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Unformatted text preview: 1 2 3 4 5 6 Explanation: Since x2  16 = 0 when x = 4, the graph of f will have vertical asymptotes at x = 4; on the other hand, since 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 x lim x2 = 1, x2  16 2 4 the graph will have a horizontal asymptote at y = 1. This already eliminates some of the possible graphs. On the other hand, f (0) = 0, so the graph of f must also pass through the origin. This eliminates another graph. To decide which of the remaining graphs is that of f we look at the sign of f to determine where f is increasing or decreasing. Now, by the Quotient Rule, f (x) = 2x(x2  16)  2x3 (x2  16)2 =  Thus (x2 32x .  16)2 6 5 5.4 3 2 1 0 1 2 3 4 5 6 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 2 4 while f (x) > 0, x < 0, f (x) < 0, x > 0, so the graph of f is increasing to the left of the origin and decreasing to the right of the origin. The only graph having all these properties is 6 5 4 3 2 1 0 1 2 3 4 5 6 Garcia, Ilse Homework 11 Due: Nov 6 2007, 3:00 am Inst: Fonken 4 2 4 2 2 4
6 5 4 3 2 1 0 1 2 3 4 5 6 2 4 Consequently, this must be the graph of f . keywords: graph, ration...
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This note was uploaded on 01/16/2009 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.
 Spring '08
 schultz
 Differential Calculus

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