to accompany
T
HOMAS
’ C
ALCULUS
E
LEVENTH
E
DITION
B
ASED ON THE
O
RIGINAL
W
ORK BY
George B. Thomas, Jr.
Massachusetts Institute of Technology
AS
REVISED BY
Maurice D. Weir
Naval Postgraduate School
J
oel Hass
University of California, Davis
Frank R. Giordano
Naval Postgraduate School
I
NSTRUCTOR
’
S
S
OLUTIONS
M
ANUAL
P
ART
O
NE
Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors.
Copyright © 2005 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.
ISBN 0-321-22653-4
This Instructor's Solutions Manual contains the solutions to every exercise in the 11th Edition of THOMAS' CALCULUS
by Maurice Weir, Joel Hass and Frank Giordano, including the Computer Algebra System (CAS) exercises. The
corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away).
In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution
conforms exactly to the methods, procedures and steps presented in the textì
is mathematically correctì
includes all of the steps necessary so a typical calculus student can follow the logical argument and algebraì
includes a graph or figure whenever called for by the exercise, or if needed to help with the explanationì
is formatted in an appropriate style to aid in its understandingì
Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing
an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations).
Acknowledgments
Solutions Writers
William Ardis, Collin County Community College-Preston Ridge Campus Joseph Borzellino, California Polytechnic State University
Linda Buchanana, Howard College
Tim Mogill
Patricia Nelson, University of Wisconsin-La Crosse
Accuracy Checkers
Karl Kattchee, University of Wisconsin-La Crosse Marie Vanisko, California State University, Stanislaus Tom Weigleitner, VISTA Information Technologies
CHAPTER 1 PRELIMINARIES
1.1 REAL NUMBERS AND THE REAL LINE
1. Executing long division, 0.1, 0.2, 0.3, 0.8, 0.9"
9 9 9 9 9
2 3 8 9
œ œ œ œ œ
2. Executing long division, 0.09, 0.18, 0.27, 0.81, 0.99"
11. (x45 2)3"(x6) Ê 12(x2)5(x6)
solution interval [0 10] ß
26. 2 Ÿ 3z# Ÿ1 2 Ê Ÿ1 3z# Ÿ3 Ê 23 Ÿ Ÿz 2;
x ; solution interval
Section 1.1 Real Numbers and the Real Line
3
solution intervals s
7/2 5/2
solution intervals ( 1) x
1 7/3
solution intervals ( ) r
1 7/3
ß qqqqqqðïïïïïïðqqqqqqp
# #
solution intervals x
41. x# x 0 Ê x#x + < 1 1 Ê x 1 < 1 Ê x 1 < 1 Ê 1 < x 1 < 1 Ê 0 < x < 1.
4 4 2 4 2 2 2 2 2 2
ˆ ‰ ¹ ¹
So the solution is the interval (0 1)ß
42. x# x 2 0 Ê x#x + 1 9 Ê x 1 3 Ê x 1 3 or x 1 3 Ê x 2 or x Ÿ 1.
of “diamond-shaped" region.
Section 1.2 Lines, Circles and Parabolas
5
is true for all n positive integers.1.2 LINES, CIRCLES, AND PARABOLAS
1. ?xœ 1 ( 3)œ2, y? œ œ 2 2 4; dœÈ( x)? #( y)? # œÈ416œ2È5
11. mœ?? œ œ0 12. mœ?? œ ; no slope
y y
x 1 2 x ( )
33 # 0
# #
perpendicular slope does not exist perpendicular slopeœ0
Section 1.2 Lines, Circles and Parabolas
7
31. x-interceptœ4, y-interceptœ3 32. x-interceptœ 4, y-interceptœ 2
33. x-interceptœÈ3, y-interceptœ È2 34. x-interceptœ 2, y-interceptœ3
35. AxByœC " Í yœ BAx B and BxAyœC # Í yœ BAx A. Since BA AB œ 1 is the
C" C# ˆ ‰ ˆ ‰
product of the slopes, the lines are perpendicular.
36. AxByœC " Í yœ BAxCB" and AxByœC # Í yœ ABxCB#. Since the lines have the same
slope AB, they are parallel.
37. New positionœaxold?x yß old?ybœ # &ß ( 3 ( 6))œ $ß ( 3).
38. New positionœaxold?x yß old?ybœ(6 ß ( 6) 0 0)œ(0 0).ß
39. ?xœ5, y? œ6, B(3ß 3). Let Aœ(x y). Then xß ? œx#x " Ê 5œ3x Ê xœ 2 and
y y y 6 3 y y 9. Therefore, A ( 9).
? œ # " Ê œ Ê œ œ #ß
41. C( 2), a!ß œ2 Ê x#(y2)#œ4 42. C($ß0), aœ3 Ê (x3)#y#œ9
43. C( 1 5), a ß œÈ10 Ê(x1)#(y5)# œ10
44. C("ß "), aœÈ2 Ê (x1)#(y1)#œ2
xœ0 Ê (01)#(y1)#œ2 Ê (y1)#œ1
y 1 1 y 0 or y 2.
Ê œ „ Ê œ œ
Similarly, yœ0 Ê xœ0 or xœ2
45. CŠÈ3ß 2 , a‹ œ2 Ê xŠ È3‹#(y2)# œ4,
xœ0 Ê Š0È3‹#(y2)#œ4 Ê (y2)# œ1
y 2 1 y 1 or y 3. Also, y 0
Ê œ „ Ê œ œ œ
x 3 (0 2) 4 x 3 0
Ê Š È ‹# #œ Ê Š È ‹#œ
x 3
Ê œ È
46. C 3ˆ ß#"‰, aœ5 Ê (x3)#ˆy"#‰# œ25, so
xœ0 Ê (03)# y" œ25 #
#
ˆ ‰
y 16 y 4 y
Ê ˆ "#‰# œ Ê "# œ „ Ê œ #9
or yœ #7. Also, yœ0 Ê (x3)#ˆ0#" #‰ œ25
(x 3) x 3
Ê #œ Ê œ „
#
99 4
3È11
x 3
Section 1.2 Lines, Circles and Parabolas
11
63. The points that lie on or inside the circle with center ( 0) and radius 2."ß
64. The points lying on or outside the circle with center ( 2) and radius 2.!ß
65. The points lying outside the circle with center ( 0) and radius 1, but inside the circle with center ( 0),!ß !ß
66. The points on or inside the circle centered
at (!ß !) with radius 2 and on or inside the
circle centered at ( 2 0) with radius 2. ß
67. x#y#6y0 Ê x#(y3)#9.
The interior points of the circle centered at
(!ß 3) with radius 3, but above the line
yœ 3.
68. x#y#4x2y4 Ê (x2)#(y1)#9.
The points exterior to the circle centered at
(2ß 1) with radius 3 and to the right of the
line xœ2.
69. (x2)#(y1)#6 70. (x4)#(y2)#16
71. x#y#Ÿ2, x 1 72. x#y#4, (x1)#(y3)#10
73. x#y#œ1 and yœ2x Ê 1œx#4x#œ5x#
x and y or x and y .
Ê Š œ " œ ‹ Š œ " œ ‹
È5 È5 È5 È5
2 2
Thus, AŠ " ‹, BŠ " ‹ are the
È5 È5 È5 È5
2 2
ß ß
Section 1.2 Lines, Circles, and Parabolas
13
are intersection points.
