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