Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises: 26

Answer

$G(x)$ is continuous on $(-\infty,\frac{-1}{2})\cup(\frac{-1}{2},1)\cup(1,\infty)$

Work Step by Step

$G(x)=\frac{x^2+1}{2x^2-x-1}$ $G(x)=\frac{x^2+1}{(2x+1)(x-1)}$ (factorize the denominator) 1) Find the domain of $G(x)$ $G(x)$ is defined only when $(2x+1)(x-1)\ne0$ $G(x)$ is defined only when $(2x+1)\ne0$ and $(x-1)\ne0$ $G(x)$ is defined only when $x\ne\frac{-1}{2}$ and $x\ne1$ Therefore, the domain of $G(x)$ is $(-\infty,\frac{-1}{2})\cup(\frac{-1}{2},1)\cup(1,\infty)$ 2) Since $G(x)$ is a rational function, according to Theorem 5, $G(x)$ is continuous on its demain. Therefore, $G(x)$ is continuous on $(-\infty,\frac{-1}{2})\cup(\frac{-1}{2},1)\cup(1,\infty)$
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