, Solution manual For Calculus Single and
Multivariable, 8th Edition Deborah Hughes-
Hallett
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, 1.1 SOLUTIONS 1
CHAPTER ONE
Solutions for Section 1.1
EXERCISES
1. Since 𝑡 represents the number of years since 2020, we see that 𝑓 (5) represents the population of the city in 2025. In 2025,
the city’s population is 8 million.
2. Since 𝑇 = 𝑓 (𝑃 ), we see that 𝑓 (200) is the value of 𝑇 when 𝑃 = 200; that is, the thickness of pelican eggs when the
concentration of PCBs is 200 ppm.
3. If there are no workers, there is no productivity, so the graph goes through the origin. At first, as the number of workers
increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest
level, after which it goes downward; in other words, as the number of workers increases beyond that point, productivity
decreases. This might, for example, be due either to the inefficiency inherent in large organizations or simply to workers
getting in each other’s way as too many are crammed on the same line. Many other reasons are possible.
4. (a) is (V), because slope is positive, vertical intercept is negative
(b) is (IV), because slope is negative, vertical intercept is positive
(c) is (I), because slope is 0, vertical intercept is positive
(d) is (VI), because slope and vertical intercept are both negative
(e) is (II), because slope and vertical intercept are both positive
(f) is (III), because slope is positive, vertical intercept is 0
5. (a) is (V), because slope is negative, vertical intercept is 0
(b) is (VI), because slope and vertical intercept are both positive
(c) is (I), because slope is negative, vertical intercept is positive
(d) is (IV), because slope is positive, vertical intercept is negative
(e) is (III), because slope and vertical intercept are both negative
(f) is (II), because slope is positive, vertical intercept is 0
6. The slope is (1 − 0)∕(1 − 0) = 1. So the equation of the line is 𝑦 = 𝑥.
7. The slope is (3 − 2)∕(2 − 0) = 1∕2. So the equation of the line is 𝑦 = (1∕2)𝑥 + 2.
8. The slope is
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Now we know that 𝑦 = (1∕2)𝑥 + 𝑏. Using the point (−2, 1), we have 1 = −2∕2 + 𝑏, which yields 𝑏 = 2. Thus, the equation
of the line is 𝑦 = (1∕2)𝑥 + 2.
6−0
9. The slope is = 2 so the equation of the line is 𝑦 − 6 = 2(𝑥 − 2) or 𝑦 = 2𝑥 + 2.
2 − (−1)
5 5
10. Rewriting the equation as 𝑦 = − 𝑥 + 4 shows that the slope is − and the vertical intercept is 4.
2 2
11. Rewriting the equation as
12 2
𝑦=− 𝑥+
7 7
shows that the line has slope −12∕7 and vertical intercept 2∕7.
12. Rewriting the equation of the line as
−2
−𝑦 = 𝑥−2
4
1
𝑦 = 𝑥 + 2,
2
we see the line has slope 1∕2 and vertical intercept 2.
,2 Chapter One /SOLUTIONS
13. Rewriting the equation of the line as
12 4
𝑦= 𝑥−
6 6
2
𝑦 = 2𝑥 − ,
3
we see that the line has slope 2 and vertical intercept −2∕3.
14. The intercepts appear to be (0, 3) and (7.5, 0), giving
−3 6 2
Slope = =− =− .
7.5 15 5
The 𝑦-intercept is at (0, 3), so a possible equation for the line is
2
𝑦 = − 𝑥 + 3.
5
(Answers may vary.)
15. 𝑦 − 𝑐 = 𝑚(𝑥 − 𝑎)
16. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then
29.2 − 27.8 1.4
Slope = = = 14.
5.3 − 5.2 0.1
Using the point-slope formula, with the point (5.2, 27.8), we get the equation
𝑦 − 27.8 = 14(𝑥 − 5.2)
which is equivalent to
𝑦 = 14𝑥 − 45.
17. 𝑦 = 5𝑥 − 3. Since the slope of this line is 5, we want a line with slope − 15 passing through the point (2, 1). The equation is
(𝑦 − 1) = − 15 (𝑥 − 2), or 𝑦 = − 15 𝑥 + 75 .
18. The line 𝑦 + 4𝑥 = 7 has slope −4. Therefore the parallel line has slope −4 and equation 𝑦 − 5 = −4(𝑥 − 1) or 𝑦 = −4𝑥 + 9.
−1
The perpendicular line has slope (−4) = 14 and equation 𝑦 − 5 = 14 (𝑥 − 1) or 𝑦 = 0.25𝑥 + 4.75.
19. The line parallel to 𝑦 = 𝑚𝑥 + 𝑐 also has slope 𝑚, so its equation is
𝑦 = 𝑚(𝑥 − 𝑎) + 𝑏.
The line perpendicular to 𝑦 = 𝑚𝑥 + 𝑐 has slope −1∕𝑚, so its equation will be
1
𝑦 = − (𝑥 − 𝑎) + 𝑏.
𝑚
20. Since the function goes from 𝑥 = 0 to 𝑥 = 4 and between 𝑦 = 0 and 𝑦 = 2, the domain is 0 ≤ 𝑥 ≤ 4 and the range is
0 ≤ 𝑦 ≤ 2.
