#### 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 \[f\left( x \right) = \sqrt{x^2 - 4} \text { on }[2, 4]\] ?

#### Solution

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

Consequently, there exists some

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

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

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

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

\[ \Rightarrow x^2 = 6\]

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

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.