Thomas Calculus 12th ed solution ch1

CHAPTER 1 FUNCTIONS

1.1 FUNCTIONS AND THEIR GRAPHS

1. domain ( ); range [1 ) 2. domain [0 ); range ( 1]œ �_ß_ œ ß_ œ ß_ œ �_ß

3. domain 2 ); y in range and y 5x 10 y can be any positive real number range ).œ Ò� ß_ œ � ! Ê Ê œ Ò!ß_È 4. domain ( 0 3, ); y in range and y x 3x y can be any positive real number range ).œ �_ß Ó � Ò _ œ � ! Ê Ê œ Ò!ß_È 2

5. domain ( 3 3, ); y in range and y , now if t 3 3 t , or if t 3œ �_ß Ñ � Ð _ œ � Ê � � ! Ê � ! �4 43 t 3 t� �

3 t y can be any nonzero real number range 0 ).Ê � � ! Ê � ! Ê Ê œ Ð�_ß Ñ � Ð!ß_43 t�

6. domain ( 4, 4 4, ); y in range and y , now if t t 16 , or ifœ �_ß�%Ñ � Ð� Ñ � Ð _ œ � �% Ê � � ! Ê � !2 2t 16 t 16

22 2� �

t 4 16 t 16 , or if t t 16 y can be any�% � � Ê � Ÿ � � ! Ê � Ÿ � ! � % Ê � � ! Ê � ! Ê2 22 2t 16 t 16

#"' � �2 2

nonzero real number range ).Ê œ Ð�_ß � Ó � Ð!ß_18

7. (a) Not the graph of a function of x since it fails the vertical line test. (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.

9. base x; (height) x height x; area is a(x) (base)(height) (x) x x ;œ � œ Ê œ œ œ œ# # ## # # # #

# " "ˆ ‰ Š ‹x 3 3 34

È È È

perimeter is p(x) x x x 3x.œ � � œ

10. s side length s s d s ; and area is a s a dœ Ê � œ Ê œ œ Ê œ# # # # #"#

d2È

11. Let D diagonal length of a face of the cube and the length of an edge. Then D d andœ j œ j � œ# # #

D 2 3 d . The surface area is 6 2d and the volume is .# # # # # # $$Î#

œ j Ê j œ Ê j œ j œ œ j œ œd 6d d d3 3 33 3È È

# # $Š ‹12. The coordinates of P are x x so the slope of the line joining P to the origin is m (x 0). Thus,ˆ ‰Èß œ œ �

ÈÈ

xx x

"

x, x , .ˆ ‰ ˆ ‰È œ " "m m#

13. 2x 4y 5 y x ; L x 0 y 0 x x x x x� œ Ê œ � � œ Ð � Ñ � Ð � Ñ œ � Ð� � Ñ œ � � �" " "# #

5 5 5 254 4 4 4 16

2 2 2 2 2 2È É É x xœ � � œ œÉ É5 5 25 20x 20x 25

4 4 16 16 42 20x 20x 252 2� � � �È

14. y x 3 y 3 x; L x 4 y 0 y 3 4 y y 1 yœ � Ê � œ œ Ð � Ñ � Ð � Ñ œ Ð � � Ñ � œ Ð � Ñ �È È È È2 2 2 2 2 2 2 2 2

y 2y 1 y y y 1œ � � � œ � �È È4 2 2 4 2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

2 Chapter 1 Functions

15. The domain is . 16. The domain is .a b a b�_ß _ �_ß _

17. The domain is . 18. The domain is .a b�_ß _ Ð�_ß !Ó

19. The domain is . 20. The domain is .a b a b a b a b�_ß ! � !ß _ �_ß ! � !ß _

21. The domain is 5 5 3 3, 5 5, 22. The range is 2, 3 .a b a b�_ß � � Ð� ß � Ó � Ò Ñ � _ Ò Ñ

23. Neither graph passes the vertical line test (a) (b)


Section 1.1 Functions and Their Graphs 3

24. Neither graph passes the vertical line test (a) (b)

x y 1 x y y 1 x

or orx y y x

k kÚ Þ Ú ÞÛ ß Û ßÜ à Ü à� œ Í Í

� œ " œ �

� œ �" œ �" �

25. x 0 1 2 26. x 0 1 2 y 0 1 0 y 1 0 0

27. F x 28. G x4 x , x 1x 2x, x 1

, x 0x, 0 x

a b a bœ œœ œ� Ÿ

� �

�

Ÿ

2

2x"

29. (a) Line through and : y x; Line through and : y x 2a b a b a b a b!ß ! "ß " œ "ß " #ß ! œ � �

f(x) x, 0 x 1

x 2, 1 x 2œ

Ÿ Ÿ� � � Ÿœ

(b) f(x)

2, xx

2 xx

œ

! Ÿ � "!ß " Ÿ � #ß # Ÿ � $!ß $ Ÿ Ÿ %

ÚÝÝÛÝÝÜ30. (a) Line through 2 and : y x 2 a b a b!ß #ß ! œ � �

Line through 2 and : m , so y x 2 xa b a b a bß " &ß ! œ œ œ � œ � � � " œ � �! � " �" " " " && � # $ $ $ $ $

f(x) x , 0 xx , x

œ� � # � Ÿ #

� � # � Ÿ &œ " &$ $

(b) Line through and : m , so y x a b a b�"ß ! !ß �$ œ œ �$ œ �$ � $�$�!!� Ð�"Ñ

Line through and : m , so y xa b a b!ß $ #ß �" œ œ œ �# œ �# � $�"�$ �%#�! #

f(x) x , xx , x

œ�$ � $ �" � Ÿ !�# � $ ! � Ÿ #œ



31. (a) Line through and : y x a b a b�"ß " !ß ! œ �

Line through and : ya b a b!ß " "ß " œ "

Line through and : m , so y x xa b a b a b"ß " $ß ! œ œ œ � œ � � " � " œ � �!�" �" " " " $$�" # # # # #

f(x)x x

xx x

œ

� �" Ÿ � !" ! � Ÿ "

� � " � � $

ÚÛÜ " $

# #

(b) Line through 2 1 and 0 0 : y x a b a b� ß � ß œ 12

Line through 0 2 and 1 0 : y 2x 2a b a bß ß œ � �

Line through 1 1 and 3 1 : y 1a b a bß � ß � œ �

f xx 2 x 0

2x 2 0 x 11 1 x 3

a bÚÛÜœ

� Ÿ Ÿ

� � � Ÿ� � Ÿ

12

32. (a) Line through and T : m , so y x 0 x ˆ ‰ ˆ ‰a bT TT T T T T# � Î# #

"�! # # #ß ! ß " œ œ œ � � œ � "a b

f x , 0 x

x , x Ta b �œ

! Ÿ Ÿ

� " � Ÿ

T

TT

##

#

(b) f x

A, x

A x T

A T x

A x T

a bÚÝÝÝÛÝÝÝÜ

œ

! Ÿ �

� ß Ÿ �

ß Ÿ �

� ß Ÿ Ÿ #

T

T

T

T

#

#$#

$#

33. (a) x 0 for x [0 1) (b) x 0 for x ( 1 0]Ú Û œ − ß Ü Ý œ − � ß

34. x x only when x is an integer.Ú Û œ Ü Ý

35. For any real number x, n x n , where n is an integer. Now: n x n n x n. ByŸ Ÿ � " Ÿ Ÿ � " Ê �Ð � "Ñ Ÿ � Ÿ �

definition: x n and x n x n. So x x for all x .Ü� Ý œ � Ú Û œ Ê �Ú Û œ � Ü� Ý œ �Ú Û − d

36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.



37. Symmetric about the origin 38. Symmetric about the y-axis Dec: x Dec: x�_ � � _ �_ � � !

Inc: nowhere Inc: x! � � _

39. Symmetric about the origin 40. Symmetric about the y-axis Dec: nowhere Dec: x! � � _

Inc: x Inc: x�_ � � ! �_ � � !

x! � � _

41. Symmetric about the y-axis 42. No symmetry Dec: x Dec: x�_ � Ÿ ! �_ � Ÿ !

Inc: x Inc: nowhere! � � _



43. Symmetric about the origin 44. No symmetry Dec: nowhere Dec: x! Ÿ � _

Inc: x Inc: nowhere�_ � � _

45. No symmetry 46. Symmetric about the y-axis Dec: x Dec: x! Ÿ � _ �_ � Ÿ !

Inc: nowhere Inc: x! � � _

47. 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.

