f09exam3solutions

f09exam3solutions - Fall 2009 Math 113 Exam #3 Solutions 1....

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Fall 2009 Math 113 Exam #3 Solutions 1. Evaluate the limit lim x 0 1 - cos x x 2 + x . Answer: Notice that both numerator and denominator go to zero as x 0. Hence, we can apply L’Hopital’s Rule: lim x 0 1 - cos x x 2 + x = lim x 0 sin x 2 x + 1 = 0 , since sin(0) = 0. 2. Find the maximum and minimum values, inflection points and asymptotes of y = ln( x 2 + 1) and use this information to sketch the graph. Answer: Notice that y 0 = 1 x 2 + 1 · 2 x = 2 x x 2 + 1 and, by the Quotient Rule, y 00 = ( x 2 + 1)(2) - 2 x (2 x ) ( x 2 + 1) 2 = 2 x 2 + 2 - 4 x 2 ( x 2 + 1) 2 = 2 - 2 x 2 ( x 2 + 1) 2 . Now, the critical points occur when y 0 = 0, which is to say when 2 x x 2 + 1 = 0 . The only happens when x = 0, so 0 is the only critical point. Notice that y 00 (0) = 2, which is greater than zero, so the second derivative test implies that 0 is a local minimum. y 00 = 0 when 2 - 2 x 2 = 0, meaning when x = ± 1, so there are inflection points at x = ± 1. Finally, lim x →-∞ ln( x 2 + 1) = = lim x + ln( x 2 + 1) , so there are no horizontal asymptotes. Putting all this together, we see that
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at UGA.

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f09exam3solutions - Fall 2009 Math 113 Exam #3 Solutions 1....

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