53 1 3 3 but 153 is not within the interval so c 153 f

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online