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) = 2x − x² on [0, 1] ?

Answer 1

We have,

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

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

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

Now, \[f\left( x \right) = 2x - x^2\]

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

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

\[\Rightarrow 2 - 2x = \frac{1 - 0}{1}\]

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

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

Thus, 

\[c = \frac{1}{2} \in \left( 0, 1 \right)\] such that 

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

Hence, Lagrange's theorem is verified.