75. y œx 1 and yœx # Ê x# œx 1
are the intersection points.
76. yœ x and C œ (x1) # Ê (x1)# œx
are the intersection points.
77. yœ2x# œ 1 x # Ê 3x#œ1
78. yœx4# œ(x1) # Ê 0œ 3x4# 2x1
Ê 0œ3x#8x œ4 (3x2)(x2)
x 2 and y 1, or x and
Ê œ œx4# œ œ 23
yœx4# œ9". Thus, A(2 1) and Bß ˆ23ß9"‰
are the intersection points.
79. x#y#œ1œ(x1)#y#
intersection points.
80. x#y#œ1œx#y Ê y#œy
intersection points.
81. (a) A¸(69° 0 in), Bß ¸(68° .4 in) ß Ê mœ 68° 69° ¸ 2.5°/in.
82. The time rate of heat transfer across a material, ? , is directly proportional to the cross-sectional area, A, of the mate
?
U
> rial,
to the temperature gradient across the material, ? (the slopes from the previous problem), and to a constant characteristic
?
temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good
insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are
not changing), we may define another constant, K, characteristics of the material: K œ ?" ÞUsing the values of XB from
?XB
? ?
the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the
poorest insulator, with Kœ0.4.
Section 1.2 Lines, Circles, and Parabolas
15
D (" "ß4) be located vertically upward from A and D (# "ß 2) be located vertically downward from A so
rotation gives a segment with slope mwœ " œ . If this segment has length equal to the original segment, its endpoint
m y x
will be ay, x or y, x , the first of these corresponds to a counter-clockwise rotation, the latter to a clockwiseb a b
rotation.
(e) ( y 0); ß (f) ( y x); ß (g) (3ß 10)
shown in the figure are congruent, the value a must lie midway between x and x , so a" # œ x"#x#.
1.3 FUNCTIONS AND THEIR GRAPHS
1. domainœ _ß _( ); rangeœ[1ß _) 2. domainœ[0ß _); rangeœ _ß( 1]
3. domainœ !ß _( ); y in range Ê yœ " , t0 Ê y#œ " and y ! Ê y can be any positive real number
Èt t
Section 1.3 Functions and Their Graphs
17
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test.
17. The domain is a_ß _b. 18. The domain is Ð_ß !Ó.
19. The domain is a_ß ! !ß _b a b. 20. The domain is a_ß ! !ß _b a b.
21. Neither graph passes the vertical line test
(a) (b)
22. Neither graph passes the vertical line test
(a) (b)
x y 1
x y y 1 x
or or
x y y x
k k
Ú Þ Ú Þ
Û ß Û ß
Ü à Ü à
œ Í Í
œ " œ
Section 1.3 Functions and Their Graphs
21
36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.
37. vœf(x)œ Ð"% x 2x 22ÑÐ 2xÑ œ %x$72x# $!)x; ! x Þ7
38. (a) Let hœheight of the triangle. Since the triangle is isosceles, AB# AB#œ2#ÊAB œÈ2 So,Þ
h# " œ# ŠÈ2‹#Êh œ " ÊB is at a!ß " Êb slope of ABœ " ÊThe equation of AB is
yœf(x)œ B "; x− Ò!ß "Ó.
(b) A xÐ Ñ œ2x yœ2xÐ "Ñ œ x 2x# #x; x− Ò!ß "Ó.
39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed.
(b) rœ ) œ %
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the
1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS
1. (a) linear, polynomial of degree 1, algebraic. (b) power, algebraic.
(c) rational, algebraic. (d) exponential.
2. (a) polynomial of degree 4, algebraic. (b) exponential.
(c) algebraic. (d) power, algebraic.
3. (a) rational, algebraic. (b) algebraic.
(c) trigonometric. (d) logarithmic.
4. (a) logarithmic. (b) algebraic.
(c) exponential. (d) trigonometric.
5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
6. (a) Graph f because it is linear.
(b) Graph g because it contains a!ß "b.
(c) Graph h because it is a nonlinear odd function.
7. Symmetric about the origin 8. Symmetric about the y-axis
Dec: x_ _ Dec: x_ !
Inc: nowhere Inc: x! _
9. Symmetric about the origin 10. Symmetric about the y-axis
Dec: nowhere Dec: x! _
Inc: x_ ! Inc: x_ !
Section 1.4 Identifying Functions; Mathematical Models
23
11. Symmetric about the y-axis 12. No symmetry
Dec: x_ Ÿ ! Dec: x_ Ÿ !
Inc: x! _ Inc: nowhere
13. Symmetric about the origin 14. No symmetry
Dec: nowhere Dec: x! Ÿ _
Inc: _ x _ Inc: nowhere
15. No symmetry 16. No symmetry
Dec: x! Ÿ _ Dec: _ xŸ !
17. Symmetric about the y-axis 18. Symmetric about the y-axis
Dec: _ xŸ ! Dec: x! Ÿ _
Inc: x! _ Inc: _ x !
19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even.
20. f xa bœx& œ " and fa b a bx œ x &œ " œ ˆ ‰" œ f x . Thus the function is odd.a b
x& a xb& x&
21. Since f xa bœx# " œ a bx # " œ f x . The function is even.a b
22. Since f xÒa bœx# Ó Á Ò x fa b a bx œ x # Óx and f xÒa bœx# Ó Á Òx f xa bœ a bx # Óx the function is neither even nor
odd.
23. Since g xa bœx$x, ga bx œ x$ œ x ax$xbœ g x . So the function is odd.a b
24. g xa bœx% $x# " œ a bx % $ Ba b# " œga b ßx thus the function is even.
25. g xa bœ " œ " œga bx . Thus the function is even.
" "
x # a xb#
26. g xa bœx #x "; ga bx œ x#x" œga bx . So the function is odd.
27. h ta bœ t "" ; ha b œt "t " ; h t a bœ" "t. Since h ta bÁ h t and h ta b a bÁha bt , the function is neither even nor odd.
28. Since t |l $ œ l a bt |, h t$ a bœha bt and the function is even.
29. h ta bœ2t ", ha b œ "t 2t . So h ta bÁha bt . h t a bœ "2t , so h ta bÁ h t . The function is neither even nora b
odd.
30. h ta bœ l l "2 t and ha b œ l l " œ l l "t 2 t 2 t . So h ta bœha bt and the function is even.
31. (a) The graph support the assumption that y is proportional to x.=
Section 1.4 Identifying Functions; Mathematical Models
25
(b) The graph support the assumption that y is proportional to x= "Î#.
The constant of proportionality is estimated from the slope of the regression line, which is 2.03.
32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the
regression line.
The graphs support the assumption that y is proportional to . The constant of proportionality is estimated from the$x
slope of the regression line, which is 5.00.
(b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from the slope of the regression line, which is 2.99.
33. (a) The scatterplot of yœreaction distance versus xœspeed is
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is
(b) Calculate xwœspeed squared. The scatterplot of x versus yw œbraking distance is:
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which
is approximately 0.059.
34. Kepler's 3rd Law is T daysa bœ !Þ%"R$Î#, R in millions of miles. "Quaoar" is 4‚ "!* miles from Earth, or about
4‚ "! *$ ‚ "! ¸ % ‚ "!* ' * miles from the sun. Let Rœ4000 (millions of miles) and Tœ !Þ%" %!!!a ba b$Î# days ¸ "!$ß (#$ days.
