Summer Mathematics Prep
Chesterfield County Public Schools
Department of Mathematics
Entering Calculus
SOLUTIONS
Domain and Range
1. Domain: All Real Numbers Range: {y: y ≥ 12}
2. Domain: x : x  4 Range: y : y  0
3. Domain: {x: x ≠ 1} Range: {y: y ≠ 0}
4. Domain: All Real Numbers Range: {y: y > 0}
5. Domain: All Real Numbers Range: {y: y ≥ 8}
6. Domain: {x: x > 3} Range: All Real Numbers
7. Domain: All Real Numbers Range: {y: 4 ≤ y ≤ 6}
8. Domain: All Real Numbers Range: {y: 0 ≤ y ≤ 1}
9. Domain: x : x  2 or x  2Range: y : y  0
Asymptotes (vertical, horizontal, oblique)
A( x) an x n  ...
Let f be the rational function f(x) =
where N(x) and D(x) have no common factors.

D( x) bm x m  ...
a. The graph of f has vertical asymptotes at the zeros of D(x).
b. The graph of f has at most one horizontal asymptote determined by comparing the degrees of N(x) and D(x).
.
1. If n< m, the line y = 0 (the x-axis) is a horizontal asymptote.
a
2. If n = m, the line y = n is a horizontal asymptote.
bm
3. If n > m, the graph of f has no horizontal asymptote.
c. If the degree of the numerator of a rational function is exactly one degree more than than the degree of the
denominator, the graph has a slant(oblique) asymptote.
10. Find all asymptotes
Vertical Asymptote
x–2=0
x = 2.
Horizontal Asymptote: y = 0
11. x 2  1  0
x 2 = -1, which is non-real
x 2 + 1  0 there is NO vertical asymptote
3x 2
 3 , hence y = 3 is a horizontal
x2
asymptote
12. x – 5 = 0
x = 5 is a vertical asymptote
x 5
x +2
x 2  3x  7
x2  5x
2x - 7
2x -10
3
y = x + 2 is the oblique asymptote. There is no
horizontal asymptote.
Chesterfield County Public Schools
Department of Mathematics
Page 1 of 9
May 2006
13. 8 – x = 0
8 = x, is a vertical asymptote.
3x
x
y = -3, horizontal asymptote
y=
Zeros
14. y = 4 x 2 – 12x + 4
17. y = x4 – 8x2 – 9
= (x2 + 1)(x2 – 9)
x2 + 1 = 0;
x2 – 9 = 0
2
x = -1
x2 = 9
no real solution
x =  3 zeros
b  b 2  4ac
2a
a = 4, b = -12, c = 4
x=
x=
12  (12)2  4(4)(4)
2(4)
12  144  16
8
12  80
x=
8
x=

x=
12  4 5 4 3  5

8
2(4)
x=
3 5
2
16. y = x4 + 2x3 + x2
= x2(x2 + 2x + 1)
= x2(x + 1)(x + 1)
x2 = 0; x + 1 = 0; x + 1 = 0
x = 0, x = -1, x = -1
x = 0, double zeros
x = -1, double zeros

15. y = 2x3 + 11x2 – 6x
= x(2x2 + 11x – 6)
x = 0; 2x2 + 11x – 6 = 0
(x + 6)(2x – 1) = 0
x + 6 = 0; 2x – 1 = 0
x = -6; 2x = 1
1
x=
2
1
x = -6, 0,
2
Period/Amplitude
2
18. Period =
; Amplitude = A
b
2
Period =
; Amplitude = 5 = 5
3
19. No amplitude; Period =

