Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.8 Continuity - 1.8 Exercises: 26

Answer

continuous on $(-\infty, -\displaystyle \frac{1}{2})\cup(-\frac{1}{2},1)\cup(1, \infty)$.

Work Step by Step

$G(x)=\displaystyle \frac{x^{2}+1}{2x^{2}-x-1}=\frac{x^{2}+1}{(2x+1)(x-1)}$ (not defined for $x=-\displaystyle \frac{1}{2}$ and $x=1$ domain of G: $(-\infty, -\displaystyle \frac{1}{2})\cup(-\frac{1}{2},1)\cup(1, \infty)$. G(x) is a rational function, so by Th. 5, [(b) Any rational function is continuous wherever it is defined.] it is continuous on its domain. continuous on $(-\infty, -\displaystyle \frac{1}{2})\cup(-\frac{1}{2},1)\cup(1, \infty)$.
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