Find the point with positive \(x\) value on the parabola \(y=x^2\) that is nearest to the point \((0,1)\).

The distance, D, between the point \((0,1)\) and a point on the parabola \((x,x^2)\) is given by

\(D(x)=\sqrt {(x-0)^2+(x^2-1)^2}\). We have to find a global minimum of D on its domain \([0,\infty )\).

We have to find the critical points of \(D\).
\begin{align*} x &= \answer {1/\sqrt {2}}\\ y &= \answer {1/2} \end{align*}