• Tidak ada hasil yang ditemukan

(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.

(b) line slope

AB 10281 œ 96 œ 3# BC 210( 4)6 œ œ46 23 CD 6% ( 3)2 œ 96 œ 3# DA 18( 3)2 œ œ64 32

CE % 6614 0

3 œ

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 143 gives yœ 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 53, 72 and 65, respectively. The slopes of AB, BC and AC are , 1 and , respectively.

È È È 7

# 8"

12. P(x 3xß 1) is a point on the line yœ3x1. If the distance from P to (!ß0) equals the distance from P to ($ß %), then x#(3x1)#œ(x3)# (3 3x) # Ê x#9x#6x œ1 x#6x 9 9 18x9x#

18x 17 or x y 3x 1 3 1 . Thus the point is P .

Ê œ œ 1718 Ê œ œ ˆ ‰1718 œ 236 ˆ1718ß236

13. yœ $ " ' Ê œ $ *ax b a b y x 14. yœ "#ax " # Ê œ b y "#x$# 15. xœ !

16. mœ" $# 'a b œ)% œ # Ê œ # $ ' Ê œ #y ax b y x

17. yœ #

18. mœ# $& $ œ&# œ Ê œ #& y #&ax $ $ Ê œ b y #&x#"&

19. yœ $ $x

20. Since x# œ #y is equivalent to yœ # #x , the slope of the given line (and hence the slope of the desired line) is 2.

yœ # " " Ê œ # &ax b y x

21. Since x% $ œ "#y is equivalent to yœ %$x %, the slope of the given line (and hence the slope of the desired line) is %$. yœ %$ax4b " Ê œ 2 y %$x#!$

22. Since x$ & œ "y is equivalent to yœ $&x"&, the slope of the given line is and the slope of the perpendicular line is$&

5$. yœ 5$ax # $ Ê œ b y 5$x"*$

23. Since x"# "$yœ " is equivalent to yœ $#x $, the slope of the given line is $# and the slope of the perpendicular line is . y#$ œ #$ax " # Ê œb y #$x)$

24. The line passes through a!ß &b and a$ß !b. mœ! &$ !a b œ Ê œ$& y $&x &

25. The area is Aœ1r and the circumference is C# œ #1r. Thus, rœ #C1 Ê œA 1ˆ ‰#C1 # œ%C1#.

26. The surface area is Sœ %1r#Ê œr ˆ ‰%S1 "Î#. The volume is Vœ$%1 r$Ê œr É$ $%V1. Substitution into the formula for surface area gives Sœ %1r#œ %1ˆ ‰$%V1 #Î$.

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 )œxx# œx. Thus the point has coordinates x xa ß #bœatan )ßtan#)b.

28. tan )œ riserun œ &!!h Ê œ &!!h tan ft.)

29. 30.

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

31. 32.

Neither Symmetric about the y-axis.

33. ya b œ x a bx # " œx# " œy x . Even.a b

34. ya b œ x a bx & a bx $ œ a bx x& x$ œ x y x . Odd.a b 35. ya b œ " x cosa b œ " x cos xœy x . Even.a b

36. ya b œx seca bx tana b œx cossin#a ba bxx œ cos xsin x# œ sec x tan xœ y x . Odd.a b

37. ya b œx a ba bx$x# %"a bx œ #xx"x œ xx#"x œ y x . Odd.a b

% %

$ $

38. ya b œ " x sina b œ " x sin x. Neither even nor odd.

39. ya b œ x x cosa b œ x x cos x. Neither even nor odd.

40. ya b œx Éa bx % " œÈx% " œy x . Even.a b

41. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since x attains all nonnegative values, the range is l l Ò#ß _Ñ. 42. (a) Since the square root requires " x#  !, the domain is Ð_ß "Ó.

(b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ. 43. (a) Since the square root requires "' x#   !, the domain is Ò%ß %Ó.

(b) For values of x in the domain, ! Ÿ "' x#Ÿ "', so ! Ÿ "' È x#Ÿ %. The range is Ò!ß %Ó. 44. (a) The function is defined for all values of x, so the domain is a_ß _b.

(b) Since $#x attains all positive values, the range is a"ß _b.

45. (a) The function is defined for all values of x, so the domain is a_ß _b. (b) Since e# x attains all positive values, the range is a$ß _b.

46. (a) The function is equivalent to yœtan x, so we require x# # Á k#1 for odd integers k. The domain is given by xÁ k%1for odd integers k.

(b) Since the tangent function attains all values, the range is a_ß _b. 47. (a) The function is defined for all values of x, so the domain is a_ß _b.

(b) The sine function attains values from " " to , so # Ÿ #sin xa$ 1bŸ # and hence $ Ÿ #sin xa$ 1b " Ÿ ". The range is Ò ß Ó3 1 .

48. (a) The function is defined for all values of x, so the domain is a_ß _b.

(b) The function is equivalent to yœÈ&x , which attains all nonnegative values. The range is # Ò!ß _Ñ. 49. (a) The logarithm requires x $ !, so the domain is a$ß _b.

(b) The logarithm attains all real values, so the range is a_ß _b. 50. (a) The function is defined for all values of x, so the domain is a_ß _b.

