4.1 SOLUTIONS
87
CHAPTER FOUR
Solutions for Section 4.1
EXERCISES
1. w = f (c).
2. m = f (v).
3. The dependent variable is N , the number of napkins used, the output of the function. The independent variable is C, the
number of customers, the input of the function.
4. The dependent variable is P , the output of the function. The independent variable is s, the input of the function.
9
7
5. To evaluate when x = −7, we substitute −7 for x in the function, giving f (−7) = − − 1 = − .
2
2
6. To evaluate when x = −7, we substitute −7 for x in the function, giving f (−7) = (−7)2 − 3 = 49 − 3 = 46.
7. We have
(a)
g(2) = (12 − 2)2 − (2 − 1)3
= 102 − 13
= 100 − 1
= 99.
(b)
g(5) = (12 − 5)2 − (5 − 1)3
= 72 − 43
= 49 − 64
= −15.
(c)
g(0) = (12 − 0)2 − (0 − 1)3
= 122 − (−1)3
= 144 − (−1)
= 145.
(d)
g(−1) = (12 − (−1))2 − (−1 − 1)3
= 132 − (−2)3
= 169 − (−8)
= 177.
88
Chapter Four /SOLUTIONS
8. We have To evaluate, we substitute the input value of the function for x in f (x) = 2x2 + 7x + 5.
(a) We substitute 3 for x:
(b) We substitute a for x:
f (3) = 2(3)2 + 7(3) + 5 = 2 · 9 + 21 + 5 = 18 + 26 = 44.
f (a) = 2(a)2 + 7(a) + 5 = 2a2 + 7a + 5.
(c) We substitute 2a for x:
f (2a) = 2(2a)2 + 7(2a) + 5 = 2 · (4a2 ) + 14a + 5 = 8a2 + 14a + 5.
(d) We substitute −2 for x:
f (−2) = 2(−2)2 + 7(−2) + 5 = 2 · 4 − 14 + 5 = 8 − 9 = −1.
9. We have
f (0) =
10. We have
g(0) = √
11. We have
1
2·0+1
= .
3−5·0
3
1
02
1
= √ = 1.
+1
1
1
1
1
g(−1) = p
= √
= √ .
1+1
2
(−1)2 + 1
This can be rewritten as
√
√
2
2
1
√ ·√ =
.
2
2
2
12. We have
f (10) =
13. We have
f (1/2) =
14. We have
g
21
21
2 · 10 + 1
=
=− .
3 − 5 · 10
−47
47
2 · 12 + 1
1+1
=
=
3 − 5 · 21
3 − 52
6
2
2
−
5
2
=
2
1
2
= 4.
√ 1
1
1
1
8 = q √ = √
= √ = .
3
2
8+1
9
8 +1
15. We have h(r) = 10 − 3r.
16. We have
h(2u) = 10 − 3 · 2u
= 10 − 6u.
17. We have
h(k − 3) = 10 − 3(k − 3)
= 10 − 3k + 9
= 19 − 3k.
4.1 SOLUTIONS
89
18. We have
h(4 − n) = 10 − 3(4 − n)
= 10 − 12 + 3n
= −2 + 3n.
19. We have
h(5t2 ) = 10 − 3 · 5t2
= 10 − 15t2 .
20. We have
h(4 − t3 ) = 10 − 3(4 − t3 )
= 10 − 12 + 3t3
= −2 + 3t3 .
21. We have
f (1 − x) = 1 − (1 − x)2 − (1 − x)
Applying formula for f
2
= 1 − (1 − 2x + x ) − (1 − x) Expanding
= 1 − 1 + 2x − x2 − 1 + x
Clearing parentheses
= −x2 + 3x − 1.
22. Since the tax is 0.06P , the total cost would be the price of the item plus the tax,or
C = P + 0.06P = 1.06P.
23. (a) Since f (T ) gives the volume at temperature T , the 40 is the temperature, so its units are degrees Celsius (◦ C).
Similarly, the 3 is the volume, so its units are liters.
(b) Since f (40) = 3, we know that when T = 40 that f (T ) = 3. Since T = 40 at 40◦ C, the volume is 3 liters at 40◦ C.
24. (a)
(b)
(c)
(d)
25. (a)
(b)
(c)
(d)
Since g(y) is 5 when y = 0, we have g(0) = 5.
Since y = −5 when g(y) = 0, we have g(−5) = 0.
g(−5) = 0
Since y = −10 when g(y) = −5, we have g(−10) = −5.
Since h(t) is −2 when t = 0, we have h(0) = −2.
There are two t values leading to h(t) = 0, namely t = −2 and t = 3. So h(−2) = 0 and h(3) = 0.
h(−2) = 0
There are two t values leading to h(t) = −2, namely t = 0 and t = 2. So h(0) = −2 and h(2) = −2.
26. (a) Since h(t) is −2 when t = 0, we have 2h(0) = −4. Thus, we need to find the value(s) of t for which h(t) = −4.
There is only one t value leads to this, namely t = −1.
(b) Since h(t) is −1 when t = −3 and h(t) is −2 when t = 2, we have 2h(−3) + h(2) = −4. Thus, we need to find
the value(s) of t for which h(t) = −4. There is only one t value that leads to this, namely t = −1.
(c) Since h(t) is 0 when t = −2, we have h(−2) = 0. Thus, we need to find the value(s) of t for which h(t) = 0. There
are two t values leading to h(t) = 0, namely t = −2 and t = 3.
(d) Since h(t) is −1 when t = 1 and h(t) is −2 when t = 2, we have h(1) + h(2) = −3. Thus, we need to find the
value(s) of t for which h(t) = −3. There are no t values that lead to this.
27. (a) Since the vertical intercept is (0, 2), we have f (0) = 2.
(b) Since the horizontal intercepts are at (−2, 0) and (1, 0), we have f (−2) = 0 and f (1) = 0.
28. (a) Since the vertical intercept is (0, 1), we have h(0) = 1.
(b) Since there are no horizontal intercepts there are no values of r for which h(r) = 0.
90
Chapter Four /SOLUTIONS
29. (a) Since the vertical intercept is (0, 4), we have g(0) = 4.
(b) Since the horizontal intercepts are at (−2, 0) and (2, 0), we have g(−2) = 0 and g(2) = 0.
PROBLEMS
30. (a) The number of games won is a function of the team since for each team there is a unique number of games won.
(b) The NBA team is not a function of games won. This is because two teams correspond to 66 games won.
31. (a)
(b)
(c)
(d)
When the company charges 15 dollars for one of its products, the revenue is 112,500 dollars.
When the company charges a dollars for one of its products, the revenue is 0 dollars.
When the company charges 1 dollar for one of its products, the revenue is b dollars.
The number c is the revenue in dollars that the company receives when it charges a price of p dollars for one of its
products.
32. (a)
(b)
(c)
(d)
The car takes 111 feet to stop when traveling at 30 mph.
The car takes 10 feet to stop when traveling at a mph.
The car takes b feet to stop when traveling at 10 mph.
The number s is the distance it takes the car to stop when traveling at v mph.
33. The expression r(99) gives the average number of songs downloaded daily at a price of $0.99 per song.
34. The current price per song is p0 , so r(p0 ) is the average number of daily downloads at the current price.
35. The expression p0 − 10 means 10 cents less than the current price per song, so r(p0 − 10) means the average number of
daily downloads at a price 10 cents lower than the current price.
36. The expression p0 − 10 means 10 cents less than the current price per song, so r(p0 − 10) means the average number of
daily downloads at a price 10 cents lower than the current price. So
r(p0 − 10) − r(p0 )
means
Number of daily downloads
at lower price
−
Current number
of daily downloads
,
or the change in the average number of daily downloads if the price drops 10 cents from the current price.
37. The expression r(p0 ) is the average number of daily downloads at the current price. Since there are 365 days in a year,
365r(p0 ) is the number of downloads in one year at current price.
38. The expression 0.80p0 is 80% of the current price, or 20% less than the current price. Thus, r(0.80p0 ) is the average
number of daily downloads after the price drops 20%.
