, Solution manual for Functions Modeling
Change A Preparation for Calculus, 6th Edition
Eric Connally
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, 1.1 SOLUTIONS 1
CHAPTER ONE
Solutions for Section 1.1
Skill Refresher
S1. Combining like terms, we see that 4𝑥 + 8𝑥 = 12𝑥.
S2. Combining like terms, we see that 8𝑤 − 5𝑤 = 3𝑤.
S3. Finding the common denominator we get 𝑐 + 12 𝑐 = 2𝑐+𝑐
2
= 3𝑐
2
= 32 𝑐.
S4. 𝑃 (1 + 0.07 + 0.02) = 1.09𝑃 .
S5. 2𝜋𝑟2 + 2𝜋𝑟 ⋅ 2𝑟 = 2𝜋𝑟2 + 4𝜋𝑟2 = 6𝜋𝑟2 .
12𝜋 − 2𝜋 10𝜋 5
S6. = = .
6𝜋 6𝜋 3
S7. We substitute 𝑥 = 3 and evaluate: 5𝑥 − 2 = 5(3) − 2 = 15 − 2 = 13.
S8. We substitute 𝑥 = 2 and evaluate: 3𝑥2 + 5 = 3(22 ) + 5 = 3(4) + 5 = 12 + 5 = 17.
( )
S9. 12 − 5(−5) = 12 + 25 = 51 2
.
S10. 1 − 12(3) + (3)2 = 1 − 36 + 9 = −26.
3 3 3
S11. = = = 1.
2 − (−1)3 2 − (−1) 2+1
4 4 4
S12. = = 1 = −12.
1 + 13 1 − 34 −3
−4
S13. The figure is a parallelogram, so 𝐴 = (−2, 8).
S14. The figure is a parallelogram, so 𝐴 = (3, 21).
S15. We see that 𝑤 is a function of 𝑟, so on a graph of 𝑓 , the variable 𝑟 is on the horizontal axis and the variable 𝑤 is on the
vertical axis.
S16. We see that 𝑇 is a function of 𝐴, so on a graph of 𝑓 , the variable 𝐴 is on the horizontal axis and the variable 𝑇 is on the
vertical axis.
S17. The variable corresponding to 2 is on the horizontal axis and the variable corresponding to 6 is on the vertical axis, so the
coordinates are (2, 6).
S18. The variable corresponding to -8 is on the horizontal axis and the variable corresponding to 5 is on the vertical axis, so
the coordinates are (−8, 5).
EXERCISES
1. 𝑚 = 𝑓 (𝑣).
2. 𝑤 = 𝑓 (𝑐).
3. Appropriate axes are shown in Figure 1.1.
𝑃 (millions)
𝑡 (years)
Figure 1.1
,2 Chapter One /SOLUTIONS
4. Appropriate axes are shown in Figure 1.2.
Cost ($)
𝑞 (items)
Figure 1.2
5. Appropriate axes are shown in Figure 1.3.
𝑝, pressure (lbs∕in2 )
𝑣, volume (in3 )
Figure 1.3
6. Appropriate axes are shown in Figure 1.4.
𝐷
𝑦
Figure 1.4
7. (a) On the graph, the high tides occur when the graph is at its highest points. On this particular day, there were two high
tides.
(b) The low tides occur when the graph is at its lowest points. There were two low tides on this day.
(c) To find the amount of time elapsed between high tides, find the distance between the two highest points on the graph.
It is about 12 hours.
8. These data are plotted in Figure 1.5. The independent variable is 𝐴 and the dependent variable is 𝑛.
𝑛 (gallons)
6
5
4
3
2
1
0 𝐴, area (ft2 )
500 1000 1500
Figure 1.5
9. (a) Since the vertical intercept is (0, 40), we have 𝑓 (0) = 40.
(b) Since the horizontal intercept is (2, 0), we have 𝑓 (2) = 0.
, 1.1 SOLUTIONS 3
10. (a) Since 𝑓 (𝑥) is 4 when 𝑥 = 0, we have 𝑓 (0) = 4.
(b) Since 𝑥 = 3 when 𝑓 (𝑥) = 0, we have 𝑓 (3) = 0.
(c) 𝑓 (1) = 2
(d) There are two 𝑥 values leading to 𝑓 (𝑥) = 1, namely 𝑥 = 2 and 𝑥 = 4. So 𝑓 (2) = 1 and 𝑓 (4) = 1.
11. 𝑓 (6.9) = 2.9.
12. (2.2, 2.9); (6.1, 4.9)
13. Since 𝑓 (0) = 𝑓 (4) = 𝑓 (8) = 0, the solutions are 𝑥 = 0, 4, 8.
14. Since the graphs touch at 𝑥 = 2.2 and 𝑥 = 6.1, these are the solutions.
15. (a) 𝑤 goes on the horizontal axis
(b) (−4, 10)
(c) (6, 1)
16. We substitute the given value for the input 𝑥, and evaluate.
(a) 𝑓 (0) = 3(0) + 8 = 0 + 8 = 8.
