liggett_quizsolutions2,3

liggett_quizsolutions2,3 - point at x = e 8 / 3 . Quiz for...

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Mathematics 31B/4 – Fall 2004 – Quiz Solutions Quiz for Tuesday, October 26 Find the inverse function to f ( x ) = 1 + e x 1 - e x . Solution: Write y = 1 + e x 1 - e x and solve for x : e x = y - 1 y + 1 , and then x = ln ± y - 1 y + 1 . Therefore, f - 1 ( y ) = ln ± y - 1 y + 1 . Note, f has domain ( -∞ , 0) (0 , ) and range ( -∞ , - 1) (1 , ) so f - 1 has domain ( -∞ , - 1) (1 , ) and range ( -∞ , 0) (0 , ). 2. Find the intervals of concavity (up and down) and the inflection point(s) of f ( x ) = ln x x . Solution: f 0 ( x ) = 2 - ln x 2 x 3 / 2 and f 00 ( x ) = 3ln x - 8 4 x 5 / 2 , so the graph of f is concave down on (0 ,e 8 / 3 ), concave up on ( e 8 / 3 , ), and has an inflection
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Unformatted text preview: point at x = e 8 / 3 . Quiz for Thursday, October 28 1. Evaluate Z 9 4 x + 1 x 2 dx. Solution: Z 9 4 x + 1 x 2 dx = Z 9 4 ( x + 2 + x-1 ) dx = x 2 2 + 2 x + ln x 9 4 = 85 2 + ln 9 4 . 2. Dierentiate the following function, and simplify your answer if possible: f ( x ) = x cos-1 x-p 1-x 2 . Solution: f ( x ) =-x 1-x 2 + cos-1 x-1 2 (1-x 2 )-1 / 2 (-2 x ) = cos-1 x....
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This note was uploaded on 04/29/2010 for the course MATH 31B 31B taught by Professor Liggett during the Spring '10 term at UCLA.

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