39. The expression r(p0 ) is the average number of daily downloads at the current price. Since there are 24 hours in one day,
r(p0 )/24 is the average number of downloads every hour.
40. The expression r(p0 ) is the average number of daily downloads at a price of p0 cents per song, so
p0 · r(p0 ) = Price per song · Number of downloaded songs
= Average daily revenue for music store.
We see that multiplying the number of downloaded songs by the price per song gives the average daily revenue of the
music store—that is, the average amount of money it brings in each day.
41. We have
1
· 100(100 + 1)
2
= 50(101)
f (100) =
= 5050.
4.2 SOLUTIONS
91
42. Working this in steps, we first find f (n + 1):
1
(n + 1)((n + 1) + 1)
2
1
= (n + 1)(n + 2).
2
f (n + 1) =
Now we subtract:
f (n + 1) − f (n) =
1
1
(n + 1)(n + 2) − n(n + 1)
2
2
{z
} | {z }
|
f (n+1)
f (n)
1
= (n + 1) ((n + 2) − n)
2
1
= (n + 1) · 2
2
= n + 1.
factor out 12 (n + 1)
43. This is the number of people (in 1000s) infected by a strain whose open-air survival time is 3 minutes longer than the
most common strain.
44. h(t0 ) is the number eventually infected by the most common strain, and h(2t0 ) is the number eventually infected by a
strain that survives twice as long in open air. The ratio of these numbers gives how many times more people are infected
by the longer-living strain.
45. This is the gas mileage of the car if its tire pressure is 90% of the recommended pressure, that is, if the tires are 10%
underinflated.
46. g(P0 ) is the mileage at the recommended pressure, and g(P0 − 5) is the mileage if the tires are underinflated by 5 lbs/in2 .
So g(P0 ) − g(P0 − 5) is the difference in gas mileage between driving on properly inflated tires and tires underinflated
by 5 lbs/in2 .
Solutions for Section 4.2
EXERCISES
1. (a) We have f (1) = 8 and f (3) = 6, so f (1) is larger than f (3).
(b) Since the input value x is subtracted from 9 to get the output value, a greater input value subtracts a greater amount,
and so give a lesser output value.
2. (a) We have g(−5) = −5 − 2 = −7 and g(−2) = −2 − 2 = −4. Since −4 > −7, the greater output is g(−2).
(b) The output value is obtained by subtracting 2 from the input value, so a greater input gives a greater output. Since −2
is greater than −5, it gives a greater output value.
3. (a) We have C(100) = −100/5 = −20 and C(200) = −200/5 = −40, so C(100) is larger, since −20 > −40.
(b) By rewriting −p/5 as
−1
p,
5
we see that the output value is obtained by multiplying the input value by −1/5. When you multiply two numbers
by a negative number you change their order, so a lesser input gives a greater output. Since 100 < 200, the greater
output is C(100). This might seem strange, but keep in mind that we find the output of C by multiplying the input by
a negative number. This means that the larger the positive input value, the smaller (the “more negative”) the output
value. Since 200 is a bigger input value than 100, it leads to a smaller (“more negative”) output value.
4. (a) We have h(4) = 4/5 = 0.8, and h(6) = 6/5 = 1.2, so h(6) is greater.
(b) The output is obtained by dividing the input by 5, so a greater input gives a greater output.
92
Chapter Four /SOLUTIONS
5. These expressions are equivalent. In each case, we divide f (t) by 3.
6. These two expressions do not appear to be equivalent, so we check by trying t = 0:
f (02 ) = 02 /2 + 7 = 7,
and
(f (0))2 =
Since 7 6= 49, they are not equivalent.
2
0
+7
2
= 72 = 49.
7. These two expressions do not appear to be equivalent, so we check:
2f (t) = 2
t
+ 7 = t + 14,
2
and
2t
+ 7 = t + 7.
2
Indeed, they are not equivalent (for example, if t = 1, the first is 15 and the second is 8).
f (2t) =
8. These expressions are equivalent. We note that (2t)2 = (2t)(2t) = 4t2 , so they are the same.
9. The same, since the expressions for f and g are equivalent. We have (x − 2)2 + 1 = x2 − 4x + 5 for all x.
10. Different, since the two expressions are not equivalent. We have 2(x + 1)(x − 3) = 2x2 − 4x − 6, which is 2 times the
expression for g(x).
11. The same. The expressions are the same except that a different letter is used for the independent variable. This does not
change the function.
12. The same, since the expressions are equivalent.
13. Different. We have 5(x − 2) + 3 = 5x − 7, which is not equivalent ot 5x + 7.
14. The same, since the two expressions are equivalent.
15. Different. The first function is half the second one.
16. The same, since the two expressions are equivalent.
17. The variable is r and the constant is π.
18. The variable is x and the constants are b and m.
19. The variable is t and the constants are r and A.
20. The variable is r and the constants are A and t.
PROBLEMS
21. (a) We have G(50) = 17 − 0.05 · 50 = 14.5, and G(100) = 17 − 0.05 · 100 = 12. So G(50) > G(100).
(b) Since the term 0.05d is subtracted from 17 in the expression for G(d), the value of G(d) is smaller when d is larger.
This makes sense, since there is less gas left when you have driven further.
22. (a) We have B(35) = 30 − 480/35 = 16 minutes, and B(45) = 30 − 480/45 = 19 minutes, so B(45) > B(35).
(b) Since v is in the denominator, the fraction 480/v is smaller when v is larger. So when v is larger, a smaller amount
is being subtracted from 30, giving a larger answer. This makes sense, since if you drive faster you have longer for
breakfast.
23. (a) Abby’s expression is 325 − 65t and Leah’s is 65(5 − t).
(b) The two expressions are equivalent and define the same function.
24. (a) Sharif’s calculation starts with P , then adds 0.05P to get P +0.05P . Then he takes 5% of this to get 0.05(P +0.05P ),
and adds that to P + 0.05P to get
P + 0.05P + 0.05(P + 0.05P ).
Donald starts with P , multiplies it by 1.05, giving 1.05P . Then he multiplies again, giving 1.05(1.05P ) = 1.052 P .
4.3 SOLUTIONS
93
(b) Yes. Distributing and collecting like terms, we see that Sharif’s function is
f (P ) = P + 0.05P + 0.05P + 0.0025P = (1 + 0.05 + 0.05 + 0.0025)P = 1.1025P.
Donald’s is g(P ) = (1.05)2 P = 1.1025P .
25. Writing this as Q = (1/4)t we have k = 1/4.
26. Writing this as Q = (n + 1)t we have k = n + 1.
27. Factoring, we can write this as Q = (b + r)t, so k = b + r.
28. Writing this as
Q=
we have k =
√
√
3/2 or k = 0.5 3.
√
3
t,
2
29. Writing this as
Q=
α−β
· t,
γ
we have k = (α − β)/γ.
30. Expanding parentheses gives
Q = (t − 3)(t + 3) − (t + 9)(t − 1)
= t2 − 9 − (t2 + 8t − 9)
= −8t,
so Q = −8t, and k = −8.
31. (a) Since your possibilities are all apartments of 1000 square feet, A is the constant (always 1000), and d is the variable
(since you can have the same-sized apartment at different distances from the station).
(b) Since your possibilities are all apartments 1 mile from the station, d is the constant (always 1), and A is the variable
(since you can have apartments of different sizes that are the same distance from the station).
(c) Here, both d and A are variables. In choosing the apartment, you must decide whether you want to be farther from
the station or to have a smaller area.
Solutions for Section 4.3
EXERCISES
1. We begin by setting f (x) = 0:
√
x−2−4 = 0
√
x − 2 = 4.
Since x − 2 must equal 16 in order for the square root to be 4, x − 2 = 16, so x = 18.
2. We begin by setting f (x) = 0:
6 − 3x = 0
−3x = −6
x = 2.
94
Chapter Four /SOLUTIONS
3. We begin by setting f (x) = 0:
4x − 9 = 0
4x = 9
9
x= .
