Sample Test Answers for Chapter 5 1 SAMPLE TEST ANSWERS FOR CHAPTER 5 1. (a) The following graph is illustrative. Answer Key 1. (a) Since the population is believed to be growing exponentially, we calculate the regression line for the natural logarithm of the data: ln N = 0.139t + 2.996. Since r is the slope of this regression line, r = 0.139. (b) The doubling time t rd = = = ln ln . . 2 2 0 139 4 99 days. (c) The data in the table doubles exactly every 5 days. 2. (a) Answers will vary. The following graph is illustrative. (b) In practical terms, N(5) is the number of inhabitants (in thousands) 5 years after the state was settled. N(5) = 47.24 thousand inhabitants. (c) i. The carrying capacity for the population in this environment is the limiting value for N. Using a graph or table, we see that the limiting value is 62 thousand inhabitants. ii. The population growing the fastest when the population equals one-half of the carrying capacity (since it is logistic). Using the graph, N = = 62 2 31 when t = 3.32 years after the state was settled. iii. The graph is concave up from t = 0 until t = 3.32. In practical terms, the population was increasing at an increasing rate during that interval. 3. (a) Answers will vary. The following graph is illustrative. Page 1 (b) In practical terms, N (5) is the number of inhabitants (in thousands) 5 years after the state was settled. N (5) = 67.2 0.1 + 0.55 = 47.24 thousand inhabitants. (c) i. The carrying capacity for the population in this environment is the limiting value for N . Using a graph or table, we see that the limiting value is 62 thou- sand inhabitants. ii. The population growing the fastest when the population equals one-half of the carrying capacity (since it is logistic). Using the graph, N = 62 2 = 31, when t = 3.32 years after the state was settled. iii. The graph is concave up from t = 0 until t = 3.32. In practical terms, the population was increasing at an increasing rate during that interval. 2. (a) The following graph is illustrative (b) According to a graph or table, the carrying capacity for the population is 184 million people. Since the current population is over 280 million, the model does not seem accurate. (c) The U.S. population was growing the fastest when it reaches one-half of 184 million, so when N = 92. This occurred when t = 140, so in the year 1780 + 140 = 1920. (d) If another logistic model has the population growing the fastest at 150 million, then the carrying capacity for the population under this new model is twice 150, so 300 million people. 4. The graph below shows the survivorship curve. Since the graph is more or less a straight line, it is a Type II curve and represents a nearly constant mortality rate. Page 2 (b) According to a graph or table, the carrying capacity for the population is 184 mil- lion people. Since the current population is over 300 million, the model does not seem acc.

1. Sample Test Answers for Chapter 5 1
SAMPLE TEST ANSWERS FOR CHAPTER 5
1. (a) The following graph is illustrative.
Answer Key
1. (a) Since the population is believed to be growing
exponentially, we calculate the
regression line for the natural logarithm of the data: ln N =
0.139t + 2.996. Since r is the
slope of this regression line, r = 0.139.
(b) The doubling time
t
rd
= = =
ln ln
.
.
2 2
0 139
4 99
days.

2. (c) The data in the table doubles exactly every 5 days.
2. (a) Answers will vary. The following graph is illustrative.
(b) In practical terms, N(5) is the number of inhabitants (in
thousands) 5 years after the
state was settled. N(5) = 47.24 thousand inhabitants.
(c) i. The carrying capacity for the population in this
environment is the limiting value
for N. Using a graph or table, we see that the limiting value is
62 thousand inhabitants.
ii. The population growing the fastest when the population
equals one-half of the
carrying capacity (since it is logistic). Using the graph,
N = =
62
2
31
when t = 3.32
years after the state was settled.
iii. The graph is concave up from t = 0 until t = 3.32. In
practical terms, the population
was increasing at an increasing rate during that interval.
3. (a) Answers will vary. The following graph is illustrative.
Page 1
(b) In practical terms, N (5) is the number of inhabitants (in
thousands) 5 years after

3. the state was settled. N (5) =
67.2
0.1 + 0.55
= 47.24 thousand inhabitants.
(c) i. The carrying capacity for the population in this
environment is the limiting
value for N . Using a graph or table, we see that the limiting
value is 62 thou-
sand inhabitants.
ii. The population growing the fastest when the population
equals one-half of
the carrying capacity (since it is logistic). Using the graph, N =
62
2
= 31,
when t = 3.32 years after the state was settled.
iii. The graph is concave up from t = 0 until t = 3.32. In
practical terms, the
population was increasing at an increasing rate during that
interval.
2. (a) The following graph is illustrative
(b) According to a graph or table, the carrying capacity for the

