College Algebra Final Exam Review Key, Fall 2012
1. Since an absolute value can’t be <0, there is no solution.
2. Either: 1) 5x+1 = x-6 or 2) 5x+1 = -(x-6). Thus the solutions are x = -7/4 or x = 5/6.
3. Either: 2 − 3 x ≥ 11 or 2 − 3 x ≤ −11 . This leads to x ≤ −3 or x ≥ 13 / 3 . In interval notation:
(−∞, −3] ∪ [13 / 3, ∞ ) .
4. First isolate the absolute value: 5 x − 4 < 8 , or −8 < 5 x − 4 < 8 . In interval notation:
(-4/5,12/5).
5. No, since 3 is associated with two different values.
6. No, since it does not satisfy the vertical line test.
7. Yes, since it does satisfy the vertical line test.
8. Domain: [-3,0], range: [-5,-1]
9. Domain: ( −∞, ∞ ) , range: {8}.
10. y 2 = 5 x does not define y as a single function of x, since each x (except 0) would lead to two different y
values.
11. y =
5
does define y as a function of x. Each x value would result in only one y value.
x −1
12. Domain and range of y = ( x + 4 ) − 4 . This is just a parabola (opening upward) with vertex at (-4,-4). The
2
domain is ( −∞, ∞) and the range is [ −4, ∞) .
13. Domain and range of y =
4 x − 5 . Domain: 4 x − 5 ≥ 0, or x ≥ 5 / 4, or [5 / 4, ∞) . The range is a little harder,
but note that y can’t be <0. So the range is [0, ∞ ) . Look at a graph.
−7
. Domain: any x except 6 will work. So the domain is ( −∞, 6) ∪ (6, ∞ ) . To
x−6
see the range, look at the graph: y ≠ 0 . Thus the range is ( −∞, 0) ∪ (0, ∞ ) .
14. Domain and range of y =
15. If f ( x) = 2 x 2 + 3 x + 5, then f ( −3) = 2( −3) 2 + 3( −3) + 5 = 14
16. If f ( x) = 3 x 2 − 5 x + 4, then f ( − x) = 3( − x) 2 − 5( − x) + 4 = 3 x 2 + 5 x + 4
17. a) f(-3) = 0; b) When f(x) = -2, the x values are 3 and approximately -2.5.
18. To get the average rate of change, we just need the slope of the line:
50 − 35
= 0.25 dollars per minute
60 − 0
=25 cents per minute
LSC-Tomball Math Dept.
Page 1
College Algebra Final Exam Review Key, Fall 2012
1900 − 600
19. Rate of change =
= 32.5 ft/min
50 − 10
20. f(6)
6) = 6(6) = 36. The middle ‘piece’ of the function is used since 6 is in the middle interval.
21. The slope is m = (0-2)/(4-0) = -1/2,
1/2, xx-intercept is (3,0) and y-intercept is (0,2)
22. This horizontal line is y = -3
23. A line parallel to x = 6 must also be
e a vertical line, thus its equation is x = 7.
24. A line perpendicular to -5x
5x + y = 5 must have slope -1/5. Thus its equation: y – 5 = -1/5(x-5),
or in standard
form: x + 5y = 26 (after clearing fractions)
25.
axis, and 2) shifting left 5 units.
26. f(x) = -(x + 5)2 is just y = x2 after 1) reflecting in the xx-axis,
27. g ( x ) = −3 f ( x + 2) − 3, or g ( x ) = −3 x + 2 − 3
28. The original graph needs to be reflected in the xx-axis:
29. x-axis symmetry: no; y-axis
axis symmetry: no; origin symmetry: yes
