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Chapter 7: Algebra for College Mathematics Courses Lesson 7.1.1 7-1. a. See graph at right.
Increasing: x > 2 ; Decreasing: x < 2 b. As the x-values get larger, the y-values get larger. Or, the slope
of the tangent line is positive. 7-2. a. First x-value is less than the 2nd and both are in the interval [a, b]. b. First y-value is less than the 2nd. c. See graph at right. d. Yes, f (x1) < f (x2 ) . 7-3. a. x1 = 2!!!!!x2 = 3
22 < 23
4 < 8
x1 = 1.7!!!!!x2 = 1.821.7 < 21.8
3.25 < 3.48
b. g(x) is an increasing function on the interval [a, b] if, for every two points x and x + h with a ! x < x + h ! b, h > 0 , g(x) < g(x + h) .
c. 2x+h = 2h !2x . Since h > 0, 2h > 1 ; therefore 2h !2x > 2x . 7-4. a. Something like, “as x gets larger, y gets smaller.” b. If x2 > x1 , then f (x2 ) < f (x1) . c. f (x) is a decreasing function on the interval [a, b] if, for every two points 1x and 2x with
a ! x1 < x2 ! b , then f (x2 ) < f (x1) . d. Given h > 0 , 5 ! (x + h)2 = 5 ! h(2x + h) .
If !" < x < x + h < 0 , then 2x + h < 0 ; therefore !h(2x + h) > 0 and 5 ! (x + h)2 > 5 ! x2 . If 0 < x < x + h < ! , then 2x + h > 0 ; therefore !h(2x + h) < 0 and 5 ! (x + h)2 < 5 ! x2 .
7-6. A graph is concave down over an interval [a, b] if a line segment joining any two points on
the graph over that interval lies completely below the graph. 7-7. See graph at right. Increasing on (!", !1) and (1,!) , decreasing on (!1, 0) and
(0,1) ; concave up on (0,!) , concave down on (!", 0) .
7-8. Any line or odd function that passes through the origin, for example:
y = 5x,!y = x5,!y = sin x . Review and Preview 7.1.1 7-9. a. Increasing: (–∞, –3) ∪ (1, 5), Decreasing: (–3, 1) ∪ (5, ∞), Concave Up: (–1, 3),
Concave Down: (–∞, –1) ∪ (3, ∞); b. Decreasing: (–∞, ∞), Concave Up (–∞, 0), Concave down (0, ∞) 7-10. See sample graph at right. 7-11. Brittany; segments connecting any two points on the graph
are above the graph. 7-12. a. b(x) = a(x ! 2) + 5 = (x ! 2)3 ! 3(x ! 2) + 5
d = r ! t m2 +1 Lesson 7.1.2 7-17. a. b. Changing of sign of x does not affect f(x). c. y = (a)2 = a2
y = (!a)2 = a2 They are equal.
d. f (a) = f (!a) e. Symmetric about the y-axis. 7-18. a. These functions have even power exponents. b. f (!x) = f (x) c. They are symmetric about the y-axis. d. y = cos x or y = x are good choices.
7-19. a. b. Changing the sign of x changes only the sign of f (x) .
c. y = (a)3 = a3
y = (!a)3 = !a3
d. f (!a) = ! f (a) e. Symmetric around the origin. 7-20. a. These functions have odd exponents. b. f (!x) = ! f (x) c. They are symmetric about the origin. d. y = sin x is a good choice. 7-21. a. (–2, 5) b. (3, –5) c. unknown 7-23. a. Even g(!x) = (!x)2 + cos2(!x)
g(!x) = x2 + cos2 x
b. Neither f (!x) = (!x)2 + 3(!x)3
f (!x) = x2 ! 3x3
c. Odd h(!x) = (!x)!1 + 2 sin(!x)h(!x) = !x!1 ! 2 sin x
Review and Preview 7.1.2 7-24. a. cos(!x) = cos(x) and sin(!x) = ! sin x b. Sine is odd, cosine is even. 7-25. a. Any parabola with a vertex on the y-axis. Example: f (x) = x2 ! 3 b. Impossible c. Any parabola with a vertex not on the y-axis. Example: f (x) = 2x2 ! 8x + 5 7-26. Tangent is odd. tan(!x) = sin(!x)
cos(!x) =! sin xcos x = ! sin x
cos x = ! tan x 7-27. Increasing: (–2, 4); Decreasing: (–∞, –2) ∪ (4, ∞);
7-28. a. 2.9 – 1.7 = 1.2 seconds b. 46 – 20 = 26 c. y = 13 cos 2!
