1. No category

Pre-Calculus Math 260
Pierce College
Homework Solutions
Part I
Table of Contents
Section 1.5..................................................................................................................................................... 1
Section 1.7..................................................................................................................................................... 4
Section 1.8..................................................................................................................................................... 8
Section 1.10................................................................................................................................................. 13
Section 2.1................................................................................................................................................... 16
Section 2.3................................................................................................................................................... 28
Section 2.4................................................................................................................................................... 30
Section 2.5................................................................................................................................................... 32
Section 2.6................................................................................................................................................... 38
Section 2.7................................................................................................................................................... 42
Section 3.1................................................................................................................................................... 47
Section 3.2................................................................................................................................................... 51
Section 3.7................................................................................................................................................... 55
Section 4.1................................................................................................................................................... 59
Section 4.2................................................................................................................................................... 63
Section 4.3................................................................................................................................................... 65
Section 4.4................................................................................................................................................... 69
Section 4.5................................................................................................................................................... 70
Section 4.6................................................................................................................................................... 73
Section 1.5
20. The given equation is either linear of equivalent to a linear equation. Solve the equation.
2
1
y 1
y   y  3 
3
2
4
2
1
y 1
12 y  12  y  3  12
3
2
4
8 y  6  y  3  3  y  1
8 y  6 y  18  3 y  3
14 y  3 y  3  18
14 y  3 y  21
11 y  21
21
y
11
26. The given equation is either linear of equivalent to a linear equation. Solve the equation.
4

x 1
4

x 1
2
35
 2
x 1 x 1
2
35

x  1  x  1 x  1
4
2
  x  1 x  1
 35
x 1
x 1
4  x  1  2  x  1  35
 x  1 x  1
4 x  4  2 x  2  35
6 x  2  35
6 x  35  2  33
33 11
x

6
2
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54. Solve the equation by factoring.
 2 x  1
2
8
4x2  4 x  1  8
4x2  4 x  7  0
x
1 2 2
2
62. Solve the equation by completing the square.
2
3
1
x  x
4
8
3
1

0
x  
8  64

3
1
x2  x   0
4
8
3
9
9 1
x2  x     0
4
64 64 8
3
1

x  
8  64

3
1
x 
8
8
2
2
3
8
9

0
x   
8  64 64

2
3 1 1 1
x   ,
8 8 2 4
75. Find all real solutions to the quadratic equation.
w2  3  w  1
w2  3w  3  0
a  1, b  3, c  3
w
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3
 3
2
 4 1  3
2 1

3  9  12 3  3

 no real solutions
2
2
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89. Find all real solutions of the equation.
x5
5
28
5
28

 2


x  2 x  2 x  4 x  2  x  2  x  2 
x5
5
  x  2  x  2 
 28
x2
x2
 x  2  x  5  5  x  2   28
 x  2  x  2 
x 2  2 x  5 x  10  5 x  10  28
x 2  7 x  10  5 x  18
x2  2 x  8  0
 x  4  x  2   0
x  4, 2
Note: only x  4 is a solution since when x  2 the original equation is undefined.
91. Find all real solutions of the equation.
2x 1 1  x
2x 1  x 1
2 x  1   x  1  x 2  2 x  1
2
x2  4 x  0
x  x  4  0
x  0, 4
105. Find all real solutions of the equation.
3x  5  1
3x  5  1
3 x  5  1
3 x  1  5  4
3 x  6
4
x
x  2
3
108. Find all real solutions of the equation.
x  6  1 This equation has no solutions since the absolute value must always be positive or zero.
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Section 1.7
19. Solve the linear inequality. Express the solution using interval notation and graph the solution set.
3 x  11  6 x  8
3  3x
1 x
1,  
-5
-4
-3
-2
-1
0
1
2
3
4
5
29. Solve the linear inequality. Express the solution using interval notation and graph the solution set.
1  2 x  5  7
4  2 x  12
2 x6
-4
-3
-2
-1
0
1
2
3
4
5
6
 2, 6 
37. Solve the linear inequality. Express the solution using interval notation and graph the solution set.
x  2x  7  0
In this case we must have x  0 and 2 x  7  0 or x  0 and 2 x  7  0 . Now
2x  7  0
2 x  7
7
x
2
Thus 2 x  7  0 when x  
7
7
and 2 x  7  0 when x   . Therefore the solution set is:
2
2
7

 ,     0,  
2

-5
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-2
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1
2
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45. Solve the linear inequality. Express the solution using interval notation and graph the solution set.
x2  3  x  6
x2  3  x  6
x 2  3 x  18  0
 x  6  x  3  0
We note that  x  6  x  3  0 when x  6 and x  3 . When x  3 , then  x  6  x  3  0
since both factors are negative. When x  6 then  x  6  x  3  0 since both factors are positive.
When 3  x  6 then  x  6  x  3  0 since the first factor is negative and the second positive.
Therefore the solution set is:
 , 3   6,  
-10
-8
-6
-2
-4
2
0
4
6
8
10
61. Solve the linear inequality. Express the solution using interval notation and graph the solution set.
2x 1
3
x 5
2x 1
3 0
x 5
2 x  1  3  x  5
0
x 5
2 x  1  3 x  15
0
x 5
16  x
0
x 5
Note: numerator is zero when x  16 and he denominator is zero when x  5 . When x  5 the
denominator is negative and the numerator is positive and their quotient is therefore negative.
When x  16 the denominator is positive and the numerator is negative and their quotient is
therefore negative. Finally when 5  x  16 the denominator is positive and then numerator is also
positive, so their quotient is also positive. The solution set is therefore  , 5  16,  
-25
MAP
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-15
-10
-5
0
5
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77. Solve the absolute value inequality. Express the answer using interval notation and graph the
solution set.
x  5  3 We are looking for solution where x  5  3 and x  5  3 . For x  5  3 we have
x  8 and for x  5  3 we have x  2 . The solution set is  2,8 .
-3
-2
-1
0
1
2
1
4
3
5
6
7
8
81. Solve the absolute value inequality. Express the answer using interval notation and graph the
solution set.
3x  2  5
We are looking for solutions where 3x  2  5 and 3x  2  5 . We therefore must have x 
7
3
7
3


