Answer
$(-\infty,-1]\cup[1,4]$
Work Step by Step
$\sqrt{4-x}$ is defined for x such that
$4-x\geq 0$
$4\geq x,\qquad x\in(-\infty,4]$
$\sqrt{x^{2}-1}$ is defined for x such that
$x^{2}-1\geq 0$
$x^{2}\geq 1,\qquad x\in(-\infty,-1]\cup[1,\infty)$
$h$ is defined where BOTH these fuctions are defined, the intersection of their domains.
Domain of h is
$(-\infty,-1]\cup[1,4]$
