The domain of the function

$f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$

for $\sqrt{4-x}$

4 − x ≥ 0

i.e 4 ≥ x

i.e  x ≤ 4

and $\frac{1}{\sqrt{x^{2}-1}}$

Since $x^{2}-1>0 \quad\left(\because x^{2}-1=0\right.$ is not defined for $\left.\frac{1}{\sqrt{x^{2}-1}}\right)$

i.e   x2 > 1 

i.e   x < − 1 or x > 1 

for $\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$, Domain is $(-\infty,-1) \cup(1,4]$