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209-2-sample-sol

# 209-2-sample-sol - DEPARTMENT OF MATHEMATICS AND STATISTICS...

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Unformatted text preview: DEPARTMENT OF MATHEMATICS AND STATISTICS MATH 209 Second class test duration 1h20m 1. [15pts] Find dy dx of the following (a) y = ( x 2 + 4 x ) 5 ( x 5 + 2) 3 , (b) y = £ x 2 + ln( x- 2) / 3 , (c) y = ln x e x +1 . Solution (a) y = 5( x 2 + 4 x ) 4 (2 x + 4)( x 5 + 2) 3 + ( x 2 + 4 x ) 5 · 3( x 5 + 2) 2 (5 x 4 ) (apply product rule, then general power rule) (b) y = 3[ x 2 + ln( x- 2)] 2 ( 2 x + 1 x- 2 ) (apply general power rule, then differ- entiate x 2 +ln( x- 2). When differentiating the logarithm, use the chain rule again remembering that ( x- 2) = 1.) (c) y = (ln x ) ( e x + 1)- ln x ( e x + 1) ( e x + 1) 2 = e x +1 x- ln xe x ( e x + 1) 2 , no further simplifications are possible. 2. [30pts] For the function f ( x ) = x +1 x 2 (a) locate horizontal and vertical asymptotes and x- and y-intercepts (if any) domain are all x except x = 0. At x = 0 f (0) = 1 / 0, that is, infinite, so x = 0 is a vertical asymptote. lim x →∞ f ( x ) = 0 because the power in denominator is bigger than that of the numerator. y = 0 is the horizontal asymptote therefore. No y-intercept because x = 0 is outside the domain. x-intercept at x + 1 = 0 or x =- 1....
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209-2-sample-sol - DEPARTMENT OF MATHEMATICS AND STATISTICS...

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