The function f(x)=x/2+2/x, x∈ R- 0, :

The correct answer is Option (3) → has a local minimum at $x=2$ and a local maximum at $x= -2$

$f(x)=\frac{x}{2}+\frac{2}{x};x∉R-\{0\}$

$f'(x)=\frac{1}{2}-\frac{2}{x^2}$

for critical points,

$f'(c)=0$

$⇒\frac{2}{x^2}=\frac{1}{2}$

$⇒x^2=4⇒x=±2$

Now,

$f''(x)=\frac{4}{x^3}$

$f''(2)=\frac{2}{3}>0$ ⇒ local minimum at $x=2$

$f''(-2)=-\frac{2}{3}<0$ ⇒ local maximum at $x=-2$