and x  1 . The solution set is therefore:  , 1   ,   .
-5
-4
-3
-2
-1
0
1
2
3
4
5
88. Solve the absolute value inequality. Express the answer using interval notation and graph the
solution set.
7 x2 5  4
In this case we note that the right side of the inequality is always 5 or greater. In this case, the
solution is set all the real numbers:  ,  
5
-4
-1
2
-2 company
3 renting
4 a car.
0offers1two plans
107. Rental-5Car Cost.
A-3car rental
for
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Plan A: $30 per day and 10¢ per mile.
Plan B: $50 per day will unlimited mileage.
For what range of miles will Plan B save you money?
If x is the number of miles driven, then under Plan A, the cost will be 30  0.1x . Under plan B the
cost is just 50. The value of x where the cost is the same is:
30  0.1x  50
x
20
 200
0.1
0.1x  50  30  20
Therefore Plan B saves money whenever more than 200 miles is driven.
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Section 1.8
13. A pair of points is graphed.
(a) Plot the points in a coordinate plane.
(b) Find the distance between them.
(c) Find the midpoint of the segment that joins them.
 0,8 ,  6,16 
y
(a)
x
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(b) Using the distance formula we have: d 
 6  0  16  8
2
2
 36  64  100  10
 6  0 16  8 
,
   3,12 
2 
 2
(c) Using the mid-point formula:  x, y   
33. Which of the points A  6, 7  or B  5,8 is closer to the origin.
Using the distance formula we have:
d A  62  7 2  36  49  85
dB 
 5
2
 8  25  64  89
Therefore point A is
2
closer.
66. Make a table of values and sketch the graph if the equation. Find the x and y intercepts and test for
symmetry.
y  9  x2
x
-5
-4
-3
-2
-1
0
1
2
3
4
5
y
-16
-7
0
5
8
9
8
5
0
-7
-16
x-intercepts are at -3 and 3. Y-intercept is at 9. The graph is symmetrical about the y-axis.
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68. Make a table of values and sketch the graph if the equation. Find the x and y intercepts and test for
symmetry.
y  x4
x
-4
-3
-2
-1
0
1
2
3
4
5
y
0
1
1.4
1.7
2
2.2
2.4
2.6
2.8
3
The x-intercept is a -4. The y-intercept is at 2. There is no symmetry exhibited in the graph.
91. Find the center of the circle and sketch the graph.
 x  3   y  4 
2
MAP
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 25 The center of the circle is at  3, 4  and it has radius 5.
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93. Find the equation of the circle that satisfies the given conditions: Center at
 2, 1 with radius 3.
Using the standard equation for a circle we have:  x  2    y  1  9
2
2
95. Find the equation of the circle that satisfies the given conditions: Center at origin and passes
through  4, 7  . Since the point  4, 7  lies on the perimeter of the circle, the distance of this point
to the origin is the radius of the circle:
r  42  72  16  49  65
The equation of the circle is therefore: x 2  y 2  65
97. Find the equation of the circle that satisfies the given conditions: the endpoints of the diameter are
P  1,1 and Q  5,9  . The distance between these points is the diameter:
d
5   1  9 1
2
2
 62  82  36  64  100  10
The radius is therefore 5. The center of the circle is at the midpoint between P and Q. Using the
 1  5 9  1 
,
   2,5  . The equation of the circle is:
2 
 2
midpoint formula we have:  x, y   
 x  2    y  5
2
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103. Show that the equation represents a circle, and find the center and radius of the circle.
x2  y 2  4 x  10 y  13  0
To show that this is the equation of a circle put the equation in the standard form for a circle. We do
this by completing the squares:
x 2  y 2  4 x  10 y  13  0
x
x
x
2
 4 x    y 2  10 y   13  0
2
 4 x  4  4    y 2  10 y  25  25   13  0
2
 4 x  4   4   y 2  10 y  25   25  13  0
 x  2    y  5  4  25  13  0
2
2
 x  2    y  5  16
2
2
 x  2    y  5   42
2
2
The center of the circle is at  2, 5  and its radius is 4.
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Section 1.10
19. Find the equation of the line that satisfies the given conditions: through  2,3 with slope 5.
Using the point slope form of the line we have:  y  3  5  x  2 
27. Find the equation of the line that satisfies the given conditions: x-intercept 1, y-intercept -3.
The line passes though the points 1, 0  and  0, 3 . The slope is therefore m 
3  0
 3 . Using
0 1
the slope intercept for the line we have: y  3x  3
29. Find the equation of the line that satisfies the given conditions: though  4,5  and parallel to the xaxis.
If the line is parallel to the x-axis then it is a horizontal line with slope zero. The equation of the line
is therefore y  5
32. Find the equation of the line that satisfies the given conditions: y-intercept 6 and parallel to the line
2x  3 y  4  0 .
If the line is parallel to the given line, then it has the same slope. Putting the given line into slope
intercept form:
2x  3y  4  0
3 y  2 x  4
2
4
y  x
3
3
The slope is therefore 
2
. Using the slope intercept form of the line we have:
3
2
y   x6
3
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35. Find the equation of the line that satisfies the given conditions: though  1, 2  perpendicular to
the line 2 x  5 y  8  0 .
If the line is perpendicular to the given line, then it has the negative reciprocal of the slope. Putting
the given line into slope intercept form:
2x  5 y  8  0
5 y  2 x  8
2 8
y 
5 5
The slope is therefore
5
. Using the point slope form of the line we have:
2
5
2
 y  2    x  1
71. Crickets and Temperature. Biologists have observed that the chirping of crickets of a certain species
is related to temperature, and the relationship appears to be very nearly linear. A cricket produces
120 chirps per minute at 70°F and 168 chirps per minute at 80°F.
(a) Find the linear equation that relates the temperature t and the number of chirps per minutes n .
(b) If the crickets are chirping at 150 chirps per minute, estimate the temperature.
(a) From the problem information, we have two points on the linear equation relating chirps per
minutes and temperature: 120, 70  and 168,80  . The slope of this line is:
m
80  70
10 5