35. (a)
The hypothesis is reasonable.
(b) The constant of proportionality is the slope of the line ¸ )Þ(%" !"! ! in./unit massœ !Þ)(%in./unit mass.
(c) y(in.)œ !Þ)(a in./unit massba"$unit massbœ ""Þ$" in.
36. (a) (b)
Graph (b) suggests that yœk x is the better model. This graph is more linear than is graph (a).$
1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. D : f _ x _, D : xg 1 Ê Df gb œD : xfg 1. R : f _ y _, R : yg 0, R : yf gb 1, R : yfg 0
2. D : xf 1 0 Ê x 1, D : xg 1 0 Ê x 1. Therefore Df gb œD : xfg 1.
11. g(x) f(x) (f g)(x)‰
The completed table is shown. Note that the absolute value sign in part (d) is optional.
Section 1.5 Combining Functions; Shifting and Scaling Graphs
29
19. 20.
21. 22.
23. 24.
27. 28.
29. 30.
31. 32.
33. 34.
Section 1.5 Combining Functions; Shifting and Scaling Graphs
31
37. 38.
39. 40.
41. 42.
45. 46.
47. 48.
49. (a) domain: [0 2]; range: [ß #ß $] (b) domain: [0 2]; range: [ 1 0]ß ß
(c) domain: [0 2]; range: [0 2] ß ß (d) domain: [0 2]; range: [ 1 0]ß ß
Section 1.5 Combining Functions; Shifting and Scaling Graphs
33
(g) domain: [ 2 0]; range: [ ß !ß "] (h) domain: [ 1 1]; range: [ ß !ß "]
50. (a) domain: [0 4]; range: [ 3 0] ß ß (b) domain: [ 4 0]; range: [ ß !ß $]
(c) domain: [ 4 0]; range: [ ß !ß $] (d) domain: [ 4 0]; range: [ ß "ß %]
(g) domain: [ 5]; range: [ 3 0] "ß ß (h) domain: [0 4]; range: [0 3]ß ß
51. yœ3x#3
52. yœa b2x #1œ %x#1
53. yœ"#ˆ" x"#‰œ "# #"x#
54. yœ 1 axÎ$"b# œ 1 x*#
55. yœÈ% x 1
56. yœ3Èx 1
57. yœÉ% ˆ ‰x œ È16x
# #
# " #
58. yœ" % x
$È #
59. yœ " a b3x $œ " 27x$
60. yœ " ˆ ‰x œ " x
# ) $ $
61. Let yœ È# " œx f x and let g xa b a bœx"Î#, h xa bœˆx "‰ , i xa bœÈ#ˆx "‰ , and
# #
"Î# "Î#
j xa bœ ’È ˆ# x "‰ “œ Bfa b. The graph of h x is the graph of g x shifted left unit; the graph of i x is the grapha b a b " a b
# #
"Î#
Section 1.5 Combining Functions; Shifting and Scaling Graphs
35
62. Let yœÈ" x œf x Let g xa bÞ a b a bœ x , h xa b aœ #x b , and i xa bœ a #x b œÈ" x œf xa bÞ
# #
"Î# "Î# " "Î# #
È
The graph of g x is the graph of ya b œÈx reflected across the x-axis. The graph of h x is the graph of g x shifted righta b a b
two units. And the graph of i x is the graph of h x compressed vertically by a factor of a b a b È#.
63. yœf xa bœx . Shift f x one unit right followed by a shift two units up to get g x$ a b a b aœ x "b3 #.
64. yœ " Ba b$ # œ Òax "b$ # Ó œa b f x . Let g xa b a bœx , h x$ a b aœ x "b a b a$, i x œ x "b$ #a b, and
j xa bœ Òax "b$ # Óa b. The graph of h x is the graph of g x shifted right one unit; the graph of i x is the graph ofa b a b a b
65. Compress the graph of f xa bœ " horizontally by a factor of 2 to get g xa bœ ". Then shift g x vertically down 1 unit toa b
#
x x
get h xa bœ " ".
#x
66. Let f xa bœ " and g xa bœ # " œ " " œ " " œ " "Þ Since È# ¸ "Þ%, we see that the graph of
Î # "Î # B
x x
x
# # B#
#
# #
Š ‹ Š È ‹ ’Š È ‹ “
f x stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g x .a b a b
67. Reflect the graph of yœf xa bœÈ$ x across the x-axis to get g xa bœ È$ x.
68. yœf xa b aœ #xb#Î$œ Ò " # Óa ba bx#Î$ œ "a b a b#Î$ #x #Î$œ #a bx #Î$. So the graph of f x is the graph of g xa b a bœx#Î$
77. x# y# has its center at . Shiftinig 4 units left and 3 units up gives the center at h, k . So the
"' * œ " a!ß !b a b aœ %ß $b
equation is <x 44a b‘# ay 33b ax %4 b ay $3 b . Center, C, is , and major axis, AB, is the segment
# # # #
# # #
œ " Ê œ " a%ß $b
from a)ß $b a bto .!ß $
78. The ellipse x# y# has center h, k . Shifting the ellipse 3 units right and 2 units down produces an ellipse
% #&œ " a b aœ !ß !b
with center at h, ka b aœ $ß #b and an equation ax %3b#<y ##&a b‘ œ ". Center, C, is 3a ß #b, and AB, the segment from
#
to is the major axis.
a$ß $b a$ß (b
79. (a) (fg)( x) œf( x)g( x) œf(x)( g(x)) œ (fg)(x), odd
(b) ( x)Š ‹f Š ‹f (x), odd
g g( x) g(x) g f( x) f(x)
œ œ œ
(c) ( x)ˆ ‰fg œ g( x)f( x) œf(x)g(x) œ ˆ ‰gf (x), odd
(d) f ( x)# œf( x)f( x) œf(x)f(x)œf (x), # even
(e) g ( x)# (g( x))# ( g(x))# g (x), # even
œ œ œ
(f) (f g)( x)‰ œf(g( x)) œf( g(x)) œf(g(x))œ(f g)(x), ‰ even
(g) (g f)( x)‰ œg(f( x)) œg(f(x))œ(g f)(x), ‰ even
(h) (f f)( x)‰ œf(f( x)) œf(f(x))œ(f f)(x), ‰ even
(i) (g g)( x)‰ œg(g( x)) œg( g(x)) œ g(g(x))œ (g g)(x), ‰ odd
Section 1.6 Trigonometric Functions
39
81. (a) (b)
(c) (d)
82.
1.6 TRIGONOMETRIC FUNCTIONS
1. (a) sœr)œ(10)ˆ ‰4 œ8 m 1 (b) sœr)œ(10)(110°)ˆ ‰œ 110 œ 55 m
51 180°1 181 91
2. )œ sr œ 1081 œ 541 radians and 541ˆ180°1 ‰œ225°
Section 1.6 Trigonometric Functions
41
19. 20.
periodœ2 1 periodœ21
21. 22.
periodœ2 1 periodœ21
23. periodœ 1#, symmetric about the origin 24. periodœ1, symmetric about the origin
25. periodœ4, symmetric about the y-axis 26. periodœ4 , symmetric about the origin1
27. (a) Cos x and sec x are positive in QI and QIV and negative in QII and QIII. Sec x is undefined when
cos x is 0. The range of sec x is (_ß 1] "ß _[ );
(b) Sin x and csc x are positive in QI and QII and negative in QIII and QIV. Csc x is undefined when
sin x is 0. The range of csc x is (_ß 1][1ß _);
the range of sin x is ["ß "].
