The function $f(x)=\frac{x}{2}+\frac{2}{x}, x\in R-$ {0}, :

has a local minimum at x=2 and a local maximum at x= -2

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 $x=-2$