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 3 œ
_ß Ñ  Ð _ œ  Ê
 ! Ê  !  4 4 3 t 3 t

3 t y can be any nonzero real number range 0 ). Ê
 ! Ê  ! Ê Ê œ Ð
_ß Ñ  Ð!ß _ 4 3 t
6. domain ( 4, 4 4, ); y in range and y , now if t t 16 , or if œ
_ß
%Ñ  Ð
Ñ  Ð _ œ 
% Ê
 ! Ê  ! 2 2 t 16 t 16 2 2 2

t 4 16 t 16 , or if t t 16 y can be any
%   Ê
Ÿ
 ! Ê
Ÿ  !  % Ê
 ! Ê  ! Ê 2 2 2 2 t 16 t 16 # "'

2 2 nonzero real number range ). Ê œ Ð
_ß
Ó  Ð!ß _ 1 8 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 3 4 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 d œ Ê  œ Ê œ œ Ê œ # # # # # " # d 2 È 11. Let D diagonal length of a face of the cube and the length of an edge. Then D d and œ 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 œ d 6d d d 3 3 3 3 3 È 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 ‰ È ß œ œ  È È x x 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 25 4 4 4 4 16 2 2 2 2 2 2 È É É x x œ
 œ œ É É 5 5 25 20x 20x 25 4 4 16 16 4 2 20x 20x 25 2 2

 È 14. y x 3 y 3 x; L x 4 y 0 y 3 4 y y 1 y œ
Ê  œ œ Ð
Ñ  Ð
Ñ œ Ð 
Ñ  œ Ð
Ñ  È È È È 2 2 2 2 2 2 2 2 2 y 2y 1 y y y 1 œ
  œ
 È È 4 2 2 4 2 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 or x y y x k k Ú Þ Ú Þ Û ß Û ß Ü 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 x 4 x , x 1 x 2x, x 1 , x 0 x, 0 x a b a b œ œ œ œ
Ÿ    Ÿ 2 2 x " 29. (a) Line through and : y x; Line through and : y x 2 a b a b a b a b !ß ! "ß " œ "ß " #ß ! œ
 f(x) x, 0 x 1 x 2, 1 x 2 œ Ÿ Ÿ
  Ÿ œ (b) f(x) 2, x x 2 x x œ ! Ÿ  " !ß " Ÿ  # ß # Ÿ  $ !ß $ Ÿ Ÿ % Ú Ý Ý Û Ý Ý Ü 30. (a) Line through 2 and : y x 2 a b a b !ß #ß ! œ
 Line through 2 and : m , so y x 2 x a b a b a b ß " &ß ! œ œ œ
œ

 " œ
 !
"
" " " " & &
# $ $ $ $ $ f(x) x , 0 x x , 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 x a b a b !ß $ #ß
" œ œ œ
# œ
#  $
"
$
% #
! # f(x) x , x x , x œ
$
$
"  Ÿ !
#  $ !  Ÿ # œ 4 Chapter 1 Functions 31. (a) Line through and : y x a b a b
"ß " !ß ! œ
Line through and : y a b a b !ß " "ß " œ " Line through and : m , so y x x a b a b a b "ß " $ß ! œ œ œ
œ

"  " œ
 !
"
" " " " $ $
" # # # # # f(x) x x x x x œ

" Ÿ  ! " !  Ÿ "
 "   $ Ú Û Ü " $ # # (b) Line through 2 1 and 0 0 : y x a b a b
ß
ß œ 1 2 Line through 0 2 and 1 0 : y 2x 2 a b a b ß ß œ
 Line through 1 1 and 3 1 : y 1 a b a b ß
ß
œ
f x x 2 x 0 2x 2 0 x 1 1 1 x 3 a b Ú Û Ü œ
Ÿ Ÿ
  Ÿ
 Ÿ 1 2 32. (a) Line through and T : m , so y x 0 x ˆ ‰ ˆ ‰ a b T T T T T T T #
Î# # "
! # # # ß ! ß " œ œ œ
 œ
" a b f x , 0 x x , x T a b
œ ! Ÿ Ÿ
"  Ÿ T T T # # # (b) f x A, x A x T A T x A x T a b Ú Ý Ý Ý Û Ý Ý Ý Ü 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. By Ÿ 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. Section 1.1 Functions and Their Graphs 5 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 !   _ 6 Chapter 1 Functions 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 b ˆ ‰ œ œ
œ
œ œ
œ