30. f(-x) = (-x)4 -4(-x)2 + 1 = x4 – 4x2 +1, so f is even.
31. f(-x)
x) does not change every sign, so f is not odd. (and not even because some signs are changed)
32. y =
x has been shifted left 7 units and up 6 units, so the new equation is y = x + 7 + 6 .
33. y=x3 is the basic graph. It has been shrunk vertically by a factor of 1/3, moved left 5 units and up 4.
34. ( fg )( x) = f ( x) g ( x) = (4 x + 8)(9 x + 7) = 36 x 2 + 100 x + 56
LSC-Tomball Math Dept.
Page 2
College Algebra Final Exam Review Key, Fall 2012
4
The only restricted value is x = 2, so the domain of f + g is
35. ( f + g )( x ) = f ( x) + g ( x ) = 2 x + 5 +
x−2
( −∞, 2) ∪ (2, ∞) .
 10 − 1 
 = g (3) = 2(3) + 3 = 9 .
 3 
36. ( g f )(10) = g[ f (10] = g 
37. ( f g )( x) = f [ g ( x)] = f (8 x − 8) = 8 x − 8 + 4 = 4(2 x − 1) = 2 2 x − 1 .
38. The function is increasing from ( −∞, 2] and decreasing from [2, ∞ ) .
39. The vertex of y = x 2 − 20 x + 105 is (10,5). Use −
b
or complete the square to find it.
2a
40. The minimum value of c(x) will be at its vertex: x = −
41. t =
−368
= 46 watches.
2(4)
39
seconds
4
42. Letting x=the width and 100-2x the length, we have to solve the quadratic equation:
x (100 − 2 x ) = 1250 . The solution is width = 25 ft. and length = 50 ft.
43. Yes, -2 is a zero (the synthetic division remainder = 0).
44. No, 2+i is NOT a zero (the synthetic division remainder is 8+21i ).
45. f(-2) = 80
46. The other two zeros are 3 + 2i and 3 – 2i .
() (
)(
)(
47. f x = x+6 x−6 x −2
)
48. ± 16 , ± 13 , ± 12 , ± 1, ± 23 , ± 3
49. {-1, 2, 7}
50.
51.
52.
f ( x ) = ( x +1)( x − 2 )( x − 7 )
f (x)= x 3 − 7x 2 −14x + 48
f (x)= x 3 − 7x 2 + 17x − 15
f (x)=
1 3
(x − 4x 2 − 3x + 18)
6
LSC-Tomball Math Dept.
Page 3
College Algebra Final Exam Review Key, Fall 2012
53.
54.
f (x)= − 8x 4 +16x 2 − 8
55. Horizontal: y = 0; Vertical: none
56. Horizontal: y = 1; Vertical: x = -3, x = 3
57.
58. f(x)
=
௫ሺ௫ି7ሻ
ሺ௫ି4ሻమ
59. f −1 (x) = 3
60. f −1 (x)=
x+7
3
8x + 3
x
61. Yes, they are inverses. Their graphs are symmetric about the line y = x.
 tu 4 
62. log a 
 s 
x+3
63. y = 2( ) − 4
64. x = 3
65. Amount is approximately $1801.05.
LSC-Tomball Math Dept.
Page 4
College Algebra Final Exam Review Key, Fall 2012
66. -2
67. log 8 32 =
5
3
68. x = -2
69.
Domain (1, ∞ ) ; Range
70. log 5 ( 0.497) =
( −∞, ∞ )
ln(0.497)
≈ − 0.4344
ln(5)
71. x is approximately 0.920
72. x = 16/5
73. x = 14
74. P(10) is approximately 2326.90 people
75. t is approximately 3.71 years
76. t ≈ 94.06 years
 1 −1 2 −9 


77.  2 0 1 −3  augmented matrix Solutions are (0, 3, -3)
 1 4 1 9 
 4 0 2 50 


78.  0 9 −2 63  augmented matrix
 2 2 −2 16 
 6 −1 4 25 


79.  9 8 −9 108  augmented matrix Solutions are (5, 9, 1)
0 
 7 −4 1

80. Vertical: x = 7; horizontal: none; oblique (or slant): y = x + 9
LSC-Tomball Math Dept.
Page 5