4
4. We begin by setting f (x) = 0:
2x2 − 18 = 0
2x2 = 18
x2 = 9.
This means that x = 3 and x = −3 are solutions.
5. We begin by setting f (x) = 0:
√
2 x − 10 = 0
√
2 x = 10
√
x=5
x = 25.
6. We begin by setting f (x) = 0:
2(2x − 3) + 2 = 0
2(2x − 3) = −2
2x − 3 = −1
2x = 2.
x = 1.
7. We begin by setting g(t) = 1:
6−t = 1
−t = −5
t = 5.
8. We begin by setting g(t) = 10:
2
t + 6 = 10
3
2
t=4
3
t = 6.
9. We begin by setting g(t) = −3:
3(2t − 1) = −3
2t − 1 = −1
2t = 0
t = 0.
4.3 SOLUTIONS
95
10. We begin by setting g(t) = 1:
t−1
=1
3
t−1 = 3
t = 4.
11. We begin by setting g(t) = 0:
2(t − 1) + 4(2t + 3) = 0
2t − 2 + 8t + 12 = 0
10t + 10 = 0
10t = −10
t = −1.
12. The graph contains the points (20, 60), (100, 60), so the solutions are x = 20, 100.
13. The graph contains the point (60, 20), so v(60) = 20.
14. (a) It crosses the y-axis at −2.
(b) It crosses the x-axis at approximately −1.5, 1.5.
15. (a) It crosses the y-axis at 75.
(b) It crosses the x-axis at 5.
16. (a) From the graph we see that f (0) = 1.
(b) The function is equal to zero when x = −1 and x = 3.
17. (a) The table shows f (6) = 3.7, so t = 6. In a typical June, Chicago has 3.7 inches of rain.
(b) First evaluate f (2) = 1.8. Solving f (t) = 1.8 gives t = 1 or t = 2. Chicago has 1.8 inches of rain in January and
in February.
PROBLEMS
18. (a) We want to know when 4800 − 400m is equal to 2000, so we want to solve
4800 − 400m = 2000.
(b) Table 4.1 shows the balance, and Figure 4.1 shows the plotted points and the solution at m = 7.
Table 4.1 Balance in an account 1–11 months after its
establishment
m (months)
1
3
5
7
9
11
Balance $4800 − 400m
4400
3600
2800
2000
1200
400
account balance ($)
5000
$2000
when m = 7
2000
3
6
9
Figure 4.1
12
m(months)
96
Chapter Four /SOLUTIONS
19. (a) We want to know when 15 − d/20 is equal to 10, so we want to solve
15 − d/20 = 10.
(b) Table 4.2 shows the number of gallons, and Figure 4.2 shows the plotted points and the solution at d = 100.
Table 4.2 Gallons of gas left 40–140 miles after leaving a
gas station
d (miles)
40
60
80
100
120
140
Gallons of gas (15 − d/20)
13
12
11
10
9
8
gallons of gas
15
10 gallons
when d = 100
10
5
100
200
d(miles)
Figure 4.2
20. (a)
(b)
(c)
(d)
(e)
When t = 0 the car’s position is expressed as s(0) = 112 (0) + 0 + 40 = 40, so the car is 40 miles from home.
The car’s position after one hour is expressed as s(1). To find s(1), we have s(1) = 11(1)2 + 1 + 40 = 52 miles.
In function notation, we have s(t) = 142.
To express when the car is 142 miles from home, we write 142 = 11t2 + t + 40.
A little trial and error shows that when t = 3, s(3) = 142.
21. We have
f (b) = 20
b
= 20
2 − 3b
| {z }
f (b)
b = 20(2 − 3b) Multiply by denominator
b = 40 − 60b
61b = 40
40
b=
.
61
Distribute
Collect b-terms
22. We have
3h(x) + 1 = 7
3h(x) = 6
h(x) = 2
2
3− = 2
x
because h(x) = 3 − 2/x
4.3 SOLUTIONS
−
2
= −1
x
2
=1
x
x = 2.
Checking our answer, we have
3h(2) + 1 = 3(2) + 1
= 6 + 1 = 7,
because h(2) = 3 − 2/2 = 2
as required.
23. We have
w(t) = 3t + 5
w(t − 1) = 3(t − 1) + 5 = 3t + 2
w(2t) = 3(2t) + 5
= 6t + 5.
Solving, we see that
w(t − 1) = w(2t)
3t + 2 = 6t + 5
−3t = 3
t = −1.
Checking, we see that at t = −1,
w(t − 1) = w(−1 − 1) = w(−2) = 3(−2) + 5 = −1
w(2t) = w (2(−1)) = w(−2) = 3(−2) + 5 = −1,
as required.
24. We have
w(0.2x + 1) = 0.5 − 0.25(0.2x + 1)
= 0.5 − 0.05x − 0.25
= 0.25 − 0.05x
0.2w(1 − x) = 0.2 (0.5 − 0.25(1 − x))
= 0.2 (0.5 − 0.25 + 0.25x)
= 0.2 (0.25 + 0.25x)
= 0.05 + 0.05x.
Thus, solving gives
w(0.2x + 1) = 0.2w(1 − x)
0.25 − 0.05x = 0.05 + 0.05x
−0.1x = −0.2
x = 2.
Checking our answer, we see that
w(0.2 · 2 + 1) = w(1.4) = 0.5 − 0.25(1.4)
= 0.15
0.2w(1 − 2) = 0.2(−1) = 0.2(0.5 − 0.25(−1)) = 0.2(0.75) = 0.15,
as required.
97
98
Chapter Four /SOLUTIONS
25. (a)
(b)
(c)
(d)
The function f (x) is greater than g(x) for x = −2, x = −1 and x = 3.
The functions are equal at x = 2 and x = 4.
When x = 0, we see that f (x) = 0.
There are no values for which g(x) = 0.
26. (a) Reading from the table, we see that 3x + 2 < 8 when x = 0 and when x = 1.
(b) Reading from the table, we see that 3x + 2 is greater than 8 when x = 3 and when x = 4.
(c) Reading from the table, we see that 3x + 2 equals 8 when x = 2.
27. (a) Reading from the table, we see that 10 + 2s − 9.8s2 < 9.89 when s = 0.5, when s = 0.75, and when s = 1.
(b) Reading from the table, we see that his height is greater than 9.89 meters when s = 0.
(c) Reading from the table, we see that height = 9.89 meters when s = 0.25.
28. (a) Reading from the table, we see that 12 − 3v is less than −3 + 2v when v = 4, when v = 5, and when v = 6.
(b) Reading from the table, we see that 12 − 3v > −3 + 2v when v = 0, when v = 1, and when v = 2.
(c) Reading from the table, we see that 12 − 3v = −3 + 2v when v = 3.
29. (a) From the table, we see that the monthly premium a 65 year old male pays is f (65) = $218.
(b) From the table, we see that the monthly premium a 50 year old female pays is g(50) = $39.
(c) We seek the value of a such that f (a) = 102. This is the age at which males pay a monthly premium of $102. This
occurs when a = 60.
(d) We seek the value of a such that g(a) = 57. This is the age at which females pay a monthly premium of $57. This
occurs when a = 55.
(e) We are looking for the ages at which males and females pay the same monthly premium. This occurs at three ages,
35, 40, and 45.
(f) Women pay smaller monthly premiums at ages 50, 55, 60, 65, and 70.
30. (a) Looking at the graph, we see that p is close to 50 when profits per package are $8000.
(b) We substitute p = 50 into the equation:
10,000 −
100,000
= 8000.
p
The left-hand side becomes
10,000 −
100,000
100,000
= 10,000 −
= 10,000 − 2000 = 8000.
p
50
Since the two sides are equal, p = 50 is a solution.
31. (a) Looking at the graph, we see that t is close to 7 years when tuition is $3700.
(b) We substitute t = 7 into the equation:
3000 + 100t = 3700.
The left-hand side becomes
3000 + 100t = 3000 + 100 · 7 = 3000 + 700 = 3700.