48. f x x and f x x f x . Thus the function is odd.a b a b a b a bˆ ‰œ œ � œ � œ œ � œ ��& " " "�&

�x xx& &&a b

49. Since f x x x f x . The function is even.a b a b a bœ � " œ � � " œ �# #

50. Since f x x x f x x x and f x x x f x x x the function is neither even norÒ œ � Ó Á Ò � œ � � Ó Ò œ � Ó Á Ò� œ � � Óa b a b a b a b a b a b# ## #

odd.

51. Since g x x x, g x x x x x g x . So the function is odd.a b a b a b a bœ � � œ � � œ � � œ �$ $ $

52. g x x x x x g x thus the function is even.a b a b a b a bœ � $ � " œ � � $ � � " œ � ß% # % #

53. g x g x . Thus the function is even.a b a bœ œ œ �" "� " � �"x x# #a b

54. g x ; g x g x . So the function is odd.a b a b a bœ � œ � œ �x xx x# #� " �"

55. h t ; h t ; h t . Since h t h t and h t h t , the function is neither even nor odd.a b a b a b a b a b a b a bœ � œ � œ Á � Á �" " "� " � � " " �t t t

56. Since t | t |, h t h t and the function is even.l œ l � œ �$ $a b a b a bCopyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.


57. h t 2t , h t 2t . So h t h t . h t 2t , so h t h t . The function is neither even nora b a b a b a b a b a b a bœ � " � œ � � " Á � � œ � � " Á �

odd.

58. h t 2 t and h t 2 t 2 t . So h t h t and the function is even.a b a b a b a bœ l l � " � œ l� l � " œ l l � " œ �

59. s kt 25 k 75 k s t; 60 t t 180œ Ê œ Ð Ñ Ê œ Ê œ œ Ê œ" " "3 3 3

60. K c v 12960 c 18 c 40 K 40v ; K 40 10 4000 joulesœ Ê œ Ê œ Ê œ œ œ# # #a b a b2

61. r 6 k 24 r ; 10 sœ Ê œ Ê œ Ê œ œ Ê œk k 24 24 12s 4 s s 5

62. P 14.7 k 14700 P ; 23.4 v 628.2 inœ Ê œ Ê œ Ê œ œ Ê œ ¸k k 14700 14700 24500v 1000 v v 39

3

63. v f(x) x 2x 22 2x x 72x x; x 7œ œ Ð"% � ÑÐ � Ñ œ % � � $!) ! � � Þ$ #

64. (a) Let h height of the triangle. Since the triangle is isosceles, AB AB 2 AB 2 So,œ � œ Ê œ Þ# # # È h 2 h B is at slope of AB The equation of AB is# #

#

� " œ Ê œ " Ê !ß " Ê œ �" ÊŠ ‹È a b y f(x) x ; x .œ œ � � " − Ò!ß "Ó

(b) A x 2x y 2x x 2x x; x .Ð Ñ œ œ Ð� � "Ñ œ � � # − Ò!ß "Ó#

65. (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.

66. (a) Graph f because it is linear. (b) Graph g because it contains .a b!ß "

(c) Graph h because it is a nonlinear odd function.

67. (a) From the graph, 1 x ( 2 0) ( )x 4x# � � Ê − � ß � %ß_

(b) 1 1 0x 4 x 4x x# #� � Ê � � �

x 0: 1 0 0 0� � � � Ê � Ê �x 4 x 2x 8x 2x x

(x 4)(x 2)# #

� � � �#

x 4 since x is positive;Ê �

x 0: 1 0 0 0� � � � Ê � Ê �x 4 x 2x 82 x 2x x

(x 4)(x 2)#� � � �#

x 2 since x is negative;Ê � �

sign of (x 4)(x 2)� �

2

ïïïïïðïïïïïðïïïïî� ��

�%

Solution interval: ( 0) ( )�#ß � %ß_



68. (a) From the graph, x ( 5) ( 1 1)3 2x 1 x 1� �� Ê − �_ß� � � ß

(b) x 1: 2Case � � � Ê �3 2x 1 x 1 x 1

3(x 1)� � �

�

3x 3 2x 2 x 5.Ê � � � Ê � �

Thus, x ( 5) solves the inequality.− �_ß�

1 x 1: 2Case � � � � Ê �3 2x 1 x 1 x 1

3(x 1)� � �

�

3x 3 2x 2 x 5 which is trueÊ � � � Ê � �

if x 1. Thus, x ( 1 1) solves the� � − � ß

inequality.

1 x: 3x 3 2x 2 x 5Case � � Ê � � � Ê � �3 2x 1 x 1� �

which is never true if 1 x, so no solution here.�

In conclusion, x ( 5) ( 1 1).− �_ß� � � ß

69. A curve symmetric about the x-axis will not pass the vertical line test because the points x, y and x, y lie on the sama b a b� e vertical line. The graph of the function y f x is the x-axis, a horizontal line for which there is a single y-value, ,œ œ ! !a b for any x.

70. price 40 5x, quantity 300 25x R x 40 5x 300 25xœ � œ � Ê œ � �a b a ba b71. x x h x ; cost 5 2x 10h C h 10 10h 5h 2 22 2 2 h

22 h 2 h2 2� œ Ê œ œ œ � Ê œ � œ �È

È Èa b a b Š ‹ Š ‹È72. (a) Note that 2 mi = 10,560 ft, so there are 800 x feet of river cable at $180 per foot and 10,560 x feet of landÈ a b# #� �

cable at $100 per foot. The cost is C x 180 800 x 100 10,560 x .a b a bÈœ � � �# #

(b) C $a b! œ "ß #!!ß !!!

C $a b&!! ¸ "ß "(&ß )"#

C $a b"!!! ¸ "ß ")'ß &"#

C $a b"&!! ¸ "ß #"#ß !!!

C $a b#!!! ¸ "ß #%$ß ($#

C $a b#&!! ¸ "ß #()ß %(*

C $a b$!!! ¸ "ß $"%ß )(!

Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P.

1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS

1. D : x , D : x 1 D D : x 1. R : y , R : y 0, R : y 1, R : y 0f g f g fg f g f g fg�_ � � _ Ê œ �_ � � _ � �

2. D : x 1 0 x 1, D : x 1 0 x 1. Therefore D D : x 1.f g f g fg� Ê � � Ê œ �

R R : y 0, R : y 2, R : y 0f g f g fgœ �È

3. D : x , D : x , D : x , D : x , R : y 2, R : y 1,f g f g g f f g�_ � � _ �_ � � _ �_ � � _ �_ � � _ œ Î Î

R : 0 y 2, R : yf gÎ � Ÿ Ÿ � _g fÎ"#

4. D : x , D : x 0 , D : x 0, D : x 0; R : y 1, R : y 1, R : 0 y 1, R : 1 yf g f g g f f g f g�_ � � _ œ � Ÿ Ÿ � _Î Î Î g fÎ

5. (a) 2 (b) 22 (c) x 2# �

(d) (x 5) 3 x 10x 22 (e) 5 (f) 2� � œ � � �# #

(g) x 10 (h) (x 3) 3 x 6x 6� � � œ � �# # % #


Section 1.2 Combining Functions; Shifting and Scaling Graphs 9

6. (a) (b) 2 (c) 1� � œ" " �� 3 x 1 x 1

x

(d) (e) 0 (f) "x 4

3

(g) x 2 (h) � œ œ" " �"� �#" �#

� �x 1 x 1x1

xx

7. f g h x f g h x f g 4 x f 3 4 x f 12 3x 12 3x 1 13 3xa ba b a b a b a b a b a ba b a b a ba b‰ ‰ œ œ � œ � œ � œ � � œ �

8. f g h x f g h x f g x f 2 x 1 f 2x 1 3 2x 1 4 6x 1a ba b a b a b a b a b a ba b a b a ba b‰ ‰ œ œ œ � œ � œ � � œ �2 2 2 2 2

9. f g h x f g h x f g f fa ba b a b Éa ba b ˆ ‰ ˆ ‰ˆ ‰ Š ‹ É‰ ‰ œ œ œ œ œ � " œ1 1 x x 5xx 1 4x 1 4x 1 4x1

x �% � � ��"

10. f g h x f g h x f g 2 x f fa ba b a ba ba b Š ‹Š ‹È � � ˆ ‰‰ ‰ œ œ � œ œ œ œŠ ‹È