1.2 (x "1.7)( ) + 33 (other answers are possible) d. Because the period of the graph is less than the horizontal shift, one solution is x = 2.0445 !1.2 = 0.8445 . The graph will have another maximum at x = 1.7 !1.2 = 0.5 . Due to the symmetry of the graph, the second solution is x = 0.5 ! (0.8445 ! 0.5) = 0.1555 . 7-29.
Review and Preview 7.2.1 7-36. Graph or average the x-intercepts (x = 0 and x = 60) to find that the product will be a maximum when x = 30. 30 + y = 60!!!!!y = 30
7-61. a. b. c. The graph in Y1 follows the graph in Y2 except that it has an asymptote at x = 2. d. The quotient tells you about the general or global shape of the graph.
7-71. a. Decrease by 1 each time. b. Increase by 1 each time. c. Each time the sum is the same as the exponent of expansion. 7-73. (x + y)0 = 1 It goes in “Row 0.” 7-74. a. Row 9 b. x9 + 9x8y c. x6 + 6x5y +15x4y2 + 20x3y3 +15x2y4 + 6xy5 + y6
Lesson 7.3.1 7-86. a. Each pair equals 101. b. 50 pairs c. 101 x 50 = 5050 7-87. a. 12 x 4 = 48 b. It is twice as large as A. c. 1000 by 999 7-88. a. 1100 by 899 b. 1100!899
2 = 494, 450 7-89. n = number of terms, a
1= first term of the sequence, and an = nth term of the sequence.
7-90. a. 10.2 !10 = 0.2 b. 49 times c. 49 !0.2 = 9.8
Lesson 7.3.2 7-100. a. 1+ 3+ 9 + 27 + 81+ 243+ 729 = 1093 b. 3 ! 729 = 2187 , 3 is the multiplier. c. 3S = 3 + 9 + 27 + 81 + 243 + 729 + 2187 d. 2S = 2186 S = 1 + 3 + 9 + 27 + 81 + 243 + 729 S = 1093 3S – S = 2S = 3279 – 1093 = 2186 This is twice as much as what was found in part (a). 7-101. a. S = 1+ 5 + 25 +125 + 625 + 3125 +15625
Review and Preview 7.3.2 7-107. S = 1! 3+ 9 ! 27 + 81! 243+ 729 ! 2187 = !1640 . The method still works when r < 0. 7-108. a. 4(5+x)
2 = 2004(5 + x) = 4005 + x = 100
x = 9595!53 = 30
Series = 5 + 35 + 65 + 95
b. 200 = 5 + 5r + 5r2 + 5r3
195 = 5r3 + 5r2 + 5r39 = r3 + r2 + rr = 3
Series = 5 +15 + 45 +135
7-109.
a. S = 3(211!1)2!1 = 3(2047) = 6141
b. 20(800+1560)2 = 23, 600
c. S = 0.02(311!34 )3!1 = 0.02(177,066)
2 = 3541.322 = 1770.66
d. 12 +
22 +
32 +
42…+ 20
2 =20 1
2+10!
"#$%&
2 =20 21
2( )2
= 10(21)2
= 105
7-110.
S = 100(1.00512 !1)1.005!1 = 100(0.0617)
0.005 = 1233.56 7-111.
2x ! 3 4x4 ! 2x3 + 0x2 ! 7x ! 54x4 ! 6x3
!!4x3 ! 0x2
!!!! 4x3 ! 6x2
!!!6x2 ! 7x!!!6x2 ! 9x
!!!!!2x ! 5 !!!!!!!2x ! 3 !!!! ! 2
2x3 + 2x2 + 3x +1! 22x!3
7-112. 500 = 4x + 2y!!!!!y = 250 " 2x A = xy = x(250 ! 2x) This is a maximum when x = 62.5 ft. Therefore y = 125 ft and the maximum area is 7812.5 ft2.
7-120. a. sin2 u ! sin u + 0.24 = 0 b. Let v = sin u . c. v2 ! v + 0.24 = 0 d. v = sin(3x ! 5) 7-121. (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 Let x = a and y = bc.
