Question

How do you find the domain of $f\left(x\right)=\sqrt{x^2-16}$?

Answer 1

Find the domain of the function:

The domain of a function $f\left(x\right)=\sqrt{g\left(x\right)}$can be found by $g\left(x\right)\ge 0$.

Write the greater than equal to type inequality by taking the function under square root on the left side and $0$on the right side:

.$x^2-16\ge 0$

Add $16$ on both sides:

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

So for the values of $x\ge 4\mathrm{and}x\le -4$, $x^2-16\ge 0$
The remaining values $-4<x<4$, would give you a negative term for $x^2-16$ which would not be valid under square root making the function invalid.

Hence, the domain is $x\ge 4\mathrm{and}x\le -4$.