Since the two sides are equal, t = 7 is a solution.
32. (a) For each hour, we subtract 2 gallons from Antonio’s total and 6 from Lucia’s total, giving Table 4.3. However, we
note that in hour 5, Lucia runs out of gas and fills up to 30 gallons again.
Table 4.3
Hours driven
0
1
2
3
4
5
6
Antonio (gallons)
14
12
10
8
6
4
2
0
Lucia (gallons)
30
24
18
12
6
0
24
18
(b) We see that after 4 hours, they each have 6 gallons of gas.
7
4.4 SOLUTIONS
99
(c) Antonio gets to San Diego first, since Lucia has to stop for gas.
(d) Antonio arrives in San Diego, since he does not need to get gas. Lucia, on the other hand, needs gas after 5 hours, so
she gets stuck in the desert.
(e) (i) The two vehicles have the same amount of gas when 14 − 2t = 30 − 6t, so t = 4, as we read in the table.
(ii) Antonio runs out of gas when 14 − 2t = 0, which is when t = 7.
(iii) Lucia runs out of gas when 30 − 6t = 0, which is when t = 5.
Solutions for Section 4.4
EXERCISES
1. We have
f (4) − f (2)
28 − 10
18
∆f
=
=
=
= 9.
∆x
4−2
4−2
2
2. We have
f (4) − f (−2)
28 − (−2)
30
∆f
=
=
=
= 5.
∆x
4 − (−2)
4 − (−2)
6
3. We have
f (−4) − f (−2)
4 − (−2)
∆f
6
=
=
=
= −3.
∆x
−4 − (−2)
−4 − (−2)
−2
4. We have
f (3) − f (1)
∆f
18 − 4
14
=
=
=
= 7.
∆x
3−1
3−1
2
5. We have
g(1) = 2 · 13 − 3 · 12 = −1
g(3) = 2 · 33 − 3 · 32 = 27,
so
Average rate
of change
g(3) − g(1)
3−1
27 − (−1)
=
= 14.
2
=
6. We have
g(−1) = 2(−1)3 − 3(−1)2 = −5
g(4) = 2 · 43 − 3 · 42
so
Average rate
of change
g(4) − g(−1)
4 − (−1)
80 − (−5)
= 17.
=
5
=
= 80,
100
Chapter Four /SOLUTIONS
7. We have
g(0) = 2 · 03 − 3 · 02
3
=0
2
g(10) = 2 · 10 − 3 · 10 = 1700,
so
Average rate
of change
g(10) − g(0)
10 − 0
1700 − 0
=
= 170.
10
=
8. We have
g(−0.1) = 2(−0.1)3 − 3(0.1)2 = −0.032
g(0.1) = 2(0.1)3 − 3(0.1)2
= −0.028,
so
Average rate
of change
g(0.1) − g(0.1)
0.1 − (−0.1)
−0.028 − (−0.032)
= 0.02.
=
0.2
=
9. Year 2004 is t = 1, and year 2007 is t = 3. We have
Value in 2004 = 1000 · 21/6 = 1122.46
Value in 2007 = 1000 · 24/6 = 1587.40,
so
Average rate
of change
1587.40 − 1122.46
2007 − 2004
464.94
=
3
= 154.98.
=
This means the investment’s value climbs at an average rate of $154.98/year.
10. We have
Average rate
of change
382 − 336
2007 − 1979
46
=
= 1.643.
28
=
This means that CO2 levels have risen at an average rate of 1.643 ppm per year.
11. We have
Average rate
of change
Change in sea level
Time elapsed
130 m
=
22,000 years
= 0.00591 meters/year.
=
4.4 SOLUTIONS
101
12. The time period in question is 2100 − 1990 = 110 years. We have:
Average rate of change =
10◦ F
= 0.0909◦ F/year.
110 years
PROBLEMS
13. We have
∆f
f (5) − f (1)
7932.7 − 5807.5
2125.2
=
=
=
= 531.3.
∆x
5−1
5−1
4
This tells us that between the years 2001 and 2005, the total US debt increased at an average rate of $531.3 billion/year.
14. We have
7932.7 − 6783.2
1149.5
f (5) − f (3)
=
=
= 574.75
5−3
5−3
2
f (3) − f (0)
6783.2 − 5674.2
1109
=
=
= 369.67.
3−0
3−0
3
We see that the US debt increases at a higher average rate between 2003 and 2005 than it does between 2000 and 2003.
15. We have
f (6) − f (5)
8494.1 − 7932.7
561.4
=
=
= 561.4
6−5
6−5
1
f (5) − f (4)
7932.7 − 7379.1
553.6
=
=
= 553.6.
5−4
5−4
1
The US debt is not going down; it just did not climb as much between 2005 and 2006 as it did between 2004 and 2005.
16. We have
f (6) − f (0)
8494.1 − 5674.2
2819.9
=
=
= 469.98.
6−0
6−0
6
This tells us the US debt grew at an average rate of $469.98 billion/year between 2006 and 2007. Letting
f (10) − f (6)
= 469.98
10 − 6
f (10) − f (6)
= 469.98
4
f (10) − f (6) = 4 · 469.98
= 10,374.04.
Here, we assume that the average rate of change between 2006 and 2010 is the same as between 2000 and 2006. Given
this assumption, we predict that the US debt rises to over $10 trillion by 2010.
17. The expression g(5) − g(0) represents the change in the market value of the house between years t = 0 and t = 5. Thus,
the statement g(5) − g(0) = 30 tells us that the house’s market value increased by $30, 000.
18. The expression (g(10) − g(4))/6 represents the average rate of change in the market value of the house over the six-year
period from year t = 4 to t = 10. Thus, the statement
g(10) − g(4)
=3
6
tells us that the house’s market value increased at an average rate of $3000/year.
102
Chapter Four /SOLUTIONS
19. The expression (g(20) − g(12))/(20 − 12) represents the average rate of change in the market value of the house over
the eight-year period from year t = 12 to t = 20. Thus, the statement
g(20) − g(12)
= −1
20 − 12
tells us that the house’s market value decreased at an average rate of $1000/year.
20. This expression is the change in the town’s population from year 3 to year a. Since the town is growing, f (a) − f (3) is
positive so long as a > 3.
21. This expression is the average rate of change in the town’s population from year 3 to year a. Since the town is growing,
this expression is positive provided a 6= 3.
22. This expression is the change in the town’s population from year t + a to year t + b, that is, the change from b years after
year t to a years after year t. Since the town is growing, this change will be positive so long as a > b.
23. This tells us the area of wetlands decreases by 25,000 acres between year 0 and year 25.
24. This tells us that between year t = 10 and t = 20, the area of wetlands decreased at an average rate of 520 acres per year.
25. This tells us that the change in the area of wetlands during the 10-year period from t = 20 to t = 30 is less than (more
negative than) the change from t = 10 to t = 20. In other words, a larger area is lost between years 20 and 30 than
between years 10 and 20.
26. This is the difference between the amount of electricity used in year 10 and year h. The amount used goes down over
time, so for this difference to be negative, year h must come after year 10, so h > 10.
27. This is the average rate of change of the amount of electricity used in year 5 and year 5 + h. The amount used goes down
over time, so for this rate to be negative, year 5 + h must come after year 5, so h > 0.
28. This is the difference between the amount of electricity used in year h − 1 and year h + 1. The amount used goes down
over time, and since year h + 1 comes after year h − 1, we see that h can take any value for which the expression is
defined.
29. We have
w(10) − w(5)
1715 − 1660
55
=
=
= 11,
10 − 5
5
5
which means that methane levels increased at an average rate of 11 ppb per year between 1985 and 1990.
30. We have
w(10) − w(0)
1715 − 1575
140
=
=
= 14
10 − 0
10
10
w(25) − w(10)
1775 − 1715
60
=
=
=4
25 − 10
15
4
Larger
Smaller.
This tells us that the methane level rises at a slower average rate from 1990 to 2005 than from 1980 to 1990 (4 ppb versus
11 ppb). In other words, this suggests that the methane level is increasing, but a slower and slower pace.