4. population is 184 million
people. Since the current population is over 280 million, the
model does not seem
accurate.
(c) The U.S. population was growing the fastest when it reaches
one-half of 184 million,
so when N = 92. This occurred when t = 140, so in the year
1780 + 140 = 1920.
(d) If another logistic model has the population growing the
fastest at 150 million, then
the carrying capacity for the population under this new model is
twice 150, so 300
million people.
4. The graph below shows the survivorship curve. Since the
graph is more or less a straight
line, it is a Type II curve and represents a nearly constant
mortality rate.
Page 2
(b) According to a graph or table, the carrying capacity for the
population is 184 mil-
lion people. Since the current population is over 300 million,
the model does not
seem accurate.
2 Sample Test Answers for Chapter 5
(c) The U.S. population was growing the fastest when it reaches
one-half of 184 mil-

5. lion, so when N = 92. This occured when t = 140, so in the year
1780+140 = 1920.
(d) If another logistic model has the population growing the
fastest is 150 million,
then the carrying capacity for the population under this new
model is twice 150,
so 300 million people.
3. We calculate differences to identify linear data and ratios to
identify exponential data.
Table A The ratios are all 1.16, to two decimal places, so this is
exponential data and so
f = P at. The initial data, P , is 6.7, and the growth factor, a, is
1.16, so the formula
is f = 6.7 × 1.16t.
Table B Neither differences nor ratios are constant, so we hope
this is power data. Taking
the natural logarithms of both t and g, we find the regression
line ln g = 1.160 ln t+
2.041 is a excellent fit, so it is power data for which k = 1.16
and c = e2.041 = 7.7.
The formula is g = 7.7t1.16.
Table C The differences are all 1.73, so this is linear and so h =
mt + b. The initial data, b,

6. is 5.8, and the slope, m, is 1.73, so the formula is h = 1.73t +
5.8.
4. We calculate differences to identify linear data and ratios to
identify exponential data.
(a) i. The function f is exponential since all the ratios are 1.1.
ii. Since the ratios are 1.1, a = 1.1. The initial value is 4.55,
calculated as 5/1.1, so
f = 4.55 × 1.1t.
(b) i. The function g is power. Neither differences nor ratios are
constant, so it is
neither linear nor exponential. It is power since its natural
logarithms lie on
the line ln g = 0.138 ln t + 1.609.
ii. A formula for g is ctk where k = 0.138 and c = e1.609 = 5, so
g = 5t0.138.
(c) i. The function h is linear since all the differences are 0.5.
ii. The formula for h is h = 0.5t + 4.5 since the slope is 0.5 and
the initial value is
h(1) − m = 5 − 0.5 = 4.5.
5. (a) Since the tax is a linear function of income with slope
0.12, for each additional
dollar of income, the tax you owe increases by 0.12 dollar. If

7. your income increases
by $150, you will owe 150 × 0.12 = 18 dollars more tax.
(b) The decade growth factor is 1.0510 = 1.63, so that is the
factor be which the circu-
lation will increase over a decade.
Sample Test Answers for Chapter 5 3
(c) If one bird is twice as long as another, then it flies 20.3 =
1.23 times as fast, since
the speed is a power function with power k = 0.3.
6. Since weight W is a power function of its length L, W = cLk
for some values of c and
k.
(a) If one lizard is twice as long as a second and the first weighs
3 times as much as
the second, then 3 = 2k. Solving, we find that the power is k =
1.58.
(b) If one lizard is 4 times as long as a second, then their
weights are related by a factor
of 4k. Using the value of k from Part (a), the longer lizard is 4k
= 41.58 = 8.94 times
as heavy as the shorter.

