2 {2 + 4{4
i | 2 = u2 3 {2
i u2 {2 3 {4 = u4 3 4u2 {2 + 4{4
This is a quadratic equation in {2 . By the quadratic formula, {2 =
I
5± 5 2
u ,
10
i
i 5{4 3 5u2 {2 + u4 = 0.
but we reject the root with the + sign since it
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.7
t
doesn’t satisfy (B). [The right side is negative and the left side is positive.] So { =
maximum surface area occurs at the critical number and {2 =
the surface area is
t I t I
I 2 5 +10 5 u2 + 4 5 310 5 5 +10 5 u2 = u2 2 ·
= u2
32.
k
I
5+ 5
10
I
53 5 2
u
10
+4
I
I
5+ 5 + 2·2 5
5
Perimeter = 30 i 2| + { + t
l
2
I
5)(5+ 5)
10
= u2
{
I
53 5
10
i | 2 = u2 3
I
(53
k
I
5+5 5
5
= u2
l
OPTIMIZATION PROBLEMS
401
u. Since V(0) = V(u) = 0, the
I
53 5 2
u
10
k
¤
I
5+ 5
5
+
=
I
5+ 5 2
u
10
i
I l
2 20
5
I = u2 1 + 5 .
= 30 i
{ {
{
1
30 3 { 3
= 15 3 3
. The area is the area of the rectangle plus the area of
2
2
2
4
{ 2
{ { 1 2
the semicircle, or {| + 12 , so D({) = { 15 3 3
+ 8 { = 15{ 3 12 {2 3 8 {2 .
2
2
4
|=
D0 ({) = 15 3 1 + 4 { = 0 i { =
The dimensions are { =
15
60
=
. D00 ({) = 3 1 +
? 0, so this gives a maximum.
1 + @4
4+
4
60
30
15
60 + 15 3 30 3 15
30
ft and | = 15 3
3
=
=
ft, so the height of the
4+
4+
4+
4+
4+
rectangle is half the base.
33.
{| = 384 i | = 384@{. Total area is
D({) = (8 + {)(12 + 384@{) = 12(40 + { + 256@{), so
D0 ({) = 12(1 3 256@{2 ) = 0 i { = 16. There is an absolute minimum
when { = 16 since D0 ({) ? 0 for 0 ? { ? 16 and D0 ({) A 0 for { A 16.
When { = 16, | = 384@16 = 24, so the dimensions are 24 cm and 36 cm.
34.
{| = 180, so | = 180@{. The printed area is
({ 3 2)(| 3 3) = ({ 3 2)(180@{ 3 3) = 186 3 3{ 3 360@{ = D({).
I
D0 ({) = 33 + 360@{2 = 0 when {2 = 120 i { = 2 30. This gives an absolute
I
I
maximum since D0 ({) A 0 for 0 ? { ? 2 30 and D0 ({) ? 0 for { A 2 30. When
I
I
I
I
{ = 2 30, | = 180@(2 30 ), so the dimensions are 2 30 in. and 90@ 30 in.
35.
Let { be the length of the wire used for the square. The total area is
{ 2 1 10 3 { I3 10 3 { +
D({) =
4
2
3
2
3
=
I
I
1 2
{
16
+
I
3
(10
36
I
3 {)2 , 0 $ { $ 10
I
9
3
I
{ + 4723 { 3 4072 3 = 0 C { = 9 40
.
D0 ({) = 18 { 3 183 (10 3 {) = 0 C 72
+4 3
I I 3
I
Now D(0) = 363 100 E 4=81, D(10) = 100
= 6=25 and D 9 40
E 2=72, so
16
+4 3
(a) The maximum area occurs when { = 10 m, and all the wire is used for the square.
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°
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402
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(b) The minimum area occurs when { =
I
40 I
3
9+4 3
E 4=35 m.
2
10 3 {
(10 3 {)2
{2
+
, 0 $ { $ 10.
=
4
2
16
4
10 3 {
1
1
5
{
D0 ({) = 3
=
+
{ 3 = 0 i { = 40@(4 + ).
8
2
2
8
36.
Total area is D({) =
{ 2
+
D(0) = 25@ E 7=96, D(10) = 6=25, and D(40@(4 + )) E 3=5, so the maximum
occurs when { = 0 m and the minimum occurs when { = 40@(4 + ) m.
37.
The volume is Y = u2 k and the surface area is
Y
2Y
= u2 +
V(u) = u2 + 2uk = u2 + 2u
.
u2
u
u
Y
2Y
cm.
V 0 (u) = 2u 3 2 = 0 i 2u3 = 2Y i u = 3
u
u
u
Y
3 Y
0
0
This gives an absolute minimum since V (u) ? 0 for 0 ? u ?
and V (u) A 0 for u A 3
.
u
u
Y
Y
Y
3 Y
=
=
When u = 3
,k=
cm.
u2
(Y@)2@3
gO
= 38 csc cot + 4 sec tan = 0 when
g
I
I
sec tan = 2 csc cot C tan3 = 2 C tan = 3 2 C = tan31 3 2.
I
I
gO@g ? 0 when 0 ? ? tan31 3 2, gO@g A 0 when tan31 3 2 ? ? 2 , so O has
I
an absolute minimum when = tan31 3 2, and the shortest ladder has length
I
I
1 + 2 2@3
O=8
+ 4 1 + 2 2@3 E 16=65 ft.
1@3
2
38.
O = 8 csc + 4 sec , 0 ? ?
,
2
Another method: Minimize O2 = {2 + (4 + |)2 , where
k2 + u2 = U2
39.
Y 0 (k) =
2
3 (U
i Y =
2
u k
3
=
(U2
3
3 3k2 ) = 0 when k =
3 k2 )k =
I1 U.
3
40. The volume and surface area of a cone with radius u and height k are given by Y =
D = 2 u2 (u2 + k2 ) = 2
81
k
81
+ k2
k
=
1
u2 k
3
I1 U.
3
1
u2 k
3
= 27 i u2 =
812
+ 81k, so D0 = 0 i
k2
(U2 k
3
3 k3 ).
This gives an absolute maximum, since
Y 0 (k) A 0 for 0 ? k ? I13 U and Y 0 (k) ? 0 for k A
3
1
2
Y I13 U = 3 I13 U3 3 3I
U
U3 .
= 9I
3
3
We’ll minimize D = V 2 subject to Y = 27. Y = 27 i
8
{
= .
4+|
|
The maximum volume is
and V = u
81
k
I
u2 + k2 .
(1).
32 · 812
+ 81 = 0 i
k3
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.7
u
OPTIMIZATION PROBLEMS
¤
u
162
6
27
81
2 · 812
162
81
s
= I
i k= 3
=3 3
E 3=722. From (1), u2 =
=
i k3 =
3
k3
k
· 3 3 6@
62
I
3 3
E 2=632. D00 = 6 · 812 @k4 A 0, so D and hence V has an absolute minimum at these values of u and k.
u= I
6
62
81 =
41.
By similar triangles,
K 3k
K
=
U
u
so we’ll solve (1) for k.
k=K3
Ku
=K 3k i
U
K
2 K
u · (U 3 u) =
(Uu2 3 u3 ) i
3
U
3U
K
K
(2Uu 3 3u2 ) =
u(2U 3 3u).
3U
3U
K 1 1
K
U 3 23 U =
i u = 23 U and from (2), k =
U = 3 K.
U
U 3
Y 0 (u) =
Y 0 (u) = 0 i u = 0 or 2U = 3u
Y 0 (u) changes from positive to negative at u = 23 U, so the inner cone has a maximum volume of
2 1 4
K = 27
· 13 U2 K, which is approximately 15% of the volume of the larger cone.
Y = 13 u2 k = 13 23 U
3
42. We need to minimize I for 0 $ ? @2.
I () =
Z
sin + cos i I 0 () =
3Z ( cos 3 sin )
[by the
( sin + cos )2
Reciprocal Rule]. I 0 () A 0 i cos 3 sin ? 0 i cos ? sin i ? tan i A tan31 .
So I is decreasing on 0> tan31 and increasing on tan31 > 2 . Thus, I attains its minimum value at = tan31 .
Z
This maximum value is I (tan31 ) = s
.
2 + 1
43. S (U) =
H2U
(U + u)2
i
S 0 (U) =
=
S 0 (U) = 0 i U = u
(U + u)2 · H 2 3 H 2 U · 2(U + u)
(U2 + 2Uu + u2 )H 2 3 2H 2 U2 3 2H 2 Uu
=
[(U + u)2 ]2
(U + u)4
H 2 u2 3 H 2 U2
H 2 (u2 3 U2 )
H 2 (u + U)(u 3 U)
H 2 (u 3 U)
=
=
=
4
4
4
(U + u)
(U + u)
(U + u)
(U + u)3
i S (u) =
H2 u
H2u
H2
=
=
.
2
2
(u + u)
4u
4u
The expression for S 0 (U) shows that S 0 (U) A 0 for U ? u and S 0 (U) ? 0 for U A u. Thus, the maximum value of the
power is H 2 @(4u), and this occurs when U = u.
44. (a) H(y) =
dOy 3
y3x
2y 3 = 3xy 2
i H 0 (y) = dO
i
(1). The volume of the inner cone is Y = 13 u2 k,
KU 3 Ku
K
Ku
=
= (U 3 u) (2).
U
U
U
Thus, Y(u) =
403
(y 3 x)3y 2 3 y 3
= 0 when
(y 3 x)2
(b)
i 2y = 3x i y = 32 x.
The First Derivative Test shows that this value of y gives the minimum
value of H.
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404
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
45. V = 6vk 3 32 v2 cot + 3v2
I
3
2
I
csc I
csc cot or 32 v2 csc csc 3 3 cot .
(a)
gV
= 32 v2 csc2 3 3v2
g
(b)
I
gV
= 0 when csc 3 3 cot = 0 i
g
3
2
I cos 1
3 3
= 0 i cos =
sin sin that the minimum surface area occurs when = cos31 I13 E 55 .
If cos =
(c)
I1 ,
3
then cot =
V = 6vk 3 32 v2 I12 +
I1
2
and csc =
I
I3 ,
2
The First Derivative Test shows
so the surface area is
I
I
3v2 23 I32
= 6vk 3
6
1
= 6vk + 2I
v2 = 6v k + 2I
v
2
2
46.
I1 .
3
3
I
v2
2 2
+
9
I
v2
2 2
Let w be the time, in hours, after 2:00 PM. The position of the boat heading south
at time w is (0> 320w). The position of the boat heading east at time w is
(315 + 15w> 0). If G(w) is the distance between the boats at time w, we
minimize i (w) = [G(w)]2 = 202 w2 + 152 (w 3 1)2 .
i 0 (w) = 800w + 450(w 3 1) = 1250w 3 450 = 0 when w =
0=36 h ×
60 min
h
450
1250
= 0=36 h.
= 21=6 min = 21 min 36 s. Since i 00 (w) A 0, this gives a minimum, so the boats are closest together
at 2:21:36 PM.
I
I
1
{
{2 + 25
53{
3 = 0 C 8{ = 6 {2 + 25 C
47. Here W ({) =
+
, 0 $ { $ 5 i W 0 ({) = I
6
8
8
6 {2 + 25
16{2 = 9({2 + 25) C { =
15
I
.
7
But
15
I
7
A 5, so W has no critical number. Since W (0) E 1=46 and W (5) E 1=18, he
should row directly to E.
48.
In isosceles triangle DRE, _R = 180 3 3 , so _ERF = 2. The distance rowed is
4 cos while the distance walked is the length of arc EF = 2(2) = 4. The time taken
is given by W () =
4
4 cos +
= 2 cos + , 0 $ $
2
4
W 0 () = 32 sin + 1 = 0 C sin =
1
2
i =
2.
.
6
Check the value of W at = 6 and at the endpoints of the domain of W ; that is, = 0 and = 2 .
I
W (0) = 2, W 6 = 3 + 6 E 2=26, and W 2 = 2 E 1=57. Therefore, the minimum value of W is
the woman should walk all the way. Note that W 00 () = 32 cos ? 0 for 0 $ ?
,
2
so =
6
2
when =
that is,
gives a maximum time.
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;
2
NOT FOR SALE
SECTION 4.7
49. There are (6 3 {) km over land and
OPTIMIZATION PROBLEMS
¤
405
I
{2 + 4 km under the river.
We need to minimize the cost F (measured in $100,000) of the pipeline.
I
{2 + 4 (8) i
F({) = (6 3 {)(4) +
8{
.
F 0 ({) = 34 + 8 · 12 ({2 + 4)31@2 (2{) = 34 + I
{2 + 4
I
8{
F 0 ({) = 0 i 4 = I
{2 + 4 = 2{ i {2 + 4 = 4{2 i 4 = 3{2 i {2 = 43 i
i
2
{ +4
I
I
{ = 2@ 3 [0 $ { $ 6]. Compare the costs for { = 0, 2@ 3, and 6. F(0) = 24 + 16 = 40,
I
I
I
I
I F 2@ 3 = 24 3 8@ 3 + 32@ 3 = 24 + 24@ 3 E 37=9, and F(6) = 0 + 8 40 E 50=6. So the minimum cost is about
I
$3=79 million when S is 6 3 2@ 3 E 4=85 km east of the re¿nery.
50. The distance from the re¿nery to S is now
Thus, F({) = 4
s
I
(6 3 {)2 + 12 = {2 3 12{ + 37.
I
I
{2 3 12{ + 37 + 8 {2 + 4 i
8{
4({ 3 6)
+I
.
F 0 ({) = 4 · 12 ({2 3 12{ + 37)31@2 (2{ 3 12) + 8 · 12 ({2 + 4)31@2 (2{) = I
{2 3 12{ + 37
{2 + 4
F 0 ({) = 0 i { E 1=12 [from a graph of F 0 or a numerical root¿nder]. F(0) E 40=3, F(1=12) E 38=3, and
F(6) E 54=6. So the minimum cost is slightly higher (than in the previous exercise) at about $3=83 million when S is
approximately 4=88 km from the point on the bank 1 km south of the re¿nery.
51.
The total illumination is L({) =
L 0 ({) =
3n
n
+
, 0 ? { ? 10. Then
{2
(10 3 {)2
36n
2n
+
= 0 i 6n(10 3 {)3 = 2n{3
{3
(10 3 {)3
i
I
I
I
I
I
3
3 (10 3 {) = { i 10 3 3 3 3 3 { = { i 10 3 3 = { + 3 3 { i
I
I
I 10 3 3
I E 5=9 ft. This gives a minimum since L 00 ({) A 0 for 0 ? { ? 10.
10 3 3 = 1 + 3 3 { i { =
1+ 33
3(10 3 {)3 = {3
52.
i
The line with slope p (where p ? 0) through (3> 5) has equation | 3 5 = p({ 3 3) or
| = p{ + (5 3 3p). The |-intercept is 5 3 3p and the {-intercept is 35@p + 3. So the
triangle has area D(p) = 12 (5 3 3p)(35@p + 3) = 15 3 25@(2p) 3 92 p. Now
D0 (p) =
D00 (p) = 3
25
9
3 = 0 C p2 =
2p2
2
25
9
i
p = 3 53 (since p ? 0).
25
A 0, so there is an absolute minimum when p = 3 53 . Thus, an equation of the line is | 3 5 = 3 53 ({ 3 3)
p3
or | = 3 53 { + 10.
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406
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
Every line segment in the ¿rst quadrant passing through (d> e) with endpoints on the {-
53.
and |-axes satis¿es an equation of the form | 3 e = p({ 3 d), where p ? 0. By setting
e
> 0 . The
{ = 0 and then | = 0, we ¿nd its endpoints, D(0> e 3 dp) and E d 3 p
t
e
3 0]2 + [0 3 (e 3 dp)]2 .
distance g from D to E is given by g = [ d 3 p
It follows that the square of the length of the line segment, as a function of p, is given by
2
e2
2de
e
+ 2 + d2 p2 3 2dep + e2 . Thus,
+ (dp 3 e)2 = d2 3
V(p) = d 3
p
p
p
2de
2e2
2
3 3 + 2d2 p 3 2de = 3 (dep 3 e2 + d2 p4 3 dep3 )
2
p
p
p
2
2
= 3 [e(dp 3 e) + dp3 (dp 3 e)] = 3 (dp 3 e)(e + dp3 )
p
p
t
t
2
Thus, V 0 (p) = 0 C p = e@d or p = 3 3 de . Since e@d A 0 and p ? 0, p must equal 3 3 de . Since 3 ? 0, we see
p
t
t
t
that V 0 (p) ? 0 for p ? 3 3 de and V 0 (p) A 0 for p A 3 3 de . Thus, V has its absolute minimum value when p = 3 3 de .
V 0 (p) =
That value is
2 t t 2 t
2 I
2
I
3
3 2
3 d
3 e
3 e
+ 3d d 3 e = d + de2 +
d e+e
V 3 d = d+e e
= d2 + 2d4@3 e2@3 + d2@3 e4@3 + d4@3 e2@3 + 2d2@3 e4@3 + e2 = d2 + 3d4@3 e2@3 + 3d2@3 e4@3 + e2
The last expression is of the form {3 + 3{2 | + 3{| 2 + | 3
[= ({ + |)3 ] with { = d2@3 and | = e2@3 ,
I
so we can write it as (d2@3 + e2@3 )3 and the shortest such line segment has length V = (d2@3 + e2@3 )3@2 .
54. | = 1 + 40{3 3 3{5
i | 0 = 120{2 3 15{4 , so the tangent line to the curve at { = d has slope p(d) = 120d2 3 15d4 .
Now p0 (d) = 240d 3 60d3 = 360d(d2 3 4) = 360d(d + 2)(d 3 2), so p0 (d) A 0 for d ? 32, and 0 ? d ? 2, and
p0 (d) ? 0 for 32 ? d ? 0 and d A 2. Thus, p is increasing on (3"> 32), decreasing on (32> 0), increasing on (0> 2), and
decreasing on (2> ") = Clearly, p(d) < 3" as d < ±", so the maximum value of p(d) must be one of the two local
maxima, p(32) or p(2). But both p(32) and p(2) equal 120 · 22 3 15 · 24 = 480 3 240 = 240. So 240 is the largest
slope, and it occurs at the points (32> 3223) and (2> 225). Note: d = 0 corresponds to a local minimum of p.
55. | =
|3
3
{
i |0 = 3
3
, so an equation of the tangent line at the point (d> d3 ) is
{2
3
6
3
3
= 3 2 ({ 3 d), or | = 3 2 { + . The |-intercept [{ = 0] is 6@d. The
d
d
d
d
{-intercept [| = 0] is 2d. The distance g of the line segment that has endpoints at the
intercepts is g =
s
36
(2d 3 0)2 + (0 3 6@d)2 . Let V = g2 , so V = 4d2 + 2
d
i
I
72
= 8d C d4 = 9 C d2 = 3 i d = 3.
d3
I
216
V 00 = 8 + 4 A 0, so there is an absolute minimum at d = 3= Thus, V = 4(3) + 36
= 12 + 12 = 24 and
3
d
I
I
hence, g = 24 = 2 6.
V 0 = 8d 3
72
. V0 = 0 C
d3
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.7
56. | = 4 3 {2
OPTIMIZATION PROBLEMS
¤
407
i | 0 = 32{, so an equation of the tangent line at (d> 4 3 d2 ) is
| 3 (4 3 d2 ) = 32d({ 3 d), or | = 32d{ + d2 + 4. The |-intercept [{ = 0]
d2 + 4
. The area D of the triangle is
2d
1 d2 + 4 2
1
1 d4 + 8d2 + 16
16
1
(d +4) =
=
d3 + 8d +
.
D = (base)(height) = ·
2
2
2d
4
d
4
d
1
16
D0 = 0 i
3d2 + 8 3 2 = 0 i 3d4 + 8d2 3 16 = 0 i
4
d
4
1
2
32
(3d2 3 4)(d2 + 4) = 0 i d2 =
i d = I . D00 =
6d + 3 A 0, so there is an absolute minimum at
3
4
d
3
I
1 4 3 16
32 I
1 4@3 + 4 4
2
I
+4 = ·
·
=
3.
d = I . Thus, D = ·
2
3
2
3
3
9
3
2(2@ 3)
is d2 + 4. The {-intercept [| = 0] is
57. (a) If f({) =
F({)
{F 0 ({) 3 F({)
. Now f0 ({) = 0 when
, then, by the Quotient Rule, we have f0 ({) =
{
{2
F({)
= f({). Therefore, the marginal cost equals the average cost.
{
I
(b) (i) F({) = 16,000 + 200{ + 4{3@2 , F(1000) = 16,000 + 200,000 + 40,000 10 E 216,000 + 126,491, so
{F 0 ({) 3 F({) = 0 and this gives F 0 ({) =
F(1000) E $342,491. f({) = F({)@{ =
F 0 (1000) = 200 + 60
16,000
+ 200 + 4{1@2 , f(1000) E $342=49@unit. F 0 ({) = 200 + 6{1@2 ,
{
I
10 E $389=74@unit.
(ii) We must have F 0 ({) = f({) C 200 + 6{1@2 =
16,000
+ 200 + 4{1@2
{
C 2{3@2 = 16,000 C
{ = (8,000)2@3 = 400 units. To check that this is a minimum, we calculate
f0 ({) =
316,000
2
2
+ I = 2 ({3@2 3 8000). This is negative for { ? (8000)2@3 = 400, zero at { = 400,
{2
{
{
and positive for { A 400, so f is decreasing on (0> 400) and increasing on (400> "). Thus, f has an absolute minimum
at { = 400. [Note: f00 ({) is not positive for all { A 0.]
(iii) The minimum average cost is f(400) = 40 + 200 + 80 = $320@unit.
58. (a) The total pro¿t is S ({) = U({) 3 F({). In order to maximize pro¿t we look for the critical numbers of S , that is, the
numbers where the marginal pro¿t is 0. But if S 0 ({) = U0 ({) 3 F 0 ({) = 0, then U0 ({) = F 0 ({). Therefore, if the pro¿t
is a maximum, then the marginal revenue equals the marginal cost.
(b) F({) = 16,000 + 500{ 3 1=6{2 + 0=004{3 , s({) = 1700 3 7{. Then U({) = {s({) = 1700{ 3 7{2 . If the pro¿t is
maximum, then U0 ({) = F 0 ({) C 1700 3 14{ = 500 3 3=2{ + 0=012{2
C 0=012{2 + 10=8{ 3 1200 = 0 C
{2 + 900{ 3 100,000 = 0 C ({ + 1000)({ 3 100) = 0 C { = 100 (since { A 0). The pro¿t is maximized if
S 00 ({) ? 0, but since S 00 ({) = U00 ({) 3 F 00 ({), we can just check the condition U00 ({) ? F 00 ({). Now
U00 ({) = 314 ? 33=2 + 0=024{ = F 00 ({) for { A 0, so there is a maximum at { = 100.
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°
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408
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
59. (a) We are given that the demand function s is linear and s(27,000) = 10, s(33,000) = 8, so the slope is
10 3 8
27,000 3 33,000
1
1
({ 3 27,000) i
= 3 3000
and an equation of the line is | 3 10 = 3 3000
1
| = s({) = 3 3000
{ + 19 = 19 3 ({@3000).
(b) The revenue is U({) = {s({) = 19{ 3 ({2@3000) i U0 ({) = 19 3 ({@1500) = 0 when { = 28,500. Since
U00 ({) = 31@1500 ? 0, the maximum revenue occurs when { = 28,500 i the price is s(28,500) = $9=50.
60. (a) Let s({) be the demand function. Then s({) is linear and | = s({) passes through (20> 10) and (18> 11), so the slope is
3 12 and an equation of the line is | 3 10 = 3 12 ({ 3 20) C | = 3 12 { + 20. Thus, the demand is s({) = 3 12 { + 20
and the revenue is U({) = {s({) = 3 12 {2 + 20{.
(b) The cost is F({) = 6{, so the pro¿t is S ({) = U({) 3 F({) = 3 12 {2 + 14{. Then 0 = S 0 ({) = 3{ + 14 i
{ = 14. Since S 00 ({) = 31 ? 0, the selling price for maximum pro¿t is s(14) = 3 12 (14) + 20 = $13.
61. (a) As in Example 6, we see that the demand function s is linear. We are given that s(1000) = 450 and deduce that
s(1100) = 440, since a $10 reduction in price increases sales by 100 per week. The slope for s is
440 3 450
1100 3 1000
1
= 3 10
,
1
1
so an equation is s 3 450 = 3 10
({ 3 1000) or s({) = 3 10
{ + 550.
1 2
{ + 550{. U0 ({) = 3 15 { + 550 = 0 when { = 5(550) = 2750.
(b) U({) = {s({) = 3 10
s(2750) = 275, so the rebate should be 450 3 275 = $175.
1 2
1 2
{ + 550{ 3 68,000 3 150{ = 3 10
{ + 400{ 3 68,000,
(c) F({) = 68,000 + 150{ i S ({) = U({) 3 F({) = 3 10
S 0 ({) = 3 15 { + 400 = 0 when { = 2000. s(2000) = 350. Therefore, the rebate to maximize pro¿ts should be
450 3 350 = $100.
62. Let { denote the number of $10 increases in rent. Then the price is s({) = 800 + 10{, and the number of units occupied is
100 3 {. Now the revenue is
U({) = (rental price per unit) × (number of units rented)
= (800 + 10{)(100 3 {) = 310{2 + 200{ + 80,000 for 0 $ { $ 100
i
U0 ({) = 320{ + 200 = 0 C { = 10. This is a maximum since U00 ({) = 320 ? 0 for all {. Now we must check the
value of U({) = (800 + 10{) (100 3 {) at { = 10 and at the endpoints of the domain to see which value of { gives the
maximum value of U. U(0) = 80,000, U(10) = (900)(90) = 81,000, and U(100) = (1800)(0) = 0. Thus, the maximum
revenue of $81,000@week occurs when 90 units are occupied at a rent of $900@week.
63.
Here v2 = k2 + e2@4, so k2 = v2 3 e2@4. The area is D = 12 e
Let the perimeter be s, so 2v + e = s or v = (s 3 e)@2 i
s
s
D(e) = 12 e (s 3 e)2@4 3 e2@4 = e s2 3 2se@4. Now
s
s2 3 2se
es@4
33se + s2
0
3s
= s
.
D (e) =
2
4
s 3 2se
4 s2 3 2se
s
v2 3 e2@4.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.7
OPTIMIZATION PROBLEMS
¤
409
Therefore, D0 (e) = 0 i 33se + s2 = 0 i e = s@3. Since D0 (e) A 0 for e ? s@3 and D0 (e) ? 0 for e A s@3, there
is an absolute maximum when e = s@3. But then 2v + s@3 = s, so v = s@3 i v = e i the triangle is equilateral.
64. See the ¿gure. The area is given by
I
I
I
I
I
{2 + e2 3 d2 = d2 3 {2 { + {2 + e2 3 d2 for 0 $ { $ d.
2 d2 3 {2 { + 12 2 d2 3 {2
I
I
{
3{
+ { + {2 + e2 3 d2 I
=0 C
Now D0 ({) = d2 3 {2 1 + I
2
2
2
{ +e 3d
d2 3 {2
I
I
I
{
{ + {2 + e2 3 d2
2
2
2
2
2
I
I
.
