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 theore f(x) = (x − 1)(x − 2)(x − 3) on [0, 4] ?

Answer 1

 We have,

\[f\left( x \right) = \left( x - 1 \right)\left( x - 2 \right)\left( x - 3 \right)\] which can be rewritten as

\[f\left( x \right) = x^3 - 6 x^2 + 11x - 6\]

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

\[f\left( x \right)\] is continuous on \[\left[ 0, 4 \right]\] and differentiable on \[\left( 0, 4 \right)\] .

Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ \[c \in \left( 0, 4 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0} = \frac{f\left( 4 \right) - f\left( 0 \right)}{4}\]

Now, \[f\left( x \right) = x^3 - 6 x^2 + 11x - 6\]

\[\Rightarrow f'\left( x \right) = 3 x^2 - 12x + 11\] ,\[f\left( 0 \right) = - 6\] ,\[f\left( 4 \right) = 64 - 96 + 44 - 6 = 6\]

∴ \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\]

\[\Rightarrow 3 x^2 - 12x + 11 = \frac{6 + 6}{4}\]

\[ \Rightarrow 3 x^2 - 12x + 8 = 0\]

\[ \Rightarrow x = 2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}\]

Thus,\[c = 2 \pm \frac{2}{\sqrt{3}} \in \left( 0, 4 \right)\] such that\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\] .

Hence, Lagrange's theorem is verified.