 Interval subdivision

Interval subdivision covers $X$ by smaller closed intervals $X_j$, computes a valid interval enclosure $F(X_j)$ on each, and takes their union. Every allowed input belongs to a piece, so its output belongs to that union. If one closed interval is wanted, take the smallest one containing the union; gaps between separate output pieces can then be lost. Splitting narrows the freedom given to repeated inputs, while increasing the number of evaluations.

For $f(x)=x(1-x)$, split $[0,1]$ at $1/2$. The natural interval extensions areTheir union is $[0,1/2]$. Four equal pieces sharpen the upper bound again:The union is now $[0,3/8]=[0,0.375]$, still wider than the actual image of a function $[0,1/4]$. These are exact rational number bounds: the excess width comes from the interval dependency problem, rather than roundoff error.