& " " "
&
x x x & & & a b 49. Since f x x x f x . The function is even. a b a b a b œ  " 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 nor Ò œ  Ó Á Ò
œ

Ó Ò œ  Ó Á Ò
œ

Ó a 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 b œ 
œ

œ
 œ
$ $ $ 52. g x x x x x g x thus the function is even. a b a b a b a b œ  $
" œ
 $

" œ
ß % # % # 53. g x g x . Thus the function is even. a b a b œ œ œ
" "
"

" x x # # a b 54. g x ; g x g x . So the function is odd. a b a b a b œ
œ
œ
x x x 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 b œ
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 b Section 1.1 Functions and Their Graphs 7 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 nor a b a b a b a b a b a b a b œ  "
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 b œ l l  "
œ l
l  " œ l l  " œ
59. s kt 25 k 75 k s t; 60 t t 180 œ 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 joules œ Ê œ Ê œ Ê œ œ œ # # # a b a b 2 61. r 6 k 24 r ; 10 s œ Ê œ Ê œ Ê œ œ Ê œ k k 24 24 12 s 4 s s 5 62. P 14.7 k 14700 P ; 23.4 v 628.2 in œ Ê œ Ê œ Ê œ œ Ê œ ¸ k k 14700 14700 24500 v 1000 v v 39 3 63. v f(x) x 2x 22 2x x 72x x; x 7 œ 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
" Ê Š ‹ 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 4 x #   Ê −
ß  %ß _ (b) 1 1 0 x 4 x 4 x x # #   Ê

 x 0: 1 0 0 0 

 Ê  Ê  x 4 x 2x 8 x 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 8 2 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 _ 8 Chapter 1 Functions 68. (a) From the graph, x ( 5) ( 1 1) 3 2 x 1 x 1
  Ê −
_ß

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

 3x 3 2x 2 x 5 which is true Ê  
Ê 
if x 1. Thus, x ( 1 1) solves the 
−
ß inequality. 1 x: 3x 3 2x 2 x 5 Case   Ê  
Ê 
3 2 x 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 sam a 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 25x œ  œ
Ê œ 
a b a ba b 71. x x h x ; cost 5 2x 10h C h 10 10h 5h 2 2 2 2 2 h 2 2 h 2 h 2 2  œ Ê œ œ œ  Ê œ  œ  È È È a b a b Š ‹ Š ‹ È 72. (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 land È a b # # 
cable at $100 per foot. The cost is C x 180 800 x 100 10,560 x . a b a b È œ  
# # (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 0 f 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 0 f g f g fg œ 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 : y f g Î  Ÿ Ÿ  _ g f Î " # 4. D : x , D : x 0 , D : x 0, D : x 0; R : y 1, R : y 1, R : 0 y 1, R : 1 y f g f g g f f g f g
_   _ œ  Ÿ Ÿ  _ Î Î Î g f Î 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 1 x 1 x x 7. f g h x f g h x f g 4 x f 3 4 x f 12 3x 12 3x 1 13 3x a ba b a b a b a b a b a b a b a b a b a 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 1 a ba b a b a b a b a b a b a b a b a b a b ‰ ‰ œ œ œ
œ
œ
 œ  2 2 2 2 2 9. f g h x f g h x f g f f a ba b a b É a b a b ˆ ‰ ˆ ‰ ˆ ‰ Š ‹ É ‰ ‰ œ œ œ œ œ  " œ 1 1 x x 5x x 1 4x 1 4x 1 4x 1 x  %     " 10. f g h x f g h x f g 2 x f f a ba b a b a b a b Š ‹ Š ‹ È   ˆ ‰ ‰ ‰ œ œ
œ œ œ œ Š ‹ È Š ‹ È 2 x 2 x 1 2 x 8 3x x 7 2x 2 3



$


2 2 2 x 2 x x x $ $ 11. (a) f g x (b) j g x (c) g g x a ba b a ba b a ba b ‰ ‰ ‰ (d) j j x (e) g h f x (f) h j f x a ba b a ba b a ba b ‰ ‰ ‰ ‰ ‰ 12. (a) f j x (b) g h x (c) h h x a ba b a ba b a ba b ‰ ‰ ‰ (d) f f x (e) j g f x (f) g f h x a 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) x x x x x 1 x 1 1 x (x 1)




x x 1 x x 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
"l x (b) f g x so g x x . a ba b a b ‰ œ œ Ê "
œ Ê "
œ Ê œ ß œ  " g x g x x g x x x g x x g x x x x a b a b a b a b a b
"  "  "  "  " " " " " (c) Since f g x g x x , g x x . a ba b a b a b È ‰ 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 b ˆ ‰ 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 È È ‰ l l  " l l l l " "
" l
"l
"  " # # x x x x x x 15. (a) f g 1 f 1 1 (b) g f 0 g 2 2 (c) f f 1 f 0 2 a b a b a b a b a b a b a 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 0 a b a b a b a b a b a b a b a b a b œ œ
œ œ
œ
œ 16. (a) f g 0 f 1 2 1 3, where g 0 0 1 1 a b a b a b a b a b œ
œ

