,Solution manual for Applied Calculus 7e Hughes-Hallett
Hello all ,
We have all what you need with best price
Our email :
Our website :
testbanks-store.com
, 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.
, 1.1 SOLUTIONS 5
28. (a) Since the potato is getting hotter, the temperature is increasing. The temperature of the potato eventually levels o˙ at
the temperature of the oven, so the best answer is graph (III).
(b) The vertical intercept represents the temperature of the potato at time t = 0 or just before it is put in the oven.
29. (a) The graph shows that the yield increases as more kg of seeds are planted. Moreover the yield initially increases very
rapidly but then levels o˙ at about 2.5 tons, meaning there is little beneft from continuing to increase the quantity of
seed.
(b) We see that at 80 kg the graph is almost a horizontal line, meaning that any additional seed planted will not impact
the crop yield, so it is not worth the cost of planting additional seed.
30. (a) f (30) = 10 means that the value of f at t = 30 was 10. In other words, the temperature at time t = 30 minutes was
10 C. So, 30 minutes after the object was placed outside, it had cooled to 10 C.
(b) The intercept a measures the value of f (t) when t = 0. In other words, when the object was initially put outside, it
had a temperature of a C. The intercept b measures the value of t when f (t) = 0. In other words, at time b the object’s
temperature is 0 C.
31. (a) Since N = f (ℎ), the statement f (500) = 100, means that if ℎ = 500, then N = 100. This tells us that at an elevation
of 500 feet above sea level, there are 100 species of bats.
(b) The vertical intercept k is on the N-axis, so it represents the value of N when ℎ = 0. The intercept, k, is the number
of bat species at sea level. The horizontal intercept, c, represents the value of ℎ when N = 0. The intercept, c, is the
lowest elevation above which no bats are found.
32. The number of species of algae is low when there are few snails or lots of snails. The greatest number of species of algae
(about 10) occurs when the number of snails is at a medium level (around 125 snails per square meter.) The graph supports
the statement that diversity peaks at intermediate predation levels, when there are an intermediate number of snails.
33. (a) From the graph, we estimate f (3) = 0.14. This means that after 3 hours, the level of nicotine is about 0.14 mg.
(b) About 4 hours.
(c) The vertical intercept is 0.4. It represents the level of nicotine in the blood right after the cigarette is smoked.
(d) A horizontal intercept would represent the value of t when N = 0, or the number of hours until all nicotine is gone
from the body.
34. (a) The original deposit is the balance, B, when t = 0, which is the vertical intercept. The original deposit was $1000.
(b) It appears that f (10) 2200. The balance in the account after 10 years is about $2200.
(c) When B = 5000, it appears that t 20. It takes about 20 years for the balance in the account to reach $5000.
35. (a) The statement f (10) = 0.2 means that C = 0.2 when t = 10. In other words, CFC consumption was 0.2 million tons
in 1997.
(b) The vertical intercept is the value of C when t = 0. It represents CFC consumption in 1987.
(c) The horizontal intercept is the value of t when C = 0. It represents the year when CFC consumption is expected to be
zero.
36. (a) The graph shows maximum range is about 78 miles and it occurs at approximately 65 F.
(b) The graph is decreasing for temperatures above 65 F, so above 65 F, every increase in the temperature will reduce
the range.
37. (a) We see that the defcit is decreasing during this time period.
(b) The statement f (4) = 485 means that the defcit was 485 billion dollars in 2014.
(c) We can estimate the location of 4 on the horizontal axis, since the graph covers 2011–2015, corresponding to 1 ≤ t ≤ 5.
The value 485 on the vertical axis is then the corresponding y-coordinate. These values, and a dot at the corresponding
point, are shown in Figure 1.7.
billion $
D = f (t)
485
t (years since 2010)
4
Figure 1.7
,6 Chapter One /SOLUTIONS
(d) A vertical intercept represents the value of D when t = 0, which is the US defcit (in billions of dollars) in the year
2010.
(e) A horizontal intercept represents the value of t when D = 0, or the number of years after 2010 when the defcit is
zero: the horizontal intercept occurs at a moment when the defcit is zero.
38. See Figure 1.8.
heart
rate
time
administration
of drug
Figure 1.8
39. Figure 1.9 shows one possible graph of gas mileage increasing to a high at a speed of 45 mph and then decreasing again.
Other graphs are also possible.
miles per gallon
speed
45 (mph)
Figure 1.9
40. One possible graph is shown in Figure 1.10. Since the patient has none of the drug in the body before the injection,
the vertical intercept is 0. The peak concentration is labeled on the concentration (vertical) axis and the time until peak
concentration is labeled on the time (horizontal) axis. See Figure 1.10.
concentration
Peak
concen-
tration
time
Time of
peak concentration
Figure 1.10
41. (a) We want a graph of expected return as a function of risk, so expected return is on the vertical axis and risk is on the
horizontal axis. The return increases as the risk increases, so the graph might look like Figure 1.11. However, the
graph may not be a straight line; many other answers are also possible.
