Question

Find the value of c prescribed by Lagrange's mean value theorem for the function
$f\left(x\right)=\sqrt{x^2-4}$ defined on [2, 3].

Answer 1

We have

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

Here, $f\left(x\right)$ will exist, if
$$\begin{gathered} x^2-4\ge 0 \\ \Rightarrow x\le -2\mathrm{or}x\ge 2 \end{gathered}$$

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

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

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

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

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

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}=\frac{f\left(3\right)-f\left(2\right)}{1}$

Now,
$f\left(x\right)=\sqrt{x^2-4}$

$f\text{'}\left(x\right)=\frac{x}{\sqrt{x^2-4}}$ , $f\left(3\right)=\sqrt{5}$ , $f\left(2\right)=0$

∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$

$$\begin{gathered} \Rightarrow \frac{x}{\sqrt{x^2-4}}=\sqrt{5} \\ \Rightarrow \frac{x^2}{x^2-4}=5 \\ \Rightarrow x^2=5x^2-20 \\ \Rightarrow 4x^2=20 \\ \Rightarrow x=\pm \sqrt{5} \end{gathered}$$
Thus, $c=\sqrt{5}\in \left(2,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$.

Hence, Lagrange's theorem is verified.