21. Since 𝑥 goes from 1 to 5 and 𝑦 goes from 1 to 6, the domain is 1 ≤ 𝑥 ≤ 5 and the range is 1 ≤ 𝑦 ≤ 6.
22. Since the function goes from 𝑥 = −2 to 𝑥 = 2 and from 𝑦 = −2 to 𝑦 = 2, the domain is −2 ≤ 𝑥 ≤ 2 and the range is
−2 ≤ 𝑦 ≤ 2.
23. Since the function goes from 𝑥 = 0 to 𝑥 = 5 and between 𝑦 = 0 and 𝑦 = 4, the domain is 0 ≤ 𝑥 ≤ 5 and the range is
0 ≤ 𝑦 ≤ 4.
24. The domain is all numbers. The range is all numbers ≥ 2, since 𝑥2 ≥ 0 for all 𝑥.
1
25. The domain is all 𝑥-values, as the denominator is never zero. The range is 0 < 𝑦 ≤ .
2
, 1.1 SOLUTIONS 3
26. The value of 𝑓 (𝑡) is real provided 𝑡2 − 16 ≥ 0 or 𝑡2 ≥ 16. This occurs when either 𝑡 ≥ 4, or 𝑡 ≤ −4. Solving 𝑓 (𝑡) = 3, we
have
√
𝑡2 − 16 = 3
𝑡2 − 16 = 9
𝑡2 = 25
so
𝑡 = ±5.
4 3
27. We have 𝑉 = 𝑘𝑟3 . You may know that 𝑉 = 𝜋𝑟 .
3
𝑑
28. If distance is 𝑑, then 𝑣 = .
𝑡
29. For some constant 𝑘, we have 𝑆 = 𝑘ℎ2 .
30. We know that 𝐸 is proportional to 𝑣3 , so 𝐸 = 𝑘𝑣3 , for some constant 𝑘.
31. We know that 𝑁 is proportional to 1∕𝑙2 , so
𝑘
𝑁= , for some constant 𝑘.
𝑙2
PROBLEMS
32. (a) Each date, 𝑡, has a unique daily snowfall, 𝑆, associated with it. So snowfall is a function of date.
(b) On December 12, the snowfall was approximately 5 inches.
(c) On December 11, the snowfall was above 10 inches.
(d) Looking at the graph we see that the largest increase in the snowfall was between December 10 and December 11.
33. (a) When the car is 5 years old, it is worth $6000.
(b) Since the value of the car decreases as the car gets older, this is a decreasing function. A possible graph is in Figure 1.1:
𝑉 (thousand dollars)
(5, 6)
𝑎 (years)
Figure 1.1
(c) The vertical intercept is the value of 𝑉 when 𝑎 = 0, or the value of the car when it is new. The horizontal intercept is
the value of 𝑎 when 𝑉 = 0, or the age of the car when it is worth nothing.
34. (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 flat 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 first 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.
35. 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.
,4 Chapter One /SOLUTIONS
36. 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.
37. 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).
38. 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.
39. See Figure 1.2.
driving speed
time
Figure 1.2
40. See Figure 1.3.
distance driven
time
Figure 1.3
41. See Figure 1.4.
distance from exit
time
Figure 1.4
, 1.1 SOLUTIONS 5
42. See Figure 1.5.
distance between cars
distance driven
Figure 1.5
43. (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.
44. (a) The height of the rock decreases as time passes, so the graph falls as you move from left to right. One possibility is
shown in Figure 1.6.
𝑠 (meters)
𝑡 (sec)
Figure 1.6
(b) The statement 𝑓 (7) = 12 tells us that 7 seconds after the rock is dropped, it is 12 meters above the ground.
(c) The vertical intercept is the value of 𝑠 when 𝑡 = 0; that is, the height from which the rock is dropped. The horizontal
intercept is the value of 𝑡 when 𝑠 = 0; that is, the time it takes for the rock to hit the ground.
45. See Figure 1.7.
Distance from
Kalamazoo
155
120
Time
start in arrive in arrive in
Chicago Kalamazoo Detroit
Figure 1.7
46. (a) We select two points on the line. Using the values (1992, 6.75) and (2006, 12.25) we have
12.25 − 6.75 million barrels per day
Slope = = 0.4 .
2006 − 1992 year
(b) With 𝑡 as the year and 𝑓 (𝑡) as the quantity of imports in millions of barrels per day, we have
𝑓 (𝑡) = 6.75 + 0.4(𝑡 − 1992).
Alternatively, if 𝑡 is the number of years since 1992 and if 𝑓 (𝑡) is the quantity of imports in millions of barrels
per day
𝑓 (𝑡) = 6.75 + 0.4𝑡.
,6 Chapter One /SOLUTIONS
(c) We have
6.75 + 0.4𝑡 = 18
𝑡 = 28.125.
The model predicts imports will reach 18 million barrels a day in the year 1992 + 28.125 = 2020.125.
This prediction could serve as a guideline, but it is very risky to put too much reliance on a prediction more than
ten years into the future, especially since there has been a drop in imports since 2008. Many unexpected events could
drastically change the economic environment during that time. This prediction should be used with caution.
