Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 31

Determine the domain of the function $\displaystyle h(x) = \frac{1}{\sqrt[4]{x^2 - 5x}}$



The function is a rational function that involves a square root. In order for the function to exist, the value of the radicand should not be equal to 0. So,



$
\begin{equation}
\begin{aligned}
x^2 - 5x & \neq 0\\
x(x-5) & \neq 0\\
\end{aligned}
\end{equation}
$

$ x \neq 0 \text{ and } x \neq 5$
Also, the root function is defined only for positive x values so,

$
\begin{equation}
\begin{aligned}
x^2 - 5x & > 0\\
x(x-5) &> 0 \\
\end{aligned}
\end{equation}
$

$x< 0 \text{ and } x > 5 $

Therefore,

The domain of the function is: $(-\infty,0) \bigcup (5,\infty)$