Solving Bezier cubic derivative for t Getting the derivative of the cubic Bezier curve:

$P(t)=P_0(1-t)^3+P_13t(1-t)^2+P_23t^2(1-t)+P_3t^3$

Produces the following:

$P'(t)=3(-P_0-2P_1)+6t(P_0+P_1+P_2)+3t^2(P_3-P_0)$

Assuming P'(t)=0, is it possible to solve for t?
 A: Let’s assume that your points are 2D, say $P_0 = (x_0,y_0)$, $P_1= (x_1,y_1)$, and so on.
Then the equation $P’(t)=0$ is actually two equations that need to be satisfied:
\begin{align}
3(−x_0−2x_1)+6(x_0+x_1+x_2)+3^2(x_3−x_0) &=0 \\
3(−y_0−2y_1)+6(y_0+y_1+y_2)+3^2(y_3−y_0) &=0 \\
\end{align}
Each of these is a quadratic, which you can solve using the high-school formula, to get values of $t$. Then you can check if there’s a value of $t$ that satisfies both equations.
For a general cubic, it’s highly unlikely that you will find a point where $P’(t)=0$. The equation of $P’$ represents a parabola, and $P’(t)=0$ means that the parabola passes through the origin, which probably won’t happen, in general.
If your points are 3D, then three equations will have to be satisfied simultaneously to get $P’(t)=0$, so this is even less likely to happen.
Maybe you’re thinking that $P’(t)=0$ corresponds to places where the curve tangent is horizontal. That’s not true. The tangent is horizontal where $y’(t)=0$, and you can find such places by solving the second of the two equations shown above:
$$
3(−y_0−2y_1)+6(y_0+y_1+y_2)+3^2(y_3−y_0) =0
$$
This equation won’t always have (real) solutions, but it often will.
