This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 θ → sin( θ 2 ) 1 cos( θ ) This is the case of l’Hˆ opital’s Rule. lim θ → sin( θ 2 ) 1 cos( θ ) = lim θ → 2 θ cos( θ 2 ) sin( θ ) by l’Hˆ opital’s Rule, still case = lim θ → 2 cos( θ 2 ) 4 θ 2 sin( θ 2 ) cos( θ ) by l’Hˆ opital’s 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 = π 8 or x = 5 π 8 (in the given interval). Therefore, we must consider three intervals [ 1 , π 8 ] , [ π 8 , 5 π 8 ] and [ 5 π 8 , 2] . Evaluating at , we see that on the first interval f ( x ) is positive, evaluating at x = π 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....
View
Full Document
 Winter '08
 WAlker
 Calculus, Derivative, lim, Mathematical analysis

Click to edit the document details