(b) 𝑓 (2) = 3(2) + 8 = 6 + 8 = 14.
(c) 𝑓 (−1) = 3(−1) + 8 = −3 + 8 = 5.
17. Note that the input variable is 𝑥 in the expression 𝑓 (𝑥), so we substitute the given input value for 𝑥, and evaluate.
(a) 𝑓 (0) = 𝑝(0) + 𝑞 = 𝑞.
(b) 𝑓 (1) = 𝑝(1) + 𝑞 = 𝑝 + 𝑞.
(c) 𝑓 (2) = 𝑝(2) + 𝑞 = 2𝑝 + 𝑞.
(d) 𝑓 (−1) = 𝑝(−1) + 𝑞 = −𝑝 + 𝑞.
18. (a) We substitute 𝑥 = 2 and evaluate:
𝑓 (2) = 4(2) − 5 = 8 − 5 = 3.
(b) We see in part (a) that 𝑓 (2) = 3 so 3𝑓 (2) = 3(3) = 9.
(c) When 𝑥 = 2, we have
𝑓 (3𝑥) = 𝑓 (3 ⋅ 2) = 𝑓 (6) = 4(6) − 5 = 24 − 5 = 19.
Notice that 3𝑓 (𝑥) ≠ 𝑓 (3𝑥).
19. Any 𝑤 value can give two 𝑧 values. For example, if 𝑤 = 0,
7 ⋅ 02 + 5 = 𝑧2
√
± 5 = 𝑧,
so there are two 𝑧 values (one positive and one negative) for 𝑤 = 0. Thus, 𝑧 is not a function of 𝑤.
A similar argument shows that 𝑤 is not a function of 𝑧.
20. Here, 𝑦 is a function of 𝑥, because any particular 𝑥 value gives one and only one 𝑦 value. For example, if we input the
constant 𝑎 as the value of 𝑥, we have 𝑦 = 𝑎4 − 1, which is one particular 𝑦 value.
However, some values of 𝑦 lead to more than one value of 𝑥. For example, if 𝑦 = 15, then 15 = 𝑥4 − 1, so 𝑥4 = 16,
giving 𝑥 = ±2. Thus, 𝑥 is not a function of 𝑦.
21. Here, 𝑚 is a function of 𝑡. For any 𝑡, there is only one possible value of 𝑚. In addition, for any 𝑚, there is only one possible
value of 𝑡, given by 𝑡 = 𝑚2 . Thus, 𝑡 is a function of 𝑚.
22. Both of the relationships are functions because any quantity of gas determines the quantity of coffee that can be bought,
and vice versa. For example, if you buy 20 gallons of gas, spending $60, you buy 4 pounds of coffee.
,4 Chapter One /SOLUTIONS
23. We apply the vertical-line test. As you can see in Figure 1.6, there is no vertical line that meets the graph at more than one
point, so this graph represents 𝑦 as a function of 𝑥.
𝑦
𝑥
Figure 1.6
24. We apply the vertical-line test. As you can see in Figure 1.7, there is a vertical line meeting the graph in more than one
point. Thus, this graph fails the vertical-line test and does not represent a function.
𝑦
𝑥
Figure 1.7
25. We apply the vertical-line test. As you can see in Figure 1.8, there is a vertical line meeting the graph in more than one
point. Thus, this graph fails the vertical-line test and does not represent a function.
𝑦
𝑥
Figure 1.8
26. We apply the vertical-line test. As you can see in Figure 1.9, there is no vertical line that meets the graph at more than one
point, so this graph represents 𝑦 as a function of 𝑥.
𝑦
𝑥
Figure 1.9
, 1.1 SOLUTIONS 5
27. We apply the vertical-line test. As you can see in Figure 1.10, there is a vertical line meeting the graph in more than one
point. Thus, this graph fails the vertical-line test and does not represent a function.
𝑦
𝑥
Figure 1.10
28. We apply the vertical-line test. As you can see in Figure 1.11, there is no vertical line that meets the graph at more than
one point, so this graph represents 𝑦 as a function of 𝑥.
𝑦
𝑥
Figure 1.11
PROBLEMS
29. The value of 𝑁 is not necessarily a function of 𝐺, since each value of 𝐺 does not need to have a unique value of 𝑁
associated to it. For example, suppose we choose the value of 𝐺 to be a B. There may be more than one student who
received a B, so there may be more than one ID number corresponding to B.
