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Solution Manual For Applied Calculus 7th Edition By Deborah Hughes Hallett Andrew M. Gleason, Atti Frazer Lock, Daniel E.Flath | All Chapter (1_10)|Latest 2025

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Solution Manual For Applied Calculus 7th Edition By Deborah Hughes Hallett Andrew M. Gleason, Atti Frazer Lock, Daniel E.Flath | All Chapter (1_10)|Latest 2025

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1.1 SOLUTIONS 1


CHAPTER ONE
Solutions for Section 1.1

1. (a) The story in (a) matches Graph (IV), in which the person forgot her books and had to return home.
(b) The story in (b) matches Graph (II), the fat tire story. Note the long period of time during which the distance from
home did not change (the horizontal part).
(c) The story in (c) matches Graph (III), in which the person started calmly but sped up later.
The frst graph (I) does not match any of the given stories. In this picture, the person keeps going away from home,
but his speed decreases as time passes. So a story for this might be: I started walking to school at a good pace, but since I
stayed up all night studying calculus, I got more and more tired the farther I walked.
2. The height is going down as time goes on. A possible graph is shown in Figure 1.1. The graph is decreasing.

height




time

Figure 1.1


3. The amount of carbon dioxide is going up as time goes on. A possible graph is shown in Figure 1.2. The graph is increasing.

CO2




time

Figure 1.2

,2 Chapter One /SOLUTIONS

4. The number of air conditioning units sold is going up as temperature goes up. A possible graph is shown in Figure 1.3.
The graph is increasing.

AC units




temperature

Figure 1.3


5. The noise level is going down as distance goes up. A possible graph is shown in Figure 1.4. The graph is decreasing.

noise level




distance

Figure 1.4


6. If we let t represent the number of years since 1900, then the population increased between t = 0 and t = 40, stayed
approximately constant between t = 40 and t = 50, and decreased for t ≥ 50. Figure 1.5 shows one possible graph. Many
other answers are also possible.

population




years
20 40 60 80 100 120 since 1900

Figure 1.5

, 1.1 SOLUTIONS 3

7. Amount of grass G = f (r) increases as the amount of rainfall r increases, so f (r) is an increasing function.
8. We are given information about how atmospheric pressure P = f (ℎ) behaves when the altitude ℎ decreases: as altitude
ℎ decreases the atmospheric pressure P increases. This means that as altitude ℎ increases the atmospheric pressure P
decreases. Therefore, P = f (ℎ) is a decreasing function.
9. We are given information about how battery capacity C = f (T ) behaves when air temperature T decreases: as T decreases,
battery capacity C also decreases. This means that increasing temperature T increases battery capacity C. Therefore,
C = f (T ) is an increasing function.
10. Time T = f (m) increases as m increases, so f (m) is increasing.
11. The attendance A = f (P ) decreases as the price P increases, so f (P ) is a decreasing function.
12. The cost of manufacturing C = f (v) increases as the number of vehicles manufactured v increases, so f (v) is an increasing
function.
13. We are given information about how commuting time, T = f (c) behaves as the number of cars on the road c decreases:
as c decreases, commuting time T also decreases. This means that increasing the number of cars on the road c increases
commuting time T . Therefore, T = f (c) is an increasing function.
14. The statement f (4) = 20 tells us that W = 20 when t = 4. In other words, in 2019, Argentina produced 20 million metric
tons of wheat.
15. (a) If we consider the equation
C = 4T − 160
simply as a mathematical relationship between two variables C and T , any T value is possible. However, if we think
of it as a relationship between cricket chirps and temperature, then C cannot be less than 0. Since C = 0 leads to
0 = 4T − 160, and so T = 40 F, we see that T cannot be less than 40 F. In addition, we are told that the function is
not defned for temperatures above 134 . Thus, for the function C = f (T ) we have

Domain = All T values between 40 F and 134 F
= All T values with 40 ≤ T ≤ 134
= [40, 134].


(b) Again, if we consider C = 4T − 160 simply as a mathematical relationship, its range is all real C values. However,
when thinking of the meaning of C = f (T ) for crickets, we see that the function predicts cricket chirps per minute
between 0 (at T = 40 F) and 376 (at T = 134 F). Hence,

Range = All C values from 0 to 376
= All C values with 0 ≤ C ≤ 376
= [0, 376].


16. (a) The statement f (19) = 415 means that C = 415 when t = 19. In other words, in the year 2019, the concentration of
carbon dioxide in the atmosphere was 415 ppm.
(b) The expression f (22) represents the concentration of carbon dioxide in the year 2022.
17. (a) At p = 0, we see r = 8. At p = 3, we see r = 7.
(b) When p = 2, we see r = 10. Thus, f (2) = 10.
18. Substituting x = 5 into f (x) = 2x + 3 gives

f (5) = 2(5) + 3 = 10 + 3 = 13.

19. Substituting x = 5 into f (x) = 10x − x2 gives

f (5) = 10(5) − (5)2 = 50 − 25 = 25.

20. We want the y-coordinate of the graph at the point where its x-coordinate is 5. Looking at the graph, we see that the
y-coordinate of this point is 3. Thus
f (5) = 3.

, 4 Chapter One /SOLUTIONS

21. Looking at the graph, we see that the point on the graph with an x-coordinate of 5 has a y-coordinate of 2. Thus

f (5) = 2.

22. In the table, we must fnd the value of f (x) when x = 5. Looking at the table, we see that when x = 5 we have

f (5) = 4.1

23. (a) We are asked for the value of y when x is zero. That is, we are asked for f (0). Plugging in we get

f (0) = (0)2 + 2 = 0 + 2 = 2.

(b) Substituting we get
f (3) = (3)2 + 2 = 9 + 2 = 11.
(c) Asking what values of x give a y-value of 11 is the same as solving

y = 11 = x2 + 2
x2 = 9

x = ± 9 = ±3.

We can also solve this problem graphically. Looking at Figure 1.6, we see that the graph of f (x) intersects the
line y = 11 at x = 3 and x = −3. Thus, when x equals 3 or x equals −3 we have f (x) = 11.

y
f (x)




11

x
−6 −3 3 6

Figure 1.6


(d) No. No matter what, x2 is greater than or equal to 0, so y = x2 + 2 is greater than or equal to 2.
24. The year 2018 was 2 years before 2020 so 2018 corresponds to t = 2. Thus, an expression that represents the statement is:

f (2) = 7.088.


25. The year 2020 was 0 years before 2020 so 2020 corresponds to t = 0. Thus, an expression that represents the statement is:

f (0) meters.

26. The year 1949 was 2020 − 1949 = 71 years before 2020 so 1949 corresponds to t = 71. Similarly, we see that the year
2000 corresponds to t = 20. Thus, an expression that represents the statement is:

f (71) = f (20).


27. The year 2018 was 2 years before 2020 so 2018 corresponds to t = 2. Similarly, t = 3 corresponds to the year 2017. Thus,
f (3) and f (2) are the average annual sea level values, in meters, in 2017 and 2018, respectively. Because 11 millimeters
is the same as 0.011 meters, the average sea level in 2018, f (2), is 0.011 less than the sea level in 2017 which is f (3). An
expression that represents the statement is:
f (2) = f (3) − 0.011.
Note that there are other possible equivalent expressions, such as: f (2) − f (3) = −0.011.
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