(b) The cube root attains all real values, so the range is a_ß _b. 51. (a) The function is defined for % Ÿ Ÿ %x , so the domain is Ò%ß %Ó.

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

52. (a) The function is defined for # Ÿ Ÿ #x , so the domain is Ò#ß #Ó. (b) The range is Ò"ß "Ó.

53. First piece: Line through a!ß "b and a"ß !b. mœ! "" ! œ "" œ " Ê œ " œ " y x x

Second piece: Line through a"ß "b and a#ß !b. mœ ! "# " œ "" œ " Ê œ " " œ # œ # y ax b x x

f x x, x

x, x

a bœ " œ ! Ÿ "

# " Ÿ Ÿ #

54. First piece: Line through a!ß !b and 2 5 . ma ß b œ52 ! ! œ Ê œ52 y 52x

Second piece: Line through 2 5 and 4a ß b a ß !b. mœ ! 452 œ 25 œ Ê œ 52 y 52ax2b œ 5 52x10œ105x2 f x x, x 2 (Note: x 2 can be included on either piece.)

10 , 2 x 4

a bœ ! Ÿ œ

Ÿ Ÿ

5 2 5x

2

55. (a) f ga‰ " œba b f ga a" œbb fŠÈ" #" ‹œ " œ œ "fa b ""

(b) ag f‰ # œba b g faa b# œb gˆ ‰" œ " œ " or É#

# #Þ& &

2 É" È

#

(c) af f x‰ ba bœf f xaa bbœfˆ ‰"x œ"x œx, xÁ ! (d) ag g x‰ ba bœg g xa a bbœgŠ " #‹œ " œ

#

#

" # #

È É É

È È x

x

" x

#

% Èx

56. (a) f ga‰ " œba b f ga a" œbb fˆÈ$ " " œ ! œ # ! œ #‰ fa b (b) ag f‰ # œba b f ga a b# œ # # œ ! œ ! " œ "b ga b ga b È$ (c) af f x‰ ba bœf f xaa bbœ # fa xbœ # # a xbœx (d) ag g x‰ ba bœg g xa a bbœgˆÈ$ x " œ‰ É$ È$ x " "

57. (a) f g xa‰ ba bœf g xa a bbœfˆÈx # œ # ‰ ˆÈx # œ ‰# x, x  #.

g f x f g x g x x x

a ‰ ba bœ a a bbœ # a #bœÈa# #b # œ % È # (b) Domain of f g: ‰ Ò#ß _ÑÞ

Domain of g f: ‰ Ò#ß #ÓÞ (b) Range of f g: ‰ Ð_ß #ÓÞ

Range of g f: ‰ Ò!ß #ÓÞ

58. (a) f g xa‰ ba bœf g xa a bbœfŠÈ" x‹œÉÈ" œ " x È% x.

g f xa ‰ ba bœf g xa a bbœgˆÈx‰œÉ" Èx (b) Domain of f g: ‰ Ð_ß "ÓÞ

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

59. 60.

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.

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)"

across the x-axis.

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 13 œ œcb b# Ê bœ2 sin 13 œ2ŠÈ#3‹œÈ3. By the theorem of Pythagoras, a#b# œc # Ê œa Èc#b#œÈ4 œ3 1.

(b) sin Bœsin 13 œ œcb c2 Ê œ c sin 21 œ 2 œ 43. Thus, aœ c b œ 43 (2) œ 34 œ 23.

3 ŠÈ3 È È È

#

È # # ÊŠ ‹# # É

76. (a) sin Aœ ac Ê œ a c sin A (b) tan Aœ ab Ê œ a b tan A 77. (a) tan Bœ ba Ê œ a tan Bb (b) sin Aœ ac Ê œ c sin Aa

78. (a) sin Aœ ac (c) sin Aœ œac 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 hc, tan 35°œ hb, and b œc 10.

Thus, hœc tan 50° and hœb tan 35°œ (c 10) tan 35°

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 tan 50° tan 35°10 tan 35° Ê œ

16.98 œ 10 tan 35° tan 50° ¸ m.

tan 50° tan 35°

80. Let hœheight of balloon above ground. From the figure at the right, tan 40°œ ha, tan 70°œ hb, and a œb 2. Thus, hœb tan 70° Ê hœ(2a) tan 70° and hœa tan 40°

(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 tan 40° tan 70°2 tan 70° Ê œ 1.3 œ 2 tan 70° tan 40° ¸ km.

tan 40° tan 70°

81. (a)

(b) The period appears to be 4 .1

(c) f(x4 )1 œsin (x4 )1 cosˆx 4#1‰œsin (x2 )1 cosˆx#21‰œsin xcos x# since the period of sine and cosine is 2 . Thus, f(x) has period 4 .1 1

82. (a)

(b) Dœ _ß !ß _( 0) ( ); Rœ ß[ 1 1]

(c) f is not periodic. For suppose f has period p. Then fˆ#"1kp‰œfˆ ‰#"1 œsin 21œ0 for all integers k. Choose k so large that #"1kp1" Ê 0(1/2 ) kp1" 1. But then

#"1 kp‰œsinŠ(1/# 1") kp‹0 which is a contradiction. Thus f has no period, as claimed.