Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem \[f\left( x \right) = \sqrt{x^2 - 4} \text { on }[2, 4]\] ?

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  \[x \in \left[ 2, 4 \right]\]  the function\[f\left( x \right)\] attains a unique definite value.

So, \[f\left( x \right)\] is continuous on \[\left[ 2, 4 \right]\]

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

So, \[f\left( x \right)\] is differentiable on \[\left( 2, 4 \right)\] .

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some 

\[c \in \left( 2, 4 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 2 \right)}{4 - 2} = \frac{f\left( 4 \right) - f\left( 2 \right)}{2}\]

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

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

∴ \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 2 \right)}{4 - 2}\]

\[\Rightarrow \frac{x}{\sqrt{x^2 - 4}} = \frac{2\sqrt{3}}{2}\]

\[ \Rightarrow \frac{x}{\sqrt{x^2 - 4}} = \sqrt{3}\]

\[ \Rightarrow \frac{x^2}{x^2 - 4} = 3 \]

\[ \Rightarrow x^2 = 3 x^2 - 12\]

\[ \Rightarrow x^2 = 6\]

\[ \Rightarrow x = \pm \sqrt{6}\]

Thus, \[c = \sqrt{6} \in \left( 2, 4 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 2 \right)}{4 - 2}\] .

Hence, Lagrange's theorem is verified.