solx2f05 - SOLUTIONS TO MIDTERM 2 OF FALL 2005 1. d dx...

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Unformatted text preview: 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 ( x ) = 4- 5 e x . Setting f ( 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 ( x ) > 0 for x < ln(4 / 5), f ( x ) < f (ln(4 / 5)) for x < ln(4 / 5) and as f ( 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)) = 4ln(4 / 5)- 5 e ln(4 / 5) = 4ln(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)....
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This note was uploaded on 04/24/2010 for the course MATH 9374 taught by Professor Smith during the Spring '10 term at University of California, Berkeley.

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solx2f05 - SOLUTIONS TO MIDTERM 2 OF FALL 2005 1. d dx...

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