Exam_solutions_2_(2)

# Exam_solutions_2_(2) - Math 141, Sections 03**, T...

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Unformatted text preview: Math 141, Sections 03**, T Pilachowski, Spring 2007 SOLUTIONS , page 1 of 2 March 2, 2007 MATH 141 TEST 2 (7.1 8.3) [Pilachowski] SOLUTIONS 1. a. (12 points) Let ( ) 7 18 3 2 2 3 + + = x x x x f and Let [ a , b ] be the largest interval containing x = 0 such that f has an inverse function 1 f on this interval. Find a and b . (You must show calculus work.) ( ) 2 13 2 1 , 2 13 2 1 2 13 2 1 3 6 18 6 6 2 2 + = = = = + = + = b a x x x x x f b. (4 extra points) You know that g (1) = 2, and ( ) ( ) 3 2 1 = g . What is ( ) 1 g ? answer : ( ) ( ) ( ) 3 1 2 1 1 1 = = g g 2. (14 points) Use integration by substitution to evaluate 3 2 2 cos 1 2 sin 2 dx x x . Let u = cos 2 x , du = 2 sin 2 x dx ; ( ) ; 2 1 3 2 cos 3 ; 1 cos = = = = = = u x u x ( ) 3 2 2 6 1 sin 2 1 sin 1 sin 1 1 2 cos 1 2 sin 2 1 1 2 1 1 1 2 2 2 1 3 = + = +...
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## This note was uploaded on 09/12/2009 for the course MATH 141 taught by Professor Hamilton during the Spring '07 term at Maryland.

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Exam_solutions_2_(2) - Math 141, Sections 03**, T...

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