6

6
20. Amplitude =
Period =
1 1

2 2
2 

4
2
Writing the equation of a line given a graph
1
21. Through (6, 1) with slope m = ; Use the point-slope form of the equation of a line,
2
1
y – y1 = m(x – x1) with the slope m =
and the point (6, 1) for (x1, y1) with x1 = 6
2
and y1 = 1.
1
y – 1 = (x – 6)
2
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Department of Mathematics
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May 2006
y–1=
1
x–3
2
1
- x+y+2=0
2
1
x–y–2=0
2
or
x – 2y – 4 = 0
22. First, find the slope of the line through the points (-1, -4) and (2, 5) using the slope
y  y1
formula m = 2
with x1 = -1, y1 = -4, x2 = 2, and y2 = 5.
x2  x1
5  (4) 5  4
9
=
= = 3.
3
2 1
2  (1)
Then use the point-slope form of the equation of a line, y – y1 = m(x – x1). Using the
slope of m = 3 and the point (-1, -4, the equation is
y – (-4) = 3[x – (-1)],
y + 4 = 3(x + 1)
y + 4 = 3x + 3
-3x + y + 1 = 0; 3x – y – 1 = 0
m=
23. The line perpendicular to the line with equation y = -
2
3
x + 1 has a slope of since
3
2
2
and the slopes of perpendicular line are negative
3
reciprocals of each other. Since the line passes through the point (8, 17) and has a
3
slope of , its equation using the point-slope form of the equation of a line is
2
3
y – 17 = (x – 8)
2
3
y – 17 = x – 12
2
3
- x+y–5=0
2
-3x + 2y – 10 = 0; 3x – 2y + 10 = 0
the given line has a slope of -
24. A line with an x-intercept of -2 and a y-intercept of 4 passes through the points
(-2, 0) and (0, 4). Calculating the slope of a line through these points gives
40
4
m=
 = 2.
0  (2) 2
Using the slope-intercept form of the equation of a line, y = mx + b, with slope m = 2
and y-intercept 4 gives an equation of y = 2x + 4.
In the general form, the equation is -2x + y – 4 = 0 or 2x – y + 4 = 0.
Chesterfield County Public Schools
Department of Mathematics
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May 2006
25. A line parallel to the line with the equation 2 x  3 y  7 will have the same slope. If you put the equation in
2
2
7
slope-intercept form y  x  , you can identify the slope as = . So the parallel line has the same slope and
3
3
3
2
passes through the point (5, 1) - given. Using the point-slope form of the line, you get y  1  ( x  5) or
3
2
7
y  x  . Therefore, we see that the only line parallel to 2 x  3 y  7 is the same line, because it passes
3
3
through the point (5,1).
Composition of functions
26. (g f)(x) = g(f(x))
= g(3x – 2)
= 3x  2
(g f)(9) = g(f(9));
= g(25)
= 25
=5
29. (g h)(x) = g(h(x))
=
f(9) = 3(9) – 2
= 27 – 2 = 25
27. (f h)(x) = f(h(x))
= 3(x2 – 2x + 1) – 2
= 3x2 – 6x + 3 – 2
= 3x2 – 6x + 1
(f h)(5) = f(h(5));
= 3(16) – 2
= 48 – 2
= 46
2
2
) = h(f( ));
3
3
h(5) = 52 – 2(5) + 1
= 25 – 10 + 1
= 15 + 1
= 16
f(
2
2
) = 3( ) – 2
3
3
= h(0)
=2–2
= 02 – 2(0) + 1
=0
=1
Chesterfield County Public Schools
Department of Mathematics
= ( x  1) 2
=x–1
(g h)(-2) = g(h(-2));
= g(9)
= 9
=3
h(-2) = (-2)2 -2(-2) + 1
=4+4+1
=9
Finding the inverse of a function
2x 1
30. y =
; interchange x and y
4
28. (h f)(x) = h(f(x))
= (3x – 2)2 – 2(3x – 2) + 1
= 9x2 – 12x + 4 – 6x + 4 + 1
= 9x2 – 18x + 9
= 9(x2 – 2x + 1)
= 9(x – 1)2
(h f)(
x2  2 x  1
2 y 1
4
4x = 2y – 1
4x + 1 = 2y
4x 1
=y
2
4x 1
= f -1(x)
2
1
2x +
= f -1(x)
2
x=
3x  1
y = 3x  1
Interchange x and y,
x = 3y 1
31. f(x) =
x2 = ( 3 y  1)2
x2 = 3y + 1
x2 1 3 y