7-127. If the y-axis is a line of symmetry then there is not a horizontal shift. The line y = 15
touches either the top or the bottom of the graph. Since the point (20, 50) is on the graph, the line y = 15 must touch the bottom. If (20, 50) is the next point of symmetry and in the middle, then the period is 80. Therefore the amplitude is 50 !15 = 35 and 80b = 2! !or !b = !
40 . Hence a possible equation is y = 35 cos ! x40( ) + 50 .
7-128. a. The zeros are at x = 1, 3, and 5. x < 1 or 3 < x < 5 The intervals to check are (!",1),!(1, 3), (3, 5),!and!(5,") . Choose a point in each interval and check to see if it makes the inequality true. (!",1)!!choose!x = 0!!#!!(1! 0)(0 ! 3)(0 ! 5) = 15 > 0!!true
Therefore the solution set is x < 1!!or !!3 < x < 5 . b. x2 ! 2x !15 < 0!!"!!(x ! 5)(x + 3) < 0 The zeros are at x = –3 and 5. The intervals to check are (!", !3), (!3, 5),!and!(5,") . Choose a point in each interval and check to see if it makes the inequality true. (!", !3)!!choose!x = !4 !!#!!(!4 ! 5)(!4 + 3) = 9 /< 0!!false
Chapter 7 Closure 7-130. a. The function must be cosine because it is even. If the increasing regions repeat every 4
units, then the period is 8 units. Since amplitude = 10 and 8b = 2! !or !b = !4 , a possible
equation is y = 10 cos !4 x( ) .
b. y = 12 x
odd!# c. The given information indicates that there is a vertical asymptote at x = 2 and a horizontal
asymptote at y = –1. Therefore this is a rational function. A possible equation is y = 1
x!2 !1 . d. A function that has only one horizontal asymptote is an exponential function. Since it is
concave down it is reflected over the x-axis. Since the horizontal asymptote is y = –4, the function has been shifted down 4 units. Thus a possible equation is y = !(2)x ! 4 .
e. Since the asymptotes are at x = –2 and x = 2, a possible equation is y = 1(x+2)(x!2) =
1x2 !4
.
f. This will be an odd power function centered at x = 4. A possible equation is y = !(x ! 4)3 . CL 7-131. See graph below right. a. !2 < x < 2 b. x < !2 and x > 2 c. x < 0 d. x > 0 CL 7-132. SA = 2! r2 + 2! rh
V = ! r2h
h = V! r2
SA = 2! r2 + 2! r V! r2( )
SA = 2! r2 + 2Vr
CL 7-133. 15, 050 = 100(2+a2 )
230,100 = 100(2 + a2 )301 = 2 + a2299 = a2
CL 7-134. 100 + 90 + 80 + ...! 20 !18 !16 ! ...
10(100+10)2 ! 10(20+2)2 = 1100
2 ! 2202 = 550 !110 = 440
Separate the positive terms from the negative. Each forms an arithmetic sequence.
The graph looks like the line 3 4x + globally, but has an asymptote at 2x = . CL 7-136. (3+ 2x !1)2 + 24 = 10(3+ 2x !1) Let u = 3+ 2x !1 . u2 + 24 = 10u
u2 !10u + 24 = 0(u ! 6)(u ! 4) = 0
u = 63+ 2x !1 = 6
2x !1 = 32x !1 = 92x = 10x = 5
u = 43+ 2x !1 = 4
2x !1 = 12x !1 = 12x = 2x = 1
u = 6!or !u = 4 Both answers check. CL 7-137. The signs are alternating so this must be subtraction. By looking at the pattern on the
exponents, the powers of x are decreasing by 1 and the power of y are increasing by 2. Therefore start with (ax ! by2 )? . The missing power must be 7 because x5(y2 )2 !!!!!5 + 2 = 7 . Now use the binomial formula to find a and b. 75
!"
#$ (ax)
5 (by2 )2 = 84x5y4
21a5x5b2y4 = 84x5y4a5b2 = 4
74
!"
#$ (ax)
4 (by2 )3 = %280x4y6
%35a4x4b3y6 = %280x4y6a4b3 = 8
Dividing the 2 new equations yields: ab =12 !!or !!2a = b .