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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
# # $Š ‹12. The coordinates of P are x x so the slope of the line joining P to the origin is m (x 0). Thus,ˆ ‰Èß œ œ �
ÈÈ
xx x
"
x, x , .ˆ ‰ ˆ ‰È œ " "m m#
13. 2x 4y 5 y x ; L x 0 y 0 x x x x x� œ Ê œ � � œ Ð � Ñ � Ð � Ñ œ � Ð� � Ñ œ � � �" " "# #
5 5 5 254 4 4 4 16
2 2 2 2 2 2È É É x xœ � � œ œÉ É5 5 25 20x 20x 25
4 4 16 16 42 20x 20x 252 2� � � �È
14. y x 3 y 3 x; L x 4 y 0 y 3 4 y y 1 yœ � Ê � œ œ Ð � Ñ � Ð � Ñ œ Ð � � Ñ � œ Ð � Ñ �È È È È2 2 2 2 2 2 2 2 2
43. Symmetric about the origin 44. No symmetry Dec: nowhere Dec: x! Ÿ � _
Inc: x Inc: nowhere�_ � � _
45. No symmetry 46. Symmetric about the y-axis Dec: x Dec: x! Ÿ � _ �_ � Ÿ !
Inc: nowhere Inc: x! � � _
47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even.
48. f x x and f x x f x . Thus the function is odd.a b a b a b a bˆ ‰œ œ � œ � œ œ � œ ��& " " "�&
�x xx& &&a b
49. Since f x x x f x . The function is even.a b a b a bœ � " œ � � " œ �# #
50. Since f x x x f x x x and f x x x f x x x the function is neither even norÒ œ � Ó Á Ò � œ � � Ó Ò œ � Ó Á Ò� œ � � Óa b a b a b a b a b a b# ## #
odd.
51. Since g x x x, g x x x x x g x . So the function is odd.a b a b a b a bœ � � œ � � œ � � œ �$ $ $
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
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_
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 bȜ � � �# #
(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 $"%ß )(!
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P.
1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. D : x , D : x 1 D D : x 1. R : y , R : y 0, R : y 1, R : y 0f g f g fg f g f g fg�_ � � _ Ê œ �_ � � _ � �
2. D : x 1 0 x 1, D : x 1 0 x 1. Therefore D D : x 1.f g f g fg� Ê � � Ê œ �
R R : y 0, R : y 2, R : y 0f g f g fgœ �È
3. D : x , D : x , D : x , D : x , R : y 2, R : y 1,f g f g g f f g�_ � � _ �_ � � _ �_ � � _ �_ � � _ œ Î Î
R : 0 y 2, R : yf gÎ � Ÿ Ÿ � _g fÎ"#
4. D : x , D : x 0 , D : x 0, D : x 0; R : y 1, R : y 1, R : 0 y 1, R : 1 yf g f g g f f g f g�_ � � _ œ � Ÿ Ÿ � _Î Î Î g fÎ
Section 1.2 Combining Functions; Shifting and Scaling Graphs 15
67. Let y x f x and let g x x ,œ � # � " œ œÈ a b a b "Î#
h x x , i x x , anda b a bˆ ‰ ˆ ‰Èœ � œ # �" "# #
"Î# "Î#
j x x f . The graph ofa b a b’ “È ˆ ‰œ � # � œ B"#
"Î#
h x is the graph of g x shifted left unit; thea b a b "#
graph of i x is the graph of h x stretcheda b a b vertically by a factor of ; and the graph ofÈ#
j x f x is the graph of i x reflected acrossa b a b a bœ
the x-axis.
68. Let y f x Let g x x ,œ " � œ Þ œ �È a b a b a bx#
"Î#
h x x , and i x xa b a b a b a bœ � � # œ � � #"Î# "Î#"
#È f x The graph of g x is theœ " � œ ÞÈ a b a bx
#
graph of y x reflected across the x-axis.œ È The graph of h x is the graph of g x shifteda b a b right two units. And the graph of i x is thea b graph of h x compressed vertically by a factora b of .È#
69. y f x x . Shift f x one unit right followed by aœ œa b a b$
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
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# # # #
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 �
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.
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
26. (a) The function is defined for all values of x, so the domain is .a b�_ß _
(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 _
28. (a) The function is defined for all values of x, so the domain 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
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" "
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
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