hw12solutions - Math 2250 Written HW #12 Solutions 1....

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Math 2250 Written HW #12 Solutions 1. Consider f ( x ) = 1 4 x 2 + 7 (a) Find all critical points of f ( x ). For each one determine whether it is a local maximum, a local minimum, or neither. Answer: To find the critical points, first we need to compute f 0 ( x ) = - 1 (4 x 2 + 7) 2 · 8 x = - 8 x (4 x 2 + 7) 2 . Since the denominator is always positive, f 0 ( x ) = 0 exactly when the numerator is zero, so the only critical point occurs when x = 0, so this is the point (0 , 1 / 7). Now, again since the denominator is always positive, f 0 ( x ) has the same sign as the numerator, which is - 8 x . Therefore, f 0 ( x ) changes sign from positive to negative at x = 0, so the first derivative test implies that (0 , 1 / 7) is a local maximum. (b) If it exists, what is the absolute maximum of f ( x )? If it exists, what is the absolute minimum? Answer: Since f ( x ) has only one critical point, and that critical point is a local max- imum, it must also be an absolute maximum. Therefore, the absolute maximum value of f ( x ) is 1 / 7. On the other hand, any absolute minimum would also have to be a local
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.

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hw12solutions - Math 2250 Written HW #12 Solutions 1....

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