{+ { +e 3d = d 3{
d2 3 {2
{2 + e2 3 d2
D({) =
1
2
Except for the trivial case where { = 0, d = e and D({) = 0, we have
I
{ + {2 + e2 3 d2 A 0. Hence, cancelling this factor gives
I
I
d2 3 {2
{
I
= I
i { {2 + e2 3 d2 = d2 3 {2 i
d2 3 {2
{2 + e2 3 d2
{2 ({2 + e2 3 d2 ) = d4 3 2d2 {2 + {4
{2(e2 + d2 ) = d4
i {2 (e2 3 d2 ) = d4 3 2d2 {2
i
d2
i {= I
.
2
d + e2
Now we must check the value of D at this point as well as at the endpoints of the domain to see which gives the maximum
I
value. D(0) = d e2 3 d2 , D(d) = 0 and
5
6
v
v
2
2
2
2
2
d
d2
d
d
7I
I
+
+ e2 3 d2 8
D I
= d2 3 I
d2 + e2
d2 + e2
d2 + e2
d2 + e2
de
= I
d2 + e2
Since e D
d2
e2
I
+I
2
2
2
d +e
d + e2
=
de(d2 + e2 )
= de
d2 + e2
I
I
d2
e2 3 d2 , D d2@ d2 + e2 D D (0). So there is an absolute maximum when { = I
. In this case the
d2 + e2
I
d2 + e2
2de
and the vertical piece should be I
= d2 + e2 .
horizontal piece should be I
d2 + e2
d2 + e2
65. Note that |DG| = |DS | + |S G|
i 5 = { + |S G| i |S G| = 5 3 {.
Using the Pythagorean Theorem for {S GE and {S GF gives us
s
s
O({) = |DS | + |ES | + |FS | = { + (5 3 {)2 + 22 + (5 3 {)2 + 32
I
I
= { + {2 3 10{ + 29 + {2 3 10{ + 34 i
{35
{35
O0 ({) = 1 + I
+I
. From the graphs of O
{2 3 10{ + 29
{2 3 10{ + 34
and O0 , it seems that the minimum value of O is about O(3=59) = 9=35 m.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
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410
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
66. We note that since f is the consumption in gallons per hour, and y is the velocity in miles per hour, then
f
gallons@hour
gallons
=
=
gives us the consumption in gallons per mile, that is, the quantity J. To ¿nd the minimum,
y
miles@hour
mile
gf
gf
gy
y
3f
g f y gy 3 f gy
gJ
gy
=
=
=
.
we calculate
gy
gy y
y2
y2
This is 0 when y
gf
3f =0 C
gy
gf
f
= . This implies that the tangent line
gy
y
of f(y) passes through the origin, and this occurs when y E 53 mi@h. Note that
the slope of the secant line through the origin and a point (y> f(y)) on the graph
is equal to J(y), and it is intuitively clear that J is minimized in the case where
the secant is in fact a tangent.
67.
The total time is
W ({) = (time from D to F) + (time from F to E)
s
I
e2 + (g 3 {)2
d2 + {2
+
, 0?{?g
=
y1
y2
W 0 ({) =
{
g3{
sin 1
sin 2
I
3 s
=
3
y1
y2
y1 d2 + {2
y2 e2 + (g 3 {)2
The minimum occurs when W 0 ({) = 0 i
sin 1
sin 2
=
.
y1
y2
[Note: W 00 ({) A 0]
68.
If g = |TW |, we minimize i (1 ) = |S U| + |UV| = d csc 1 + e csc 2 .
Differentiating with respect to 1 , and setting
gi
equal to 0, we get
g1
g2
gi
= 0 = 3d csc 1 cot 1 3 e csc 2 cot 2
.
g1
g1
So we need to ¿nd an expression for
g2
. We can do this by observing that |TW | = constant = d cot 1 + e cot 2 .
g1
Differentiating this equation implicitly with respect to 1 , we get 3d csc2 1 3 e csc2 2
g2
=0 i
g1
d csc2 1
gi
g2
=3
. We substitute this into the expression for
to get
g1
e csc2 2
g1
csc2 1 cot 2
d csc2 1
3d csc 1 cot 1 3 e csc 2 cot 2 3
cot
+
d
=0 C
=
0
C
3d
csc
1
1
e csc2 2
csc 2
cot 1 csc 2 = csc 1 cot 2
C
cot 1
cot 2
=
csc 1
csc 2
C cos 1 = cos 2 . Since 1 and 2 are both acute, we
have 1 = 2 .
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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NOT FOR SALE
SECTION 4.7
OPTIMIZATION PROBLEMS
¤
411
| 2 = {2 + } 2 , but triangles FGH and EFD are similar, so
I
I
}@8 = {@ 4 { 3 4
i } = 2{@ { 3 4. Thus, we minimize
69.
i ({) = | 2 = {2 + 4{2 @({ 3 4) = {3 @({ 3 4), 4 ? { $ 8.
i 0 ({) =
({ 3 4)(3{2 ) 3 {3
{2 [3({ 3 4) 3 {]
2{2 ({ 3 6)
=
=
=0
({ 3 4)2
({ 3 4)2
({ 3 4)2
when { = 6. i 0 ({) ? 0 when { ? 6, i 0 ({) A 0 when { A 6, so the minimum
occurs when { = 6 in.
70.
Paradoxically, we solve this maximum problem by solving a minimum problem.
Let O be the length of the line DFE going from wall to wall touching the inner
corner F. As < 0 or <
,
2
we have O < " and there will be an angle that
makes O a minimum. A pipe of this length will just ¿t around the corner.
From the diagram, O = O1 + O2 = 9 csc + 6 sec i gO@g = 39 csc cot + 6 sec tan = 0 when
I
2@3
6 sec tan = 9 csc cot C tan3 = 96 = 1=5 C tan = 3 1=5. Then sec2 = 1 + 32
and
k
k
32@3
32@3 l1@2
2@3 l1@2
csc2 = 1 + 32
, so the longest pipe has length O = 9 1 + 32
+ 6 1 + 32
E 21=07 ft.
Or, use = tan31
71.
I
3
1=5 E 0=853 i O = 9 csc + 6 sec E 21=07 ft.
It suf¿ces to maximize tan . Now
tan # + tan w + tan 3w
= tan(# + ) =
=
. So
1
1 3 tan # tan 1 3 w tan 3w(1 3 w tan ) = w + tan 2w
Let i(w) = tan =
1 + 3w2
i 2w = (1 + 3w2 ) tan i tan =
2w
.
1 + 3w2
2 1 + 3w2 3 2w(6w)
2 1 3 3w2
i i (w) =
=
= 0 C 1 3 3w2 = 0 C
(1 + 3w2 )2
(1 + 3w2 )2
0
since w D 0. Now i 0 (w) A 0 for 0 $ w ? I13 and i 0 (w) ? 0 for w A I13 , so i has an absolute maximum when w =
I 2 1@ 3
1
and tan =
i = 6 . Substituting for w and in 3w = tan(# + ) gives us
I 2 = I
3
1 + 3 1@ 3
I
3 = tan # + 6
i # = 6 .
w=
I1
3
72. We maximize the cross-sectional area
D() = 10k + 2 12 gk = 10k + gk = 10(10 sin ) + (10 cos )(10 sin )
= 100(sin + sin cos ), 0 $ $
2
D0 () = 100(cos + cos2 3 sin2 ) = 100(cos + 2 cos2 3 1)
= 100(2 cos 3 1)(cos + 1) = 0 when cos =
1
2
C
=
3
[ cos 6= 31 since 0 $ $
I
Now D(0) = 0, D 2 = 100 and D 3 = 75 3 E 129=9, so the maximum occurs when =
.]
2
3.
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°
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I1
3
412
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
2
5
and tan =
. Since
{
33{
5
2
+ + = 180 = , = 3 tan31
3 tan31
i
{
33{
2
1
g
5
1
=3
2
2 3 2 3
g{
{
(3 3 {)2
5
2
1+
1+
{
33{
From the ¿gure, tan =
73.
=
Now
g
=0
g{
{2
5
2
= 2
{2 + 25
{ 3 6{ + 13
i
{2
5
2
(3 3 {)2
· 2 3
·
.
+ 25 {
(3 3 {)2 + 4 (3 3 {)2
i 2{2 + 50 = 5{2 3 30{ + 65 i
I
{2 3 10{ + 5 = 0 i { = 5 ± 2 5. We reject the root with the + sign, since it is
I
I
larger than 3. g@g{ A 0 for { ? 5 3 2 5 and g@g{ ? 0 for { A 5 3 2 5, so is maximized when
I
|DS | = { = 5 3 2 5 E 0=53.
3{2 3 30{ + 15 = 0
i
74. Let { be the distance from the observer to the wall. Then, from the given ¿gure,
= tan31
k+g
{
3 tan31
g
, {A0 i
{
g
g
k+g
1
g
k+g
1
+ 2
3
3
3
=3 2
=
g{
1 + [(k + g)@{]2
{2
1 + (g@{)2
{2
{ + (k + g)2
{ + g2
=
k2 g + kg2 3 k{2
g[{2 + (k + g)2 ] 3 (k + g)({2 + g2 )
= 2
=0 C
2
2
2
2
[{ + (k + g) ]({ + g )
[{ + (k + g)2 ]({2 + g2 )
s
s
k{2 = k2 g + kg2 C {2 = kg + g2 C { = g(k + g). Since g@g{ A 0 for all { ? g(k + g) and g@g{ ? 0
s
s
for all { A g(k + g), the absolute maximum occurs when { = g(k + g).
75.
In the small triangle with sides d and f and hypotenuse Z , sin =
d
and
Z
cos =
g
f
. In the triangle with sides e and g and hypotenuse O, sin = and
Z
O
cos =
e
. Thus, d = Z sin , f = Z cos , g = O sin , and e = O cos , so the
O
area of the circumscribed rectangle is
D() = (d + e)(f + g) = (Z sin + O cos )(Z cos + O sin )
= Z 2 sin cos + Z O sin2 + OZ cos2 + O2 sin cos = OZ sin2 + OZ cos2 + (O2 + Z 2 ) sin cos = OZ (sin2 + cos2 ) + (O2 + Z 2 ) ·
1
2
· 2 sin cos = OZ + 12 (O2 + Z 2 ) sin 2, 0 $ $
This expression shows, without calculus, that the maximum value of D() occurs when sin 2 = 1 C 2 =
= 4 . So the maximum area is D 4 = OZ + 12 (O2 + Z 2 ) = 12 (O2 + 2OZ + Z 2 ) = 12 (O + Z )2 .
2
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°
© Cengage Learning. All Rights Reserved.
2
i
NOT FOR SALE
SECTION 4.7
76. (a) Let G be the point such that d = |DG|. From the ¿gure, sin =
cos =
|EG|
d 3 |DE|
=
|EF|
|EF|
e
|EF|
OPTIMIZATION PROBLEMS
¤
413
i |EF| = e csc and
i |EF| = (d 3 |DE|) sec . Eliminating |EF| gives
(d 3 |DE|) sec = e csc i e cot = d 3 |DE| i |DE| = d 3 e cot . The total resistance is
d 3 e cot |EF|
e csc |DE|
+
.
U() = F 4 + F 4 = F
u1
u2
u14
u24
(b) U0 () = F
e csc2 e csc cot 3
u14
u24
U0 () = 0 C
csc cot = 4
u14
u2
csc cot A 4
u14
u2
when cos = u24 u14 .
U0 () A 0 C
(c) When u2 = 23 u1 , we have cos =
= eF csc csc cot 3
.
u14
u24
u24
cot =
= cos .
u14
csc C
i cos ?
2 4
3
u24
u24
0
4 and U () ? 0 when cos A 4 , so there is an absolute minimum
u1
u1
, so = cos31
77. (a)
2 4
3
E 79 .
If n = energy@km over land, then energy@km over water = 1=4n.
I
So the total energy is H = 1=4n 25 + {2 + n(13 3 {), 0 $ { $ 13,
and so
Set
gH
= 0: 1=4n{ = n(25 + {2 )1@2
g{
gH
1=4n{
3 n.
=
g{
(25 + {2 )1@2
i 1=96{2 = {2 + 25 i 0=96{2 = 25 i { =
I5
0=96
E 5=1.
Testing against the value of H at the endpoints: H(0) = 1=4n(5) + 13n = 20n, H(5=1) E 17=9n, H(13) E 19=5n.
Thus, to minimize energy, the bird should Ày to a point about 5=1 km from E.
(b) If Z@O is large, the bird would Ày to a point F that is closer to E than to G to minimize the energy used Àying over water.
If Z@O is small, the bird would Ày to a point F that is closer to G than to E to minimize the distance of the Àight.
I
I
25 + {2
Z{
gH
Z
= I
=
. By the same sort of
3 O = 0 when
H = Z 25 + {2 + O(13 3 {) i
2
g{
O
{
25 + {
argument as in part (a), this ratio will give the minimal expenditure of energy if the bird heads for the point { km from E.
(c) For Àight direct to G, { = 13, so from part (b), Z@O =
I
25 + 132
13
E 1=07. There is no value of Z@O for which the bird
should Ày directly to E. But note that lim (Z@O) = ", so if the point at which H is a minimum is close to E, then
{<0+
Z@O is large.
(d) Assuming that the birds instinctively choose the path that minimizes the energy expenditure, we can use the equation for
I
gH@g{ = 0 from part (a) with 1=4n = f, { = 4, and n = 1: f(4) = 1 · (25 + 42 )1@2 i f = 41@4 E 1=6.
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°
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414
¤
NOT FOR SALE
CHAPTER 4
78. (a) L({) x
L({) =
APPLICATIONS OF DIFFERENTIATION
strength of source
. Adding the intensities from the left and right lightbulbs,
(distance from source)2
n
n
n
n
+
= 2
+ 2
.
{2 + g 2
{ + g2
{ 3 20{ + 100 + g2
(10 3 {)2 + g2
(b) The magnitude of the constant n won’t affect the location of the point of maximum intensity, so for convenience we take
n = 1. L 0 ({) = 3
2{
2({ 3 10)
3 2
.
({2 + g2 )2
({ 3 20{ + 100 + g 2 )2
Substituting g = 5 into the equations for L({)and L 0 ({), we get
L5 ({) =
1
1
+ 2
{2 + 25
{ 3 20{ + 125
and
L50 ({) = 3
2({ 3 10)
2{
3 2
({2 + 25)2
({ 3 20{ + 125)2
From the graphs, it appears that L5 ({) has a
minimum at { = 5 m.
(c) Substituting g = 10 into the equations for L({) and L 0 ({) gives
L10 ({) =
1
1
+ 2
{2 + 100
{ 3 20{ + 200
and
0
({) = 3
L10
({2
2 ({ 3 10)
2{
2 3
2
+ 100)
({ 3 20{ + 200)2
From the graphs, it seems that for g = 10, the
intensity is minimized at the endpoints, that is,
{ = 0 and { = 10. The midpoint is now the
most brightly lit point!
(d) From the ¿rst ¿gures in parts (b) and (c), we see that the minimal illumination changes from the midpoint ({ = 5 with
g = 5) to the endpoints ({ = 0 and { = 10 with g = 10).
So we try g = 6 (see the ¿rst ¿gure) and we see that the minimum value still occurs at { = 5. Next, we let g = 8 (see the
second ¿gure) and we see that the minimum value occurs at the endpoints. It appears that for some value of g between 6
and 8, we must have minima at both the midpoint and the endpoints, that is, L(5) must equal L(0). To ¿nd this value of g,
we solve L(0) = L(5) (with n = 1):
1
1
1
2
1
+
=
+
=
g2
100 + g 2
25 + g2
25 + g 2
25 + g2
2500 + 125g2 + g4 + 25g2 + g4 = 200g2 + 2g4
i (25 + g2 )(100 + g2 ) + g2 (25 + g2 ) = 2g 2 (100 + g2 ) i
i 2500 = 50g2
i g2 = 50 i g = 5
I
2 E 7=071
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°
© Cengage Learning. All Rights Reserved.
APPLIED PROJECT
THE SHAPE OF A CAN
¤
415
[for 0 $ g $ 10]. The third ¿gure, a graph of L(0) 3 L(5) with g independent, con¿rms that L(0) 3 L(5) = 0, that is,
I
L(0) = L(5), when g = 5 2. Thus, the point of minimal illumination changes abruptly from the midpoint to the
I
endpoints when g = 5 2.
APPLIED PROJECT
The Shape of a Can
1. In this case, the amount of metal used in the making of each top or bottom is (2u)2 = 4u2 . So the quantity we want to
minimize is D = 2uk + 2(4u2 ). But Y = u2 k C k = Y@u2 . Substituting this expression for k in D gives
D = 2Y @u + 8u2 . Differentiating D with respect to u, we get gD@gu = 32Y@u2 + 16u = 0 i
16u3 = 2Y = 2u2 k C
8
g2 D
k
4Y
= E 2=55. This gives a minimum because
= 16 + 3 A 0.
u
gu2
u
We need to ¿nd the area of metal used up by each end, that is, the area of each
2.
hexagon. We subdivide the hexagon into six congruent triangles, each sharing one
side (v in the diagram) with the hexagon. We calculate the length of
v = 2u tan 6 =
I2 u,
3
so the area of each triangle is 12 vu =
area of the hexagon is 6 ·
I1 u 2
3
=2
I1 u 2 ,
3
and the total
I 2
3 u . So the quantity we want to minimize
I
I 2
2Y
gD
= 3 2 + 8 3 u.
3 u . Substituting for k as in Problem 1 and differentiating, we get
gu
u
I
I
4 3
k
Setting this equal to 0, we get 8 3 u3 = 2Y = 2u2 k i
=
E 2=21. Again this minimizes D because
u
I
g2 D
4Y
= 8 3 + 3 A 0.
2
gu
u
I
I
Y
Y
3. Let F = 4 3 u2 + 2uk + n (4u + k) = 4 3 u2 + 2u
+
n
4u
+
. Then
u2
u2
I
2Y
2nY
Y
gF
. Setting this equal to 0, dividing by 2 and substituting 2 = k and
= 8 3 u 3 2 + 4n 3
gu
u
u3
u
I
Y
k
nk
C
= in the second and fourth terms respectively, we get 0 = 4 3 u 3 k + 2n 3
3
u
u
u
u
u
I
I
3
3
I
Y
Y n
k
n 2 3 k@u
k
3 Y
I = 1. We now multiply by
= k 3 4 3 u i
, noting that
=
,
= 3
n 2 3
u
u k@u 3 4 3
n
n u
u3
u
u
I
3
Y
k
2 3 k@u
I .
and get
= 3
·
n
u k@u 3 4 3
I
I
2 3 {
I . We see from
4.
Let 3 Y @n = W and k@u = { so that W ({) = 3 { ·
{ 3 4 3
I
the graph of W that when the ratio 3 Y @n is large; that is, either the volume of
is D = 2uk + 2 · 2
the can is large or the cost of joining (proportional to n) is small, the optimum
value of k@u is about 2=21, but when
I
3
Y @n is small, indicating small volume
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated,
°
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416
¤
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
or expensive joining, the optimum value of k@u is larger. (The part of the graph for
I
3
Y @n ? 0 has no physical meaning, but
con¿rms the location of the asymptote.)
5. Our conclusion is usually true in practice. But there are exceptions, such as cans of tuna, which may have to do with the shape
of a reasonable slice of tuna. And for a comfortable grip on a soda or beer can, the geometry of the human hand is a restriction
on the radius. Other possible considerations are packaging, transportation and stocking constraints, aesthetic appeal and other
marketing concerns. Also, there may be better models than ours which prescribe a differently shaped can in special
circumstances.
4.8 Newton's Method
1. (a)
The tangent line at { = 1 intersects the {-axis at { E 2=3, so
{2 E 2=3. The tangent line at { = 2=3 intersects the {-axis at
{ E 3, so {3 E 3=0.
(b) {1 = 5 would not be a better ¿rst approximation than {1 = 1 since the tangent line is nearly horizontal. In fact, the second
approximation for {1 = 5 appears to be to the left of { = 1.
2.
The tangent line at { = 9 intersects the {-axis at { E 6=0, so
{2 E 6=0. The tangent line at { = 6=0 intersects the {-axis at
{ E 8=0, so {3 E 8=0.
3. Since the tangent line | = 9 3 2{ is tangent to the curve | = i ({) at the point (2> 5), we have {1 = 2, i({1 ) = 5, and
i 0 ({1 ) = 32 [the slope of the tangent line]. Thus, by Equation 2,
{2 = {1 3
Note that geometrically
9
2
5
9
i ({1 )
=23
=
i 0 ({1 )
32
2
represents the {-intercept of the tangent line | = 9 3 2{.
4. (a)
(b)
If {1 = 0, then {2 is negative, and {3 is even more
If {1 = 1, the tangent line is horizontal and Newton’s
negative. The sequence of approximations does not
method fails.
converge, that is, Newton’s method fails.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 4.8 NEWTON’S METHOD
¤
417
(d)
(c)
If {1 = 3, then {2 = 1 and we have the same situation
If {1 = 4, the tangent line is horizontal and Newton’s
as in part (b). Newton’s method fails again.
method fails.
If {1 = 5, then {2 is greater than 6, {3 gets closer to 6, and
(e)
the sequence of approximations converges to 6. Newton’s
method succeeds!
5. The initial approximations {1 = d> e, and f will work, resulting in a second approximation closer to the origin, and lead to the
root of the equation i ({) = 0, namely, { = 0. The initial approximation {1 = g will not work because it will result in
successive approximations farther and farther from the origin.
6. i ({) =
1 3
{
3
{2 = 33 3
+ 12 {2 + 3 i i 0 ({) = {2 + {, so {q+1 = {q 3
+ 12 {2q + 3
. Now {1 = 33 i
{2q + {q
39 + 92 + 3
= 33 3 3 14 = 32=75 i {3 = 32=75 3
933
7. i ({) = {5 3 { 3 1
{2 = 1 3
1 3
{
3 q
i i 0 ({) = 5{4 3 1, so {q+1 = {q 3
1
(32=75)3
3
+ 12 (32=75)2 + 3
E 32=7186.
(32=75)2 + (32=75)
{5q 3 {q 3 1
. Now {1 = 1 i
5{4q 3 1
13131
(1=25)5 3 1=25 3 1
= 1 3 3 14 = 1=25 i {3 = 1=25 3
E 1=1785.
531
5(1=25)4 3 1
8. i ({) = {7 + 4
{2 = 31 3
{7q + 4
. Now {1 = 31 i
7{6q
10 7
3 7 +4
10
i {3 = 3 3
6 E 31=2917.
7
7 3 10
i i 0 ({) = 7{6 , so {q+1 = {q 3
10
(31)7 + 4
3
= 31 3 = 3
7(31)6
7
7
9. i ({) = {3 + { + 3
i
i 0 ({) = 3{2 + 1, so {q+1 = {q 3
7
{3q + {q + 3
=
3{2q + 1
Now {1 = 31 i
{2 = 31 3
31 3 1 + 3
1
(31)3 + (31) + 3
= 31 3
= 31 3 = 31=25.
3(31)2 + 1
3+1
4
Newton’s method follows the tangent line at (31> 1) down to its intersection with
the {-axis at (31=25> 0), giving the second approximation {2 = 31=25.
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°
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418
¤
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
10. i ({) = {4 3 { 3 1
i i 0 ({) = 4{3 3 1, so {q+1 = {q 3
{4q 3 {q 3 1
.
4{3q 3 1
31
4
14 3 1 3 1
=13
= . Newton’s method
4 · 13 3 1
3
3
follows the tangent line at (1> 31) up to its intersection with the {-axis at 43 > 0 ,
Now {1 = 1 i {2 = 1 3
giving the second approximation {2 = 43 .
11. To approximate { =
{q+1 = {q 3
I
5
20 (so that {5 = 20), we can take i ({) = {5 3 20. So i 0 ({) = 5{4 , and thus,
I
{5q 3 20
. Since 5 32 = 2 and 32 is reasonably close to 20> we’ll use {1 = 2. We need to ¿nd approximations
5{4q
until they agree to eight decimal places. {1 = 2 i {2 = 1=85, {3 E 1=82148614, {4 E 1=82056514,
{5 E 1=82056420 E {6 . So
I
5
20 E 1=82056420, to eight decimal places.
Here is a quick and easy method for ¿nding the iterations for Newton’s method on a programmable calculator.
(The screens shown are from the TI-84 Plus, but the method is similar on other calculators.) Assign i ({) = {5 3 20
to Y1 , and i 0 ({) = 5{4 to Y2 . Now store {1 = 2 in X and then enter X 3 Y1 @Y2 < X to get {2 = 1=85. By successively
pressing the ENTER key, you get the approximations {3 , {4 , = = = .
In Derive, load the utility ¿le SOLVE. Enter NEWTON(xˆ5-20,x,2) and then APPROXIMATE to get
[2, 1.85, 1.82148614, 1.82056514, 1.82056420]. You can request a speci¿c iteration by adding a fourth argument. For
example, NEWTON(xˆ5-20,x,2,2) gives [2> 1=85> 1=82148614].
In Maple, make the assignments i := { < {ˆ5 3 20;, j := { < { 3 i ({)@G(i)({);, and { := 2=;. Repeatedly execute
the command { := j({); to generate successive approximations.
In Mathematica, make the assignments i [{_ ] := {ˆ5 3 20, j[{_ ] := { 3 i[{]@i 0 [{], and { = 2= Repeatedly execute the
command { = j[{] to generate successive approximations.
12. i ({) = {100 3 100
i i 0 ({) = 100{99 , so {q+1 = {q 3
{100
q 3 100
. We need to ¿nd approximations until they agree
100{99
q
to eight decimal places. {1 = 1=05 i {2 E 1=04748471, {3 E 1=04713448, {4 E 1=04712855 E {5 .
So
I
100 E 1=04712855, to eight decimal places.
100
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NOT FOR SALE
SECTION 4.8 NEWTON’S METHOD
13. i ({) = {4 3 2{3 + 5{2 3 6
i i 0 ({) = 4{3 3 6{2 + 10{ i {q+1 = {q 3
¤
419
{4q 3 2{3q + 5{2q 3 6
. We need to ¿nd
4{3q 3 6{2q + 10{q
approximations until they agree to six decimal places. We’ll let {1 equal the midpoint of the given interval, [1> 2].
{1 = 1=5 i {2 = 1=2625, {3 E 1=218808, {4 E 1=217563, {5 E 1=217562 E {6 . So the root is 1=217562 to six decimal
places.
14. i ({) = 2=2{5 3 4=4{3 + 1=3{2 3 0=9{ 3 4=0
{q+1 = {q 3
i i 0 ({) = 11{4 3 13=2{2 + 2=6{ 3 0=9 i
2=2{5q 3 4=4{3q + 1=3{2q 3 0=9{q 3 4=0
. {1 = 31=5 i {2 E 31=425369, {3 E 31=405499,
11{4q 3 13=2{2q + 2=6{q 3 0=9
{4 E 31=404124, {5 E 31=404118 E {6 . So the root is 31=404118 to six decimal places.
h{ = 4 3 {2 , so i ({) = h{ 3 4 + {2
15.
i {q+1 = {q 3
h{q 3 4 + {2q
.
h{q + 2{q
From the ¿gure, the negative root of h{ = 4 3 {2 is near 32.
{1 = 32 i {2 E 31=964981, {3 E 31=964636 E {4 . So the negative
root is 31=964636, to six decimal places.
3 sin { = {, so i({) = 3 sin { 3 { i i 0 ({) = 3 cos { 3 1 i
16.
{q+1 = {q 3
3 sin {q 3 {q
. From the ¿gure, the positive root of
3 cos {q 3 1
3 sin { = { is near 2. {1 = 2 i {2 E 2=323732, {3 E 2=279595,
{4 E 2=278863 E {5 . So the positive root is 2=278863, to six decimal places.