168  120 48 24
Using the point slope form of the line:
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5
 n  120 
24
5
 5 
t
n  120    70
24
 24 
5
t
n  45
24
 t  70  
(b) Using the equation we found in part (a):
t
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5
150   45  76.25
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Section 2.1
19. Evaluate the function at the indicated values. f  x   2 x  1 at
1
f 1 , f  2  , f   , f  a  , f  a  , f  a  b 
2
f 1  2 1  1  3
f  2   2  2   1   3
1
1
, f    2  1  2
2
2
f  a   2a  1
f   a   1  2a
f a  b  2 a  b 1
25. Evaluate the function at the indicated values. f  x   2 x  1 at
1
f  2  , f  0  , f   , f  2  , f  x  1 , f  x 2  2 
2
f  2   2 2  1  2
f  0  2 0  1  2
1
1
f    2 1  3
2
2
f  2  2 2  1  6
f  x  1  2 x  2
f  x 2  2   2 x 2  3  2  x 2  3
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 x2
x0
at
 x 1 x  0
27. Evaluate the piecewise defined function at the indicated values. f  x   
f  2  , f  1 , f  0  , f 1 , f  2 
f  2    2   4
2
f  1   1  1
2
f  0  0  1  1
f 1  1  1  2
f  2  2  1  3
 x2  2x
x  1

29. Evaluate the piecewise defined function at the indicated values. f  x    x
 1  x  1 at
 1
x 1

 3
f  4  , f    , f  1 , f  0  , f  25 
 2
f  4    4   2  4   8
2
2
3
 3  3
 3
f         2    
4
 2  2
 2
f  1   1  2  1  1
2
f  0  0
f  25   1
35. Find f  a  , f  a  h  and the difference quotient
f  a  h  f a
where h  0 when
h
f  x   3x  2 .
f  a   3a  2
f  a  h  3 a  h  2
f  a  h   f  a  3  a  h   2   3a  2  3a  3h  2  3a  2 3h



3
h
h
h
h
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Pierce College Math 260
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36 Find f  a  , f  a  h  and the difference quotient
f  a  h  f a
where h  0 when
h
f  x   x2  1 .
f  a   a2  1
f a  h  a  h 1
2
2
f  a  h   f  a   a  h   1   a  1 a 2  2ah  h 2  1  a 2  1 2ah  h 2



 2a  h
h
h
h
h
2
45. Find the domain of the function f  x   2 x,  1  x  5 .
Normally the domain would be all the real numbers, but since the value of x is restricted the
domain is  1,5
49. Find the domain of the function f  x  
x2
.
x2 1
The function is defined everywhere except at -1 and 1. The domain is therefore
 , 1   1,1  1,  
51. Find the domain of the function f  x  
x 5 .
The function is defined for all values of x  5 . The domain is therefore 5,  
61. Find the domain of the function f  x  
3
. The function is defined for all values of x  4
x4
. The domain is therefore  4,   .
MAP
Pierce College Math 260
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69. Production Cost. The cost C in dollars of producing x yards of a certain fabric is given by the
2
3
function C  x   1500  3x  0.02 x  0.0001x
a. Find C 10  and C 100  .
b. What do your answers in part (a) represent?
c. Find C  0  (This number represents the fixed costs.)
C 10   1500  3 10   0.02 10   0.000110   1500  30  2  0.10  1532.10
2
a.
3
C 100   1500  3 100   0.02 100   0.0001100   1500  300  200  100  2100
2
3
b.
It costs $1532.10 to produce 10 yards of fabric and $2100 to produce 100 yards of fabric.
c.
C  0   1500  3  0   0.02  0   0.0001 0   1500
2
3
77. Internet Purchases. An Internet bookstore charges $15 shipping for orders under $100 but provides
free shipping for orders of $100 or more. The cost C of an order is a function of the total price x of
the books purchased, given by:
 x  15 x  100
C  x  
x  100
x
a. Find C  75 , C  90  , C 100  , 105 
b. What do your answers in part (a) represent?
C  75   75  15  90
a.
C  90   90  15  105
C 100   100
105  105
b. The cost of purchasing and shipping the books.
MAP
Pierce College Math 260
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Section 2.2
x2
3
 x 1 x  2
35. Sketch the graph of the piecewise defined function f  x   
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Pierce College Math 260
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x0
x
 x 1 x  0
37. Sketch the graph of the piecewise defined function f  x   
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Pierce College Math 260
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x  1
 1

40. Sketch the graph of the piecewise defined function f  x    1  1  x  1
1
x 1

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Pierce College Math 260
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2
41. Sketch the graph of the piecewise defined function f  x   
 x
MAP
Pierce College Math 260
2
x  1
x  1
Page 23
51. Use the vertical line test to determine whether the curve is the graph of a function of x .
For (a) and (c) the curves pass the vertical line test since it is possible to draw a vertical line
anywhere without intersecting the curves more than once; therefore these curves are graphs of
functions. For (b) and (d) these curves fail the vertical line test. For example the y-axis intersects
both curves twice. These are not the graphs of functions.
53. Use the vertical line test to determine whether the curve is the graph of a function of x . If so, state
the domain and range of the function.
From the graph, it looks the curve passes the vertical line test, since any vertical line intersects the
curve only once. The domain of the function is  3, 2 and the range is  2, 2
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Pierce College Math 260
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57. Determine whether the equation defines y as a function of x . (See Example 9.)
x2  2 y  4
Solving the equation for y : y  2 
x2
. The figure below shows a plot of this equation.
2
Since the curve passes the vertical line test, the equation defines y as a function of x .
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59. Determine whether the equation defines y as a function of x . (See Example 9.)
x  y2
The figure below shows a graph of the equation.
The curve shown in the graph clearly fails the vertical line test so that the equation does not define
y as a function of x .
MAP
Pierce College Math 260
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65. Determine whether the equation defines y as a function of x . (See Example 9.)
2 x  y0
Solving the equation for y : y  2 x . The figure below shows a plot of this equation.
Since the curve passes the vertical line test, the equation defines y as a function of x .
MAP
Pierce College Math 260
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Section 2.3
5. The graph of a function h is given.
a. Find h  2  , h  0  , h  2  and h  3 .
b. Find the domain and range of h .
c. Find the values of x for which h  x   3 .
d. Find the values of x for which h  x   3 .
a.
h  2   2
h  2  3
h  0   1
h  3  4
b. Domain:  3, 4 , Range:  1, 4
c.
x  3, 2, 4
d.
 3, 2  4
19. The graph of a function is given. Determine the intervals on which the function is (a) increasing and
(b) decreasing.
a. Increasing:  1,1 ,  2, 4 
b. Decreasing 1, 2 
31. The graph of a function is given. (a) Find the local maximum and minimum values of the function and
the values of x at which each occurs. (b) Find the intervals on which the function is increasing and
on which the function is decreasing.
a. Local minimums are at x  2 and x  2 with values 1 and 0 ,
respectively. There is a local maximum at x  0 with a value of 2 .
b. The function is decreasing on the intervals  , 2  and  0, 2  and
increasing on the intervals  2, 0  and  2,  
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Pierce College Math 260
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33. The graph of a function is given. (a) Find the local maximum and minimum values of the function and
the values of x at which each occurs. (b) Find the intervals on which the function is increasing and
on which the function is decreasing.
a. Local minimums are at x  2 and x  1 with values 2 and 1 ,
respectively. There is a local maximum at x  3 with a value of 1 .
b. The function is decreasing on the intervals  , 2  ,  0,1 and  3,  
and increasing on the intervals  2, 0  and 1,3 .
MAP
Pierce College Math 260
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Section 2.4
8. The graph of a function is given. Determine the average rate of change of the function between the
indicated points on the graph.
From the graph, the indicated points are  1, 0  and  5, 4  so that the average rate of change
between these points is
40
4 2
 
