$f(x) = \sqrt {25 - {x^2}}$ in [1,5]

We have, $f(x) = \sqrt {25 - {x^2}}$ in [1,5]

(i) Since, $f(x) = {\left( {25 - {x^2}} \right)^{1/2}},$ where $25 - {x^2} \ge 0$

$\Rightarrow$ ${x^2} \le \pm 5 \Rightarrow - 5 \le x \le 5$

Hence, $f(x)$ is continuous in [1,5] .

(ii) ${f^\prime }(x) = \frac{1}{2}{\left( {25 - {x^2}} \right)^{ - 1/2}} \cdot - 2x = \frac{{ - x}}{{\sqrt {25 - {x^2}} }},$ which exists in (1,5) .

Hence, ${f^\prime }(x)$ is differentiable in (1,5) .

Since, conditions of mean value theorem are satisfied.

By mean value theorem $\exists c \in (1,5)$ such that
$f'(c) = \frac{{f(5) - f(1)}}{{5 - 1}} \Rightarrow \frac{{ - c}}{{\sqrt {25 - {c^2}} }} = \frac{{0 - \sqrt {24} }}{4}$
$\Rightarrow$ $\frac{{{c^2}}}{{25 - {c^2}}} = \frac{{24}}{{16}}$
$\Rightarrow$ $16{c^2} = 600 - 24{c^2}$

$\Rightarrow$ ${c^2} = \frac{{600}}{{40}} = 15$

therefore,$c = \pm \sqrt {15}$

Also, $c = \sqrt {15} \in (1,5)$

Hence, the mean value theorem has been verified.