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

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

œ 10 Chapter 1 Functions (d) f f 2 f 0 2 0 2, where f 2 2 2 0 a b a b a b a b œ œ
œ œ
œ (e) g f 0 g 2 2 1 1, where f 0 2 0 2 a b a b a b a b œ œ
œ œ
œ (f) f g f 2 , where g 1 ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ " " " " " " # # # # # # # œ
œ

œ œ
œ
5 17. (a) f g x f g x 1 a ba b a b a b É É ‰ œ œ  œ 1 1 x x x  g f x g f x a ba b a b a b ‰ œ œ 1 x 1 È  (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 x a ba b a b a b È ‰ œ œ
 g f x g f x 1 x a ba b a b k k a 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 : , 1 a 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 2x a ba b a b a b a b a b a b a b ‰ œ Ê œ Ê œ Ê œ
œ †
g x g x 2 a b a b
g x x g x 2x g x Ê
† œ
Ê œ
œ a b a b a b 2x 2x 1 x x 1

20. f g x x 2 f g x x 2 2 g x 4 x 2 g x g x a ba b a b a b a b a b a b a b a b É ‰ œ  Ê œ  Ê
œ  Ê œ Ê œ 3 3 x 6 x 6 2 2   3 21. (a) y (x 7) (b) y (x 4) œ
 œ

# # 22. (a) y x 3 (b) y x 5 œ  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. Section 1.2 Combining Functions; Shifting and Scaling Graphs 11 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 12 Chapter 1 Functions 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. Section 1.2 Combining Functions; Shifting and Scaling Graphs 13 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 " 14 Chapter 1 Functions 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 1 œ
œ
œ %
# # # a b 59. y 60. y 1 1 œ "  œ  œ  œ  " " " " " * # # # Î$ ˆ ‰ x x x x # # # # a b 61. y x 1 62. y 3 x 1 œ %  œ  È È 63. y 16 x 64. y x œ %
œ
œ %
É ˆ ‰ È È x # # $ # " " # # 65. y 3x 27x 66. y œ "
œ "
œ "
œ "
a b ˆ ‰ $ $ # ) $ x x $ Section 1.2 Combining Functions; Shifting and Scaling Graphs 15 67. Let y x f x and let g x x , œ
#  " œ œ È a b a b "Î# h x x , i x x , and a b a b ˆ ‰ ˆ ‰ È œ  œ #  " " # # "Î# "Î# j x x f . The graph of a b a b ’ “ È ˆ ‰ œ
#  œ B " # "Î# h x is the graph of g x shifted left unit; the a b a b " # graph of i x is the graph of h x stretched a b a b vertically by a factor of ; and the graph of È # j x f x is the graph of i x reflected across a b a b a b œ the x-axis. 68. Let y f x Let g x x , œ "
œ Þ œ
È a b a b a b x # "Î# h x x , and i x x a b a b a b a b œ
 # œ
 # "Î# "Î# " # È f x The graph of g x is the œ "
œ Þ È a b a b x # graph of y x reflected across the x-axis. œ È The graph of h x is the graph of g x shifted a b a b right two units. And the graph of i x is the a b graph of h x compressed vertically by a factor a b of . È # 69. y f x x . Shift f x one unit right followed by a œ œ a b a b $ shift two units up to get g x x . a b a b œ
"  # 3 70. y x f x . œ "
B  # œ
Ò
" 
# Ó œ a 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 b œ œ
" œ
" 
# $ $ $ and j x x . The graph of h x is the a b a b a b a b œ
Ò
" 
# Ó $ graph of g x shifted right one unit; the graph of i x is a b a b the graph of h x shifted down two units; and the graph a 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 factor a b œ " x of 2 to get g x . Then shift g x vertically down 1 a b a b œ " # x unit to get h x . a b œ
" " # x