Why does a distance and its square reach their minimum at the same point?

Question

There is a question in my calculus textbook that asks to find a point on the parabola $y^2 = 2x$ that is closest to point $(1,4)$. They want us to first use the distance formula, but then proceeded to square it leaving us with $d^2 = (y^2/2 - 1)^2 + (y-4)^2$. I am following the math up to this point, where I get lost is when the textbook states *"You should convince yourself that the minimum of $d^2$ occurs at the same point as the minimum of $d$, but $d^2$ is easier to work with"*. I don't understand how that can be true at all, $d$ is a distance between a fixed point and another point on the graph $y^2 = 2x$. $d^2$ must be that distance scaled by itself, how can the minimum be the same then? I've worked out the math for $d$ and $d^2$ and I do get the same point but I don't understand why.

Answer 1

The ultimate reason that d and d² achieve the maximum or minimum simultaneously is because the function f(x)=x² is increasing when x is non-negative. If max(d) = d(x₀), this is equivalent to d(x₀) ≥ d(x) for any x in the domain of d(x) (here x is not necessarily a scalar). But then d²(x₀) ≥ d²(x) for any x as well. In other words, max(d²)=d²(x₀).