8. (c) If one lizard weighs twice as much as a second, then their
lengths are related by a
factor of t where 2 = tk. Using the value of k from Part (a), 2 =
t1.58. Solving, we
see that the heavier lizard is t = 1.55 times as long as the
lighter.
7. Since terminal velocity T is proportional to the square root of
length L, T is a power
function of L with power k = 0.5.
(a) The length of a 4-foot monkey is 4/3 = 1.33 times that of a
3-foot monkey, so the
terminal velocity of a 4-foot monkey is 1.33k = 1.330.5 = 1.15
times that of a 3-foot
monkey.
(b) Since the terminal velocity of a 4-foot monkey is twice that
of a certain smaller
mammal with a similar shape, the length of the monkey is t
times that of the
smaller mammal where 2 = tk = t0.5, so t = 4. Since the monkey
is 4 feet long, the
smaller mammal is 4/4 = 1 foot long.
(c) For some c, T = cL0.5. Since the terminal velocity of a 4-
foot monkey is about 98

9. miles per hour, 98 = c × 40.5, so c = 49 and so the formula is T
= 49L0.5. For the
cat, L = 1, so T = 49. This also follows immediately from Part
(b).
8. Since the distance D is a power function of the period P , D =
cP k.
(a) If the period of one satellite is twice that of another, then its
distance is 2k times as
far, so if the distance is 1.59 times, then 2k = 1.59 and so the
power is k = 0.67.
(b) If the period of one satellite is 3 times that of a second, then
their distances from
the center of the Earth differ by a factor of 3k. Using the value
of k from Part (a),
they differ by a factor of 30.67 = 2.09.
(c) If the distance from the center of the Earth of one satellite is
twice that of a second,
then their periods differ by a factor of t where tk = 2. Using the
value of k from
Part (a), t0.67 = 2 and so t = 2.81. Their periods differ by a
factor of 2.81.
4 Sample Test Answers for Chapter 5

10. 9. (a) To find a formula that models W as a power function of
L, we take the natural
logarithms of L and W and find the linear regression line ln W =
3.092 ln L−5.094.
Thus W = cLk where k = 3.09 and c = e−5.094 = 0.0061, so W =
0.0061L3.09.
(b) If one fish is twice as long as another, then their weights
differ by a factor of 2k =
23.09 = 8.51.
(c) If one fish is twice as heavy as another, then their lengths
differ by a factor of t
where 2 = tk = t3.09, so a factor of t = 1.25.
10. (a) The graph is show below. (The horizontal span is 1.45 to
3.38 and the vertical span
is −0.24 to 6.28.) Since the graph of ln P versus ln v is very
close to linear, it is
reasonable to model P as a power function of v.
5. Since terminal velocity T is proportional to the square root of
length L, T is a power
function of L with power k = 0.5.
(a) The length of a 4-foot monkey is 4/3 times that of a 3-foot
monkey, so the terminal
velocity of a 4-foot monkey is (4/3)k = (4/3)0.5 = 1.15 times

11. that of a 3-foot monkey.
(b) Since the terminal velocity of a 4-foot monkey is twice that
of a certain smaller
mammal with a similar shape, the length of the monkey is t
times that of the smaller
mammal where 2 = tk = t0.5. Thus t = 4. Since the monkey is 4
feet long, the smaller
mammal is 4/4 = 1 foot long.
(c) For some c, we have T = cL0.5. Since the terminal velocity
of a 4-foot monkey is
about 98 miles per hour, 98 = c × 40.5, so c = 49. Thus the
formula is T = 49L0.5. For the
cat, L = 1, so T = 49. This also follows immediately from Part
(b).
6. Since the distance D is a power function of the period P, we
have D = cPk.
(a) If the period of one satellite is twice that of another, then its
distance is 2k times as
far; so, if the distance is 1.59 times, then 2k = 1.59. Thus the
power is k = 0.67.
(b) If the period of one satellite is 3 times that of a second, then
their distances from the
center of the Earth differ by a factor of 3k. Using the value of k
from Part (a), they differ
by a factor of 30.67 = 2.09.
(c) If the distance from the center of the Earth of one satellite is
twice that of a second,
then their periods differ by a factor of t where tk = 2. Using the
value of k from Part (a),
we have t0.67 = 2, and so t = 2.81. Their periods differ by a
factor of 2.81.
7. (a) To find a formula that models W as a power function of
L, we take the natural

