Question

What is the domain of the function$f\left(x\right)=\sqrt{\frac{x-2}{x+4}}$?

Short answer

The domain of the function is $\left\{x\right|x<-4,x\ge 2\}$and the interval notation is$(-\infty ,-4)\cup [2,\infty ).$

Step-by-step solution

Step 1. Given Information 

The given function is$f\left(x\right)=\sqrt{\frac{x-2}{x+4}}$

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,

$\sqrt{\frac{x-2}{x+4}}=0$

First, find the real zeros of the numerator

$$\begin{gathered} \sqrt{x-2}=0 \\ x-2=0 \\ x=2 \end{gathered}$$

Now, the real zeros of denomiantor

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

Step 3. Determining interval

The three intervals we get by the real zeros are

$(-\infty ,-4),(-4,2),(2,\infty ).$

Step 4. Check the points

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

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

So, it is true.

Let $x=-2$substitute in the given function

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

So, it is not true.

Let $x=4$substitute in the given function

$$\begin{gathered} f\left(4\right)=\sqrt{\frac{4-2}{4+4}} \\ =\sqrt{\frac{2}{8}} \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

So, it is true.

Therefore, the domain of the function isrole="math" localid="1646279976919" $\left\{x\right|x<-4,x\ge 2\}$or the interval notation is$(-\infty ,-4)\cup [2,\infty ).$