College Algebra, 2013a

2.1 Linear Functions 171
35. (a) F (C) = 9 5 C +32 (b) C(F )= 5 9 (F − 32) = 5 9 F − 160
9
(c) F (−40) = −40 = C(−40).
36. N(T )=− 2
15 T + 43
3
and N(20) =
35
3
≈ 12 howls per hour.
Having a negative number of howls makes no sense and since N(107.5) = 0 we can put an
upper bound of 107.5 ◦ F on the domain. The lower bound is trickier because there’s nothing
other than common sense to go on. As it gets colder, he howls more often. At some point
it will either be so cold that he freezes to death or he’s howling non-stop. So we’re going to
say that he can withstand temperatures no lower than −60 ◦ F so that the applied domain is
[−60, 107.5].
{
39. C(p) =
6p +1.5 if 1 ≤ p ≤ 5
5.5p if p ≥ 6
{
15n if 1 ≤ n ≤ 9
40. T (n) =
12.5n if n ≥ 10
{
10 if 0 ≤ m ≤ 500
41. C(m) =
10 + 0.15(m − 500) if m>500
{
0.12c if 1 ≤ c ≤ 100
42. P (c) =
12 + 0.1(c − 100) if c>100
43. (a)
(b)
(c)
⎧
⎪⎨ 8 if 0 ≤ d ≤ 15
D(d) = −
⎪ 1 2 d + 31
2
if 15 ≤ d ≤ 27
⎩
2 if 27 ≤ d ≤ 37
⎧
⎪⎨
D(s) =
⎪⎩
2 if 0 ≤ s ≤ 10
1
2
s − 3 if 10 ≤ s ≤ 22
8 if 22 ≤ s ≤ 37
8
2
8
2
15 27 37
y = D(d)
10 22 37
y = D(s)