The domain of the real function

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

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

$16-x^{2} \geq 0$

$\Rightarrow 16 \geq x^{2}$

$\Rightarrow x^{2} \leq 16$

$\Rightarrow x \leq 4$ and $x \geq-4$

$\Rightarrow-4 \leq x \leq 4$

$\Rightarrow x \in[-4,4]$

Hence, the domain of the real function $f(x)=\sqrt{16-x^{2}}$ is $[-4.4]$.