## College Algebra 7th Edition

$(-\infty,\ -1)\cup (-1,1)\cup (1,\ \infty)$
We are given: $f(x)= \frac{x+2}{x^{2}-1}$ The domain of a function consists of all the values that $x$ is allowed to have. In this case, we can not have division by $0$: $x^{2}-1\neq 0$ $x^{2}\neq 1$ $x\neq\pm 1$ Therefore, the domain is: $(-\infty,\ -1)\cup (-1,1)\cup (1,\ \infty)$