Š ‹È2 x

2 x 1

2 x 8 3xx 7 2x

2

3

�

� �

� �$� �

�

�

2

2 2 x

2 xx

x

�

$�

�

$�

11. (a) f g x (b) j g x (c) g g xa ba b a ba b a ba b‰ ‰ ‰

(d) j j x (e) g h f x (f) h j f xa ba b a ba b a ba b‰ ‰ ‰ ‰ ‰

12. (a) f j x (b) g h x (c) h h xa ba b a ba b a ba b‰ ‰ ‰

(d) f f x (e) j g f x (f) g f h xa ba b a ba b a ba b‰ ‰ ‰ ‰ ‰

13. g(x) f(x) (f g)(x)‰

(a) x 7 x x 7 � �È È (b) x 2 3x 3(x 2) 3x 6� � œ �

(c) x x 5 x 5# #È È� �

(d) xx x xx 1 x 1 1 x (x 1)� � � � �

xx 1

xx 1

�

�

œ œ

(e) 1 x" "�x 1 x�

(f) x" "x x

14. (a) f g x g x .a ba b a b‰ œ l l œ "l � "lx

(b) f g x so g x x .a ba b a b‰ œ œ Ê " � œ Ê " � œ Ê œ ß œ � "g xg x x g x x x g x x g x

x x xa ba b a b a b a b�"

�" �" �" �"" " " "

(c) Since f g x g x x , g x x .a ba b a b a bÈ‰ œ œ l l œ #

(d) Since f g x f x x , f x x . (Note that the domain of the composite is .)a ba b a bˆ ‰È‰ œ œ l l œ Ò!ß _Ñ#

The completed table is shown. Note that the absolute value sign in part (d) is optional.

g x f x f g x x

x

x x x

x x x

a b a b a ba b

ÈÈ

‰

l l

� "

l l

l l

" "�" l � "l

� "�"

#

#

x xx x

x x

15. (a) f g 1 f 1 1 (b) g f 0 g 2 2 (c) f f 1 f 0 2a b a b a b a b a b a ba b a b a b� œ œ œ � œ � œ œ �

(d) g g 2 g 0 0 (e) g f 2 g 1 1 (f) f g 1 f 1 0a b a b a b a b a b a ba b a b a bœ œ � œ œ � œ � œ

16. (a) f g 0 f 1 2 1 3, where g 0 0 1 1a b a b a b a ba b œ � œ � � œ œ � œ �

(b) g f 3 g 1 1 1, where f 3 2 3 1a b a b a b a ba b œ � œ � � œ œ � œ �

(c) g g 1 g 1 1 1 0, where g 1 1 1a b a b a b a ba b� œ œ � œ � œ � � œ



(d) f f 2 f 0 2 0 2, where f 2 2 2 0a b a b a ba b œ œ � œ œ � œ

(e) g f 0 g 2 2 1 1, where f 0 2 0 2a b a b a ba b œ œ � œ œ � œ

(f) f g f 2 , where g 1ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ˆ ‰" " " " " "# # # # # # #œ � œ � � œ œ � œ �5

17. (a) f g x f g x 1a ba b a ba b É É‰ œ œ � œ1 1 xx x

�

g f x g f x a ba b a ba b‰ œ œ 1x 1È �

(b) Domain f g : , 1 0, , domain g f : 1, a b a b‰ Ð�_ � Ó � Ð _Ñ ‰ Ð� _Ñ

(c) Range f g : 1, , range g f : 0, a b a b‰ Ð _Ñ ‰ Ð _Ñ

18. (a) f g x f g x 1 2 x xa ba b a ba b È‰ œ œ � �

g f x g f x 1 x a ba b a b k ka b‰ œ œ �

(b) Domain f g : 0, , domain g f : , a b a b‰ Ò _Ñ ‰ Ð�_ _Ñ

(c) Range f g : 0, , range g f : , 1a b a b‰ Ð _Ñ ‰ Ð�_ Ó

19. f g x x f g x x x g x g x 2 x x g x 2xa ba b a b a b a b a ba b a b‰ œ Ê œ Ê œ Ê œ � œ † �g xg x 2

a ba b�

g x x g x 2x g xÊ � † œ � Ê œ � œa b a b a b 2x 2x1 x x 1� �

20. f g x x 2 f g x x 2 2 g x 4 x 2 g x g xa ba b a b a b a b a ba b a b a b É‰ œ � Ê œ � Ê � œ � Ê œ Ê œ3 3 x 6 x 62 2� �3

21. (a) y (x 7) (b) y (x 4)œ � � œ � �# #

22. (a) y x 3 (b) y x 5œ � œ �# #

23. (a) Position 4 (b) Position 1 (c) Position 2 (d) Position 3

24. (a) y (x 1) 4 (b) y (x 2) 3 (c) y (x 4) 1 (d) y (x 2)œ � � � œ � � � œ � � � œ � �# # # #

25. 26.

27. 28.



29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.



41. 42.

43. 44.

45. 46.

47. 48.

49. 50.

51. 52.



53. 54.

55. (a) domain: [0 2]; range: [ ] (b) domain: [0 2]; range: [ 1 0]ß #ß $ ß � ß

(c) domain: [0 2]; range: [0 2] (d) domain: [0 2]; range: [ 1 0]ß ß ß � ß

(e) domain: [ 2 0]; range: [ 1] (f) domain: [1 3]; range: [ ]� ß !ß ß !ß "

(g) domain: [ 2 0]; range: [ ] (h) domain: [ 1 1]; range: [ ]� ß !ß " � ß !ß "



56. (a) domain: [0 4]; range: [ 3 0] (b) domain: [ 4 0]; range: [ ]ß � ß � ß !ß $

(c) domain: [ 4 0]; range: [ ] (d) domain: [ 4 0]; range: [ ]� ß !ß $ � ß "ß %

(e) domain: [ 4]; range: [ 3 0] (f) domain: [ 2 2]; range: [ 3 0]#ß � ß � ß � ß

(g) domain: [ 5]; range: [ 3 0] (h) domain: [0 4]; range: [0 3]"ß � ß ß ß

57. y 3x 3 58. y 2x 1 x 1œ � œ � œ % �# ##a b

59. y 60. y 1 1œ " � œ � œ � œ �" " " " " *# # # Î$ˆ ‰

x x xx# # ##a b

61. y x 1 62. y 3 x 1œ % � œ �È È

63. y 16 x 64. y xœ % � œ � œ % �É ˆ ‰ È Èx# # $

# " "# #

65. y 3x 27x 66. yœ " � œ " � œ " � œ " �a b ˆ ‰$ $# )

$x x$



67. Let y x f x and let g x x ,œ � # � " œ œÈ a b a b "Î#

h x x , i x x , anda b a bˆ ‰ ˆ ‰Èœ � œ # �" "# #

"Î# "Î#

j x x f . The graph ofa b a b’ “È ˆ ‰œ � # � œ B"#

"Î#

h x is the graph of g x shifted left unit; thea b a b "#

graph of i x is the graph of h x stretcheda b a b vertically by a factor of ; and the graph ofÈ#

j x f x is the graph of i x reflected acrossa b a b a bœ

the x-axis.

68. Let y f x Let g x x ,œ " � œ Þ œ �È a b a b a bx#

"Î#

h x x , and i x xa b a b a b a bœ � � # œ � � #"Î# "Î#"

#È f x The graph of g x is theœ " � œ ÞÈ a b a bx

#

graph of y x reflected across the x-axis.œ È The graph of h x is the graph of g x shifteda b a b right two units. And the graph of i x is thea b graph of h x compressed vertically by a factora b of .È#

69. y f x x . Shift f x one unit right followed by aœ œa b a b$

shift two units up to get g x x .a b a bœ � " � #3

70. y x f x .œ " � B � # œ �Ò � " � �# Ó œa b a b a b a b$ $

Let g x x , h x x , i x x ,a b a b a b a b a b a bœ œ � " œ � " � �#$ $ $

and j x x . The graph of h x is thea b a b a b a bœ �Ò � " � �# Ó$

graph of g x shifted right one unit; the graph of i x isa b a b the graph of h x shifted down two units; and the grapha b of f x is the graph of i x reflected across the x-axis.a b a b

71. Compress the graph of f x horizontally by a factora b œ "x

of 2 to get g x . Then shift g x vertically down 1a b a bœ "#x

unit to get h x .a b œ � ""#x



72. Let f x and g xa b a bœ œ � " œ � "" # "x x# #

B#

#Š ‹

Sinceœ � " œ � "Þ" "

Î # "Î # BŠ ‹ ’Š ‹ “È Èx# #

, we see that the graph of f x stretchedÈ a b# ¸ "Þ%

horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g x .a b

73. Reflect the graph of y f x x across the x-axisœ œa b È$ to get g x x.a b Èœ � $

74. y f x x xœ œ �# œ Ò �" # Óa b a b a ba b#Î$ #Î$

x x . So the graphœ �" # œ #a b a b a b#Î$ #Î$ #Î$

of f x is the graph of g x x compresseda b a b œ #Î$

horizontally by a factor of 2.