17.
From the graph, we see that there appear to be points of intersection near
{ = 34, { = 32, and { = 1. Solving 3 cos { = { + 1 is the same as solving
i({) = 3 cos { 3 { 3 1 = 0. i 0 ({) = 33 sin { 3 1, so
{q+1 = {q 3
3 cos {q 3 {q 3 1
.
33 sin {q 3 1
{1 = 34
{1 = 32
{1 = 1
{2 E 33=682281
{2 E 31=856218
{2 E 0=892438
{3 E 33=638960
{3 E 31=862356
{3 E 0=889473
{4 E 33=637959
{4 E 31=862365 E {5
{4 E 0=889470 E {5
{5 E 33=637958 E {6
To six decimal places, the roots of the equation are 33=637958, 31=862365, and 0=889470.
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18.
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
From the graph, we see that there appear to be points of intersection near
I
{ = 30=5 and { = 2. Solving { + 1 = {2 3 { is the same as solving
I
1
3 2{ + 1, so
{ + 1 3 {2 + { = 0. i 0 ({) = I
2 {+1
I
{q + 1 3 {2q + {q
.
= {q 3
1
I
3 2{q + 1
2 {q + 1
i ({) =
{q+1
{1 = 30=5
{1 = 2
{2 E 30=484155
{2 E 1=901174
{3 E 30=484028 E {4
{3 E 1=897186
{4 E 1=897179 E {5
To six decimal places, the roots of the equation are 30=484028 and 0=897179.
19.
From the graph, we see that there appear to be points of intersection near
{ = 1=5 and { = 3. Solving ({ 3 2)2 = ln { is the same as solving
i ({) = ({ 3 2)2 3 ln { = 0. i ({) = ({ 3 2)2 3 ln { i
i 0 ({) = 2({ 3 2) 3 1@{, so {q+1 = {q 3
({q 3 2)2 3 ln {q
.
2({q 3 2) 3 1@{q
{1 = 1=5
{1 = 3
{2 E 1=406721
{2 E 3=059167
{3 E 1=412370
{3 E 3=057106
{4 E 1=412391 E {5
{4 E 3=057104 E {5
To six decimal places, the roots of the equation are 1=412391 and 3=057104.
20.
From the graph, we see that there appear to be points of intersection near
{ = 31=2 and { = 0=8. Solving
i ({) =
1
= 1 + {3 is the same as solving
{
1
1
3 1 3 {3 = 0. i ({) = 3 1 3 {3
{
{
{q+1 = {q 3
i i 0 ({) = 3
1
3 3{2 , so
{2
1@{q 3 1 3 {3q
.
31@{2q 3 3{2q
{1 = 31=2
{1 = 0=8
{2 E 31=221006
{2 E 0=724767
{3 E 31=220744 E {4
{3 E 0=724492 E {4
To six decimal places, the roots of the equation are 31=220744 and 0=724492.
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NOT FOR SALE
SECTION 4.8 NEWTON’S METHOD
21.
¤
From the graph, we see that the graphs intersect at 0 and at { = ± d, where
d E 1. [Both functions are odd, so the roots are negatives of each other.]
Solving {3 = tan31 { is the same as solving i ({) = {3 3 tan31 { = 0.
i 0 ({) = 3{2 3
1
{3q 3 tan31 {q
,
so
{
=
{
3
.
q+1
q
1
1 + {2
3{2q 3
2
1 + {q
Now {1 = 1 i {2 E 0=914159, {3 E 0=902251, {4 E 0=902026, {5 E 0=902025 E {6 . To six decimal places, the
nonzero roots of the equation are ±0=902025.
22.
From the graph, we see that there appear to be points of intersection near
{ = 31 and { = 2. Solving sin { = {2 3 2 is the same as solving
i ({) = sin { 3 {2 + 2 = 0. i 0 ({) = cos { 3 2{, so
{q+1 = {q 3
sin {q 3 {2q + 2
.
cos {q 3 2{q
{1 = 31
{1 = 2
{2 E 31=062406
{2 E 1=753019
{3 E 31=061550 E {4
{3 E 1=728710
{4 E 1=728466 E {5
To six decimal places, the roots of the equation are 31=061550 and 1=728466.
i({) = {6 3 {5 3 6{4 3 {2 + { + 10 i
23.
i 0 ({) = 6{5 3 5{4 3 24{3 3 2{ + 1 i
{q+1 = {q 3
{6q 3 {5q 3 6{4q 3 {2q + {q + 10
.
6{5q 3 5{4q 3 24{3q 3 2{q + 1
From the graph of i , there appear to be roots near 31=9, 31=2, 1=1, and 3.
{1 = 31=9
{1 = 31=2
{1 = 1=1
{1 = 3
{2 E 31=94278290
{2 E 31=22006245
{2 E 1=14111662
{2 E 2=99
{3 E 31=21997997 E {4
{3 E 1=13929741
{3 E 2=98984106
{4 E 1=13929375 E {5
{4 E 2=98984102 E {5
{3 E 31=93828380
{4 E 31=93822884
{5 E 31=93822883 E {6
To eight decimal places, the roots of the equation are 31=93822883, 31=21997997, 1=13929375, and 2=98984102.
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422
¤
NOT FOR SALE
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
i ({) = {5 3 3{4 + {3 3 {2 3 { + 6 i
24.
i 0 ({) = 5{4 3 12{3 + 3{2 3 2{ 3 1 i
{q+1 = {q 3
{5q 3 3{4q + {3q 3 {2q 3 {q + 6
. From the graph of i , there
5{4q 3 12{3q + 3{2q 3 2{q 3 1
appear to be roots near 31, 1=3, and 2=7.
{1 = 31
{1 = 1=3
{1 = 2=7
{2 E 31=04761905
{2 E 1=33313045
{2 E 2=70556135
{3 E 31=04451724
{3 E 1=33258330
{4 E 31=04450307 E {5
{3 E 2=70551210
{4 E 1=33258316 E {5
{4 E 2=70551209 E {5
To eight decimal places, the roots of the equation are 31=04450307, 1=33258316, and 2=70551209.
25.
Solving
I
1 3 {2
1
{
3 1 3 { = 0. i 0 ({) = 2
+ I
+1
({ + 1)2
2 13{
I
{q
3 1 3 {q
{2q + 1
= {q 3
.
1 3 {2q
1
I
+
2 1 3 {q
({2q + 1)2
i ({) =
{q+1
I
{
= 1 3 { is the same as solving
{2 + 1
{2
i
From the graph, we see that the curves intersect at about 0=8. {1 = 0=8 i {2 E 0=76757581, {3 E 0=76682610,
{4 E 0=76682579 E {5 . To eight decimal places, the root of the equation is 0=76682579.
26.
Solving cos({2 3 {) = {4 is the same as solving
i ({) = cos({2 3 {) 3 {4 = 0. i 0 ({) = 3(2{ 3 1) sin({2 3 {) 3 4{3
{q+1 = {q 3
cos({2q
i
{4q
3 {q ) 3
. From the equations
3(2{q 3 1) sin({2q 3 {q ) 3 4{3q
| = cos({2 3 {) and | = {4 and the graph, we deduce that one root of the
equation cos({2 3 {) = {4 is { = 1. We also see that the graphs intersect at
approximately { = 30=7. {1 = 30=7 i {2 E 30=73654354, {3 E 30=73486274, {4 E 30=73485910 E {5 .
To eight decimal places, one root of the equation is 30=73485910; the other root is 1.
27.
2
Solving 4h3{ sin { = {2 3 { + 1 is the same as solving
2
i ({) = 4h3{ sin { 3 {2 + { 3 1 = 0.
2
i 0 ({) = 4h3{ (cos { 3 2{ sin {) 3 2{ + 1 i
2
{q+1 = {q 3
4h3{q sin {q 3 {2q + {q 3 1
.
(cos {q 3 2{q sin {q ) 3 2{q + 1
3{2
q
4h
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SECTION 4.8 NEWTON’S METHOD
¤
423
From the ¿gure, we see that the graphs intersect at approximately { = 0=2 and { = 1=1.
{1 = 0=2
{1 = 1=1
{2 E 0=21883273
{2 E 1=08432830
{3 E 0=21916357
{3 E 1=08422462 E {4
{4 E 0=21916368 E {5
To eight decimal places, the roots of the equation are 0=21916368 and 1=08422462.
I
From the equations | = harctan { and | = {3 + 1 and the graph, we deduce
I
that one root of the equation harctan { = {3 + 1 is { = 0. We also see that the
28.
graphs intersect at approximately { = 30=9 and { = 2.
I
harctan {
3{2
{3 + 1 i i 0 ({) =
3 I
, so
1 + {2
2 {3 + 1
I
harctan {q 3 {3q + 1
.
= {q 3 arctan {q
h
3{2q
I
3
1 + {2q
2 {3q + 1
i ({) = harctan { 3
{q+1
{1 = 30=9
{1 = 2
{2 E 30=91934894
{2 E 2=01843839
{3 E 30=91798835
{3 E 2=01831502
{4 E 30=91798012 E {5
{4 E 2=01831501 E {5
To eight decimal places, the nonzero roots of the equation are 30=91798012 and 2=01831501.
29. (a) i ({) = {2 3 d
{q+1
i i 0 ({) = 2{, so Newton’s method gives
{2q 3 d
1
d
1
d
1
d
= {q 3
= {q 3 {q +
= {q +
=
.
{q +
2{q
2
2{q
2
2{q
2
{q
(b) Using (a) with d = 1000 and {1 =
I
So 1000 E 31=622777.
30. (a) i ({) =
I
900 = 30, we get {2 E 31=666667, {3 E 31=622807, and {4 E 31=622777 E {5 .
1
1
1@{q 3 d
3 d i i 0 ({) = 3 2 , so {q+1 = {q 3
= {q + {q 3 d{2q = 2{q 3 d{2q .
{
{
31@{2q
(b) Using (a) with d = 1=6894 and {1 =
1
2
= 0=5, we get {2 = 0=5754, {3 E 0=588485, and {4 E 0=588789 E {5 .
So 1@1=6984 E 0=588789.
31. i ({) = {3 3 3{ + 6
i i 0 ({) = 3{2 3 3. If {1 = 1, then i 0 ({1 ) = 0 and the tangent line used for approximating {2 is
horizontal. Attempting to ¿nd {2 results in trying to divide by zero.
32. {3 3 { = 1
C {3 3 { 3 1 = 0. i ({) = {3 3 { 3 1 i i 0 ({) = 3{2 3 1, so {q+1 = {q 3
{3q 3 {q 3 1
.
3{2q 3 1
(a) {1 = 1, {2 = 1=5, {3 E 1=347826, {4 E 1=325200, {5 E 1=324718 E {6
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NOT FOR SALE
¤
CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
(b) {1 = 0=6, {2 = 17=9, {3 E 11=946802, {4 E 7=985520, {5 E 5=356909, {6 E 3=624996, {7 E 2=505589,
{8 E 1=820129, {9 E 1=461044, {10 E 1=339323, {11 E 1=324913, {12 E 1=324718 E {13
(c) {1 = 0=57, {2 E 354=165455, {3 E 336=114293, {4 E 324=082094, {5 E 316=063387, {6 E 310=721483,
{7 E 37=165534, {8 E 34=801704, {9 E 33=233425, {10 E 32=193674, {11 E 31=496867, {12 E 30=997546,
{13 E 30=496305, {14 E 32=894162, {15 E 31=967962, {16 E 31=341355, {17 E 30=870187, {18 E 30=249949,
{19 E 31=192219, {20 E 30=731952, {21 E 0=355213, {22 E 31=753322, {23 E 31=189420, {24 E 30=729123,
{25 E 0=377844, {26 E 31=937872, {27 E 31=320350, {28 E 30=851919, {29 E 30=200959, {30 E 31=119386,
{31 E 30=654291, {32 E 1=547010, {33 E 1=360051, {34 E 1=325828, {35 E 1=324719, {36 E 1=324718 E {37 .
From the ¿gure, we see that the tangent line corresponding to {1 = 1 results
(d)
in a sequence of approximations that converges quite quickly ({5 E {6 ).
The tangent line corresponding to {1 = 0=6 is close to being horizontal, so
{2 is quite far from the root. But the sequence still converges — just a little
more slowly ({12 E {13 ). Lastly, the tangent line corresponding to
{1 = 0=57 is very nearly horizontal, {2 is farther away from the root, and
the sequence takes more iterations to converge ({36 E {37 )=
33. For i ({) = {1@3 , i 0 ({) =
1 32@3
{
3
and
1@3
{q+1 = {q 3
i ({q )
{q
= {q 3 1 32@3 = {q 3 3{q = 32{q .
i 0 ({q )
{
3 q
Therefore, each successive approximation becomes twice as large as the
previous one in absolute value, so the sequence of approximations fails to
converge to the root, which is 0. In the ¿gure, we have {1 = 0=5,
{2 = 32(0=5) = 31, and {3 = 32(31) = 2.
34. According to Newton’s Method, for {q A 0,
{q+1
I
{q
= {q 3
I = {q 3 2{q = 3{q and for {q ? 0,
1@ 2 {q
{q+1 = {q 3
I
3 3 {q
= {q 3 [32(3{q )] = 3{q . So we can see that
I
1@ 2 3{q
after choosing any value {1 the subsequent values will alternate between 3{1
and {1 and never approach the root.
35. (a) i ({) = {6 3 {4 + 3{3 3 2{
i i 0 ({) = 6{5 3 4{3 + 9{2 3 2 i
i 00 ({) = 30{4 3 12{2 + 18{. To ¿nd the critical numbers of i , we’ll ¿nd the
zeros of i 0 . From the graph of i 0 , it appears there are zeros at approximately
{ = 31=3, 30=4, and 0=5. Try {1 = 31=3 i
{2 = {1 3
i 0 ({1 )
E 31=293344 i {3 E 31=293227 E {4 .
i 00 ({1 )
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SECTION 4.8 NEWTON’S METHOD
¤
425
Now try {1 = 30=4 i {2 E 30=443755 i {3 E 30=441735 i {4 E 30=441731 E {5 . Finally try
{1 = 0=5 i {2 E 0=507937 i {3 E 0=507854 E {4 . Therefore, { = 31=293227, 30=441731, and 0=507854 are
all the critical numbers correct to six decimal places.
(b) There are two critical numbers where i 0 changes from negative to positive, so i changes from decreasing to increasing.
i (31=293227) E 32=0212 and i (0=507854) E 30=6721, so 32=0212 is the absolute minimum value of i correct to four
decimal places.
36. i ({) = { cos {
i i 0 ({) = cos { 3 { sin {. i 0 ({) exists for all {, so to ¿nd
the maximum of i , we can examine the zeros of i 0 . From the graph of i 0 , we
see that a good choice for {1 is {1 = 0=9. Use j({) = cos { 3 { sin { and
j 0 ({) = 32 sin { 3 { cos { to obtain {2 E 0=860781, {3 E 0=860334 E {4 .
Now we have i (0) = 0, i () = 3, and i (0=860334) E 0=561096, so
0=561096 is the absolute maximum value of i correct to six decimal places.
37.
| = {2 sin { i | 0 = {2 cos { + (sin {)(2{) i
| 00 = {2 (3 sin {) + (cos {)(2{) + (sin {)(2) + 2{ cos {
= 3{2 sin { + 4{ cos { + 2 sin {
i
| 000 = 3{2 cos { + (sin {)(32{) + 4{(3 sin {) + (cos {)(4) + 2 cos {
= 3{2 cos { 3 6{ sin { + 6 cos {.
From the graph of | = {2 sin {, we see that { = 1=5 is a reasonable guess for the {-coordinate of the inÀection point. Using
Newton’s method with j({) = | 00 and j0 ({) = | 000 , we get {1 = 1=5 i {2 E 1=520092, {3 E 1=519855 E {4 .
The inÀection point is about (1=519855> 2=306964).
38.
i ({) = 3 sin { i i 0 ({) = 3 cos {. At { = d, the slope of the tangent
line is i 0 (d) = 3 cos d. The line through the origin and (d> i (d)) is
|=
3 sin d 3 0
{. If this line is to be tangent to i at { = d, then its slope
d30
must equal i 0 (d). Thus,
3 sin d
= 3 cos d i tan d = d.
d
To solve this equation using Newton’s method, let j({) = tan { 3 {, j0 ({) = sec2 { 3 1, and {q+1 = {q 3
tan {q 3 {q
sec2 {q 3 1
with {1 = 4=5 (estimated from the ¿gure). {2 E 4=493614, {3 E 4=493410, {4 E 4=493409 E {5 . Thus, the slope of the
line that has the largest slope is i 0 ({5 ) E 0=217234.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
39. We need to minimize the distance from (0> 0) to an arbitrary point ({> |) on the
s
curve | = ({ 3 1)2 . g = {2 + | 2 i
s
s
g({) = {2 + [({ 3 1)2 ]2 = {2 + ({ 3 1)4 . When g0 = 0, g will be
minimized and equivalently, v = g2 will be minimized, so we will use Newton’s
method with i = v0 and i 0 = v00 .
i ({) = 2{ + 4({ 3 1)3
i i 0 ({) = 2 + 12({ 3 1)2 , so {q+1 = {q 3
2{q + 4({q 3 1)3
. Try {1 = 0=5 i
2 + 12({q 3 1)2
{2 = 0=4, {3 E 0=410127, {4 E 0=410245 E {5 . Now g(0=410245) E 0=537841 is the minimum distance and the point on
the parabola is (0=410245> 0=347810), correct to six decimal places.
40. Let the radius of the circle be u. Using v = u, we have 5 = u and so u = 5@. From the Law of Cosines we get
42 = u2 + u2 3 2 · u · u · cos C 16 = 2u2 (1 3 cos ) = 2(5@)2 (1 3 cos ). Multiplying by 2 gives
162 = 50(1 3 cos ), so we take i () = 162 + 50 cos 3 50 and i 0 () = 32 3 50 sin . The formula
for Newton’s method is q+1 = q 3
162q + 50 cos q 3 50
. From the graph
32q 3 50 sin q
of i , we can use 1 = 2=2, giving us 2 E 2=2662, 3 E 2=2622 E 4 . So
correct to four decimal places, the angle is 2=2622 radians E 130 .
41. In this case, D = 18,000, U = 375, and q = 5(12) = 60. So the formula D =
18,000 =
375
[1 3 (1 + {)360 ] C 48{ = 1 3 (1 + {)360
{
U
[1 3 (1 + l)3q ] becomes
l
[multiply each term by (1 + {)60 ] C
48{(1 + {)60 3 (1 + {)60 + 1 = 0. Let the LHS be called i ({), so that
i 0 ({) = 48{(60)(1 + {)59 + 48(1 + {)60 3 60(1 + {)59
= 12(1 + {)59 [4{(60) + 4(1 + {) 3 5] = 12(1 + {)59 (244{ 3 1)
{q+1 = {q 3
48{q (1 + {q )60 3 (1 + {q )60 + 1
. An interest rate of 1% per month seems like a reasonable estimate for
12(1 + {q )59 (244{q 3 1)
{ = l. So let {1 = 1% = 0=01, and we get {2 E 0=0082202, {3 E 0=0076802, {4 E 0=0076291, {5 E 0=0076286 E {6 .
Thus, the dealer is charging a monthly interest rate of 0=76286% (or 9=55% per year, compounded monthly).
42. (a) s({) = {5 3 (2 + u){4 + (1 + 2u){3 3 (1 3 u){2 + 2(1 3 u){ + u 3 1
i
s0 ({) = 5{4 3 4(2 + u){3 + 3(1 + 2u){2 3 2(1 3 u){ + 2(1 3 u). So we use
{q+1 = {q 3
{5q 3 (2 + u){4q + (1 + 2u){3q 3 (1 3 u){2q + 2(1 3 u){q + u 3 1
.
5{4q 3 4(2 + u){3q + 3(1 + 2u){2q 3 2(1 3 u){q + 2(1 3 u)
We substitute in the value u E 3=04042 × 1036 in order to evaluate the approximations numerically. The libration point
O1 is slightly less than 1 AU from the sun, so we take {1 = 0=95 as our ¿rst approximation, and get {2 E 0=96682,
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SECTION 4.9
ANTIDERIVATIVES
¤
427
{3 E 0=97770, {4 E 0=98451, {5 E 0=98830, {6 E 0=98976, {7 E 0=98998, {8 E 0=98999 E {9 . So, to ¿ve decimal
places, O1 is located 0=98999 AU from the sun (or 0=01001 AU from the earth).
(b) In this case we use Newton’s method with the function
s({) 3 2u{2 = {5 3 (2 + u){4 + (1 + 2u){3 3 (1 + u){2 + 2(1 3 u){ + u 3 1 i
0
s({) 3 2u{2 = 5{4 3 4(2 + u){3 + 3(1 + 2u){2 3 2(1 + u){ + 2(1 3 u). So
{5q 3 (2 + u){4q + (1 + 2u){3q 3 (1 + u){2q + 2(1 3 u){q + u 3 1
. Again, we substitute
5{4q 3 4(2 + u){3q + 3(1 + 2u){2q 3 2(1 + u){q + 2(1 3 u)
{q+1 = {q 3
u E 3=04042 × 1036 . O2 is slightly more than 1 AU from the sun and, judging from the result of part (a), probably less
than 0=02 AU from earth. So we take {1 = 1=02 and get {2 E 1=01422, {3 E 1=01118, {4 E 1=01018,
{5 E 1=01008 E {6 . So, to ¿ve decimal places, O2 is located 1=01008 AU from the sun (or 0=01008 AU from the earth).
4.9 Antiderivatives
1. i ({) = { 3 3 = {1 3 3
i I ({) =
{1+1
3 3{ + F = 12 {2 3 3{ + F
1+1
Check: I 0 ({) = 12 (2{) 3 3 + 0 = { 3 3 = i({)
2. i ({) =
1 2
2{
3. i ({) =
1
2
3 2{ + 6 i I ({) =
+ 34 {2 3 45 {3
Check: I 0 ({) =
1
2
{2
1 {3
32
+ 6{ + F = 16 {3 3 {2 + 6{ + F
2 3
2
i I ({) = 12 { +
3 {2+1
4 {3+1
3
+ F = 12 { + 14 {3 3 15 {4 + F
4 2+1
5 3+1
+ 14 (3{2 ) 3 15 (4{3 ) + 0 =
4. i ({) = 8{9 3 3{6 + 12{3
i I ({) = 8
5. i ({) = ({ + 1)(2{ 3 1) = 2{2 + { 3 1
2
1
2
+ 34 {2 3 45 {3 = i ({)
1 10
{
10
33
1
i I ({) = 2
7
1
3
6. i ({) = { (2 3 {) = { 4 3 4{ + {2 = 4{ 3 4{2 + {3
I ({) = 4
1
2
2
{
34
1
3
{
7. i ({) = 7{2@5 + 8{34@5
3=4
8. i ({) = {
9. i ({) =
I
231
3 2{
3
4
2
3
{7 + 12 14 {4 + F = 45 {10 3 37 {7 + 3{4 + F
{3 + 12 {2 3 { + F = 23 {3 + 12 {2 3 { + F
i
4
+ 14 { + F = 2{ 3 43 { + 14 { + F
i I ({) = 7 57 {7@5 + 8(5{1@5 ) + F = 5{7@5 + 40{1@5 + F
{4=4
32
i I ({) =
4=4
#
I
{ 2
I
2
$
+F =
5 4=4 I I2
{ 3 2{ +F
22
I
I
2 is a constant function, so I ({) = 2 { + F.
10. i ({) = h2 is a constant function, so I ({) = h2 { + F.
I
I
11. i ({) = 3 { 3 2 3 { = 3{1@2 3 2{1@3
12. i ({) =
I
I
3
{2 + { { = {2@3 + {3@2
i I ({) = 3 23 {3@2 3 2 34 {4@3 + F = 2{3@2 3 32 {4@3 + F
i I ({) = 35 {5@3 + 25 {5@2 + F
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
+1
{ 3 2 ln |{| + F1
1
2
1
1
5
has domain (3"> 0) (0> "), so I ({) = 1
13. i ({) = 3 = 3 2
5
{
5
{
{ 3 2 ln |{| + F2
5
if { ? 0
if { A 0
See Example 1(b) for a similar problem.
;
6
A
?3w 3 ln |w| 3 + F1
3w4 3 w3 + 6w2
1
6
w
14. i (w) =
= 3 3 + 2 has domain (3"> 0) (0> "), so I (w) =
A
w4
w
w
=3w 3 ln |w| 3 6 + F2
w
See Example 1(b) for a similar problem.
15. j(w) =
1 + w + w2
I
= w31@2 + w1@2 + w3@2
w
16. u() = sec tan 3 2h
17. k() = 2 sin 3 sec2 if w A 0
i J(w) = 2w1@2 + 23 w3@2 + 25 w5@2 + F
i U() = sec 3 2h + Fq on the interval q 3 2 > q + 2 .
i K() = 32 cos 3 tan + Fq on the interval q 3 2 > q + 2 .
18. i (w) = sin w + 2 sinh w
i I (w) = 3 cos w + 2 cosh w + F
19. i ({) = 5h{ 3 3 cosh {
i I ({) = 5h{ 3 3 sinh { + F
20. i ({) = 2
if w ? 0
3@2 I
{
+ 6 sin { + F = 43 {3@2 + 6 sin { + F
{ + 6 cos { = 2{1@2 + 6 cos { i I ({) = 2
3@2
2
{5 3 {3 + 2{
1
1
= { 3 + 3 = { 3 + 2{33 i
{4
{
{
{
33+1 {
{2
1
I ({) =
3 ln |{| + 2
+ F = 12 {2 3 ln |{| 3 2 + F
2
33 + 1
{
21. i ({) =
22. i ({) =
2 + {2
1 + (1 + {2 )
1
=
=
+ 1 i I ({) = tan31 { + { + F
2
1+{
1 + {2
1 + {2
23. i ({) = 5{4 3 2{5
i I ({) = 5 ·
I (0) = 4 i 05 3
1
3
{6
{5
32·
+ F = {5 3 13 {6 + F.
5
6
· 06 + F = 4
i
F = 4, so I ({) = {5 3 13 {6 + 4.
The graph con¿rms our answer since i ({) = 0 when I has a local maximum, i is
positive when I is increasing, and i is negative when I is decreasing.
24. i ({) = 4 3 3 1 + {2
31
=43
3
1 + {2
i I ({) = 4{ 3 3 tan31 { + F.
I (1) = 0 i 4 3 3 4 + F = 0 i F =
I ({) = 4{ 3 3 tan31 { +
3
4
3
4
3 4, so
3 4. Note that i is positive and I is increasing on R.
Also, i has smaller values where the slopes of the tangent lines of I are smaller.
3
2
4
{
{
{
3 12
+6
+ F = 5{4 3 4{3 + 3{2 + F
i i 0 ({) = 20
4
3
2
4
3
5
{
{
{
34
+3
+ F{ + G = {5 3 {4 + {3 + F{ + G
i ({) = 5
5
4
3
25. i 00 ({) = 20{3 3 12{2 + 6{
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SECTION 4.9
26. i 00 ({) = {6 3 4{4 + { + 1
i ({) =
1 8
56 {
27. i 00 ({) =
3
2 6
15 {
i i 0 ({) =
28. i 00 ({) = 6{ + sin {
3 45 {5 + 12 {2 + { + F
¤
i
2
3
{5@3
5@3
+ F = 25 {5@3 + F
i i ({) =
2
5
2
{
i i 0 ({) = 6
3 cos { + F = 3{2 3 cos { + F
2
{8@3
8@3
+ F{ + G =
3 8@3
20 {
+ F{ + G
i
3
{
i ({) = 3
3 sin { + F{ + G = {3 3 sin { + F{ + G
3
29. i 000 (w) = cos w
i i 00 (w) = sin w + F1
i i 0 (w) = 3 cos w + F1 w + G
i i (w) = 3 sin w + Fw2 + Gw + H,
where F = 12 F1 .
