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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 theore f(x) = tan−1 x on [0, 1] ?
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Solution
We have,
\[f\left( x \right) = \tan^{- 1} x\]
Clearly. \[f\left( x \right)\] is continuous on \[\left[ 0, 1 \right]\] and derivable on \[\left( 0, 1 \right)\]
Consequently, there exists some \[c \in \left( - 3, 4 \right)\] such that \[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0} = \frac{f\left( 1 \right) - f\left( 0 \right)}{1}\]
\[\Rightarrow \frac{1}{1 + x^2} = \frac{\pi}{4} - 0\]
\[ \Rightarrow \left( \frac{4}{\pi} - 1 \right) = x^2 \]
\[ \Rightarrow x = \pm \sqrt{\frac{4 - \pi}{\pi}}\]
Thus, \[c = \sqrt{\frac{4 - \pi}{\pi}} \in \left( 0, 1 \right)\] such that
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