31. We have
1770 − 1750
20
w(20) − w(15)
=
=
=4
20 − 15
5
5
w(25) − w(20)
1775 − 1770
5
=
=
= 1.
25 − 20
5
5
Thus, the average rate of change goes down from 4 ppb to 1 ppb. This does not mean the methane level is going down.
The methane level is still increasing, but not as fast as it once was.
32. We see that
1715 − 1575
140
w(10) − w(0)
=
=
= 14.
10 − 0
10
10
This means the methane level rose at an average rate of 14 ppb per year from 1980 to 1990. Therefore, by assumption,
w(0) − w(−20)
= 14
20
4.5 SOLUTIONS
w(0) − w(−20) = 14 · 20
103
Multiply both sides by 20
w(0) = 280 + w(−20)
w(−20) = w(0) − 280
Solve for w(−20)
= 1575 − 280 = 1295.
This means that if methane levels rose at the same rate from t = −20 (that is, 1960) to t = 0 (or 1980) as they did from
1980 to 1990, then the 1960 level would have been 1295 ppb.
Solutions for Section 4.5
EXERCISES
1. Since a distance of 55 miles is traveled for every hour, d = 55t. Thus, d is proportional to t, with k = 55.
2. Since each barrel costs $70, p = 70b. Thus P is proportional to b, with k = 70.
3. Since the sale takes 30% off, the sale price is 70% of the original price, so p = 0.7p0 . Thus p is proportional to p0 , with
k = 0.7.
4. Although the cost of drinks alone is 20n dollars, the total cost of having the drinks is C = 20 + 5n. So C is not
proportional to n.
5. Yes. We have y = kx with k = 5.
6. Yes. We have y = kx with k = 7.
7. No. This can be written as y = kx2 with k = 1, but not as y = kx.
√
8. Yes. We have y = kx with k = 5.
9. Yes. We have y = kx with k = 1/9.
10. No. This can be written as y = k/x with k = 9, but not as y = kx.
11. No. This can be written as y = mx + b with m = 1 and b = 2, but it can not be written as y = kx.
12. No. This can be written (after expanding the parentheses) as y = mx + b with m = 3 and b = 6, but it can not be written
as y = kx.
13. Yes. We have y = kx with k = 42. This is because y = 6z and z = 7x and so y = 6(7x) = 42x.
14. Since r is directly proportional to s, we know that r = ks for some constant k. Filling in values for r and s, we solve for
k:
r = ks
36 = 4k
k = 9.
Now, use the values of k and s to solve for r:
r = ks
r = (9)(5)
r = 45.
15. Since p is directly proportional to q, we know that p = kq for some constant k. Filling in values for p and q, we solve for
k:
p = kq
24 = 6k
k = 4.
104
Chapter Four /SOLUTIONS
Now, use the values of k and p to solve for q:
p = kq
32 = 4q
q = 8.
16. Since y is directly proportional to x, we know that y = kx for some constant k. Filling in values for x and y, we solve
for k:
y = kx
16 = 12k
16
k=
12
4
k= .
3
Now, use the values of k and x to solve for y:
y = kx
4
y = (9)
3
y = 12.
17. Since s is directly proportional to t, we know that s = kt for some constant k. Filling in values for s and t, we solve for
k:
s = kt
35 = 25k
35
k=
25
7
k= .
5
Now, use the values of k and s to solve for t:
s = kt
7
14 = t
5
5
t = (14)
7
t = 10.
18. Interest earned is proportional to the starting balance, so
Interest earned = k × Starting balance.
Since k = 0.06, we have
I = 0.06B.
(a) If B = 500, we have I = 0.06(500) = $30.
(b) If B = 1000, we have I = 0.06(1000) = $60.
(c) If B = 5000, we have I = 0.06(5000) = $300.
19. Substituting into the general formula y = kx, we have 6 = k(4) or k = 32 . So the formula for y is
y=
When y = 8, we get 8 = 23 x, so x =
2
3
·8=
16
3
3
x.
2
= 5.33.
20. Since graph has the form y = kx, the function it represents is a direct proportion. The graph goes through the point
15 − 0
5
(6, 15) and the origin. Therefore the rate of change is k =
= .
6−0
2
4.5 SOLUTIONS
105
21. Since graph is not in the form y = kx, the function does not represents a direct proportion.
22. Since graph is not in the form y = kx, the function does not represents a direct proportion.
23. Since graph has the form y = kx, the function it represents is a direct proportion. The graph goes through the point
2
1−0
= .
(1.5, 1) and the origin. Therefore the rate of change is k =
1.5 − 0
3
24. Since graph is not in the form y = kx, the function does not represents a direct proportion.
25. Since graph has the form y = kx, the function it represents is a direct proportion. The graph goes through the point
4−0
= − 34 .
(−3, 4) and the origin. Therefore the rate of change is k = −3−0
26. Since graph is not in the form y = kx, the function does not represents a direct proportion.
27. Since graph curves, it is not a direct proportion.
28. Since graph curves, it is not a direct proportion.
PROBLEMS
29. Each week 50 vehicles are produced, so we have v = 50t. Thus
t=
v
=
50
1
50
v,
so t is proportional to v, with k = 1/50.
30. Each week 120 pounds of apples are sold. Thus we have a = 120w, so a is proportional to w, with k = 120.
31. The perimeter of a square can be expressed as P = 4s where s is the length of one side of the square. Therefore, k = 4.
32. (a) The revenue can be expressed as R = kx where x is the number of bicycles sold. Since the revenue is $20,000 when
50 bicycles are sold, we have $20,000 = k · 50, so k = 400. Notice that for the units on each side of the equal sign
to agree, the units of k are dollars per bicycle.
(b) Now we can write R = 400x. If x = 75 then R = 400 · 75 = $30,000.
33. (a) Since the total cost is proportional to the number of gallons pumped, we have C = f (n) = kn. Solving for k, we
have $36.63 = k(11) gallons, so k = $3.33, and it units are dollars per gallon.
(b) To find the cost of 15 gallons, C = (3.33)(15) = $49.95.
34. (a) Since the capacity is proportional to area, C = f (A) = kA. Solving for k, 5600 = k · 280. Therefore, k = 20. The
units of k are BTUs per square foot.
(b) Solving C = f (A) = 20A for A, we have 10,000 = 20A, so A = 500 square feet.
35. (a) Since the interest rate is 2%, we have r = 0.02; also t = 0. Thus
B = P (1 + 0.02)10 = P (1.219) = 1.219P.
Thus B is proportional to P , with k = 1.219.
(b) We solve for P in terms of B:
B = P (1.02)10
B
P =
= 0.820B.
(1.02)10
Thus P is proportional to B with k = 0.820.
36. (a) We have m = kr for some constant k.
(b) We have k = 1/87. We set m = 10.5 and solve for r:
m = kr
1
r
10.5 =
87
r = 87 · 10.5
r = 913.5.
106
Chapter Four /SOLUTIONS
The real locomotive is 913.5 inches long. To convert this answer to feet, divide by 12 to get 76.125 feet. The
locomotive is about 76 feet long.
(c) We have k = 1/220. We convert 75 feet to 75 · 12 = 900 inches. Thus
1
· 900 = 4.091.
220
m = kr =
The scale model measures about 4 inches long.
37. (a) The cost C is proportion to the amount bought x, so
C = kx.
(b) We substitute C = 28.50 dollars and x = 3 yards and solve for k:
28.50 = k(3)
28.50 dollars
= 9.50 dollars per yard.
k=
3 yards
So the formula is
C = 9.50x.
(c) The cost for 5.5 yards is C = 9.50(5.5) = $52.25.
38. (a) We have M = kd. Using M = 1/2 and d = 5 we get
M = kd
0.5 = k(5)
0.5 inches
k=
= 0.1 inches per mile.
5 miles
(b) The formula is
M = 0.1d.
(c) When M = 3.25, we have 3.25 = 0.1d so d = 3.25/0.1 = 32.5. These two towns are 32.5 miles apart.
