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Exam_solutions_2_

# Exam_solutions_2_ - MATH 141 2nd Examination-Solution Key...

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MATH 141, 2nd Examination—Solution Key Prof. Jonathan Rosenberg Monday, October 8, 2007 1. (30 points, 10 points per part) Let f ( x ) = x 4 - 4 x - 1 . (a) What is the biggest interval I containing 0 such that the restriction g of f to I is invertible? (Note: I is allowed to extend out to + , to -∞ , or both, as appropriate.) Solution. The function f has derivative f ( x ) = 4 x 3 - 4 = 4( x 3 - 1) , so it is increasing when x 3 > 1 and decreasing when x 3 < 1 or x < 1 . Since 0 < 1 , f is decreasing near x = 0 and so I should be the largest interval on which f is decreasing, i.e., I = ( -∞ , 1] . (b) What is the domain of the inverse function g 1 ? Solution. dom( g 1 ) = range( g ) = f ( I ) . Now as x → -∞ , f ( x ) + and f is decreasing on I , so this is [ f (1) , ) = [ - 4 , ) . (c) Compute ( g 1 ) ( - 1) . Solution. Since f (0) = - 1 , g 1 ( - 1) = 0 ( g 1 ) ( - 1) = 1 g ( g 1 ( - 1)) = 1 g (0) = 1 - 4 . 2. (20 points) Compute lim x 0 + (sin x )(ln x ) . using L’Hˆopital’s Rule. Solution. Since sin(0) = 0 and ln x → -∞ as x 0 + , this is indeterminate of type 0 · ∞ . It’s more convenient to

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