3
3
Page 4 of 9
May 2006
x2 1
y
3
x2 1
 f 1 ( x) ,
3
40. tan
x>0
32. f(x) = (x – 6)2
Not one-one, NO INVERSE
33. f(x) = ln (x – 1)
y = ln(x – 1)
Replace x by y and y by x and solve for y.
x = ln(y – 1)
Write in exponential form to obtain ex = y – 1.
ex + 1 = y
So, f -1(x) = ex + 1, for all x.
Basic Trigonometry (Unit circle)
5
34.  =
. The point (x, y) on the unit circle that
6
5   3 1 
,  .
corresponds to  =
is 
6
 2 2
41. tan
2
 3
3

2
undefined
1
2
1
)
42. cos( Sin
43. Sin1 (sin
7

)
6
6
cos 2 x  1  sin 2 x
44.

  sin x
sin x
sin x
(Use the identity: sin 2 x  cos 2 x  1 and
sin 2 x  1  cos 2 x ;  sin 2 x  cos 2 x  1 )
45.  sin x  cos x   sin 2 x  2sin x cos x  cos 2 x
2
=  sin 2 x  cos 2 x  + 2 sin x cos x
5
. The point (x, y) on the unit circle that
4
 2  2
5
,
corresponds to  =
is 
.
4
2 
 2
35.  =
3
. The point (x, y) on the unit circle
2
3
that corresponds to  =
is (0, 1).
2
36.  =
11
37.  =
. The point (x, y) on the unit circle
3
1  3
11
that corresponds to  =
is  ,
 .
3
2 2 
38.  = 5 . The point (x, y) on the unit circle that
corresponds to  = 5 is (-1, 0).
39. tan

6

3
2
46.
=
1
+ 2 sin x cos x
=
1
+ sin 2x
csc2 x
csc2 x

1  tan 2 x sec2 x
=
1
sin 2 x
1
cos 2 x
=
1
cos 2 x
sin 2 x
=
cos 2 x
= cot 2 x
2
sin x
3
3
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Department of Mathematics
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May 2006
 cos x 
47. sin x + (cos x )(cot x) = sin x + cos x 

 sin x 
= sin x +
cos 2 x
sin x
2 x 2  10 x  12
2( x 2  5 x  6)
2(x + 1)(x – 6)
2
1
=
= csc x
sin x
48.
49.
sin x  cos x
sin x
2
=
Factor completely.
(tan x + cot x)(sin x + cos x)
50. x 4  81   x 2  9  x 2  9 
x
sec 2 x
(sin x + cos x)
tan x
 9  (x + 3)(x – 3)
51. 2 x 3  10 x 2  x  5
2x2 (x + 5) - (x + 5)
1 

 tan x 
 (sin x + cos x)
tan x 

 tan 2 x  1 

 (sin x + cos x)
 tan x 
2
(x + 5) (2x2 – 1)
52.
5x4 – 40x =
5x(x3 – 8)
5x(x – 2)(x2 + 2x + 4)
 1  cos x 


 (sin x + cos x)
2
 cos x  sin x 
1



 (sin x + cos x)
 cos x sin x 
1
1

= sec x + csc x
cos x sin x
Simplifying rational expressions
53. 8 -
8  x  3  6
6

x3
x3
=
8 x  24  6
8 x  18
=
x3
x3
=
2(4 x  9)
x3
Chesterfield County Public Schools
Department of Mathematics
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May 2006
54.
6x  8

3x  2

=
 6 x  8 
3x  2
3x  2

 6 x  8 

3x  2
3x  2
3x  4

  2  3x  4  
3x  2
 3x  4 
55.
1
1
x  ( x  h)
h
 

.
xh x
x ( x  h)
x ( x  h)
56.
2
5
3
x2  2  x  x
.
10
5
x 2 10
x5
2

3x  2

1
 1
1
1
 3(3  x) 


33 x
1
3 x 3   3 x 3


57.
.
x
x(3(3  x)
3x(3  x)
3(3  x)
2x
1
8

 2

x  6 x  9 x  1 x  2x  3
2 x( x  1)
( x  3)2
8( x  3)



