Minimizing distance from a point to a curve

To avoid differentiating a square root, minimize $f(x) = D(x)^2 = (x-3)^2 + x$ instead, since $D(x)\ge 0$ and squaring preserves the location of minima. Expanding gives $f(x) = x^2 - 6x + 9 + x = x^2 - 5x + 9$. Differentiating, $f'(x) = 2x - 5$, and setting $f'(x) = 0$ gives $x = 5/2$, which lies in the domain $x \ge 0$.

Since $f''(x) = 2 > 0$ for all $x$, the function $f$ is concave up everywhere, so $x = 5/2$ is a global minimum on $x\ge 0$. Plugging back in, $f(5/2) = (5/2)^2 - 5(5/2) + 9 = 25/4 - 25/2 + 9 = 11/4$. Therefore the minimum distance is $D(5/2) = \sqrt{11/4} = \dfrac{\sqrt{11}}{2}$.