Question:
The domain of the real function $f(x)=\sqrt{16-x^{2}}$ is _________.
Solution:
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 $[-4.4]$.