000
34
w
30. i (w) = h + w
0
i w) =
00
has domain (3"> 0) (0> ") i i (w) =
+ w 1 32
h + 6 w + F1 w + G1
w
h +
1 32
6w
I
31. i 0 ({) = 1 + 3 {
+ F2 w + G2
if w ? 0
if w A 0
i i (w) =
+ w 1 33
h 3 3 w + F1
w
h 3
1 33
w
3
+ F2
if w ? 0
if w A 0
+ w 1 31 1
h 3 6 w + 2 F1 w2 + G1 w + H1
w
h 3
1 31
6w
+
2
1
2 F2 w
+ G2 w + H2
i
if w ? 0
if w A 0
i i ({) = { + 3 23 {3@2 + F = { + 2{3@2 + F. i (4) = 4 + 2(8) + F and i(4) = 25 i
20 + F = 25 i F = 5, so i ({) = { + 2{3@2 + 5.
32. i 0 ({) = 5{4 3 3{2 + 4
i i({) = {5 3 {3 + 4{ + F. i (31) = 31 + 1 3 4 + F and i (31) = 2 i
34 + F = 2 i F = 6, so i ({) = {5 3 {3 + 4{ + 6.
33. i 0 (w) =
4
1 + w2
+ F and i (1) = 0 i + F = 0
i i(w) = 4 arctan w + F. i (1) = 4
4
i
F = 3,
so i (w) = 4 arctan w 3 .
34. i 0 (w) = w +
i (w) =
1
1
1
1
1
, w A 0 i i (w) = w2 3 2 + F. i (1) = 3 + F and i (1) = 6 i F = 6, so
w3
2
2w
2
2
1 2
1
w 3 2 + 6.
2
2w
35. i 0 (w) = 2 cos w + sec2 w
i
3
i i (w) = 2 sin w + tan w + F because 3@2 ? w ? @2.
I
I
I
I
I
= 2 3@2 + 3 + F = 2 3 + F and i 3 = 4 i F = 4 3 2 3, so i (w) = 2 sin w + tan w + 4 3 2 3.
1
{2 3 1
= { 3 has domain (3"> 0) (0> ") i i ({) =
36. i ({) =
{
{
0
i (1) =
1
2
3 ln 1 + F1 =
1
2
+ F1 and i (1) =
1
2
+1
2
2{
1 2
2{
3 ln { + F1
if { A 0
3 ln(3{) + F2
if { ? 0
i F1 = 0.
3 ln 1 + F2 = 12 + F2 and i (31) = 0 i F2 = 3 12 .
+1 2
{ 3 ln {
if { A 0
2
Thus, i({) = 1 2
1
{ 3 ln(3{) 3 2 if { ? 0
2
i (31) =
429
+ 16 {3 + 12 {2 + F{ + G
i i 0 ({) =
2 2@3
3{
1 7
7{
ANTIDERIVATIVES
1
2
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CHAPTER 4
0
31@3
37. i ({) = {
APPLICATIONS OF DIFFERENTIATION
has domain (3"> 0) (0> ") i i ({) =
+ F1 and i (1) = 1 i F1 = 3 12 . i(31) =
+3 2@3 1
3 2 if { A 0
2{
Thus, i({) =
3 2@3
{
3 52 if { ? 0
2
i (1) =
3
2
I
38. i 0 ({) = 4@ 1 3 {2
2
3
i i ({) = 4 sin31 { + F. i
+F =1 i F =13
39. i 00 ({) = 32 + 12{ 3 12{2
2
,
3
31
so i ({) = 4 sin
1
2
3
2
+3
2@3
2{
+ F1
if { A 0
3 2@3
{
2
+ F2
if { ? 0
+ F2 and i (31) = 31 i F2 = 3 52 .
= 4 sin31
{+13
2
.
3
1
2
+F = 4·
6
+ F and i
1
2
=1 i
i i 0 ({) = 32{ + 6{2 3 4{3 + F. i 0 (0) = F and i 0 (0) = 12 i F = 12, so
i 0 ({) = 32{ + 6{2 3 4{3 + 12 and hence, i ({) = 3{2 + 2{3 3 {4 + 12{ + G. i (0) = G and i(0) = 4 i G = 4,
so i ({) = 3{2 + 2{3 3 {4 + 12{ + 4.
40. i 00 ({) = 8{3 + 5
i i 0 ({) = 2{4 + 5{ + F. i 0 (1) = 2 + 5 + F and i 0 (1) = 8 i F = 1, so
i 0 ({) = 2{4 + 5{ + 1. i({) = 25 {5 + 52 {2 + { + G. i(1) =
so i ({) = 25 {5 + 52 {2 + { 3
41. i 00 () = sin + cos 2
5
+
5
2
+1+G =G+
39
10
and i(1) = 0 i G = 3 39
10 ,
39
.
10
i i 0 () = 3 cos + sin + F. i 0 (0) = 31 + F and i 0 (0) = 4 i F = 5, so
i 0 () = 3 cos + sin + 5 and hence, i() = 3 sin 3 cos + 5 + G. i (0) = 31 + G and i (0) = 3 i G = 4,
so i () = 3 sin 3 cos + 5 + 4.
I
42. i 00 (w) = 3@ w = 3w31@2
i i 0 (w) = 6w1@2 + F. i 0 (4) = 12 + F and i 0 (4) = 7 i F = 35, so i 0 (w) = 6w1@2 3 5
and hence, i (w) = 4w3@2 3 5w + G. i(4) = 32 3 20 + G and i(4) = 20 i G = 8, so i (w) = 4w3@2 3 5w + 8.
43. i 00 ({) = 4 + 6{ + 24{2
i i 0 ({) = 4{ + 3{2 + 8{3 + F
i i ({) = 2{2 + {3 + 2{4 + F{ + G. i (0) = G and
i (0) = 3 i G = 3, so i ({) = 2{2 + {3 + 2{4 + F{ + 3. i(1) = 8 + F and i (1) = 10 i F = 2,
so i ({) = 2{2 + {3 + 2{4 + 2{ + 3.
44. i 00 ({) = {3 + sinh {
i i 0 ({) = 14 {4 + cosh { + F
i (0) = 1 i G = 1, so i({) =
1 5
{
20
i i ({) =
+ sinh { + F{ + 1. i (2) =
sinh 2 + 2F = 0 i F = 3 12 sinh 2, so i ({) =
1 5
{
20
1 5
{
20
32
20
+ sinh { + F{ + G. i(0) = G and
+ sinh 2 + 2F + 1 and i (2) = 2=6 i
+ sinh { 3 12 (sinh 2){ + 1.
45. i 00 ({) = 2 + cos {
i i 0 ({) = 2{ + sin { + F i i ({) = {2 3 cos { + F{ + G.
F = 3 2 @4 i
i (0) = 31 + G and i (0) = 31 i G = 0. i 2 = 2 @4 + 2 F and i 2 = 0 i
2
F = 3 2 , so i ({) = {2 3 cos { 3 2 {.
46. i 00 (w) = 2hw + 3 sin w
i i 0 (w) = 2hw 3 3 cos w + F
i i(w) = 2hw 3 3 sin w + Fw + G. i (0) = 2 + G and
i (0) = 0 i G = 32. i () = 2h + F 3 2 and i () = 0 i F = 2 3 2h
so i (w) = 2hw 3 3 sin w +
i F=
2 3 2h
,
2 3 2h
w 3 2.
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SECTION 4.9
47. i 00 ({) = {32 , { A 0
i i 0 ({) = 31@{ + F
ANTIDERIVATIVES
¤
431
i i ({) = 3 ln |{| + F{ + G = 3 ln { + F{ + G [since { A 0].
i (1) = 0 i F + G = 0 and i(2) = 0 i 3 ln 2 + 2F + G = 0 i 3 ln 2 + 2F 3 F = 0 [since G = 3F] i
3 ln 2 + F = 0 i F = ln 2 and G = 3 ln 2. So i ({) = 3 ln { + (ln 2){ 3 ln 2.
48. i 000 ({) = cos {
i i 00 ({) = sin { + F. i 00 (0) = F and i 00 (0) = 3 i F = 3. i 00 ({) = sin { + 3 i
i 0 ({) = 3 cos { + 3{ + G. i 0 (0) = 31 + G and i 0 (0) = 2 i G = 3. i 0 ({) = 3 cos { + 3{ + 3 i
i ({) = 3 sin { + 32 {2 + 3{ + H. i (0) = H and i (0) = 1 i H = 1. Thus, i ({) = 3 sin { + 32 {2 + 3{ + 1.
49. Given i 0 ({) = 2{ + 1, we have i ({) = {2 + { + F. Since i passes through (1> 6), i (1) = 6
i
12 + 1 + F = 6 i
F = 4. Therefore, i ({) = {2 + { + 4 and i(2) = 22 + 2 + 4 = 10.
50. i 0 ({) = {3
i i ({) = 14 {4 + F. { + | = 0 i | = 3{ i p = 31. Now p = i 0 ({) i 31 = {3
i
{ = 31 i | = 1 (from the equation of the tangent line), so (31> 1) is a point on the graph of i. From i ,
1 = 14 (31)4 + F
i F = 34 . Therefore, the function is i ({) = 14 {4 + 34 .
51. e is the antiderivative of i . For small {, i is negative, so the graph of its antiderivative must be decreasing. But both d and f
are increasing for small {, so only e can be i ’s antiderivative. Also, i is positive where e is increasing, which supports our
conclusion.
52. We know right away that f cannot be i ’s antiderivative, since the slope of f is not zero at the {-value where i = 0. Now i is
positive when d is increasing and negative when d is decreasing, so d is the antiderivative of i .
53.
The graph of I must start at (0> 1). Where the given graph, | = i ({), has a
local minimum or maximum, the graph of I will have an inÀection point.
Where i is negative (positive), I is decreasing (increasing).
Where i changes from negative to positive, I will have a minimum.
Where i changes from positive to negative, I will have a maximum.
Where i is decreasing (increasing), I is concave downward (upward).
54.
Where y is positive (negative), v is increasing (decreasing).
Where y is increasing (decreasing), v is concave upward (downward).
Where y is horizontal (a steady velocity), v is linear.
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CHAPTER 4
APPLICATIONS OF DIFFERENTIATION
55.
;
2
A
?
i 0 ({) = 1
A
=
31
if 0 $ { ? 1
if 1 ? { ? 2
if 2 ? { $ 3
;
2{ + F
A
?
i i ({) = { + G
A
=
3{ + H
if 0 $ { ? 1
if 1 ? { ? 2
if 2 ? { $ 3
i (0) = 31 i 2(0) + F = 31 i F = 31. Starting at the point
(0> 31) and moving to the right on a line with slope 2 gets us to the point (1> 1).
The slope for 1 ? { ? 2 is 1, so we get to the point (2> 2). Here we have used the fact that i is continuous. We can include the
point { = 1 on either the ¿rst or the second part of i . The line connecting (1> 1) to (2> 2) is | = {, so G = 0. The slope for
2 ? { $ 3 is 31, so we get to (3> 1).
i(3) = 1 i 33 + H = 1 i H = 4. Thus
;
2{ 3 1 if 0 $ { $ 1
A
?
if 1 ? { ? 2
i ({) = {
A
=
3{ + 4 if 2 $ { $ 3
Note that i 0 ({) does not exist at { = 1 or at { = 2.
56. (a)
(b) Since I (0) = 1, we can start our graph at (0> 1). i has a minimum at about
{ = 0=5, so its derivative is zero there. i is decreasing on (0> 0=5), so its
derivative is negative and hence, I is CD on (0> 0=5) and has an IP at
{ E 0=5. On (0=5> 2=2), i is negative and increasing (i 0 is positive), so I is
decreasing and CU. On (2=2> "), i is positive and increasing, so I is
increasing and CU.
(c) i ({) = 2{ 3 3
I
{ i I ({) = {2 3 3 · 23 {3@2 + F.
(d)
I (0) = F and I (0) = 1 i F = 1, so I ({) = {2 3 2{3@2 + 1.
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NOT FOR SALE
SECTION 4.9
57. i ({) =
ANTIDERIVATIVES
¤
433
sin {
, 32 $ { $ 2
1 + {2
Note that the graph of i is one of an odd function, so the graph of I will
be one of an even function.
58. i ({) =
I
{4 3 2{2 + 2 3 2, 33 $ { $ 3
Note that the graph of i is one of an even
function, so the graph of I will be one of an
odd function.
59. y(w) = v0 (w) = sin w 3 cos w
i v(w) = 3 cos w 3 sin w + F. v(0) = 31 + F and v(0) = 0 i F = 1, so
v(w) = 3 cos w 3 sin w + 1.
60. y(w) = v0 (w) = 1=5
I
w i v(w) = w3@2 + F. v(4) = 8 + F and v(4) = 10 i F = 2, so v(w) = w3@2 + 2.
61. d(w) = y 0 (w) = 2w + 1
i y(w) = w2 + w + F. y(0) = F and y(0) = 32 i F = 32, so y(w) = w2 + w 3 2 and
v(w) = 13 w3 + 12 w2 3 2w + G. v(0) = G and v(0) = 3 i G = 3, so v(w) = 13 w3 + 12 w2 3 2w + 3.
62. d(w) = y 0 (w) = 3 cos w 3 2 sin w
i y(w) = 3 sin w + 2 cos w + F. y(0) = 2 + F and y(0) = 4 i F = 2, so
y(w) = 3 sin w + 2 cos w + 2 and v(w) = 33 cos w + 2 sin w + 2w + G. v(0) = 33 + G and v(0) = 0 i G = 3,
so v(w) = 33 cos w + 2 sin w + 2w + 3.
63. d(w) = y 0 (w) = 10 sin w + 3 cos w
i y(w) = 310 cos w + 3 sin w + F
i v(w) = 310 sin w 3 3 cos w + Fw + G.
v(0) = 33 + G = 0 and v(2) = 33 + 2F + G = 12 i G = 3 and F =
v(w) = 310 sin w 3 3 cos w +
64. d(w) = w2 3 4w + 6
6
w
Thus,
+ 3.
i y(w) = 13 w3 3 2w2 + 6w + F
v(0) = 0 i G = 0. v(1) =
6
.
29
12
i v(w) =
+ F and v(1) = 20 i F =
1 4
12 w
211
.
12
3 23 w3 + 3w2 + Fw + G. v(0) = G and
Thus, v(w) =
1 4
w
12
3 23 w3 + 3w2 +
211
w.
12
65. (a) We ¿rst observe that since the stone is dropped 450 m above the ground, y(0) = 0 and v(0) = 450.
y0 (w) = d(w) = 39=8 i y(w) = 39=8w + F. Now y(0) = 0 i F = 0, so y(w) = 39=8w i
v(w) = 34=9w2 + G. Last, v(0) = 450 i G = 450 i v(w) = 450 3 4=9w2 .
(b) The stone reaches the ground when v(w) = 0. 450 3 4=9w2 = 0 i w2 = 450@4=9 i w1 =
s
(c) The velocity with which the stone strikes the ground is y(w1 ) = 39=8 450@4=9 E 393=9 m@s.
s
450@4=9 E 9=58 s.
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APPLICATIONS OF DIFFERENTIATION
(d) This is just reworking parts (a) and (b) with y(0) = 35. Using y(w) = 39=8w + F, y(0) = 35 i 0 + F = 35 i
y(w) = 39=8w 3 5. So v(w) = 34=9w2 3 5w + G and v(0) = 450 i G = 450 i v(w) = 34=9w2 3 5w + 450.
I
Solving v(w) = 0 by using the quadratic formula gives us w = 5 ± 8845 (39=8) i w1 E 9=09 s.
66. y 0 (w) = d(w) = d
i y(w) = dw + F and y0 = y(0) = F
v(w) = 12 dw2 + y0 w + G
i v0 = v(0) = G
i y(w) = dw + y0
i
i v(w) = 12 dw2 + y0 w + v0
67. By Exercise 66 with d = 39=8, v(w) = 34=9w2 + y0 w + v0 and y(w) = v0 (w) = 39=8w + y0 . So
[y(w)]2 = (39=8w + y0 )2 = (9=8)2 w2 3 19=6y0 w + y02 = y02 + 96=04w2 3 19=6y0 w = y02 3 19=6 34=9w2 + y0 w .
But 34=9w2 + y0 w is just v(w) without the v0 term; that is, v(w) 3 v0 . Thus, [y(w)]2 = y02 3 19=6 [v(w) 3 v0 ].
68. For the ¿rst ball, v1 (w) = 316w2 + 48w + 432 from Example 7. For the second ball, d(w) = 332
i y(w) = 332w + F, but
2
y(1) = 332(1) + F = 24 i F = 56, so y(w) = 332w + 56 i v(w) = 316w + 56w + G, but
v(1) = 316(1)2 + 56(1) + G = 432 i G = 392, and v2 (w) = 316w2 + 56w + 392. The balls pass each other
when v1 (w) = v2 (w) i 316w2 + 48w + 432 = 316w2 + 56w + 392 C 8w = 40 C w = 5 s.
Another solution: From Exercise 66, we have v1 (w) = 316w2 + 48w + 432 and v2 (w) = 316w2 + 24w + 432.
We now want to solve v1 (w) = v2 (w 3 1) i 316w2 + 48w + 432 = 316(w 3 1)2 + 24(w 3 1) + 432 i
48w = 32w 3 16 + 24w 3 24 i 40 = 8w i w = 5 s.
69. Using Exercise 66 with d = 332, y0 = 0, and v0 = k (the height of the cliff ), we know that the height at time w is
v(w) = 316w2 + k. y(w) = v0 (w) = 332w and y(w) = 3120 i 332w = 3120 i w = 3=75, so
0 = v(3=75) = 316(3=75)2 + k i k = 16(3=75)2 = 225 ft.
70. (a) HL| 00 = pj(O 3 {) + 12 j(O 3 {)2
3
HL| = 16 pj(O 3 {) +
1
24 j(O
i HL| 0 = 3 12 pj(O 3 {)2 3 16 j(O 3 {)3 + F
i
3 {) + F{ + G. Since the left end of the board is ¿xed, we must have | = | 0 = 0
4
1
when { = 0. Thus, 0 = 3 12 pjO2 3 16 jO3 + F and 0 = 16 pjO3 + 24
jO4 + G. It follows that
1
1
j(O 3 {)4 + 12 pjO2 + 16 jO3 { 3 16 pjO3 + 24
jO4 and
HL| = 16 pj(O 3 {)3 + 24
i ({) = | =
1 1
pj(O 3 {)3 +
HL 6
1
j(O
24
3 {)4 +
1
2
pjO2 + 16 jO3 { 3 16 pjO3 +
1
jO4
24
(b) i (O) ? 0, so the end of the board is a distance approximately 3i (O) below the horizontal. From our result in (a), we
calculate
3i (O) =
31 1
pjO3 + 16 jO4 3 16 pjO3 3
HL 2
Note: This is positive because j is negative.
71. Marginal cost = 1=92 3 0=002{ = F 0 ({)
1
jO4
24
=
31 1
jO3
pjO3 + 18 jO4 = 3
3
HL
HL
p O
+
3
8
i F({) = 1=92{ 3 0=001{2 + N. But F(1) = 1=92 3 0=001 + N = 562 i
N = 560=081. Therefore, F({) = 1=92{ 3 0=001{2 + 560=081 i F(100) = 742=081, so the cost of producing
100 items is $742=08.
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SECTION 4.9
72. Let the mass, measured from one end, be p({). Then p(0) = 0 and =
p(0) = F = 0, so p({) = 2
gp
= {31@2
g{
ANTIDERIVATIVES
¤
435
i p({) = 2{1@2 + F and
I
I
{. Thus, the mass of the 100-centimeter rod is p(100) = 2 100 = 20 g.
73. Taking the upward direction to be positive we have that for 0 $ w $ 10 (using the subscript 1 to refer to 0 $ w $ 10),
d1 (w) = 3 (9 3 0=9w) = y10 (w) i y1 (w) = 39w + 0=45w2 + y0 , but y1 (0) = y0 = 310 i
y1 (w) = 39w + 0=45w2 3 10 = v01 (w) i v1 (w) = 3 92 w2 + 0=15w3 3 10w + v0 . But v1 (0) = 500 = v0
i
v1 (w) = 3 92 w2 + 0=15w3 3 10w + 500. v1 (10) = 3450 + 150 3 100 + 500 = 100, so it takes
more than 10 seconds for the raindrop to fall. Now for w A 10, d(w) = 0 = y 0 (w) i
y(w) = constant = y1 (10) = 39(10) + 0=45(10)2 3 10 = 355 i y(w) = 355.
At 55 m@s, it will take 100@55 E 1=8 s to fall the last 100 m. Hence, the total time is 10 +
74. y 0 (w) = d(w) = 322. The initial velocity is 50 mi@h =
The car stops when y(w) = 0 C w =
2
+
v 10
= 311 10
3
3
220
3
10
3
·
=
1100
9
220
3 · 22
=
5280
3600
=
220
3
220
3
ft@s, so y(w) = 322w +
Since v(w) = 311w2 +
220
w,
3
130
11
E 11=8 s.
220
.
3
the distance covered is
5280
3600
= 44 ft@s, and the ¿nal velocity (after 5 seconds) is
ft@s. So y(w) = nw + F and y(0) = 44 i F = 44. Thus, y(w) = nw + 44 i
y(5) = 5n + 44. But y(5) =
76. d(w) = 316
=
=
= 122=2 ft.
75. d(w) = n, the initial velocity is 30 mi@h = 30 ·
50 mi@h = 50 ·
10
.
3
50 · 5280
3600
100
55
220
,
3
so 5n + 44 =
220
3
i
5n =
88
3
i
n=
88
15
E 5=87 ft@s2 .
i y(w) = 316w + y0 where y0 is the car’s speed (in ft@s) when the brakes were applied. The car stops when
316w + y0 = 0 C w =
1
y .
16 0
Now v(w) = 12 (316)w2 + y0 w = 38w2 + y0 w. The car travels 200 ft in the time that it takes
1 2
1 1 y0 = 200 i 200 = 38 16
y0 + y0 16
y0 =
to stop, so v 16
1 2
y
32 0
i y02 = 32 · 200 = 6400 i
y0 = 80 ft@s [54=54 mi@h].
77. Let the acceleration be d(w) = n km@h2 . We have y(0) = 100 km@h and we can take the initial position v(0) to be 0.
We want the time wi for which y(w) = 0 to satisfy v(w) ? 0=08 km. In general, y 0 (w) = d(w) = n, so y(w) = nw + F,
where F = y(0) = 100. Now v0 (w) = y(w) = nw + 100, so v(w) = 12 nw2 + 100w + G, where G = v(0) = 0.
Thus, v(w) = 12 nw2 + 100w. Since y(wi ) = 0, we have nwi + 100 = 0 or wi = 3100@n, so
2
1
100
100
1
1
5,000
+ 100 3
v(wi ) = n 3
= 10,000
3
=3
. The condition v(wi ) must satisfy is
2
n
n
2n
n
n
3
5,000
5,000
? 0=08 i 3
An
n
0=08
[n is negative] i n ? 362,500 km@h2 , or equivalently,
n ? 3 3125
E 34=82 m@s2 .
648
78. (a) For 0 $ w $ 3 we have d(w) = 60w
v(w) = 10w3 + F
i y(w) = 30w2 + F
i v(0) = 0 = F
i y(0) = 0 = F
i y(w) = 30w2 , so
i v(w) = 10w3 . Note that y(3) = 270 and v(3) = 270.
For 3 ? w $ 17: d(w) = 3j = 332 ft@s i y(w) = 332(w 3 3) + F
i y(3) = 270 = F
i
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APPLICATIONS OF DIFFERENTIATION
y(w) = 332(w 3 3) + 270 i v(w) = 316(w 3 3)2 + 270(w 3 3) + F
i v(3) = 270 = F
i
v(w) = 316(w 3 3)2 + 270(w 3 3) + 270. Note that y(17) = 3178 and v(17) = 914.
For 17 ? w $ 22: The velocity increases linearly from 3178 ft@s to 318 ft@s during this period, so
{y
318 3 (3178)
160
=
=
= 32. Thus, y(w) = 32(w 3 17) 3 178 i
{w
22 3 17
5
v(w) = 16(w 3 17)2 3 178(w 3 17) + 914 and v(22) = 424 ft.
For w A 22: y(w) = 318 i v(w) = 318(w 3 22) + F. But v(22) = 424 = F
i v(w) = 318(w 3 22) + 424.
Therefore, until the rocket lands, we have
; 2
30w
A
A
A
A
?332 (w 3 3) + 270
y(w) =
A32(w 3 17) 3 178
A
A
A
=
318
and
if 0 $ w $ 3
if 3 ? w $ 17
if 17 ? w $ 22
if w A 22
; 3
10w
A
A
A
A
?316(w 3 3)2 + 270(w 3 3) + 270
v(w) =
A
16(w 3 17)2 3 178 (w 3 17) + 914
A
A
A
=
318(w 3 22) + 424
if 0 $ w $ 3
if 3 ? w $ 17
if 17 ? w $ 22
if w A 22
(b) To ¿nd the maximum height, set y(w) on 3 ? w $ 17 equal to 0. 332(w 3 3) + 270 = 0 i w1 = 11=4375 s and the
maximum height is v(w1 ) = 316(w1 3 3)2 + 270(w1 3 3) + 270 = 1409=0625 ft.
(c) To ¿nd the time to land, set v(w) = 318(w 3 22) + 424 = 0. Then w 3 22 =
79. (a) First note that 90 mi@h = 90 ×
5280
3600
F = 0. Now 4w = 132 when w =
ft@s = 132 ft@s. Then d(w) = 4 ft@s2
132
4
424
18
= 23=5, so w E 45=6 s.
i y(w) = 4w + F, but y(0) = 0 i
= 33 s, so it takes 33 s to reach 132 ft@s. Therefore, taking v(0) = 0, we have
v(w) = 2w2 , 0 $ w $ 33. So v(33) = 2178 ft. 15 minutes = 15(60) = 900 s, so for 33 ? w $ 933 we have
y(w) = 132 ft@s i v(933) = 132(900) + 2178 = 120,978 ft = 22=9125 mi.
(b) As in part (a), the train accelerates for 33 s and travels 2178 ft while doing so. Similarly, it decelerates for 33 s and travels
2178 ft at the end of its trip. During the remaining 900 3 66 = 834 s it travels at 132 ft@s, so the distance traveled is
132 · 834 = 110,088 ft. Thus, the total distance is 2178 + 110,088 + 2178 = 114,444 ft = 21=675 mi.
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CHAPTER 4 REVIEW
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437
(c) 45 mi = 45(5280) = 237,600 ft. Subtract 2(2178) to take care of the speeding up and slowing down, and we have
233,244 ft at 132 ft@s for a trip of 233,244@132 = 1767 s at 90 mi@h. The total time is
1767 + 2(33) = 1833 s = 30 min 33 s = 30=55 min.
(d) 37=5(60) = 2250 s. 2250 3 2(33) = 2184 s at maximum speed. 2184(132) + 2(2178) = 292,644 total feet or
292,644@5280 = 55=425 mi.
