How to find a superstable period-$2$ orbit of the logistic map.

Question

Suppose that $G\colon R \to R$ such that $G(x)=rx(1-x)$.

I need to find the value for $r$ at which the super stable period-$1$ and period-$2$ points exist. I think I know how to get the super stable period-$1$ point but I'm having trouble finding how to do the same for period-$2$.

Answer 1

An orbit is super-stable if and only if there is a critical point in that orbit. Now, $G_r(x)=rx(1-x)$ has exactly one critical point, namely $1/2$, which is independent of $r$. So, to find an $r$ parameter such that $G_r$ has a super-stable orbit of period 2, you simply solve the equation $G_r(G_r(1/2))=1/2$ for $r$.

Answer 2

Hint: for the period-2 points, look at the second iterate of the map: $$ G^2(x) = - r^2x(x-1)(1-rx+rx^2).$$