111_quiz2_spr08_key

111_quiz2_spr08_key - ⇒ 1 = 2( x-4) ⇒ 1 = 2 x-8 ⇒ 9 =...

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APMA 111 Quiz 2 January 29, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. [3 pts] Simplify: log 3 100 - log 3 18 - log 3 50 Solution: log 3 100 - log 3 18 - log 3 50 = log 3 ± 100 18 · 50 ² = log 3 ± 2 18 ² = log 3 ± 1 9 ² = log 3 ( 3 - 2 ) = - 2 2. [3 pts] Find the derivative: f ( x ) = 10 x · x 10 Solution: f 0 ( x ) = (10 x ) 0 · x 10 + 10 x · ( x 10 ) 0 = (10 x ln 10) · x 10 + 10 x · (10 x 9 ) = 10 x x 9 ( x ln 10 + 10) 3. [3 pts] Solve for x : 10 1 / ( x - 4) = 100 Solution: 10 1 / ( x - 4) = 100 10 1 / ( x - 4) = 10 2 1 / ( x - 4) = 2
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Unformatted text preview: ⇒ 1 = 2( x-4) ⇒ 1 = 2 x-8 ⇒ 9 = 2 x ⇒ x = 9 / 2 Quiz 2, Page 2 of 2 January 29, 2008 4. [3 pts] Z e x sin( e x ) dx Solution: u = e x ⇒ du = e x dx ( u-substitution) ⇒ Z sin( e x ) e x dx = Z sin udu =-cos u + C =-cos( e x ) + C 5. [3 pts] lim x → 2-ln ± 2-x 3 ² Solution: Since x → 2-, 2-x 3 → + , and so ln ± 2-x 3 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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111_quiz2_spr08_key - ⇒ 1 = 2( x-4) ⇒ 1 = 2 x-8 ⇒ 9 =...

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