Find the value of c prescribed by Lagrange's mean value theorem for the function

Solution:

We have

$f(x)=\sqrt{x^{2}-4}$

Here, $f(x)$ will exist, if

$x^{2}-4 \geq 0$

$\Rightarrow x \leq-2$ or $x \geq 2$

Since for each $x \in[2,3]$, the function $f(x)$ attains a unique definite value, $f(x)$ is continuous on $[2,3]$.

Also, $f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}}(2 x)=\frac{x}{\sqrt{x^{2}-4}}$ exists for all $x \in(2,3)$.

So, $f(x)$ is differentiable on $(2,3)$.

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists $c \in(2,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}=\frac{f(3)-f(2)}{1}$

Now,

$f(x)=\sqrt{x^{2}-4}$

$f^{\prime}(x)=\frac{x}{\sqrt{x^{2}-4}}, f(3)=\sqrt{5}, f(2)=0$

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

$\Rightarrow \frac{x}{\sqrt{x^{2}-4}}=\sqrt{5}$

$\Rightarrow \frac{x^{2}}{x^{2}-4}=5$

$\Rightarrow x^{2}=5 x^{2}-20$

$\Rightarrow 4 x^{2}=20$

 

$\Rightarrow x=\pm \sqrt{5}$

Thus, $c=\sqrt{5} \in(2,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}$

Hence, Lagrange's theorem is verified.