22. (a)
Since
f
9
(
x
)
5
h
,
f
is not differentiable at
x
5
1. (If
f
were differentiable
at
x
5
1, it would violate the Intermediate Value
Theorem for Derivatives.)
(b)
[0, 3] by [
2
1, 3]
(c)
We require
f
9
(
c
)
5
}
f
(3
3
)
2
2
f
0
(0)
}5}
2
2
3
1
} 5 }
1
3
}
, but
f
9
(
x
)
56
1 for all
x
where
f
9
(
x
) is defined. Therefore,
there is no such value of
c
.
23. (a)
Since
f
9
(
x
)
5
h
,
f
is not differentiable at
x
5
0. (If
f
were differentiable
at
x
5
1, it would violate the Intermediate Value
Theorem for Derivatives.)
(b)
[
2
1, 1] by [
2
1, 2]
(c)
We require
f
9
(
c
)
5
}
f
(1
1
)
2
2
(
f
2
(
2
1)
1)
0
2
2
0
} 5
0, but
f
9
(
x
)
1 for all
x
where
f
9
(
x
) is defined. Therefore,
there is no such value of
c
.
24. (a)
We test for differentiability at
x
5
0, using the limits
given in Section 3.5.
Left-hand derivative:
lim
h
→
0
2
}
f
(0
1
h
h
)
2
f
(0)
}5
lim
h
→
0
2
}
cos
h
h
2
1
0
Right-hand derivative:
lim
h
→
0
1
}
f
(0
1
h
h
)
2
f
(0)
lim
h
→
0
1
}
(1
1
sin
h
h
)
2
1
}
5
lim
h
→
0
1
}
sin
h
h
}
5
1
Since the left- and right-hand derivatives are not equal,
f
is not differentiable at
x
5
0.
(b)
[
2
p
,
p
] by [
2
1, 2]
(c)
Note that
f
9
(
x
)
5
h
.
We require
f
9
(
c
)
5
}
f
(
p
p
)
2
2
(
f
2
(
2
p
)
p
)
}
1
2
2
p
(
2
1)
p
1
}
.
For
2
p
,
c
,
0, this occurs when
2
sin
c
5 }
p
1
}
, so
c
52
sin
2
1
1
}
p
1
}
2
<
2
0.324 or
c
p
1
sin
2
1
1
}
p
1
}
2
<
2
2.818. For 0
,
c
,
p
, this
occurs when cos
c
5 }
p
1
}
, so
c
5
cos
2
1
1
}
p
1
}
2
<
1.247. The
possible values of
c
are approximately
2
2.818,
2
0.324, and 1.247.
25.
f
(
x
)
5 }
x
2
2
} 1
C
26.
f
(
x
)
5
2
x
1
C
27.
f
(
x
)
5
x
3
2
x
2
1
x
1
C
28.
f
(
x
)
cos
x
1
C
29.
f
(
x
)
5
e
x
1
C
30.
f
(
x
)
5
ln (
x
2
1)
1
C
31.
f
(
x
)
5 }
1
x
}
1
C
,
x
.
0
f
(2)
5
1
}
1
2
}
1
C
5
1
C
5 }
1
2
}
f
(
x
)
5 }
1
x
}
1 }
1
2
}
,
x
.
0
32.
f
(
x
)
5
x
1/4
1
C
f
(1)
2
1
1/4
1
C
2
1
1
C
2
C
3
f
(
x
)
5
x
1/4
2
3
33.
f
(
x
)
5
ln (
x
1
2)
1
C
f
(
2
1)
5
3
ln (
2
1
1
2)
1
C
5
3
0
1
C
5
3
C
5
3
f
(
x
)
5
ln (
x
1
2)
1
3
34.
f
(
x
)
5
x
2
1
x
2
sin
x
1
C
f
(0)
5
3
0
1
C
5
3
C
5
3
f
(
x
)
5
x
2
1
x
2
sin
x
1
3
2
p
#
x
,
0
0
,
x
#
p
2
sin
x
,
cos
x
,
2
1
#
x
,
0
1
$
x
.
0
1,
2
1,
0
#
x
,
1
3
$
x
.
1
2
1,
1,
136
Section 4.2