, 1.1 SOLUTIONS 7
expected
return
risk
Figure 1.11
(b) High expected return and low risk is in the top left corner. See Figure 1.11.
42. (a) The statement f (1000) = 3500 means that when a = 1000, we have S = 3500. In other words, when $1000 is spent
on advertising, the number of sales per month is 3500.
(b) Graph I, because we expect that as advertising expenditures go up, sales will go up (not down).
(c) The vertical intercept represents the value of S when a = 0, or the sales per month if no money is spent on advertising.
43. (a) In 1970, fertilizer use in the US was about 13 million tons, in India about 2 million tons, and in the former Soviet
Union about 10 million tons.
(b) Fertilizer use in the US rose steadily between 1950 and 1980 and has stayed relatively constant at about 18 million
tons since then. Fertilizer use in India has risen steadily throughout this 50-year period. Fertilizer use in the former
Soviet Union rose rapidly between 1950 and 1985 and then declined very rapidly between 1985 and 2000.
44. Since P (t) is the total cases, it can’t decrease. So (I) is the graph of the function N(t), and (II) is the graph of P (t).
45. (a) P (100) is the total number of confrmed cases of the disease up to day 100, which is the last day of the outbreak, so
it is the total number of people who contract the disease during the outbreak. Therefore P (100) = 20,000.
N(100) is the number of new cases on day 100. Since the outbreak is over on day 100 there are no new cases, so
N(100) = 0.
(b) Since the outbreak is over on day 100, there are no new cases after day 100, so P (t) = P (100) for t > 100.
46. (a) On January 8, 2021, there were 1419 new Covid-19 cases reported in Iowa.
(b) On January 9, 2021, there were 898 fewer Covid-19 cases reported than on January 8, 2021.
(c) On January 10 there were 581 more cases reported than on January 9.
47. (a) On February 3, 2021, there were 2296 new Covid-19 cases reported in Arizona.
(b) On February 4, 2021, there were 2121 more Covid-19 cases reported than on February 3, 2021.
(c) On February 5 there were 591 fewer cases reported than on February 4.
48. (a) (I) The incidence of cancer increase with age, but the rate of increase slows down slightly. The graph is nearly linear.
This type of cancer is closely related to the aging process.
(II) In this case a peak is reached at about age 55, after which the incidence decreases.
(III) This type of cancer has an increased incidence until the age of about 48, then a slight decrease, followed by a
gradual increase.
(IV) In this case the incidence rises steeply until the age of 30, after which it levels out completely.
(V) This type of cancer is relatively frequent in young children, and its incidence increases gradually from about the
age of 20.
(VI) This type of cancer is not age-related – all age-groups are equally vulnerable, although the overall incidence is
low (assuming each graph has the same vertical scale).
(b) Graph (V) shows a relatively high incidence rate for children. Leukemia behaves in this way.
(c) Graph (III) could represent cancer in women with menopause as a signifcant factor. Breast cancer is a possibility
here.
(d) Graph (I) shows a cancer which might be caused by toxins building up in the body. Lung cancer is a good example of
this.
49. (a) Since 2014 corresponds to t = 0, the average annual sea level in Aberdeen in 2014 was 7.071 meters.
(b) Looking at the table, we see that the average annual sea level was 7.083 twenty fve years before 2014, or in the year
1989. Similar reasoning shows that the average sea level was 6.985 meters 100 years before 2014, or in 1914.
(c) Because 1889 was 125 years before 2014, we see that the sea level value corresponding to the year 1889 is 6.900 (this
is the sea level value corresponding to t = 125). Similar reasoning yields Table 1.1.
, 8 Chapter One /SOLUTIONS
Table 1.1
Year 1889 1914 1939 1964 1989 2014
S 6.900 6.985 6.964 6.990 7.083 7.071
50. (a) When the temperature is 14 F the graph is at 60%, so on average an electrical vehicle will achieve 60% of its advertised
range.
(b) The temperatures that give 80% of advertised range are approximately 32 F and 104 F.
(c) If the vehicle has a shorter range than the advertised range, then the percentage of the advertised range achieved will
be less than 100%. All temperatures where the percentage is less than 100% correspond to temperatures where the
vehicle will not achieve its advertised range. From the graph we see that these correspond to temperatures below 50 F
and above approximately 90 F.
51. See Figure 1.12.
driving speed
time
Figure 1.12
52. See Figure 1.13.
distance driven
time
Figure 1.13
53. See Figure 1.14.
distance from exit
time
Figure 1.14
54. See Figure 1.15.
distance between cars
distance driven
Figure 1.15
Hello all ,
We have all what you need with best price
Our email :
Our website :
testbanks-store.com
, 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.