4 Review
1. A function i has an absolute maximum at { = f if i (f) is the largest function value on the entire domain of i , whereas i has
a local maximum at f if i(f) is the largest function value when { is near f. See Figure 6 in Section 4.1.
2. (a) See the Extreme Value Theorem on page 275.
(b) See the Closed Interval Method on page 278.
3. (a) See Fermat’s Theorem on page 276.
(b) See the de¿nition of a critical number on page 277.
4. (a) See Rolle’s Theorem on page 284.
(b) See the Mean Value Theorem on page 285. Geometric interpretation—there is some point S on the graph of a function i
[on the interval (d> e)] where the tangent line is parallel to the secant line that connects (d> i (d)) and (e> i (e)).
5. (a) See the Increasing/Decreasing (I/D) Test on page 290.
(b) If the graph of i lies above all of its tangents on an interval L, then it is called concave upward on L.
(c) See the Concavity Test on page 293.
(d) An inÀection point is a point where a curve changes its direction of concavity. They can be found by determining the points
at which the second derivative changes sign.
6. (a) See the First Derivative Test on page 291.
(b) See the Second Derivative Test on page 295.
(c) See the note before Example 7 in Section 4.3.
7. (a) See l’Hospital’s Rule and the three notes that follow it in Section 4.4.
(b) Write i j as
i
j
or
.
1@j
1@i
(c) Convert the difference into a quotient using a common denominator, rationalizing, factoring, or some other method.
(d) Convert the power to a product by taking the natural logarithm of both sides of | = i j or by writing i j as hj ln i .
8. Without calculus you could get misleading graphs that fail to show the most interesting features of a function.
See Example 1 in Section 4.6.
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
9. (a) See Figure 3 in Section 4.8.
(b) {2 = {1 3
i({1 )
i 0 ({1 )
(c) {q+1 = {q 3
i ({q )
i 0 ({q )
(d) Newton’s method is likely to fail or to work very slowly when i 0 ({1 ) is close to 0. It also fails when i 0 ({l ) is unde¿ned,
such as with i ({) = 1@{ 3 2 and {1 = 1.
10. (a) See the de¿nition at the beginning of Section 4.9.
(b) If I1 and I2 are both antiderivatives of i on an interval L, then they differ by a constant.
1. False.
For example, take i ({) = {3 , then i 0 ({) = 3{2 and i 0 (0) = 0, but i (0) = 0 is not a maximum or minimum;
(0> 0) is an inÀection point.
2. False.
For example, i({) = |{| has an absolute minimum at 0, but i 0 (0) does not exist.
3. False.
For example, i({) = { is continuous on (0> 1) but attains neither a maximum nor a minimum value on (0> 1).
Don’t confuse this with i being continuous on the closed interval [d> e], which would make the statement true.
0
i(1) 3 i (31)
= = 0. Note that |f| ? 1 C f M (31> 1).
1 3 (31)
2
4. True.
By the Mean Value Theorem, i 0 (f) =
5. True.
This is an example of part (b) of the I/D Test.
6. False.
For example, the curve | = i ({) = 1 has no inÀection points but i 00 (f) = 0 for all f.
7. False.
i 0 ({) = j 0 ({) i i ({) = j({) + F. For example, if i({) = { + 2 and j({) = { + 1, then i 0 ({) = j 0 ({) = 1,
but i({) 6= j({).
8. False.
Assume there is a function i such that i (1) = 32 and i (3) = 0. Then by the Mean Value Theorem there exists a
number f M (1> 3) such that i 0 (f) =
9. True.
10. False.
i (3) 3 i (1)
0 3 (32)
=
= 1. But i 0 ({) A 1 for all {, a contradiction.
331
2
The graph of one such function is sketched.
At any point (d> i (d)), we know that i 0 (d) ? 0. So since the tangent line at (d> i (d)) is not horizontal, it must cross
the {-axis—at { = e, say. But since i 00 ({) A 0 for all {, the graph of i must lie above all of its tangents;
in particular, i (e) A 0. But this is a contradiction, since we are given that i({) ? 0 for all {.
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CHAPTER 4 REVIEW
11. True.
¤
439
Let {1 ? {2 where {1 > {2 M L. Then i ({1 ) ? i({2 ) and j({1 ) ? j({2 ) [since i and j are increasing on L ],
so (i + j)({1 ) = i ({1 ) + j({1 ) ? i ({2 ) + j({2 ) = (i + j)({2 ).
12. False.
i ({) = { and j({) = 2{ are both increasing on (0> 1), but i({) 3 j({) = 3{ is not increasing on (0> 1).
13. False.
Take i ({) = { and j({) = { 3 1. Then both i and j are increasing on (0> 1). But i ({) j({) = {({ 3 1) is not
increasing on (0> 1).
14. True.
Let {1 ? {2 where {1 > {2 M L. Then 0 ? i ({1 ) ? i ({2 ) and 0 ? j({1 ) ? j({2 ) [since i and j are both positive
and increasing]. Hence, i ({1 ) j({1 ) ? i({2 ) j({1 ) ? i ({2 ) j({2 ). So i j is increasing on L.
15. True.
Let {1 > {2 M L and {1 ? {2 . Then i ({1 ) ? i ({2 ) [i is increasing] i
1
1
A
[i is positive] i
i ({1 )
i ({2 )
j({1 ) A j({2 ) i j({) = 1@i ({) is decreasing on L.
16. False.
If i is even, then i ({) = i (3{). Using the Chain Rule to differentiate this equation, we get
i 0 ({) = i 0 (3{)
17. True.
g
(3{) = 3i 0 (3{). Thus, i 0 (3{) = 3i 0 ({), so i 0 is odd.
g{
If i is periodic, then there is a number s such that i ({ + s) = i (s) for all {. Differentiating gives
i 0 ({) = i 0 ({ + s) · ({ + s)0 = i 0 ({ + s) · 1 = i 0 ({ + s), so i 0 is periodic.
The most general antiderivative of i({) = {32 is I ({) = 31@{ + F1 for { ? 0 and I ({) = 31@{ + F2
18. False.
for { A 0 [see Example 1(b) in Section 4.9].
By the Mean Value Theorem, there exists a number f in (0> 1) such that i (1) 3 i (0) = i 0 (f)(1 3 0) = i 0 (f).
19. True.
Since i 0 (f) is nonzero, i(1) 3 i (0) 6= 0, so i (1) 6= i (0).
20. False.
lim
{<0
lim {
0
{
{<0
=
= = 0, not 1.
{
h
lim h{
1
{<0
1. i ({) = {3 3 6{2 + 9{ + 1, [2> 4].
i 0 ({) = 3{2 3 12{ + 9 = 3({2 3 4{ + 3) = 3({ 3 1)({ 3 3). i 0 ({) = 0 i
{ = 1 or { = 3, but 1 is not in the interval. i 0 ({) A 0 for 3 ? { ? 4 and i 0 ({) ? 0 for 2 ? { ? 3, so i (3) = 1 is a local
minimum value. Checking the endpoints, we ¿nd i (2) = 3 and i (4) = 5. Thus, i(3) = 1 is the absolute minimum value and
i (4) = 5 is the absolute maximum value.
2. i ({) = {
I
1 3 3{
1 3 {, [31> 1]. i 0 ({) = { · 12 (1 3 {)31@2 (31) + (1 3 {)1@2 (1) = (1 3 {)31@2 3 12 { + (1 3 {) = I 2 .
13{
i 0 ({) = 0 i { = 23 . i 0 ({) does not exist C { = 1. i 0 ({) A 0 for 31 ? { ? 23 and i 0 ({) ? 0 for 23 ? { ? 1, so
t
I
I
i 23 = 23 13 = 29 3 [E 0=38] is a local maximum value. Checking the endpoints, we ¿nd i(31) = 3 2 and i (1) = 0.
I
Thus, i(31) = 3 2 is the absolute minimum value and i 23 =
2
9
I
3 is the absolute maximum value.
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
3. i ({) =
3{ 3 4
({2 + 1)(3) 3 (3{ 3 4)(2{)
3(3{2 3 8{ 3 3)
3(3{ + 1)({ 3 3)
, [32> 2]. i 0 ({) =
=
=
.
2
2
2
{ +1
({ + 1)
({2 + 1)2
({2 + 1)2
i 0 ({) = 0 i { = 3 13 or { = 3, but 3 is not in the interval. i 0 ({) A 0 for 3 13 ? { ? 2 and i 0 ({) ? 0 for
35
= 3 92 is a local minimum value. Checking the endpoints, we ¿nd i (32) = 32 and
32 ? { ? 3 13 , so i 3 13 = 10@9
i (2) = 25 . Thus, i 3 13 = 3 92 is the absolute minimum value and i(2) =
2
5
is the absolute maximum value.
I
2{ + 1
1
{2 + { + 1, [32> 1]. i 0 ({) = 12 ({2 + { + 1)31@2 (2{ + 1) = I
. i 0 ({) = 0 i { = 3 .
2
2 {2 + { + 1
I
i 0 ({) A 0 for 3 12 ? { ? 1 and i 0 ({) ? 0 for 32 ? { ? 3 12 , so i 3 12 = 3@2 is a local minimum value. Checking the
I
I
I
endpoints, we ¿nd i (32) = i (1) = 3. Thus, i 3 12 = 3@2 is the absolute minimum value and i (32) = i(1) = 3 is
4. i ({) =
the absolute maximum value.
i 0 ({) = 1 3 2 sin {. i 0 ({) = 0 i sin { = 12 i { = 6 , 5
. i 0 ({) A 0 for
6
I
5 0
, so i 6 = 6 + 3 E 2=26 is a local maximum value and
3> 6 and 5
6 > , and i ({) ? 0 for 6 > 6
5 I
i 5
= 6 3 3 E 0=89 is a local minimum value. Checking the endpoints, we ¿nd i (3) = 3 3 2 E 35=14 and
6
I
i () = 3 2 E 1=14. Thus, i (3) = 3 3 2 is the absolute minimum value and i 6 = 6 + 3 is the absolute
5. i ({) = { + 2 cos {, [3> ].
maximum value.
6. i ({) = {2 h3{ , [31> 3].
i 0 ({) = {2 (3h3{ ) + h3{ (2{) = {h3{ (3{ + 2). i 0 ({) = 0 i { = 0 or { = 2.
i 0 ({) A 0 for 0 ? { ? 2 and i 0 ({) ? 0 for 31 ? { ? 0 and 2 ? { ? 3, so i(0) = 0 is a local minimum value and
i (2) = 4h32 E 0=54 is a local maximum value. Checking the endpoints, we ¿nd i (31) = h E 2=72 and
i (3) = 9h33 E 0=45. Thus, i (0) = 0 is the absolute minimum value and i (31) = h is the absolute maximum value.
7. This limit has the form 00 .
lim
{<0
h{ 3 1 H
h{
1
= =1
= lim
{<0 sec2 {
tan {
1
8. This limit has the form 00 .
lim
{<0
tan 4{ H
4 sec2 4{
4(1)
4
=
=
= lim
{ + sin 2{ {<0 1 + 2 cos 2{
1 + 2(1)
3
9. This limit has the form 00 .
lim
h4{ 3 1 3 4{ H
4h4{ 3 4 H
16h4{
=
lim
=
lim
= lim 8h4{ = 8 · 1 = 8
{<0
{<0
{<0
{2
2{
2
10. This limit has the form
"
.
"
{<0
h4{ 3 1 3 4{ H
4h4{ 3 4 H
16h4{
= lim 8h4{ = "
= lim
= lim
2
{<"
{<"
{<"
{<"
{
2{
2
lim
11. This limit has the form " · 0.
H
{2 3 {3 "
2{ 3 3{2 "
form = lim
form
"
"
32{
{<3" h
{<3" 32h32{
H
2 3 6{ "
36
H
= lim
form = lim
=0
"
{<3" 4h32{
{<3" 38h32{
lim ({2 3 {3 )h2{ = lim
{<3"
12. This limit has the form 0 · ".
lim ({ 3 ) csc { = lim
{< {< H
{ 3 0
1
1
form = lim
=
= 31
sin { 0
31
{< cos {
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
CHAPTER 4 REVIEW
13. This limit has the form " 3 ".
lim
{<1+
{
1
3
{31
ln {
= lim
{<1+
H
= lim
{<1+
14. | = (tan {)cos {
lim
{<(@2)
so
lim
ln | =
{<(@2)
{ ln { 3 { + 1
({ 3 1) ln {
H
= lim
{<1+
¤
441
{ · (1@{) + ln { 3 1
ln {
= lim
({ 3 1) · (1@{) + ln { {<1+ 1 3 1@{ + ln {
1@{
1
1
=
=
1@{2 + 1@{
1+1
2
i ln | = cos { ln tan {, so
lim
{<(@2)
(tan {)cos { =
(1@ tan {) sec2 {
0
ln tan { H
sec {
cos {
=
lim
= 2 = 0,
=
lim
=
lim
2
2
sec {
sec { tan {
1
{<(@2)
{<(@2) tan {
{<(@2) sin {
lim
{<(@2)
hln | = h0 = 1.
15. i (0) = 0, i 0 (32) = i 0 (1) = i 0 (9) = 0, lim i ({) = 0, lim i({) = 3",
{<"
{<6
i 0 ({) ? 0 on (3"> 32), (1> 6), and (9> "), i 0 ({) A 0 on (32> 1) and (6> 9),
i 00 ({) A 0 on (3"> 0) and (12> "), i 00 ({) ? 0 on (0> 6) and (6> 12)
16. For 0 ? { ? 1, i 0 ({) = 2{, so i ({) = {2 + F. Since i (0) = 0,
i ({) = {2 on [0> 1]. For 1 ? { ? 3, i 0 ({) = 31, so i ({) = 3{ + G.
1 = i(1) = 31 + G
i G = 2, so i ({) = 2 3 {. For { A 3, i 0 ({) = 1,
so i ({) = { + H. 31 = i (3) = 3 + H
i H = 34, so i({) = { 3 4.
Since i is even, its graph is symmetric about the |-axis.
17. i is odd, i 0 ({) ? 0 for 0 ? { ? 2,
i 00 ({) A 0 for 0 ? { ? 3,
i 0 ({) A 0 for { A 2,
i 00 ({) ? 0 for { A 3, lim{<" i ({) = 32
18. (a) Using the Test for Monotonic Functions we know that i is increasing on (32> 0) and (4> ") because i 0 A 0 on (32> 0)
and (4> "), and that i is decreasing on (3"> 32) and (0> 4) because i 0 ? 0 on (3"> 32) and (0> 4).
(b) Using the First Derivative Test, we know that i has a local maximum at { = 0 because i 0 changes from positive to
negative at { = 0, and that i has a local minimum at { = 4 because i 0 changes from negative to positive at { = 4.
(c)
(d)
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°
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442
¤
NOT FOR SALE
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
19. | = i ({) = 2 3 2{ 3 {3
A. G = R B. |-intercept: i (0) = 2.
H.
The {-intercept (approximately 0=770917) can be found using Newton’s
Method. C. No symmetry D. No asymptote
E. i 0 ({) = 32 3 3{2 = 3(3{2 + 2) ? 0, so i is decreasing on R.
F. No extreme value G. i 00 ({) = 36{ ? 0 on (0> ") and i 00 ({) A 0 on
(3"> 0), so i is CD on (0> ") and CU on (3"> 0). There is an IP at (0> 2).
20. | = i ({) = {3 3 6{2 3 15{ + 4
A. G = R B. |-intercept: i(0) = 4; {-intercepts: i ({) = 0 i
{ E 32=09, 0=24, 7=85 C. No symmetry D. No asymptote
H.
E. i 0 ({) = 3{2 3 12{ 3 15 = 3({2 3 4{ 3 5) = 3({ + 1)({ 3 5),
so i is increasing on (3"> 31), decreasing on (31> 5), and increasing
on (5> "). F. Local maximum value i (31) = 12, local minimum
value i (5) = 396. G. i 00 ({) = 6{ 3 12 = 6({ 3 2), so i is CD
on (3"> 2) and CU on (2> "). There is an IP at (2> 342).
21. | = i({) = {4 3 3{3 + 3{2 3 { = {({ 3 1)3
A. G = R B. |-intercept: i (0) = 0; {-intercepts: i({) = 0 C
{ = 0 or { = 1 C. No symmetry D. i is a polynomial function and hence, it has no asymptote.
E. i 0 ({) = 4{3 3 9{2 + 6{ 3 1. Since the sum of the coef¿cients is 0, 1 is a root of i 0 , so
i 0 ({) = ({ 3 1) 4{2 3 5{ + 1 = ({ 3 1)2 (4{ 3 1). i 0 ({) ? 0 i { ? 14 , so i is decreasing on 3"> 14
and i is increasing on 14 > " . F. i 0 ({) does not change sign at { = 1, so
H.
1
27
is a local minimum value.
there is not a local extremum there. i 4 = 3 256
G. i 00 ({) = 12{2 3 18{ + 6 = 6(2{ 3 1)({ 3 1). i 00 ({) = 0 C { =
or 1. i 00 ({) ? 0 C 12 ? { ? 1 i i is CD on 12 > 1 and CU on
1
3"> 12 and (1> "). There are inÀection points at 12 > 3 16
and (1> 0).
22. | = i ({) =
{
1 3 {2
1
2
A. G = (3"> 31) (31> 1) (1> ") B. |-intercept: i (0) = 0; {-intercept: 0
C. i (3{) = 3i ({), so i is odd and the graph is symmetric about the origin. D.
lim
{<31 1
lim
{<±" 1
{
= 0, so | = 0 is a HA.
3 {2
{
{
{
= " and lim
= 3", so { = 31 is a VA. Similarly, lim
= " and
3 {2
{<31+ 1 3 {2
{<1 1 3 {2
(1 3 {2 )(1) 3 {(32{)
1 + {2
{
0
=
3",
so
{
=
1
is
a
VA.
E.
i
({)
=
=
A 0 for { 6= ±1, so i is
(1 3 {2 )2
(1 3 {2 )2
{<1+ 1 3 {2
lim
increasing on (3"> 31), (31> 1), and (1> "). F. No local extrema
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
CHAPTER 4 REVIEW
G. i 00 ({) =
=
(1 3 {2 )2 (2{) 3 (1 + {2 )2(1 3 {2 )(32{)
[(1 3 {2 )2 ]2
¤
H.
2{(1 3 {2 )[(1 3 {2 ) + 2(1 + {2 )]
2{(3 + {2 )
=
(1 3 {2 )4
(1 3 {2 )3
i 00 ({) A 0 for { ? 31 and 0 ? { ? 1, and i 00 ({) ? 0 for 31 ? { ? 0 and
{ A 1, so i is CU on (3"> 31) and (0> 1), and i is CD on (31> 0) and (1> ").
(0> 0) is an IP.
23. | = i({) =
D.
lim
{<±"
1
{({ 3 3)2
A. G = {{ | { 6= 0> 3} = (3"> 0) (0> 3) (3> ") B. No intercepts. C. No symmetry.
1
= 0, so | = 0 is a HA.
{({ 3 3)2
so { = 0 and { = 3 are VA. E. i 0 ({) = 3
lim
{<0+
1
1
1
= ", lim
= 3", lim
= ",
2
{<3 {({ 3 3)2
{({ 3 3)2
{<0 {({ 3 3)
({ 3 3)2 + 2{({ 3 3)
3(1 3 {)
= 2
{2 ({ 3 3)4
{ ({ 3 3)3
so i is increasing on (1> 3) and decreasing on (3"> 0), (0> 1), and (3> ").
F. Local minimum value i(1) =
1
4
G. i 00 ({) =
i i 0 ({) A 0 C 1 ? { ? 3,
H.
6(2{2 3 4{ + 3)
.
{3 ({ 3 3)4
Note that 2{2 3 4{ + 3 A 0 for all { since it has negative discriminant.
So i 00 ({) A 0 C { A 0 i i is CU on (0> 3) and (3> ") and
CD on (3"> 0). No IP
24. | = i ({) =
1
1
3
{2
({ 3 2)2
1
1
=
{2
({ 3 2)2
A. G = {{ | { 6= 0> 2} B. |-intercept: none; {-intercept: i ({) = 0 i
C ({ 3 2)2 = {2
C {2 3 4{ + 4 = {2
C 4{ = 4 C { = 1 C. No symmetry
D. lim i({) = " and lim i ({) = 3", so { = 0 and { = 2 are VA; lim i ({) = 0, so | = 0 is a HA
{<0
{<2
E. i 0 ({) = 3
{<±"
2
2
+
A0 i
{3
({ 3 2)3
3({ 3 2)3 + {3
A0 C
{3 ({ 3 2)3
3{3 + 6{2 3 12{ + 8 + {3
A0 C
{3 ({ 3 2)3
2(3{2 3 6{ + 4)
A 0. The numerator is positive (the discriminant of the quadratic is negative), so i 0 ({) A 0 if { ? 0 or
{3 ({ 3 2)3
{ A 2, and hence, i is increasing on (3"> 0) and (2> ") and decreasing on (0> 2).
F. No local extreme values G. i 00 ({) =
({ 3 2)4 3 {4
A0 C
{4 ({ 3 2)4
6
6
3
A0 i
{4
({ 3 2)4
H.
{4 3 8{3 + 24{2 3 32{ + 16 3 {4
A0 C
{4 ({ 3 2)4
38({3 3 3{2 + 4{ 3 2)
A0 C
{4 ({ 3 2)4
38({ 3 1)({2 3 2{ + 2)
A 0. So i 00 is
{4 ({ 3 2)4
positive for { ? 1 [{ 6= 0] and negative for { A 1 [{ 6= 2]. Thus, i is CU on
(3"> 0) and (0> 1) and i is CD on (1> 2) and (2> "). IP at (1> 0)
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°
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443
444
NOT FOR SALE
¤
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
25. | = i ({) =
D. lim
{<"
lim
{<38+
64
{2
= {38+
{+8
{+8
A. G = {{ | { 6= 38} B. Intercepts are 0 C. No symmetry
{2
64
= ", but i ({) 3 ({ 3 8) =
< 0 as { < ", so | = { 3 8 is a slant asymptote.
{+8
{+8
{2
{2
64
{({ + 16)
= " and lim
= 3", so { = 38 is a VA. E. i 0 ({) = 1 3
=
A0 C
{+8
({ + 8)2
({ + 8)2
{<38 { + 8
{ A 0 or { ? 316, so i is increasing on (3"> 316) and (0> ") and
H.
decreasing on (316> 38) and (38> 0) =
F. Local maximum value i (316) = 332, local minimum value i (0) = 0
G. i 00 ({) = 128@({ + 8)3 A 0 C { A 38, so i is CU on (38> ") and
CD on (3"> 38). No IP
26. | = i ({) =
I
I
1 3 { + 1 + { A. 1 3 { D 0 and 1 + { D 0 i { $ 1 and { D 31, so G = [31> 1].
B. |-intercept: i (0) = 1 + 1 = 2; no {-intercept because i ({) A 0 for all {.
C. i (3{) = i({), so the curve is symmetric about the |-axis= D. No asymptote
E. i 0 ({) = 12 (1 3 {)31@2 (31) + 12 (1 + {)31@2 =
I
I
1
3 1+{+ 13{
31
I
I
I
+ I
=
A0 i
2 13{
2 1+{
2 13{ 1+{
I
I
I
I
3 1+{+ 13{A0 i
1 3 { A 1 + { i 1 3 { A 1 + { i 32{ A 0 i { ? 0, so i 0 ({) A 0 for
31 ? { ? 0 and i 0 ({) ? 0 for 0 ? { ? 1. Thus, i is increasing on (31> 0)
H.
and decreasing on (0> 1). F. Local maximum value i (0) = 2
G. i 00 ({) = 3 12 3 12 (1 3 {)33@2 (31) + 12 3 12 (1 + {)33@2
=
31
31
+
?0
4(1 3 {)3@2
4(1 + {)3@2
for all { in the domain, so i is CD on (31> 1). No IP
27. | = i ({) = {
I
2 + { A. G = [32> ") B. |-intercept: i(0) = 0; {-intercepts: 32 and 0 C. No symmetry
I
3{ + 4
{
1
I
+ 2+{= I
[{ + 2(2 + {)] = I
= 0 when { = 3 43 , so i is
2+{
2 2+{
2 2+{
t
I
decreasing on 32> 3 43 and increasing on 3 43 > " . F. Local minimum value i 3 43 = 3 43 23 = 3 4 9 6 E 31=09,
D. No asymptote E. i 0 ({) =
2
no local maximum
H.
I
1
2 2 + { · 3 3 (3{ + 4) I
6(2 + {) 3 (3{ + 4)
2
+{
=
G. i 00 ({) =
4(2 + {)
4(2 + {)3@2
=
3{ + 8
4(2 + {)3@2
i 00 ({) A 0 for { A 32, so i is CU on (32> "). No IP
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© Cengage Learning. All Rights Reserved.
CHAPTER 4 REVIEW
28. | = i ({) =
¤
445
I
3
{2 + 1 A. G = R B. |-intercept: i (0) = 1; i ({) A 0, so there is no {-intercept C. i (3{) = i ({), so
the curve is symmetric about the |-axis. D. No asymptote E. i 0 ({) = 13 ({2 + 1)32@3 (2{) =
2{
A0 C
3({2 + 1)2@3
{ A 0 and i 0 ({) ? 0 C { ? 0, so i is increasing on (0> ") and decreasing on (3"> 0).
F. Local minimum value i (0) = 1
2({2 + 1)31@3 3({2 + 1) 3 4{2
3({2 + 1)2@3 (2) 3 2{(3) 23 ({2 + 1)31@3 (2{)
2(3 3 {2 )
=
=
.
G. i ({) =
2
2@3
2
2
4@3
[3({ + 1) ]
9({ + 1)
9({2 + 1)5@3
I
I
I
i 00 ({) A 0 C 3 3 ? { ? 3 and i 00 ({) ? 0 C { ? 3 3 and
H.
I I
I I I
3> " .
{ A 3, so i is CU on 3 3> 3 and CD on 3"> 3 3 and
00
I
I I I There are inÀection points at ± 3> i ± 3 = ± 3> 3 4 .
29. | = i ({) = h{ sin {, 3 $ { $ A. G = [3> ] B. |-intercept: i (0) = 0; i ({) = 0 C sin { = 0 i
{ = 3> 0> . C. No symmetry D. No asymptote E. i 0 ({) = h{ cos { + sin { · h{ = h{ (cos { + sin {).
i 0 ({) = 0 C 3 cos { = sin { C 31 = tan { i { = 3 4 > 3
. i 0 ({) A 0 for 3 4 ? { ? 3
and i 0 ({) ? 0
4
4
for 3 ? { ? 3 4 and 3
? { ? , so i is increasing on 3 4 > 3
> .
and i is decreasing on 3> 3 4 and 3
4
4
4
I
F. Local minimum value i 3 4 = (3 2@2)h3@4 E 30=32 and
I
=
local maximum value i 3
2@2 h3@4 E 7=46
4
H.
G. i 00 ({) = h{ (3 sin { + cos {) + (cos { + sin {)h{ = h{ (2 cos {) A 0 i
3 2 ? { ? 2 and i 00 ({) ? 0 i 3 ? { ? 3 2 and 2 ? { ? , so i is
CU on 3 2 > 2 , and i is CD on 3> 3 2 and 2 > . There are inÀection
points at 3 2 > 3h3@2 and 2 > h@2 .
30. | = i ({) = 4{ 3 tan {, 3 2 ? { ?
curve is symmetric about (0> 0) = D.