5   1 6 3
9. A function is given. Determine the average rate of change of the function between the given values of
the variable. f  x   3x  2 ; x  2 , x  3
Average rate of change 
f  3  f  2  3  3  2   3  2   2 

 9262  3
3 2
1
16. A function is given. Determine the average rate of change of the function between the given values
2
of the variable. f  x   4  x ; x  1 , x  1  h
Average rate of change
2
f 1  h   f 1 4  1  h    4  1  4  1  2h  h2  4  1 2h  h 2




 2  h
1  h 1
h
h
h
2
19. A function is given. Determine the average rate of change of the function between the given values
of the variable. f  t  
2
; t  a, t  ah
t
Average rate of change
2
2
f a  h  f a a  h  a
2
2 2a  2  a  h 
2h
2







aha
h
 a  h  h ha  a  h  ha  a  h  ha a  a  h 
MAP
Pierce College Math 260
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23. Changing Water Levels. The graph shows the depth of water W in a reservoir over a one year
period as a function of the number of days x since the beginning of the year. What was the average
rate of change of W between x  100 and x  200 .
From the graph we see that at 100 days the water level was about 75 feet and at 200
days the water level was about 50 feet. The average rate of change is therefore
50  75
25

 0.25 feet per day.
200  100 100
MAP
Pierce College Math 260
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Section 2.5
19. Use the graph of y  x 2 in Figure 4 to graph the following.
2
a. g  x   x  1
b. g  x    x  1
2
2
c. g  x    x
d. g  x    x  1  3
2
a. Shift up by one.
b. Shift right by one.
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Pierce College Math 260
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c. Reflect about x-axis.
d. Shift right by one and up by three.
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Pierce College Math 260
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23. Sketch the graph of the function not by plotting points but by starting with the graph of a standard
function and applying transformations. f  x  
Shit f  x  
x 1
x up by one.
24. Sketch the graph of the function not by plotting points but by starting with the graph of a standard
function and applying transformations. f  x   x  1
Shift f  x   x down by one.
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Pierce College Math 260
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27. Sketch the graph of the function not by plotting points but by starting with the graph of a standard
function and applying transformations. f  x  
Shift f  x  
x4
x to the left by 4.
28. Sketch the graph of the function not by plotting points but by starting with the graph of a standard
function and applying transformations. f  x   x  3
Shift f  x   x to the right by 3.
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Pierce College Math 260
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29. Sketch the graph of the function not by plotting points but by starting with the graph of a standard
3
function and applying transformations. f  x    x
3
Reflect f  x   x about the x-axis.
38. Sketch the graph of the function not by plotting points but by starting with the graph of a standard
function and applying transformations. f  x  
Shift f  x  
MAP
x  4 3
x to the left by 4 and down by 3.
Pierce College Math 260
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49. A function f is given and the indicated transformations applied to its graph (in the given order).
Write the equation for the final transformed graph. f  x   x ; shift three units to the right and
shift upward by one unit.
f  x  x  3 1
51. A function f is given and the indicated transformations applied to its graph (in the given order).
Write the equation for the final transformed graph. f  x  
4
x ; reflect in the y-axis an shift
upward by one unit.
f  x  4 x 1
52. A function f is given and the indicated transformations applied to its graph (in the given order).
2
Write the equation for the final transformed graph. f  x   x ; shift 2 units to the left and reflect
about the x-axis.
f  x     x  2
2
63a. The graph of f is given. Sketch the graphs of the following functions: y  f  x  2  .
Shift the graph to the right by 2:
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Pierce College Math 260
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Section 2.6
2
5. Find f  g , f  g , fg , and f g and their domains. f  x   x  3 , g  x   x
Note that the domain of f is  ,   and the domain of g is  ,   . The intersection of these
domains is  ,   .
f  g  x 2  x  3 Domain:  ,  
f  g   x 2  x  3 Domain:  ,  
fg  x 2  x  3 Domain:  ,  
f g
x 3
Domain:  ,0    0,  
x2
2
7. Find f  g , f  g , fg , and f g and their domains. f  x   4  x , g  x  
x 1
Note: the domain of f is  2, 2 and the domain of g is  1,   . The intersection of these
domains is:  1, 2
f  g  4  x 2  x  1 Domain:  1, 2
f  g  4  x 2  x  1 Domain:  1, 2
fg 
f g
MAP
 4  x   x  1 Domain:  1, 2
2
4  x2
Domain:  1, 2
x 1
Pierce College Math 260
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9. Find f  g , f  g , fg , and f g and their domains. f  x  
2
4
; g  x 
x
x4
Note: the domain of f is  ,0    0,   and the domain of g is  , 4    4,   . The
intersection of these domains is  , 4    4,0    0,   .
f g
2
4
6x  8


Domain:  , 4    4,0    0,  
x x  4 x  x  4
f g 
2
4
8  2x


Domain:  , 4    4,0    0,  
x x  4 x  x  4
8
2 4
fg   

Domain:  , 4    4,0    0,  
 x  x  4 x  x  4
2 x4 x4
Domain:  , 4    4,0    0,  
f g  

2x
 x 4
2
21. Use f  x   3x  5 and g  x   2  x to evaluate the expression.