12. logarithms of L and W and find the linear regression line ln W =
3.092 ln L – 5.094.
Thus W = cLk where k = 3.09 and c = e–5.094 = 0.0061, so W =
0.0061L3.09
(b) If one fish is twice as long as another, then their weights
differ by a factor of 2k =
23.09 = 8.51.
(c) If one fish is twice as heavy as another, then their lengths
differ by a factor of t
where 2 = tk = t3.09, so a factor of t = 1.25.
8. (a) The graph is shown below. (The horizontal span is 1.45 to
3.38 and the vertical span
is from –0.24 to 6.28.) Since the graph of ln P versus lnυ is
very close to linear, it is
reasonable to model P as a power function of υ.
(b) Using linear regression on the natural logarithms, we find
that
ln . ln . .P = −3 024 4 269υ A power model of P as a function
of υ is P = cvk where k =
3.024 and c = e–4.269 = 0.014, so the formula is P = 0 014 3 02.
..υ
(c) If υ = 35, then P = 0.014 × 353.02 = 644.49 watts.
Page 2
(b) Using linear regression on the natural logarithms, we find
that ln P = 3.024 ln v −
4.269. A power model of P as a function of v is P = cvk where k
= 3.024 and

13. c = e−4.269 = 0.014, so the formula is P = 0.014v3.02.
(c) If v = 35, then P = 0.014 × 353.02 = 644.49 watts.
(d) To generate 41 watts, we need a wind velocity v for which
41 = P , that is, 41 =
0.014v3.02. Solving, we find that v = 14.06 miles per hour.
(e) If wind speed increases by a factor of 3, then the power
generated increases by a
factor of 3k = 33.02 = 27.60.
11. (a) Since S = 4πr2, √
S
4π
=
√
4πr2
4π
=
√
r2 = r.
(b) Using the formula for V and the expression for r from Part
(a),
V =
4
3
πr3 =

14. 4
3
π
(√
S
4π
)3
.
12. (a) Since W is proportional to the cube of L, W = cL3.
Sample Test Answers for Chapter 5 5
(b) Using the formula for W from Part (a) and the given
expression for L, function
composition gives
W = cL3 = c(8(1 − e−t))3,
which is a formula expressing weight as a function of time.
(c) Since W = 17, when t = 3, the formula from Part (b) gives 17
= c(8(1 − e−3))3,
and so c = 17/((8(1 − e−3))3) = 0.039.
13. (a) The limiting value is the maximum length, 13 inches.
(b) Since L = 13 − D and L(0) = 0.5, the initial value of D is

15. D(0) = 13 − L(0) =
13 − 0.5 = 12.5.
(c) D is an exponential function with initial value 12.5, so D =
12.5at for some growth
or decay factor a, and so L = 13 − D = 13 − 12.5at. Now L = 8
when t = 3, so
8 = 13 − 12.5a3. Solving, we find that a = 0.74, so D = 12.5 ×
0.74t, which is a
formula for D in terms of t.
(d) To find a formula for L in terms of t, we use our formulas
above: L = 13 − D =
13 − 12.5 × 0.74t.
(e) Since W = 0.01L3 and L = 13−D = 13−12.5×0.74t, function
composition gives
W = 0.01L3 = 0.01(13 − 12.5 × 0.74t)3.
14. (a) The limiting value of R is the total voting-age
population, which is 40,000.
(b) By Part (a), R = 40,000 − D, so R(0) = 40,000 − D(0). Since
the initial value of R
is 28,000, we have 28,000 = 40,000 − D(0) and so the initial
value of D is 12,000.
(c) Since D is an exponential function with initial value 12,000,
D = 12,000at for some

16. growth or decay factor a. When t = 1, R = 33,000 so D = 7,000,
and therefore
7,000 = D = 12,000a1. Solving, a = 0.58 and so D = 12,000 ×
0.58t.
(d) Since R = 40,000−D and D = 12,000×0.58t, we use function
composition to find
a formula for R in terms of t:
R = 40,000 − D = 40,000 − 12,000 × 0.58t.
15. To determine if data are quadratic, we see if the second-
order differences are constant.
Table A The first-order differences are −2, 4, 12, and 16 and so
the second-order differences
are 6, 8 , and 4, so the data in Table A are not quadratic.
6 Sample Test Answers for Chapter 5
Table B The first-order differences are −2, 4, 10, and 16 and so
the second-order differences
are 6, 6 , and 6, so the data in Table A are quadratic, moreover
the second-order
difference is 6, the initial first-order difference is −2, and the
initial value is 2.
The formula is g = ax2 + bx + c where a = 1

