The value of c prescribed by Lagrange's mean value theorem, when $f(x)=\sqrt{x^2-4}, a=2$ and $b=3$, is

$\sqrt{5}$

Clearly, $f(x)=\sqrt{x^2-4}$ is continuous on $[2,3]$ and differentiable on $(2,3)$.

So, by Lagrange's mean value theorem there exists $c \in(2,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}$

$\Rightarrow \frac{c}{\sqrt{c^2-4}}=\sqrt{5}-0$        $\left[∵ f(x)=\sqrt{x^2-4} \Rightarrow f^{\prime}(x)=\frac{x}{\sqrt{x^2-4}}\right]$

$\Rightarrow c^2=5\left(c^2-4\right)$

$\Rightarrow 4 c^2=20$

$\Rightarrow c=\sqrt{5}$