91影视

We have given the following function :-

$h\left(x\right)=\sqrt{\frac{6}{x+3}}$

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

In the following given function :-

$h\left(x\right)=\sqrt{\frac{6}{x+3}}$,

we can see that radical is 2, that is even.

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

Then we have :-

$\frac{6}{x+3}鈮�0$

Now we have :-

$\frac{6}{x+3}鈮�0$.

Now $\frac{6}{x+3}$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 :-

$$\begin{gathered} x+3>0 \\ 鈬�x>-3 \end{gathered}$$

Also as radicand is a fraction so denominator cannot be zero.

That is :-

$$\begin{gathered} x+3鈮�0 \\ 鈬�x鈮�-3 \end{gathered}$$

From above step, we have :-

$$\begin{gathered} x>-3 \\ andx鈮�-3 \end{gathered}$$

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

$x>-3$

Required interval notation of domain of given function is :-

$\left(-3,鈭�\right)$.