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) = x² − 2x + 4 on [1, 5] ?

Answer 1

We have,

\[f\left( x \right) = x^2 - 2x + 4\]

Since a polynomial function is everywhere continuous and differentiable.
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 1, 5 \right]\] and differentiable on \[\left( 1, 5 \right)\] .

Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number

\[c \in \left( 1, 5 \right)\] such that \[f'\left( c \right) = \frac{f\left( 5 \right) - f\left( 1 \right)}{5 - 1} = \frac{f\left( 5 \right) - f\left( 1 \right)}{4}\]

Now, 

\[f\left( x \right) = x^2 - 2x + 4\]

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

\[f\left( 5 \right) = 25 - 10 + 4 = 19\] ,

\[f\left( 1 \right) = 1 - 2 + 4 = 3\]

∴  \[f'\left( x \right) = \frac{f\left( 5 \right) - f\left( 1 \right)}{4}\]

\[\Rightarrow 2x - 2 = \frac{19 - 3}{4}\]

\[ \Rightarrow 2x - 2 - 4 = 0\]

\[ \Rightarrow x = \frac{6}{2} = 3\]

Thus, \[c = 3 \in \left( 1, 5 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 5 \right) - f\left( 1 \right)}{5 - 1}\] .

Hence, Lagrange's theorem is verified.