Function f that has infinitely many superstable points

Question

Suppose $f: R \rightarrow R$. Point $x^*$ is called superstable when it's fixed $(f(x^*)=x^*)$ and $f'(x)\vert_x* = 0$

Is this possible to find $f$ that has an infinite number of supestable points?

Definitely, it's possible to find some functions, that have infinitely many fixed points (e.g. $x$ itself, $x\sin x, x\cos x$...). But I can't figure anything for supestability case.

Answer 1

From a geometric point of view, you are asking for a function whose graph intersects (infinitely many times) the graph of $g(x)=x$ and the tangent in the intersection points has slope 0: it's not difficult to figure out a lot of examples.

There are also function with infinite "$\infty$-superstable" points: $f(x)=\lfloor x\rfloor+\frac{1}{2}$ has infinite stable points in which all the derivatives are $0$.

Answer 2

Of course. There are basically two constraints on $f$. The first constraint require that the graph of $f$ intersect with the graph of $x$. Then at those intersections, let the slop of tangent line on $f$ be 0.