a. f g  0  Since g  0   2 , f g  0   f  2   1


b. g f  0  Since f  0   5 , g f  0   g  5  23
2
22. Use f  x   3x  5 and g  x   2  x to evaluate the expression.
a. f
 f  4  Since f  4  7 , f  f  4  f  7   16




b. g g  3 Since g  3  7 , g g  3  g  7   47
2
23. Use f  x   3x  5 and g  x   2  x to evaluate the expression.
a.
f
b.  g
MAP
g  2  Since g  2   2 ,  f g  2   f  2   11
f  2  Since f  2   11 ,  g f  2  g  11  119
Pierce College Math 260
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2
24. Use f  x   3x  5 and g  x   2  x to evaluate the expression.
a.
f
f  1 Since f  1  8 ,  f
b.  g g  x 
f  1  f  8  29
g  x   2   2  x 2   2   4  4 x 2  x 4    x 4  4 x 2  2
g
2
2
25. Use f  x   3x  5 and g  x   2  x to evaluate the expression.
a.
f
g  x 
b.  g
g  x   3  2  x 2   5  6  3x 2  5  1  3x 2
f
f  x   g f  x   2   3x  5  2   9 x 2  30 x  25  9 x 2  30 x  23
2
38. Find the functions f
g, g f , f
f , and g g and their domains.
f  x   x2 , g  x   x  3
Note: From page 193: The domain of f
g is the set of all x in the domain of g such that g  x  is
in the domain of f .
f g

x 3

2
 x  3 ; Domain: Since the domain of f is  ,   and the domain of g is
3,   , the domain of
g f 
x 3 
2
f g is 3,   .
x 2  3 ; Domain: Since the domain of f is  ,   and the domain of g is
3,   , the domain of g
f


f is ,  3    3,  .
f   x 2   x 4 ; Domain: Since the domain of f is  ,   , then the domain of f
2
f is also
 ,   .
g g
x  3  3 ; Domain: Since the domain of g is 3,   , then the domain of g g is
12,  
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61. Area of a Ripple. A stone is dropped in a lake, creating a circular ripple that travels outward at a
speed of 60 cm/s.
a. Find a function g that models the radius of the circle as function of the time.
b. Find a function f that models the area of the circle as a function of the radius.
c. Find f
g . What does this function represent?
a. Using the formula Distance = speed X time, we have r  60t where r is the radius of the circle
and t is the time in seconds.
b. The area f of a circle is f   r 2 , where r is the radius.
c.
MAP
f g  3600 t 2 is the area of the circle as a function of time.
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Page 41
Section 2.7
5. The graph of the function f is given. Determine whether f is one-to-one.
Since the graph passes the horizontal line test, f is one-to-one.
8. The graph of the function f is given. Determine whether f is one-to-one.
Since the graph fails the horizontal line test, f is not one-to-one.
13. Determine whether the function is one-to-one. g  x  
The graph of g  x  
x .
x , shows that g  x  passes the horizontal line test and is therefore one-to-
one.
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4
17. Determine whether the function is one-to-one. g  x   x  5
4
The graph of g  x   x  5 , shows that g  x  fails the horizontal line test and is therefore not one-
to-one.
21. Assume that f is a one-to-one function.
1
a. If f  2   7 find f  7 
1
b. If f  3  1 find f  1
1
a. Since f  2   7 we must have that f  7   2
1
b. B. Since f  3  1 we must have f  1  3
25. Use the Inverse Function Property to show that f and g are inverses of each other.
f  x  x  6 ; g  x  x  6
f g   x  6  6  x  6  6  x
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27. Use the Inverse Function Property to show that f and g are inverses of each other.
f  x   2x  5 ; g  x  
x5
2
 x5
f g  2
5  x 55  x
 2 
39. Find the inverse of f . f  x   4 x  7
f  x  4x  7
y  4x  7
y  7  4x
y7
x
4
x7
y
4
x7
f 1 
4
43. Find the inverse of f . f  x  
f  x 
1
x2
1
x2
1
x2
1
x2
y
1
1 2 y
x  2
y
y
y
1 2x
x
1 2x
f 1  x  
x
y
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2
53. Find the inverse of f . f  x   4  x ; x  0
f  x   4  x2
y  4  x2
x2  4  y
x  4 y
y  4 x
f 1 ( x)  4  x
Note: f  x  passes the horizontal line test because of its restricted domain. When solving for x we
take the positive square root because x  0 .
79. Use the graph of f to sketch the graph of f 1 .
From the graph, we note that the points  1, 2  , 1, 1 ,  2, 2 
, and  3,3 appear to be on the graph. Between these points, the
graph appear linear. Therefore the points  2, 1 ,  1,1 ,
 2, 2  , and  3,3 will be on the graph of
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85. Temperature Scales. The relationship between the Fahrenheit  F  and Celsius  C  scales is given
by:
9
F  C   C  32
5
a. Find F 1 . What does F 1 represent?
1
b. Find F  86  . What does your answer represent?
9
5
F  C   C  32
 F  32   C
5
9
9
5
a.
F  C  32
F 1   F  32  F 1 is the conversion from Celsius to Fahrenheit.
5
9
9
F  32  C
5
5
5
b. F 1  86    86  32    54   5  6   30 . Thus 86°F corresponds to 30°C
9
9
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Section 3.1
23. A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its
2
graph. (c) Find its maximum or minimum value. f  x   x  2 x  1 .
a. We complete the square to put the equation in standard form:
f  x   x2  2x 1
 x2  2x  1  1  1
 x  1
2
1
b.
c. The vertex is located at  1, 1 and since the leading coefficient is positive, the quadratic has a
minimum with value -1.
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39. Find the minimum or maximum value of the function.
1 2
x  2x  6 .
2
Put the quadratic in standard form by completing the square:
1 2
x  2x  6
2
1 2
 x  4x  6
2
1 2
 x  4x  4  4  6
2
1 2
 x  4x  4  2  6
2
1
2
 x  2  8
2
The vertex is located at  2, 8 and since the leading term in the quadratic is positive, the curve
opens upward. The vertex is therefore the location of the minimum, which is -8.
2
45. Find the domain and range of the function. f  x    x  4 x  3
Put the quadratic in standard form by completing the square:
 x2  4x  3
  x2  4x   3
  x2  4x  4  4  3
  x2  4x  4  4  3
  x  2  1
2
The vertex is located at  2,1 and since the leading term in the quadratic is negative, the curve
opens downward. The vertex is therefore the location of the maximum, which is 1. The domain is
 ,   and the range is  ,1 .
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2
47. Find the domain and range of the function. f  x   2 x  6 x  7
Put the quadratic in standard form by completing the square:
2 x2  6 x  7
2  x 2  3x   7
9 9