47. (a) Reading coordinates from the graph, we see that rainfall 𝑟 = 100 mm corresponds to about 𝑄 = 600 kg∕hectare, and
𝑟 = 600 mm corresponds to about 𝑄 = 5800 kg∕hectare. Using a difference quotient we have
Δ𝑄 5800 − 600
Slope = = = 10.4 kg∕hectare per mm.
Δ𝑟 600 − 100
(b) Every additional 1 mm of annual rainfall corresponds to an additional 10.4 kg of grass per hectare.
(c) Using the slope, we see that the equation has the form
𝑄 = 𝑏 + 10.4𝑟.
Substituting 𝑟 = 100 and 𝑄 = 600 we can solve for 𝑏.
𝑏 + 10.4(100) = 600
𝑏 = −440.
The equation of the line is
𝑄 = −440 + 10.4𝑟.
48. (a) Reading coordinates from the graph, we see that rainfall 𝑟 = 100 mm corresponds to about 𝑄 = 300 kg∕hectare, and
𝑟 = 600 mm corresponds to about 𝑄 = 3200 kg∕hectare. Using a difference quotient we have
Δ𝑄 3200 − 300
Slope = = = 5.8 kg∕hectare per mm.
Δ𝑟 600 − 100
(b) Every additional 1 mm of annual rainfall corresponds to an additional 5.8 kg of grass per hectare.
(c) Using the slope, we see that the equation has the form
𝑄 = 𝑏 + 5.8𝑟.
Substituting 𝑟 = 100 and 𝑄 = 300 we can solve for 𝑏.
𝑏 + 5.8(100) = 300
𝑏 = −280.
The equation of the line is
𝑄 = −280 + 5.8𝑟.
49. The difference quotient Δ𝑄∕Δ𝑟 equals the slope of the line and represents the increase in the quantity of grass per millimeter
of rainfall. We see from the graph that the slope of the line for 1939 is larger than the slope of the line for 1997. Thus, each
additional 1 mm of rainfall in 1939 led to a larger increase in the quantity of grass than in 1997.
50. (a) Substituting into the function 𝑔(𝑡):
𝑔(10) = 300 + 1.4 ⋅ (10) = 314 ppm
𝑔(20) = 300 + 1.4 ⋅ (20) = 328 ppm
𝑔(100) = 300 + 1.4 ⋅ (100) = 440 ppm
(b) Substituting into ℎ(𝑡):
ℎ(10) = 300 + 1.5 ⋅ (10) = 315 ppm
ℎ(20) = 300 + 1.5 ⋅ (20) = 330 ppm
ℎ(100) = 300 + 1.5 ⋅ (100) = 450 ppm
(c) They differ in slope, 1.4 vs. 1.5 ppm per year. This creates a larger difference between 𝑔(𝑡) and ℎ(𝑡) for larger values
of 𝑡:
For 𝑡 = 10, the difference is ℎ(10) − 𝑔(10) = 315 − 314 = 1 ppm.
For 𝑡 = 100, the difference is ℎ(100) − 𝑔(100) = 450 − 440 = 10 ppm.
, 1.1 SOLUTIONS 7
51. (a) Let 𝑥 be the average minimum daily temperature (◦ C), and let 𝑦 be the date the marmot is first sighted. Reading
coordinates from the graph, we see that temperature 𝑥 = 12◦ C corresponds to date 𝑦 = 137 days after Jan 1, and
𝑥 = 22◦ C corresponds to 𝑦 = 109 days. Using a difference quotient, we have
Δ𝑦 109 − 137 days
Slope = = = −2.8 ◦ .
Δ𝑥 22 − 12 C
(b) The slope is negative. An increase in the temperatures corresponds to an earlier sighting of a marmot. Marmots come
out of hibernation earlier in years with warmer average daily minimum temperature.
(c) We have
Δ𝑦 = Slope × Δ𝑥 = (−2.8)(6) = 16.8 days.
◦
If temperatures are 6 C higher, then marmots come out of hibernations about 17 days earlier.
(d) Using the slope, we see that the equation has the form
𝑦 = 𝑏 − 2.8𝑥.
Substituting 𝑥 = 12 and 𝑦 = 137 we solve for 𝑏.
𝑏 − 2.8(12) = 137
𝑏 = 171.
The equation of the line is
𝑦 = 171 − 2.8𝑥.
52. (a) If the first date of bare ground is 140, then, according to the figure, the first bluebell flower is sighted about day 150,
that is 150 − 140 = 10 days later.
(b) Let 𝑥 be the first date of bare ground, and let 𝑦 be the date the first bluebell flower is sighted. Reading coordinates
from the graph, we see that date 𝑥 = 130 days after Jan 1 corresponds to date 𝑦 = 142 days after Jan 1, and 𝑥 = 170
days corresponds to 𝑦 = 173 days. Using a difference quotient we have
Δ𝑦 173 − 142
Slope = = = 0.775 days per day.
