#### 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 $(-\infty,\frac{-1}{2})\cup(\frac{-1}{2},1)\cup(1,\infty)$