The value of 𝐺 must be a function of 𝑁, because each ID number (each student) receives exactly one grade. Therefore
each value of 𝑁 has a unique value of 𝐺 associated with it. Writing 𝐺 = 𝑓 (𝑁) indicates that the ID number is the input
which uniquely determines the grade, the output.
30. Figure 1.12 shows a possible graph of blood sugar level as a function of time over one day. Note that the actual curve is
smooth, and does not have any sharp corners.
blood-sugar level
time
8 am Noon 6 pm Midnight
Figure 1.12
,6 Chapter One /SOLUTIONS
31. A possible graph is shown in Figure 1.13
𝐻
𝑡
Figure 1.13
32. A possible graph is shown in Figure 1.14
Height
Time
Figure 1.14
33. (a) In 2010, or year 𝑡 = 0, the ranking for Olivia was 4, making it most popular, and the ranking for Charlotte was 46,
making it least popular.
(b) In 2015, or year 𝑡 = 5, the ranking for Olivia was 2, making it still the most popular, and the ranking for Madison was
11, making it least popular.
34. (a) We have 𝑟𝑐 (0) − 𝑟𝑜 (0) = 46 − 4 = 42. This tells us that in 2010 (year 𝑡 = 0), the name Olivia was ranked 42 places
higher than Charlotte on the list of most popular names. (Recall that the lower the ranking, the higher a name’s position
on the list.)
(b) We have 𝑟𝑚 (5) − 𝑟𝑐 (5) = 11 − 9 = 2. This tells us that in 2015 (year 𝑡 = 5), the name Madison was ranked 2 places
lower than Charlotte on the list of most popular names.
(c) We have 𝑟𝑚 (𝑡) > 𝑟𝑐 (𝑡) for 𝑡 = 5. This tells us that the name Madison was ranked lower than the name Charlotte on the
list of most popular names in the year 2015.
35. (a) Since the person starts out 5 miles from home, the vertical intercept on the graph must be 5. Thus, (i) and (ii) are
possibilities. However, since the person rides 5 mph away from home, after 1 hour the person is 10 miles from home.
Thus, (ii) is the correct graph.
(b) Since this person also starts out 5 miles from home, (i) and (ii) are again possibilities. This time, however, the person
is moving at 10 mph and so is 15 miles from home after 1 hour. Thus, (i) is correct.
(c) The person starts out 10 miles from home so the vertical intercept must be 10. The fact that the person reaches home
after 1 hour means that the horizontal intercept is 1. Thus, (v) is correct.
(d) Starting out 10 miles from home means that the vertical intercept is 10. Being half way home after 1 hour means that
the distance from home is 5 miles after 1 hour. Thus, (iv) is correct.
(e) We are looking for a graph with vertical intercept of 5 and where the distance is 10 after 1 hour. This is graph (ii).
Notice that graph (iii), which depicts a bicyclist stopped 10 miles from home, does not match any of the stories.
, 1.1 SOLUTIONS 7
36. (a) At 40 mph, fuel consumption is about 28 mpg, so the fuel used is 300∕28 = 10.71 gallons.
(b) At 60 mph, fuel consumption is about 29 mpg. At 70 mph, fuel consumption is about 28 mpg. Therefore, on a 200
mile trip
200 200
Fuel saved = − = 0.25 gallons.
28 29
(c) The most fuel-efficient speed is where mpg is a maximum, which is about 55 mph.
37. (a) (i) The mass of water in 1 kg of air that has 100% relative humidity at 30◦ C is approximately 30 g.
(ii) The mass of water in 1 kg of air that has 50% relative humidity at 30◦ C is approximately 15 g. We can either
read this from the graph of or use the answer to the previous part.
(iii) The mass of water in 1 kg of air that has 75% relative humidity at 30◦ C is approximately 22.5 g since it will be
half way between the amount of water we found at 100% and 50% humidity.
(b) At 50% relative humidity, 1 kg of 20◦ air contains approximately 7.5 g of water. Therefore, 300 kg of air contains
7.5 ⋅ 300 = 2250 g of water. Since 1 kg = 1 liter of water, this is 2.25 liters.
(c) The answer will depend on the dimensions of the classroom. A classroom that is 8 meters by 10 meters with a height
of 3 meters has volume 8 ⋅ 10 ⋅ 3 = 240 m3 . Given that the density of air is approximately 1.2 kg/m3 , the mass of air
in the classroom is 240 m3 ⋅ 1.2 kg/m3 = 288 kg. At 50% relative humidity and 20◦ C, 1 kg of air contains 7.5 g of
water. Therefore the classroom contains 288 ⋅ 7.5 = 2160 g of water. Since 1000 g = 1 l of water, 2160 g = 2.16 l of
water, which is a little over half a gallon.
38. (a) The number of people who own cell phones in the year 2000 is 100,300,000.
(b) There are 20,000,000 people who own cell phones 𝑎 years after 1990.
