## Calculus 8th Edition

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