2
A. G = 3 2 > 2 . B. |-intercept = i (0) = 0 C. i (3{) = 3i ({), so the
lim (4{ 3 tan {) = 3",
{<@2
lim
{<3@2+
are VA. E. i 0 ({) = 4 3 sec2 { A 0 C sec { ? 2 C cos { A
1
2
(4{ 3 tan {) = ", so { =
C 3 3 ? { ?
I
3 3 > 3 and decreasing on 3 2 > 3 3 and 3 > 2 = F. i 3 = 4
3 3 is
3
I
a local maximum value, i 3 3 = 3 3 4
is a local minimum value.
3
,
3
2
and { = 3 2
so i is increasing on
H.
G. i 00 ({) = 32 sec2 { tan { A 0 C tan { ? 0 C 3 2 ? { ? 0,
so i is CU on 3 2 > 0 and CD on 0> 2 . IP at (0> 0)
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446
¤
CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
31. | = i ({) = sin31 (1@{)
A. G = {{ | 31 $ 1@{ $ 1} = (3"> 31] [1> ") = B. No intercept
C. i (3{) = 3i({), symmetric about the origin D.
lim sin31 (1@{) = sin31 (0) = 0, so | = 0 is a HA.
{<±"
1
31
? 0, so i is decreasing on (3"> 31) and (1> ") =
3 2 = I
{
{4 3 {2
1
E. i 0 ({) = s
1 3 (1@{)2
F. No local extreme value, but i (1) =
2
is the absolute maximum value
H.
and i (31) = 3 2 is the absolute minimum value.
{ 2{2 3 1
4{3 3 2{
=
A 0 for { A 1 and i 00 ({) ? 0
G. i 00 ({) =
2({4 3 {2 )3@2
({4 3 {2 )3@2
for { ? 31, so i is CU on (1> ") and CD on (3"> 31). No IP
2
32. | = i ({) = h2{3{
A. G = R B. |-intercept 1; no {-intercept C. No symmetry D.
2
is a HA. E. | = i ({) = h2{3{
2
lim h2{3{ = 0, so | = 0
{<±"
2
i i 0 ({) = 2(1 3 {)h2{3{ A 0 C { ? 1, so i is increasing on (3"> 1) and
decreasing on (1> "). F. i (1) = h is a local and absolute maximum value.
2
G. i 00 ({) = 2 2{2 3 4{ + 1 h2{3{ = 0 C { = 1 ±
I
2
.
2
so i is CU on 3"> 1 3
i 00 ({) A 0 C { ? 1 3 22 or { A 1 +
I
I
I I I and 1 + 22 > " , and CD on 1 3 22 > 1 + 22 . IP at 1 ± 22 > h
I
33. | = i ({) = ({ 3 2)h3{
I
2
,
2
H.
I 2
2
A. G = R B. |-intercept: i (0) = 32; {-intercept: i({) = 0 C { = 2
C. No symmetry D. lim
{<"
{32 H
1
= lim { = 0, so | = 0 is a HA. No VA
{<" h
h{
E. i 0 ({) = ({ 3 2)(3h3{ ) + h3{ (1) = h3{ [3({ 3 2) + 1] = (3 3 {)h3{ .
H.
i 0 ({) A 0 for { ? 3, so i is increasing on (3"> 3) and decreasing on (3> ").
F. Local maximum value i (3) = h33 , no local minimum value
G. i 00 ({) = (3 3 {)(3h3{ ) + h3{ (31) = h3{ [3(3 3 {) + (31)]
= ({ 3 4)h3{ A 0
for { A 4, so i is CU on (4> ") and CD on (3"> 4). IP at (4> 2h34 )
34. | = i({) = { + ln({2 + 1)
A. G = R B. |-intercept: i (0) = 0 + ln 1 = 0; {-intercept: i ({) = 0 C
ln({2 + 1) = 3{ C {2 + 1 = h3{
i { = 0 since the graphs of | = {2 + 1 and | = h3{ intersect only at { = 0.
C. No symmetry D. No asymptote E. i 0 ({) = 1 +
{2 + 2{ + 1
({ + 1)2
2{
=
= 2
. i 0 ({) A 0 if { 6= 31 and
2
+1
{ +1
{ +1
{2
i is increasing on R. F. No local extreme values
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
CHAPTER 4 REVIEW
G. i 00 ({) =
({2 + 1)2 3 2{(2{)
2[({2 + 1) 3 2{2 ]
2(1 3 {2 )
=
= 2
.
2
2
2
2
({ + 1)
({ + 1)
({ + 1)2
H.
i 00 ({) A 0 C 31 ? { ? 1 and i 00 ({) ? 0 C { ? 31 or { A 1, so i is
CU on (31> 1) and i is CD on (3"> 31) and (1> "). IP at (31> 31 + ln 2)
and (1> 1 + ln 2)
35. i ({) =
{2 3 1
{3
i 00 ({) =
i i 0 ({) =
{3 (2{) 3 {2 3 1 3{2
3 3 {2
=
6
{
{4
{4 (32{) 3 3 3 {2 4{3
2{2 3 12
=
8
{
{5
i
Estimates: From the graphs of i 0 and i 00 , it appears that i is increasing on
(31=73> 0) and (0> 1=73) and decreasing on (3"> 31=73) and (1=73> ");
i has a local maximum of about i (1=73) = 0=38 and a local minimum of about
i (31=7) = 30=38; i is CU on (32=45> 0) and (2=45> "), and CD on
(3"> 32=45) and (0> 2=45); and i has inÀection points at about
(32=45> 30=34) and (2=45> 0=34).
3 3 {2
is positive for 0 ? {2 ? 3, that is, i is increasing
{4
I I on 3 3> 0 and 0> 3 ; and i 0 ({) is negative (and so i is decreasing) on
I
I I
3> " . i 0 ({) = 0 when { = ± 3.
3"> 3 3 and
I
i 0 goes from positive to negative at { = 3, so i has a local maximum of
Exact: Now i 0 ({) =
i
I (I3 )2 3 1
3 = I 3 =
( 3)
I
2 3
9 ;
and since i is odd, we know that maxima on the
interval (0> ") correspond to minima on (3"> 0), so i has a local minimum of
I
I 2{2 3 12
is positive (so i is CU) on
i 3 3 = 3 2 9 3 . Also, i 00 ({) =
{5
I I
I 6> " , and negative (so i is CD) on 3"> 3 6 and
3 6> 0 and
I I I
I I 0> 6 . There are IP at
6> 5366 and 3 6> 3 5366 .
36. i ({) =
{3 3 {
+{+3
{2
i i 0 ({) =
{4 + 2{3 + 10{2 3 3
({2 + { + 3)2
i i 00 ({) =
36({3 3 3{2 3 12{ 3 1)
.
({2 + { + 3)3
i ({) = 0 C { = ±1; i 0 ({) = 0 C { E 30=57, 0=52; i 00 ({) = 0 C { E 32=21, 30=09, 5=30.
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
From the graphs of i 0 and i 00 , it appears that i is increasing on (3"> 30=57) and (0=52> ") and decreasing on
(30=57> 0=52); i has a local maximum of about i (30=57) = 0=14 and a local minimum of about i (0=52) = 30=10;
i is CU on (3"> 32=21) and (30=09> 5=30), and CD on (32=21> 30=09) and (5=30> "); and i has inÀection points at about
(32=21> 31=52), (30=09> 0=03), and (5=30> 3=95).
37. i ({) = 3{6 3 5{5 + {4 3 5{3 3 2{2 + 2
i i 0 ({) = 18{5 3 25{4 + 4{3 3 15{2 3 4{ i
i 00 ({) = 90{4 3 100{3 + 12{2 3 30{ 3 4
From the graphs of i 0 and i 00 , it appears that i is increasing on (30=23> 0) and (1=62> ") and decreasing on (3"> 30=23)
and (0> 1=62); i has a local maximum of i(0) = 2 and local minima of about i (30=23) = 1=96 and i (1=62) = 319=2;
i is CU on (3"> 30=12) and (1=24> ") and CD on (30=12> 1=24); and i has inÀection points at about (30=12> 1=98) and
(1=24> 312=1).
38. i ({) = {2 + 6=5 sin {, 35 $ { $ 5
i i 0 ({) = 2{ + 6=5 cos { i i 00 ({) = 2 3 6=5 sin {. i ({) = 0 C
{ E 32=25 and { = 0; i 0 ({) = 0 C { E 31=19, 2=40, 3=24; i 00 ({) = 0 C { E 33=45, 0=31, 2=83.
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CHAPTER 4 REVIEW
¤
From the graphs of i 0 and i 00 , it appears that i is decreasing on (35> 31=19) and (2=40> 3=24) and increasing on
(31=19> 2=40) and (3=24> 5); i has a local maximum of about i (2=40) = 10=15 and local minima of about
i (31=19) = 34=62 and i(3=24) = 9=86; i is CU on (33=45> 0=31) and
(2=83> 5) and CD on (35> 33=45) and (0=31> 2=83); and i has inÀection points
at about (33=45> 13=93), (0=31> 2=10), and (2=83> 10=00).
39.
From the graph, we estimate the points of inÀection to be about (±0=82> 0=22).
2
2
i ({) = h31@{
i i 0 ({) = 2{33 h31@{
i
2 (33{ )] = 2{36 h31@{ 2 3 3{2 .
t
This is 0 when 2 3 3{2 = 0 C { = ± 23 , so the inÀection points
t
are ± 23 > h33@2 .
00
33
i ({) = 2[{
40. (a)
(b) i ({) =
33
(2{
31@{2
)h
31@{2
+h
34
1
.
1 + h1@{
lim i ({) =
{<"
1
1
1
1
= , lim i ({) =
= ,
1+1
2 {<3"
1+1
2
as { < 0+ , 1@{ < ", so h1@{ < " i
as { < 03 , 1@{ < 3", so h1@{ < 0 i
lim i ({) = 0,
{<0+
lim i ({) =
{<0
1
=1
1+0
(c) From the graph of i , estimates for the IP are (30=4> 0=9) and (0=4> 0=08).
(d) i 00 ({) = 3
h1@{ [h1@{ (2{ 3 1) + 2{ + 1]
{4 (h1@{ + 1)3
(e) From the graph, we see that i 00 changes sign at { = ±0=417
({ = 0 is not in the domain of i ). IP are approximately (0=417> 0=083)
and (30=417> 0=917).
cos2 {
, 3 $ { $ {2 + { + 1
41. i ({) = I
i 00 ({) = 3
i i 0 ({) = 3
cos { [(2{ + 1) cos { + 4({2 + { + 1) sin {]
2({2 + { + 1)3@2
i
(8{4 + 16{3 + 16{2 + 8{ + 9) cos2 { 3 8({2 + { + 1)(2{ + 1) sin { cos { 3 8({2 + { + 1)2 sin2 {
4({2 + { + 1)5@2
i ({) = 0 C { = ± 2 ; i 0 ({) = 0 C { E 32=96, 31=57, 30=18, 1=57, 3=01;
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
i 00 ({) = 0 C { E 32=16, 30=75, 0=46, and 2=21.
The {-coordinates of the maximum points are the values at which i 0 changes from positive to negative, that is, 32=96,
30=18, and 3=01. The {-coordinates of the minimum points are the values at which i 0 changes from negative to positive, that
is, 31=57 and 1=57. The {-coordinates of the inÀection points are the values at which i 00 changes sign, that is, 32=16, 30=75,
0=46, and 2=21.
h30=1{ ({2 3 1) ln({2 3 1) 3 20{
i
42. i ({) = h
ln({ 3 1) i i ({) =
10(1 3 {2 )
h30=1{ ({2 3 1)2 ln({2 3 1) 3 40({3 + 5{2 3 { + 5)
i 00 ({) =
.
100({2 3 1)2
I
The domain of i is (3"> 31) (1> "). i ({) = 0 C { = ± 2; i 0 ({) = 0 C { E 5=87;
30=1{
0
2
i 00 ({) = 0 C { E 34=31 and 11=74.
i 0 changes from positive to negative at { E 5=87, so 5=87 is the {-coordinate of the maximum point. There is no minimum
point. The {-coordinates of the inÀection points are the values at which i 00 changes sign, that is, 34=31 and 11=74.
43. The family of functions i ({) = ln(sin { + F) all have the same period and all
have maximum values at { =
2
+ 2q. Since the domain of ln is (0> "), i has
a graph only if sin { + F A 0 somewhere. Since 31 $ sin { $ 1, this happens
if F A 31, that is, i has no graph if F $ 31. Similarly, if F A 1, then
sin { + F A 0 and i is continuous on (3"> "). As F increases, the graph of
i is shifted vertically upward and Àattens out. If 31 ? F $ 1, i is de¿ned where sin { + F A 0 C
sin { A 3F C sin31 (3F) ? { ? 3 sin31 (3F). Since the period is 2, the domain of i is
2q + sin31 (3F)> (2q + 1) 3 sin31 (3F) , q an integer.
44. We exclude the case f = 0, since in that case i ({) = 0 for all {. To ¿nd the maxima and minima, we differentiate:
i({) = f{h3f{
2
k
l
2
2
2
i i 0 ({) = f {h3f{ (32f{) + h3f{ (1) = fh3f{ (32f{2 + 1)
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CHAPTER 4 REVIEW
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451
I
This is 0 where 32f{2 + 1 = 0 C { = ±1@ 2f. So if f A 0, there are two maxima or minima, whose {-coordinates
approach 0 as f increases. The negative root gives a minimum and the positive root gives a maximum, by the First Derivative
I
s
2
I I Test. By substituting back into the equation, we see that i ±1@ 2f = f ±1@ 2f h3f(±1@ 2f ) = ± f@2h. So as f
increases, the extreme points become more pronounced. Note that if f A 0, then lim i ({) = 0. If f ? 0, then there are no
{<±"
extreme values, and lim i ({) = ~".
{<±"
2
To ¿nd the points of inÀection, we differentiate again: i 0 ({) = fh3f{ 32f{2 + 1 i
k
l
2
2
2
i 00 ({) = f h3f{ (34f{) + (32f{2 + 1)(32f{h3f{ ) = 32f2 {h3f{ (3 3 2f{2 ). This is 0 at { = 0 and where
s
s
s
3 3 2f{2 = 0 C { = ± 3@(2f) i IP at ± 3@(2f)> ± 3f@2 h33@2 . If f A 0 there are three inÀection points,
and as f increases, the {-coordinates of the nonzero inÀection points approach 0. If f ? 0, there is only one inÀection point,
the origin.
45. Let i ({) = 3{ + 2 cos { + 5. Then i(0) = 7 A 0 and i (3) = 33 3 2 + 5 = 33 + 3 = 33( 3 1) ? 0> and since i is
continuous on R (hence on [3> 0]), the Intermediate Value Theorem assures us that there is at least one zero of i in [3> 0].
Now i 0 ({) = 3 3 2 sin { A 0 implies that i is increasing on R, so there is exactly one zero of i, and hence, exactly one real
root of the equation 3{ + 2 cos { + 5 = 0.
46. By the Mean Value Theorem, i 0 (f) =
i (4) 3 i (0)
430
C 4i 0 (f) = i (4) 3 1 for some f with 0 ? f ? 4. Since
2 $ i 0 (f) $ 5, we have 4(2) $ 4i 0 (f) $ 4(5) C 4(2) $ i (4) 3 1 $ 4(5) C 8 $ i (4) 3 1 $ 20 C
9 $ i(4) $ 21.
47. Since i is continuous on [32> 33] and differentiable on (32> 33), then by the Mean Value Theorem there exists a number f in
0
(32> 33) such that i (f) =
1 34@5
f
5
I
I
5
I
33 3 5 32
=
= 5 33 3 2, but 15 f34@5 A 0 i
33 3 32
I
5
33 3 2 A 0 i
i 0 is decreasing, so that i 0 (f) ? i 0 (32) = 15 (32)34@5 = 0=0125 i 0=0125 A i 0 (f) =
I
Therefore, 2 ? 5 33 ? 2=0125.
48. Since the point (1> 3) is on the curve | = d{3 + e{2 , we have 3 = d(1)3 + e(1)2
I
5
33 3 2 i
I
5
33 A 2. Also
I
5
33 ? 2=0125.
i 3 = d + e (1).
| 0 = 3d{2 + 2e{ i |00 = 6d{ + 2e. | 00 = 0 [for inÀection points] C { =
e
32e
= 3 . Since we want { = 1,
6d
3d
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
e
i e = 33d. Combining with (1) gives us 3 = d 3 3d C 3 = 32d C d = 3 32 . Hence,
3d
e = 33 3 32 = 92 and the curve is | = 3 32 {3 + 92 {2 .
1=3
49. (a) j({) = i ({2 )
i j0 ({) = 2{i 0 ({2 ) by the Chain Rule. Since i 0 ({) A 0 for all { 6= 0, we must have i 0 ({2 ) A 0 for
{ 6= 0, so j 0 ({) = 0 C { = 0. Now j 0 ({) changes sign (from negative to positive) at { = 0, since one of its factors,
i 0 ({2 ), is positive for all {, and its other factor, 2{, changes from negative to positive at this point, so by the First
Derivative Test, i has a local and absolute minimum at { = 0.
(b) j 0 ({) = 2{i 0 ({2 ) i j 00 ({) = 2[{i 00 ({2 )(2{) + i 0 ({2 )] = 4{2 i 00 ({2 ) + 2i 0 ({2 ) by the Product Rule and the Chain
Rule. But {2 A 0 for all { 6= 0, i 00 ({2 ) A 0 [since i is CU for { A 0], and i 0 ({2 ) A 0 for all { 6= 0, so since all of its
factors are positive, j 00 ({) A 0 for { 6= 0. Whether j 00 (0) is positive or 0 doesn’t matter [since the sign of j00 does not
change there]; j is concave upward on R.
50. Call the two integers { and |. Then { + 4| = 1000, so { = 1000 3 4|. Their product is S = {| = (1000 3 4|)|, so our
problem is to maximize the function S (|) = 1000| 3 4| 2 , where 0 ? | ? 250 and | is an integer. S 0 (|) = 1000 3 8|, so
S 0 (|) = 0 C | = 125. S 00 (|) = 38 ? 0, so S (125) = 62,500 is an absolute maximum. Since the optimal | turned
out to be an integer, we have found the desired pair of numbers, namely { = 1000 3 4(125) = 500 and | = 125.
F
F |D{1 + E|1 + F|
I
51. If E = 0, the line is vertical and the distance from { = 3 to ({1 > |1 ) is {1 + =
, so assume
D
D
D2 + E 2
E 6= 0. The square of the distance from ({1 > |1 ) to the line is i ({) = ({ 3 {1 )2 + (| 3 |1 )2 where D{ + E| + F = 0, so
2
F
D
we minimize i ({) = ({ 3 {1 )2 + 3 { 3
3 |1
E
E
D
F
D
3
3 |1
.
i i 0 ({) = 2 ({ 3 {1 ) + 2 3 { 3
E
E
E
E 2 {1 3 DE|1 3 DF
D2
00
i ({) = 0 i { =
and this gives a minimum since i ({) = 2 1 + 2 A 0. Substituting
D2 + E 2
E
0
this value of { into i ({) and simplifying gives i ({) =
(D{1 + E|1 + F)2
, so the minimum distance is
D2 + E 2
s
|D{1 + E|1 + F|
I
i({) =
.
D2 + E 2
52. On the hyperbola {| = 8, if g({) is the distance from the point ({> |) = ({> 8@{) to the point (3> 0), then
[g({)]2 = ({ 3 3)2 + 64@{2 = i ({). i 0 ({) = 2({ 3 3) 3 128@{3 = 0 i {4 3 3{3 3 64 = 0 i
({ 3 4) {3 + {2 + 4{ + 16 = 0 i { = 4 since the solution must have { A 0. Then | =
8
4
= 2, so the point is (4> 2).
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CHAPTER 4 REVIEW
53.
By similar triangles,
453
u
|
= I
, so the area of the triangle is
2
{
{ 3 2u{
u{2
i
D({) = 12 (2|){ = {| = I
{2 3 2u{
I
I
2u{ {2 3 2u{ 3 u{2 ({ 3 u)@ {2 3 2u{
u{2 ({ 3 3u)
D0 ({) =
=0
=
2
{ 3 2u{
({2 3 2u{)3@2
when { = 3u=
I
u(9u2 )
D0 ({) ? 0 when 2u ? { ? 3u, D0 ({) A 0 when { A 3u. So { = 3u gives a minimum and D(3u) = I
= 3 3 u2 .
3u
The volume of the cone is Y = 13 | 2 (u + {) = 13 (u2 3 {2 )(u + {), 3u $ { $ u.
54.
Y 0 ({) =
=
2
3 [(u
3 (u
3 {2 ) (1) + (u + {)(32{)] =
3 [(u
+ {)(u 3 { 3 2{)]
+ {)(u 3 3{) = 0 when { = 3u or { = u@3.
Now Y (u) = 0 = Y (3u), so the maximum occurs at { = u@3 and the volume is
u 4u
32u3
u2
=
u2 3
=
.
Y
3
3
9
3
81
I
We minimize O({) = |S D| + |S E| + |S F| = 2 {2 + 16 + (5 3 {),
I
I
0 $ { $ 5. O0 ({) = 2{ {2 + 16 3 1 = 0 C 2{ = {2 + 16 C
4{2 = {2 + 16 C { = I43 . O(0) = 13, O I43 E 11=9, O(5) E 12=8, so the
55.
minimum occurs when { =
56.
4
I
3
E 2=3.
If |FG| = 2, the last part of O({) changes from (5 3 {) to (2 3 {) with
0 $ { $ 2. But we still get O0 ({) = 0 C { =
[0> 2]. Now O(0) = 10 and O(2) = 2
I4 ,
3
which isn’t in the interval
I
I
20 = 4 5 E 8=9. The minimum occurs
when S = F.
57. y = N
u
O
F
+
F
O
gy
N
= s
gO
2 (O@F) + (F@O)
i
1
F
3 2
F
O
=0 C
1
F
= 2
F
O
C O2 = F 2
C O = F.
This gives the minimum velocity since y 0 ? 0 for 0 ? O ? F and y 0 A 0 for O A F.
58.
We minimize the surface area V = u2 + 2uk + 12 (4u2 ) = 3u2 + 2uk.
Y 3 23 u3
Y
Solving Y = u2 k + 23 u3 for k, we get k =
=
3 23 u, so
u2
u2
Y
2Y
2
3
u
= 53 u2 +
V(u) = 3u2 + 2u
.
3
u2
u
2Y
V (u) = 3 2 +
u
0
10
u
3
=
10
u3
3
3 2Y
u2
=0 C
10
u3
3
3Y
= 2Y C u =
5
3
C u=
u
3
3Y
.
5
[continued]
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
This gives an absolute minimum since V 0 (u) ? 0 for 0 ? u ?
u
3
3Y
and V 0 (u) A 0 for u A
5
u
3
3Y
. Thus,
5
3Y
s
s
u
Y 3 23 ·
Y 3 25 Y 3 (5)2
3Y 3 (5)2
3Y
5
v
s
s
=u
=
=
= 3
k=
3
3
2
2
5
2
(3Y )
5 (3Y )
3 (3Y )
(5)2
59. Let { denote the number of $1 decreases in ticket price. Then the ticket price is $12 3 $1({), and the average attendance is
11,000 + 1000({). Now the revenue per game is
U({) = (price per person) × (number of people per game)
= (12 3 {)(11,000 + 1000{) = 31000{2 + 1000{ + 132,000
for 0 $ { $ 4 [since the seating capacity is 15,000] i U0 ({) = 32000{ + 1000 = 0 C { = 0=5. This is a
maximum since U00 ({) = 32000 ? 0 for all {. Now we must check the value of U({) = (12 3 {)(11,000 + 1000{) at
{ = 0=5 and at the endpoints of the domain to see which value of { gives the maximum value of U.
U(0) = (12)(11,000) = 132,000, U(0=5) = (11=5)(11,500) = 132,250, and U(4) = (8)(15,000) = 120,000. Thus, the
maximum revenue of $132,250 per game occurs when the average attendance is 11,500 and the ticket price is $11=50.
60. (a) F({) = 1800 + 25{ 3 0=2{2 + 0=001{3 and
U({) = {s({) = 48=2{ 3 0=03{2 .
The pro¿t is maximized when F 0 ({) = U0 ({).
From the ¿gure, we estimate that the tangents are parallel when { E 160.
(b) F 0 ({) = 25 3 0=4{ + 0=003{2 and U0 ({) = 48=2 3 0=06{. F 0 ({) = U0 ({) i 0=003{2 3 0=34{ 3 23=2 = 0 i
{1 E 161=3 ({ A 0). U00 ({) = 30=06 and F 00 ({) = 30=4 + 0=006{, so U00 ({1 ) = 30=06 ? F 00 ({1 ) E 0=57 i
pro¿t is maximized by producing 161 units.
(c) f({) =
1800
F({)
=
+ 25 3 0=2{ + 0=001{2 is the average cost. Since
{
{
the average cost is minimized when the marginal cost equals the average
cost, we graph f({) and F 0 ({) and estimate the point of intersection.
From the ¿gure, F 0 ({) = f({) C { E 144.
61. i ({) = {5 3 {4 + 3{2 3 3{ 3 2
i i 0 ({) = 5{4 3 4{3 + 6{ 3 3, so {q+1 = {q 3
{5q 3 {4q + 3{2q 3 3{q 3 2
.
5{4q 3 4{3q + 6{q 3 3
Now {1 = 1 i {2 = 1=5 i {3 E 1=343860 i {4 E 1=300320 i {5 E 1=297396 i
{6 E 1=297383 E {7 , so the root in [1> 2] is 1=297383, to six decimal places.
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CHAPTER 4 REVIEW
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455
62. Graphing | = sin { and | = {2 3 3{ + 1 shows that there are two roots,
one about 0=3 and the other about 2=8. i ({) = sin { 3 {2 + 3{ 3 1 i
i 0 ({) = cos { 3 2{ + 3 i {q+1 = {q 3
sin {q 3 {2q + 3{q 3 1
.
cos {q 3 2{q + 3
Now {1 = 0=3 i {2 E 0=268552 i {3 E 0=268881 E {4 and
{1 = 2=8 i {2 E 2=770354 i {3 E 2=770058 E {4 , so to six
decimal places, the roots are 0=268881 and 2=770058.
63. i (w) = cos w + w 3 w2
i i 0 (w) = 3 sin w + 1 3 2w. i 0 (w) exists for all
w, so to ¿nd the maximum of i , we can examine the zeros of i 0 .
From the graph of i 0 , we see that a good choice for w1 is w1 = 0=3.
Use j(w) = 3 sin w + 1 3 2w and j0 (w) = 3 cos w 3 2 to obtain
w2 E 0=33535293, w3 E 0=33541803 E w4 . Since i 00 (w) = 3 cos w 3 2 ? 0
for all w, i(0=33541803) E 1=16718557 is the absolute maximum.
64. | = i ({) = { sin {, 0 $ { $ 2.
A. G = [0> 2] B. |-intercept: i (0) = 0; {-intercepts: i({) = 0 C { = 0 or
sin { = 0 C { = 0, , or 2. C. There is no symmetry on G, but if i is de¿ned for all real numbers {, then i is an even
function. D. No asymptote E. i 0 ({) = { cos { + sin {. To ¿nd critical numbers in (0> 2), we graph i 0 and see that there
are two critical numbers, about 2 and 4=9. To ¿nd them more precisely, we use Newton’s method, setting
j({) = i 0 ({) = { cos { + sin {, so that j 0 ({) = i 00 ({) = 2 cos { 3 { sin { and {q+1 = {q 3
{q cos {q + sin {q
.
