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) = x³ − 2x² − x + 3 on [0, 1] ?

Answer 1

We have,

\[f\left( x \right) = x^3 - 2 x^2 - x + 3 = 0\]

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

\[f\left( x \right)\] is continuous on \[\left[ 0, 1 \right]\]  and differentiable on \[\left( 0, 1 \right)\]
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ \[c \in \left( 0, 1 \right)\] such that \[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\]

Now, \[f\left( x \right) = x^3 - 2 x^2 - x + 3 = 0\]

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

\[f\left( 1 \right) = 1\] ,\[f\left( 0 \right) = 3\]

∴ \[f'\left( x \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\]

\[\Rightarrow 3 x^2 - 4x - 1 = \frac{1 - 3}{1}\]

\[ \Rightarrow 3 x^2 - 4x - 1 + 2 = 0\]

\[ \Rightarrow 3 x^2 - 4x + 1 = 0\]

\[ \Rightarrow 3 x^2 - 3x - x + 1 = 0\]

\[ \Rightarrow \left( 3x - 1 \right)\left( x - 1 \right) = 0\]

\[ \Rightarrow x = \frac{1}{3}, 1\]

Thus, 

\[c = \frac{1}{3} \in \left( 0, 1 \right)\] such that \[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\] .Hence, Lagrange's theorem is verified.