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Unformatted text preview: we have a relative minimum and f (1) = 0. So 0 si a relative min and occurs for x = 1 A f ( x ) = ( x1) 3 B f ( x ) = ( x1) 2 3 C f ( x ) = √ x 21 D f ( x ) = sin 2 ( x ) E f ( x ) = x x +1 4. Find the absolute extrema of the following fcts in the given interval example f ( x ) = x 2 [1 , 3]. We need to look for abs extrema in the critical point and on the boundary. f ( x ) = 2 x = 0 → x = 0 and we have f (0) = 0. On the boundary f (1) = 1 and f (3) = 9. So 9 is the abs max occurring in x = 3 while the abs min is 0 and occurs in x = 0 A f ( x ) =  x 24  [4 , 5] B f ( x ) = ( x5) 2 / 3 [2 , 7] C f ( x ) = x 2 x 2 +3 [1 , 1] D f ( x ) = ( x1) 2 3 [5 , 2] E f ( x ) = cos ( x ) sin ( x ) [0 ,π ] 2...
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This note was uploaded on 10/08/2010 for the course MATH 16A taught by Professor Sabalka during the Spring '08 term at UC Davis.
 Spring '08
 Sabalka
 Calculus, Derivative

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