# How do you find the domain of (x^2-x-12)^-(1/4)?

The domain is D=(-oo;-3)uu(4;+oo)
This function can be written as $y = \frac{1}{\sqrt[4]{{x}^{2} - x - 12}}$.
so the domain is this subset of $\mathbb{R}$ where ${x}^{2} - x - 12 > 0$