, 1.1 SOLUTIONS 5
28. (a) Since the potato is getting hotter, the temperature is increasing. The temperature of the potato eventually levels o˙ at
the temperature of the oven, so the best answer is graph (III).
(b) The vertical intercept represents the temperature of the potato at time t = 0 or just before it is put in the oven.
29. (a) The graph shows that the yield increases as more kg of seeds are planted. Moreover the yield initially increases very
rapidly but then levels o˙ at about 2.5 tons, meaning there is little beneft from continuing to increase the quantity of
seed.
(b) We see that at 80 kg the graph is almost a horizontal line, meaning that any additional seed planted will not impact
the crop yield, so it is not worth the cost of planting additional seed.
30. (a) f (30) = 10 means that the value of f at t = 30 was 10. In other words, the temperature at time t = 30 minutes was
10 C. So, 30 minutes after the object was placed outside, it had cooled to 10 C.
(b) The intercept a measures the value of f (t) when t = 0. In other words, when the object was initially put outside, it
had a temperature of a C. The intercept b measures the value of t when f (t) = 0. In other words, at time b the object’s
temperature is 0 C.
31. (a) Since N = f (ℎ), the statement f (500) = 100, means that if ℎ = 500, then N = 100. This tells us that at an elevation
of 500 feet above sea level, there are 100 species of bats.
(b) The vertical intercept k is on the N-axis, so it represents the value of N when ℎ = 0. The intercept, k, is the number
of bat species at sea level. The horizontal intercept, c, represents the value of ℎ when N = 0. The intercept, c, is the
lowest elevation above which no bats are found.
32. The number of species of algae is low when there are few snails or lots of snails. The greatest number of species of algae
(about 10) occurs when the number of snails is at a medium level (around 125 snails per square meter.) The graph supports
the statement that diversity peaks at intermediate predation levels, when there are an intermediate number of snails.
33. (a) From the graph, we estimate f (3) = 0.14. This means that after 3 hours, the level of nicotine is about 0.14 mg.
(b) About 4 hours.
(c) The vertical intercept is 0.4. It represents the level of nicotine in the blood right after the cigarette is smoked.
(d) A horizontal intercept would represent the value of t when N = 0, or the number of hours until all nicotine is gone
from the body.
34. (a) The original deposit is the balance, B, when t = 0, which is the vertical intercept. The original deposit was $1000.
(b) It appears that f (10) 2200. The balance in the account after 10 years is about $2200.
(c) When B = 5000, it appears that t 20. It takes about 20 years for the balance in the account to reach $5000.
35. (a) The statement f (10) = 0.2 means that C = 0.2 when t = 10. In other words, CFC consumption was 0.2 million tons
in 1997.
(b) The vertical intercept is the value of C when t = 0. It represents CFC consumption in 1987.
(c) The horizontal intercept is the value of t when C = 0. It represents the year when CFC consumption is expected to be
zero.
36. (a) The graph shows maximum range is about 78 miles and it occurs at approximately 65 F.
(b) The graph is decreasing for temperatures above 65 F, so above 65 F, every increase in the temperature will reduce
the range.
37. (a) We see that the defcit is decreasing during this time period.
(b) The statement f (4) = 485 means that the defcit was 485 billion dollars in 2014.
(c) We can estimate the location of 4 on the horizontal axis, since the graph covers 2011–2015, corresponding to 1 ≤ t ≤ 5.
The value 485 on the vertical axis is then the corresponding y-coordinate. These values, and a dot at the corresponding
point, are shown in Figure 1.7.
billion $
D = f (t)
485
t (years since 2010)
4
Figure 1.7
,6 Chapter One /SOLUTIONS
(d) A vertical intercept represents the value of D when t = 0, which is the US defcit (in billions of dollars) in the year
2010.
(e) A horizontal intercept represents the value of t when D = 0, or the number of years after 2010 when the defcit is
zero: the horizontal intercept occurs at a moment when the defcit is zero.
38. See Figure 1.8.
heart
rate
time
administration
of drug
Figure 1.8
39. Figure 1.9 shows one possible graph of gas mileage increasing to a high at a speed of 45 mph and then decreasing again.
Other graphs are also possible.
miles per gallon
speed
45 (mph)
Figure 1.9
40. One possible graph is shown in Figure 1.10. Since the patient has none of the drug in the body before the injection,
the vertical intercept is 0. The peak concentration is labeled on the concentration (vertical) axis and the time until peak
concentration is labeled on the time (horizontal) axis. See Figure 1.10.
concentration
Peak
concen-
tration
time
Time of
peak concentration
Figure 1.10
41. (a) We want a graph of expected return as a function of risk, so expected return is on the vertical axis and risk is on the
horizontal axis. The return increases as the risk increases, so the graph might look like Figure 1.11. However, the
graph may not be a straight line; many other answers are also possible.
