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Example
Consider the function
f
(
x
) =
x
3

x,
x
∈
[1
,
5]
.
Verify that the Mean Value Theorem holds in this case, and show that there is a
c
∈
(1
,
5) for which
f
0
(
c
) =
f
(5)

f
(1)
5

1
.
(1)
Solution:
The function
f
(
x
) =
x
3

x
is diﬀerentiable on (
∞
,
∞
), so it is continuous on the
closed interval [1
,
5] and diﬀerentiable on the open interval (1
,
5), the two conditions required
for the results of the Mean Value Theorem to hold.
From the Mean Value Theorem we are guaranteed the existence of at least one
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Unformatted text preview: c ∈ (1 , 5) so that equation (1) holds. To ﬁnd c we note that f (5)f (1) 51 = 1204 = 30 , and that f ( x ) = 3 x 21 . So we need to ﬁnd c ∈ (1 , 5) for which f ( c ) = 30, or 3 c 21 = 30 = ⇒ 3 c 2 = 31 = ⇒ c = ± s 31 3 . Clearly c =s 13 3 is not in the interval (1 , 5). The other solution satisﬁes c = s 31 3 ≈ 3 . 21 ∈ (1 , 5) . So the answer is c = s 13 3 ....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus, Mean Value Theorem

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