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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 \[f\left( x \right) = x + \frac{1}{x} \text { on }[1, 3]\] ?
Solution
We have,
\[f\left( x \right) = x + \frac{1}{x} = \frac{x^2 + 1}{x}\]
Clearly, \[f\left( x \right)\] is continuous on \[\left[ 1, 3 \right]\] and derivable on \[\left( 1, 3 \right)\]
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some \[c \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1} = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]
Now, \[f\left( x \right) = \frac{x^2 + 1}{x}\]\[f'\left( x \right) = \frac{x^2 - 1}{x^2}\]\[f\left( 1 \right) = 2\],\[f\left( 3 \right) = \frac{10}{3}\]
\[\Rightarrow \frac{x^2 - 1}{x^2} = \frac{4}{6}\]
\[ \Rightarrow \frac{x^2 - 1}{x^2} = \frac{2}{3}\]
\[ \Rightarrow 3 x^2 - 3 = 2 x^2 \]
\[ \Rightarrow x = \pm \sqrt{3}\]
Thus, \[c = \sqrt{3} \in \left( 1, 3 \right)\] such that
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