, 1.1 SOLUTIONS 7
expected
return
risk
Figure 1.11
(b) High expected return and low risk is in the top left corner. See Figure 1.11.
42. (a) The statement f (1000) = 3500 means that when a = 1000, we have S = 3500. In other words, when $1000 is spent
on advertising, the number of sales per month is 3500.
(b) Graph I, because we expect that as advertising expenditures go up, sales will go up (not down).
(c) The vertical intercept represents the value of S when a = 0, or the sales per month if no money is spent on advertising.
43. (a) In 1970, fertilizer use in the US was about 13 million tons, in India about 2 million tons, and in the former Soviet
Union about 10 million tons.
(b) Fertilizer use in the US rose steadily between 1950 and 1980 and has stayed relatively constant at about 18 million
tons since then. Fertilizer use in India has risen steadily throughout this 50-year period. Fertilizer use in the former
Soviet Union rose rapidly between 1950 and 1985 and then declined very rapidly between 1985 and 2000.
44. Since P (t) is the total cases, it can’t decrease. So (I) is the graph of the function N(t), and (II) is the graph of P (t).
45. (a) P (100) is the total number of confrmed cases of the disease up to day 100, which is the last day of the outbreak, so
it is the total number of people who contract the disease during the outbreak. Therefore P (100) = 20,000.
N(100) is the number of new cases on day 100. Since the outbreak is over on day 100 there are no new cases, so
N(100) = 0.
(b) Since the outbreak is over on day 100, there are no new cases after day 100, so P (t) = P (100) for t > 100.
46. (a) On January 8, 2021, there were 1419 new Covid-19 cases reported in Iowa.
(b) On January 9, 2021, there were 898 fewer Covid-19 cases reported than on January 8, 2021.
(c) On January 10 there were 581 more cases reported than on January 9.
47. (a) On February 3, 2021, there were 2296 new Covid-19 cases reported in Arizona.
(b) On February 4, 2021, there were 2121 more Covid-19 cases reported than on February 3, 2021.
(c) On February 5 there were 591 fewer cases reported than on February 4.
48. (a) (I) The incidence of cancer increase with age, but the rate of increase slows down slightly. The graph is nearly linear.
This type of cancer is closely related to the aging process.
(II) In this case a peak is reached at about age 55, after which the incidence decreases.
(III) This type of cancer has an increased incidence until the age of about 48, then a slight decrease, followed by a
gradual increase.
(IV) In this case the incidence rises steeply until the age of 30, after which it levels out completely.
(V) This type of cancer is relatively frequent in young children, and its incidence increases gradually from about the
age of 20.
(VI) This type of cancer is not age-related – all age-groups are equally vulnerable, although the overall incidence is
low (assuming each graph has the same vertical scale).
(b) Graph (V) shows a relatively high incidence rate for children. Leukemia behaves in this way.
(c) Graph (III) could represent cancer in women with menopause as a signifcant factor. Breast cancer is a possibility
here.
(d) Graph (I) shows a cancer which might be caused by toxins building up in the body. Lung cancer is a good example of
this.
49. (a) Since 2014 corresponds to t = 0, the average annual sea level in Aberdeen in 2014 was 7.071 meters.
(b) Looking at the table, we see that the average annual sea level was 7.083 twenty fve years before 2014, or in the year
1989. Similar reasoning shows that the average sea level was 6.985 meters 100 years before 2014, or in 1914.
(c) Because 1889 was 125 years before 2014, we see that the sea level value corresponding to the year 1889 is 6.900 (this
is the sea level value corresponding to t = 125). Similar reasoning yields Table 1.1.
, 8 Chapter One /SOLUTIONS
Table 1.1
Year 1889 1914 1939 1964 1989 2014
S 6.900 6.985 6.964 6.990 7.083 7.071
50. (a) When the temperature is 14 F the graph is at 60%, so on average an electrical vehicle will achieve 60% of its advertised
range.
(b) The temperatures that give 80% of advertised range are approximately 32 F and 104 F.
(c) If the vehicle has a shorter range than the advertised range, then the percentage of the advertised range achieved will
be less than 100%. All temperatures where the percentage is less than 100% correspond to temperatures where the
vehicle will not achieve its advertised range. From the graph we see that these correspond to temperatures below 50 F
and above approximately 90 F.
51. See Figure 1.12.
driving speed
time
Figure 1.12
52. See Figure 1.13.
distance driven
time
Figure 1.13
53. See Figure 1.14.
distance from exit
time
Figure 1.14
54. See Figure 1.15.
distance between cars
distance driven
Figure 1.15
