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Unformatted text preview: Fall 2010 Math 113 Final Exam Solutions 1. If f ( x ) = x ln x , find f ( e 3 ). Answer: Using the quotient rule, f ( x ) = ln x d dx ( x ) x d dx (ln x ) (ln x ) 2 = ln x 1 x 1 x (ln x ) 2 = ln x 1 (ln x ) 2 . Therefore, f ( e 3 ) = ln( e 3 ) 1 (ln( e 3 )) 2 = 3 1 3 2 = 2 9 . 2. At what value(s) of x (if any) is the function f ( x ) defined below discontinuous? f ( x ) = x 2 + 4 x + 5 if x < 2 x 2 if 2 x 2 1 + x 2 if x > 2 For x < 2, f ( x ) = x 2 + 4 x + 5, which is a polynomial and, hence, continuous. Also, for 2 < x < 2, f ( x ) = x 2 , which is also a polynomial and hence continuous. Finally, for x > 2, f ( x ) = 1 + x 2, which is continuous since its the composition of continuous functions (specifically, f ( x ) = g h ( x ) where g ( u ) = 1 + u and h ( x ) = x 2). Therefore, the only possible points of discontinuity for f are at x = 2 and x = 2. At x = 2, f ( x ) will be continuous if lim x  2 f ( x ) = f ( 2) = 2 2 = 1 . Clearly, lim x  2 + f ( x ) = lim x  2 + x 2 = 1. On the other hand, lim x  2 f ( x ) = lim x  2 ( x 2 + 4 x + 5 ) = ( 2) 2 + 4( 2) + 5 = 4 8 + 5 = 1 . Therefore, since lim x  2 + f ( x ) = 1 and lim x  2 f ( x ) = 1, we can see that f ( x ) is not continuous at x = 2. Turning to x = 2, f ( x ) will be continuous if lim x 2 f ( x ) = f (2) = 2 2 = 1 . Now lim x 2 f ( x ) = lim x 2 x 2 = 1. On the other hand lim x 2 + f ( x ) = lim x 2 + ( 1 + x 2 ) = 1 . Since the limits from the left and the right are both equal to 1, we conclude that lim x 2 f ( x ) = 1 , so f ( x ) is continuous at x = 2. Therefore, f ( x ) is continuous at all real numbers except x = 2. 1 3. Find the equation of the tangent line to the curve x cos y = 1 at the point ( 2 , 3 ) . Answer: We will find the slope of the tangent line using implicit differentiation. Differentiating both sides yields 1 cos y + x ( sin y y ) = 0 or cos y xy sin y = 0 . Therefore, cos y = xy sin y and so we have that y = cos y x sin y ....
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 Fall '08
 CHESTKOFSKY
 Calculus, Quotient Rule

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