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) = 2x² − 3x + 1 on [1, 3] ?

Answer 1

 We have,

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

Since a polynomial function is everywhere continuous and differentiable.
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 1, 3 \right]\] and differentiable on \[\left( 1, 3 \right)\].

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

Now, \[f\left( x \right) = 2 x^2 - 3x + 1\]

\[\Rightarrow f'\left( x \right) = 4x - 3\],\[f\left( 3 \right) = 10\] ,\[f\left( 1 \right) = 2 \left( 1 \right)^2 - 3\left( 1 \right) + 1 = 0\]

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

\[\Rightarrow 4x - 3 = \frac{10 - 0}{2}\]

\[ \Rightarrow 4x - 3 - 5 = 0\]

\[ \Rightarrow x = 2\]

Thus, 

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

Hence, Lagrange's theorem is verified.