75. 76.

77. x y 78. x y* � #& œ ##& Ê � œ " "' � ( œ ""# Ê � œ "# # # #& $ %

(

x xy y# #

# # #

# #

#Š ‹È



79. x y 80. x y$ � � # œ $ Ê � œ " � " � # œ % Ê � œ "# ## � # #" #

$ #

� �"a b a bx y x y#

# #

#

# #

##a b

Š ‹ Š ‹È È� ‘a b

81. x y 82. x y$ � " � # � # œ ' ' � � * � œ &%a b a b ˆ ‰ ˆ ‰# # $ "# #

# #

Ê � œ " Ê � œ "a bŠ ‹ Š ‹ Š ‹È È

� ‘a b ’ “ˆ ‰ ˆ ‰È

x y x y � "

# $ '

� �# � �

$

�#

# # #

# $

#

#

#

"

#

#

83. has its center at . Shiftinig 4 unitsx y# #

"' *� œ " !ß !a b left and 3 units up gives the center at h, k .a b a bœ �%ß $

So the equation is � ‘a b a bx 4

4 3y 3� � �

#

# #

#

� œ "

. Center, C, is , andÊ � œ " �%ß $a b a bx y 4 3� % � $# #

# # a b major axis, AB, is the segment from to .a b a b�)ß $ !ß $

84. The ellipse has center h, k .x y# #

% #&� œ " œ !ß !a b a b Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at h, ka b a bœ $ß �#

and an equation . Center,a b � ‘a bx 3 y �% #&

� �## #

� œ "

C, is 3 , and AB, the segment from toa b a bß �# $ß $

is the major axis.a b$ß�(

85. (a) (fg)( x) f( x)g( x) f(x)( g(x)) (fg)(x), odd� œ � � œ � œ �

(b) ( x) (x), oddŠ ‹ Š ‹f fg g( x) g(x) g

f( x) f(x)� œ œ œ ��



(c) ( x) (x), oddˆ ‰ ˆ ‰g g( x) g(x) gf f( x) f(x) f� œ œ œ ��

�

(d) f ( x) f( x)f( x) f(x)f(x) f (x), even# #� œ � � œ œ

(e) g ( x) (g( x)) ( g(x)) g (x), even# # # #� œ � œ � œ

(f) (f g)( x) f(g( x)) f( g(x)) f(g(x)) (f g)(x), even‰ � œ � œ � œ œ ‰

(g) (g f)( x) g(f( x)) g(f(x)) (g f)(x), even‰ � œ � œ œ ‰

(h) (f f)( x) f(f( x)) f(f(x)) (f f)(x), even‰ � œ � œ œ ‰

(i) (g g)( x) g(g( x)) g( g(x)) g(g(x)) (g g)(x), odd‰ � œ � œ � œ � œ � ‰

86. Yes, f(x) 0 is both even and odd since f( x) 0 f(x) and f( x) 0 f(x).œ � œ œ � œ œ �

87. (a) (b)

(c) (d)

88.


Section 1.3 Trigonometric Functions 19

1.3 TRIGONOMETRIC FUNCTIONS

1. (a) s r (10) 8 m (b) s r (10)(110°) mœ œ œ œ œ œ œ) 1 )ˆ ‰ ˆ ‰4 110 555 180° 18 91 1 1 1

2. radians and 225°) œ œ œ œs 10 5 5 180°r 8 4 4

1 1 1

1ˆ ‰

3. 80° 80° s (6) 8.4 in. (since the diameter 12 in. radius 6 in.)) )œ Ê œ œ Ê œ œ œ Ê œˆ ‰ ˆ ‰1 1 1

180° 9 94 4

4. d 1 meter r 50 cm 0.6 rad or 0.6 34°œ Ê œ Ê œ œ œ ¸)s 30 180°r 50

ˆ ‰1

5. 0 6

sin 0 0

cos 1 0

tan 0 3 0 und.

cot und. und. 0 1

sec 1 und. 2

csc und. und. 2

) 1

)

)

)

)

)

)

� �

� "

� � " �

�"

�

� �# " �

� "

2 33 43

2

2

3

23

1 1 1

#

#

"

" "

#

"

ÈÈ

È

È

È

È

ÈÈ

.

sin

cos

tan und. 3

cot 3 3

sec und. 2

csc

)

)

)

)

)

)

)

� � �

" � �

! �

� � " �

! � � " �

# �

" � �#

33

32

3 32

3 3

3

2 23 3

23

1 1 1 1 1

# ' % '

&

# # #

" " "

" "

# # #

" "

"

ÈÈ

È ÈÈ

È È

È

È È

È

ÈÈ È

ÈÈ2 #

7. cos x , tan x 8. sin x , cos xœ � œ � œ œ4 3 25 4 5 5È È

"

9. sin x , tan x 8 10. sin x , tan xœ � œ � œ œ �È8

3 13 512 12È

11. sin x , cos x 12. cos x , tan xœ � œ � œ � œ" "

#È È ÈÈ

5 5 32 3

13. 14.

period period 4œ œ1 1

15. 16.

period 2 period 4œ œ



17. 18.


19. 20.

period 2 period 2œ œ1 1

21. 22.


23. period , symmetric about the origin 24. period 1, symmetric about the originœ œ1

#

25. period 4, symmetric about the s-axis 26. period 4 , symmetric about the originœ œ 1



27. (a) Cos x and sec x are positive for x in the interval

, ; and cos x and sec x are negative for x in theˆ ‰� 1 1

2 2

intervals , and , . Sec x is undefinedˆ ‰ ˆ ‰� �3 32 2 2 21 1 1 1

when cos x is 0. The range of sec x is ( 1] [ ); the range of cos x is [ 1].�_ß� � "ß_ �"ß

(b) Sin x and csc x are positive for x in the intervals

, and , ; and sin x and csc x are negativeˆ ‰ a b� � !321

1 1

for x in the intervals , and , . Csc x isa b ˆ ‰� !1 1321

undefined when sin x is 0. The range of csc x is ( 1] [1 ); the range of sin x is [ ].�_ß� � ß_ �"ß "

28. Since cot x , cot x is undefined when tan x 0œ œ"

tan x

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: y 1, 0, 1 30. D: x ; R: y 1, 0, 1�_ � � _ œ � �_ � � _ œ �

31. cos x cos x cos sin x sin (cos x)(0) (sin x)( 1) sin xˆ ‰ ˆ ‰ ˆ ‰� œ � � � œ � � œ1 1 1

# # #

32. cos x cos x cos sin x sin (cos x)(0) (sin x)(1) sin xˆ ‰ ˆ ‰ ˆ ‰� œ � œ � œ �1 1 1

# # #

33. sin x sin x cos cos x sin (sin x)(0) (cos x)(1) cos xˆ ‰ ˆ ‰ ˆ ‰� œ � œ � œ1 1 1

# # #

34. sin x sin x cos cos x sin (sin x)(0) (cos x)( 1) cos xˆ ‰ ˆ ‰ ˆ ‰� œ � � � œ � � œ �1 1 1

# # #

35. cos (A B) cos (A ( B)) cos A cos ( B) sin A sin ( B) cos A cos B sin A ( sin B)� œ � � œ � � � œ � �

cos A cos B sin A sin Bœ �

36. sin (A B) sin (A ( B)) sin A cos ( B) cos A sin ( B) sin A cos B cos A ( sin B)� œ � � œ � � � œ � �

sin A cos B cos A sin Bœ �

37. If B A, A B 0 cos (A B) cos 0 1. Also cos (A B) cos (A A) cos A cos A sin A sin Aœ � œ Ê � œ œ � œ � œ �

cos A sin A. Therefore, cos A sin A 1.œ � � œ# # # #



38. If B 2 , then cos (A 2 ) cos A cos 2 sin A sin 2 (cos A)(1) (sin A)(0) cos A andœ � œ � œ � œ1 1 1 1

sin (A 2 ) sin A cos 2 cos A sin 2 (sin A)(1) (cos A)(0) sin A. The result agrees with the� œ � œ � œ1 1 1

fact that the cosine and sine functions have period 2 .1

39. cos ( x) cos cos sin sin x ( 1)(cos x) (0)(sin x) cos x1 1 1� œ B � œ � � œ �

