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 − 3x + 2 on [−1, 2] ?

Answer 1

We have,

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

Since a polynomial function is everywhere continuous and differentiable.
Therefore,

\[f\left( x \right)\] is continuous on \[\left[ - 1, 2 \right]\] and differentiable on \[\left( - 1, 2 \right)\]

Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​\[c \in \left( - 1, 2 \right)\] such that \[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{2 + 1} = \frac{f\left( 2 \right) - f\left( - 1 \right)}{3}\]

Now, 

\[f\left( x \right) = x^2 - 3x + 2\] \[\Rightarrow f'\left( x \right) = 2x - 3\] ,\[f\left( 2 \right) = 0\] ,\[f\left( - 1 \right) = \left( - 1 \right)^2 - 3\left( - 1 \right) + 2 = 6\]

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

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

\[ \Rightarrow 2x - 1 = 0\]

\[ \Rightarrow x = \frac{1}{2}\]

Thus, \[c = \frac{1}{2} \in \left( - 1, 2 \right)\] such that \[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{2 - \left( - 1 \right)}\] .

Hence, Lagrange's theorem is verified.