(c) There are 𝑏 million people who own cell phones in the year 2010.
(d) The number 𝑛 is the number of people (in millions) who own cell phones 𝑡 years after 1990.
39. (a) When there is no snow, it is equivalent to no rain. Thus, the vertical intercept is 0. Since every ten inches of snow is
equivalent to one inch of rain, we can specify the slope as
Δrain 1
= = 0.1
Δsnow 10
We have a vertical intercept of 0 and a slope of 0.1. Thus, the equation is: 𝑟 = 𝑓 (𝑠) = 0.1𝑠.
(b) By substituting 5 in for 𝑠, we get 𝑓 (5) = 0.1(5) = 0.5 This tells us that five inches of snow is equivalent to approxi-
mately 1∕2 inch of rain.
(c) Substitute 5 inches for 𝑟 = 𝑓 (𝑠) in the equation: 5 = 0.1𝑠. Solving gives 𝑠 = 50. Five inches of rain is equivalent to
approximately 50 inches of snow.
40. Figure 1.15 shows the tank.
8 ft
3 ft
Figure 1.15: Cylindrical water tank
(a) The volume of a cylinder is equal to the area of the base times the height, where the area of the base is 𝜋𝑟2 . Here, the
radius of the base is (1∕2)(6) = 3 ft, so the area is 𝜋 ⋅ 32 = 9𝜋 ft2 . Therefore, the capacity of this tank is (9𝜋)8 = 72𝜋
ft3 .
(b) If the height of the water is 5 ft, the volume becomes (9𝜋)5 = 45𝜋 ft3 .
, 8 Chapter One /SOLUTIONS
(c) In general, if the height of water is ℎ ft, the volume of the water is (9𝜋)ℎ. If we let 𝑉 (ℎ) be the volume of water in the
tank as a function of its height, then
𝑉 (ℎ) = 9𝜋ℎ.
Note that this function only makes sense for a non-negative value of ℎ, which does not exceed 8 feet, the height of the
tank.
41. (a) The number 𝑓 (5) = 0.35 is the amount of rain, in inches, that Tucson received in the month of May.
(b) From the table, we see that 𝑡 = 1 is the only solution to the equation 𝑓 (𝑡) = 0. This solution indicates that January is
the only month among our data during which Tucson received no rainfall.
(c) From the table, we see that 𝑡 = 2 and 𝑡 = 4 are the only solutions to 𝑓 (𝑡) = 0.1. These solutions indicate that in
February and April, Tucson received 0.1 inches of rain.
42. (a) From the table, we see that 𝑓 (100) = 1154.8. This means that there is approximately $1154.8 billion worth of $100
bills in circulation in the United States.
(b) To determine the number of $5 bills, we divide 14.2 by 5. Thus, we have 2.84 billion $5 bills in circulation. The
number of $1 bills is the same as the value, so there are 11.7 billion $1 bills in circulation. There are more $1 bills.
43. (a) 69◦ F
(b) July 17th and 20th
(c) Yes. For each date, there is exactly one low temperature.
(d) No, it is not true that for each low temperature, there is exactly one date: for example, 73◦ corresponds to both the
17th and 20th .
44. (a) Figure 1.16 shows the plot of 𝑅 versus 𝑡. 𝑅 is a function of 𝑡 because no vertical line intersects the graph in more than
one place.
(b) Figure 1.17 shows the plot of 𝐹 versus 𝑡. 𝐹 is a function of 𝑡 because no vertical line intersects the graph in more than
one place.
Rabbits Foxes
1500 150
1300 130
1100 110
900 90
700 70
500 50
300 30
100 10
Time Time
1 2 3 4 5 6 7 8 9 1011 1 2 3 4 5 6 7 8 9 1011
Figure 1.16: The graph of 𝑅 versus 𝑡 Figure 1.17: The graph of 𝐹 versus 𝑡
(c) Figure 1.18 shows the plot of 𝐹 versus 𝑅. We have also drawn the vertical line corresponding to 𝑅 = 567. This tells
us that 𝐹 is not a function of 𝑅 because there is a vertical line that intersects the graph twice. In fact the lines 𝑅 = 567,
𝑅 = 750, 𝑅 = 1000, 𝑅 = 1250, and 𝑅 = 1433 all intersect the graph twice. However, the existence of any one of
them is enough to guarantee that 𝐹 is not a function of 𝑅.
(d) Figure 1.19 shows the plot of 𝑅 versus 𝐹 . We have drawn the vertical line corresponding to 𝐹 = 57. This tells us
that 𝑅 is not a function of 𝐹 because there is a vertical line that intersects the graph twice. In fact the lines 𝐹 = 57,
𝐹 = 75, 𝐹 = 100, 𝐹 = 125, and 𝐹 = 143 all intersect the graph twice. However, the existence of any one of them is
enough to guarantee that 𝑅 is not a function of 𝐹 .
