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 \[f\left( x \right) = x + \frac{1}{x} \text { on }[1, 3]\] ?

Answer 1

 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, \[f\left( x \right) = \frac{x^2 + 1}{x}\]\[f'\left( x \right) = \frac{x^2 - 1}{x^2}\]\[f\left( 1 \right) = 2\],\[f\left( 3 \right) = \frac{10}{3}\]

∴\[f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]

\[\Rightarrow \frac{x^2 - 1}{x^2} = \frac{4}{6}\]

\[ \Rightarrow \frac{x^2 - 1}{x^2} = \frac{2}{3}\]

\[ \Rightarrow 3 x^2 - 3 = 2 x^2 \]

\[ \Rightarrow x = \pm \sqrt{3}\]

Thus, \[c = \sqrt{3} \in \left( 1, 3 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\] .

Hence, Lagrange's theorem is verified.