17. 2
× 6 = 3, b = (−2) − 3 = −5, and
c = 2, so the final formula is g = 3x2 − 5x + 2.
16. (a) Using quadratic regression, the equation of the parabola
approximately followed
by the cannonball is h = −0.00066d2 + 0.80d + 0.027.
(b) Using the equation from Part (a), 0 = h = −0.00066d2 +
0.80d + 0.027 where
d = 1212.16, so the cannonball will strike the ground 1212.16
feet downrange.
(c) i. Since the slope of inclination of the cannon, s is the
coefficient of x, the analo-
gous number in the formula from Part (a) is 0.80, so the slope s
= 0.8.
ii. Comparing the coefficient of x2 and the analogous number in
the formula
from Part (a), −16
1 + s2
v20
= −0.00066. Since s = 0.8, −16
1 + 0.82
v20
= −0.00066.

18. Solving, we find an initial velocity of v0 = 199.39 feet per
second.
17. (a) A quadratic model for profit as a function of advertising
expenditure is, using
regression, P = −0.035A2 + 46.04A + 221.17.
(b) Graphing the profit function from Part (a), we see that the
maximum profit is
P = 15,361.75 dollars and occurs when A = 657.71 is spent on
advertising.
18. (a) Graphing the profit function, we see that its maximum
occurs when a = 250.00
dollars per month is spent on advertising.
(b) P = 5000 when 5000 = 1 + 50a − 0.1a2. Solving, we find
that a = 138.15 or
a = 361.85. Since we’d rather pay less than more, we should
spend $138.15 per
month on advertising.
(c) The graph of profit is decreasing when a = 350, so profit
will increase as the man-
ufacturer spends less on advertising. Hence the manufacturer
should decrease the
amount spent on advertising.
19. (a) Using the quadratic formula, the solutions are

19. −k ±
√
k2 − 4k
2
.
(b) The equation has exactly one solution when k2 − 4k = 0,
that is, when k = 0 or
k = 4.
Sample Test Answers for Chapter 5 7
(c) There are two real roots when k2−4k > 0. Graphing k2−4k,
we see that k2−4k > 0
when k < 0 or k > 4.
20. (a) The graph is shown below using a horizontal span of 0 to
5 and a vertical span of
0 to 11.
formula is g = ax2 + bx + c where a = × =
1
2 6 3, b = (–2) – 3 = –5, and c = 2, so the final
formula is g = 3x2 – 5x + 2.
14. (a) Using quadratic regression, we find that the equation of
the parabola approximately

20. followed by the cannonball is h = –0.00066d2 + 0.80d + 0.027.
(b) Using the equation from Part (a), we have 0 = h = –
0.00066d2 + 0.80d + 0.027. We
find the solution d = 1212.16, so the cannonball will strike the
ground 1212.16 feet
downrange.
(c) i. Since the slope of inclination of the cannon, s, is the
coefficient of x, the analogous
number in the formula from Part (a) is 0.80, so the slope s =
0.8.
ii. Comparing the coefficient of x2 and the analogous number in
the formula from Part
(a), we have
−
+
= −16
1
0 00066
2
0
2
s
υ
. .
Since s = 0.8, we have
−
+

21. = −16
1 0 8
0 00066
2
0
2
.
. .
υ
Solving, we find an initial velocity of υ 0 199 39= . feet per
second.
15. (a) A quadratic model for profit as a function of advertising
expenditure is, using
regression, P = –0.035A2 + 46.04A + 221.17.
(b) Graphing the profit function from Part (a), we see that the
maximum profit is P =
15,361.75 dollars and occurs when A = 657.71 is spent on
advertising.
16. (a) Graphing the profit function, we see that its maximum
occurs when a = 250.00
dollars per month is spent on advertising.
(b) P = 5000 when 5000 = 1 + 50a – 0.1a2. Solving, we find that
a = 138.15 or a =
361.85. Since we'd rather pay less than more, we should spend
$138.15 per month on
advertising.
(c) The graph of profit is decreasing when a = 350, so profit
will increase as the
manufacturer spends less on advertising. Hence the