Δ𝑥 170 − 130
(c) The slope is positive, so an increase in the 𝑥-variable corresponds to an increase in the 𝑦-variable. These variables
represents days in the year, and larger values indicate days that are later in the year. Thus if bare ground first occurs
later in the year, then bluebells first flower later in the year. Positive slope means that bluebells flower later when the
snow cover lasts longer.
(d) Using the slope, we see that the equation has the form
𝑦 = 𝑏 + 0.775𝑥.
Substituting 𝑥 = 130 and 𝑦 = 142 we can solve for 𝑏.
𝑏 + 0.775(130) = 142
𝑏 = 41.25.
The equation of the line is
𝑦 = 41.25 + 0.775𝑥.
256
53. (a) We have 𝑚 = = 14.222 cm per hour. When the snow started, there were 100 cm on the ground, so
18
𝑓 (𝑡) = 100 + 14.222𝑡.
(b) The domain of 𝑓 is 0 ≤ 𝑡 ≤ 18 hours. The range is 100 ≤ 𝑓 (𝑡) ≤ 356 cm.
54. (a) We find the slope 𝑚 and intercept 𝑏 in the linear equation 𝐶 = 𝑏 + 𝑚𝑤. To find the slope 𝑚, we use
Δ𝐶 1098.32 − 694.55
𝑚= = = 134.59 dollars per ton.
Δ𝑤 8−5
We substitute to find 𝑏:
𝐶 = 𝑏 + 𝑚𝑤
694.55 = 𝑏 + (134.59)(5)
𝑏 = 21.6 dollars.
The linear formula is 𝐶 = 21.6 + 134.59𝑤.
(b) The slope is 134.59 dollars per ton. Each additional ton of waste collected costs $134.59.
(c) The intercept is $21.60. The flat fee for waste collection is $21.60. This is the amount charged even if there is no waste.
, 8 Chapter One /SOLUTIONS
55. We are looking for a linear function 𝑦 = 𝑓 (𝑥) that, given a time 𝑥 in years, gives a value 𝑦 in dollars for the value of the
refrigerator. We know that when 𝑥 = 0, that is, when the refrigerator is new, 𝑦 = 950, and when 𝑥 = 7, the refrigerator is
worthless, so 𝑦 = 0. Thus (0, 950) and (7, 0) are on the line that we are looking for. The slope is then given by
950
𝑚=
−7
It is negative, indicating that the value decreases as time passes. Having found the slope, we can take the point (7, 0) and
use the point-slope formula:
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ).
So,
950
𝑦−0 = − (𝑥 − 7)
7
950
𝑦=− 𝑥 + 950.
7
Δ$ 55 − 40
56. (a) Charge per cubic foot = = = $0.025/cu. ft.
Δ cu. ft. 1600 − 1000
Alternatively, if we let 𝑐 = cost, 𝑤 = cubic feet of water, 𝑏 = fixed charge, and 𝑚 = cost/cubic feet, we obtain
𝑐 = 𝑏 + 𝑚𝑤. Substituting the information given in the problem, we have
40 = 𝑏 + 1000𝑚
55 = 𝑏 + 1600𝑚.
Subtracting the first equation from the second yields 15 = 600𝑚, so 𝑚 = 0.025.
(b) The equation is 𝑐 = 𝑏 + 0.025𝑤, so 40 = 𝑏 + 0.025(1000), which yields 𝑏 = 15. Thus the equation is 𝑐 = 15 + 0.025𝑤.
(c) We need to solve the equation 100 = 15 + 0.025𝑤, which yields 𝑤 = 3400. It costs $100 to use 3400 cubic feet of
water.
57. (a) The first company’s price for a day’s rental with 𝑚 miles on it is 𝐶1 (𝑚) = 40 + 0.15𝑚. Its competitor’s price for a day’s
rental with 𝑚 miles on it is 𝐶2 (𝑚) = 50 + 0.10𝑚.
(b) See Figure 1.8.
𝐶 (cost in dollars)
𝐶1 (𝑚) = 40 + 0.15𝑚
150
100
50 𝐶2 (𝑚) = 50 + 0.10𝑚
0 𝑚 (miles)
200 400 600 800
Figure 1.8
(c) To find which company is cheaper, we need to determine where the two lines intersect. We let 𝐶1 = 𝐶2 , and thus
40 + 0.15𝑚 = 50 + 0.10𝑚
0.05𝑚 = 10
𝑚 = 200.
If you are going more than 200 miles a day, the competitor is cheaper. If you are going less than 200 miles a day, the
first company is cheaper.
58. (a) We find the slope 𝑚 and intercept 𝑏 in the linear equation 𝑆 = 𝑏 + 𝑚𝑡. To find the slope 𝑚, we use
Δ𝑆 66 − 113
𝑚= = = −0.94.
Δ𝑡 50 − 0
When 𝑡 = 0, we have 𝑆 = 113, so the intercept 𝑏 is 113. The linear formula is
𝑆 = 113 − 0.94𝑡.
(b) We use the formula 𝑆 = 113 − 0.94𝑡. When 𝑆 = 20, we have 20 = 113 − 0.94𝑡 and so 𝑡 = 98.9. If this linear model
were correct, the average male sperm count would drop below the fertility level during the year 2038.
