39.
5422(1.018)
19
<
7609.7 million
40. (a)
[
2
5, 5] by [
2
2, 10]
In this window, it appears they cross twice, although a
third crossing off-screen appears likely.
(b)
It happens by the time
x
5
4.
(c)
Solving graphically,
x
<
2
0.7667,
x
5
2,
x
5
4.
(d)
The solution set is approximately
(
2
0.7667, 2)
<
(4,
‘
).
41.
Since
f
(1)
5
4.5 we have
ka
5
4.5, and since
f
(
2
1)
5
0.5
we have
ka
2
1
5
0.5.
Dividing, we have
}
ka
ka
2
1
} 5 }
4
0
.
.
5
5
}
a
2
5
9
a
56
3
Since
f
(
x
)
5
k
?
a
x
is an exponential function, we require
a
.
0, so
a
5
3. Then
ka
5
4.5 gives 3
k
5
4.5, so
k
5
1.5.
The values are
a
5
3 and
k
5
1.5.
42.
Since
f
(1)
5
1.5 we have
ka
5
1.5, and since
f
(
2
1)
5
6
we have
ka
2
1
5
6. Dividing, we have
}
ka
ka
2
1
} 5 }
1
6
.5
}
a
2
5
0.25
a
56
0.5
Since
f
(
x
)
5
k
?
a
x
is an exponential function, we require
a
.
0, so
a
5
0.5. Then
ka
5
1.5 gives 0.5
k
5
1.5, so
k
5
3. The values are
a
5
0.5 and
k
5
3.
■
Section 1.4
Parametric Equations
(pp. 26–31)
Exploration 1
Parametrizing Circles
1.
Each is a circle with radius
)
a
)
. As
)
a
)
increases, the radius
of the circle increases.
[
2
4.7, 4.7] by [
2
3.1, 3.1]
2.
0
#
t
# }
p
2
}
:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
0
#
t
#
p
:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
0
#
t
# }
3
2
p
}
:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
2
p
#
t
#
4
p
:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
0
#
t
#
4
p
:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
Let
d
be the length of the parametric interval. If
d
,
2
p
,
you get
}
2
d
p
}
of a complete circle. If
d
5
2
p
, you get the
complete circle. If
d
.
2
p
, you get the complete circle but
portions of the circle will be traced out more than once. For
example, if
d
5
4
p
the entire circle is traced twice.
16
Section 1.4
x
change in
Y
1
change in
Y
2
1
3
2
2
5
4
3
7
8
4