Q. 10 - Precalculus Enhanced with Graphing Utilities

Question

Find the domain of each function. 

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

Step-by-Step Solution

Answer

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

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 $\left\{x\right|x\ge -1;x\ne 2\}$.