## Calculus: Early Transcendentals 8th Edition

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