asst5-f10-solns

asst5-f10-solns - MATH 137 Solutions to Assignment #5, Part...

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MATH 137 Fall 2010 Solutions to Assignment #5, Part II 1. For the function y = x - 1 x + 1 , dy dx = ( x + 1) - ( x - 1) ( x + 1) 2 = 2 ( x + 1) 2 . Since the slope of the line x - 2 y = 2 is 1 2 , we want to find the equation of all tangent lines to the curve for which dy dx = 1 2 . Thus we solve the equation 2 ( x + 1) 2 = 1 2 , hence ( x + 1) 2 = 4 or x + 1 = ± 2, so x = 1 or x = - 3. At the point (1 , 0), the equation of the tangent line to the curve is y = 1 2 ( x - 1). At the point ( - 3 , 2), the equation of the tangent line to the curve is y - 2 = 1 2 ( x + 3). 2. We first find the points where the graphs meet. That is, we solve x 3 - 3 x + 4 = 3 x 2 - 3 x , or x 3 - 3 x 2 + 4 = 0. Let f ( x ) = x 3 - 3 x 2 + 4. Then f ( - 1) = 0, so x + 1 is a factor of f ( x ). Using synthetic (or long) division we find the complete factorization of f ( x ) to be ( x + 1)( x - 2) 2 . Therefore, the curves meet where x = - 1 and where x = 2, or more specifically, the curves meet at the points ( - 1 , 6) and (2
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This note was uploaded on 01/18/2012 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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asst5-f10-solns - MATH 137 Solutions to Assignment #5, Part...

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