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Unformatted text preview: Math 124, Autumn 2006 Final Exam Solutions Page 1 of 9 1. [ 12 points total ] Calculate the derivatives of the following functions. You need not simplfy your answers. (a) [ 4 points ] y = 5 x 7 sin(3 x ) + e 2 + ln x . y = 35 x 6 3 cos(3 x ) + 0 + 1 x (b) [ 4 points ] f ( r ) = 1 + r 3 1 1 r ! 3 f ( r ) = 3(1 + r ) 2 1 2 r ! 1 1 r ! 3 + (1 + r ) 3 3 1 1 r ! 2 1 2 r 3 / 2 OR if they simplified first f ( r ) = ( r 1 / 2 r 1 / 2 ) 3 f ( r ) = 3 2 ( r 1 / 2 r 1 / 2 ) 2 ( r 1 / 2 + r 3 / 2 ) (c) [ 4 points ] g ( t ) = (ln t ) ln t u = (ln t ) ln t ln u = (ln t )(ln (ln t )) u u = 1 t (ln (ln t )) + (ln t ) 1 t (ln t ) = ln(ln t ) + 1 t g ( t ) = ln(ln t ) + 1 t (ln t ) ln t Math 124, Autumn 2006 Final Exam Solutions Page 2 of 9 2. [ 10 points total ] Compute the following limits. You must show your work or adequately explain your answers. No credit will be given for an unsupported answer. (a) [ 5 points ] lim x 2 2 x 2 1 2 x 2 lim x 2 2 x 2 1 2 x 2 = lim x 2 4 x 2 2 x 2 ( x 2) = lim x 2 ( x + 2) 2 x 2 = 1 2 (b) [ 5 points ] lim sin( 2 ) 1 cos( ) This is the case of lH opitals Rule. lim sin( 2 ) 1 cos( ) = lim 2 cos( 2 ) sin( ) by lH opitals Rule, still case = lim 2 cos( 2 ) 4 2 sin( 2 ) cos( ) by lH opitals Rule = 2 using continuity Math 124, Autumn 2006 Final Exam Solutions Page 3 of 9 3. [ 12 points total ] Let f ( x ) = e 2 x cos(2 x ) on the domain D = [ 1 , 2]. (a) [ 3 points ] Find the subintervals of D on which f is increasing, and the subintervals on which f is decreasing. To find the intervals on which f is increasing or decreasing, we compute the derivative f and determine the intervals on which it is positive or negative. To compute the derivative we use the product rule and chain rule to obtain f ( x ) = e 2 x  sin(2 x ) 2 + cos(2 x ) e 2 x 2 = 2 e 2 x (cos(2 x ) sin(2 x )) . Since e 2 x is never zero, the derivative can be zero only where cos(2 x ) sin(2 x ) = 0 . Now, cos(2 x ) = sin(2 x ) when x = 8 or x = 5 8 (in the given interval). Therefore, we must consider three intervals [ 1 , 8 ] , [ 8 , 5 8 ] and [ 5 8 , 2] . Evaluating at , we see that on the first interval f ( x ) is positive, evaluating at x = 4 , we see that f ( x ) is negative on the second interval and evaluating at 2 , we see that f ( x ) is positive on the third interval....
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This note was uploaded on 04/28/2008 for the course MATH 124 taught by Professor Walker during the Winter '08 term at University of Washington.
 Winter '08
 WAlker
 Derivative

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