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solx2f05

# solx2f05 - SOLUTIONS TO MIDTERM 2 OF FALL 2005 1 d(ln(7x 1...

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SOLUTIONS TO MIDTERM 2 OF FALL 2005 1. d dx (ln(7 x + 1) ln(5 x - 1)) = d dx (ln(7 x + 1)) · ln(5 x - 1) + ln(7 x + 1) · d dx (ln(5 x - 1)) = 7 7 x + 1 · ln(5 x - 1) + ln(7 x + 1) · 5 5 x - 1 2. We compute f 0 ( x ) = 4 - 5 e x . Setting f 0 ( x ) = 0 we have 0 = 4 - 5 e x , or e x = 4 / 5. Applying ln, we conclude x = ln(4 / 5). As f 00 ( x ) = - 5 e x is always negative, we have a relative maximum at a = ln(4 / 5). As f 0 ( x ) > 0 for x < ln(4 / 5), f ( x ) < f (ln(4 / 5)) for x < ln(4 / 5) and as f 0 ( x ) < 0 for x > ln(4 / 5), f ( x ) < f (ln(4 / 5)) for x > ln(4 / 5). Hence, this relative maximum is actually an absolute maximum. Evaluating, f (ln(4 / 5)) = 4 ln(4 / 5) - 5 e ln(4 / 5) = 4 ln(4 / 5) - 4. Thus, the point ( a, b ) on the graph of y = f ( x ) where f attains its maximum is (ln(4 / 5) , 4(ln(4 / 5) - 1)). 3. e < 3, so e 2 < 3 2 = 9 < 13. As x < y = ln( x ) < ln( y ), we conclude that 2 = ln( e 2 ) < ln(13). ln(16 / 81) = ln(2 4 / 9 2 ) = 4 ln(2) - 2 ln(9) > 2 ln(2) - 2 ln(9) since ln(2) > 0. ln( e e 3 ) = ln( e 1 2 e 3 ) = ln( e - 2 . 5 ) = - 2 . 5 < - 2 Since 1 < 7 5 , we know 0 = ln(1) < ln( 7 5 ) e 3 ln(2) = ( e ln(2) ) 3 = 2 3 = 8 4. d dx (ln( f ( x ))) = d dx (ln( 5 s ( x 8 + 9 x ) 2 (6 x 2 - 1) 3 )) = d dx (ln(( ( x 8 + 9 x ) 2 (6 x 2 - 1) 3 ) 1 5 )) = 1 5 d dx (2 ln( x 8 + 9 x ) - 3 ln(6 x 2 - 1)) = 1 5 (2 8 x

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solx2f05 - SOLUTIONS TO MIDTERM 2 OF FALL 2005 1 d(ln(7x 1...

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