40. sin (2 x) sin 2 cos ( x) cos (2 ) sin ( x) (0)(cos ( x)) (1)(sin ( x)) sin x1 1 1� œ � � � œ � � � œ �

41. sin x sin cos ( x) cos sin ( x) ( 1)(cos x) (0)(sin ( x)) cos xˆ ‰ ˆ ‰ ˆ ‰3 3 31 1 1

# # #� œ � � � œ � � � œ �

42. cos x cos cos x sin sin x (0)(cos x) ( 1)(sin x) sin xˆ ‰ ˆ ‰ ˆ ‰3 3 31 1 1

# # #� œ � œ � � œ

43. sin sin sin cos cos sin 71 4 3 4 3 4 3 4

2 2 3 6 21 1 1 1 1 1 1

# # # # #

" �œ � œ � œ � œˆ ‰ ˆ ‰Š ‹ Š ‹Š ‹È È È È È

44. cos cos cos cos sin sin 11 2 2 21 4 3 4 3 4 3 4

2 2 3 2 61 1 1 1 1 1 1

# # # # #

" �œ � œ � œ � � œ �ˆ ‰ ˆ ‰Š ‹ Š ‹Š ‹È È È ÈÈ

45. cos cos cos cos sin sin1 1 1 1 1 1 1

12 3 4 3 4 3 42 23 1 3

2 2œ � œ � � � œ � � œˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ Š ‹Š ‹"

# # # #

�È È È ÈÈ

46. sin sin sin cos cos sin5 2 2 21 3 4 3 4 3 4

3 1 32 22 2

1 1 1 1 1 1 1

# # # # #

" �œ � œ � � � œ � � � œˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰Š ‹Š ‹ Š ‹È È È ÈÈ

47. cos 48. cos # #� � �

# # # # #

��

�1 1

8 4 1 41 cos 1 1 cos2 2 5 1 2 3œ œ œ œ œ œ

ˆ ‰ ˆ ‰È ÈŠ ‹2 108 1

2 31 1

È È

# ##

49. sin 50. sin # #

# # # # #

� � ��

�1 1

1 4 8 41 cos 1 1 cos2 3 3 1 2 2œ œ œ œ œ œ

ˆ ‰ ˆ ‰È ÈŠ ‹2 61 8

3 21 1

# ##

È È

51. sin sin , , , 2 3 2 4 54 2 3 3 3 3

3) ) )œ Ê œ „ Ê œ

È1 1 1 1

52. sin cos tan 1 tan 1 , , , 2 2 2sin cos 3 5 7cos cos 4 4 4 4) ) ) ) )œ Ê œ Ê œ Ê œ „ Ê œ

2 2

2 2) ) 1 1 1 1

) )

53. sin 2 cos 0 2sin cos cos 0 cos 2sin 1 0 cos 0 or 2sin 1 0 cos 0 or) ) ) ) ) ) ) ) ) )� œ Ê � œ Ê � œ Ê œ � œ Ê œa b sin , , or , , , , ) ) ) )œ Ê œ œ Ê œ"

#

1 1 1 1 1 1 1 1

2 2 6 6 6 2 6 23 5 5 3

54. cos 2 cos 0 2cos 1 cos 0 2cos cos 1 0 cos 1 2cos 1 0) ) ) ) ) ) ) )� œ Ê � � œ Ê � � œ Ê � � œ2 2 a ba b cos 1 0 or 2cos 1 0 cos 1 or cos or , , , Ê � œ � œ Ê œ � œ Ê œ œ Ê œ) ) ) ) ) 1 ) ) 1

"

#

1 1 1 1

3 3 3 35 5

55. tan (A B)� œ œ œsin (A B)cos (A B) cos A cos B sin A sin B

sin A cos B cos A cos B�

� �

� �sin A cos B cos A sin Bcos A cos B cos A cos Bcos A cos B sin A sin Bcos A cos B cos A cos B�

�

�œ tan A tan B

1 tan A tan B

56. tan (A B)� œ œ œsin (A B)cos (A B) cos A cos B sin A sin B

sin A cos B cos A cos B�

� �

� �sin A cos B cos A sin Bcos A cos B cos A cos Bcos A cos B sin A sin Bcos A cos B cos A cos B�

�

�œ tan A tan B

1 tan A tan B

57. According to the figure in the text, we have the following: By the law of cosines, c a b 2ab cos # # #œ � � )

1 1 2 cos (A B) 2 2 cos (A B). By distance formula, c (cos A cos B) (sin A sin B)œ � � � œ � � œ � � �# # # # #

cos A 2 cos A cos B cos B sin A 2 sin A sin B sin B 2 2(cos A cos B sin A sin B). Thusœ � � � � � œ � �# # # #

c 2 2 cos (A B) 2 2(cos A cos B sin A sin B) cos (A B) cos A cos B sin A sin B.# œ � � œ � � Ê � œ �



58. (a) cos A B cos A cos B sin A sin Ba b� œ �

sin cos and cos sin) ) ) )œ � œ �ˆ ‰ ˆ ‰1 1

# #

Let A B) œ �

sin A B cos A B cos A B cos A cos B sin A sin Ba b a b’ “ ’ “ˆ ‰ ˆ ‰ ˆ ‰� œ � � œ � � œ � � �1 1 1 1

# # # #

sin A cos B cos A sin Bœ �

(b) cos A B cos A cos B sin A sin Ba b� œ �

cos A B cos A cos B sin A sin Ba b a b a ba b� � œ � � �

cos A B cos A cos B sin A sin B cos A cos B sin A sin BÊ � œ � � � œ � �a b a b a b a b cos A cos B sin A sin Bœ �

Because the cosine function is even and the sine functions is odd.

59. 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 � œˆ ‰

Thus, c 7 2.65.œ ¸È

60. c a b 2ab cos C 2 3 2(2)(3) cos (40°) 13 12 cos (40°). Thus, c 13 12 cos 40° 1.951.# # # # #œ � � œ � � œ � œ � ¸È

61. From the figures in the text, we see that sin B . If C is an acute angle, then sin C . On the other hand,œ œh hc b

if C is obtuse (as in the figure on the right), then sin C sin ( C) . Thus, in either case,œ � œ1hb

h b sin C c sin B ah ab sin C ac sin B.œ œ Ê œ œ

By the law of cosines, cos C and cos B . Moreover, since the sum of theœ œa b c a c b2ab 2ac

# # # # # #� � � �

interior angles of a triangle is , we have sin A sin ( (B C)) sin (B C) sin B cos C cos B sin C1 1œ � � œ � œ �

2a b c c b ah bc sin A.œ � œ � � � � œ Ê œˆ ‰ ˆ ‰ ˆ ‰’ “ ’ “ a bh a b c a c b h h ahc 2ab 2ac b 2abc bc

# # # # # #� � � � # # # # #

Combining our results we have ah ab sin C, ah ac sin B, and ah bc sin A. Dividing by abc givesœ œ œ

.h sin A sin C sin Bbc a c bœ œ œðóóóóóóóñóóóóóóóò

law of sines

62. By the law of sines, . By Exercise 61 we know that c 7. Thus sin B 0.982.sin A sin B3 c

3/2 3 32 7#

œ œ œ œ ¶È È

ÈÈ

63. From the figure at the right and the law of cosines, b a 2 2(2a) cos B# # #œ � �

a 4 4a a 2a 4.œ � � œ � �# #"

#ˆ ‰

Applying the law of sines to the figure, sin A sin Ba bœ

b a. Thus, combining results,Ê œ Ê œÈ È2/2

a b3/2 3É

#

a 2a 4 b a 0 a 2a 4# # # #

# #

"� � œ œ Ê œ � �3

0 a 4a 8. From the quadratic formula and the fact that a 0, we haveÊ œ � � �#

a 1.464.œ œ ¶� � � �

# #

�4 4 4(1)( 8) 4 3 4È È#

64. (a) The graphs of y sin x and y x nearly coincide when x is near the origin (when the calculatorœ œ

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.



