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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) = 2x2 − 3x + 1 on [1, 3] ?
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Solution
We have,
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 1, 3 \right]\] and differentiable on \[\left( 1, 3 \right)\].
So, there must exist at least one real number \[c \in \left( 1, 3 \right)\] such that\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1} = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]
\[ \Rightarrow 4x - 3 - 5 = 0\]
\[ \Rightarrow x = 2\]
Thus,
\[c = 2 \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\]
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