Explanation

(d)
Since, $f(x) = \frac{x}{2} + \frac{2}{x}$
$\therefore\ \ \ \ \ \ f^{'}(x) = \frac{1}{2} - \frac{2}{x^{2}}$
For maxima or minima, put $f^{'}(x) = 0$
$\therefore\ \ \ \ \ \frac{1}{2} - \frac{2}{x^{2}} = 0\ \ \ \ \Rightarrow \ \ \ \ \ x = \pm 2$
Now, $f^{''}(x) = \frac{4}{x^{3}}$
$\Rightarrow \ \ \ \ \ \ f^{''}(2) = \frac{4}{8} = \frac{1}{2} > 0$, minima
And $f^{''}( - 2) = - \frac{4}{8} = - \frac{1}{2} < 0$, maxima
Hence, $f(x)$ has local minimum at $x = 2$