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 

\[x^2 - 4 \geq 0\]
\[ \Rightarrow x \leq - 2 \text { or } x \geq 2\]

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, \[f'\left( x \right) = \frac{1}{2\sqrt{x^2 - 4}}\left( 2x \right) = \frac{x}{\sqrt{x^2 - 4}}\] 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

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

Now,

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

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

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

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

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

\[ \Rightarrow x^2 = 5 x^2 - 20\]

\[ \Rightarrow 4 x^2 = 20\]

\[ \Rightarrow x = \pm \sqrt{5}\]

Thus, \[c = \sqrt{5} \in \left( 2, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}\].

Hence, Lagrange's theorem is verified.