$f(x) = \sqrt {4 - {x^2}}$ in [-2,2]

We have, $f(x) = \sqrt {4 - {x^2}} = {\left( {4 - {x^2}} \right)^{1/2}}$

(i) $f(x) = \sqrt {4 - {x^2}}$ is a continuous function.

[since every polynomial function is a continuous function]
Hence, $f(x)$ is continuous in [-2,2] .

(ii) ${f^\prime }(x) = \frac{1}{2}{\left( {4 - {x^2}} \right)^{ - 1/2}} \cdot ( - 2x)$

$= - x \cdot \frac{1}{{\sqrt {4 - {x^2}} }},$ which exists everywhere except at $x = \pm 2$.

Hence, $f(x)$ is differentiable in (-2,2) .

(iii) $f( - 2) = \sqrt {(4 - 4)} = 0$ and $f(2) = \sqrt {(4 - 4)} = 0$

$\Rightarrow$ $f( - 2) = f(2)$

conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that ${f^\prime }(c) = 0$.

$\Rightarrow$ $- c\frac{1}{{\sqrt {4 - {c^2}} }} = 0$

$\Rightarrow$ $c = 0 \in ( - 2,2)$

Hence, Rolle's theorem has been verified.