Question

f(x) =\[\frac{x}{2} + \frac{2}{x} , x > 0\] .

Answer 1

\[\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{x}{2} + \frac{2}{x}\]

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

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

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

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

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

\[ \Rightarrow x^2 = \pm 2\]

Since x > 0,  f '(x) changes from negative to positive when x increases through 2. So, x = 2 is a point of local minima.

The local minimum value of  f (x) at x = 2 is given by \[\frac{2}{2} + \frac{2}{2} = 2\]