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Unformatted text preview: EXAM 3 Section XXX November 4, 2005 NAME: Make sure to read the question carefully and answer the question that is asked. Show ALL relevant work so that partial credit may be given and indicate where the solution is. Lack of sufficient work may result in a loss of credit, even if a correct answer is given. Good luck!! “On my honor, as a University of Colorado at Boulder student, I have neither given nor received unautho rized assistance on this work.” YOUR SIGNATURE: 1. Absolute Extrema (5 points) Find the absolute maximum value and the absolute minimum value of the function on the indicated interval. f ( x ) = x 3 12 x + 4 on the interval [ 3 , 3] Solution: To find the absolute extrema of a continuous function on a closed interval, we can simply find the critical points and compare the values of the end points and critical points in the original function. f ′ ( x ) = 3 x 2 12 = 3( x 2 4) = 3( x 2)( x + 2) The critical points are the points in the domain where the derivative is equal to 0 or does not exist. Since f ′ ( x ) is a polynomial, we only need consider where f ′ ( x ) = 0....
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This note was uploaded on 05/03/2008 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.
 Spring '08
 JOHANSON
 Calculus

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