Number of Curvature Maxima of a 2D Cubic Bezier curve I am trying to prove that a standard cubic Bezier curve can only have at most 2 curvature maxima over $t \in [0,1]$. Assuming that no 3 adjacent control points are colinear, the curvature will either have 2 true local maxima, and the curvature at the endpoints will not be locally maximum, or else the curvature will have 0 or 1 true local maxima, but 2 or 1 endpoints will be a local maximum. 
Intuitively this appears to be true, and experimentally this holds, but I cannot figure out how to go about proving this. Any direction would be of great help

Have you tried writing out the formula for the curvature directly in terms of the polynomials? Note that a standard Bezier curve is really just a way of writing a general cubic curve, so "Bezier" is a red herring here: you're really asking if an arbitrary  cubic curve can have more than two curvature extreme on its whole domain. 
The curvature formula is something like $(\ddot{x}\dot{y} - \ddot{y} \dot{x})/(\dot{x}^2 + \dot{y}^2)^\frac{3}{2}$. The numerator is therefore a quadratic, and the denominator's the $3/2$ power of a quadratic. I'm not certain whether there's anyting useful to drag out of that, but it might be worth writing out in terms of the actual coeffs of $x$ and $y$. 
