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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] ?
Solution
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.
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