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Unformatted text preview: , 3]. 1. abs. min. value = 2 2. abs. min. value = 4 3. abs. min. value = 3 correct 4. none oF the other answers 5. abs. min. value = 1 Explanation: The absolute minimum value oF f on [0 , 3] occurs either at an endpoint oF [0 , 3] or at a critical point oF f in (0 , 3). Now f ( x ) = x 2 4 x + 3 = ( x 1)( x 3) , so the critical points oF f occur at x = 1 , 3. But only x = 1 lies in (0 , 3). On the other hand, f (0) = 3 , f (1) = 13 3 , f (3) = 3 . Consequently, abs. min. value = 3 . 010 10.0 points ind all the critical values oF f ( x ) = x (5 x ) 6 . 1. x = 5 , 5 7 correct 2. x = 5 , 1...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.
 Spring '08
 CLARK,C.W./HOY,R.R

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