Change A Preparation for Calculus, 6th Edition
Eric Connally
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
Skill Refresher
S1. Combining like terms, we see that 4𝑥 + 8𝑥 = 12𝑥.
S2. Combining like terms, we see that 8𝑤 − 5𝑤 = 3𝑤.
S3. Finding the common denominator we get 𝑐 + 12 𝑐 = 2𝑐+𝑐
2
= 3𝑐
2
= 32 𝑐.
S4. 𝑃 (1 + 0.07 + 0.02) = 1.09𝑃 .
S5. 2𝜋𝑟2 + 2𝜋𝑟 ⋅ 2𝑟 = 2𝜋𝑟2 + 4𝜋𝑟2 = 6𝜋𝑟2 .
12𝜋 − 2𝜋 10𝜋 5
S6. = = .
6𝜋 6𝜋 3
S7. We substitute 𝑥 = 3 and evaluate: 5𝑥 − 2 = 5(3) − 2 = 15 − 2 = 13.
S8. We substitute 𝑥 = 2 and evaluate: 3𝑥2 + 5 = 3(22 ) + 5 = 3(4) + 5 = 12 + 5 = 17.
( )
S9. 12 − 5(−5) = 12 + 25 = 51 2
.
S10. 1 − 12(3) + (3)2 = 1 − 36 + 9 = −26.
3 3 3
S11. = = = 1.
2 − (−1)3 2 − (−1) 2+1
4 4 4
S12. = = 1 = −12.
1 + 13 1 − 34 −3
−4
S13. The figure is a parallelogram, so 𝐴 = (−2, 8).
S14. The figure is a parallelogram, so 𝐴 = (3, 21).
S15. We see that 𝑤 is a function of 𝑟, so on a graph of 𝑓 , the variable 𝑟 is on the horizontal axis and the variable 𝑤 is on the
vertical axis.
S16. We see that 𝑇 is a function of 𝐴, so on a graph of 𝑓 , the variable 𝐴 is on the horizontal axis and the variable 𝑇 is on the
vertical axis.
S17. The variable corresponding to 2 is on the horizontal axis and the variable corresponding to 6 is on the vertical axis, so the
coordinates are (2, 6).
S18. The variable corresponding to -8 is on the horizontal axis and the variable corresponding to 5 is on the vertical axis, so
the coordinates are (−8, 5).
EXERCISES
1. 𝑚 = 𝑓 (𝑣).
2. 𝑤 = 𝑓 (𝑐).
3. Appropriate axes are shown in Figure 1.1.
𝑃 (millions)
𝑡 (years)
Figure 1.1
,2 Chapter One /SOLUTIONS
4. Appropriate axes are shown in Figure 1.2.
Cost ($)
𝑞 (items)
Figure 1.2
5. Appropriate axes are shown in Figure 1.3.
𝑝, pressure (lbs∕in2 )
𝑣, volume (in3 )
Figure 1.3
6. Appropriate axes are shown in Figure 1.4.
𝐷
𝑦
Figure 1.4
7. (a) On the graph, the high tides occur when the graph is at its highest points. On this particular day, there were two high
tides.
(b) The low tides occur when the graph is at its lowest points. There were two low tides on this day.
(c) To find the amount of time elapsed between high tides, find the distance between the two highest points on the graph.
It is about 12 hours.
8. These data are plotted in Figure 1.5. The independent variable is 𝐴 and the dependent variable is 𝑛.
𝑛 (gallons)
6
5
4
3
2
1
0 𝐴, area (ft2 )
500 1000 1500
Figure 1.5
9. (a) Since the vertical intercept is (0, 40), we have 𝑓 (0) = 40.
(b) Since the horizontal intercept is (2, 0), we have 𝑓 (2) = 0.
, 1.1 SOLUTIONS 3
10. (a) Since 𝑓 (𝑥) is 4 when 𝑥 = 0, we have 𝑓 (0) = 4.
(b) Since 𝑥 = 3 when 𝑓 (𝑥) = 0, we have 𝑓 (3) = 0.
(c) 𝑓 (1) = 2
(d) There are two 𝑥 values leading to 𝑓 (𝑥) = 1, namely 𝑥 = 2 and 𝑥 = 4. So 𝑓 (2) = 1 and 𝑓 (4) = 1.
11. 𝑓 (6.9) = 2.9.
12. (2.2, 2.9); (6.1, 4.9)
13. Since 𝑓 (0) = 𝑓 (4) = 𝑓 (8) = 0, the solutions are 𝑥 = 0, 4, 8.
14. Since the graphs touch at 𝑥 = 2.2 and 𝑥 = 6.1, these are the solutions.
15. (a) 𝑤 goes on the horizontal axis
(b) (−4, 10)
(c) (6, 1)
16. We substitute the given value for the input 𝑥, and evaluate.
(a) 𝑓 (0) = 3(0) + 8 = 0 + 8 = 8.
