The function $f(x)=\dfrac{x}{2}+\dfrac{2}{x}$ has a local minimum at

Given the function $f(x) = \dfrac{x}{2} + \dfrac{2}{x}$.

To find the local extrema, calculate the first derivative $f'(x)$:

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

Set $f'(x) = 0$ to find the critical points:

$\dfrac{1}{2} - \dfrac{2}{x^2} = 0 \Rightarrow \dfrac{1}{2} = \dfrac{2}{x^2} \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.

Calculate the second derivative $f''(x)$ to determine the nature of the critical points:

$f''(x) = \dfrac{d}{dx} \left( \dfrac{1}{2} - 2x^{-2} \right) = 0 - 2(-2)x^{-3} = \dfrac{4}{x^3}$.

Evaluate $f''(x)$ at the critical points:

For $x = 2$, $f''(2) = \dfrac{4}{2^3} = \dfrac{4}{8} = \dfrac{1}{2} > 0$. Since $f''(2) > 0$, the function has a local minimum at $x = 2$.

For $x = -2$, $f''(-2) = \dfrac{4}{(-2)^3} = \dfrac{4}{-8} = -\dfrac{1}{2} < 0$. Since $f''(-2) < 0$, the function has a local maximum at $x = -2$.

Answer: $x=2$