2
2
( x  3) ( x  1) ( x  3) ( x  1) ( x  3) 2 ( x  1)
2
58.
2 x2  2 x  x 2  6 x  9  8 x  24

( x  3)2 ( x  1)
.
x2  15
( x  3)2 ( x  1)
59. Evaluate
f ( x  h)  f ( x )
2
and simplify if f ( x)  x  2 x .
h
( x  h) 2  2( x  h)  ( x 2  2 x)
x 2  2 xh  h 2  2 x  2h  x 2  2 x
2 xh  h 2  2h


 2x  h  2
h
h
h
60.
( x  y) 3  x 3  3x 2 y  3xy2  y 3 .
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May 2006
2
61.
5
5
19
8
x 3 (x  x 2  x2 )  x 3  x 6  x 3 .
n2
1
9
 0   2   8  15 .

2
2
n 0 2
4
62.
Solving Trigonometric Equations
63. sin x tan x = sin x
sin x tan x – sin x = 0
sin x(tan x – 1) = 0
sin x = 0 or tan x – 1 = 0
tan x = 1
x = 0  , 180  ; x = 45  , 225 
64. 2 sec2 x – 5 sec x + 2 = 0
(2 sec x – 1)(sec x – 2) = 0
2 sec x – 1 =0 or sec x – 2 = 0
2 sec x = 1
sec x = 2
1
sec x =
x = 60  , x = 300 
2

Solution: x = 60  , x = 300 
65. tan2 x cos x – 3 cos x = 0
cos x (tan2x – 3) = 0
cos x = 0
or
tan2 x – 3 = 0
x = 90  , x = 270 
tan2 x = 3
tan x =  3
x = 60  , 120  , 240  ,
300 
Solving Equations
68. log10 (x2 – 15x) = 2
Express the equation in exponential form and
solve for x.
x2 – 15x = 102
x2 – 15x = 100
x2 – 15x – 100 = 0
(x + 5)(x – 20) = 0
x + 5 = 0 or x – 20 = 0
x = -5
x = 20
Solution: x = -5, x = 20
x 2  3x  4  14
x 2  3x  18  0
69.
( x  6)( x  3)  0
x  6; x  3
x4  1
0
x3
70. x 4  1  0
( x 2  1)( x 2  1)  0
x  1,  i
66. 2 cos x + 3 = 0
2 cos x = - 3
3
2
x = 150  , 210 
( x  5) 2  9
x  5  3
cos x = -
67. 8 sin x + 13 sin x – 6 = 0
(8 sin x – 3)(sin x + 2) = 0
8 sin x – 3 = 0 or
sin x + 2 = 0
8 sin x = 3
sin x = -2
3
sin x = = 0.375

8
x  22  , x = 180  - 22  = 158 
2
Chesterfield County Public Schools
Department of Mathematics
71.
x  53
x  8; x  2
2 x2  5 x  8
72. 2 x 2  5 x  8  0
x
5  89
4
Page 8 of 9
May 2006
12 x 2  3 x
12 x 3  3 x  0
73. 3 x ( 4 x  1)  0
x  0;
ln 3
77. e  3
78.
e1ln x  e  e ln x  ex
79. ln 1  0
1
4
7
80. ln e  7
( x  1) 2 ( x  2)  ( x  1)( x  2) 2  0
( x  1)( x  2)( x  1  x  2)  0
81. log 3
74. ( x  1)( x  2)( 2 x  1)  0
x  1;2;
27
2x
9
3 2x
(3 )
1
2
x 3
 (32 ) x 3
36 x  32 x 6
75. 6 x  2 x  6
4 x  6
3
x
2
e3x  5
3 x  ln 5
76.
1
x  ln 5
3
x  .536
1
 1
3
82. log 1 8  3
2
83. ln
1
  ln 2  .6931
2
3 ln x
 e ln x  x 3
84. e
3
85.
4 xy

2

1
3
12 x y 5
2
3
3
2
4
3
x y3
3
86. (5a )( 4a )  20a
87.
2 3

3 2
 20a
13
6
3(n  1)! 3(n  1)

5n!
5
Exponents and Logarithms
Chesterfield County Public Schools
Department of Mathematics
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May 2006