Question

What is the domain of the function$f\left(x\right)=\sqrt{x^4-16}$?

Short answer

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

Step-by-step solution

Step 1. Given Information

The given function is $f\left(x\right)=\sqrt{x^4-16}$

We have to find the domain.

Step 2. Finding the domain

As the given function is a square root function.

To find the domain of a square root function, set the radicand to greater than or equal to zero.

So,

$$\begin{gathered} x^4-16\ge 0 \\ \left(x^2-4\right)\left(x^2+4\right)\ge 0 \end{gathered}$$

Step 3. Check the points

Let $x=-6$and substitute in $(x^2+4)(x^2-4)\ge 0$

$$\begin{gathered} \left({\left(-6\right)}^2+4\right)\left({\left(-6\right)}^2-4\right)\ge 0 \\ \left(40\right)\left(32\right)\ge 0 \end{gathered}$$

So, it is true.

Let $x=0$and substitute in $(x^2+4)(x^2-4)\ge 0$

$$\begin{gathered} \left(0+4\right)\left(0-4\right)\ge 0 \\ -16\ge 0 \end{gathered}$$

So, it is false.

Let $x=6$and substitute in $(x^2+4)(x^2-4)\ge 0$

$$\begin{gathered} \left({\left(6\right)}^2+4\right)\left({\left(6\right)}^2-4\right)\ge 0 \\ \left(40\right)\left(32\right)\ge 0 \end{gathered}$$

So, it is true.

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