2 cos {q 3 {q sin {q
{1 = 2 i {2 E 2=029048, {3 E 2=028758 E {4 and {1 = 4=9 i {2 E 4=913214, {3 E 4=913180 E {4 , so the
critical numbers, to six decimal places, are u1 = 2=028758 and u2 = 4=913180. By checking sample values of i 0 in (0> u1 ),
(u1 > u2 ), and (u2 > 2), we see that i is increasing on (0> u1 ), decreasing on (u1 > u2 ), and increasing on (u2 > 2). F. Local
maximum value i (u1 ) E 1=819706, local minimum value i (u2 ) E 34=814470. G. i 00 ({) = 2 cos { 3 { sin {. To ¿nd
points where i 00 ({) = 0, we graph i 00 and ¿nd that i 00 ({) = 0 at about 1 and 3=6. To ¿nd the values more precisely, we use
Newton’s method. Set k({) = i 00 ({) = 2 cos { 3 { sin {. Then k0 ({) = 33 sin { 3 { cos {, so
{q+1 = {q 3
2 cos {q 3 {q sin {q
. {1 = 1 i {2 E 1=078028, {3 E 1=076874 E {4 and {1 = 3=6 i
33 sin {q 3 {q cos {q
{2 E 3=643996> {3 E 3=643597 E {4 , so the zeros of i 00 , to six decimal places, are u3 = 1=076874 and u4 = 3=643597.
By checking sample values of i 00 in (0> u3 ), (u3 > u4 ), and (u4 > 2), we see that i
H.
is CU on (0> u3 ) > CD on (u3 > u4 ), and CU on (u4 > 2). i has inÀection points at
(u3 > i(u3 ) E 0=948166) and (u4 > i (u4 ) E 31=753240).
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
1
1 3 {2
65. i 0 ({) = cos { 3 (1 3 {2 )31@2 = cos { 3 I
i i({) = 2h{ + sec { + Fq on the interval q 3 2 > q + 2 .
66. i 0 ({) = 2h{ + sec { tan {
67. i 0 ({) =
i i({) = sin { 3 sin31 { + F
I
I
3
{3 + {2 = {3@2 + {2@3
68. i 0 ({) = sinh { + 2 cosh {
i i ({) =
{5@3
{5@2
+
+ F = 25 {5@2 + 35 {5@3 + F
5@2
5@3
i i({) = cosh { + 2 sinh { + F.
i (0) = 1 + 0 + F and i (0) = 2 i F = 1, so i({) = cosh { + 2 sinh { + 1.
69. i 0 (w) = 2w 3 3 sin w
i i (w) = w2 + 3 cos w + F.
i (0) = 3 + F and i (0) = 5 i F = 2, so i (w) = w2 + 3 cos w + 2.
70. i 0 (x) =
i (1) =
I
x2 + x
= x + x31@2
x
1
2
i i (x) = 12 x2 + 2x1@2 + F.
+ 2 + F and i(1) = 3 i F = 12 , so i (x) = 12 x2 + 2
71. i 00 ({) = 1 3 6{ + 48{2
I
x + 12 .
i i 0 ({) = { 3 3{2 + 16{3 + F. i 0 (0) = F and i 0 (0) = 2 i F = 2, so
i 0 ({) = { 3 3{2 + 16{3 + 2 and hence, i({) = 12 {2 3 {3 + 4{4 + 2{ + G.
i (0) = G and i (0) = 1 i G = 1, so i ({) = 12 {2 3 {3 + 4{4 + 2{ + 1.
72. i 00 ({) = 2{3 + 3{2 3 4{ + 5
i ({) =
1 5
{
10
+
i (1) =
1
10
1
4
i ({) =
1 5
{
10
+
1 4
{
4
3
2
3
3
2 3
{
3
+
5
2
i i 0 ({) = 12 {4 + {3 3 2{2 + 5{ + F
+
5 2
{
2
+ F{ + G. i(0) = G and i (0) = 2 i G = 2.
6
+ F + 2 and i (1) = 0 i F = 3 60
3
+ 14 {4 3 23 {3 + 52 {2 3
73. y(w) = v0 (w) = 2w 3
1
1 + w2
i
251
{
60
15
60
+
40
60
3
150
60
3
120
60
= 3 251
, so
60
+ 2.
i v(w) = w2 3 tan31 w + F.
v(0) = 0 3 0 + F = F and v(0) = 1 i F = 1, so v(w) = w2 3 tan31 w + 1.
74. d(w) = y 0 (w) = sin w + 3 cos w
i y(w) = 3 cos w + 3 sin w + F.
y(0) = 31 + 0 + F and y(0) = 2 i F = 3, so y(w) = 3 cos w + 3 sin w + 3 and v(w) = 3 sin w 3 3 cos w + 3w + G.
v(0) = 33 + G and v(0) = 0 i G = 3, and v(w) = 3 sin w 3 3 cos w + 3w + 3.
75. (a) Since i is 0 just to the left of the |-axis, we must have a minimum of I at the same place since we are increasing through
(0> 0) on I . There must be a local maximum to the left of { = 33, since i changes from positive to negative there.
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
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NOT FOR SALE
CHAPTER 4 REVIEW
(b) i ({) = 0=1h{ + sin { i
¤
457
(c)
I ({) = 0=1h{ 3 cos { + F. I (0) = 0 i
0=1 3 1 + F = 0 i F = 0=9, so
I ({) = 0=1h{ 3 cos { + 0=9.
i i 0 ({) = 4{3 + 3{2 + 2f{. This is 0 when { 4{2 + 3{ + 2f = 0 C { = 0
I
33 ± 9 3 32f
or 4{2 + 3{ + 2f = 0. Using the quadratic formula, we ¿nd that the roots of this last equation are { =
.
8
76. i ({) = {4 + {3 + f{2
Now if 9 3 32f ? 0 C f A
9
,
32
then (0> 0) is the only critical point, a minimum. If f =
9
,
32
then there are two critical
points (a minimum at { = 0, and a horizontal tangent with no maximum or minimum at { = 3 38 ) and if f ?
9
,
32
then there are
three critical points except when f = 0, in which case the root with the + sign coincides with the critical point at { = 0. For
I
I
3
9 3 32f
9 3 32f
3
9
0 ? f ? 32
, there is a minimum at { = 3 3
, a maximum at { = 3 +
, and a minimum at { = 0.
8
8
8
8
For f = 0, there is a minimum at { = 3 34 and a horizontal tangent with no extremum at { = 0, and for f ? 0, there is a
I
9 3 32f
3
maximum at { = 0, and there are minima at { = 3 ±
. Now we calculate i 00 ({) = 12{2 + 6{ + 2f.
8
8
I
36 ± 36 3 4 · 12 · 2f
. So if 36 3 96f $ 0 C f D 38 , then there is no inÀection
The roots of this equation are { =
24
I
1
9 3 24f
point. If f ? 38 , then there are two inÀection points at { = 3 ±
.
4
12
Value of f
No. of CP No. of IP
f?0
f=0
0?f?
f=
9
32
9
32
9
32
?f?
fD
3
8
3
8
3
2
2
2
3
2
2
2
1
2
1
0
77. Choosing the positive direction to be upward, we have d(w) = 39=8
0
i y(w) = 39=8w + y0 , but y(0) = 0 = y0
2
i
2
y(w) = 39=8w = v (w) i v(w) = 34=9w + v0 , but v(0) = v0 = 500 i v(w) = 34=9w + 500. When v = 0,
t
t
500
E
10=1
i
y(w
)
=
39=8
E 398=995 m@s. Since the canister has been
34=9w2 + 500 = 0 i w1 = 500
1
4=9
4=9
designed to withstand an impact velocity of 100 m@s, the canister will not burst.
78. Let vD (w) and vE (w) be the position functions for cars D and E and let i (w) = vD (w) 3 v(w). Since D passed E twice, there
must be three values of w such that i (w) = 0. Then by three applications of Rolle’s Theorem (see Exercise 4.2.22), there is a
number f such that i 00 (f) = 0. So v00D (f) = v00E (f); that is, D and E had equal accelerations at w = f. We assume that i is
continuous on [0> W ] and twice differentiable on (0> W ), where W is the total time of the race.
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
79. (a)
The cross-sectional area of the rectangular beam is
I
D = 2{ · 2| = 4{| = 4{ 100 3 {2 , 0 $ { $ 10, so
gD
= 4{ 12 (100 3 {2 )31@2 (32{) + (100 3 {2 )1@2 · 4
g{
4[3{2 + 100 3 {2 ]
34{2
2 1@2
=
+
4(100
3
{
)
=
.
(100 3 {2 )1@2
(100 3 {2 )1@2
gD
= 0 when 3{2 + 100 3 {2 = 0 i
g{
{2 = 50 i { =
t
I
I 2 I
50 E 7=07 i | = 100 3
50 = 50.
Since D(0) = D(10) = 0, the rectangle of maximum area is a square.
The cross-sectional area of each rectangular plank (shaded in the ¿gure) is
I I
I I
D = 2{ | 3 50 = 2{ 100 3 {2 3 50 , 0 $ { $ 50, so
(b)
I I
gD
= 2 100 3 {2 3 50 + 2{ 12 (100 3 {2 )31@2 (32{)
g{
= 2(100 3 {2 )1@2 3 2
Set
I
50 3
2{2
(100 3 {2 )1@2
I
I
gD
= 0: (100 3 {2 ) 3 50 (100 3 {2 )1@2 3 {2 = 0 i 100 3 2{2 = 50 (100 3 {2 )1@2
g{
i
10,000 3 400{2 + 4{4 = 50(100 3 {2 ) i 4{4 3 350{2 + 5000 = 0 i 2{4 3 175{2 + 2500 = 0 i
I
I
175 ± 10,625
E 69=52 or 17=98 i { E 8=34 or 4=24. But 8=34 A 50, so {1 E 4=24 i
{2 =
4
s
I
I
| 3 50 = 100 3 {21 3 50 E 1=99. Each plank should have dimensions about 8 12 inches by 2 inches.
(c) From the ¿gure in part (a), the width is 2{ and the depth is 2|, so the strength is
V = n(2{)(2|)2 = 8n{| 2 = 8n{(100 3 {2 ) = 800n{ 3 8n{3 , 0 $ { $ 10. gV@g{ = 800n 3 24n{2 = 0 when
t
I
I
10
i {= I
i | = 200
= 10I3 2 = 2 {. Since V(0) = V(10) = 0, the
24n{2 = 800n i {2 = 100
3
3
3
maximum strength occurs when { =
10
I
.
3
The dimensions should be
20
I
3
E 11=55 inches by
I
20
I 2
3
E 16=33 inches.
j
{2 . The parabola intersects the line when
2y 2 cos2 j
{2 i
(tan ){ = (tan ){ 3 2
2y cos2 80. (a)
| = (tan ){ 3
{=
U() =
{
=
cos (tan 3 tan )2y2 cos2 j
sin sin 3
cos cos = (sin cos 3 sin cos )
2y 2 cos2 =
j cos i
sin sin 2y 2 cos 3
(cos cos )
cos cos j cos2 2y2 cos 2y 2 cos = sin( 3 )
2
j cos j cos2 c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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NOT FOR SALE
CHAPTER 4 REVIEW
(b) U0 () =
=
¤
459
2y2
2y 2
[cos · cos( 3 ) + sin( 3 )(3 sin )] =
cos[ + ( 3 )]
2
j cos j cos2 2y 2
cos(2 3 ) = 0
j cos2 when cos(2 3 ) = 0 i 2 3 =
2
i =
@2 + = + . The First Derivative Test shows that this
2
4
2
gives a maximum value for U(). [This could be done without calculus by applying the formula for sin { cos | to U().]
Replacing by 3 in part (a), we get U() =
(c)
2y 2 cos sin( + )
.
j cos2 Proceeding as in part (b), or simply by replacing by 3 in the result of
part (b), we see that U() is maximized when =
81. We ¿rst show that
i 0 ({) =
3 .
4
2
{
{
? tan31 { for { A 0. Let i ({) = tan31 { 3
. Then
1 + {2
1 + {2
1
1(1 + {2 ) 3 {(2{)
(1 + {2 ) 3 (1 3 {2 )
2{2
3
=
=
A 0 for { A 0. So i ({) is increasing
1 + {2
(1 + {2 )2
(1 + {2 )2
(1 + {2 )2
on (0> "). Hence, 0 ? { i 0 = i (0) ? i ({) = tan31 { 3
{
{
. So
? tan31 { for 0 ? {. We next show
1 + {2
1 + {2
that tan31 { ? { for { A 0. Let k({) = { 3 tan31 {. Then k0 ({) = 1 3
1
{2
=
A 0. Hence, k({) is increasing
1 + {2
1 + {2
on (0> "). So for 0 ? {, 0 = k(0) ? k({) = { 3 tan31 {. Hence, tan31 { ? { for { A 0, and we conclude that
{
? tan31 { ? { for { A 0.
1 + {2
82. If i 0 ({) ? 0 for all {, i 00 ({) A 0 for |{| A 1, i 00 ({) ? 0 for |{| ? 1, and
lim [i ({) + {] = 0, then i is decreasing everywhere, concave up on
{<±"
(3"> 31) and (1> "), concave down on (31> 1), and approaches the line
| = 3{ as { < ±". An example of such a graph is sketched.
83. (a) L =
n cos n(k@g)
k
k
k
=
= n 3 = n I
3 = n
g2
g2
g
(1600 + k2 )3@2
402 + k2
i
(1600 + k2 )3@2 3 k 32 (1600 + k2 )1@2 · 2k
gL
n(1600 + k2 )1@2 (1600 + k2 3 3k2 )
=n
=
2
3@2
2
gk
(1600 + k2 )3
[(1600 + k ) ]
=
n(1600 3 2k2 )
(1600 + k2 )5@2
[n is the constant of proportionality]
Set gL@gk = 0: 1600 3 2k2 = 0 i k2 = 800 i k =
I
maximum at k = 20 2 E 28 ft.
I
I
800 = 20 2. By the First Derivative Test, L has a local
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CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
n[(k 3 4)@g]
n(k 3 4)
n cos =
=
g2
g2
g3
33@2
n(k 3 4)
=
= n(k 3 4) (k 3 4)2 + {2
3@2
2
2
[(k 3 4) + { ]
L =
(b)
g{
= 4 ft@s
gw
35@2
gL
gL g{
g{
=
·
= n(k 3 4) 3 32 (k 3 4)2 + {2
· 2{ ·
gw
g{ gw
gw
312{n(k
3 4)
35@2
·4=
= n(k 3 4)(33{) (k 3 4)2 + {2
2
[(k 3 4) + {2 ]5@2
gL 480n(k 3 4)
=3
gw { = 40
[(k 3 4)2 + 1600]5@2
84. (a) Y 0 (w) is the rate of change of the volume of the water with respect to time. K 0 (w) is the rate of change of the height of the
water with respect to time. Since the volume and the height are increasing, Y 0 (w) and K 0 (w) are positive.
(b) Y 0 (w) is constant, so Y 00 (w) is zero (the slope of a constant function is 0).
(c) At ¿rst, the height K of the water increases quickly because the tank is narrow. But as the sphere widens, the rate of
increase of the height slows down, reaching a minimum at w = w2 . Thus, the height is increasing at a decreasing rate on
(0> w2 ), so its graph is concave downward and K 00 (w1 ) ? 0. As the sphere narrows for w A w2 , the rate of increase of the
height begins to increase, and the graph of K is concave upward. Therefore, K 00 (w2 ) = 0 and K 00 (w3 ) A 0.
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°
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NOT FOR SALE
PROBLEMS PLUS
2
2
1. Let | = i ({) = h3{ . The area of the rectangle under the curve from 3{ to { is D({) = 2{h3{ where { D 0. We maximize
2
2
2
D({): D0 ({) = 2h3{ 3 4{2 h3{ = 2h3{ 1 3 2{2 = 0 i { =
I1
2
for 0 $ { ?
and D0 ({) ? 0 for { A
1
I
.
2
I1 .
2
This gives a maximum since D0 ({) A 0
We next determine the points of inÀection of i ({). Notice that
2
i 0 ({) = 32{h3{ = 3D({). So i 00 ({) = 3D0 ({) and hence, i 00 ({) ? 0 for 3 I12 ? { ?
and { A
I1 .
2
I1
2
and i 00 ({) A 0 for { ? 3 I12
So i({) changes concavity at { = ± I12 , and the two vertices of the rectangle of largest area are at the inÀection
points.
2. Let i({) = sin { 3 cos { on [0> 2] since i has period 2. i 0 ({) = cos { + sin { = 0
C cos { = 3 sin { C
I
= 2,
or
Evaluating i at its critical numbers and endpoints, we get i (0) = 31, i 3
tan { = 31 C { =
4
I
I
I
7 i 4 = 3 2, and i (2) = 31. So i has absolute maximum value 2 and absolute minimum value 3 2. Thus,
I
I
I
3 2 $ sin { 3 cos { $ 2 i |sin { 3 cos {| $ 2.
3
4
7
.
4
3. i ({) has the form hj({) , so it will have an absolute maximum (minimum) where j has an absolute maximum (minimum).
2
j({) = 10|{ 3 2| 3 { =
0
j ({) =
+
+
10({ 3 2) 3 {2
if { 3 2 A 0
10[3({ 3 2)] 3 {2 if { 3 2 ? 0
=
+
3{2 + 10{ 3 20 if { A 2
3{2 3 10{ + 20 if { ? 2
i
32{ + 10 if { A 2
32{ 3 10 if { ? 2
0
j ({) = 0 if { = 35 or { = 5, and j0 (2) does not exist, so the critical numbers of j are 35, 2, and 5. Since j00 ({) = 32 for
all { 6= 2, j is concave downward on (3"> 2) and (2> "), and j will attain its absolute maximum at one of the critical
numbers. Since j(35) = 45, j(2) = 34, and j(5) = 5, we see that i (35) = h45 is the absolute maximum value of i . Also,
lim j({) = 3", so lim i ({) = lim hj({) = 0= But i ({) A 0 for all {, so there is no absolute minimum value of i .
{<"
{<"
{<"
4. {2 | 2 4 3 {2 4 3 | 2 = {2 4 3 {2 | 2 4 3 | 2 = i ({)i (|), where i (w) = w2 4 3 w2 . We will show that 0 $ i (w) $ 4
for |w| $ 2, which gives 0 $ i ({)i (|) $ 16 for |{| $ 2 and ||| $ 2.
I
i (w) = 4w2 3 w4 i i 0 (w) = 8w 3 4w3 = 4w 2 3 w2 = 0 i w = 0 or ± 2.
I i (0) = 0, i ± 2 = 2(4 3 2) = 4, and i (2) = 0. So 0 is the absolute minimum value of i(w) on [32> 2] and 4 is the
absolute maximum value of i (w) on [32> 2]. We conclude that 0 $ i (w) $ 4 for |w| $ 2 and hence, 0 $ i ({)i (|) $ 42 or
0 $ {2 4 3 {2 |2 4 3 |2 $ 16.
5. | =
sin {
{
i |0 =
{ cos { 3 sin {
{2
i | 00 =
3{2 sin { 3 2{ cos { + 2 sin {
. If ({> |) is an inÀection point,
{3
then |00 = 0 i (2 3 {2 ) sin { = 2{ cos { i (2 3 {2 )2 sin2 { = 4{2 cos2 { i
(2 3 {2 )2 sin2 { = 4{2 (1 3 sin2 {) i (4 3 4{2 + {4 ) sin2 { = 4{2 3 4{2 sin2 { i
(4 + {4 ) sin2 { = 4{2
i ({4 + 4)
sin2 {
sin {
.
= 4 i |2 ({4 + 4) = 4 since | =
{2
{
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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462
NOT FOR SALE
¤
CHAPTER 4 PROBLEMS PLUS
6. Let S d> 1 3 d2 be the point of contact. The equation of the tangent line at S is | 3 1 3 d2 = (32d)({ 3 d)
| 3 1 + d2 = 32d{ + 2d2
i
i | = 32d{ + d2 + 1. To ¿nd the {-intercept, put | = 0: 2d{ = d2 + 1 i
d2 + 1
. To ¿nd the |-intercept, put { = 0: | = d2 + 1. Therefore, the area of the triangle is
2d
2
2
2
2
d +1
d +1
1 d2 + 1 2
d +1 =
. Therefore, we minimize the function D(d) =
, d A 0.
2
2d
4d
4d
2
2
2
(4d)2 d2 + 1 (2d) 3 d2 + 1 (4)
d + 1 [4d2 3 d2 + 1 ]
d + 1 3d2 3 1
0
D (d) =
=
=
.
16d2
4d2
4d2
{=
D0 (d) = 0 when 3d2 3 1 = 0 i d =
D0 (d) A 0 for d A I13 . So by the First Derivative
Test, there is an absolute minimum when d = I13 . The required point is I13 > 23 and the corresponding minimum area
is D I13 =
7. Let O = lim
{<0
I1 .
3
D0 (d) ? 0 for d ?
1
I
,
3
I
4 3
9 .
d{2 + sin e{ + sin f{ + sin g{
. Now O has the indeterminate form of type 00 , so we can apply l’Hospital’s
3{2 + 5{4 + 7{6
Rule. O = lim
{<0
2d{ + e cos e{ + f cos f{ + g cos g{
. The denominator approaches 0 as { < 0, so the numerator must also
6{ + 20{3 + 42{5
approach 0 (because the limit exists). But the numerator approaches 0 + e + f + g, so e + f + g = 0. Apply l’Hospital’s Rule
again. O = lim
{<0
2d 3 e2 sin e{ 3 f2 sin f{ 3 g 2 sin g{
2d
2d 3 0
=
, which must equal 8.
=
6 + 60{2 + 210{4
6+0
6
2d
= 8 i d = 24. Thus, d + e + f + g = d + (e + f + g) = 24 + 0 = 24.
6
8. Case (i) (¿rst graph): For { + | D 0, that is, | D 3{, |{ + || = { + | $ h{
i | $ h{ 3 {.
Note that | = h{ 3 { is always above the line | = 3{ and that | = 3{ is a slant asymptote.
Case (ii) (second graph): For { + | ? 0, that is, | ? 3{, |{ + || = 3{ 3 | $ h{
Note that 3{ 3 h{ is always below the line | = 3{ and | = 3{ is a slant asymptote.
i | D 3{ 3 h{ .
Putting the two pieces together gives the third graph.
9. Differentiating {2 + {| + | 2 = 12 implicitly with respect to { gives 2{ + | + {
At a highest or lowest point,
g|
g|
g|
2{ + |
+ 2|
= 0, so
=3
.
g{
g{
g{
{ + 2|
g|
= 0 C | = 32{. Substituting 32{ for | in the original equation gives
g{
{2 + {(32{) + (32{)2 = 12, so 3{2 = 12 and { = ±2. If { = 2, then | = 32{ = 34, and if { = 32 then | = 4. Thus,
the highest and lowest points are (32> 4) and (2> 34).
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 4 PROBLEMS PLUS
| = f{3 + h{
10.
i | 0 = 3f{2 + h{
¤
463
i | 00 = 6f{ + h{ . The curve will have
inÀection points when | 00 changes sign. | 00 = 0 i 36f{ = h{ , so |00 will change
sign when the line | = 36f{ intersects the curve | = h{ (but is not tangent to it).
Note that if f = 0, the curve is just | = h{ , which has no inÀection point.
The ¿rst ¿gure shows that for f A 0, | = 36f{ will intersect | = h{ once, so
| = f{3 + h{ will have one inÀection point.
The second ¿gure shows that for f ? 0, the line | = 36f{ can intersect the curve
| = h{ in two points (two inÀection points), be tangent to it (no inÀection point), or not
intersect it (no inÀection point). The tangent line at (d> hd ) has slope hd , but from the
diagram we see that the slope is
hd
hd
. So
= hd
d
d
i d = 1. Thus, the slope is h.
The line | = 36f{ must have slope greater than h, so 36f A h i f ? 3h@6.
3
{
Therefore, the curve | = f{ + h will have one inÀection point if f A 0 and two inÀection points if f ? 3h@6.
11. | = {2
i | 0 = 2{, so the slope of the tangent line at S (d> d2 ) is 2d and the slope of the normal line is 3
An equation of the normal line is | 3 d2 = 3
1
({ 3 d). Substitute {2 for | to ¿nd the {-coordinates of the two points of
2d
intersection of the parabola and the normal line. {2 3 d2 = 3
{=
31 ±
1
for d 6= 0.
2d
{
1
+
2d 2
i 2d{2 + { 3 2d3 3 d = 0 i
s
s
I
1 3 4(2d) (32d3 3 d)
31 ± (4d2 + 1)2
31 ± 1 + 16d4 + 8d2
31 ± (4d2 + 1)
=
=
=
2(2d)
4d
4d
4d
34d2 3 2
1
4d2
or
, or equivalently, d or 3d 3
.
4d
4d
2d
#
2 $
1
1
So the point T has coordinates 3d 3
. The square V of the distance from S to T is given by
> 3d 3
2d
2d
=
&2 2
2 %
2
2 1
1
1
1
2
3d 3
3d +
3d
= 32d 3
+
d2 + 1 + 2 3 d2
V = 3d 3
2d
2d
2d
4d
=
V 0 = 8d 3
2 1
1
1
3
1
1
2
4d2 + 2 + 2 + 1 + 2 = 4d2 + 2 + 2 + 1 + 2 +
= 4d2 + 3 + 2 +
4d
4d
4d
4d
16d4
4d
16d4
6
4
3
1
32d6 3 6d2 3 1
3
=
8d
3
3
=
. The only real positive zero of the equation V 0 = 0 is
4d3
16d5
2d3
4d5
4d5
1
9
5
1
d = I . Since V 00 = 8 + 4 + 6 A 0, d = I corresponds to the shortest possible length of the line segment S T.
2d
4d
2
2
12. To sketch the region ({> |) | 2{| $ |{ 3 || $ {2 + | 2 , we consider two cases.
Case 1: { D |
This is the case in which ({> |) lies on or below the line | = {. The double inequality becomes
2{| $ { 3 | $ {2 + | 2 . The right-hand inequality holds if and only if {2 3 { + | 2 + | D 0 C
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464
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CHAPTER 4 PROBLEMS PLUS
2 2
{ 3 12 + | + 12 D
1
2
C ({> |) lies on or outside the circle with radius
I1
2
centered at
1
1
2>32
.
The left-hand inequality holds if and only if 2{| 3 { + | $ 0 C {| 3 12 { + 12 | $ 0 C
{ + 12 | 3 12 $ 3 14 C ({> |) lies on or below the hyperbola { + 12 | 3 12 = 3 14 , which passes through the
origin and approaches the lines | =
Case 2: | D {
1
2
and { = 3 12 asymptotically.
This is the case in which ({> |) lies on or above the line | = {. The double inequality becomes
2{| $ | 3 { $ {2 + | 2 . The right-hand inequality holds if and only if {2 + { + | 2 3 | D 0 C
2 2
{ + 12 + | 3 12 D
1
2
C ({> |) lies on or outside the circle of radius
I1
2
centered at 3 12 > 12 . The left-hand
inequality holds if and only if 2{| + { 3 | $ 0 C {| + 12 { 3 12 | $ 0 C { 3 12 | + 12 $ 3 14 C ({> |) lies
on or above the left-hand branch of the hyperbola { 3 12 | + 12 = 3 14 , which passes through the origin and approaches the
lines | = 3 12 and { =
1
2
asymptotically. Therefore, the region of interest consists of the points on or above the left branch of
the hyperbola { 3 12 | + 12 = 3 14 that are on or outside the circle
2 2
{ + 12 + | 3 12 = 12 , together with the points on or below the right
branch of the hyperbola { + 12 | 3 12 = 3 14 that are on or outside the circle
2 2
{ 3 12 + | + 12 = 12 . Note that the inequalities are unchanged when {
and | are interchanged, so the region is symmetric about the line | = {. So we
need only have analyzed case 1 and then reÀected that region about the line
| = {, instead of considering case 2.
