39.
[
2
2.35, 2.35] by [
2
3.5, 3.5]
y
95
x
?
}
2
ˇ
4
w
1
2
w
x
w
2
w
}
(
2
2
x
)
1
(1)
ˇ
4
w
2
w
x
w
2
w
5
}
2
x
ˇ
2
1
4
w
(
2
w
4
2
x
w
2
w
x
2
)
}5
}
ˇ
4
4
w
2
2
w
2
x
x
w
2
2
w
}
40.
[
2
4.7, 4.7] by [
2
1, 5]
y
5
x
2
?
}
2
ˇ
3
w
1
2
w
x
w
}
(
2
1)
1
2
x
ˇ
3
w
2
w
x
w
5
}
2
x
2
2
1
ˇ
3
w
4
x
2
w
(3
x
w
2
x
)
}
2
2
5
ˇ
x
2
3
w
1
2
w
1
x
w
2
x
}
41.
[
2
4.7, 4.7] by [0, 6.2]
y
h
42.
[
2
4, 4] by [
2
1, 6]
y
h
43.
[
2
4, 6] by [
2
2, 6]
y
h
44.
[
2
4, 6] by [
2
5, 5]
We begin by determining whether
f
9
(
x
) is defined at
x
5
1,
where
f
(
x
)
5
h
Lefthand derivative:
lim
h
→
0
2
}
f
(1
1
h
h
)
2
f
(1)
lim
h
→
0
2
5
lim
h
→
0
2
}
2
h
4
2
h
2
h
}
5
lim
h
→
0
2
}
1
4
}
(
2
h
2
4
h
)
52
1
Righthand derivative:
lim
h
→
0
1
}
f
(1
1
h
h
)
2
f
(1)
}
5
lim
h
→
0
1
5
lim
h
→
0
1
}
h
3
2
3
h
h
2
2
h
}
5
lim
h
→
0
1
(
h
2
2
3
h
2
1)
1
Thus
f
9
(
x
)
5
h
x
#
1
x
.
1
2}
1
2
}
x
2 }
1
2
}
,
3
x
2
2
12
x
1
8,
(1
1
h
)
3
2
6(1
1
h
)
2
1
8(1
1
h
)
2
3
}}}}
h
2}
1
4
}
(1
1
h
)
2
2 }
1
2
}
(1
1
h
)
1 }
1
4
5
} 2
3
}}}}
h
x
#
1
x
.
1
2}
1
4
}
x
2
2 }
1
2
}
x
1 }
1
4
5
}
,
x
3
2
6
x
2
1
8
x
,
x
,
1
x
.
1
2
2
x
2
2,
2
2
x
1
6,
x
,
0
x
.
0
2
1,
2
2
2
x
,
x
,
1
x
.
1
2
2,
1,
Section 4.1
131
crit. pt.
derivative
extremum
value
x
5
1
undefined
minimum
2
crit. pt.
derivative
extremum
value
x
5
0
0
minimum
0
x
5 }
1
5
2
}
0
local max
}
1
1
4
2
4
5
}
15
1/2
<
4.462
x
5
3
undefined
minimum
0
crit. pt.
derivative
extremum
value
x
2
undefined
local max
0
x
ˇ
2
w
0
minimum
2
2
x
5
ˇ
2
w
0
maximum
2
x
5
2
undefined
local min
0
crit. pt.
derivative
extremum
value
x
1
0
maximum
5
x
5
1
undefined
local min
1
x
5
3
0
maximum
5
crit. pt.
derivative
extremum
value
x
5
0
undefined
local min
3
x
5
1
0
local max
4
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View Full Document44. continued
Note that
2}
1
2
}
x
2 }
1
2
}
5
0 when
x
52
1, and
3
x
2
2
12
x
1
8
5
0 when
x
5
5
}
12
6
6
ˇ
4
w
8
w
}5
2
6 }
2
ˇ
3
3
w
}
.
But 2
2 }
2
ˇ
3
3
w
}
<
0.845
,
1, so the only critical points
occur at
x
1 and
x
5
2
1 }
2
ˇ
3
3
w
}
<
3.155.
45.
Graph (c), since this is the only graph that has positive
slope at
c
.
46.
Graph (b), since this is the only graph that represents a
differentiable function at
a
and
b
and has negative slope
at
c
.
47.
Graph (d), since this is the only graph representing a
function that is differentiable at
b
but not at
a
.
48.
Graph (a), since this is the only graph that represents a
function that is not differentiable at
a
or
b
.
49. (a)
V
(
x
)
5
160
x
2
52
x
2
1
4
x
3
V
9
(
x
)
5
160
2
104
x
1
12
x
2
5
4(
x
2
2)(3
x
2
20)
The only critical point in the interval (0, 5) is at
x
5
2.
The maximum value of
V
(
x
) is 144 at
x
5
2.
(b)
The largest possible volume of the box is 144 cubic
units, and it occurs when
x
5
2.
50. (a)
P
9
(
x
)
5
2
2
200
x
2
2
The only critical point in the interval (0,
‘
) is at
x
5
10. The minimum value of
P
(
x
) is 40 at
x
5
10.
(b)
The smallest possible perimeter of the rectangle is
40 units and it occurs at
x
5
10, which makes the
rectangle a 10 by 10 square.
51. (a)
f
9
(
x
)
5
3
ax
2
1
2
bx
1
c
is a quadratic, so it can have
0, 1, or 2 zeros, which would be the critical points of
f
.
Examples:
[
2
3, 3] by [
2
5, 5]
The function
f
(
x
)
5
x
3
2
3
x
has two critical points at
x
1 and
x
5
1.
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 Spring '08
 ALL
 Calculus, Critical Point, maximum

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