## Calculus: Early Transcendentals 8th Edition

$x\lt0$ or $x\gt5$
$\because h(x)$ is defined for all real x where the denominator is not equal to zero $[\sqrt[4] (x^{2}-5x)\ne0, x^{2}-5x\ne0$], and only the fourth root of non-negative numbers is real $[x^{2}-5x\geq0]$ $\therefore x^{2}-5x\gt0,$ $x(x-5)\gt0,$ $x\lt0$ or $x\gt5$