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Chapter 1 – Suggested Solution(1.1-1.4)
§1
9, 14, 22, 50, 56, 65, 68, 70, 72
9. If f (t )  (2t  1) 3 / 2 ; f (1), f (5), f (13), .
Solution
f (1) 
f (5) 
f (13) 
1
(2  1  1) 3

1
1
1
3
1
1

(2  5  1) 3
(2  13  1)
1
93

1

25
3
1
27

1
125
3
if t  5

14. If f (t )  t  1 if  5  t  5 ; f (6), f (5), f (0), f (16) .

 t if t  5
Solution (piecewise-defined function:)
Since t  6 satisfies t  5 , use the top part of the formula to find
f (6)  3
However, t  5 and t  0 satisfy  5  t  5 , so both f (5) and f (0) are found by using the
middle part of the formula:
f (5)  5  1  4
f (0)  0  1  1
and
And t  16 satisfies t  5 , so use the bottom part of the formula to find
f (16)  16  4
22. Find the range and domain of function f (t ) 
t 1
.
t t 2
2
Solution (example 1.1.4)
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Rewrite the function as f (t ) 
possible, the domain of
t 1
. Since division by any number other than 0 is
(t  2)(t  1)
f is the set of all number t except t  2 and t  1 . The range of
f is the set of all numbers y except 0 and 
1
1
, since for t   2 , and t  1.
3
y
50 Find the indicated composite function f ( x 2  2 x  9) where f ( x)  2 x  20
Solution (Composition of functions: Example 1.1.8)
Write the formula for f in more neutral terms, say as
f ( )  2( )  20
to find f ( x 2  2 x  9) , we simply insert the expression x 2  2 x  9 inside each box, getting
f ( x 2  2 x  9)  2  ( x 2  2 x  9)  20
 2x 2  4x  2
56. If f ( x)  x  4 
1
find the functions h(x) and g (u ) such that f ( x)  g (h( x))
( x  4) 3
Solution (Composition of functions: Example 1.1.9)
The form of the given function is
f ( x)  ( ) 
1
( ) 3
where each box contains the expression x  4 . Thus, f ( x)  g (h( x)) , where
g (u )  u 
1
and h( x)  x  4
u3
65. IMMUNIZATION Suppose that during a nationwide program to immunize the population
against a certain form of influenza, public health officials found that the cost of inoculating x% of
the population was approximately C ( x) 
150 x
million dollars.
200  x
a. What is the domain of the function C?
b. For what values of x does C(x) have a practical interpretation in this context?
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e. What percentage of the population had been inoculated by the time 37.5 million dollars had
been spent?
Solution ( Example 1.1.4; Example 1.1.6)
a. Since division by any number other than 0 is possible, the domain of C is the set of all real
numbers x  200 .
b. C(x) has a practical interpretation for 0  x  100 since x means the number of percent of
the population.
e. Set C (x) equal to 37.5 and solve for x to get
150 x
 37.5
200  x
x  40
That is, 40% of the population had been inoculated by the time 37.5 million dollars had been
spent.
68. EXPERIMENTAL PSYCHOLOGY To study the rate at which animals learn, a psychology
student performed an experiment in which a rat was sent repeatedly through a laboratory maze.
Suppose that the time required for the rat to traverse the maze on the nth trial was approximately
12
T ( n)  3 
minutes.
n
a. what is the domain of the function T?
b. For what values of n does T(n) have meaning in the context of the psychology experiment?
c. How long did it take the rat to traverse the maze on the 3rd trial?
d. On which trial did the rat first traverse the maze in 4 minutes or less?
e. According to the function T, what will happen to the time required for the rat to traverse the
maze as the number of trials increases? Will the rat ever be able to traverse the maze in less
than 3 minutes?
Solution
a. Since division by any number other than 0 is possible, the domain of T is the set of all real
numbers n  0 .
b. T(n) has a practical interpretation for all positive integer since n means the number of trials.
c. On the 3rd trial, it took the rat approximately
12
T (3)  3   7 minutes.
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d. Since
T ( n)  4
12
3  4
n
n  12
the rat first traverse the maze in 4 minutes on 12th trial.
e. It is obviously that the time required for the rat to traverse the maze will decrease as the
number of trials increases. But the time can not be less than 3 minutes, since 12/n is always
larger than 0.
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70. POSITION OF A MOVING OBJECT A ball has been dropped from the top of a building.
Its height (in feet) after t seconds is given by the function H (t )  16t 2  256 .
a. What is the height of the ball after 2 seconds?
b. How far will the ball travel during the third second?
c. How tall is the building?
d. When will the ball hit the ground?
Solution
a. After 2 seconds the height of the ball is
H (2)  16(2) 2  2 5 6 1 9 2feet
b. During the third second the distance that the ball will travel is the difference between the
height of the ball after 2 second and the height after 3 second. That is
H (2)  H (3)  192  (16(3) 2  256)  80 feet
c. The height of the building is the height of the ball when t  0 . That is
H (0)  16(0) 2  2 5 6 2 5 6feet
d. Set H (t ) equal to 0 and solve for t to get
 16t 2  2 5 6 0
t 2  16
t  16  4
That is, the ball will hit the ground after 4 seconds.
72. POPULATION DENSITY Observations suggest that for herbivorous mammals, the number
91.2
of animals N per square kilometer can be estimated by the formula N  0.73 , where m is the
m
average mass of the animal in kilograms.
a. Assuming that the average elk on a particular reserve has mass 300 kg, approximately how
many elk would you expect to find per square kilometer in the reserve?
b. Using this formula, it is estimated that there is less than one animal of a certain species per
square kilometer. How large can the average animal of this species be?
c. One species of large mammal has twice the average mass as a second species. If a particular
reserve contains 100 animals of the large species, how many animals of the smaller species
would you expect to find there?
Solution
a. There is approximately N 
91.2
 1.4 elk per square kilometer in the reserve.
300 0.73
4
N
b.
91.2
 1 , which means
m 0.73
91.2  m 0.73
m  0.73 91.2  484.1
So the average animal of this species should has mass at less 484.1 kg.
c. Supposed the average mass and number for large animals are m1 and N1 respectively, and
for the smaller ones are m2 and N 2 respectively. According the context, m1  2m2 and
N1  100 . The number of smaller species is
N2 
§2
91.2
91.2
91.2

