45.
Because the function is not continuous on [0, 1]. The
function does not satisfy the hypotheses of the Mean Value
Theorem, and so it need not satisfy the conclusion of the
Mean Value Theorem.
46.
Because the Mean Value Theorem applies to the function
y
5
sin
x
on any interval, and
y
5
cos
x
is the derivative of
sin
x
. So, between any two zeros of sin
x
, its derivative,
cos
x
, must be zero at least once.
47.
f
(
x
) must be zero at least once between
a
and
b
by the
Intermediate Value Theorem. Now suppose that
f
(
x
) is zero
twice between
a
and
b
. Then by the Mean Value Theorem,
f
9
(
x
) would have to be zero at least once between the two
zeros of
f
(
x
), but this can’t be true since we are given that
f
9
(
x
)
±
0 on this interval. Therefore,
f
(
x
) is zero once and
only once between
a
and
b
.
48.
Let
f
(
x
)
5
x
4
1
3
x
1
1. Then
f
(
x
) is continuous and
differentiable everywhere.
f
9
(
x
)
5
4
x
3
1
3, which is never
zero between
x
52
2 and
x
1. Since
f
(
2
2)
5
11 and
f
(
2
1)
1, exercise 47 applies, and
f
(
x
) has exactly one
zero between
x
2 and
x
1.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.
 Fall '08
 GERMAN
 Mean Value Theorem

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