65. A 2, B 2 , C , D 1œ œ œ � œ �1 1

66. A , B 2, C 1, Dœ œ œ œ" "

# #

67. A , B 4, C 0, Dœ � œ œ œ21 1

"

68. A , B L, C 0, D 0œ œ œ œL21

69-72. 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="#69 (Section 1.3)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[2 /b (x c)] + d1 �

Plot[f[x]/.{a 3, b 1, c 0, d 0}, {x, 4 , 4 }]Ä Ä Ä Ä � 1 1



69. (a) The graph stretches horizontally.

(b) The period remains the same: period B . The graph has a horizontal shift of period.œ l l "

#

70. (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 '

71. (a) The graph shifts upwards D units for Dl l � !

(b) The graph shifts down D units for Dl l � !Þ

72. (a) The graph stretches A units. (b) For A , the graph is inverted.l l � !



1.4 GRAPHING WITH CALCULATORS AND COMPUTERS

1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.

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. 2 5 by 15 40 6. 4 4 by 4 4Ò� ß Ó Ò� ß Ó Ò� ß Ó Ò� ß Ó



Section 1.4 Graphing with Calculators and Computers 27




15. 3 3 by 0 10 16. 1 2 by 0 1Ò� ß Ó Ò ß Ó Ò� ß Ó Ò ß Ó




19. 4 4 by 0 3 20. 5 5 by 2 2Ò� ß Ó Ò ß Ó Ò� ß Ó Ò� ß Ó

21. by 22. by Ò�"!ß "!Ó Ò�'ß 'Ó Ò�&ß &Ó Ò�#ß #Ó

23. by 24. by Ò�'ß "!Ó Ò�'ß 'Ó Ò�$ß &Ó Ò�#ß "!Ó


Section 1.4 Graphing with Calculators and Computers 29

25. 0 03 0 03 by 1 25 1 25 26. 0 1 0 1 by 3 3Ò� Þ ß Þ Ó Ò� Þ ß Þ Ó Ò� Þ ß Þ Ó Ò� ß Ó

27. 300 300 by 1 25 1 25 28. 50 50 by 0 1 0 1Ò� ß Ó Ò� Þ ß Þ Ó Ò� ß Ó Ò� Þ ß Þ Ó

29. 0 25 0 25 by 0 3 0 3 30. 0 15 0 15 by 0 02 0 05Ò� Þ ß Þ Ó Ò� Þ ß Þ Ó Ò� Þ ß Þ Ó Ò� Þ ß Þ Ó

31. x x y y y x x .# # #� # œ % � % � Ê œ # „ � � # � )È The lower half is produced by graphing

y x x .œ # � � � # � )È #

32. y x y x . The upper branch# # #� "' œ " Ê œ „ " � "'È is produced by graphing y x .œ " � "'È #



33. 34.

35. 36.

37. 38 Þ

39. 40.

CHAPTER 1 PRACTICE EXERCISES

1. The area is A r and the circumference is C r. Thus, r A .œ œ # œ Ê œ œ1 1 1## # %

#C C C1 1 1

ˆ ‰ #

2. The surface area is S r r . The volume is V r r . Substitution into the formula forœ % Ê œ œ Ê œ1 1# $% $ %

"Î# % $ˆ ‰ ÉS V1 1

$

surface area gives S r .œ % œ %1 1# $%

#Î$ˆ ‰V1


Chapter 1 Practice Exercises 31

3. The coordinates of a point on the parabola are x x . The angle of inclination joining this point to the origin satisfiesa bß # )

the equation tan x. Thus the point has coordinates x x tan tan .) ) )œ œ ß œ ßxx

# a b a b# #

4. tan h tan ft.) )œ œ Ê œ &!!rise hrun &!!

5. 6.

Symmetric about the origin. Symmetric about the y-axis.

7. 8.

Neither Symmetric about the y-axis.

9. y x x x y x . Even.a b a b a b� œ � � " œ � " œ# #

10. y x x x x x x x y x . Odd.a b a b a b a b a b� œ � � � � � œ � � � œ �& $ & $

11. y x cos x cos x y x . Even.a b a b a b� œ " � � œ " � œ

12. y x sec x tan x sec x tan x y x . Odd.a b a b a b a b� œ � � œ œ œ � œ �sin xcos x cos x

sin xa ba b��

�# #

13. y x y x . Odd.a b a b� œ œ œ � œ �a ba b a b

� �"

� �# ��" �"

� �# �#x

x xx xx x x x

%

$

% %

$ $

14. y x x sin x x sin x x sin x y x . Odd.a b a b a b a b a b a b� œ � � � œ � � œ � � œ �

15. y x x cos x x cos x. Neither even nor odd.a b a b� œ � � � œ � �

16. y x x cos x x cos x y x . Odd.a b a b a b a b� œ � � œ � œ �

17. Since f and g are odd f x f x and g x g x .Ê � œ � � œ �a b a b a b a b (a) f g x f x g x f x g x f x g x f g x f g is evena ba b a b a b a b a b a b a b a ba b† � œ � � œ Ò� ÓÒ� Ó œ œ † Ê †

(b) f x f x f x f x f x f x f x f x f x f x f x f is odd.3 3 3a b a b a b a b a b a b a b a b a b a b a b� œ � � � œ Ò� ÓÒ� ÓÒ� Ó œ � † † œ � Ê

(c) f sin x f sin x f sin x f sin x is odd.a b a b a b a ba b a b a b a b� œ � œ � Ê

(d) g sec x g sec x g sec x is even.a b a b a ba b a b a b� œ Ê

(e) g x g x g x g is evenl � l œ l� l œ l l Ê l l Þa b a b a b



18. Let f a x f a x and define g x f x a . Then g x f x a f a x f a x f x a g xa b a b a b a b a b a b a b a b a b a ba b� œ � œ � � œ � � œ � œ � œ � œ

g x f x a is even.Ê œ �a b a b19. (a) The function is defined for all values of x, so the domain is .a b�_ß _

(b) Since x attains all nonnegative values, the range is .l l Ò�#ß _Ñ

20. (a) Since the square root requires x , the domain is ." � ! Ð�_ß "Ó

(b) Since x attains all nonnegative values, the range is .È" � Ò�#ß _Ñ

21. (a) Since the square root requires x , the domain is ."' � ! Ò�%ß %Ó#

(b) For values of x in the domain, x , so x . The range is .! Ÿ "' � Ÿ "' ! Ÿ "' � Ÿ % Ò!ß %Ó# #È22. (a) The function is defined for all values of x, so the domain is .a b�_ß _

(b) Since attains all positive values, the range is .$ "ß _#�x a b23. (a) The function is defined for all values of x, so the domain is .a b�_ß _

(b) Since e attains all positive values, the range is .# �$ß _�x a b24. (a) The function is equivalent to y tan x, so we require x for odd integers k. The domain is given by x forœ # # Á Ák k1 1

# %

odd integers k. (b) Since the tangent function attains all values, the range is .a b�_ß _

25. (a) The function is defined for all values of x, so the domain is .a b�_ß _

(b) The sine function attains values from to , so sin x and hence sin x . The�" " �# Ÿ # $ � Ÿ # �$ Ÿ # $ � � " Ÿ "a b a b1 1

range is 3 1 .Ò� ß Ó


(b) The function is equivalent to y x , which attains all nonnegative values. The range is .œ Ò!ß _ÑÈ& #

27. (a) The logarithm requires x , so the domain is .� $ � ! $ß _a b (b) The logarithm attains all real values, so the range is .a b�_ß _


(b) The cube root attains all real values, so the range is .a b�_ß _

29. (a) Increasing because volume increases as radius increases (b) Neither, since the greatest integer function is composed of horizontal (constant) line segments (c) Decreasing because as the height increases, the atmospheric pressure decreases. (d) Increasing because the kinetic (motion) energy increases as the particles velocity increases.