 0.73  20.73  N1  20.73  100  20.73  165.9
0.73
0.73
(m1 / 2)
m2
m1
53, 68,
53. AIR POLLUTION Lead emissions are a major source of air pollution. Using data gathered
by the U.S. Environmental Protection Agency in the 1990s, it can be shown that the formula
N (t )  35t 2  2 9 t9 3 3 4 7
estimates the total amount of lead emission N (in thousands of tons) occurring in the United
States t years after the base year 1990.
a. Sketch the graph of the pollution function N(t).
b. Approximately how much lead emission did the formula predict for the year 1995? (The actual
amount was about 3924 thousand tons.)
c. Based on this formula, when during the decade 1990-2000 would you expect the maximum
lead emission to have occurred?
d. Can this formula be used to predict the current level of lead emission? Explain.
Solution
N
5,000
2,500
0
10
20
t
b. In year 1995, t=5, so the lead emission is about
5
N (5)  35  52  299  5  3347  3,967 thousand tons.
c. The maximum lead emission occurs when 
299
 4.27 years after 1990, or in March
 2 * 35
1994.
d. No, the formula predicts negative emissions after December 2004.
8x 2  9 x  3
54. Graph f ( x)  2
. Determine the values of x for which the function is defined.
x  x 1
Solution
The function is defined for all real numbers except 
§3
1
5 1
5

