Draw an example where a critical point for the curvature does not correspond to a local maximum/minimum
Does the curve for infinity sign satisfy this? I am having trouble seeing why it's true, if of course it's true! I know that at the intersection point the curvature is not defined. I am not sure about the derivative of the curvature at that point
Thanks
If I read you right, you're looking to sketch a plane curve $C$ for which the curvature has a critical point at some point $p$, but $p$ is not an extremum for the $y$ coordinate along $C$. Not sure what you have in mind for "the curve for infinity" (a specific figure eight on its side?)....
If you take the parabola $y = x^{2}$ with $-1 \leq x \leq 1$ (which has maximum curvature at the origin) and rotate it slightly about the origin, you'll get a curve having a curvature maximum at the origin, but for which the tangent line at the origin is not horizontal. Particularly, the origin is not a local minimum or maximum of height. (If necessary, interpolate at the ends to get a graph defined "everywhere".)