Pre-Calc Homework Solutions 107

A when x 1 y3 2y 2 we have y 3 1 0 clearly y 2y 1 or

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Unformatted text preview: we have y 3 1 0. Clearly, y 2y 1, or 1 is one solution, and 1)(y 2 y 1). The 3 2 3 3 4 At 3 1 , : 4 2 3 4 1 2 3 4 1 3 2 1 2 1 4 we may factor y 3 solutions of y 2 4 4 3 1 2y + 1 as (y y 1 = 0 are y 3 1 2 1 Slope (b) (1) 4(1)( 1) 2(1) 2 1 2 1 2 5 . Hence, 5 2 there are three possible y-values: 1, 5 , and . y3 (b) [ 1.8, 1.8] by [ 1.2, 1.2] xy 1 d ( 1) dx d 3 (y ) dx d (xy) dx Parameter interval: 38. (a) 1 y (2 2 t x) x)] dy dx dy x) dx dy dx 1 x 3 3y 2y d 3 (x ) dx xy (3y 2 (y)(1) x)y y 0 y y 3y 2 x y d dx 3y 2 x (3y 2 x)(y ) (y)(6y y (3y 2 x)2 y xy 3y 2 y 2 (3y x)2 d 2 [y (2 dx (y2)( 1) (2 x)(2y) 3x 2 3x 2 3x 2y(2 4 2 2 2y(2 y2 y x) 2 y 1) Slope at (1, 1): Tangent: y Normal: y 3(1)2 (1)2 2(1)(2 1) 2 2x 1 1 x 2 3 2 x3 2 x 2(x 1 (x 2 1) 1) 1 or y 1 or y Since we are working with numerical information, there is no need to write a general expression for y in terms of x and y. . To evaluate f (2), evaluate the expression for y using x 2 and y 1 3(1)2 2 (b) One way is to graph the equations y 1: 1 1 f (2) 1 To evaluate f (2), evaluate the expression for y using x 2, y 1, and y (1) 2(1) [3(1)2 1: 3(1)2(1) 2]2 4 1 f z (2) 4...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.

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