I have a polynomial $P(x) = Ax^3 + Bx^2 + Cx + D$, with $P(0) = 1$ and $P(1) = 0$. This means that $D=1$ and $A + B + C + D = 0$. Suppose that $P'(0) = d_0 \leq 0$ and $P'(1) = d_1 \leq 0$. Depending on $d_0$ and $d_1$, $P(x)$ may or may not be monotonic. Is there a way to clamp $d_0$ and $d_1$ such that $P(x)$ is monotonic on $x\in]0, 1[$, with minimal change in $d_0$ and $d_1$?
It is easy to find any local extrema for $P(x)$:
Assuming $A = 0$ and $B = 0$, that is $C = -D$, no one exists. Thus, $P(x)$ is always monotonic in this case. Indeed, it is a straight line.
Assuming $A = 0$ and $B \neq 0$, the derivative has one zero at $x = -\frac{C}{2B}$, which gives two options $-\frac{C}{2B} < 0$ ($B$ and $C$ must have the same sign), or $C < -2B$.
Assuming $A \neq 0$, we have a complete second degree polynomial. Then $\frac{-2B - \sqrt{4B^2 - 12AC}}{6A} < 0 \land\frac{-2B + \sqrt{4B^2 - 12AC}}{6A} > 1$, or alternatively $4B^2 - 12AC < 0$.
My initial idea was to try to find an explicit forumula for all allowed ranges and then substitute a clamped value back into the original equation. I am not really sure how to do it and where to go from there. At least, it is easy to verify that any given combination of $d_0$ and $d_1$ is within the constraints.