1
CHAPTER 1
Functions
EXERCISE SET 1.1
1.
(a)
around 1943
(b)
1960; 4200
(c)
no; you need the year’s population
(d)
war; marketing techniques
(e)
news of health risk; social pressure, antismoking campaigns, increased taxation
2.
(a)
1989; $35,600
(b)
1975, 1983; $32,000
(c)
the first two years; the curve is steeper (downhill)
3.
(a)
−
2
.
9
,
−
2
.
0
,
2
.
35
,
2
.
9
(b)
none
(c)
y
= 0
(d)
−
1
.
75
≤
x
≤
2
.
15
(e)
y
max
= 2
.
8 at
x
=
−
2
.
6;
y
min
=
−
2
.
2 at
x
= 1
.
2
4.
(a)
x
=
−
1
,
4
(b)
none
(c)
y
=
−
1
(d)
x
= 0
,
3
,
5
(e)
y
max
= 9 at
x
= 6;
y
min
=
−
2 at
x
= 0
5.
(a)
x
= 2
,
4
(b)
none
(c)
x
≤
2; 4
≤
x
(d)
y
min
=
−
1; no maximum value
6.
(a)
x
= 9
(b)
none
(c)
x
≥
25
(d)
y
min
= 1; no maximum value
7.
(a)
Breaks could be caused by war, pestilence, ﬂood, earthquakes, for example.
(b)
C
decreases for eight hours, takes a jump upwards, and then repeats.
8.
(a)
Yes, if the thermometer is not near a window or door or other source of sudden temperature
change.
(b)
No; the number is always an integer, so the changes are in movements (jumps) of at least
one unit.
9.
(a)
The side adjacent to the building has length
x
, so
L
=
x
+ 2
y
. Since
A
=
xy
= 1000
,
L
=
x
+ 2000
/x
.
(b)
x >
0 and
x
must be smaller than the width of the building, which was not given.
(c)
120
80
20
80
(d)
L
min
≈
89
.
44 ft
10.
(a)
V
=
lwh
= (6
−
2
x
)(6
−
2
x
)
x
(b)
From the figure it is clear that 0
< x <
3.
(c)
20
0
0
3
(d)
V
max
≈
16 in
3
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2
Chapter 1
11.
(a)
V
= 500 =
πr
2
h
so
h
=
500
πr
2
. Then
C
= (0
.
02)(2)
πr
2
+ (0
.
01)2
πrh
= 0
.
04
πr
2
+ 0
.
02
πr
500
πr
2
= 0
.
04
πr
2
+
10
r
;
C
min
≈
4
.
39 at
r
≈
3
.
4
, h
≈
13
.
8
.
7
4
1.5
6
(b)
C
= (0
.
02)(2)(2
r
)
2
+ (0
.
01)2
πrh
= 0
.
16
r
2
+
10
r
. Since
0
.
04
π <
0
.
16, the top and bottom now get more weight.
Since they cost more, we diminish their sizes in the
solution, and the cans become taller.
7
4
1.5
5.5
(c)
r
≈
3
.
1 cm,
h
≈
16
.
0 cm,
C
≈
4
.
76 cents
12.
(a)
The length of a track with straightaways of length
L
and semicircles of radius
r
is
P
= (2)
L
+ (2)(
πr
) ft. Let
L
= 360 and
r
= 80 to get
P
= 720 + 160
π
= 1222
.
65 ft.
Since this is less than 1320 ft (a quartermile), a solution is possible.
(b)
P
= 2
L
+ 2
πr
= 1320 and 2
r
= 2
x
+ 160, so
L
=
1
2
(1320
−
2
πr
) =
1
2
(1320
−
2
π
(80 +
x
))
= 660
−
80
π
−
πx.
450
0
0
100
(c)
The shortest straightaway is
L
= 360, so
x
= 15
.
49 ft.
(d)
The longest straightaway occurs when
x
= 0, so
L
= 660
−
80
π
= 408
.
67 ft.
EXERCISE SET 1.2
1.
(a)
f
(0) = 3(0)
2
−
2 =
−
2;
f
(2) = 3(2)
2
−
2 = 10;
f
(
−
2) = 3(
−
2)
2
−
2 = 10;
f
(3) = 3(3)
2
−
2 = 25;
f
(
√
2) = 3(
√
2)
2
−
2 = 4;
f
(3
t
) = 3(3
t
)
2
−
2 = 27
t
2
−
2
(b)
f
(0) = 2(0) = 0;
f
(2) = 2(2) = 4;
f
(
−
2) = 2(
−
2) =
−
4;
f
(3) = 2(3) = 6;
f
(
√
2) = 2
√
2;
f
(3
t
) = 1
/
3
t
for
t >
1 and
f
(3
t
) = 6
t
for
t
≤
1.
2.
(a)
g
(3) =
3 + 1
3
−
1
= 2;
g
(
−
1) =
−
1 + 1
−
1
−
1
= 0;
g
(
π
) =
π
+ 1
π
−
1
;
g
(
−
1
.
1) =
−
1
.
1 + 1
−
1
.
1
−
1
=
−
0
.
1
−
2
.
1
=
1
21
;
g
(
t
2
−
1) =
t
2
−
1 + 1
t
2
−
1
−
1
=
t
2
t
2
−
2
(b)
g
(3) =
√
3 + 1 = 2;
g
(
−
1) = 3;
g
(
π
) =
√
π
+ 1;
g
(
−
1
.
1) = 3;
g
(
t
2
−
1) = 3 if
t
2
<
2 and
g
(
t
2
−
1) =
√
t
2
−
1 + 1 =

t

if
t
2
≥
2.
3.
(a)
x
= 3
(b)
x
≤ −
√
3 or
x
≥
√
3
(c)
x
2
−
2
x
+ 5 = 0 has no real solutions so
x
2
−
2
x
+ 5 is always positive or always negative. If
x
= 0, then
x
2
−
2
x
+ 5 = 5
>
0; domain: (
−∞
,
+
∞
).
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