1. Given function

   $f(x)=\frac{x}{2}+\frac{2}{x}, \qquad x\neq 0$

   We need to find where this function has a local minimum.

2. Find the first derivative

   $f'(x)=\frac{d}{dx}\left(\frac{x}{2}+\frac{2}{x}\right)=\frac{1}{2}-\frac{2}{x^2}$

3. Find critical points

   Set $f'(x)=0$:

   $\frac{1}{2}-\frac{2}{x^2}=0$

   $\frac{1}{2}=\frac{2}{x^2}$

   $x^2=4$

   $x=\pm 2$

4. Use the second derivative test

   $f''(x)=\frac{d}{dx}\left(\frac{1}{2}-\frac{2}{x^2}\right)=\frac{4}{x^3}$

   • At $x=2$: $f''(2)=\frac{4}{8}=\frac{1}{2}>0$ So, $x=2$ is a local minimum.

   • At $x=-2$: $f''(-2)=\frac{4}{-8}=-\frac{1}{2}<0$ So, $x=-2$ is a local maximum.

5. Check the options

   • A: $x=2$ ✅ local minimum
   • B: $x=-2$ ❌ local maximum
   • C: $x=0$ ❌ not in domain
   • D: $x=1$ ❌ not a critical point

Therefore, the function has a local minimum at:

$\boxed{x=2}$