Question

`f(x) = x/2+2/x, x>0 `.

Answer 1

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

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

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

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

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

\[ \Rightarrow x^2 = 4\]

\[ \Rightarrow x = 2\text { and } - 2\]

\[\text { Thus, x = 2 and x = - 2 are the possible points of local maxima or a local minima } . \]

\[\text { Since }x > 0, x = 2\]

\[\text { Now,} \]

\[f''\left( x \right) = \frac{4}{x^3}\]

\[\text { At }x = 2: \]

\[ f''\left( 2 \right) = \frac{4}{\left( 2 \right)^3} = \frac{1}{2} > 0\]

\[\text { So, x = 2 is the point of local minimum } . \]

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

\[f\left( 2 \right) = \frac{x}{2} + \frac{2}{x} = 1 + 1 = 2\]