For the function f(x) = x + 1/x

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\text { maximum value < minimum value} \text { Given:} f\left( x \right) = x + 1/x \Rightarrow f'\left( x \right) = 1 - 1/x^2 \text { For a local maxima or a local minima, we must have} f'\left( x \right) = 0 \Rightarrow 1 - 1/x^2 = 0 \Rightarrow x^2 - 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = $\pm$ 1 \text { Now }, f''\left( x \right) = 2/x^3 \Rightarrow f''\left( 1 \right) = $\frac{2}{1}$ = 2 > 0 \text { So, x = 1 is a local minima.} \text { Also }, f''\left( - 1 \right) = - 2 < 0 \text {So, x = - 1 is a localmaxima }. \text { The local minimum value is given by } f\left( 1 \right) = 2 \text { The local maximum value is given by } f\left( - 1 \right) = - 2 \therefore \text { Maximum value < Minimum value }

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