2  x 2  3x     7
4 4

9 9

2  x 2  3x     7
4 2

2
3  23

2 x   
2
2

 3
 2
The vertex is located at   , 
23 
 and since the leading term in the quadratic is positive, the
2 
curve opens upward. The vertex is therefore the location of the minimum, which is 
23
. The
2
 23

,   .
 2

domain is  ,   and the range is  
65. Revenue. A manufacturer finds that the revenue generated by selling x units of a certain
2
commodity is given by the function R  x   80 x  0.4 x where R  x  is measured in dollars. What
is the maximum revenue, and how many units should be manufactured to obtain this maximum?
Since the equation for the revenue is quadratic all that’s required is to find the vertex. Again, we
complete the square to place the equation in standard form:
R  x   80 x  0.4 x 2
 0.4  x 2  200 x 
 0.4  x 2  200 x  10000  10000 
 0.4  x 2  200 x  10000   4000
 0.4  x  100   4000
2
The vertex is at 100, 4000  and since the leading coefficient is negative this is the maximum.
Therefore the maximum revenue is $4000 and is obtained when 100 units are produced.
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75. Fencing a Horse Corral. Carol has 2400 ft. of fencing to fence in a rectangular horse corral.
a. Find a function that models the area of the corral in terms of the width x of the corral.
b. Find the dimensions of the rectangle that maximizes the area of the corral.
a. The area, A , of the corral is A  xy where y is the length of the corral. We note that the
perimeter of the corral is 2 x  2 y and that this equals 2400, the length of the available fencing.
As such we have:
2 x  2 y  2400
2 y  2400  2 x
y  1200  x
The area is therefore:
A  xy  x 1200  x 
b. Since the area is a quadratic function of the width, we place the equation for A into standard
form:
A  x 1200  x 
  x 2  1200 x
   x 2  1200 x 
   x 2  1200 x  360000  360000 
   x 2  1200 x  360000   360000
   x  600   360000
2
The vertex is  600,360000  which is the location of the maximum since the leading coefficient
is negative. The width is there 600 feet, and using the formula from part (a) the length is also
600. The maximum area is 360,000 square feet.
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Section 3.2
17. Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits
the proper end behavior. P  x   x  x  3 x  2 
Note that there are zeros at x  0, x  3, x  2 . The y-intercept is at 0. As x becomes large and
positive, P  x  becomes large and positive and as x becomes large and negative, P  x  becomes
large and negative.
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21. Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits
the proper end behavior. P  x    x  1
2
 x  3
Note that there are zeros at x  1, x  3 . The y-intercept is at -3. As x becomes large and positive,
P  x  becomes large and positive and as x becomes large and negative, P  x  becomes large and
negative.
27. Factor the polynomial and use the factored form to find the zeros. Then sketch the graph.
P  x   x3  x 2  6 x
P  x   x3  x 2  6 x
 x  x2  x  6
 x  x  2  x  3
Zeros are at x  0, x  3, x  2
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31. Factor the polynomial and use the factored form to find the zeros. Then sketch the graph.
P  x   x 4  3x 3  2 x 2
P  x   x 4  3x3  2 x 2
 x 2  x 2  3x  2 
 x 2  x  1 x  2 
Zeros are at x  0, x  1, x  2
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34. Factor the polynomial and use the factored form to find the zeros. Then sketch the graph.
P  x   x3  3x 2  4 x  12
P  x   x 3  3x 2  4 x  12
  x  2  x2  5x  6
  x  2  x  2  x  3
Zeros are at x  3, x  2, x  2
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Section 3.7
22. Find all the horizontal and vertical asymptotes (if any). r  x  
2x  3
x2 1
Vertical asymptotes are where the denominator is zero, in this case at x  1 and x  1 . Since the
degree of the polynomial in the numerator is less than the degree of the polynomial in the
denominator, the horizontal asymptote is the x-axis.
6 x2  1
25. Find all the horizontal and vertical asymptotes (if any). s  x   2
2x  x 1
We factor the rational function: s  x  
are at x  1 and x 
6x2  1
6 x2  1

. The vertical asymptotes
1
2 x2  x  1

2  x    x  1
2

1
. Since the degrees of the polynomials in the numerator and denominator
2
are equal and the leading coefficients are 6 and 2, respectively, the horizontal asymptote is at
y  3.
41. Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the
domain and range. r  x  
Factor: r  x  
4x  4
x2
4 x  4 4  x  1

x2
x2
x-intercepts: are the zeros of the numerator which in this case is x  1
y-intercept: is at r  0  
4  0  1 4

 2
02
2
Vertical asymptotes: is where the denominator is zero, in this case at x  2
Behavior near vertical asymptotes: examine the function on each side of x  2 :
r  1.99  
4   1.99   1

4  2.99 
 1196; y   as x  2
0.01
 1.99   2
4   2.01  1 4  3.01
r  2.01 

 1204 y   as x  2
0.01
 2.01  2
Horizontal Asymptote: since the degrees of the polynomials in the numerator and denominator are
equal and the leading coefficients are 4 and 1, respectively, y  4 is the horizontal asymptote.
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Graph:
Domain:
 , 2   2,  
Range:  , 4    4,  
45. Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the
domain and range. r  x  
Factor: r  x  
18
 x  2
2
18
 x  2
2
x-intercepts: are the zeros of the numerator and in this case there are none.
y-intercept: is at r  0  
18
 0  2
2

18 9

4 2
Vertical asymptotes: is where the denominator is zero, in this case at x  2
Behavior near vertical asymptotes: examine the function on each side of x  2 :
r 1.99  
r  2.01 
18
1.99  2 
2
18
 2.01  2 
2


18
 0.01
18
 0.01
2
2
 180,000; y   as x  2
 180,000 y   as x  2
Horizontal Asymptote: since the degrees of the polynomial in the numerator is less than the degree
of the polynomial in the denominator y  0 is the horizontal asymptote.
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Graph:
Domain:
 , 2   2,  
Range:  0,  
56. Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the
domain and range. r  x  
Factor: r  x  
4 x2
x2  2 x  3
4 x2
4 x2

x 2  2 x  3  x  1 x  3
x-intercepts: are the zeros of the numerator which in this case is x  0 .
y-intercept: is at r  0  
4  0
 0
2
2
 2  0  3