22. manufacturer should decrease the
amount paid on advertising.
17. (a) Using the quadratic formula, we find that the solutions
are
− ± −k k k2 4
2
.
(b) The equation has exactly one solution when k – 4k = 0, that
is, when k = 0 or k = 4. 2
(c) There are two real roots when k2 – 4k > 0. Graphing k2 –
4k, we see that k2 – 4k > 0
when k < 0 or k > 4.
18.
(a)
(b) As D increases, it's easier to catch the prey, so more are
eaten, so the graph is
increasing; on the other hand, the predator will only eat so
many prey, so even if the
Page 4
(b) As D increases, it’s easier to catch the prey, so more are
eaten, so the graph is
increasing; on the other hand, the predator will only eat so
many prey, so even
if the prey are very easy to catch, the additional number eaten
will be fewer and

23. fewer.
(c) Using the graph or a table, we see that the horizontal
asymptote is P = 10.67 prey
eaten per day.
(d) The physical significance of the horizontal asymptote is that
it indicates that, on
average, the amount consumed by predators will increase to a
level of 10.67 prey
per day as the density of prey increases.
21. (a) F = k
1
d2
.
(b) Near the pole at d = 0, F increases without bound.
(c) In practical terms, the meaning of the pole at d = 0 is that as
the two bodies get
closer and closer together, the gravitational attraction increases
without bound.
22. (a) Using regression, we find that a quartic model for
patronage as a function of time
since 10 a.m. is p = −0.068t4 + 1.644t3 − 12.226t2 + 29.220t +
4.118.

24. (b) Graphing p versus t and the line p = 10, we see that the
graph dips below the
p = 10 line between t = 4.49 and t = 6.97, that is, between about
2:30 p.m. and 5
p.m. (since t = 0 is 10 a.m.).
(c) The graph of p versus t has peaks at t = 1.76 and t = 10.62,
so the peak business
times are at about 11:45 a.m. and 8:30 p.m.
Sample Test Questions for Chapter 5 1
SAMPLE TEST QUESTIONS FOR CHAPTER 5
1. The number N (in thousands) of inhabitants in the state of
Gamma is a function of time
t, measured in years since the state was settled. The formula is
N =
6.2
0.1 + 0.5t
.
(a) Make a graph of N versus t and draw that graph on your
paper. Include times up
to 10 years, and be sure to indicate the window you use.

25. (b) Explain in practical terms what N(5) means, and then
calculate it.
(c) As you can see from the graph in Part (a), this population
growth is logistic. Use
your graph or a table of values to answer the following
questions.
i. What is the carrying capacity for the population in this
environment? (Note:
The carrying capacity does not equal the numerator, 6.2, in the
formula for N ,
since this formula is not in standard form for a logistic function.
Instead, use
your graph or a table to find the carrying capacity.)
ii. At what population size N is the population growing the
fastest? Explain how
you got your answer.
iii. What portion of the graph is concave up? Explain in
practical terms what this
means.
2. In an attempt to predict the growth of the population of the
U.S., biologists studied
census records from 1790 through 1940. They developed the
logistic formula

26. N =
184
1 + 66.7e−0.03t
.
Here N is the U.S. population in millions and t is time
(measured in years since 1780).
(a) Make a graph of N versus t and draw that graph on your
paper. Include times up
to 250 years (corresponding to dates up to 2030).
(b) Use your graph or a table of values to determine the
carrying capacity for the
population. Does this make the model look accurate?
(c) According to this model, on what date was the U.S.
population growing the fastest?
(d) Suppose another biologist has proposed an alternative
logistic formula to model
the U.S. population. Under this model, the population level at
which the popu-
lation grows the fastest is 150 million. What would the carrying
capacity for the
population be under this new model?
2 Sample Test Questions for Chapter 5

27. 3. The tables below show linear, exponential, or power data.
Determine which is which,
explain your reasoning, and write a formula for the function in
each case.
Table A
t 0 1 2 3 4 5
f(t) 6.7 7.77 9.02 10.46 12.13 14.07
Table B
t 1 2 3 4 5
g(t) 7.7 17.2 27.54 38.45 49.81
Table C
t 0 1 2 3 4 5
h(t) 5.8 7.53 9.26 10.99 12.72 14.45
4. The tables below show data modeled by a linear function, an
exponential function, and
a power function.
(a)
t 1 2 3 4
f(t) 5 5.5 6.05 6.66
i. What type of function is f : linear, exponential, or power?
Why?
ii. Give a formula for f .

