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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 − 1 on [2, 3] ?
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Solution
We have
\[f\left( x \right) = x^2 - 1\]
Since a polynomial function is everywhere continuous and differentiable, \[f\left( x \right)\] is continuous on \[\left[ 2, 3 \right]\] and differentiable on \[\left( 2, 3 \right)\].
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number \[c \in \left( 2, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}\]
Now,
\[\Rightarrow 2x = \frac{8 - 3}{1}\]
\[ \Rightarrow x = \frac{5}{2}\]
Thus, \[c = \frac{5}{2} \in \left( 2, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}\].
Hence, Lagrange's theorem is verified.
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