Problem Set 10s

T2 5 3 f x x 1 x solution t2 c f

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Unformatted text preview: 1 ⎤ྏ ⎡ྎ So the critical points are : x c = ⎢ྎ0, 1, ⎥ྏ 4 ⎦ྏ ⎣ྏ (t − 2) ⇒ t − 2 = 0 ⇒ t = 2 f. t−2 5/ 3 f ( x ) = x (1 + x ) Solution: t=2 c. f ( x ) = 3 x (2 − 2 x ) Solution: f ( x ) = 3 x (2 − 2 x ) 1 f (' x ) = 3x −1/2 ( 2 − 2 x ) + 3x1/2 (−2) 2 3 3− 9x f (' x ) = −9 x ⇒ x x Therefore the critical number is x = 0 and 1 3 − 9 x = 0 ⇒ x = 3 d. t −2 t2 +1 Solution: t−2 f (x) = 2 t +1 1(t 2 + 1) − ( 2t ) (t − 2) t 2 + 1 − 2t 2 + 4t f '( x ) = = 2 2 t 2 + 1) ( (t 2 + 1) f ( x) = = 2. Find...
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This document was uploaded on 02/27/2014 for the course M 408c at University of Texas.

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