Answer
$\pm 2$
Work Step by Step
We have: $y=\dfrac{x}{2}+\dfrac{2}{x}$
We differentiate both sides with respect to $x$.
$y^{\prime}=\dfrac{d}{dx} [\dfrac{x}{2}+\dfrac{2}{x}] \\=\dfrac{1}{2}-\dfrac{2}{x^2}$
The tangent line to the graph will be horizontal when $ \dfrac{dy}{dx} =0$
Therefore, $\dfrac{1}{2}-\dfrac{2}{x^2}=0 \implies x=\pm 2$