30. (a) Increasing on 2, (b) Increasing on 1, Ò _Ñ Ò� _Ñ

(c) Increasing on , (d) Increasing on , a b�_ _ Ò _Ñ"#

31. (a) The function is defined for x , so the domain is .�% Ÿ Ÿ % Ò�%ß %Ó

(b) The function is equivalent to y x , x , which attains values from to for x in the domain. Theœ l l �% Ÿ Ÿ % ! #È range is .Ò!ß #Ó



32. (a) The function is defined for x , so the domain is .�# Ÿ Ÿ # Ò�#ß #Ó

(b) The range is .Ò�"ß "Ó

33. First piece: Line through and . m y x xa b a b!ß " "ß ! œ œ œ �" Ê œ � � " œ " �!�" �""�! "

Second piece: Line through and . m y x x xa b a b a b"ß " #ß ! œ œ œ �" Ê œ � � " � " œ � � # œ # �!�" �"#�" "

f xx, xx, x

a b œœ" � ! Ÿ � "# � " Ÿ Ÿ #

34. First piece: Line through and 2 5 . m y xa b a b!ß ! ß œ œ Ê œ5 5 52 2 2�!�!

Second piece: Line through 2 5 and 4 . m y x 2 5 x 10 10a b a b a bß ß ! œ œ œ � Ê œ � � � œ � � œ �!� ��

5 5 5 5 5 5x4 2 2 2 2 2 2

f x (Note: x 2 can be included on either piece.) x, x 2

10 , 2 x 4a b �œ œ

! Ÿ �

� Ÿ Ÿ

525x2

35. (a) f g f g f fa ba b a b a ba b Š ‹‰ �" œ �" œ œ " œ œ "" "�"�# "È

(b) g f g f g or a ba b a ba b ˆ ‰ É‰ # œ # œ œ œ" " " #

�# #Þ& &2 É È"

#

(c) f f x f f x f x, xa ba b a ba b ˆ ‰‰ œ œ œ œ Á !" ""Îx x

(d) g g x g g x ga ba b a ba b Š ‹‰ œ œ œ œ" "�# �#

�#

"�# �#È É É

ÈÈx

x

x"

�#

%

Èx

36. (a) f g f g f fa ba b a b a ba b ˆ ‰È‰ �" œ �" œ �" � " œ ! œ # � ! œ #$

(b) g f f g g ga ba b a b a b a ba b È‰ # œ # œ # � # œ ! œ ! � " œ "$

(c) f f x f f x f x x xa ba b a b a b a ba b‰ œ œ # � œ # � # � œ

(d) g g x g g x g x xa ba b a ba b ˆ ‰È ÈÉ‰ œ œ � " œ � " � "$ $$

37. (a) f g x f g x f x x x, x .a ba b a ba b ˆ ‰ ˆ ‰È È‰ œ œ � # œ # � � # œ � �##

g f x f g x g x x xa ba b a b a b a ba b È È‰ œ œ # � œ # � � # œ % �# # #

(b) Domain of f g: (c) Range of f g: ‰ Ò�#ß _ÑÞ ‰ Ð�_ß #ÓÞ

Domain of g f: Range of g f: ‰ Ò�#ß #ÓÞ ‰ Ò!ß #ÓÞ

38. (a) f g x f g x f x x x.a ba b a ba b Š ‹È È ÈÉ‰ œ œ " � œ " � œ " �%

g f x f g x g x xa ba b a ba b ˆ ‰È ÈÉ‰ œ œ œ " �

(b) Domain of f g: (c) Range of f g: ‰ Ð�_ß "ÓÞ ‰ Ò!ß _ÑÞ

Domain of g f: Range of g f: ‰ Ò!ß "ÓÞ ‰ Ò!ß "ÓÞ

39. y f x y f f xœ œ ‰a b a ba b



40.

41. 42.

The graph of f (x) f x is the same as the It does not change the graph.# "œ a bk k 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 y f (x), x 0 across the y-axis.œ "

43. 44.

Whenever g (x) is positive, the graph of y g (x) Whenever g (x) is positive, the graph of y g (x) g (x)" # " # "œ œ œ k k g (x) is the same as the graph of y g (x). is the same as the graph of y g (x). When g (x) isœ œ œk k" " " "

When g (x) is negative, the graph of y g (x) is negative, the graph of y g (x) is the reflection of the" # #œ œ

the reflection of the graph of y g (x) across the graph of y g (x) across the x-axis.œ œ" "

x-axis.



45. 46.

Whenever g (x) is positive, the graph of The graph of f (x) f x is the same as the" # "œ a bk k y g (x) g (x) is the same as the graph of graph of f (x) to the right of the y-axis. The œ œ# " "k k y g (x). When g (x) is negative, the graph of graph of f (x) to the left of the y-axis is the œ " " #

y g (x) is the reflection of the graph of reflection of y f (x), x 0 across the y-axis. œ œ # "

y g (x) across the x-axis.œ "

47. 48.

The graph of f (x) f x is the same as the The graph of f (x) f x is the same as the# " # "œ œa b a bk k k k 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 y f (x), x 0 across the y-axis. reflection of y f (x), x 0 across the y-axis.œ œ " "

49. (a) y g x 3 (b) y g x 2œ � � œ � �a b ˆ ‰" ## 3

(c) y g x (d) y g xœ � œ �a b a b (e) y 5 g x (f) y g 5xœ † œa b a b50. (a) Shift the graph of f right 5 units (b) Horizontally compress the graph of f by a factor of 4 (c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis

(d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left unit."#

(e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the

graph up unit."4

51. Reflection of the grpah of y x about the x-axisœ È followed by a horizontal compression by a factor of

then a shift left 2 units.12



52. Reflect the graph of y x about the x-axis, followedœ

by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit.

53. Vertical compression of the graph of y by aœ 1x2

factor of 2, then shift the graph up 1 unit.

54. Reflect the graph of y x about the y-axis, thenœ 1 3Î

compress the graph horizontally by a factor of 5.

55. 56.

period period 4œ œ1 1

57. 58.




59. 60.


61. (a) sin B sin b 2 sin 2 3. By the theorem of Pythagoras,œ œ œ Ê œ œ œ1 1

3 c 3b b 3

# #Š ‹ ÈÈ

a b c a c b 4 3 1.# # # # #� œ Ê œ � œ � œÈ È (b) sin B sin c . Thus, a c b (2) .œ œ œ Ê œ œ œ œ � œ � œ œ1

3 c c sin 3b 2 2 2 4 4 4 2

3 3 31

33Š ‹ È È ÈÈ

#

È ÊŠ ‹ É# # ##

62. (a) sin A a c sin A (b) tan A a b tan Aœ Ê œ œ Ê œa ac b

63. (a) tan B a (b) sin A cœ Ê œ œ Ê œb b a aa tan B c sin A

64. (a) sin A (c) sin Aœ œ œa ac c c

c bÈ # #�

65. Let h height of vertical pole, and let b and c denote theœ

distances of points B and C from the base of the pole, measured along the flatground, respectively. Then,

tan 50° , tan 35° , and b c 10.œ œ � œh hc b

Thus, h c tan 50° and h b tan 35° (c 10) tan 35°œ œ œ �

c tan 50° (c 10) tan 35°Ê œ �

c (tan 50° tan 35°) 10 tan 35°Ê � œ

c h c tan 50°Ê œ Ê œ10 tan 35°tan 50° tan 35°�

16.98 m.œ ¸10 tan 35° tan 50°tan 50° tan 35°�

66. Let h height of balloon above ground. From the figure atœ

the right, tan 40° , tan 70° , and a b 2. Thus,œ œ � œh ha b

h b tan 70° h (2 a) tan 70° and h a tan 40°œ Ê œ � œ

(2 a) tan 70° a tan 40° a(tan 40° tan 70°)Ê � œ Ê �

2 tan 70° a h a tan 40°œ Ê œ Ê œ2 tan 70°tan 40° tan 70°�

1.3 km.œ ¸2 tan 70° tan 40°tan 40° tan 70°�

67. (a)

(b) The period appears to be 4 .1



(c) f(x 4 ) sin (x 4 ) cos sin (x 2 ) cos 2 sin x cos � œ � � œ � � � œ �1 1 1 1ˆ ‰ ˆ ‰x 4 x x�# # #1

since the period of sine and cosine is 2 . Thus, f(x) has period 4 .1 1

68. (a)

(b) D ( 0) ( ); R [ 1 1]œ �_ß � !ß_ œ � ß

(c) f is not periodic. For suppose f has period p. Then f kp f sin 2 0 for allˆ ‰ ˆ ‰" "# #1 1

� œ œ œ1

integers k. Choose k so large that kp 0 . But then" " "# �1 1 1

� � Ê � �(1/2 ) kp 1

f kp sin 0 which is a contradiction. Thus f has no period, as claimed.ˆ ‰ Š ‹" "# # �1 1

