I'm trying to find out for what values of $x_0$ and $x_1$, $\underset{x\in I}{\max}|(x-x_0)(x-x_1)|$ becomes minimum for $I=[-1,1]$. Note that we only look at the function diagram in $[-1, 1]$. I looked at the diagram and guessed value of function at $-1$ and $1$ and vertex of the parabola should be equal. So $-x_0=x_1=\frac{1}{\sqrt{2}}$. Is there a proof or string reasoning to this, if the answer is correct? If it's not, what's the correct answer?
Any help is so much appreciated!
Assuming you mean:
$\operatorname{argmin}_{x_0, x_1} \max_{x \in (-1,1)} \left|(x - x_0)(x - x_1)\right|$.
Your reasoning often does work, and while it isn't directly valid for this problem, it does give the correct answer.
Usually with smooth conditions if you have different possible maxima and trying to find the point that minimizes the largest of them, they indeed must be equal -- if they weren't you could probably lower the biggest possible maximum by raising one of the currently irrelevant lower ones. This reasoning doesn't directly hold for this problem, because the interval constraint makes it not so smooth. The third value might not be relevant -- if that value is from outside the interval, we don't need to make it equal.
But setting the boundary values must always be accounted for, which puts the possible maximum values inside the interval, which is why it does actually work for this problem.
The possible values of $x$ that maximize the inner expression are of course the boundaries $x = -1$ and $x = 1$, and where the derivative $2x - x_0 - x_1$ is zero ($x* = (x_0 + x_1)/2$). This local maximum will need to be evaluated so long as that $x$ is in the interval.
At the boundaries the function attains values $y_- = \left|(1+x_0)(1+x_1)\right|$ and $y_+ = \left|(1-x_0)(1-x_1)\right|$ for $x = -1$ and $x = 1$ respectively, while the possible zero derivative value at $x*$ is $y* = (x_1 - x_0)^2/4$.
Due to symmetry, we need only consider $x_0 \leq x_1$. Also due to symmetry, for $y_0 = y_1$, $x_0$ and $x_1$ must interchange their roles with respect to the boundary, and thus be symmetric around $0$. This reduces us to a 1-dimensional problem.
Rewriting in terms of the single variable $z = x_1 = - x_0$ $y_- = y_+ = z(1-z)$ and $y* = (x_1 - x_0)^2 / 4 = z^2$.
Setting these equal gives $z^2 = 1 - z^2$; $z^2 = 1/2$; $z = \sqrt{1/2}$, and x_0 = -\sqrt{1/2}$ and $x_1 = \sqrt{1/2}$.
It's still instructive to "case it out" without this knowledge, but it's nine different cases to worry about, and I'm lazy.
However, here's a plot of the function at the optimal x0 and x1: