For the function f (x) = x + 1/x ∈ [1, 3], the value of c for the Lagrange's mean value theorem is

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$\sqrt$(3) We have f\left( x \right) = x + 1/x = 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 existsc \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) = x^2 + 1/x f'\left( x \right) = x^2 - 1/x^2,f\left( 1 \right) = 2 ,f\left( 3 \right) = $\frac{10}{3}$ $\therefore$ f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2} \Rightarrow x^2 - 1/x^2 = $\frac{4}{6}$ \Rightarrow 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} .

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