28. Since cot xœtan x" , cot x is undefined when tan xœ0
and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.
29. D: _ x _; R: yœ 1, 0, 1 30. D: _ x _; R: yœ 1, 0, 1
31. cos xˆ #1‰œcos x cosˆ1#‰sin x sinˆ#1‰œ(cos x)(0)(sin x)( 1) œsin x
32. cos xˆ #1‰œcos x cosˆ ‰1# sin x sinˆ ‰1# œ(cos x)(0)(sin x)(1)œ sin x
33. sin xˆ 1#‰œsin x cosˆ ‰1# cos x sinˆ ‰1# œ(sin x)(0)(cos x)(1)œcos x
34. sin xˆ 1#‰œsin x cosˆ1#‰cos x sinˆ1#‰œ(sin x)(0)(cos x)( 1) œ cos x
35. cos (AB) œcos (A ( B))œcos A cos ( B) sin A sin ( B) œcos A cos Bsin A ( sin B)
cos A cos B sin A sin B
œ
36. sin (AB) œsin (A ( B))œsin A cos ( B) cos A sin ( B) œsin A cos Bcos A ( sin B)
sin A cos B cos A sin B
œ
37. If BœA, ABœ0 Ê cos (AB)œcos 0œ1. Also cos (AB)œcos (AA)œcos A cos Asin A sin A
cos A sin A. Therefore, cos A sin A 1.
œ # # # # œ
38. If Bœ2 , then cos (A1 2 )1 œcos A cos 21sin A sin 21œ(cos A)(1)(sin A)(0)œcos A and
sin (A2 )1 œsin A cos 21cos A sin 21œ(sin A)(1)(cos A)(0)œsin A. The result agrees with the
fact that the cosine and sine functions have period 2 .1
Section 1.6 Trigonometric Functions
43
Because the cosine function is even and the sine functions is odd.
55. c# a# b# 2ab cos C 2# 3# 2(2)(3) cos (60°) 4 9 12 cos (60°) 13 12 " 7.
#
œ œ œ œ ˆ ‰œ
Thus, cœÈ7¸2.65.
57. From the figures in the text, we see that sin Bœ h. If C is an acute angle, then sin Cœh. On the other hand,
law of sines
58. By the law of sines, sin A sin B . By Exercise 55 we know that c 7.
59. From the figure at the right and the law of cosines,
b# œa#2#2(2a) cos B
b a. Thus, combining results,
Ê Èa2/2 œÈb3/2 Ê œÉ3#
is in radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode.
61. Aœ2, Bœ2 , C1 œ 1, Dœ 1
62. Aœ ", Bœ2, Cœ1, Dœ"
Section 1.6 Trigonometric Functions
45
63. Aœ 21, Bœ4, Cœ0, Dœ1
"
64. Aœ 21L, BœL, Cœ0, Dœ0
65. (a) amplitudeœk kA œ37 (b) periodœk kB œ365
(c) right horizontal shiftœCœ101 (d) upward vertical shiftœDœ25
66. (a) It is highest when the value of the sine is 1 at f(101)œ37 sin (0)25œ62° F.
The lowest mean daily temp is 37( 1) 25œ 12° F.
(b) The average of the highest and lowest mean daily temperaturesœ62° ( 12)°b c# œ25° F.
The average of the sine function is its horizontal axis, yœ25.
67-70. Example CAS commands: Maple
f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0;
f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#67 (Section 1.6)" );
Mathematica
Clear[a, b, c, d, f, x]
f[x_]:=a Sin[2 /b (x1 c)] + d
Plot[f[x]/.{aÄ3, bÄ1, cÄ0, dÄ0}, {x, 4 , 4 }] 1 1
(b) The period remains the same: periodœ l lB . The graph has a horizontal shift of period." #
68. (a) The graph is shifted right C units.
(b) The graph is shifted left C units.
(c) A shift of „one period will produce no apparent shift. C l l œ '
69. The graph shifts upwards D units for Dl l ! and down D units for Dl l !Þ
70. (a) The graph stretches A units.l l
(b) For A !, the graph is inverted.
1.7 GRAPHING WITH CALCULATORS AND COMPUTERS
1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and
Section 1.7 Graphing with Calculators and Computers
47
1. d. 2. c.
3. d. 4. b.
5-30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5 30
are not unique in appearance.
5. Ò ß Ó2 5 by Ò ß15 40 Ó 6. Ò ß Ó4 4 by Ò ß Ó4 4
9. 4 4 by 5 5 Ò ß Ó Ò ß Ó 10. 2 2 by 2 8Ò ß Ó Ò ß Ó
11. 2 6 by 5 4 Ò ß Ó Ò ß Ó 12. 4 4 by 8 8Ò ß Ó Ò ß Ó
13. by Ò"ß 'Ó Ò"ß %Ó 14. by Ò"ß 'Ó Ò"ß &Ó
Section 1.7 Graphing with Calculators and Computers
49
17. by Ò&ß "Ó Ò&ß &Ó 18. by Ò&ß "Ó Ò#ß %Ó
19. by Ò%ß %Ó Ò!ß $Ó 20. by Ò&ß &Ó Ò#ß #Ó
21. by Ò"!ß "!Ó Ò'ß 'Ó 22. Ò&ß &Óby Ò#ß #Ó
25. by Ò!Þ!$ß !Þ!$Ó Ò"Þ#&ß "Þ#&Ó 26. Ò!Þ"ß !Þ"Óby Ò$ß $Ó
27. by Ò$!!ß $!!Ó Ò"Þ#&ß "Þ#&Ó 28. Ò&!ß &!Óby Ò!Þ"ß !Þ"Ó
29. by Ò!Þ#&ß !Þ#&Ó Ò!Þ$ß !Þ$Ó 30. by Ò!Þ"&ß !Þ"&Ó Ò!Þ!#ß !Þ!&Ó
31. x# # œ % % x y y#Êyœ # „Èx# # )x .
The lower half is produced by graphing yœ # Èx# # )x .
32. y# "'x#œ " Êyœ „È" "'x . The upper branch#
Section 1.7 Graphing with Calculators and Computers
51
33. 34.
35. 36.
37. 38 Þ
39. 40.
41. (a) yœ "!&*Þ"% #!(%*(#x
(c)
(d) Answers may vary slightly. yœ "!&*Þa 14ba#!"! #!(%*(# œ &$ßb $ 899
42. (a) Let Cœcost and xœyear.
C œ (*'!Þ("a bx "Þ' ‚ "!(
(b) Slope represents increase in cost per year
(c) C œ #'$(Þ"%a bx &Þ# ‚ "!'
(d) The median price is rising faster in the northease (the slope is larger).
43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is dœ !Þ!)''x# "Þ*( &!Þ"x .
(b)
(c) From the graph in part (b), the stopping distance is about $(! feet when the vehicle is mph and it is about (# &#& feet
when the speed is mph.)&
Algebraically: dquadratica b(# œ !Þ!)'' (#a b# "Þ*( (# &!Þ" œ $'(Þ'a b ft.
dquadratica b)& œ !Þ!)'' )&a b# "Þ*( )& &!Þ" œ &##Þ)a b ft.