39. Let M = blood mass and B = body mass. Then M = k · B. Using the fact that M = 150 when B = 3000, we have
M = k·B
150 = k · 3000
k = 150/3000 = 0.05.
We have M = 0.05B. For a human with B = 70, we have M = 0.05(70) = 3.5 kilograms of blood.
40. Distance traveled is proportion to quantity of gas consumed, so
Distance traveled = k × Quantity of gas.
If we let D represent distance traveled in miles and G represent the quantity of gas consumed in gallons, we have
D = kG.
When D = 225 and G = 5, we have
225 = k(5)
225 miles
k=
= 45 miles per gallon.
5 gallons
The constant k gives the rate of gas consumption in miles per gallon, or the fuel efficiency of the car.
SOLUTIONS to Review Problems for Chapter Four
107
Solutions for Chapter 4 Review
EXERCISES
1. (a)
(b)
(c)
(d)
At 4000 miles from the earth’s center the astronaut weighs 180 pounds.
At a thousand miles from the earth’s center the astronaut weighs 36 pounds.
At 36,000 miles from the earth’s center the astronaut weighs b pounds.
The number w is the astronaut’s weight in pounds at r thousand miles from the earth’s center.
2. Since w(c) represents the number of cars without parking spaces when there are c cars, w(8000) represents the number
of cars without spaces when there are 8000 cars. We evaluate w(8000) = −2000 + 8000 = 6000. There are 6000 cars
without parking spaces when there are 8000 cars.
3.
0+1
2·0+1
= 1.
f (0) =
4.
−1 + 1
2(−1) + 1
0
=
−2 + 1
= 0.
f (−1) =
5.
0.5 + 1
2 · 0.5 + 1
1.5
=
2
= 0.75.
f (0.5) =
6.
−0.5 + 1
2(−0.5) + 1
0.5
=
−1 + 1
0.5
.
=
0
f (−0.5) =
Division by 0 is undefined, so the expression f (−0.5) is undefined.
7.
f
1
3
=
=
1
3
2·
1
3
2
3
+1
+1
1
3
+
+
4/3
5/3
4
= .
5
=
3
3
3
3
108
Chapter Four /SOLUTIONS
8.
f (π) =
π+1
.
2π + 1
9. (a) To evaluate g(3), we evaluate the expression for g at t = 3:
g(3) =
(3)2 + 1
10
=
= 1.25.
5 + (3)
8
(b) To evaluate g(−1), we evaluate the expression at t = −1
g(−1) =
(−1)2 + 1
2
= = 0.5.
5 + (−1)
4
(c) To evaluate g(a), we evaluate the expression at t = a:
g(a) =
a2 + 1
.
5+a
10. To evaluate, we substitute the input value of the function for s in g(s) =
(a) We substitute 4 for s:
g(4) =
(b) We substitute a for s:
5s + 3
.
2s − 1
20 + 3
23
5(4) + 3
=
=
.
2(4) − 1
8−1
7
g(a) =
(c) We substitute a for s and add 4 to our result:
g(a) + 4 =
5(a) + 3
5a + 3
=
.
2(a) − 1
2a − 1
5(a) + 3
5a + 3
+4 =
+ 4.
2(a) − 1
2a − 1
(d) We substitute a + 4 for s:
g(a + 4) =
11. (a)
(b)
(c)
(d)
5(a + 4) + 3
5a + 20 + 3
5a + 23
=
=
.
2(a + 4) − 1
2a + 8 − 1
2a + 7
Since g(y) is −10 when y = 10, we have g(10) − 10 = −10 − 10 = −20.
Since g(y) is 5 when y = 0, we have 6g(0) = 6 · 5 = 30.
Since g(y) is −10 when y = 10 and g(y) is −5 when y = −10, we have g(10) − g(−10) = −10 − (−5) = −5.
Since g(y) is 10 when y = 5, we have g(5) + 7g(5) = 8g(5) = 80.
12. (a) The value of the function at t = 1 is C(1) = 128.3. Therefore, in 1991 there were 128.3 million cars.
(b) This is asking for the t value where C(t) = 127.3 This occurs when t = 3. Therefore in 1993 there were 127.3
million cars.
13. (a) The expression P (20) asks us to find the population when t = 20. From the table, P (20) = 205.0. Therefore, in
1970 there were 205.0 million people.
(b) The equation P (t) = 281.4 asks us to find the year in which the population was 281.4 million people. From the
table, this occurs when t = 50. Therefore in 2000 there were 281.4 million people.
14. (a)
(b)
(c)
(d)
15. (a)
(b)
(c)
(d)
From the graph, we see that f (2) = 5, while f (1) = 4. Thus, f (2) − f (1) = 5 − 4 = 1.
From the graph, we see that f (2) = 5, while f (−1) = −4. Thus, f (2) − f (−1) = 5 − (−4) = 9.
From the graph, we see that f (−1) = −4, while f (2) = 5. Thus, 2f (−1) + f (2) = 2(−4) + 5 = −3.
From the graph, we see that f (−2) = −5. Thus, 1/f (−2) = 1/(−5) = −0.2.
From the graph, we see that d(2) = 1, while d(1) = 4. Thus, d(2)/d(1) = 1/4 = 0.25.
From the graph, we see that d(1) = 4. Thus, d(1) · d(1) = 4 · 4 = 16.
From the graph, we see that d(−2) = 1, while d(−1) = 4. Thus, 4d(−2) − d(−1) = 4(1) − 4 = 0.
From the graph, we see that d(−2) = 1. Thus, 4d(−2) − (−1) = 4(1) + 1 = 5.
SOLUTIONS to Review Problems for Chapter Four
16. The expressions are equivalent because 22 = 4, so 22 g(z) = 4g(z).
17. These two expressions do not appear to be equivalent, so we check:
4g(z) = 4(4z 2 − 3) = 16z 2 − 12,
and
g(4z) = 4(4z)2 − 3 = 4(16z 2 ) − 3 = 64z 2 − 3.
Indeed, they are not equivalent (for example, if z = 1, the first is 4 and the second is 61).
18. These two expressions do not appear to be equivalent, so we check:
√
√
g( z) = 4( z)2 − 3 = 4z − 3,
and
p
g(z) =
p
4z 2 − 3.
Indeed, they are not equivalent (for example, if z = 2, the first is 5 and the second is
2
2
19. The expressions are equivalent because z = z · z, so g(z ) = g(zz).
20. We are trying to find the value of x at which f (x) = 3. Since
f (x) = 5x − 2,
we have
f (x) = 3 = 5x − 2
3 + 2 = 5x
5 = 5x
1 = x.
21. We are trying to find the value of x at which f (x) = 3. Since
f (x) = 5x − 5,
we have
f (x) = 3 = 5x − 5
3 + 5 = 5x
8 = 5x
8
= x.
5
In decimal form, x = 1.6.
22. We are trying to find the value of x at which f (x) = 3. Since
f (x) =
2x
,
5
we have
f (x) = 3 =
2x
5
5·3 = 5·
15 = 2x
15
= x.
2
In decimal form, x = 7.5.
2x
5
√
13).
109
110
Chapter Four /SOLUTIONS
23. We are trying to find the value of x at which f (x) = 3. Since
f (x) =
5
,
2x
f (x) = 3 =
5
2x
we have
5
2x
2x · 3 = 2x ·
6x = 5
5
x= .
6
In decimal form, x = 0.833.
24. (a) We want to know when 5000 + 350t is equal to 12,000, so we want to solve
5000 + 350t = 12,000.
(b) Table 4.4 shows the population values, and Figure 4.3 shows the plotted points and the solution at t = 20.
Table 4.4
Population of town from 14 to 24 years after it was founded
t (years)
14
16
18
20
22
24
Population 5000 + 350t
9900
10,600
11,300
12,000
12,700
13,400
Pop. 12,000 when t = 20
population
14,000
?
10,000
6000
2000
10
t (years)
20
Figure 4.3
25. (a) We want to know when 8 + 4t is equal to 24, so we want to solve
8 + 4t = 24.
