91Ó°ÊÓ

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

We have to find the domain.

As the given function is a square root function.

To find the domain first, find the real zeros of the numerator and denominator inside the radical of the equation.

So, $\sqrt{\frac{x-1}{x+4}}=0$

First, find the real zeros of the numerator

$$\begin{gathered} \sqrt{x-1}=0 \\ x-1=0 \\ x=1 \end{gathered}$$

Now, the real zeros of the denominator

$$\begin{gathered} \sqrt{x+4}=0 \\ x+4=0 \\ x=-4 \end{gathered}$$

The three intervals we get by the real zeros are

$(-∞,-4),(-4,1),(1,∞).$

Let $x=-6$substitute in the given function

$$\begin{gathered} f(-6)=\sqrt{\frac{-6-1}{-6+4}} \\ =\sqrt{\frac{-7}{-2}} \\ =\sqrt{\frac{7}{2}} \end{gathered}$$

So, it is true.

Let $x=0$substitute in the given function

$$\begin{gathered} f\left(0\right)=\sqrt{\frac{0-1}{0+4}} \\ =\sqrt{\frac{-1}{4}} \end{gathered}$$

So, it is not true.

Let $x=2$substitute in the given function

$$\begin{gathered} f\left(2\right)=\sqrt{\frac{2-1}{2+4}} \\ f\left(2\right)=\sqrt{\frac{1}{6}} \end{gathered}$$

So, it is true.

Therefore, the domain of the function is$\left\{x\right|x<-4,xâ‰\yen 1\}$or the interval notation is$(-∞,-4)∪[1,∞).$