Multivariable, 8th Edition Deborah Hughes-
Hallett
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, 1.1 SOLUTIONS 1
CHAPTER ONE
Solutions for Section 1.1
EXERCISES
1. Since 𝑡 represents the number of years since 2020, we see that 𝑓 (5) represents the population of the city in 2025. In 2025,
the city’s population is 8 million.
2. Since 𝑇 = 𝑓 (𝑃 ), we see that 𝑓 (200) is the value of 𝑇 when 𝑃 = 200; that is, the thickness of pelican eggs when the
concentration of PCBs is 200 ppm.
3. If there are no workers, there is no productivity, so the graph goes through the origin. At first, as the number of workers
increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest
level, after which it goes downward; in other words, as the number of workers increases beyond that point, productivity
decreases. This might, for example, be due either to the inefficiency inherent in large organizations or simply to workers
getting in each other’s way as too many are crammed on the same line. Many other reasons are possible.
4. (a) is (V), because slope is positive, vertical intercept is negative
(b) is (IV), because slope is negative, vertical intercept is positive
(c) is (I), because slope is 0, vertical intercept is positive
(d) is (VI), because slope and vertical intercept are both negative
(e) is (II), because slope and vertical intercept are both positive
(f) is (III), because slope is positive, vertical intercept is 0
5. (a) is (V), because slope is negative, vertical intercept is 0
(b) is (VI), because slope and vertical intercept are both positive
(c) is (I), because slope is negative, vertical intercept is positive
(d) is (IV), because slope is positive, vertical intercept is negative
(e) is (III), because slope and vertical intercept are both negative
(f) is (II), because slope is positive, vertical intercept is 0
6. The slope is (1 − 0)∕(1 − 0) = 1. So the equation of the line is 𝑦 = 𝑥.
7. The slope is (3 − 2)∕(2 − 0) = 1∕2. So the equation of the line is 𝑦 = (1∕2)𝑥 + 2.
8. The slope is
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Now we know that 𝑦 = (1∕2)𝑥 + 𝑏. Using the point (−2, 1), we have 1 = −2∕2 + 𝑏, which yields 𝑏 = 2. Thus, the equation
of the line is 𝑦 = (1∕2)𝑥 + 2.
6−0
9. The slope is = 2 so the equation of the line is 𝑦 − 6 = 2(𝑥 − 2) or 𝑦 = 2𝑥 + 2.
2 − (−1)
5 5
10. Rewriting the equation as 𝑦 = − 𝑥 + 4 shows that the slope is − and the vertical intercept is 4.
2 2
11. Rewriting the equation as
12 2
𝑦=− 𝑥+
7 7
shows that the line has slope −12∕7 and vertical intercept 2∕7.
12. Rewriting the equation of the line as
−2
−𝑦 = 𝑥−2
4
1
𝑦 = 𝑥 + 2,
2
we see the line has slope 1∕2 and vertical intercept 2.
,2 Chapter One /SOLUTIONS
13. Rewriting the equation of the line as
12 4
𝑦= 𝑥−
6 6
2
𝑦 = 2𝑥 − ,
3
we see that the line has slope 2 and vertical intercept −2∕3.
14. The intercepts appear to be (0, 3) and (7.5, 0), giving
−3 6 2
Slope = =− =− .
7.5 15 5
The 𝑦-intercept is at (0, 3), so a possible equation for the line is
2
𝑦 = − 𝑥 + 3.
5
(Answers may vary.)
15. 𝑦 − 𝑐 = 𝑚(𝑥 − 𝑎)
16. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then
29.2 − 27.8 1.4
Slope = = = 14.
5.3 − 5.2 0.1
Using the point-slope formula, with the point (5.2, 27.8), we get the equation
𝑦 − 27.8 = 14(𝑥 − 5.2)
which is equivalent to
𝑦 = 14𝑥 − 45.
17. 𝑦 = 5𝑥 − 3. Since the slope of this line is 5, we want a line with slope − 15 passing through the point (2, 1). The equation is
(𝑦 − 1) = − 15 (𝑥 − 2), or 𝑦 = − 15 𝑥 + 75 .
18. The line 𝑦 + 4𝑥 = 7 has slope −4. Therefore the parallel line has slope −4 and equation 𝑦 − 5 = −4(𝑥 − 1) or 𝑦 = −4𝑥 + 9.
−1
The perpendicular line has slope (−4) = 14 and equation 𝑦 − 5 = 14 (𝑥 − 1) or 𝑦 = 0.25𝑥 + 4.75.
19. The line parallel to 𝑦 = 𝑚𝑥 + 𝑐 also has slope 𝑚, so its equation is
𝑦 = 𝑚(𝑥 − 𝑎) + 𝑏.
The line perpendicular to 𝑦 = 𝑚𝑥 + 𝑐 has slope −1∕𝑚, so its equation will be
1
𝑦 = − (𝑥 − 𝑎) + 𝑏.
𝑚
20. Since the function goes from 𝑥 = 0 to 𝑥 = 4 and between 𝑦 = 0 and 𝑦 = 2, the domain is 0 ≤ 𝑥 ≤ 4 and the range is
0 ≤ 𝑦 ≤ 2.
