91Ó°ÊÓ

We have given the following function :-

$h\left(x\right)=\sqrt{\frac{5}{x-2}}$

We will apply the concept that if radical is even, then radicand must be greater than or equal to zero.

In the given function :-

$h\left(x\right)=\sqrt{\frac{5}{x-2}}$,

We can see that radical is $2$that is even.

So that the radicand must be greater than or equal to zero.

Then we have:-

localid="1645625043294" $\frac{5}{x-2}â‰\yen 0$

Now we have :-

$\frac{5}{x-2}â‰\yen 0$

Now $\frac{5}{x-2}$is positive if denominator is positive as numerator is already positive and divide of two positives is a positive.

So we take denominator as positive.

That is :-

role="math" localid="1645625182817" $x-2>0$

As radicand is a fraction so denominator cannot be zero.

That is :-

$x-2≠0$

From above step, we have :-

$$\begin{gathered} x-2>0 \\ ⇒x>2and \\ x-2≠0 \\ ⇒x≠2 \end{gathered}$$

From these two inequalities we have domain of the given function is :-

$x>2$.

Required interval notation of domain of given function is :-

$\left(2,∞\right)$.