 Solution

For $f(x)=rx(1-x)$, the fixed points are $x=0$ and $x_{\ast }=1-1/r$ when the latter is defined. On the physical interval the nonzero branch exists for $r>1$, meeting zero at $r=1$. Their multipliers are $r$ and $2-r$, respectively. Thus zero is linearly attracting for $0\le r<1$ and repelling for $r>1$; the positive point is linearly attracting for $1<r<3$ and repelling for $r>3$. At $r=1$ and $r=3$ the relevant multiplier has magnitude one, so linearization is inconclusive. At $r=1$, $x_{n+1}=x_n-x_n^2$ attracts small positive points to zero nonlinearly.

Factoring $f^2(x)-x$ and removing its fixed-point factors gives $r^2{x}^2-r(r+1)x+(r+1)=0$. Its distinct roots areThey map to each other, giving a genuine two-cycle only beyond $r_0=3$. The cycle multiplier isSolving $|M|<1$ in $r>3$ givesAt $r=3$, the two roots coalesce with $x_{\ast }=2/3$, where the fixed-point multiplier is $-1$: a supercritical period-doubling bifurcation creates the attracting two-cycle. At the upper endpoint its multiplier reaches $-1$, marking the next loss of linear stability. The two-cycle is superstable at $r=1+\sqrt{5}$, where $M=0$. At exactly $r=3$ the fixed point remains weakly attracting through higher-order terms: for $\eta =x-2/3$, the second iterate is $\eta \mapsto \eta -18{\eta }^3-27{\eta }^4$, showing alternating perturbations decay slowly.