Answer
$(3,\ \infty)$
Work Step by Step
We are given:
$\displaystyle h(x)=(x-3)^{-1/4}=\frac{1}{(x-3)^{1/4}}=\frac{1}{\sqrt[4]{x-3}}$
We can not have a negative root or division by 0, so:
$x-3\geq 0$
$x\geq 3$
And:
$x-3\ne 0$
$x\ne 3$
Thus overall we have the domain: $x\gt3$ or $(3,\ \infty)$
