91Ó°ÊÓ

We have given the following function :-

$f\left(x\right)=\sqrt{\frac{x-1}{x+4}}$

We have to find domain of this function.

We will use the concept that if radical is even, then radicand is greater than or equals to zero.

Given function is :-

$f\left(x\right)=\sqrt{\frac{x-1}{x+4}}$.

Here radical is 2. That is even.

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

So that we have :-

$\frac{x-1}{x+4}â‰\yen 0$

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 :-

localid="1645066227223" $x-1â‰\yen 0andx+4>0$

CASE 2 :-

$x-1≤0andx+4<0$

We have :-

$x-1â‰\yen 0andx+4>0$.

From these two inequalities we have :-

$$\begin{gathered} x-1â‰\yen 0andx+4>0 \\ ⇒xâ‰\yen 1andx>4 \end{gathered}$$

We can change this to interval form as following :-

$$\begin{gathered} x∈[1,∞)and\left(4,∞\right) \\ ⇒x∈[1,∞)âˆ\copyright \left(4,∞\right) \\ ⇒x∈\left(4,∞\right) \end{gathered}$$

This is the domain for case 1.

Now consider the following case 2 :-

$x-1≤0andx+4<0$

From these inequalities we have :-

$$\begin{gathered} x-1≤0andx+4<0 \\ ⇒x≤1andx<-4 \\ ⇒x∈(-∞,1]and\left(-∞,-4\right) \\ ⇒x∈(-∞,1]âˆ\copyright \left(-∞,-4\right) \\ ⇒x∈\left(-∞,-4\right) \end{gathered}$$

This is domain for case 2.

From case 1, we have :-

$x∈\left(4,∞\right)$and

From case 2, we have :-

$x∈\left(-∞,-4\right)$.

By combining both of these intervals, we have:-

localid="1645066200154" $\left(-∞,-4\right)∪\left(4,∞\right)$.

This is the required domain of the given function.