, 
.
2 2
2 2
22, 30, 33, 35, 49, 55.
22.Through (-1, 2) with slope 2/3.
Solution
Use the formula y  y0  m( x  x0 ) with ( x0 , y0 )  (1,2) and m  2 / 3 to get
2
( x  1)
3
2
8
y  x
3
3
y2
Or
30. Through (-2,3) and (0,5).
Solution (The Point-Slope Form of the Equation of a Line; Example 1.3.5)
The slope is
6
35
1
20
Use the point-slope formula with ( x0 , y0 )  (1,2) and m  1 to get
y 5 x
m
or
y  x5
33. Through (4,1) and parallel to the line 2 x  y  3 .
Solution (Linear Function; Example 1.3.5)
By rewriting the equation 2 x  y  3 in the slope-intercept form y  2 x  3 , we see that this
line has the slope ml  2 . A line parallel to line 2 x  y  3 must have slope m  ml  2 .
Since the required line contains point (4,1), we have
y  1  2( x  4)
y  2 x  9
or
35. Through (3,5) and perpendicular to the line x  y  4 .
Solution (Example 1.3.8)
By rewriting the equation x  y  4 in the slope-intercept form y   x  4 , we see that this
line has the slope ml  1 . A line perpendicular to line x  y  4 must have slope
m
1
 1 . Since the required line contains point (3,5), we have
ml
y 5  x3
y  x2
49. GROWTH OF A CHILD The average height H is certimeters of a child of age A years can
be estimated by the linear function H  6.5 A  50 . Use this formula to answer questions.
a. What is the average height of a 7-year-old child?
b. How old must a child be to have an average heigh of 150 cm?
c. What is tha average height of a nowborn baby? Does this answer seem reaonable?
d. What is th aaverage height of a 20-year-old? Does this answer seem reaonable?
Solution
a. The average height of a 7-year-old child is
H (7)  6.5  7  50  95.5 cm
b. Set H ( A) equal to 150 and solve for A to get
6.5 A  50  1 5 0
A  15.4
It follows that a child have an average height of 150 cm by 15.4 years old.
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c. The average height of a nowborn baby is the height when A  0 . That is
H (0)  6.5  0  50  50 cm
So, the answer is reasonable.
d. The average height of a 20-year-old is the height when A  20 . That is
H (20)  6.5  20  50  1 8 0cm
So, the answer is reasonable.
55. COLLEGE ADMISSIONS The average scores of incoming students at an eastern liberal
arts college in the SAT mathematics examination have been declining at a constant rate in recent
years. In 1995, the average SAT score was 575, while in 2000 it was 545.
a. Express the average SAT score as a function of time.
b. If the trend continues, what will the average SAT score of incoming students be in 2005?
c. If the trend continues, when will the average SAT score be 527?
Solution (Example 1.3.6)
a. Let t denote the number of year and y the average SAT score. Since y changes at a constant rate
with respect to t, the function relating y to x must be linear. The line is through the points (1995,
575) and (2000, 545), so the slope of the line will be
m
545  575
 6
2000  1995
y  6(t  1995)  575
To get
b. If the trend continues, in 2005, the average SAT score of incoming students will be
f (2005)  6(2005  1995)  575  515
c. Set y equal to 527 and solve for t to get
 6(t  1995)  575  527
t  2003
That is, the average SAT score will be 527 in 2003 if the trend continues.
§4
31, 38.
31. INCOME TAX The accompanying table represents the 2004 federal income tax rate
schedule for single taxpayers.
a. Express an individual’s income tax as a function of the taxable income x for 0  x  146750
and draw the graph.
b. The graph in part(a) should consist of four line segments. Compute the slope of each segment.
What happens to these slopes as the taxable income increase? Explain the behavior of the slopes
in practical terms.
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If the Taxable Income Is
The Income Tax Is
Over
But not Over
Of the Amout Over
0
$7,150
10%
0
$7,150
$29,050
$715+15%
$7,150
$29,050
$70,350
$4,000+25%
$29,050
$70,350
$146,750
$14,325+28%
$70,350
Solution
10% x
0  x  7150

 715  ( x  7150) 15%
7150  x  29050

a. f ( x)  
 4000  ( x  29050)  25% 29050  x  70350
14325  ( x  70350)  28% 70350  x  146750
y
x
0
b.
7150
29050
70350
146750
0.1x
0  x  7150

 0.15 x  357.5
7150  x  29050

f ( x)  
0.25 x  1237.5 29050  x  70350
 0.28 x  5373 70350  x  146750
From this function, we can find each slope is larger than the preceding one. It reflects that an
individual’s income tax rate is increasing with the income. In practical terms, the higher
income one earns, the effective tax-rate is also higher.
38. CONSTRUCTION COST An open box is to be made from a square piece of cardboard,
18 inches by 18 inches, by removing a small square from each corner and folding up the flaps to
form the sides. Express the volume of the resulting box moved squares.Draw the graph and
estimate the value of x for which the volume of the resulting box is greatest.
9
x
18
Solution
We denote x as the lengths of small square and y as the lengths of big square. Expressing the
volume of V of the resulting box moved squares in terms of these two variables, we get
V
y
2
 x Using the fact that the length of the cardboard is to be 18 inches that is
2 x  y  18
 y  18  2 x and substitute the resulting expression for y into the formula for V to get
V ( x)  (18  2 x)  x  4 x  72 x  324 x
2
3
2
Since x  0, y  0 , we find that 0  x  9
The graph of this function is shown the greatest volume of the resulting box is x=3.
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