30.
f
(
x
)
5
e
x
1
sin
x
f
9
(
x
)
5
e
x
1
cos
x
L
(
x
)
5
f
(0)
1
f
9
(0)(
x
2
0)
5
1
1
2(
x
2
0)
5
2
x
1
1
31.
The global minimum value of
}
1
2
}
occurs at
x
5
2.
32. (a)
The values of
y
9
and
y
0
are both negative where the
graph is decreasing and concave down, at
T
.
(b)
The value of
y
9
is negative and the value of
y
0
is posi
tive where the graph is decreasing and concave up, at
P
.
33. (a)
The function is increasing on the interval (0, 2].
(b)
The function is decreasing on the interval [
2
3, 0).
(c)
The local extreme values occur only at the endpoints of
the domain. A local maximum value of 1 occurs at
x
52
3, and a local maximum value of 3 occurs at
x
5
2.
34.
The 24th day
35.
36. (a)
We know that
f
is decreasing on [0, 1] and increasing
on [1, 3], the absolute minimum value occurs at
x
5
1
and the absolute maximum value occurs at an endpoint.
Since
f
(0)
5
0,
f
(1)
2, and
f
(3)
5
3, the absolute
minimum value is
2
2 at
x
5
1 and the absolute maxi
mum value is 3 at
x
5
3.
(b)
The concavity of the graph does not change. There are
no points of inflection.
(c)
37. (a)
f
(
x
) is continuous on [0.5, 3] and differentiable on
(0.5, 3).
(b)
f
9
(
x
)
5
(
x
)
1
}
1
x
}
2
1
(ln
x
)(1)
5
1
1
ln
x
Using
a
5
0.5 and
b
5
3, we solve as follows.
f
9
(
c
)
5
}
f
(3
3
)
2
2
f
0
(
.5
0.5)
}
1
1
ln
c
5
}
3ln3
2
2.
0
5
.5 ln 0.5
}
ln
c
1
ln
c
5
0.4 ln(27
ˇ
2
w
)
2
1
c
5
e
2
1
(27
ˇ
2
w
)
0.4
c
5
e
2
1
ˇ
5
1
w
4
w
5
w
8
w
<
1.579
(c)
The slope of the line is
m
5
}
f
(
b
b
)
2
2
f
a
(
a
)
}5
0.4 ln (27
ˇ
2
w
)
5
0.2 ln 1458, and
the line passes through (3, 3 ln 3). Its equation is
y
5
0.2(ln 1458)(
x
2
3)
1
3 ln 3, or approximately
y
5
1.457
x
2
1.075.
(d)
The slope of the line is
m
5
0.2 ln 1458, and the line
passes through
(
c
,
f
(
c
))
5
(
e
2
1
ˇ
5
1
w
4
w
5
w
8
w
,
e
2
1
ˇ
5
1
w
4
w
5
w
8
w
(
2
1
1
0.2 ln 1458))
<
(1.579, 0.722).
Its equation is
y
5
0.2(ln 1458)(
x
2
c
)
1
f
(
c
),
y
5
0.2 ln 1458(
x
2
e
2
1
ˇ
5
1
w
4
w
5
w
8
w
)
1
e
2
1
ˇ
5
1
w
4
w
5
w
8
w
(
2
1
1
0.2 ln 1458),
y
5
0.2(ln 1458)
x
2
e
2
1
ˇ
5
1
w
4
w
5
w
8
w
,
or approximately
y
5
1.457
x
2
1.579.
38. (a)
v
(
t
)
5
s
9
(
t
)
5
4
2
6
t
2
3
t
2
(b)
a
(
t
)
5
v
9
(
t
)
6
2
6
t
(c)
The particle starts at position 3 moving in the positive
direction, but decelerating. At approximately
t
5
0.528,
it reaches position 4.128 and changes direction, begin
ning to move in the negative direction. After that, it
continues to accelerate while moving in the negative
direction.
39. (a)
L
(
x
)
5
f
(0)
1
f
9
(0)(
x
2
0)
1
1
0(
x
2
0)
1
(b)
f
(0.1)
<
L
(0.1)
1
(c)
Greater than the approximation in (b), since
f
9
(
x
) is
actually positive over the interval (0, 0.1) and the
estimate is based on the derivative being 0.
40. (a)
Since
}
d
d
y
x
} 5
(
x
2
)(
2
e
2
x
)
1
(
e
2
x
)(2
x
)
5
(2
x
2
x
2
)
e
2
x
,
dy
5
(2
x
2
x
2
)
e
2
x
dx
.
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 Spring '08
 ALL
 Critical Point, Cos

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