(b) Table 4.5 shows the number of stamps, and Figure 4.4 shows the plotted points and the solution at t = 4.
Table 4.5 Number of stamps in a passport
1–6 years after getting a new job
t (years)
1
2
3
4
5
6
Stamps 8 + 4t
12
16
20
24
28
32
SOLUTIONS to Review Problems for Chapter Four
number of stamps
24 stamps when t = 4
32
?
24
16
8
2
4
t (years)
6
Figure 4.4
26. (a) We want to know when 10 + 2t is equal to 26, so we solve
10 + 2t = 26.
(b) Table 4.6 shows the number of dirty socks, and Figure 4.5 shows the plotted points and the solution at t = 8.
Table 4.6 Number of dirty socks on floor 2–12 days
after the start of exams
t (days)
2
4
6
8
10
12
Dirty socks (10 + 2t)
14
18
22
26
30
34
socks
34
26
I26 socks
when t = 8
18
10
2
2
4
6
8
10 12 14
t
Figure 4.5
27. (a) We want to know when $100,000 − 10,000t is equal to $70,000, so we solve
100,000 − 10,000t = 70,000.
(b) Table 4.7 shows the machine’s value, and Figure 4.6 shows the plotted points and the solution at t = 3.
Table 4.7
Value of machine 1–6 years after its purchase
t (years)
1
2
3
4
5
6
Machine’s value ($100,000 − 10,000t)
90,000
80,000
70,000
60,000
50,000
40,000
111
112
Chapter Four /SOLUTIONS
value of machine ($)
$70, 000
100,000
when t = 3
70,000
1
3
5
7
9
t(years)
Figure 4.6
28. We have
∆f
f (4) − f (2)
48 − 4
44
=
=
=
= 22.
∆x
4−2
4−2
2
29. We have
f (4) − f (−2)
48 − (−12)
60
∆f
=
=
=
= 10.
∆x
4 − (−2)
4 − (−2)
6
30. We have
f (−4) − f (−2)
−80 − (−12)
∆f
−68
=
=
=
= 34.
∆x
−4 − (−2)
−4 − (−2)
−2
31. (a) Since f (N ) = 3N gives the number of glasses, and since N = 50 clients per hour, we have f (50) = 3 · 50 = 150.
So the cafe should have 150 glasses.
(b) The expression tells us to multiply the average number of clients per hour by 3 to determine the number of glasses
that should be kept on hand. The 3 tells us that the cafe should have 3 glasses per client per hour on average.
32. This can be rewritten as y = (1/5)x, so y is proportional to x with constant of proportionality 1/5.
33. This can be rewritten as y = x, so y is proportional to x with constant of proportionality 1.
34. This can be rewritten as y = (−3/2)x, so y is proportional to x with constant of proportionality −3/2.
35. This can be rewritten as y = 3x + 2, but not in the form y = kx for some constant k, so y is not proportional to x.
√
√
36. This can be rewritten as y = (3 5)x, so y is proportional to x with constant of proportionality 3 5.
37. This can be rewritten as y = (a − 2)x, so y is proportional to x with constant of proportionality a − 2.
PROBLEMS
38. f (p0 ) is the average number of days a house stays on the market before being sold for the average price.
39. f (p0 ) + 10 is ten days more than the average time a house stays on the market before being sold at the average price.
40. f (p0 ) is the average number of days a house stays on the market before being sold for the average price, and f (0.9p0 )
is the average number of days a house stays on the market before being sold at 10% below the average sale price. So
f (p0 ) − f (0.9p0 ) is the difference in how long it takes to sell an average-priced house and a house selling for 10% below
average. The difference will be positive if the less expensive house sells faster, and negative if it takes longer to sell.
41. (a) Substituting t = 0 in the formula H(t) = (−0.08t + 6)2 , we obtain H(0) = 36. The initial depth of the water is 36
inches.
(b) The height of the water at t = 75 is zero. This means it takes 75 hours for all the water to leak out of the cauldron.
SOLUTIONS to Review Problems for Chapter Four
42. (a) The graph shows that f (4) = 2.
(b) From the graph, we see that the x that produce f (x) = 4 is x = −2.
(c) The graph shows that there are two xs that produce f (x) = 1, x = −1 and x = 1.
43. (a)
(b)
(c)
(d)
(e)
From the graph, we see that f (0) = 0 and g(0) = 15.
The function f (x) = 0 when x = 0.
The function g(x) = 0 when x = 7.5.
The functions are equal when x = −5 and x = 3.
The function g(x) is greater than or equal to f (x) between x = −5 and x = 3.
44. We have
5+1
5−2
6
=
3
f (5) =
= 2.
45. We have
√
g(3) = 3 3 + 1 − (3 − 5)3
√
= 3 4 − (−2)3
= 3 · 2 − (−8)
= 6+8
= 14.
46. We have
q
p
16 − 25 − 42
p
√
= p16 − 25 − 16
√
g(4) =
=
16 − 9
√
16 − 3
√
= 13.
=
47. We have
q
p
25 − 4 10 − 3(−2)
p
√
= 2 25 − 4 16
√
= 2 25 − 4 · 4
√
=2 9
g(−2) = 2
= 6.
48. We have
f (4) =
=
=
=
√
4 4−7
√
5− 4·4−7
4·2−7
√
5 − 16 − 7
8−7
√
5− 9
1
1
= .
5−3
2
113
114
Chapter Four /SOLUTIONS
49. We have
g(t) = 4t − 2
so g (2t + 1) = 4(2t + 1) − 2 substitute 2t + 1 for t
= 8t + 2
simplify.
50. We have
h 4t2 =
We can rewrite this as
p
=
p
9 · 4t2 − 4
p
36t2 − 4.
36t2 − 4 =
p
2
p4 (9t − 1)
9t2 − 1.
=2
51. We first find g(−x/2):
−x 2
22 x
= −2
4
x2
=− .
2
g(−x/2) = −2
This means
x2
−2g(−x/2) = −2 −
2
| {z }
g(−x/2)
2
=x .
52. We have
h(2x − b − 3) = 2(2x − b − 3) − b − 2
= (4x − 2b − 6) − b − 2
= 4x − 3b − 8.
53. Note that in v(r) = 2z − 3r − 4s, the independent variable is r. We have
v(3z − 2r − s) = 2z − 3 (3z − 2r − s) −4s
|
{z
r
}
= 2z − (9z − 6r − 3s) − 4s
= 2z − 9z + 6r + 3s − 4s
= −7z + 6r − s.
54. We have
h(r) = −4
2 − 3r
= −4
4 − 5r
| {z }
h(r)
SOLUTIONS to Review Problems for Chapter Four
2 − 3r = −4(4 − 5r)
2 − 3r = −16 + 20r
18 = 23r
18
r=
.
23
Multiply by denominator
Distribute
Collect r-terms
55. We have
w(2v + 7) = 3(2v + 7) + 2 because w(v) = 3v + 2
= 6v + 23
w(v − 1) = 3(v − 1) + 2
because w(v) = 3v + 2
= 3v − 1
so 3w(v − 1) = 3(3v − 1)
= 9v − 3.
Solving gives
w(2v + 7) = 3w(v − 1)
6v + 23 = 9v − 3
−3v = −26
26
−26
=
.
v=
−3
3
Checking our answer, we see that
26
+7
3
52
21
+
3
3
73
3
73
w
3
73
3·
+2
3
75,
26
3
−
3
3
23
3
23
3w
3
23
3 3·
+2
3
2v + 7 = 2 ·
=
=
so w(2v + 7) =
=
=
and
v−1 =
=
so 3w(v − 1) =
=
= 3 · 25
= 75
find common denominator
because w(v) = 3v + 2
find common denominator
because w(v) = 3v + 2
as required.
56. We have f (3) = 0.935. This means that on November 4, 2007, $1 could be traded for 0.935 CAD.
57. We have
50
50
=
= 53.482.
f (5)
0.9349
This tells us that on day 5 (November 6, 2007), $53.48 USD could be traded for 50 CAD.
