131f11x3ANS - Math 131 Exam 3 Solutions 1[20 Find...

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Math 131 Exam 3 Solutions 11/17/2011 1. [20%] Find derivative. f ( x ) = x ln x - x f 0 ( x ) = x 1 x + (ln x )(1) - 1 = ln x. Find critical number(s). f 0 ( x ) = 0 ⇐⇒ ln x = 0 ⇐⇒ x = 1 . Test endpoints and critical number. x type f ( x ) 1 / 2 end 1 2 ln ± 1 2 ² - 1 2 = - 1 2 (ln 2 + 1) ≈ - 0 . 847 1 critical - 1 min 2 end 2 ln 2 - 2 = 2(ln 2 - 1) ≈ - 0 . 614 max Thus in [1 / 2 , 2]: f has maximum value 2 (ln 2 - 1) at x = 2 ; f has minimum value - 1 at x = 1 . 1
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2. [25%] (a) The domain of f consists of those x for which x + 3 0, that is, the closed ray [ - 3 , ) . (b) f ( x ) = x x + 3 f 0 ( x ) = x 1 2 x + 3 + x + 3 (1) = x + 2 ( x + 3) 2 x + 3 = 3 ( x + 2) 2 x + 3 . f 0 ( x ) = 0 ⇐⇒ 3 ( x + 2) ⇐⇒ x = - 2 . Hence the one and only critical number of f is c = - 2 . ( Note: Is x = - 3 also a critical number? It depends upon whether you even want to speak about the derivative at an endpoint of the domain!) (c) f 00 ( x ) = 2 x + 3 (3) - ± 3 ( x + 2) ² (2) · 1 2 x +3 (2 x + 3) 2 = 3 2 x + 3 - x +2 x +3 4 ( x + 3) = 3 2 ( x + 3) - ( x + 2) 4 ( x + 3) 3 / 2 = 3 x + 4 4 ( x + 3) 3 / 2 . f
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This note was uploaded on 11/22/2011 for the course MATH 131 taught by Professor Hall-seelig during the Fall '08 term at UMass (Amherst).

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131f11x3ANS - Math 131 Exam 3 Solutions 1[20 Find...

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