math137-f11-a5sols

math137-f11-a5sols - MATH 137 (Fall 2011) Assignment 5...

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Unformatted text preview: MATH 137 (Fall 2011) Assignment 5 Solutions 1. The rational function y is continuous on < because its denominator is never equal to zero, and we note that y = 2 > 0 when x =- 1 and that y =- 2 3 < 0 when x = 1. So by the Intermediate Value Theorem it follows that the function crosses the x-axis somewhere between- 1 and 1. 2. (a) For f ( x ) = 1 x we have f (- 1) =- 1 and f (- 1) =- 1 (- 1) 2 =- 1. Thus the linear approximation to f ( x ) at a =- 1 is L ( x ) = (- 1) + (- 1)( x- (- 1)) =- x- 2. (b) For y = x we have y (4) = 2 and dy dx | x =4 = 1 2 4 = 1 4 . Thus the linearization of y at a = 4 is L ( x ) = 2 + 1 4 ( x- 4) = 1 4 x + 1. (c) For y = x 3 we have y (0) = 0 and dy dx | x =0 = 3(0) 2 = 0. Thus the equation of the tangent line of y at a = 0 is L ( x ) = 0 + 0( x- 0) = 0. L ( x ) is a horizontal line and has the point of contact (0 , 0) with the curve. We note that (- 1) 3 =- 1 < 0 and 1 3 = 1 > 0 so by the IVT, the tangent line does cross the curve at its point of contact.0 so by the IVT, the tangent line does cross the curve at its point of contact....
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This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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math137-f11-a5sols - MATH 137 (Fall 2011) Assignment 5...

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