## College Algebra 7th Edition

$(3,\ \infty)$
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 $(3,\ \infty)$