This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 xaxis 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....
View
Full
Document
This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Math, Algebra, Intermediate Value Theorem

Click to edit the document details