Answer
$(-\infty,0)\cup(0,4)\cup(4,\infty)$
Work Step by Step
We are given the function:
$h(x)=\dfrac{6}{x^2-4x}$
Factor the denominator:
$h(x)=\dfrac{6}{x(x-4)}$
The domain of the function contains all real numbers $x$ except the zeros of the denominator.
Determine the zeros of the denominator:
$x(x-4)=0$
$x=0$ or $x-4=0$
$x=0$ or $x=4$
The domain is:
$(-\infty,0)\cup(0,4)\cup(4,\infty)$