21. Since 𝑥 goes from 1 to 5 and 𝑦 goes from 1 to 6, the domain is 1 ≤ 𝑥 ≤ 5 and the range is 1 ≤ 𝑦 ≤ 6.
22. Since the function goes from 𝑥 = −2 to 𝑥 = 2 and from 𝑦 = −2 to 𝑦 = 2, the domain is −2 ≤ 𝑥 ≤ 2 and the range is
−2 ≤ 𝑦 ≤ 2.
23. Since the function goes from 𝑥 = 0 to 𝑥 = 5 and between 𝑦 = 0 and 𝑦 = 4, the domain is 0 ≤ 𝑥 ≤ 5 and the range is
0 ≤ 𝑦 ≤ 4.
24. The domain is all numbers. The range is all numbers ≥ 2, since 𝑥2 ≥ 0 for all 𝑥.
1
25. The domain is all 𝑥-values, as the denominator is never zero. The range is 0 < 𝑦 ≤ .
2
, 1.1 SOLUTIONS 3
26. The value of 𝑓 (𝑡) is real provided 𝑡2 − 16 ≥ 0 or 𝑡2 ≥ 16. This occurs when either 𝑡 ≥ 4, or 𝑡 ≤ −4. Solving 𝑓 (𝑡) = 3, we
have
√
𝑡2 − 16 = 3
𝑡2 − 16 = 9
𝑡2 = 25
so
𝑡 = ±5.
4 3
27. We have 𝑉 = 𝑘𝑟3 . You may know that 𝑉 = 𝜋𝑟 .
3
𝑑
28. If distance is 𝑑, then 𝑣 = .
𝑡
29. For some constant 𝑘, we have 𝑆 = 𝑘ℎ2 .
30. We know that 𝐸 is proportional to 𝑣3 , so 𝐸 = 𝑘𝑣3 , for some constant 𝑘.
31. We know that 𝑁 is proportional to 1∕𝑙2 , so
𝑘
𝑁= , for some constant 𝑘.
𝑙2
PROBLEMS
32. (a) Each date, 𝑡, has a unique daily snowfall, 𝑆, associated with it. So snowfall is a function of date.
(b) On December 12, the snowfall was approximately 5 inches.
(c) On December 11, the snowfall was above 10 inches.
(d) Looking at the graph we see that the largest increase in the snowfall was between December 10 and December 11.
33. (a) When the car is 5 years old, it is worth $6000.
(b) Since the value of the car decreases as the car gets older, this is a decreasing function. A possible graph is in Figure 1.1:
𝑉 (thousand dollars)
(5, 6)
𝑎 (years)
Figure 1.1
(c) The vertical intercept is the value of 𝑉 when 𝑎 = 0, or the value of the car when it is new. The horizontal intercept is
the value of 𝑎 when 𝑉 = 0, or the age of the car when it is worth nothing.
34. (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 flat 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 first 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.
35. 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.
,4 Chapter One /SOLUTIONS
36. 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.
37. 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).
38. 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.
39. See Figure 1.2.
driving speed
time
Figure 1.2
40. See Figure 1.3.
distance driven
time
Figure 1.3
41. See Figure 1.4.
distance from exit
time
Figure 1.4
, 1.1 SOLUTIONS 5
42. See Figure 1.5.
distance between cars
distance driven
Figure 1.5
43. (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.
44. (a) The height of the rock decreases as time passes, so the graph falls as you move from left to right. One possibility is
shown in Figure 1.6.
𝑠 (meters)
𝑡 (sec)
Figure 1.6
(b) The statement 𝑓 (7) = 12 tells us that 7 seconds after the rock is dropped, it is 12 meters above the ground.
(c) The vertical intercept is the value of 𝑠 when 𝑡 = 0; that is, the height from which the rock is dropped. The horizontal
intercept is the value of 𝑡 when 𝑠 = 0; that is, the time it takes for the rock to hit the ground.
45. See Figure 1.7.
Distance from
Kalamazoo
155
120
Time
start in arrive in arrive in
Chicago Kalamazoo Detroit
Figure 1.7
46. (a) We select two points on the line. Using the values (1992, 6.75) and (2006, 12.25) we have
12.25 − 6.75 million barrels per day
Slope = = 0.4 .
2006 − 1992 year
(b) With 𝑡 as the year and 𝑓 (𝑡) as the quantity of imports in millions of barrels per day, we have
𝑓 (𝑡) = 6.75 + 0.4(𝑡 − 1992).
Alternatively, if 𝑡 is the number of years since 1992 and if 𝑓 (𝑡) is the quantity of imports in millions of barrels
per day
𝑓 (𝑡) = 6.75 + 0.4𝑡.
,6 Chapter One /SOLUTIONS
(c) We have
6.75 + 0.4𝑡 = 18
𝑡 = 28.125.
The model predicts imports will reach 18 million barrels a day in the year 1992 + 28.125 = 2020.125.
This prediction could serve as a guideline, but it is very risky to put too much reliance on a prediction more than
ten years into the future, especially since there has been a drop in imports since 2008. Many unexpected events could
drastically change the economic environment during that time. This prediction should be used with caution.
