Domain with a radical denominator

Text

Find the natural domain of $g(x)=\dfrac{x+1}{\sqrt{x^{2}-x-6}}$.

Solution

Since the root is in the denominator, the radicand must be strictly positive: $x^{2}-x-6>0.$ Factoring, $x^{2}-x-6=(x-3)(x+2)$, which is positive outside the roots $x=-2$ and $x=3$: $x<-2\quad\lor\quad x>3.$ $\boxed{\,D=(-\infty,\,-2)\cup(3,\,+\infty)\,}$