The best constant approximating x²

Computing $E(c)$. Since $x^2$ covers exactly $[0,1]$, $E(c)=\max_{t\in[0,1]}|t-c|$. The function $t\mapsto|t-c|$ is convex, so its maximum on $[0,1]$ is at an endpoint:
$$E(c)=\max\bigl(|c|,\,|1-c|\bigr).$$

Minimising. For $c\le\frac12$, $E(c)=1-c$ (it is decreasing); for $c\ge\frac12$, $E(c)=c$ (increasing). So the minimum is at $c=\frac12$ with
$$E\left(\tfrac12\right)=\frac12.$$

Interpretation. $x^2-\frac12$ takes the values $+\frac12$ (at $x=-1$), $-\frac12$ (at $x=0$), $+\frac12$ (at $x=1$): the error reaches its extremal value three times, with alternating signs.

Remark. This minimax problem is the first step towards Chebyshev polynomials: $x^2-\frac12=\frac12T_2(x)$ with $T_2(x)=2x^2-1$. Contrast with the *least squares* approximation, which would minimise $\int_{-1}^1(x^2-c)^2dx$ and give $c=\frac13$ (the mean of $x^2$ on $[-1,1]$).