Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 139: 57

Answer

$f(x) = \frac{-(x-2)}{(x^2)(x-3)}$

Work Step by Step

$(x-3)$ is a term in the denominator since $x=3$ is a vertical asymptote. $x^2$ is a term in the denominator since $x$ is a vertical asymptote and $\lim\limits_{x \to 0^-}f(x) = \lim\limits_{x \to 0^+}f(x) = -\infty$ $(x-2)$ is a term in the numerator since $f(2) = 0$ We can write the function $f$: $f(x) = \frac{c~(x-2)}{(x^2)(x-3)}$ where $c$ is a constant $\lim\limits_{x \to 3^-}f(x) = \infty$ Thus, we can let $c = -1$ We can write the function $f$: $f(x) = \frac{-(x-2)}{(x^2)(x-3)}$ We can see that this function satisfies all the required conditions.
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