 Interval dependency problem

The interval dependency problem is the loss of relationships between repeated occurrences of the same input. For $X=[0,1]$, interval subtraction gives $X-X=[-1,1]$: it permits $u-v$ with independent choices $u,v\in X$. But the function $x\mapsto x-x$ has actual image of a function $\{0\}=[0,0]$. Giving both operands the name $X$ does not tell an interval arithmetic calculation to enforce $u=v$.

For the polynomial $f(x)=x(1-x)$ on $X=[0,1]$, the natural interval extension isThis allows the impossible combination $x=1$ in the first factor and $x=0$ in the second, producing the spurious upper bound $1$. In the actual function, both occurrences use the same $x$. Completing the square gives $x(1-x)=1/4-(x-1/2{)}^2$, which lies between $0$ and $1/4$ on $[0,1]$; every value between these bounds occurs, by continuity and the intermediate value theorem. The maximum $1/4$ occurs at $x=1/2$, while either endpoint gives the minimum $0$.