(b) 𝑓 (2) = 3(2) + 8 = 6 + 8 = 14.
(c) 𝑓 (−1) = 3(−1) + 8 = −3 + 8 = 5.
17. Note that the input variable is 𝑥 in the expression 𝑓 (𝑥), so we substitute the given input value for 𝑥, and evaluate.
(a) 𝑓 (0) = 𝑝(0) + 𝑞 = 𝑞.
(b) 𝑓 (1) = 𝑝(1) + 𝑞 = 𝑝 + 𝑞.
(c) 𝑓 (2) = 𝑝(2) + 𝑞 = 2𝑝 + 𝑞.
(d) 𝑓 (−1) = 𝑝(−1) + 𝑞 = −𝑝 + 𝑞.
18. (a) We substitute 𝑥 = 2 and evaluate:
𝑓 (2) = 4(2) − 5 = 8 − 5 = 3.
(b) We see in part (a) that 𝑓 (2) = 3 so 3𝑓 (2) = 3(3) = 9.
(c) When 𝑥 = 2, we have
𝑓 (3𝑥) = 𝑓 (3 ⋅ 2) = 𝑓 (6) = 4(6) − 5 = 24 − 5 = 19.
Notice that 3𝑓 (𝑥) ≠ 𝑓 (3𝑥).
19. Any 𝑤 value can give two 𝑧 values. For example, if 𝑤 = 0,
7 ⋅ 02 + 5 = 𝑧2
√
± 5 = 𝑧,
so there are two 𝑧 values (one positive and one negative) for 𝑤 = 0. Thus, 𝑧 is not a function of 𝑤.
A similar argument shows that 𝑤 is not a function of 𝑧.
20. Here, 𝑦 is a function of 𝑥, because any particular 𝑥 value gives one and only one 𝑦 value. For example, if we input the
constant 𝑎 as the value of 𝑥, we have 𝑦 = 𝑎4 − 1, which is one particular 𝑦 value.
However, some values of 𝑦 lead to more than one value of 𝑥. For example, if 𝑦 = 15, then 15 = 𝑥4 − 1, so 𝑥4 = 16,
giving 𝑥 = ±2. Thus, 𝑥 is not a function of 𝑦.
21. Here, 𝑚 is a function of 𝑡. For any 𝑡, there is only one possible value of 𝑚. In addition, for any 𝑚, there is only one possible
value of 𝑡, given by 𝑡 = 𝑚2 . Thus, 𝑡 is a function of 𝑚.
22. Both of the relationships are functions because any quantity of gas determines the quantity of coffee that can be bought,
and vice versa. For example, if you buy 20 gallons of gas, spending $60, you buy 4 pounds of coffee.
,4 Chapter One /SOLUTIONS
23. We apply the vertical-line test. As you can see in Figure 1.6, there is no vertical line that meets the graph at more than one
point, so this graph represents 𝑦 as a function of 𝑥.
𝑦
𝑥
Figure 1.6
24. We apply the vertical-line test. As you can see in Figure 1.7, there is a vertical line meeting the graph in more than one
point. Thus, this graph fails the vertical-line test and does not represent a function.
𝑦
𝑥
Figure 1.7
25. We apply the vertical-line test. As you can see in Figure 1.8, there is a vertical line meeting the graph in more than one
point. Thus, this graph fails the vertical-line test and does not represent a function.
𝑦
𝑥
Figure 1.8
26. We apply the vertical-line test. As you can see in Figure 1.9, there is no vertical line that meets the graph at more than one
point, so this graph represents 𝑦 as a function of 𝑥.
𝑦
𝑥
Figure 1.9
, 1.1 SOLUTIONS 5
27. We apply the vertical-line test. As you can see in Figure 1.10, there is a vertical line meeting the graph in more than one
point. Thus, this graph fails the vertical-line test and does not represent a function.
𝑦
𝑥
Figure 1.10
28. We apply the vertical-line test. As you can see in Figure 1.11, there is no vertical line that meets the graph at more than
one point, so this graph represents 𝑦 as a function of 𝑥.
𝑦
𝑥
Figure 1.11
PROBLEMS
29. The value of 𝑁 is not necessarily a function of 𝐺, since each value of 𝐺 does not need to have a unique value of 𝑁
associated to it. For example, suppose we choose the value of 𝐺 to be a B. There may be more than one student who
received a B, so there may be more than one ID number corresponding to B.
The value of 𝐺 must be a function of 𝑁, because each ID number (each student) receives exactly one grade. Therefore
each value of 𝑁 has a unique value of 𝐺 associated with it. Writing 𝐺 = 𝑓 (𝑁) indicates that the ID number is the input
which uniquely determines the grade, the output.