0
0
3
Vertical asymptotes: is where the denominator is zero, in this case at x  1 and x  3
Behavior near vertical asymptotes: examine the function on each side of x  1 :
4  0.99 
r  0.99  
 0.99  1 0.99  3
2
98; y   as x  1
4  1.01
r  1.01 
135; y   as x  1
 1.01  1 1.01  3
2
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Examine the function on each side of x  3 :
4  2.99 
r  2.99  
 2.99  1 2.99  3
2
4  3.01
r  3.01 
 3.01  1 3.01  3
896; y   as x  3
2
904; y   as x  3
Horizontal Asymptote: since the degrees of the polynomials in the numerator and denominator are
equal and the leading coefficients are 4 and 1, respectively, y  4 is the horizontal asymptote.
Graph:
Domain:
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 , 1   1,3  3,  
Range:  , 4    4,  
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Section 4.1
25. Graph the function, not by plotting points but by starting from the graphs in Figure 2. State the
x
domain, range, and asymptote. f  x   3 .
This is a reflection of the function 3x about the x-axis.
Domain:  ,   , Range:  , 0  , y  0 is the horizontal asymptote.
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27. Graph the function, not by plotting points but by starting from the graphs in Figure 2. State the
x
domain, range, and asymptote. f  x   2  3
This is a downward shift of the function 2 x by 3 units.
Domain:  ,   , Range:  3,   , y  3 is the horizontal asymptote.
33. Graph the function, not by plotting points but by starting from the graphs in Figure 2. State the
x
domain, range, and asymptote. f  x   5  1
This is a reflection of the function 5 x about the y-axis along with an upward shift by 1 unit.
Domain:  ,   , Range: 1,   , y  1 is the horizontal asymptote.
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47. Find, rounded to two decimal places, (a) the intervals on which the function is increasing or
decreasing and (b) the range of the function. y  10 x  x
2
The graph y  10 x  x of is shown below.
2
From the graph, the function has a local maximum at x 
1
. (a) The function is increasing from
2
1

1

 ,  and decreasing from  ,   . (b) from the graph the range of the function is
2

2

 0,10   0,1.78
0.25
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51. Compound Interest. If $10,000 is invested at an interest rate of 3% per year, compounded
semiannually, find the value of the investment after the given number of years.
(a) 5 years.
(b) 10 years


Since A  t   P 1 


(c) 15 years
nt
r
 where p  10000 and r  0.03 and n  2 , we have:
n
(a) A  5  10000 1 
0.03 

2 
2 5 
 0.03 
(b) A 10   10000 1 

2 

 0.03 
(c) A 15  10000 1 

2 

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 11, 605.41
210 
 13, 468.55
215
 15, 630.80
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Section 4.2
7. Graph the function, not by plotting points, but by starting from the graph of y  e x in Figure 1. State
x
the domain, range, and asymptote. f  x   e .
This is a reflection of y  e x about the x-axis.
Domain:  ,   , Range:  , 0  , y  0 is the horizontal asymptote.
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9. Graph the function, not by plotting points, but by starting from the graph of y  e x in Figure 1. State
x
the domain, range, and asymptote. f  x   e  1
This is a reflection of y  e x about the y-axis combined with a downward shift by one unit.
Domain:  ,   , Range:  1,   , y  1 is the horizontal asymptote.
21. Radioactive Decay. A radioactive substance decays in such a way that the amount of mass remaining
0.015t
after t days is given by the function m  t   13e
where m  t  is measured in grams.
a. Find the mass at time t  0 .
b. How much of the mass remains after 20 days?
a.
m  0   13e0.0150  13e0  13
b.
m  20   13e0.015 20
9.63
31a. Compound Interest. If $2000 is invested at an interest rate of 3.5% per year, compounded
continuously, find the value of the investment after the given number of years.
a. 2 years.
A  t   Pert  2000e0.035 2  2145.02
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Section 4.3
7. Express the equation in exponential form. (a) log5  25  2 (b) log5 1  0
(a) Exponentiation of each side by 5 gives: 25  52
(b) Exponentiation of each side by 5 gives: 1  50
9. Express the equation in exponential form. (a) log8  2  
1
1
(b) log 2    3
3
8
1
3
(a) Exponentiation of each side by 8 gives: 2  8  3 8
(b) Exponentiation of each side by 2 gives:
1
 23
8
11. Express the equation in exponential form. (a) ln  5  x (b) ln  y   5
(a) Exponentiation of each side by e gives: 5  e x
(b) Exponentiation of each side by e gives: y  e5
13. Express the equation in logarithmic form. (a) 53  125 (b) 104  0.0001
(a) Taking the log base 5 of each side gives: 3  log5 125
(b) Exp Taking the log base 10 of each side gives: 4  log10  0.0001
 
21. Evaluate the expression. (a) log 6  36  (b) log9  81 (c) log 7 710
 
(a) log 6  36   log 6 62  2
 
(b) log9 81  log9 92  2
 
(c) log 7 710  10
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 1 
 (b) log10
 27 
23. Evaluate the expression. (a) log 3 
 10 
(c) log5  0.2 
 1 
1
3
  log3  3   log 3  3   3
27
3
 
 
(a) log3 
(b) log10
 12  1
10  10  

 2
 10   log
1
5
 
(c) log5  0.2   log5    log5 51  1
log 2  37 
25. Evaluate the expression. (a) 2
log2  37 
(a) 2
log3 8
(b) 3
(c) e
ln
 5
log3  8
(b) 3
(c) e
ln
 5
 37
8
 5
29. Use the definition of the logarithmic function to find x . (a) log 2  x   5 (b) log 2 16   x
(a) log 2  x   5 . Exponentiation of each side by 2 gives: x  25  32
 
(b) log 2 16   x  log 2 24  4
32. Use the definition of the logarithmic function to find x . (a) log 4  2   x (b) log 4  x   2
(a) log 4  2   x  log 4
 12  1
4  log 4  4  
  2
 
