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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) = x2 − 3x + 2 on [−1, 2] ?
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Solution
We have,
\[f\left( x \right) = x^2 - 3x + 2\]
Since a polynomial function is everywhere continuous and differentiable.
Therefore,
So, there must exist at least one real number \[c \in \left( - 1, 2 \right)\] such that \[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{2 + 1} = \frac{f\left( 2 \right) - f\left( - 1 \right)}{3}\]
\[\Rightarrow 2x - 3 = - 2\]
\[ \Rightarrow 2x - 1 = 0\]
\[ \Rightarrow x = \frac{1}{2}\]
Thus, \[c = \frac{1}{2} \in \left( - 1, 2 \right)\] such that \[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{2 - \left( - 1 \right)}\] .
Hence, Lagrange's theorem is verified.
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