13. D = {1 > {21 and E = {2 > {22 , where {1 and {2 are the solutions of the quadratic equation {2 = p{ + e. Let S = {> {2
and set D1 = ({1 > 0), E1 = ({2 > 0), and S1 = ({> 0). Let i({) denote the area of triangle S DE. Then i ({) can be expressed
in terms of the areas of three trapezoids as follows:
i({) = area (D1 DEE1 ) 3 area (D1 DS S1 ) 3 area (E1 ES S1 )
= 12 {21 + {22 ({2 3 {1 ) 3 12 {21 + {2 ({ 3 {1 ) 3 12 {2 + {22 ({2 3 {)
After expanding and canceling terms, we get
i ({) = 12 {2 {21 3 {1 {22 3 {{21 + {1 {2 3 {2 {2 + {{22 = 12 {21 ({2 3 {) + {22 ({ 3 {1 ) + {2 ({1 3 {2 )
i 0 ({) = 12 3{21 + {22 + 2{({1 3 {2 ) . i 00 ({) = 12 [2({1 3 {2 )] = {1 3 {2 ? 0 since {2 A {1 .
i 0 ({) = 0 i 2{({1 3 {2 ) = {21 3 {22
i {S = 12 ({1 + {2 ).
2 1
{1 2 ({2 3 {1 ) + {22 12 ({2 3 {1 ) + 14 ({1 + {2 )2 ({1 3 {2 )
= 12 12 ({2 3 {1 ) {21 + {22 3 14 ({2 3 {1 )({1 + {2 )2 = 18 ({2 3 {1 ) 2 {21 + {22 3 {21 + 2{1 {2 + {22
= 18 ({2 3 {1 ) {21 3 2{1 {2 + {22 = 18 ({2 3 {1 )({1 3 {2 )2 = 18 ({2 3 {1 )({2 3 {1 )2 = 18 ({2 3 {1 )3
i ({S ) =
1
2
To put this in terms of p and e, we solve the system | = {21 and | = p{1 + e, giving us {21 3 p{1 3 e = 0 i
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NOT FOR SALE
CHAPTER 4 PROBLEMS PLUS
¤
465
I
I
I
3
p 3 p2 + 4e . Similarly, {2 = 12 p + p2 + 4e . The area is then 18 ({2 3 {1 )3 = 18 p2 + 4e ,
and is attained at the point S {S > {2S = S 12 p> 14 p2 .
{1 =
1
2
Note: Another way to get an expression for i ({) is to use the formula for an area of a triangle in terms of the coordinates of
the vertices: i ({) = 12 {2 {21 3 {1 {22 + {1 {2 3 {{21 + {{22 3 {2 {2 .
14. Let { = |DH|, | = |DI | as shown. The area A of the {DHI is A =
1
{|.
2
We
need to ¿nd a relationship between { and |, so that we can take the derivative
gA@g{ and then ¿nd the maximum and minimum areas. Now let D0 be the point
on which D ends up after the fold has been performed, and let S be the intersection
of DD0 and HI . Note that DD0 is perpendicular to HI since we are reÀecting D
through the line HI to get to D0 , and that |DS | = |S D0 | for the same reason.
But |DD0 | = 1, since DD0 is a radius of the circle. Since |DS | + |S D0 | = |DD0 |, we have |DS | = 12 . Another way to express
s
s
the area of the triangle is A = 12 |HI | |DS | = 12 {2 + | 2 12 = 14 {2 + | 2 . Equating the two expressions for A, we get
s
I
1
1
{2 + | 2 i 4{2 | 2 = {2 + | 2 i | 2 4{2 3 1 = {2 i | = {@ 4{2 3 1.
2 {| = 4
(Note that we could also have derived this result from the similarity of 4D0 S H and 4D0 I H; that is,
|D0 S |
|D0 I |
= 0
|S H|
|D H|
1
|
t 2 =
2
{
{2 3 12
i
gA
1
{2
calculate
:A= I
g{
2 4{2 3 1
2{
1
{
{
= I
.) Now we can substitute for | and
i |= I 2
2
4{ 3 1 @2
4{2 3 1
gA
1
=
g{
2
i
%I
&
31@2
4{2 3 1 (2{) 3 {2 12 4{2 3 1
(8{)
. This is 0 when
4{2 3 1
I
31@2
31@2 2
4{2 3 1 3 4{3 4{2 3 1
= 0 C 2{ 4{2 3 1
4{ 3 1 3 2{2 = 0
({ A 0) C 2{2 = 1 i { =
I1 .
2
i
2
4{ 3 1 3 2{2 = 0
So this is one possible value for an extremum. We must also test the endpoints of the
interval over which { ranges. The largest value that { can attain is 1, and the smallest value of { occurs when | = 1 C
I
1 = {@ 4{2 3 1 C {2 = 4{2 3 1 C 3{2 = 1 C { = I13 . This will give the same value of A as will
{ = 1, since the geometric situation is the same (reÀected through the line | = {). We calculate
I 2
1
1@ 2
1
1
1
12
= = 0=25, and A(1) = s
= I E 0=29. So the maximum area is
A I12 = t I 2
2
4
2 4(1)2 3 1
2
3
4 1@ 2 3 1
A(1) = A I13 =
1
I
2 3
and the minimum area is A I12 = 14 .
Another method: Use the angle (see diagram above) as a variable:
A = 12 {| =
1
2
1
2
sec 1
sin 2 = 1 i 2 =
2
2
csc =
1
1
=
. A is minimized when sin 2 is maximal, that is, when
8 sin cos 4 sin 2
i =
.
4
Also note that D0 H = { =
1
2
sec $ 1 i sec $ 2 i
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466
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NOT FOR SALE
CHAPTER 4 PROBLEMS PLUS
i $
3,
and minimized at =
:
4
cos D
1
2
and similarly, D0 I = | =
1
2
csc $ 1 i csc $ 2 i sin $
As above, we ¿nd that A is maximized at these endpoints: A 6 =
A 4 =
1
1
= .
4 sin 2
4
1
2
i D
6.
1
1
1
= I =
= A 3 ;
4 sin 3
4 sin 2
2 3
3
1@{0
15. Suppose that the curve | = d{ intersects the line | = {. Then d{0 = {0 for some {0 A 0, and hence d = {0
. We ¿nd the
maximum value of j({) = {1@{ , { A 0, because if d is larger than the maximum value of this function, then the curve | = d{
1
1 1
1
= {1@{
does not intersect the line | = {. j0 ({) = h(1@{) ln { 3 2 ln { + ·
(1 3 ln {). This is 0 only where
{
{ {
{2
{ = h, and for 0 ? { ? h, i 0 ({) A 0, while for { A h, i 0 ({) ? 0, so j has an absolute maximum of j(h) = h1@h . So if
| = d{ intersects | = {, we must have 0 ? d $ h1@h . Conversely, suppose that 0 ? d $ h1@h . Then dh $ h, so the graph of
| = d{ lies below or touches the graph of | = { at { = h. Also d0 = 1 A 0, so the graph of | = d{ lies above that of | = {
at { = 0. Therefore, by the Intermediate Value Theorem, the graphs of | = d{ and | = { must intersect somewhere between
{ = 0 and { = h.
16. If O = lim
{<"
{+d
{3d
{
= lim
, then O has the indeterminate form 1" , so
1
1
{
3
{+d
{+d
ln({ + d) 3 ln({ 3 d) H
{
+
d
{
3
d
= lim
ln O = lim ln
= lim { ln
= lim
{<"
{<"
{<"
{<"
{3d
{3d
1@{
31@{2
{<"
({ 3 d) 3 ({ + d) 3{2
·
({ + d)({ 3 d)
1
= lim
{<"
2d{2
2d
= lim
= 2d
{<" 1 3 d2@{2
3 d2
{2
Hence, ln O = 2d, so O = h2d . From the original equation, we want O = h1
i 2d = 1 i d = 12 .
i ({) 3 i (0) i ({) |i ({)|
| sin {|
sin {
17. Note that i (0) = 0, so for { 6= 0, = { = |{| $ |{| = { .
{30
i({) 3 i(0) i({) 3 i (0) 0
$ lim sin { = 1.
=
lim
Therefore, |i (0)| = lim
{<0 {<0 {
{<0
{30
{30
But i({) = d1 sin { + d2 sin 2{ + · · · + dq sin q{ i i 0 ({) = d1 cos { + 2d2 cos 2{ + · · · + qdq cos q{, so
|i 0 (0)| = |d1 + 2d2 + · · · + qdq | $ 1.
S
Another solution: We are given that q
n=1 dn sin n{ $ | sin {|. So for { close to 0, and { 6= 0, we have
q
q
q
S
S
S
sin n{ sin n{ sin n{ $
1
i
lim
$
1
i
$ 1. But by l’Hospital’s Rule,
d
d
d
lim
n
n
n
{<0 n=1
{<0 sin { sin { sin { n=1
n=1
q
S
sin n{
n cos n{
= lim
= n, so ndn $ 1.
lim
{<0 sin {
{<0 cos {
n=1
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CHAPTER 4 PROBLEMS PLUS
¤
467
18. Let the circle have radius u, so |RS | = |RT| = u, where R is the center of the circle. Now _S RU has measure 12 , and
_RS U is a right angle, so tan 12 =
|S U|
and the area of 4RS U is
u
1
2
|RS | |S U| = 12 u2 tan 12 . The area of the sector cut
1 1 2
|S V|
|RV|
= 4 u . Let V be the intersection of S T and RU. Then sin 12 =
and cos 12 =
, and
2
u
u
the area of 4RVS is 12 |RV| |S V| = 12 u cos 12 u sin 12 = 12 u2 sin 12 cos 12 = 14 u2 sin .
So E() = 2 12 u2 tan 12 3 14 u2 = u2 tan 12 3 12 and D() = 2 14 u2 3 14 u2 sin = 12 u2 ( 3 sin ). Thus,
by RS and RU is 12 u2
lim
<0+
1 2
u ( 3 sin )
D()
3 sin 1 3 cos H
= lim = lim 1
= lim 22
1
1
1
1
1
2 1
E()
<0+ u tan 3 <0+ 2 tan 3 <0+ 2
2
2
2
2
2 sec 2 3 2
1 3 cos 1 3 cos H
sin 1
= lim
= lim 1
1
<0+ 2 tan sec2 sec2 12 3 1 <0+ tan2 12 2
2
2
2 sin 12 cos 12 cos3 12 sin cos3 12 = lim
= 2 lim cos4 12 = 2(1)4 = 2
= lim
1
1
+
+
+
<0
<0
<0
sin 2 sin 2 = lim
<0+
G
2 |S U|
|UV|
2k sec G 3 2k tan , W2 =
+
=
+
,
f1
f1
f2
f1
f2
19. (a) Distance = rate × time, so time = distance@rate. W1 =
W3 =
(b)
2
s
I
k2 + G2 /4
4k2 + G2
=
.
f1
f1
gW2
2k
2k
· sec tan 3
sec2 = 0 when 2k sec =
g
f1
f2
1 1
1 sin 3
=0 i
f1 cos f2 cos sin 1
=
f1 cos f2 cos 1
1
tan 3
sec f1
f2
i sin =
=0 i
f1
. The First Derivative Test shows that this gives
f2
a minimum.
(c) Using part (a) with G = 1 and W1 = 0=26, we have W1 =
4k2 + G2 = W32 f21
i k=
from part (b) and W2 =
1
2
G
f1
i f1 =
1
0=26
E 3=85 km@s. W3 =
I
4k2 + G2
f1
i
s
s
f1
W 23 f21 3 G2 = 12 (0=34)2 (1@0=26)2 3 12 E 0=42 km. To ¿nd f2 , we use sin =
f2
2k sec G 3 2k tan +
from part (a). From the ¿gure,
f1
f2
f1
f2
f2
f1
i sec = s 2
and tan = s 2
, so
f2 3 f21
f2 3 f21
s
G f22 3 f21 3 2kf1
2kf2
s
W2 = s 2
+
. Using the values for W2 [given as 0.32],
f1 f2 3 f21
f2 f22 3 f21
sin =
s
G f22 3 f21 3 2kf1
2kf2
s
k, f1 , and G> we can graph Y1 = W2 and Y2 = s 2
+
and ¿nd their intersection points.
f1 f2 3 f21
f2 f22 3 f21
Doing so gives us f2 E 4=10 and 7=66, but if f2 = 4=10, then = arcsin(f1 @f2 ) E 69=6 , which implies that point V is to
the left of point U in the diagram. So f2 = 7=66 km@s.
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468
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NOT FOR SALE
CHAPTER 4 PROBLEMS PLUS
20. A straight line intersects the curve | = i ({) = {4 + f{3 + 12{2 3 5{ + 2 in four distinct points if and only if the graph of i
has two inÀection points. i 0 ({) = 4{3 + 3f{2 + 24{ 3 5 and i 00 ({) = 12{2 + 6f{ + 24.
s
36f ± (6f)2 3 4(12)(24)
00
. There are two distinct roots for i 00 ({) = 0 (and hence two inÀection
i ({) = 0 C { =
2(12)
I
points) if and only if the discriminant is positive; that is, 36f2 3 1152 A 0 C f2 A 32 C |f| A 32. Thus, the desired
I
I
values of f are f ? 34 2 or f A 4 2.
Let d = |HI | and e = |EI | as shown in the ¿gure.
21.
Since c = |EI | + |I G|, |I G| = c 3 e. Now
|HG| = |HI | + |I G| = d + c 3 e
s
I
u2 3 {2 + c 3 (g 3 {)2 + d2
t
I
2
I
= u2 3 {2 + c 3 (g 3 {)2 +
u2 3 {2
I
I
= u2 3 {2 + c 3 g2 3 2g{ + {2 + u2 3 {2
Let i({) =
i 0 ({) =
1
2
I
I
u2 3 {2 + c 3 g2 + u2 3 2g{.
2
31@2
31@2
3{
g
u 3 {2
(32{) 3 12 g2 + u2 3 2g{
(32g) = I
+I
.
2
2
2
u 3{
g + u2 3 2g{
i 0 ({) = 0 i
{
g
I
= I
u2 3 {2
g2 + u2 3 2g{
i
g2
{2
=
u2 3 {2
g2 + u2 3 2g{
i
g2 {2 + u2 {2 3 2g{3 = g2 u2 3 g2 {2 i 0 = 2g{3 3 2g2 {2 3 u2 {2 + g2 u2 i
i 0 = 2g{2 ({ 3 g) 3 u2 ({ + g)({ 3 g) i 0 = ({ 3 g) 2g{2 3 u2 ({ + g)
0 = 2g{2 ({ 3 g) 3 u2 {2 3 g2
But g A u A {, so { 6= g. Thus, we solve 2g{2 3 u2 { 3 gu2 = 0 for {:
t
I
3 3u2 ± (3u2 )2 3 4(2g)(3gu2 )
I
u2 ± u4 + 8g 2 u2
{=
=
. Because u4 + 8g2 u2 A u2 , the “negative” can be
2(2g)
4g
I I
I
I
u2 + u u2 + 8g2
u u2 + u2 u2 + 8g2
=
[u A 0] =
u + u2 + 8g2 . The maximum
discarded. Thus, { =
4g
4g
4g
value of |HG| occurs at this value of {.
22.
Let d = FG denote the distance from the center F of the base to the midpoint G of a side of the base.
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© Cengage Learning. All Rights Reserved.
CHAPTER 4 PROBLEMS PLUS
Since {S TU is similar to {GFU,
d
u
= s
k
k(k 3 2u)
¤
469
I
uk
k
i d= s
= uI
.
k 3 2u
k(k 3 2u)
Let e denote one-half the length of a side of the base. The area D of the base is
D = 8(area of {FGH) = 8 12 de = 4d d tan 4 = 4d2 .
The volume of the pyramid is Y =
Now
1
Dk
3
$2
#
I
2
k
k2
4
uI
, with domain k A 2u.
= 4d k =
k = 43 u2
3
k 3 2u
k 3 2u
1
3
(k 3 2u)(2k) 3 k2 (1)
4 k2 3 4ku
4 k(k 3 4u)
gY
4
= u2
= u2
= u2 ·
2
gk
3
(k 3 2u)
3 (k 3 2u)2
3 (k 3 2u)2
2
2
g2 Y
4 2 (k 3 2u) (2k 3 4u) 3 k 3 4ku (2)(k 3 2u)(1)
u
=
·
gk2
3
[(k 3 2u)2 ]2
2(k 3 2u) k2 3 4ku + 4u2 3 k2 3 4ku
4
= u2 ·
3
(k 3 2u)2
and
=
8 2
4u2
32 4
1
u ·
u ·
=
.
3
(k 3 2u)3
3
(k 3 2u)3
The ¿rst derivative is equal to zero for k = 4u and the second derivative is positive for k A 2u, so the volume of the pyramid
is minimized when k = 4u.
To extend our solution to a regular q-gon, we make the following changes:
(1) the number of sides of the base is q
(2) the number of triangles in the base is 2q
(3) _GFH =
q
(4) e = d tan
q
We then obtain the following results: D = qd2 tan
and
k(k 3 4u)
qu2
k2
gY
qu2
,
,Y =
· tan
·
,
=
· tan
·
q
3
q
k 3 2u gk
3
q
(k 3 2u)2
g2 Y
1
8qu4
· tan
·
=
. Notice that the answer, k = 4u, is independent of the number of sides of the base
2
gk
3
q
(k 3 2u)3
of the polygon!
gY
gu
gY
gY
= 4u2
. But
is proportional to the surface area, so
= n · 4u2 for some constant n.
gw
gw
gw
gw
gu
gu
Therefore, 4u2
= n · 4u2 C
= n = constant. An antiderivative of n with respect to w is nw, so u = nw + F.
gw
gw
When w = 0, the radius u must equal the original radius u0 , so F = u0 , and u = nw + u0 . To ¿nd n we use the fact that
23. Y =
4
u3
3
i
when w = 3, u = 3n + u0 and Y = 12 Y0 i 43 (3n + u0 )3 = 12 · 43 u03 i (3n + u0 )3 = 12 u03 i
1
1
1
1
1
3n + u0 = I
u0 i n = 3 u0 I
3 1 . Since u = nw + u0 , u = 3 u0 I
3 1 w + u0 . When the snowball
3
3
3
2
2
has melted completely we have u = 0 i
2
1
u
3 0
I
332
3
I
33= I
E 11 h 33 min longer.
3
3
231
231
1
I
3
2
I
332
3 1 w + u0 = 0 which gives w = I
. Hence, it takes
3
231
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°
licated, or posted to a publicly accessible website, in whole or in part.
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470
¤
NOT FOR SALE
CHAPTER 4 PROBLEMS PLUS
24. By ignoring the bottom hemisphere of the initial spherical bubble, we can rephrase the problem as follows: Prove that the
I
maximum height of a stack of q hemispherical bubbles is q if the radius of the bottom hemisphere is 1. We proceed by
I
induction. The case q = 1 is obvious since 1 is the height of the ¿rst hemisphere. Suppose the assertion is true for q = n
and let’s suppose we have n + 1 hemispherical bubbles forming a stack of maximum height. Suppose the second hemisphere
(counting from the bottom) has radius u= Then by our induction hypothesis (scaled to the setting of a bottom hemisphere of
I
radius u), the height of the stack formed by the top n bubbles is n u. (If it were shorter, then the total stack of n + 1 bubbles
wouldn’t have maximum height.)
The height of the whole stack is K(u) =
I
I
n u + 1 3 u2 . (See the ¿gure.)
We want to choose u so as to maximize K(u). Note that 0 ? u ? 1.
I
u
31
and K 00 (u) =
.
We calculate K 0 (u) = n 3 I
2
(1
3
u2 )3@2
13u
K 0 (u) = 0 C u2 = n(1 3 u2 ) C (n + 1)u2 = n
C u=
u
n
.
n+1
This is the only critical number in (0> 1) and it represents a local maximum
u
n
(hence an absolute maximum) since K 00 (u) ? 0 on (0> 1). When u =
,
n+1
I
u
I
I
n
n
1
n
K(u) = n I
+ 13
+I
= n + 1. Thus, the assertion is true for q = n + 1 when
= I
n+1
n+1
n+1
n+1
it is true for q = n. By induction, it is true for all positive integers q.
Note: In general, a maximally tall stack of q hemispherical bubbles consists of bubbles with radii
1>
u
q31
>
q
u
q32
>=== >
q
u
2
>
q
u
1
=
q
c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
5
INTEGRALS
5.1 Areas and Distances
1. (a) Since i is increasing, we can obtain a lower estimate by using left
endpoints. We are instructed to use four rectangles, so q = 4.
4
S
830
e3d
=
=2
O4 =
i ({l31 ) {{
{{ =
q
4
l=1
= i ({0 ) · 2 + i ({1 ) · 2 + i ({2 ) · 2 + i ({3 ) · 2
= 2[i (0) + i (2) + i (4) + i(6)]
= 2(2 + 3=75 + 5 + 5=75) = 2(16=5) = 33
Since i is increasing, we can obtain an upper estimate by using right
endpoints.
4
S
U4 =
i({l ) {{
l=1
= i({1 ) · 2 + i({2 ) · 2 + i ({3 ) · 2 + i ({4 ) · 2
= 2[i (2) + i (4) + i(6) + i (8)]
= 2(3=75 + 5 + 5=75 + 6) = 2(20=5) = 41
Comparing U4 to O4 , we see that we have added the area of the rightmost upper rectangle, i (8) · 2, to the sum and
subtracted the area of the leftmost lower rectangle, i (0) · 2, from the sum.
8
S
830
=1
i ({l31 ){{
{{ =
(b) O8 =
8
l=1
= 1[i ({0 ) + i({1 ) + · · · + i({7 )]
= i (0) + i(1) + · · · + i (7)
E 2 + 3=0 + 3=75 + 4=4 + 5 + 5=4 + 5=75 + 5=9
= 35=2
U8 =
8
S
i({l ){{ = i (1) + i (2) + · · · + i (8)
add rightmost upper rectangle,
= O8 + 1 · i(8) 3 1 · i (0)
l=1
subtract leftmost lower rectangle
= 35=2 + 6 3 2 = 39=2
2. (a) (i) O6 =
6
S
i ({l31 ){{
l=1
[{{ =
12 3 0
6
= 2]
= 2[i ({0 ) + i({1 ) + i ({2 ) + i ({3 ) + i ({4 ) + i({5 )]
= 2[i (0) + i (2) + i (4) + i (6) + i(8) + i (10)]
E 2(9 + 8=8 + 8=2 + 7=3 + 5=9 + 4=1)
= 2(43=3) = 86=6
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471
472
¤
NOT FOR SALE
CHAPTER 5
INTEGRALS
(ii) U6 = O6 + 2 · i (12) 3 2 · i (0)
E 86=6 + 2(1) 3 2(9) = 70=6
[Add area of rightmost lower rectangle
and subtract area of leftmost upper rectangle.]
(iii) P6 =
6
S
i ({Wl ) {{
l=1
= 2[i (1) + i (3) + i (5) + i(7) + i (9) + i (11)]
E 2(8=9 + 8=5 + 7=8 + 6=6 + 5=1 + 2=8)
= 2(39=7) = 79=4
(b) Since i is decreasing, we obtain an overestimate by using left endpoints; that is, O6 .
(c) Since i is decreasing, we obtain an underestimate by using right endpoints; that is, U6 .
(d) P6 gives the best estimate, since the area of each rectangle appears to be closer to the true area than the overestimates and
underestimates in O6 and U6 .
3. (a) U4 =
4
S
i({l ) {{
l=1
{{ =
@2 3 0
=
4
8
=
= [i ({1 ) + i ({2 ) + i({3 ) + i ({4 )] {{
3
4 = cos 8 + cos 2
8 + cos 8 + cos 8 8
4
S
l=1
i ({l ) {{
E (0=9239 + 0=7071 + 0=3827 + 0) 8 E 0=7908
Since i is decreasing on [0> @2], an underestimate is obtained by using the right endpoint approximation, U4 .
4
4
S
S
(b) O4 =
i ({l31 ) {{ =
i ({l31 ) {{
l=1
l=1
= [i({0 ) + i ({1 ) + i ({2 ) + i ({3 )] {{
+ cos 3
= cos 0 + cos 8 + cos 2
8
8 8
E (1 + 0=9239 + 0=7071 + 0=3827) 8 E 1=1835
O4 is an overestimate. Alternatively, we could just add the area of the leftmost upper rectangle and subtract the area of the
rightmost lower rectangle; that is, O4 = U4 + i (0) · 8 3 i 2 · 8 .
4. (a) U4 =
4
S
l=1
i({l ) {{
k
l
430
{{ =
=1
4
= i({1 ) · 1 + i({2 ) · 1 + i ({3 ) · 1 + i ({4 ) · 1
= i(1) + i (2) + i (3) + i (4)
I
I
I
I
= 1 + 2 + 3 + 4 E 6=1463
Since i is increasing on [0> 4], U4 is an overestimate.
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°
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
SECTION 5.1
(b) O4 =
4
S
l=1
AREAS AND DISTANCES
i ({l31 ) {{ = i ({0 ) · 1 + i ({1 ) · 1 + i ({2 ) · 1 + i ({3 ) · 1
= i (0) + i(1) + i (2) + i (3)
I
I
I
I
= 0 + 1 + 2 + 3 E 4=1463
Since i is increasing on [0> 4], O4 is an underestimate.
2 3 (31)
=1 i
3
U3 = 1 · i (0) + 1 · i(1) + 1 · i (2) = 1 · 1 + 1 · 2 + 1 · 5 = 8.
5. (a) i ({) = 1 + {2 and {{ =
2 3 (31)
= 0=5 i
6
U6 = 0=5[i (30=5) + i (0) + i (0=5) + i (1) + i (1=5) + i (2)]
{{ =
= 0=5(1=25 + 1 + 1=25 + 2 + 3=25 + 5)
= 0=5(13=75) = 6=875
(b) O3 = 1 · i(31) + 1 · i(0) + 1 · i (1) = 1 · 2 + 1 · 1 + 1 · 2 = 5
O6 = 0=5[i (31) + i (30=5) + i (0) + i (0=5) + i (1) + i(1=5)]
= 0=5(2 + 1=25 + 1 + 1=25 + 2 + 3=25)
= 0=5(10=75) = 5=375
(c) P3 = 1 · i (30=5) + 1 · i (0=5) + 1 · i (1=5)
= 1 · 1=25 + 1 · 1=25 + 1 · 3=25 = 5=75
P6 = 0=5[i(30=75) + i (30=25) + i(0=25)
+ i (0=75) + i (1=25) + i (1=75)]
= 0=5(1=5625 + 1=0625 + 1=0625 + 1=5625 + 2=5625 + 4=0625)
= 0=5(11=875) = 5=9375
(d) P6 appears to be the best estimate.
6. (a)
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473
474
¤
NOT FOR SALE
CHAPTER 5
INTEGRALS
(b) i ({) = { 3 2 ln { and {{ =
531
=1 i
4
(i) U4 = 1 · i (2) + 1 · i (3) + 1 · i (4) + 1 · i (5)
= (2 3 2 ln 2) + (3 3 2 ln 3) + (4 3 2 ln 4) + (5 3