30. Figure 1.12 shows a possible graph of blood sugar level as a function of time over one day. Note that the actual curve is
smooth, and does not have any sharp corners.
blood-sugar level
time
8 am Noon 6 pm Midnight
Figure 1.12
,6 Chapter One /SOLUTIONS
31. A possible graph is shown in Figure 1.13
𝐻
𝑡
Figure 1.13
32. A possible graph is shown in Figure 1.14
Height
Time
Figure 1.14
33. (a) In 2010, or year 𝑡 = 0, the ranking for Olivia was 4, making it most popular, and the ranking for Charlotte was 46,
making it least popular.
(b) In 2015, or year 𝑡 = 5, the ranking for Olivia was 2, making it still the most popular, and the ranking for Madison was
11, making it least popular.
34. (a) We have 𝑟𝑐 (0) − 𝑟𝑜 (0) = 46 − 4 = 42. This tells us that in 2010 (year 𝑡 = 0), the name Olivia was ranked 42 places
higher than Charlotte on the list of most popular names. (Recall that the lower the ranking, the higher a name’s position
on the list.)
(b) We have 𝑟𝑚 (5) − 𝑟𝑐 (5) = 11 − 9 = 2. This tells us that in 2015 (year 𝑡 = 5), the name Madison was ranked 2 places
lower than Charlotte on the list of most popular names.
(c) We have 𝑟𝑚 (𝑡) > 𝑟𝑐 (𝑡) for 𝑡 = 5. This tells us that the name Madison was ranked lower than the name Charlotte on the
list of most popular names in the year 2015.
35. (a) Since the person starts out 5 miles from home, the vertical intercept on the graph must be 5. Thus, (i) and (ii) are
possibilities. However, since the person rides 5 mph away from home, after 1 hour the person is 10 miles from home.
Thus, (ii) is the correct graph.
(b) Since this person also starts out 5 miles from home, (i) and (ii) are again possibilities. This time, however, the person
is moving at 10 mph and so is 15 miles from home after 1 hour. Thus, (i) is correct.
(c) The person starts out 10 miles from home so the vertical intercept must be 10. The fact that the person reaches home
after 1 hour means that the horizontal intercept is 1. Thus, (v) is correct.
(d) Starting out 10 miles from home means that the vertical intercept is 10. Being half way home after 1 hour means that
the distance from home is 5 miles after 1 hour. Thus, (iv) is correct.
(e) We are looking for a graph with vertical intercept of 5 and where the distance is 10 after 1 hour. This is graph (ii).
Notice that graph (iii), which depicts a bicyclist stopped 10 miles from home, does not match any of the stories.
, 1.1 SOLUTIONS 7
36. (a) At 40 mph, fuel consumption is about 28 mpg, so the fuel used is 300∕28 = 10.71 gallons.
(b) At 60 mph, fuel consumption is about 29 mpg. At 70 mph, fuel consumption is about 28 mpg. Therefore, on a 200
mile trip
200 200
Fuel saved = − = 0.25 gallons.
28 29
(c) The most fuel-efficient speed is where mpg is a maximum, which is about 55 mph.
37. (a) (i) The mass of water in 1 kg of air that has 100% relative humidity at 30◦ C is approximately 30 g.
(ii) The mass of water in 1 kg of air that has 50% relative humidity at 30◦ C is approximately 15 g. We can either
read this from the graph of or use the answer to the previous part.
(iii) The mass of water in 1 kg of air that has 75% relative humidity at 30◦ C is approximately 22.5 g since it will be
half way between the amount of water we found at 100% and 50% humidity.
(b) At 50% relative humidity, 1 kg of 20◦ air contains approximately 7.5 g of water. Therefore, 300 kg of air contains
7.5 ⋅ 300 = 2250 g of water. Since 1 kg = 1 liter of water, this is 2.25 liters.
(c) The answer will depend on the dimensions of the classroom. A classroom that is 8 meters by 10 meters with a height
of 3 meters has volume 8 ⋅ 10 ⋅ 3 = 240 m3 . Given that the density of air is approximately 1.2 kg/m3 , the mass of air
in the classroom is 240 m3 ⋅ 1.2 kg/m3 = 288 kg. At 50% relative humidity and 20◦ C, 1 kg of air contains 7.5 g of
water. Therefore the classroom contains 288 ⋅ 7.5 = 2160 g of water. Since 1000 g = 1 l of water, 2160 g = 2.16 l of
water, which is a little over half a gallon.
38. (a) The number of people who own cell phones in the year 2000 is 100,300,000.
(b) There are 20,000,000 people who own cell phones 𝑎 years after 1990.
(c) There are 𝑏 million people who own cell phones in the year 2010.
