Answer
$$
f(x)=\sqrt {4-x^{2}}
$$
the values $x$ for $f(x)$ is defined when:
$$
4-x^{2}\geq 0
$$
we can find all zeros $f(x)$
$$
4-x^{2} = (2-x)(2+x) = 0
$$
$ \Rightarrow $
$$
x=\pm 2,
$$
this numbers form the intervals $ (-\infty,-2), (-2,2) $ and $(2,\infty)$.
we observe that the values in the interval $[-2,2]$ which satisfy the inequality
So domain of the function $f (x)$ is $[-2,2]$ .
Work Step by Step
$$
f(x)=\sqrt {4-x^{2}}
$$
the values $x$ for $f(x)$ is defined when:
$$
4-x^{2}\geq 0
$$
we can find all zeros $f(x)$
$$
4-x^{2} = (2-x)(2+x) = 0
$$
$ \Rightarrow $
$$
x=\pm 2,
$$
this numbers form the intervals $ (-\infty,-2), (-2,2) $ and $(2,\infty)$.
we observe that the values in the interval $[-2,2]$ which satisfy the inequality
So domain of the function $f (x)$ is $[-2,2]$ .
