Question

Find the domain of each function.

$f\left(x\right)=\frac{\sqrt{x+1}}{x^2-4}$

Short answer

The domain of the function$f\left(x\right)=\frac{\sqrt{x+1}}{x^2-4}$ is role="math" localid="1645963482579" $\left\{x\right|x\ge -1;x\ne 2\}$.

Step-by-step solution

Step 1. Given information 

We have been given a function $f\left(x\right)=\frac{\sqrt{x+1}}{x^2-4}$.

We have to find its domain.

Step 2. Find the values that make the square root function undefined

Square root function is defined for nonnegative numbers.

Therefore, $(x+1)$should be non-negative.

$$\begin{gathered} x+1\ge 0 \\ x+1-1\ge 0-1(\mathrm{Subtract}1\mathrm{from}\mathrm{both}\mathrm{sides}) \\ x\ge -1 \end{gathered}$$

Step 3. Find the values where the denominator becomes zero 

The given function will not be defined where the denominator becomes zero.

We need to exclude those values of $x$.

$$\begin{gathered} x^2-4=0 \\ {x}^2-4+4=0+4(\mathrm{Add}4\mathrm{on}\mathrm{both}\mathrm{sides}) \\ {x}^2=4 \\ x=\pm \sqrt{4}(\mathrm{Take}\mathrm{square}\mathrm{roots}\mathrm{on}\mathrm{both}\mathrm{sides}) \\ x=\pm 2 \end{gathered}$$

Therefore, the domain will be $\left\{x\right|x\ge -1;x\ne 2\}$.