Suppose 0 = $x_0$ < $x_1$ < $x_2$ < $x_3$ = 1 and 0 = $y_0$ < $y_1$ < $y_2$ < $y_3$ = 1. There exists a unique cubic polynomial $p(x)$ with $p(x_i) = y_i$. Is $p$ necessarily increasing everywhere on [0, 1]? If not, what conditions on $x_1$, $x_2$, $y_1$ and $y_2$ would imply $p$ is increasing?
The interpolating polynomial depends continuously on its inputs; you could see that from writing out the Lagrange form of the interpolating polynomial. When $x_1 = y_1$ and $x_2 = y_2$, the interpolating polynomial is the identity and the derivative is 1. So for some sufficiently perturbation away from $x_1 = y_1$ and $x_2 = y_2$ the derivative stays positive. It would be nice to say something more quantitative about how close the $x$'s need to be to the corresponding $y$'s.
This should not be the case: Take a cubic $p(x)$ with bump and fit it into the unit square and find a counter-example after the fact.
Example: $p(x) = 3x^3-x^2-x$ and $\underline{x} = (0,0.8,0.9,1), \underline{y} = (0, 0.096, 0.477, 1)$.
Some thoughts about monotonicity: For simplicity fix the $x_i$. By Lagrange interpolation we get $$p'(x) = a (y)x^2 +b (y)x + c (y)$$ where $a,b,c $ are linear in $y_i $. We get the quadratic condition in $y $ for $p'$ to have no change of signs $$ b^2 - 4ac < 0$$ Together with $y_1 = 0, y _4=1$ this forces $p $ to be increasing (even outside [0,1]).
I haven't computed examples so I'm not sure how much potential this has.