111_quiz1_spr08_key

# 111_quiz1_spr08_key - ( x 3 ) . Solution: 2 ln x 3 + ln 3...

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APMA 111 Quiz 1 January 22, 2008 NAME: SECTION: #6 (10 am) #2 (11 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. Answer the following questions for f ( x ) = x 2 + ln(2 x ). (a) [1 pt] What is the domain of f ( x )? Solution: 2 x > 0 x > 0 (b) [2 pts] Show that f ( x ) is one-to-one. Solution: f 0 ( x ) = 2 x + 1 2 x · 2 = 2 x + 1 x > 0 because x > 0 f ( x ) is always increasing f ( x ) is one-to-one (c) [2 pts] Find f - 1 (1 / 4). Solution: x 2 + ln(2 x ) = 1 / 4 x = 1 / 2 because (1 / 2) 2 + ln[2 · (1 / 2)] = 1 / 4 + ln 1 = 1 / 4 2. [5 pts] Simplify: 2 ln x 3 + ln 3 x - ln 4 x - 1 4 ln
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Unformatted text preview: ( x 3 ) . Solution: 2 ln x 3 + ln 3 √ x-ln 4 √ x-1 4 ln ( x 3 ) = ln( x 2 / 3 ) + ln( x 1 / 3 )-ln( x 1 / 4 )-ln( x 3 / 4 ) = ln ± x 2 / 3 · x 1 / 3 x 1 / 4 · x 3 / 4 ² = ln( x/x ) = ln 1 = 0 3. [5 pts] Compute the derivative of g ( t ) = t-ln( t + 1) t + 1 . Simplify your answer. Solution: g ( t ) = (1-1 t +1 )( t + 1)-[ t-ln( t + 1)] · 1 ( t + 1) 2 = t + 1-1-t + ln( t + 1) ( t + 1) 2 = ln( t + 1) ( t + 1) 2...
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## This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.

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