f09finalsolutions - Fall 2009 Math 113 Final Exam Solutions...

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Unformatted text preview: Fall 2009 Math 113 Final Exam Solutions 1. What are the domain and range of the function f ( x ) = 1 + e x 1- e x ? Answer: The function is well-defined everywhere except when the denominator is zero, which happens when 0 = 1- e x , or, equivalently, e x = 0. This only happens when x = 0, so we see that the domain of f is the set of all real numbers 6 = 0. As for the range of f , notice that lim x - 1 + e x 1- e x = + and lim x + 1 + e x 1- e x =- . Now, lim x - 1 + e x 1- e x = 1 and lim x + 1 + e x 1- e x = lim x + e x- e x =- 1 by LH opitals Rule, so f has horizontal asymptotes at 1 and- 1. Finally, since f ( x ) = (1- e x ) e x- (1 + e x )(- e x ) (1- e x ) 2 = e x- e 2 x + e x + e 2 x (1- e x ) 2 = 2 e x (1- e x ) 2 is never zero, f has no critical points and, thus, no maxima or minima. Putting this all together, the range of f consists of those real numbers x such that- < x <- 1 or 1 < x < + . 2. Suppose g ( x ) = ln 1 x + 1 . What is g- 1 ( x )? For what values of x is g- 1 ( x ) defined? Answer: To determine g- 1 ( x ), swap x and y and solve for y : x = ln 1 y + 1 ; exponentiate both sides to get e x = 1 y + 1; subtract 1 from both sides: e x- 1 = 1 y ; finally, take the reciprocal of both sides: 1 e x- 1 = y. 1 Therefore, g- 1 ( x ) = 1 e x- 1 . Clearly, g- 1 ( x ) is only defined for x 6 = 0. 3. What is the equation of the line tangent to the graph of y 3 + 3 x 2 y 2 + 2 x 3 = 4 at the point (1 ,- 1)? Answer: The goal is to determine y by using implicit differentiation. Differentiating both sides yields: 3 y 2 y + 6 xy 2 + 3 x 2 2 y y + 6 x 2 = 0 , or, equivalently, y (3 y 2 + 6 x 2 y ) + 6 xy 2 + 6 x 2 = 0 . Hence, y =- 6 xy 2- 6 x 2 ) 3 y 2 + 6 x 2 y . Therefore, the slope of the tangent line at (1 ,- 1) will be- 6(1)(- 1) 2- 6(1) 2 3(- 1) 2 + 6(1) 2 (- 1) =- 12- 3 = 4 ....
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f09finalsolutions - Fall 2009 Math 113 Final Exam Solutions...

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