47. (a) Reading coordinates from the graph, we see that rainfall 𝑟 = 100 mm corresponds to about 𝑄 = 600 kg∕hectare, and
𝑟 = 600 mm corresponds to about 𝑄 = 5800 kg∕hectare. Using a difference quotient we have
Δ𝑄 5800 − 600
Slope = = = 10.4 kg∕hectare per mm.
Δ𝑟 600 − 100
(b) Every additional 1 mm of annual rainfall corresponds to an additional 10.4 kg of grass per hectare.
(c) Using the slope, we see that the equation has the form
𝑄 = 𝑏 + 10.4𝑟.
Substituting 𝑟 = 100 and 𝑄 = 600 we can solve for 𝑏.
𝑏 + 10.4(100) = 600
𝑏 = −440.
The equation of the line is
𝑄 = −440 + 10.4𝑟.
48. (a) Reading coordinates from the graph, we see that rainfall 𝑟 = 100 mm corresponds to about 𝑄 = 300 kg∕hectare, and
𝑟 = 600 mm corresponds to about 𝑄 = 3200 kg∕hectare. Using a difference quotient we have
Δ𝑄 3200 − 300
Slope = = = 5.8 kg∕hectare per mm.
Δ𝑟 600 − 100
(b) Every additional 1 mm of annual rainfall corresponds to an additional 5.8 kg of grass per hectare.
(c) Using the slope, we see that the equation has the form
𝑄 = 𝑏 + 5.8𝑟.
Substituting 𝑟 = 100 and 𝑄 = 300 we can solve for 𝑏.
𝑏 + 5.8(100) = 300
𝑏 = −280.
The equation of the line is
𝑄 = −280 + 5.8𝑟.
49. The difference quotient Δ𝑄∕Δ𝑟 equals the slope of the line and represents the increase in the quantity of grass per millimeter
of rainfall. We see from the graph that the slope of the line for 1939 is larger than the slope of the line for 1997. Thus, each
additional 1 mm of rainfall in 1939 led to a larger increase in the quantity of grass than in 1997.
50. (a) Substituting into the function 𝑔(𝑡):
𝑔(10) = 300 + 1.4 ⋅ (10) = 314 ppm
𝑔(20) = 300 + 1.4 ⋅ (20) = 328 ppm
𝑔(100) = 300 + 1.4 ⋅ (100) = 440 ppm
(b) Substituting into ℎ(𝑡):
ℎ(10) = 300 + 1.5 ⋅ (10) = 315 ppm
ℎ(20) = 300 + 1.5 ⋅ (20) = 330 ppm
ℎ(100) = 300 + 1.5 ⋅ (100) = 450 ppm
(c) They differ in slope, 1.4 vs. 1.5 ppm per year. This creates a larger difference between 𝑔(𝑡) and ℎ(𝑡) for larger values
of 𝑡:
For 𝑡 = 10, the difference is ℎ(10) − 𝑔(10) = 315 − 314 = 1 ppm.
For 𝑡 = 100, the difference is ℎ(100) − 𝑔(100) = 450 − 440 = 10 ppm.
, 1.1 SOLUTIONS 7
51. (a) Let 𝑥 be the average minimum daily temperature (◦ C), and let 𝑦 be the date the marmot is first sighted. Reading
coordinates from the graph, we see that temperature 𝑥 = 12◦ C corresponds to date 𝑦 = 137 days after Jan 1, and
𝑥 = 22◦ C corresponds to 𝑦 = 109 days. Using a difference quotient, we have
Δ𝑦 109 − 137 days
Slope = = = −2.8 ◦ .
Δ𝑥 22 − 12 C
(b) The slope is negative. An increase in the temperatures corresponds to an earlier sighting of a marmot. Marmots come
out of hibernation earlier in years with warmer average daily minimum temperature.
(c) We have
Δ𝑦 = Slope × Δ𝑥 = (−2.8)(6) = 16.8 days.
◦
If temperatures are 6 C higher, then marmots come out of hibernations about 17 days earlier.
(d) Using the slope, we see that the equation has the form
𝑦 = 𝑏 − 2.8𝑥.
Substituting 𝑥 = 12 and 𝑦 = 137 we solve for 𝑏.
𝑏 − 2.8(12) = 137
𝑏 = 171.
The equation of the line is
𝑦 = 171 − 2.8𝑥.
52. (a) If the first date of bare ground is 140, then, according to the figure, the first bluebell flower is sighted about day 150,
that is 150 − 140 = 10 days later.
(b) Let 𝑥 be the first date of bare ground, and let 𝑦 be the date the first bluebell flower is sighted. Reading coordinates
from the graph, we see that date 𝑥 = 130 days after Jan 1 corresponds to date 𝑦 = 142 days after Jan 1, and 𝑥 = 170
days corresponds to 𝑦 = 173 days. Using a difference quotient we have
Δ𝑦 173 − 142
Slope = = = 0.775 days per day.