(d) The linear regression function is dœ 'Þ)* "%!Þ% Êx dlineara b(# œ 'Þ)* (# "%!Þ% œ $&&Þ(a b ft and
dlineara b)& œ 'Þ)* )& "%!Þ% œ %%&Þ#a b ft. The linear regression line is shown on the graph in part (b). The quadratic regression curve clearly gives the better fit.
Chapter 1 Practice Exercises
53
(b)
(c) 15 2 Þ km/h
(d) The linear regression function is yœ !Þ*"$'(& %Þ")**('x and it is shown on the graph in part (b). The linear
regession function gives a speed of "%Þ# km/h when yœ "" m. The power regression curve in part (a) better fits the
data.
CHAPTER 1 PRACTICE EXERCISES
(b) line slope
BD is vertical and has no slope
(c) Yes; A, B, C and D form a parallelogram.
(d) Yes. The line AB has equation y œ 1 #3(x8). Replacing x by gives y143 œ 3#ˆ143 8‰ "
1 5 1 6. Thus, E 6 lies on the line AB and the points A, B and E are collinear.
œ 3#ˆ103‰ œ œ ˆ143 ß ‰
(e) The line CD has equation y œ 3 3#(x2) or yœ 3#x. Thus the line passes through the origin.
11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are
Chapter 1 Practice Exercises
55
27. The coordinates of a point on the parabola are x x . The angle of inclination joining this point to the origin satisfiesa ß #b )
the equation tan )œx œx. Thus the point has coordinates x xß œ tan )ßtan ) .
Symmetric about the origin. Symmetric about the y-axis.
31. 32.
Neither Symmetric about the y-axis.
41. (a) The function is defined for all values of x, so the domain is a_ß _b.
odd integers k.
(b) Since the tangent function attains all values, the range is a_ß _b.
61. 62.
It does not change the graph. The graph of f (x)# œf"a bk kx is the same as the
graph of f (x) to the right of the y-axis. The"
graph of f (x) to the left of the y-axis is the#
reflection of yœf (x), x" 0 across the y-axis.
63. 64.
The graph of f (x)# œf"a bk kx is the same as the The graph of f (x)# œf"a bk kx is the same as the
graph of f (x) to the right of the y-axis. The " graph of f (x) to the right of the y-axis. The"
graph of f (x) to the left of the y-axis is the # graph of f (x) to the left of the y-axis is the#
reflection of yœf (x), x" 0 across the y-axis. reflection of yœf (x), x" 0 across the y-axis.
65. 66.
Whenever g (x) is positive, the graph of y" œg (x) # It does not change the graph.
g (x) is the same as the graph of y g (x).
œk " k œ "
When g (x) is negative, the graph of y" œg (x) is#
the reflection of the graph of yœg (x) across the"
x-axis.
67. Whenever g (x) is positive, the graph of y" œg (x)# œkg (x) is" k
the same as the graph of yœg (x). When g (x) is negative, the" "
graph of yœg (x) is the reflection of the graph of y# œg (x)"
Chapter 1 Practice Exercises
59
68. Whenever g (x) is positive, the graph of y" œg (x)# œkg (x) is" k
the same as the graph of yœg (x). When g (x) is negative, the" "
graph of yœg (x) is the reflection of the graph of y# œg (x)"
across the x-axis.
69. 70.
periodœ1 periodœ41
71. 72.
periodœ2 periodœ4
73. 74.
periodœ2 1 periodœ21
75. (a) sin Bœsin 1 œ œ Ê bœ2 sin 1 œ2 œ 3. By the theorem of Pythagoras,
3 c 3
b b 3
# Š ‹ È#
È
a#b# œc # Ê aœÈc#b#œÈ4 œ3 1.
(b) sin Bœsin 13 œ bc œ 2c Ê cœ sin 21 œ 2 œ 43. Thus, aœ c b œ 43 (2) œ 34 œ 23.
3 Š ‹È3 È È È
#
È # # ÊŠ ‹# # É
76. (a) sin Aœ ac Ê aœc sin A (b) tan Aœ ab Ê aœb tan A
78. (a) sin Aœ ac (c) sin Aœ acœ Èc#cb#
79. Let hœheight of vertical pole, and let b and c denote the
distances of points B and C from the base of the pole, measured along the flatground, respectively. Then,
tan 50°œ h, tan 35°œ h, and b œc 10.
80. Let hœheight of balloon above ground. From the figure at
Chapter 1 Additional and Advanced Exercises
61
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. (a) The given graph is reflected about the y-axis. (b) The given graph is reflected about the x-axis.
(c) The given graph is shifted left 1 unit, stretched (d) The given graph is shifted right 2 units, stretched
vertically by a factor of 2, reflected about the vertically by a factor of 3, and then shifted
x-axis, and then shifted upward 1 unit. downward 2 units.
2. (a) (b)
3. There are (infinitely) many such function pairs. For example, f(x)œ3x and g(x)œ4x satisfy
f(g(x))œf(4x)œ3(4x)œ12xœ4(3x)œg(3x)œg(f(x)).
4. Yes, there are many such function pairs. For example, if g(x)œ(2x3) and f(x)$ œx"Î$, then
(f g)(x)‰ œf(g(x))œf (2xa 3)$b aœ (2x3)$ "Î$b œ2x3.
5. If f is odd and defined at x, then f( x) œ f(x). Thus g( x) œ f( x) œ 2 f(x)2 whereas
g(x) (f(x) 2) f(x) 2. Then g cannot be odd because g( x) g(x) f(x) 2 f(x) 2
œ œ œ Ê œ
4 0, which is a contradiction. Also, g(x) is not even unless f(x) 0 for all x. On the other hand, if f is
Ê œ œ
even, then g(x)œf(x)2 is also even: g( x) œ f( x) œ2 f(x) œ2 g(x).
7. For (x y) in the 1st quadrant, xß k k k k y œ1x
8. We use reasoning similar to Exercise 7.
(1) 1st quadrant: yk ky œ x k kx
satisfy the equation.
(4) 4th quadrant: yk ky œ x k kx
y ( y) 2x 0 x. Combining
Í œ Í œ
these results we have the graph given at the right:
(b) Using the definition of the tangent function and the double angle formulas, we have
Chapter 1 Additional and Advanced Exercises
63
16. The angles labeled in the accompanying figure are#
equal since both angles subtend arc CD. Similarly, the
two angles labeled are equal since they both subtend!
arc AB. Thus, triangles AED and BEC are similar which
implies a c 2a cos b
nœ2 by the triangle inequality. We now show it holds for all positive integers n, by induction.
Suppose it holds for nœk 1: ak "a# á akk k k k kŸ a" a# á k ka (this is the inductionk
hypothesis). Then ak "a# á akak 1bk kœ aa"a# á akbak 1bk kŸ a"a# á akk k ak 1bk
(by the triangle inequality)Ÿk k k ka" a# á k k kak ak 1bk (by the induction hypothesis) and the
22. The fact that ak "a# á ank k k k k a" a# á k ka holds for nn œ1 is obvious. It holds for nœ2 by Exercise 21(b), since ak "a#k kœ a" ( a )# k k k k ka" a#kk kœ k k k ka" a# k k k k k a" a .#
We now show it holds for all positive integers n by induction.