� œ �(1/ ) kp

CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES

1. There are (infinitely) many such function pairs. For example, f(x) 3x and g(x) 4x satisfyœ œ

f(g(x)) f(4x) 3(4x) 12x 4(3x) g(3x) g(f(x)).œ œ œ œ œ œ

2. Yes, there are many such function pairs. For example, if g(x) (2x 3) and f(x) x , thenœ � œ$ "Î$

(f g)(x) f(g(x)) f (2x 3) (2x 3) 2x 3.‰ œ œ � œ � œ �a b a b$ $ "Î$

3. If f is odd and defined at x, then f( x) f(x). Thus g( x) f( x) 2 f(x) 2 whereas� œ � � œ � � œ � �

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 isÊ œ œ

even, then g(x) f(x) 2 is also even: g( x) f( x) 2 f(x) 2 g(x).œ � � œ � � œ � œ

4. If g is odd and g(0) is defined, then g(0) g( 0) g(0). Therefore, 2g(0) 0 g(0) 0.œ � œ � œ Ê œ

5. For (x y) in the 1st quadrant, x y 1 xß � œ �k k k k x y 1 x y 1. For (x y) in the 2ndÍ � œ � Í œ ß

quadrant, x y x 1 x y x 1k k k k� œ � Í � � œ �

y 2x 1. In the 3rd quadrant, x y x 1Í œ � � œ �k k k k x y x 1 y 2x 1. In the 4thÍ � � œ � Í œ � �

quadrant, x y x 1 x ( y) x 1k k k k� œ � Í � � œ �

y 1. The graph is given at the right.Í œ �


Chapter 1 Additional and Advanced Exercises 39

6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y y x x� œ �k k k k 2y 2x y x.Í œ Í œ

(2) 2nd quadrant: y y x x� œ �k k k k 2y x ( x) 0 y 0.Í œ � � œ Í œ

(3) 3rd quadrant: y y x x� œ �k k k k y ( y) x ( x) 0 0Í � � œ � � Í œ

points in the 3rd quadrantÊ all satisfy the equation. (4) 4th quadrant: y y x x� œ �k k k k y ( y) 2x 0 x. CombiningÍ � � œ Í œ

these results we have the graph given at the right:

7. (a) sin x cos x 1 sin x 1 cos x (1 cos x)(1 cos x) (1 cos x)# # # #�� œ Ê œ � œ � � Ê � œ sin x

1 cos x

#

Ê œ1 cos x sin xsin x 1 cos x�

�

(b) Using the definition of the tangent function and the double angle formulas, we have

tan .## �

�ˆ ‰x 1 cos xsincos 1 cos xœ œ œ

#

# #

#

#

"�#

"�#

#

ˆ ‰ˆ ‰

x

x

cos 2 x

cos 2 x

Š ‹Š ‹

Š ‹Š ‹

8. 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 bb a c� �

�œ )

(a c)(a c) b(2a cos b)Ê � � œ �)

a c 2ab cos bÊ � œ �# # #)

c a b 2ab cos .Ê œ � �# # # )

9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah bc sin A ab sin C ac sin Bœ œ œ

the area of ABC (base)(height) ah bc sin A ab sin C ac sin B.Ê œ œ œ œ œ" " " " "# # # # #

10. As in Section 1.3, Exercise 61, (Area of ABC) (base) (height) a h a b sin C# # # # # # # #" " "œ œ œ4 4 4

a b cos C . By the law of cosines, c a b 2ab cos C cos C .œ " � œ � � Ê œ" � �# # # # # #4 2ab

a b ca b # # #

Thus, (area of ABC) a b cos C a b# # # # # #" " � �#

#� �œ " � œ " � œ " �4 4 ab 4 4a b

a b c a b a b ca b Œ �Š ‹ Š ‹# # # # # # # # #

# #

a b

4a b a b c 2ab a b c 2ab a b cœ � � � œ � � � � � �" "# # # # # # # # # # ##

16 16Š ‹a b c da b a ba b a b (a b) c c (a b) ((a b) c)((a b) c)(c (a b))(c (a b))œ � � � � œ � � � � � � � �" "# # # #

16 16c d c da b a b s(s a)(s b)(s c), where s .œ œ � � � œ� ‘ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰a b c a b c a b c a b c a b c� � � � � � � � � � �

# # # # #

Therefore, the area of ABC equals s(s a)(s b)(s c) .È � � �

11. If f is even and odd, then f( x) f(x) and f( x) f(x) f(x) f(x) for all x in the domain of f.� œ � � œ Ê œ �

Thus 2f(x) 0 f(x) 0.œ Ê œ



12. (a) As suggested, let E(x) E( x) E(x) E is anœ Ê � œ œ œ Êf(x) f( x) f( x) f( ( x)) f(x) f( x)� � � � � � � �# # #

even function. Define O(x) f(x) E(x) f(x) . Thenœ � œ � œf(x) f( x) f(x) f( x)� � � �# #

O( x) O(x) O is an odd function� œ œ œ � œ � Êf( x) f( ( x)) f( x) f(x) f(x) f( x)� � � � � � � �# # #Š ‹

f(x) E(x) O(x) is the sum of an even and an odd function.Ê œ �

(b) Part (a) shows that f(x) E(x) O(x) is the sum of an even and an odd function. If alsoœ �

f(x) E (x) O (x), where E is even and O is odd, then f(x) f(x) 0 E (x) O (x)œ � � œ œ �" " " " " "a b (E(x) O(x)). Thus, E(x) E (x) O (x) O(x) for all x in the domain of f (which is the same as the� � � œ �" "

domain of E E and O O ). Now (E E )( x) E( x) E ( x) E(x) E (x) (since E and E are� � � � œ � � � œ �" " " " " "

even) (E E )(x) E E is even. Likewise, (O O)( x) O ( x) O( x) O (x) ( O(x))œ � Ê � � � œ � � � œ � � �" " " " "

(since O and O are odd) (O (x) O(x)) (O O)(x) O O is odd. Therefore, E E and" " " " "œ � � œ � � Ê � �

O O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is," �

E E and O O, so the decomposition of f found in part (a) is unique." "œ œ

13. y ax bx c a x x c a x cœ � � œ � � � � œ � � �# # #Š ‹ ˆ ‰b b b b ba 4a 4a 2a 4a

# # #

#

(a) If a 0 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 a 0 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 a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the� �

graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the�

right. If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward� �

to the right. If b 0, 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 c 0.? �

14. (a) If a 0, the graph rises to the right of the vertical line x b and falls to the left. If a 0, the graph� œ � �

falls to the right of the line x b and rises to the left. If a 0, the graph reduces to the horizontalœ � œ

line y c. As a increases, the slope at any given point x x increases in magnitude and the graphœ œk k !

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.

15. Each of the triangles pictured has the same base b v t v(1 sec). Moreover, the height of eachœ œ?

triangle is the same value h. Thus (base)(height) bh" "# #œ

A A A . In conclusion, the object sweepsœ œ œ œ á" # $

out equal areas in each one second interval.

16. (a) Using the midpoint formula, the coordinates of P are . Thus the slopeˆ ‰ ˆ ‰a 0 b 0 a b� �# # # #ß œ ß

of OP .œ œ œ?

?

yx a/2 a

b/2 b


Chapter 1 Additional and Advanced Exercises 41

(b) The slope of AB . The line segments AB and OP are perpendicular when the productœ œ �b 0 b0 a a��

of their slopes is . Thus, b a a b (since both are positive). Therefore, AB�" œ � œ � œ Ê œˆ ‰ ˆ ‰b b ba a a

#

#

# #

is perpendicular to OP when a b.œ

17. From the figure we see that 0 and AB AD 1. From trigonometry we have the following: sin EB,Ÿ Ÿ œ œ œ œ) )1

2 ABEB

cos AE, tan CD, and tan . We can see that:) ) )œ œ œ œ œ œAE CD EB sinAB AD AE cos

)

)

area AEB area sector DB area ADC AE EB AD AD CD˜ � � ˜ Ê � �w

)" " "# # #a ba b a b a ba b2

sin cos tan sin cosÊ � " � " Ê � �" " " " " "# # # # # #) ) ) ) ) ) )a b a ba b2 sin

cos)

)

18. f g x f g x a cx d b acx ad b and g f x g f x c ax b d acx cb da ba b a b a b a ba b a b a ba b a b‰ œ œ � � œ � � ‰ œ œ � � œ � �

Thus f g x g f x acx ad b acx bc d ad b bc d. Note that f d ad b anda ba b a ba b a b‰ œ ‰ Ê � � œ � � Ê � œ � œ �

g b cb d, thus f g x g f x if f d g b .a b a ba b a ba b a b a bœ � ‰ œ ‰ œ



NOTES:


Thomas Calculus 12th ed solution ch1

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Transcript of Thomas Calculus 12th ed solution ch1