Δ𝑥 170 − 130
(c) The slope is positive, so an increase in the 𝑥-variable corresponds to an increase in the 𝑦-variable. These variables
represents days in the year, and larger values indicate days that are later in the year. Thus if bare ground first occurs
later in the year, then bluebells first flower later in the year. Positive slope means that bluebells flower later when the
snow cover lasts longer.
(d) Using the slope, we see that the equation has the form
𝑦 = 𝑏 + 0.775𝑥.
Substituting 𝑥 = 130 and 𝑦 = 142 we can solve for 𝑏.
𝑏 + 0.775(130) = 142
𝑏 = 41.25.
The equation of the line is
𝑦 = 41.25 + 0.775𝑥.
256
53. (a) We have 𝑚 = = 14.222 cm per hour. When the snow started, there were 100 cm on the ground, so
18
𝑓 (𝑡) = 100 + 14.222𝑡.
(b) The domain of 𝑓 is 0 ≤ 𝑡 ≤ 18 hours. The range is 100 ≤ 𝑓 (𝑡) ≤ 356 cm.
54. (a) We find the slope 𝑚 and intercept 𝑏 in the linear equation 𝐶 = 𝑏 + 𝑚𝑤. To find the slope 𝑚, we use
Δ𝐶 1098.32 − 694.55
𝑚= = = 134.59 dollars per ton.
Δ𝑤 8−5
We substitute to find 𝑏:
𝐶 = 𝑏 + 𝑚𝑤
694.55 = 𝑏 + (134.59)(5)
𝑏 = 21.6 dollars.
The linear formula is 𝐶 = 21.6 + 134.59𝑤.
(b) The slope is 134.59 dollars per ton. Each additional ton of waste collected costs $134.59.
(c) The intercept is $21.60. The flat fee for waste collection is $21.60. This is the amount charged even if there is no waste.
, 8 Chapter One /SOLUTIONS
55. We are looking for a linear function 𝑦 = 𝑓 (𝑥) that, given a time 𝑥 in years, gives a value 𝑦 in dollars for the value of the
refrigerator. We know that when 𝑥 = 0, that is, when the refrigerator is new, 𝑦 = 950, and when 𝑥 = 7, the refrigerator is
worthless, so 𝑦 = 0. Thus (0, 950) and (7, 0) are on the line that we are looking for. The slope is then given by
950
𝑚=
−7
It is negative, indicating that the value decreases as time passes. Having found the slope, we can take the point (7, 0) and
use the point-slope formula:
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ).
So,
950
𝑦−0 = − (𝑥 − 7)
7
950
𝑦=− 𝑥 + 950.
7
Δ$ 55 − 40
56. (a) Charge per cubic foot = = = $0.025/cu. ft.
Δ cu. ft. 1600 − 1000
Alternatively, if we let 𝑐 = cost, 𝑤 = cubic feet of water, 𝑏 = fixed charge, and 𝑚 = cost/cubic feet, we obtain
𝑐 = 𝑏 + 𝑚𝑤. Substituting the information given in the problem, we have
40 = 𝑏 + 1000𝑚
55 = 𝑏 + 1600𝑚.
Subtracting the first equation from the second yields 15 = 600𝑚, so 𝑚 = 0.025.
(b) The equation is 𝑐 = 𝑏 + 0.025𝑤, so 40 = 𝑏 + 0.025(1000), which yields 𝑏 = 15. Thus the equation is 𝑐 = 15 + 0.025𝑤.
(c) We need to solve the equation 100 = 15 + 0.025𝑤, which yields 𝑤 = 3400. It costs $100 to use 3400 cubic feet of
water.
57. (a) The first company’s price for a day’s rental with 𝑚 miles on it is 𝐶1 (𝑚) = 40 + 0.15𝑚. Its competitor’s price for a day’s
rental with 𝑚 miles on it is 𝐶2 (𝑚) = 50 + 0.10𝑚.
(b) See Figure 1.8.
𝐶 (cost in dollars)
𝐶1 (𝑚) = 40 + 0.15𝑚
150
100
50 𝐶2 (𝑚) = 50 + 0.10𝑚
0 𝑚 (miles)
200 400 600 800
Figure 1.8
(c) To find which company is cheaper, we need to determine where the two lines intersect. We let 𝐶1 = 𝐶2 , and thus
40 + 0.15𝑚 = 50 + 0.10𝑚
0.05𝑚 = 10
𝑚 = 200.
If you are going more than 200 miles a day, the competitor is cheaper. If you are going less than 200 miles a day, the
first company is cheaper.
58. (a) We find the slope 𝑚 and intercept 𝑏 in the linear equation 𝑆 = 𝑏 + 𝑚𝑡. To find the slope 𝑚, we use
Δ𝑆 66 − 113
𝑚= = = −0.94.
Δ𝑡 50 − 0
When 𝑡 = 0, we have 𝑆 = 113, so the intercept 𝑏 is 113. The linear formula is
𝑆 = 113 − 0.94𝑡.
(b) We use the formula 𝑆 = 113 − 0.94𝑡. When 𝑆 = 20, we have 20 = 113 − 0.94𝑡 and so 𝑡 = 98.9. If this linear model
were correct, the average male sperm count would drop below the fertility level during the year 2038.