(d) The number 𝑛 is the number of people (in millions) who own cell phones 𝑡 years after 1990.
39. (a) When there is no snow, it is equivalent to no rain. Thus, the vertical intercept is 0. Since every ten inches of snow is
equivalent to one inch of rain, we can specify the slope as
Δrain 1
= = 0.1
Δsnow 10
We have a vertical intercept of 0 and a slope of 0.1. Thus, the equation is: 𝑟 = 𝑓 (𝑠) = 0.1𝑠.
(b) By substituting 5 in for 𝑠, we get 𝑓 (5) = 0.1(5) = 0.5 This tells us that five inches of snow is equivalent to approxi-
mately 1∕2 inch of rain.
(c) Substitute 5 inches for 𝑟 = 𝑓 (𝑠) in the equation: 5 = 0.1𝑠. Solving gives 𝑠 = 50. Five inches of rain is equivalent to
approximately 50 inches of snow.
40. Figure 1.15 shows the tank.
8 ft
3 ft
Figure 1.15: Cylindrical water tank
(a) The volume of a cylinder is equal to the area of the base times the height, where the area of the base is 𝜋𝑟2 . Here, the
radius of the base is (1∕2)(6) = 3 ft, so the area is 𝜋 ⋅ 32 = 9𝜋 ft2 . Therefore, the capacity of this tank is (9𝜋)8 = 72𝜋
ft3 .
(b) If the height of the water is 5 ft, the volume becomes (9𝜋)5 = 45𝜋 ft3 .
, 8 Chapter One /SOLUTIONS
(c) In general, if the height of water is ℎ ft, the volume of the water is (9𝜋)ℎ. If we let 𝑉 (ℎ) be the volume of water in the
tank as a function of its height, then
𝑉 (ℎ) = 9𝜋ℎ.
Note that this function only makes sense for a non-negative value of ℎ, which does not exceed 8 feet, the height of the
tank.
41. (a) The number 𝑓 (5) = 0.35 is the amount of rain, in inches, that Tucson received in the month of May.
(b) From the table, we see that 𝑡 = 1 is the only solution to the equation 𝑓 (𝑡) = 0. This solution indicates that January is
the only month among our data during which Tucson received no rainfall.
(c) From the table, we see that 𝑡 = 2 and 𝑡 = 4 are the only solutions to 𝑓 (𝑡) = 0.1. These solutions indicate that in
February and April, Tucson received 0.1 inches of rain.
42. (a) From the table, we see that 𝑓 (100) = 1154.8. This means that there is approximately $1154.8 billion worth of $100
bills in circulation in the United States.
(b) To determine the number of $5 bills, we divide 14.2 by 5. Thus, we have 2.84 billion $5 bills in circulation. The
number of $1 bills is the same as the value, so there are 11.7 billion $1 bills in circulation. There are more $1 bills.
43. (a) 69◦ F
(b) July 17th and 20th
(c) Yes. For each date, there is exactly one low temperature.
(d) No, it is not true that for each low temperature, there is exactly one date: for example, 73◦ corresponds to both the
17th and 20th .
44. (a) Figure 1.16 shows the plot of 𝑅 versus 𝑡. 𝑅 is a function of 𝑡 because no vertical line intersects the graph in more than
one place.
(b) Figure 1.17 shows the plot of 𝐹 versus 𝑡. 𝐹 is a function of 𝑡 because no vertical line intersects the graph in more than
one place.
Rabbits Foxes
1500 150
1300 130
1100 110
900 90
700 70
500 50
300 30
100 10
Time Time
1 2 3 4 5 6 7 8 9 1011 1 2 3 4 5 6 7 8 9 1011
Figure 1.16: The graph of 𝑅 versus 𝑡 Figure 1.17: The graph of 𝐹 versus 𝑡
(c) Figure 1.18 shows the plot of 𝐹 versus 𝑅. We have also drawn the vertical line corresponding to 𝑅 = 567. This tells
us that 𝐹 is not a function of 𝑅 because there is a vertical line that intersects the graph twice. In fact the lines 𝑅 = 567,
𝑅 = 750, 𝑅 = 1000, 𝑅 = 1250, and 𝑅 = 1433 all intersect the graph twice. However, the existence of any one of
them is enough to guarantee that 𝐹 is not a function of 𝑅.
(d) Figure 1.19 shows the plot of 𝑅 versus 𝐹 . We have drawn the vertical line corresponding to 𝐹 = 57. This tells us
that 𝑅 is not a function of 𝐹 because there is a vertical line that intersects the graph twice. In fact the lines 𝐹 = 57,
𝐹 = 75, 𝐹 = 100, 𝐹 = 125, and 𝐹 = 143 all intersect the graph twice. However, the existence of any one of them is
enough to guarantee that 𝑅 is not a function of 𝐹 .
