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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) = x(x −1) on [1, 2] ?
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Solution
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,
\[f\left( x \right) = x^2 - x\]
\[\Rightarrow 2x - 1 = \frac{2 - 0}{2 - 1}\]
\[ \Rightarrow 2x - 1 - 2 = 0\]
\[ \Rightarrow 2x = 3\]
\[ \Rightarrow x = \frac{3}{2}\]
Thus, \[c = \frac{3}{2} \in \left( 1, 2 \right)\] such that
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