I need to restrict the degree of freedom of the coefficients of a polynomial, so the function is always strictly monotonic in the domain $x\in\left[0, 1\right]$ and $y\in\left[0, 1\right]$. The polynomial also has to go through the points $(0, 0)$ and $(1, 1)$.
The general formula for a polynomial of degree 3 is
$$a_3\cdot x^3 + a_2\cdot x^2 + a_1\cdot x + a_0\quad .$$
The constraints $f(0) = 0$ and $f(1) = 1$ give
$$a_0 = 0$$
$$a_3 = 1 - a_2 - a_1$$
yielding the fitted polynomial
$$(1 - a_2 - a_1)\cdot x^3 + a_2\cdot x^2 + a_1\cdot x\quad .$$
The number of parameters is already reduced to 2, but I also need the function to be strictly monotonic, so $f'(x) \geq 0$. The additional constraints $f'(0) \geq 0$ and $f'(1) \geq 0$ give
$$a_1 \geq 0$$
$$a_2 \leq 3 - 2\cdot a_1$$
but this does not imply $f'(x) \geq 0$ in general. Is there a simple way to achieve what I want, even for the general case, where the degree of the polynomial is not restricted to 3 and can be any integer number?
As a result, I want to choose the coefficients of the polynomial according to some constraints and the function is always strictly monotonic, includes the points $(0, 0)$ and $(1, 1)$ and is inside the unit square (see curves).
EDIT: I performed a Monte Carlo simulation to determine the constraints graphically (see monte carlo). The black lines correspond to
$$a_1 \geq 0$$
$$a_2 \leq 3 - 2\cdot a_1\quad .$$
The rest looks elliptic to me. All dots inside the yellow area give monotonic increasing functions (see array of curves).
EDIT2: The accepted answer is correct. See the visual proof.