Suppose the inequality holds for nœk 1. Then ak "a# á akk k k k k a" a# á k ka (this isk
the induction hypothesis). Thus ak " á akak 1bk kœ aa" á akb a ak 1bbk
a a a (by Exercise 21(b)) a a a a a a
kka " á kbk k k 1bkk œkk " á kk k k 1bkk k " á kk k k 1bk
a a a a (by the induction hypothesis). Hence the inequality holds for all
k k k k" # á k k kk k 1bk n by induction.
23. If f is even and odd, then f( x) œ f(x) and f( x) œf(x) Ê f(x)œ f(x) for all x in the domain of f.
(a) If a0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift
of the vertex toward the y-axis and upward. If a0 the graph is a parabola that opens downward.
Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a0 the graph is a parabola that opens upward. If also b0, then increasing b causes a shift of the
graph downward to the left; if b0, then decreasing b causes a shift of the graph downward and to the
right.
If a0 the graph is a parabola that opens downward. If b0, increasing b shifts the graph upward
to the right. If b0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by c shifts the graph upward c units if c? ? ? 0, and downward ?c
units if ?c0.
26. (a) If a0, the graph rises to the right of the vertical line xœ b and falls to the left. If a0, the graph
falls to the right of the line xœ b and rises to the left. If aœ0, the graph reduces to the horizontal
line yœc. As a increases, the slope at any given point xk k œx increases in magnitude and the graph!
becomes steeper. As a decreases, the slope at x decreases in magnitude and the graph rises or fallsk k !
more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward.
27. If m0, the x-intercept of yœmx2 must be negative. If m0, then the x-intercept exceeds "
#
0 mx 2 and x x 0 m 4.
Chapter 1 Additional and Advanced Exercises
65
28. Each of the triangles pictured has the same base
bœv t? œv(1 sec). Moreover, the height of each
triangle is the same value h. Thus (base)(height)" "bh
# œ #
A A A . In conclusion, the object sweeps
œ "œ #œ $ œ á
out equal areas in each one second interval.
29. (a) By Exercise #55 of Section P.2, the coordinates of P are ˆa 0 b 0 ‰ ˆa b‰. Thus the slope
# ß # œ #ß#
of OPœ ?? œ œ .
y
x a/2 a b/2 b
(b) The slope of ABœ b 00 a œ ba. The line segments AB and OP are perpendicular when the product
of their slopes is " œˆ ‰ ˆba ba‰œ ab##. Thus, b#œa # Ê aœb (since both are positive). Therefore, AB
CHAPTER 2 LIMITS AND CONTINUITY
2.1 RATE OF CHANGE AND LIMITS
1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x)
approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1.
(b) 1
(c) 0
2. (a) 0
(b) 1
(c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t)
approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0.
3. (a) True (b) True (c) False
approaches 1. As x approaches 0 from the right, approaches 1. There is no single number L that all
x x
x x
k k k k
the function values get arbitrarily close to as x Ä 0.
6. As x approaches 1 from the left, the values of " become increasingly large and negative. As x approaches 1
x 1
from the right, the values become increasingly large and positive. There is no one number L that all the
function values get arbitrarily close to as x Ä 1, so lim does not exist.
is defined at x . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when!
x is close enough to x . That is, the existence of a limit depends on the values of f(x) for x ! near x , not on the!
definition of f(x) at x itself.!
8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of xÄ0
the value f(0) itself.
9. No, the definition does not require that f be defined at xœ1 in order for a limiting value to exist there. If f(1)
is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) 5.
xÄ1 œ
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If
lim f(x) exists, its value may be some number other than f(1) 5. We can conclude nothing about lim f(x),
xÄ1 œ xÄ1
11. (a) f(x)œax# *b/(x3)
x 3.1 3.01 3.001 3.0001 3.00001 3.000001
f(x) 6.1 6.01 6.001 6.0001 6.00001 6.000001
x 2.9 2.99 2.999 2.9999 2.99999 2.999999
f(x) 5.9 5.99 5.999 5.9999 5.99999 5.999999
x 1.4 1.41 1.414 1.4142 1.41421 1.414213
g(x) 2.81421 2.82421 2.82821 2.828413 2.828423 2.828426
(b)
x 5.9 5.99 5.999 5.9999 5.99999 5.999999
G(x) .126582 .1251564 .1250156 .1250015 .1250001 .1250000
x 6.1 6.01 6.001 6.0001 6.00001 6.000001
G(x) .123456 .124843 .124984 .124998 .124999 .124999
Section 2.1 Rates of Change and Limits
69
(b)
(c) G(x)œ x x4x6 12 œ (x x6)(x6 2)œ x" if xÁ 6, and lim x"2 œ " 2 œ "8 œ 0.125.
# '
a# b xÄ c'
14. (a) h(x)œax#2x3 / xb a #4x3b
x 2.9 2.99 2.999 2.9999 2.99999 2.999999
h(x) 2.052631 2.005025 2.000500 2.000050 2.000005 2.0000005
x 3.1 3.01 3.001 3.0001 3.00001 3.000001
h(x) 1.952380 1.995024 1.999500 1.999950 1.999995 1.999999
(b)
(c) h(x)œ xx##4x2x 33 œ (x(x3)(x3)(x1)1) œ xx11 if xÁ3, and lim xx11 œ 3311 œ 4# œ2. xÄ $
15. (a) f(x)œax#1 / xb ak k1b
x 1.1 1.01 1.001 1.0001 1.00001 1.000001
f(x) 2.1 2.01 2.001 2.0001 2.00001 2.000001
x .9 .99 .999 .9999 .99999 .999999
f(x) 1.9 1.99 1.999 1.9999 1.99999 1.999999
(c) f(x) x 1, x 0 and x 1 , and lim (1 x) 1 ( 1) 2.
x 2.1 2.01 2.001 2.0001 2.00001 2.000001
F(x) 1.1 1.01 1.001 1.0001 1.00001 1.000001
x 1.9 1.99 1.999 1.9999 1.99999 1.999999
F(x) .9 .99 .999 .9999 .99999 .999999
.1 .01 .001 .0001 .00001 .000001
g( ) .998334 .999983 .999999 .999999 .999999 .999999
) )
.1 .01 .001 .0001 .00001 .000001
g( ) .998334 .999983 .999999 .999999 .999999 .999999
)
G(t) .499583 .499995 .499999 .5 .5 .5
t .1 .01 .001 .0001 .00001 .000001
G(t) .499583 .499995 .499999 .5 .5 .5
lim G(t) 0.5
Section 2.1 Rates of Change and Limits
71
(b)
Graph is NOT TO SCALE
19. (a) f(x)œx"ÎÐ" Ñx
x .9 .99 .999 .9999 .99999 .999999
f(x) .348678 .366032 .367695 .367861 .367877 .367879
x 1.1 1.01 1.001 1.0001 1.00001 1.000001
f(x) .385543 .369711 .368063 .367897 .367881 .367878
lim f(x) 0.36788
xÄ1 ¸ (b)
Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point
(1 2.71820).ß
20. (a) f(x)œa3x1 /xb
x .1 .01 .001 .0001 .00001 .000001
f(x) 1.161231 1.104669 1.099215 1.098672 1.098618 1.098612
x .1 .01 .001 .0001 .00001 .000001
f(x) 1.040415 1.092599 1.098009 1.098551 1.098606 1.098611
lim f(x) 1.0986
xÄ ! ¸
(b)
21. lim 2x 2(2) 4 22. lim 2x 2(0) 0
xÄ # œ œ xÄ ! œ œ
23. lim (3x 1) 3 1 0 24. lim
xÄ" xÄ1
$
œ ˆ ‰" œ œ " œ "
#
3 3x 1 3(1) 1
25. lim 3x(2x 1) 3( 1)(2( 1) 1) 9 26. lim 1
(b) At tœ20, the Cobra was traveling approximately 50 m/sec or 180 km/h.
