**Unformatted text preview: **1)
1
10 − 2
7
7
3c 2 + 1 =
= 8 ⇒ c2 = ⇒ c = ±
= ±1.53
1
3
3
But − 1..53 is not within the interval so
c = 1.53
f ' ( x) = 3x 2 + 1 ⇒ f ' (c) = 3c 2 + 1 = 1
x +1 c. f ( x ) = [0,2] Solution:
1
f ( x) =
x +1 f ' ( x) = −
f ' (c ) = − (c + 1)2 1 (x + 1)2
1 (c + 1) 2 = 1 / 3 −1
1
=−
2
3 3. Use the graph = 3 ⇒ c + 1 = ± 3 ⇒ c = −1 ± 3 Since c = −1 − 3 is not on the interval. c = −1 + 3 = 0.73
d. f (x) = Solution: x
x+2 [1, 4] of the derivative of f to locate the critical points x0 at
which f has either a loca...

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