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) = x(x −1) on [1, 2] ?

Answer 1

We have,

\[f\left( x \right) = x\left( x - 1 \right)\] which can be rewritten as \[f\left( x \right) = x^2 - x\]

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

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

Now, 

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

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

\[f\left( 2 \right) = 2\] ,

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

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

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

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

\[ \Rightarrow 2x = 3\]

\[ \Rightarrow x = \frac{3}{2}\]

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

\[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}\] .

Hence, Lagrange's theorem is verified.