36. (a) Q Slope of PQœ ??
(b) Approximately 16 m/sec
37. (a)
(b) ?? 56 thousand dollars per year
p
t 1994 1992 174 62 112
œ œ # œ
(c) The average rate of change from 1991 to 1992 is?? 35 thousand dollars per year.
p
t 1992 1991 62 27
œ œ
The average rate of change from 1992 to 1993 is?? 49 thousand dollars per year.
p
t 1993 1992 111 62
œ œ
So, the rate at which profits were changing in 1992 is approximatley 35" 49 42 thousand dollars per year.
Section 2.1 Rates of Change and Limits
73
1 h 1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005
1 h 1 /h 0.4880 0.4987 0.4998 0.499 0.5
f(T) 0.476190 0.497512 0.499750 0.4999750 0.499997 0.499999
f(T) f(2) / T 2 0.2381 0.2488 0.2
41-46. Example CAS commands: :
In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be
overcome in Maple by entering the function as f := x -> (surd(x+1, 3)1)/x.
36. lim lim lim
37. (a) quotient rule
(b) difference and power rules (c) sum and constant multiple rules
38. (a) quotient rule
(b) power and product rules
(c) difference and constant multiple rules
Section 2.2 Calculating Limits Using the Limit Laws
77
lies between the other two graphs in the
figure, and the graphs converge as x Ä 0.
conditions of the sandwich theorem are satisfied, lim f(x) 5 0.
56. (a) 1œ lim œ œ Ê lim f(x)œ4.
2.3 PRECISE DEFINITION OF A LIMIT
1.
Step 1: kx5k$ Ê $ x 5 $ Ê $ 5 x $ 5
Step 2: $ œ5 7 Ê $œ2, or œ$ 5 1 Ê $œ4.
Section 2.3 Precise Definition of a Limit
83
41. Step 1: For xÁ1, kx#1k% xÊ % # " % Ê " % x# " % Ê È1 % k kx È1%
Ê È" % xÈ1%near B œ ".
Step 2: kx1k$ Ê $ x 1 $ Ê " $ x "$ .
Then " œ$ È1% Ê $œ " È1%, or $ œ1 È" % Ê $œÈ" % 1. Choose
min 1 1 , that is, the smaller of the two distances.
then by the sandwich theorem, in either case, lim x sin 0.
you can see from the figure that there are also infinitely many values of x near 0 such that sin " 0. If we
x œ
endpoint was rounded down.
56. VœRI I 5Ê VR œ Ê ¸VR ¸Ÿ0.1 Ê 0.1Ÿ 120R Ÿ5 0.1 Ê 4.9Ÿ120R Ÿ5.1 Ê 1049 1 0R# 1051 Ê
R(120)(10)51 Ÿ Ÿ (120)(10)49 Ê 23.53ŸRŸ24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) $ x 1 0 Ê " $ x1 Ê f(x)œx. Then f(x)k 2k kœ x2kœ œ2 x 2 1 1. That is,
f(x) 2 1 no matter how small is taken when x 1 lim f(x) 2.
k k " " Ê Á
Section 2.3 Precise Definition of a Limit
85
61-66. Example CAS commands (values of del may vary for a specified eps): : plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01,
color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q);
plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); s in [0.1, 0.005, 0.001 ] do # (e)
for ep
q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q);
head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta,
end do:
(assigned function and values for x0, eps and del may vary): Mathematica
2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY
Section 2.4 One-Sided Limits and Limits at Infinity
87
7. (a Ñ (b) lim f(x) lim f(x)
xÄ1 œ " œxÄ1
(c) Yes, lim f(x) 1 since the right-hand and left-hand
xÄ1 œ limits exist and equal 1
8. (a) (b) lim f(x) 0 lim f(x)
xÄ1 œ œxÄ1
(c) Yes, lim f(x) 0 since the right-hand and left-hand
xÄ1 œ limits exist and equal 0
Section 2.4 One-Sided Limits and Limits at Infinity
89
Note: In these exercises we use the result lim 0 whenever 0. This result follows immediately from
49. (a) lim xÄ _ xx13 x lim Ä _ 1 0 (b) 0 (same process as part (a)) $Î& ""Î"!"
Section 2.4 One-Sided Limits and Limits at Infinity
91
67. Yes. If lim xÄ _ g(x)f(x) œ2 then the ratio of the polynomials' leading coefficients is 2, so lim xÄ c_ g(x)f(x) œ2 as well.
68. Yes, it can have a horizontal or oblique asymptote.
69. At most 1 horizontal asymptote: If lim xÄ _ g(x)f(x) œL, then the ratio of the polynomials' leading coefficients is L, so
(c) The function f has limit 0 at x!œ0 since both the right-hand and left-hand limits exist and equal 0.
81. lim lim lim , t
2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES
1. lim 2. lim
xÄ ! xÄ !
"
3x positive 2x negative positive 5 positive
œ _ Š ‹ œ _ Š ‹
3. lim 4. lim
xÄ # xÄ $
3
x 2 negative x 3 positive positive positive
x 8 positive 2x 10 negative negative negative
œ _ Š ‹ œ _ Š ‹
7. lim 8. lim
xÄ ( xÄ !
4
(x 7) positive x (x 1) positive positive positive negative
x 1 (x 1)(x 1) positive positive positive
# œ œ _ Š † ‹
(b) lim lim
xÄ " xÄ "
x x
x 1 (x 1)(x 1) positive negative positive
# œ œ _ Š † ‹
(c) lim lim
xÄ c" xÄ c"
x x
x 1 (x 1)(x 1) positive negative negative
Section 2.5 Infinite Limits and Vertical Asymptotes
93
(d) lim lim
xÄ "c xÄ "c
x x
x 1 (x 1)(x 1) negative negative negative
# œ œ _ Š † ‹
19. (a) lim 0 lim
xÄ !b xÄ !b
x
x x negative
#
x x positive
#
2x 4 positive 2x 4 negative positive positive
x 2x x (x 2) positive negative (x 2)(x 1) negative negative #
x 2x x (x 2) positive negative (x 2)(x 1) negative negative #
x(x ) positive positive negative
27. yœ " 28. yœ "
x 1 x 1
29. yœ " 30. yœ
# x 4 x3
3
31. yœxx23 œ 1 x" 32. yœx2x1 œ # x21
#
Section 2.5 Infinite Limits and Vertical Asymptotes
95
35. yœxx# % œ " x x$ 36. yœ xx " œ"x " x$
" " # % # # %
2
37. yœx#x1 œ x "x 38. yœx$x#1 œ x x"#
39. Here is one possibility. 40. Here is one possibility.
43. Here is one possibility. 44. Here is one possibility.
45. Here is one possibility. 46. Here is one possibility.
47. For every real number B 0, we must find a $0 such that for all x, 0 kx0k $ Ê " B. Now,
for every positive number B, there exists a corresponding number $0 such that for all x,
x! $ x x ! Ê f(x)B.
