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We have given the following function :-

$f\left(x\right)=\sqrt{\frac{x+3}{x-2}}$.

We have to find the domain of this function.

Given function is :-

$f\left(x\right)=\sqrt{\frac{x+3}{x-2}}$

We know that if radical is even, then the radicand of the function is greater than or equals to zero.

Here radical is 2, that is even.

So we have :-

$\frac{x+3}{x-2}鈮�0$

This inequality holds if numerator and denominator are both positive or both negative.

Also know denominator cannot be equals to zero.

Then we have following two cases :-

CASE 1 :-

$x+3鈮�0andx-2>0$.

CASE 2 :-

$x+3鈮�0andx-2<0$.

We have :-

$x+3鈮�0andx-2>0$.

From these we have :-

$$\begin{gathered} x+3鈮�0andx-2>0 \\ 鈬�x鈮�-3andx>2 \end{gathered}$$

We can change this to interval form as following :-

localid="1645030922267" $$\begin{gathered} x鈭�[-3,鈭�)and\left(2,鈭�\right) \\ 鈬�x鈭�[-3,鈭�)鈭�\left(2,鈭�\right) \\ 鈬�x鈭�\left(2,鈭�\right) \end{gathered}$$

This is the domain for case one.

Now consider the following case 2 :-

$x+3鈮�0andx-2<0$

From these inequalities we have :-

$$\begin{gathered} x+3鈮�0andx-2<0 \\ 鈬�x鈮�-3andx<2 \\ 鈬�x鈭�(-鈭�,-3]and\left(-鈭�,2\right) \\ 鈬�x鈭�(-鈭�,-3]鈭�\left(-鈭�,2\right) \\ 鈬�x鈭�(-鈭�,-3] \end{gathered}$$

This is domain for case 2.

From case 1, we have :-

$x鈭�\left(2,鈭�\right)$.

From case 2, we have :-

role="math" $x鈭�(-鈭�,-3]$.

By combining both of these intervals, we have:-

$(-鈭�,-3]鈭�\left(2,鈭�\right)$.

This is the required domain of the given function.