(b) log 4  x   2 Exponentiation of each side by 4 gives: x  42  16
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53. Graph the function, not by plotting points, but by starting from the graphs in Figures 4 and 9. State
the domain, range, and asymptote. f  x   log 2  x  4 
This is a shift to the right by 4 units of f  x   log 2  x  . The graph is as follows:
Domain:  4,   , Range:  ,   Vertical Asymptote: at x  4
55. Graph the function, not by plotting points, but by starting from the graphs in Figures 4 and 9. State
the domain, range, and asymptote. g  x   log5   x 
This is a reflection of g  x   log5  x  about the y-axis. The graph is as follows:
Domain:  , 0  , Range:  ,   Vertical Asymptote: at x  0
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MAP
Pierce College Math 260
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Section 4.4
9. Evaluate the expression. log  4   log  25 .
log  4   log  25  log  4  25  log 100   log 102   2
11. Evaluate the expression. log 4 192   log 4  3 .
 192 
3
log 4 192   log 4  3  log 4 
  log 4  64   log 4  4   3
 3 
 
23. Use the Laws of Logarithms to expand the expression. log 610
log  610   10log  6 
45. Use the Laws of Logarithms to combine the expression. log3  5  5log3  2 
log3  5  5log3  2   log3  5  log3  25   log3  5  25   log3 160 
47. Use the Laws of Logarithms to combine the expression. log 2  A  log 2  B   2log 2  C 
 AB 
log 2  A  log 2  B   2log 2  C   log 2  AB   2log 2  C   log 2  AB   log 2  C 2   log 2  2 
C 

51. Use the Laws of Logarithms to combine the expression. ln  5  2ln  x   3ln x 2  5

ln  5  2ln  x   3ln  x 2  5  ln  5   ln  x 2   ln  x 2  5
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Pierce College Math 260
3

  ln 5x  x  5 
2
2
3
Page 69
Section 4.5
3. Find the solution to the exponential equation rounded to four decimal places. 10x  25
Take the common logarithm of each side:
10 x  25
log 10 x   log  25
x  log  25  1.3979
13. Find the solution to the exponential equation rounded to four decimal places. 4  35 x  8
4  35 x  8
35 x  8  4  4
ln  35 x   ln  4 
5 x ln  3  ln  4 
x
ln  4 
5ln  3
0.2524
39. Solve the logarithmic equation for x . log  x   2
log  x   2
x  102  0.01
41. Solve the logarithmic equation for x . log  3x  5   2
log  3 x  5   2
3x  5  102  0.01
3x  0.01  5  4.99
x  4.99 3  1.6633
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45. Solve the logarithmic equation for x . log 2  3  log 2  x   log 2 5   log 2  x  2 
log 2  3  log 2  x   log 2  5   log 2  x  2 
log 2  3 x   log 2  5  x  2  
3 x  5  x  2   5 x  10
3 x  5 x  10
2 x  10
x5
47. Solve the logarithmic equation for x . log  x   log  x  1  log  4 x 
log  x   log  x  1  log  4 x 
log  x  x  1   log  4 x 
x  x  1  4 x
x2  x  4x
x2  5x  0
x  x  5  0
x  0, x  5
We note that only x  5 is a valid solution to the original equation and is therefore the only solution.
51. Solve the logarithmic equation for x . log 2  x   log 2  x  3  2
log 2  x   log 2  x  3  2
log 2  x  x  3   2
x  x  3  22
x 2  3x  4
x 2  3x  4  0
 x  1 x  4   0
x  1, x  4
We note that x  1 is not a valid solution to the original equation and therefore x  4 is the
solution.
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75. Compound Interest. A man invests o$5000 in an account that pays 8.5% interest per year,
compounded quarterly.
a. Find the amount after 3 years.
b. How long will it take for the amount to be $8000?
nt
a.
 r
 0.085 
A  P 1    5000 1 

4 
 n

 r
A  P 1  
 n
1.6  1.02125 
4t
4t
ln 1.6   ln 1.02125   4t ln 1.02125 
4t
4t 
t
MAP
 6435.09
nt
 0.085 
8000  5000 1 

4 

b.
43
ln 1.6 
ln 1.02125 
ln 1.6 
4 ln 1.02125 
5.6
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Section 4.6
1. Bacteria Culture. A certain culture of the bacterium Streptococcus A initially has 10 bacteria and is
observed to double every 1.5 hours.
t a
a. Find an exponential model n  t   no 2 for the number of bacteria in the culture after t hours.
n  t   10  2t 1.5 
b. Estimate the number of bacteria after 35 hours.
n  35  10  235 1.5  105,689,838
c. When will the bacteria count reach 10,000?
n  t   10  2t 1.5 
10, 000  10  2t 1.5 
1, 000  2t 1.5
log 1, 000   log  2t 1.5  
t
log  2 
1.5
t
log  2 
1.5
4.5  t log  2 
3
t
4.5
14.9
log  2 
3ab. Squirrel Population. A grey squirrel population was introduced in a certain county of Great Britain
30 years ago. Biologists observe that the population doubles every 6 years, and now the population
is 100,000.
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a. What was the initial size of the squirrel population?
n  t   no 2t a
100, 000  no 230 6
no 
100, 000 100, 000

 3125
25
32
b. Estimate the squirrel population 10 years from now.
n 10   3125  240 6  317,480
9ab. Population of a City. The population of a certain city was 112,000 in 2006, and the observed
doubling time of the population is 18 years.
t a
a. Find an exponential model n  t   no 2 for the population t years after 2006.
n  t   112, 000  2t 18 
rt
b. Find an exponential model n  t   no e for the population t years after 2006.
ln  2  ln  2 

0.0385
h
18
n  t   112,000e0.0385t
r
19. Radioactive Strontium. The half-life of strontium-90 is 28 years. How long will it take a 50-mg
sample to decay to a mass of 32 mg?
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n  t   no 2 t a  50  2 t 28 
32  50  2 t 28 
32 16

50 25
 16 
ln  2 t 28   ln  
 25 
t
 16 
 ln  2   ln  
28
 25 
2 t 28 
 16 
28ln  
 25  18
t
ln  2 
21. Finding the Half-life. If 250 mg of a radioactive element decays to 200 mg in 48 hours, find the halflife of the element.
n  t   no 2 t a
200  250  248 a 
200 4

250 5
4
ln  248 a   ln  
5
248 a 

48
4
ln  2   ln  
a
5
a
MAP
48ln  2 
149
4
ln  
5
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