Question

What is the domain of the function$f\left(x\right)=\sqrt{\frac{x-1}{x+4}}$?

Short answer

The domain of the function is$\left\{x\right|x<-4,x\ge 1\}$ and the interval notation is$(-\infty ,-4)\cup [1,\infty ).$

Step-by-step solution

Step 1. Given Information 

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

We have to find the domain.

Step 2. Finding 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

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

Step 3. Determining intervals

The three intervals we get by the real zeros are

$(-\infty ,-4),(-4,1),(1,\infty ).$

Step 4. Check the points 

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$\left\{x\right|x<-4,x\ge 1\}$or the interval notation is$(-\infty ,-4)\cup [1,\infty ).$