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(x) = x² − 1 on [2, 3] ?

Answer 1

 We have

\[f\left( x \right) = x^2 - 1\]

Since a polynomial function is everywhere continuous and differentiable, \[f\left( x \right)\] is continuous on \[\left[ 2, 3 \right]\] and differentiable on \[\left( 2, 3 \right)\]. 

Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ \[c \in \left( 2, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}\]

Now, 

\[f\left( x \right) = x^2 - 1\]

\[\Rightarrow f'\left( x \right) = 2x\] ,

\[f\left( 3 \right) = \left( 3 \right)^2 - 1 = 8\] ,

\[f\left( 2 \right) = \left( 2 \right)^2 - 1 = 3\]

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

\[\Rightarrow 2x = \frac{8 - 3}{1}\]

\[ \Rightarrow x = \frac{5}{2}\]

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

Hence, Lagrange's theorem is verified.