Question

The domain of the real function $f\left(x\right)=\sqrt{16-x^2}$ is _________.

Answer 1

Given: $f\left(x\right)=\sqrt{16-x^2}$

To find the domain, we find the real values of x for which the function is defined.

$$\begin{gathered} 16-x^2\ge 0 \\ \Rightarrow 16\ge x^2 \\ \Rightarrow x^2\le 16 \\ \Rightarrow x\le 4\mathrm{and}x\ge -4 \\ \Rightarrow -4\le x\le 4 \\ \Rightarrow x\in \left[-4,4\right] \end{gathered}$$

Hence, the domain of the real function $f\left(x\right)=\sqrt{16-x^2}$ is [−4, 4].