28. (b)
t 1 2 3 4
g(t) 5 5.5 5.82 6.05
i. What type of function is g: linear, exponential, or power?
Why?
ii. Give a formula for g.
(c)
t 1 2 3 4
h(t) 5 5.5 6 6.5
i. What type of function is h: linear, exponential, or power?
Why?
ii. Give a formula for h.
Sample Test Questions for Chapter 5 3
5. (a) The tax you owe (in dollars) is a linear function of your
taxable income (in dollars),
and the slope of this function is 0.12. If your income increases
by $150, how much
more tax will you owe?
(b) The circulation of a certain magazine is an exponential
function of time, with yearly

29. growth factor 1.05. By what factor will the circulation increase
over a decade?
(c) The speed S at which a bird can fly is a power function of its
length L, and the
power is k = 0.3. If one bird is twice as long as another, how
much faster can it
fly?
6. The weight W in ounces for a certain species of lizard is a
power function of its length
L in inches.
(a) If one lizard is twice as long as a second, then the first
weighs 3 times as much as
the second. What is the value of the power k in the relationship
between W and
L?
(b) If one lizard is 4 times as long as a second, how do their
weights compare?
(c) If one lizard weighs twice as much as a second, how do their
lengths compare?
7. For similarly shaped objects, terminal velocity T is
proportional to the square root of
length L.
(a) How does the terminal velocity of a 4-foot monkey compare

30. with that of a 3-foot
monkey?
(b) The terminal velocity of a 4-foot monkey is twice that of a
certain smaller mammal
with a similar shape. How long is this smaller mammal?
(c) Assume that the terminal velocity of a 4-foot monkey is
about 98 miles per hour.
What is the terminal velocity of a one-foot (neglecting the tail)
cat?
4 Sample Test Questions for Chapter 5
8. For a satellite orbiting the earth, the distance D from the
center of the Earth is a power
function of the period P . Let k denote the power.
(a) If the period of one satellite is twice that of another, its
distance from the center of
the Earth is 1.59 times larger. Find the value of the power, k.
(b) If the period of one satellite is 3 times that of a second, how
do their distances from
the center of the Earth compare?
(c) If the distance from the center of the Earth of one satellite is
twice that of a second,

31. how do their periods compare?
9. The table below shows the relationship between the length L
in centimeters and weight
W in grams of a certain species of fish.
L 31 33 35 37 39
W 250 308 363 420 520
(a) Find a formula that models W as a power function of L.
(b) If one fish is twice as long as another, how do their weights
compare?
(c) If one fish is twice as heavy as another, how do their lengths
compare?
10. The following table gives the power P in watts generated by
a windmill with winds
blowing v miles per hour.
v 5 10 15 20 25
P 1.8 15 50.7 120 234
(a) Plot the graph of lnP versus ln v. Is it reasonable to model P
as a power function
of v? Explain your reasoning.
(b) Find a model of P as a power function of v.
(c) What power is generated by 35 mile per hour winds?

32. (d) How fast must the wind blow in order to generate 41 watts
of power?
(e) If wind speed increases by a factor of 3, how much more
power is generated?
Sample Test Questions for Chapter 5 5
11. The volume V of a sphere (ball) of radius r is given by
V =
4
3
πr3.
The surface area S is given by
S = 4πr2.
(a) Show that the radius can be expressed as a function of
surface area by the formula
r =
√
S
4π
.
(b) Use function composition to find a formula expressing
volume as a function of
surface area.

33. 12. The weight W in ounces of a certain small mammal is
proportional to the cube of its
length L in inches.
(a) Express W as a function of L using c as the constant of
proportionality.
(b) The length in inches of the animal depends on its age t in
years. The relationship
is as follows:
L = 8(1− e−t).
Use function composition to find a formula expressing weight
as a function of
time.
(c) It is found that a 3-year-old animal weighs 17 ounces. Find
the value of c.
6 Sample Test Questions for Chapter 5
13. For a certain species of fish it is found that length L is a
function of age t. The maximum
length of this species is 13 inches. The youngest fish of this
species are 0.5 inch long. If
D denotes the difference between maximum length and current
length, then

