#### Question

For the function f(x) = \[x + \frac{1}{x}\]

x = 1 is a point of maximum

x = \[-\] 1 is a point of minimum

maximum value > minimum value

maximum value < minimum value

#### Solution

\[\text { maximum value < minimum value}\]

\[\text { Given:} f\left( x \right) = x + \frac{1}{x}\]

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

\[\text { For a local maxima or a local minima, we must have} \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]

\[ \Rightarrow x^2 - 1 = 0\]

\[ \Rightarrow x^2 = 1\]

\[ \Rightarrow x = \pm 1\]

\[\text { Now }, \]

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

\[ \Rightarrow f''\left( 1 \right) = \frac{2}{1} = 2 > 0\]

\[\text { So, x = 1 is a local minima.}\]

\[\text { Also }, \]

\[f''\left( - 1 \right) = - 2 < 0\]

\[\text {So, x = - 1 is a localmaxima }.\]

\[\text { The local minimum value is given by }\]

\[f\left( 1 \right) = 2\]

\[\text { The local maximum value is given by }\]

\[f\left( - 1 \right) = - 2\]

\[ \therefore \text { Maximum value < Minimum value }\]