(b) We say that f(x) approaches minus infinity as x approaches x from the right, and write lim f(x)! ,
xÄx! œ _
if for every positive number B (or negative number B) there exists a corresponding number $0 such
Section 2.5 Infinite Limits and Vertical Asymptotes
97
(c) We say that f(x) approaches minus infinity as x approaches x from the left, and write lim f(x)! xÄx ,
!
œ _
if for every positive number B (or negative number B) there exists a corresponding number $0 such
59. yœtan x " 60. yœ" tan x
x# x
61. yœ x 62. yœ
4 x 4 x
È È
"
# #
63. yœx#Î$ " 64. yœsin
x"Î$ ˆx# 1‰
1
2.6 CONTINUITY
1. No, discontinuous at xœ2, not defined at xœ2
2. No, discontinuous at xœ3, " œ lim g(x)Ág(3)œ1.5
Section 2.6 Continuity
99
Removable discontinuity at xœ0 by assigning the number lim f(x)œ0 to be the value of f(0) rather than
xÄ !
f(0)œ1.
12. Nonremovable discontinuity at xœ1 because lim f(x) fails to exist ( lim f(x)œ2 and lim f(x)œ1).
xÄ1 xÄ "c xÄ "b
Removable discontinuity at xœ2 by assigning the number lim f(x)œ1 to be the value of f(2) rather than
xÄ #
f(2)œ2.
13. Discontinuous only when x œ2 0 Ê xœ2 14. Discontinuous only when (x2)#œ0 Ê xœ 2
15. Discontinuous only when x# % $ œ ! Êx (x3)(x1)œ0 Ê xœ3 or xœ1
16. Discontinuous only when x#3x10œ0 Ê (x5)(x2)œ0 Ê xœ5 or xœ 2
17. Continuous everywhere. ( xk 1ksin x defined for all x; limits exist and are equal to function values.)
18. Continuous everywhere. ( xk k " Á0 for all x; limits exist and are equal to function values.)
19. Discontinuous only at xœ0
20. Discontinuous at odd integer multiples of , i.e., x = (2n1 ) , n an integer, but continuous at all other x.1
# " #
21. Discontinuous when 2x is an integer multiple of , i.e., 2x1 œn , n an integer 1 Ê xœ n1, n an integer, but
#
continuous at all other x.
22. Discontinuous when is an odd integer multiple of , i.e., 1x 1 1x (2n 1) , n an integer 1 x 2n 1, n an
# # # œ # Ê œ
23. Discontinuous at odd integer multiples of , i.e., x = (2n1 1) , n an integer, but continuous at all other x.1
# #
24. Continuous everywhere since x% 1 1 and " Ÿsin xŸ1 Ê 0Ÿsin x# Ÿ1 Ê 1sin x# 1; limits exist
and are equal to the function values.
25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval ß _3# ‰.
26. Discontinuous when 3x 1 0 or x "3 Ê continuous on the interval "3ß _‰.
27. Continuous everywhere: (2x1)"Î$ is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2x)"Î& is defined for all x; limits exist and are equal to function values.
Section 2.6 Continuity
101
41. The function can be extended: f(0)¸2.3. 42. The function cannot be extended to be continuous at
xœ0. If f(0)¸2.3, it will be continuous from the
right. Or if f(0)¸ 2.3, it will be continuous from the
left.
43. The function cannot be extended to be continuous 44. The function can be extended: f(0)¸7.39.
at xœ0. If f(0)œ1, it will be continuous from
the right. Or if f(0)œ 1, it will be continuous
from the left.
45. f(x) is continuous on [!ß "] and f(0)0, f(1)0
by the Intermediate Value Theorem f(x) takes
Ê
on every value between f(0) and f(1) Ê the
equation f(x)œ0 has at least one solution between
xœ0 and xœ1.
46. cos xœx Ê (cos x) œx 0. If xœ 1, cos 1 1 0. If xœ 1, cos 1 1 0. Thus cos x œx 0
# ˆ #‰ ˆ #‰ # ˆ ‰# #
for some x between 1 and according to the Intermediate Value Theorem.1
# #
47. Let f(x)œx$15x1 which is continuous on [ 4 4]. Then f( 4) ß œ 3, f( 1) œ15, f(1)œ 13, and f(4)œ5.
By the Intermediate Value Theorem, f(x)œ0 for some x in each of the intervals % x 1, " x1, and
x 4. That is, x 15x 1 0 has three solutions in [ 4]. Since a polynomial of degree 3 can have at most 3
" $ œ %ß
solutions, these are the only solutions.
48. Without loss of generality, assume that ab. Then F(x)œ(xa) (x# b)#x is continuous for all values of
x, so it is continuous on the interval [a b]. Moreover F(a)ß œa and F(b)œb. By the Intermediate Value
Theorem, since aabb b, there is a number c between a and b such that F(x)œ abb.
49. Answers may vary. Note that f is continuous for every value of x.
Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c)œ5,000,000.
50. All five statements ask for the same information because of the intermediate value property of continuous functions.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2.
52. Answers may vary. For example, g(x)œ xb"1 has a discontinuity at xœ 1 because lim g(x) does not exist. lim f(x) exist by the same arguments used in part (a).
xÄx
is discontinuous at x 0, since it is not defined there. Theorem 10 requires that f(x) be
œ (xb"1)c1 œ "x œ
continuous at g(0), which is not the case here since g(0)œ1 and f is undefined at 1.
57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to
Section 2.7 Tangents and Derivatives
103
58. Let f(x) be the new position of point x and let d(x)œf(x)x. The displacement function d is negative if x is
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d(x)œ0 for some point in between. That is, f(x)œx for some point x, which is
then in its original position.
59. If f(0)œ0 or f(1)œ1, we are done (i.e., cœ0 or cœ1 in those cases). Then let f(0)œa0 and f(1)œb1
continuous at xœc. Similarly,
lim cos c h lim cos c cos h sin c sin h cos c lim cos h sin c lim sin h cos c.
2.7 TANGENTS AND DERIVATIVES
3. P : m" " œ#5, P : m# # œ #" 4. P : m" " œ3, P : m# #œ 3
tangent line
6. mœ lim œ lim
tangent line
7. mœ lim œ lim
Section 2.7 Tangents and Derivatives
105
tangent line
20. At xœ2, yœ 3 Ê mœ lim œ lim œ lim œ 4, slope
horizontal tangent.
24. 0œmœ lim œ lim
the origin with slope 0.
Section 2.7 Tangents and Derivatives
107
does not have a vertical tangent at (!ß ") because the limit does not exist.
Section 2.7 Tangents and Derivatives
109
vertical tangent at xœ0;
xœ1: lim œ lim œ _ Ê yœx (x1) has a
45-48. Example CAS commands: : tan_line := f(x0) + L*(x-x0);
linestyle=[1,2,5,6,7], title="Section 2.7, #45(d)",
legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)", "Secant line (h=2)","Secant line (h=3)"] );
: (function and value for x0 may change) Mathematica
CHAPTER 2 PRACTICE EXERCISES
Chapter 2 Practice Exercises
113
extended to a continuous function at xœ 1.
At xœ1: lim f(x)œ lim œ lim œ lim ( x) œ 1, and
be extended to a continuous function at x 1 either.
cannot œ
so h cannot be extended to a continuous function