34. L = Limiting value −D.
(a) What is the limiting value?
(b) It is found that D is an exponential function of t. What is the
initial value of D?
(c) It is found that a 3-year-old fish of this species is 8 inches
long. Find a formula for
D in terms of t.
(d) Find a formula for L in terms of t.
(e) It is found that the weight W in ounces of this species of
fish is given by W =
0.01L3. Use function composition to find a formula expressing
weight as a func-
tion of age.
14. You begin a voter registration drive in a town with a voting-
age population of 40,000.
Let D denote the difference between the voting-age population
and voters registered.
The total number R of registrations is given by
R = Limiting value −D.
(a) What is the limiting value of R?
(b) It is found that D is an exponential function of the number t

35. of months since the
drive began. There were 28,000 registered voters when the drive
began. What is
the initial value of D?
(c) After one month, you find that a total of 33,000 voters are
registered. Find a for-
mula for D in terms of t.
(d) Find a formula for R in terms of t
Sample Test Questions for Chapter 5 7
15. One of the two tables below shows data that can be modeled
by a quadratic function,
and the other shows data that cannot be modeled by a quadratic
function. Identify
which is which, and find a model for the quadratic data.
Table A
x 0 1 2 3 4
f(x) 2 0 4 16 32
Table B
x 0 1 2 3 4
g(x) 2 0 4 14 30
16. A cannonball is fired from a cannon. Its height h in feet is
measured d feet downrange

36. and is recorded in the table below.
d 0 300 500 800 1000
h 0 181 236 220 144
(a) We know that the cannonball should follow the path of a
parabola. Use quadratic
regression to find a parabola that is approximately followed by
the cannonball.
(b) How far downrange will the cannonball strike the ground?
(c) If the cannon has a slope of inclination s and the initial
velocity is v0 feet per
second, then the path followed by the cannonball is the graph of
−161 + s
2
v20
x2 + sx.
i. Based on your answer to Part (a), what is the slope of
inclination of the can-
non?
ii. Based on your answer to Part (a), what is the initial velocity
of the cannonball?
8 Sample Test Questions for Chapter 5

37. 17. A manufacturer has recorded the profit P (in dollars) when
there is a monthly adver-
tising expenditure of A dollars. The data is recorded in the
following table.
A 200 500 800 1100 1300
P 8030 14,480 14,630 8480 860
(a) Find a quadratic model for profit as a function of advertising
expenditure.
(b) What advertising expenditure gives a maximum profit, and
what is that profit?
18. A manufacturer has determined that profit P (in dollars) is a
quadratic function of
dollars per month a spent on advertising. The relationship is
given by
P = 1 + 50a− 0.1a2
if a is at most 500 dollars per month.
(a) How much should be spent on advertising if profit is to be a
maximum?
(b) How much should be spent on advertising if a profit of
$5000 is desired?
(c) The manufacturer is spending $350 per month on
advertising. Should the manu-
facturer increase or decrease that amount?

38. 19. Consider the quadratic x2 + kx + k = 0.
(a) Solve the equation using the quadratic formula.
(b) What condition on k will assure that the equation has
exactly one solution? Note:
This occurs when the expression under the square root sign is
zero.
(c) What condition on k will assure that there are two real
roots? Note: This occurs
when the expression under the square root sign is positive.
Sample Test Questions for Chapter 5 9
20. For a certain predator population, the number P of prey
eaten per day depends on the
density D of the prey (measured as number per square foot). The
relationship is given
by
P =
32D
1 + 3D
.
(a) Make a graph of P as a function of D covering values of D
up to 5 per square foot.

39. (b) Explain why it is reasonable that the graph is increasing and
concave down.
(c) Find the equation of the horizontal asymptote.
(d) What is the physical significance of the horizontal
asymptote?
21. Newton’s law of gravity states that the gravitational
attraction F between two bodies
is proportional to 1 over the square of the distance d between
their centers.
(a) Using k as the constant of proportionality, express F as a
function of d.
(b) What happens to F near the pole at d = 0?
(c) Explain in practical terms the meaning of the pole at d = 0.
22. The number p of patrons in a restaurant should normally
reach its peak over the lunch
and dinner hours. Thus it may be appropriate to model p as a
quartic function of time.
Typical patronage of a certain restaurant is recorded below.
Here t is hours since 10
a.m.
t 0 2 3 5 8 10 11
p 5 23 20 14 9 48 34

40. (a) Make a quartic model for patronage as a function of time
since 10 a.m.
(b) You can run your restaurant with fewer staff when the
patronage is 10 or less. At
what times in the afternoon do you need less staff?
(c) What are the peak business times?