115
116
Chapter Four /SOLUTIONS
58. We have
f (3) − f (0)
0.9350 − 0.9529
=
3−0
3
−0.0179
=
3
= −0.00597
f (6) − f (3)
0.9295 − 0.9350
=
6−3
3
−0.0055
=
3
= −0.00183.
Thus, the second expression is larger (“less negative”) than the first. This means the USD dropped in value against the
CAD less rapidly during the second 3 days of November than during the first 3 days.
59. The expression p(t)q(t) is the family’s total expenditure on gasoline in week t, and p(t − 1)q(t − 1) is the total in week
t − 1, that is, the week before. So p(t)q(t) − p(t − 1)q(t − 1) is the change (either increase or decrease) in the family’s
expenditure between week t and the preceding week.
60. The fact that p(t2 ) > p(t1 ) tells us that the cost per gallon of gasoline is higher in week t2 than in week t1 . The fact that
p(t2 )q(t2 ) < p(t1 )q(t1 ) tells us that the family’s total expenditure is lower in week t2 than in week t1 . Since the total
expenditure is lower even though the cost per gallon of gas is higher, the family must have used less gas, so we conclude
that q(t2 ) < q(t1 ).
61. The total expenditure on gasoline during these four weeks is given by
p(2)q(2) + p(3)q(3) + p(4)q(4) + p(5)q(5) .
| {z }
week 2
| {z }
week 3
| {z }
week 4
Therefore, the average weekly expenditure over this time period is
| {z }
week 5
p(2)q(2) + p(3)q(3) + p(4)q(4) + p(5)q(5)
.
4
62. Let f (t) be the CO2 level in ppm t years after 1979. We have
Average rate
of change
f (28) − f (0)
28 − 0
382 − 336
=
28 − 0
46
= 1.643.
=
28
=
The level in 2020 is given by f (41). Assuming a constant rate of change,
f (41) − f (0)
= 1.643
41 − 0
f (41) − 336
= 1.643
41
Because rate of change is constant
Because f (0) = 336
f (41) − 336 = 1.643(41)
f (41) = 403.4.
63. We have
Average rate of change =
w(20) − w(15)
1770 − 1750
=
= 4.
20 − 15
5
SOLUTIONS to Review Problems for Chapter Four
117
Assuming the rate of change is constant on this interval, we see that
w(18) − w(15)
=4
18 − 15
w(18) − 1750
=4
3
w(18) − 1750 = 12
Because rate of change is 4
Because w(15) = 1750
w(18) = 1762.
Another approach is to realize that, because the rate of change is 4 ppm/year, the methane levels rises by by 3 · 4 = 12
ppm during the 3-year period from t = 15 to t = 18, from 1750 ppm to 1762 ppm.
64. The denominator v(t) + w(t) is the total value of stocks and bonds in the investment, that is, the total value of the
investment in year t. Thus, this expression gives
Value of bonds in year t
.
Total value of investment in year t
In other words, it gives the fraction of the investment’s value due to bonds.
65. Knowing a solution is t = 5, we have
w(5) = 2v(4).
The left-hand side is the value of the bonds in year 5. The right-hand side is twice the value of the stocks in year 4. This
tells us that the investment’s bond value in year 5 is twice its stock value from the year before, or year 4.
66. The stock value in year t is v(t). The preceding year is year t − 1, so the bond value in the preceding year is w(t − 1).
The difference in these values is v(t) − w(t − 1).
67. If the bond portion exceeds the stock portion by $3000, then
Bond portion = Stock portion + $3000,
so
w(t) = v(t) + 3000.
68. (a) Looking at the graph, we see that w is close to 50 cm when the area is 7000 cm2 .
(b) We substitute w = 50 into the equation:
4500 + w2 = 7000.
The left-hand side becomes
4500 + w2 = 4500 + 502 = 4500 + 2500 = 7000.
Since the two sides are equal, w = 50 is a solution.
69. We see from the table that g(14) = 37, which means the fly takes 37 days to develop at 14◦ C.
70. We see from the table that g(18) = 24 and g(19) = 22, which means the fly takes 24 days to develop at 18◦ C and 22
days to develop at 19◦ C. Thus, we estimate g(18.5) = 23, assuming that at a temperature between 18◦ C and 19◦ C, the
fly will take between 22 and 23 days to develop.
71.
• From the table, we have
g(12) − g(10)
50 − 68
=
= −9
12 − 10
2
32 − 50
g(15) − g(12)
=
= −6.
15 − 12
3
• The second expression is larger (less negative).
• This tells us that as the ambient temperature rises, the development time drops at an average rate of 9 days per degree
increase on the interval from 10◦ C to 12◦ C, but that it drops at an average rate of 6 days per degree on the interval
from 12◦ C to 15◦ C.
In other words, as the temperature rises, the development time drops, but it drops more slowly at higher temperatures.
118
Chapter Four /SOLUTIONS
72. If the savings are directly proportional to the amount spent, we can set up the equation: S = kA, where S represents the
savings and A represents the amount spent. We can fill in either set of values and solve for k:
S = kA
5 = 50k
k = 0.1,
or
S = kA
10 = 100k
k = 0.1.
Since k = 0.1, the store has a 10% off sale.
73. (a) If the car travels for 5 hours, the distance traveled is D = 30(5) = 150 miles.
(b) We have
Distance = Speed × Time
D = 30t.
Distance is proportional to time, with constant of proportionality 30. Checking units, we see that the units agree:
D = 30 × t
miles = (miles/hour) × (hours)
miles = miles.
74. The circumference, C, is proportional to the radius, r, since C can be expressed as C = kr where k = 2π.
75. The force, F is directly proportional to the distance, x, since F can be expressed as a constant times x, where the constant
is −k.
76. Since the number of tablespoons of grounds is directly proportional to the number of cups of coffee desired, T = f (x) =
kx, where x represents the number of cups of coffee. Since 18 tablespoons are needed for 12 cups, 18 = k ·12 so k = 1.5.
Therefore, T = 1.5x. If 4 12 tablespoons are used, we have 4.5 = 1.5x so x = 3. This means three cups of coffee can be
made with 4 21 tablespoons of grounds. In this case, k represents the number of tablespoons of grounds per cup of coffee
desired, so its units are tablespoons per 8 oz cup.
77. We have C = f (p) = kp where p represents the number of crackers eaten. If 3 crackers are eaten, 36 grams of carbohydrates are ingested, so 36 = k(3). Therefore k = 12. For the units on each side of the equal sign to agree, the units
of k are carbohydrates per cracker. To find the number grams of carbohydrates ingested if 8 crackers are eaten, we have
C = 12 · 8 = 96 grams.
1−0
2−0
78. The only lines that go through the origin are D and E. For D, k =
= 1/2. For E, k =
=2
2−0
1−0
79. (a) Since H is 0.6% of M , we have H = 0.006M . Solving for M gives M = (1/0.006)H = 167H.
(b) Since B is 5% of M , we have B = 0.05M . Solving for M gives M = (1/0.05)B = 20B.
(c) Equating the two expressions for M from parts (a) and (b), we get the equation 167H = 20B. Solving for B gives
B = (167/20)H = 8.35H. Thus, the blood mass is about 8 times the heart mass.
80. Number of dollars is proportion to number of pounds, so
Number of dollars = k × Number of pounds.
We use the fact that when the number of dollars is 400, the number of pounds is 250 to solve for k:
400 = k(250)
400 dollars
= 1.6 dollars per pound.
k=
250 pounds
The constant k gives the conversion factor from pounds to dollars, or the rate of exchange in US dollars per British pound.
SOLUTIONS to Review Problems for Chapter Four
81. We use the fact that C = kb and substitute C = 245 and b = 3 to find k:
C = kb
245 = k(3)
245 calories
= 81.667 calories per ounce.
k=
3 ounces
The formula is
C = 81.667b.
In four ounces of ground beef, there are C = 81.667(4) = 326.667 calories.
119