= Solution
For $f(x)=rx(1-x)$, the <fixed points> are $x=0$ and $x_*=1-1/r$ 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 $0\leq r<1$ 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 $r^2x^2-r(r+1)x+(r+1)=0$. Its distinct roots are
$$
\boxed{x_\pm=\frac{r+1\pm\sqrt{(r-3)(r+1)}}{2r},\qquad r>3.}
$$
They map to each other, giving a genuine <two-cycle> only beyond $r_0=3$. The cycle multiplier is
$$
M=f'(x_+)f'(x_-)=4+2r-r^2.
$$
Solving $|M|<1$ in $r>3$ gives
$$
\boxed{3<r<1+\sqrt6.}
$$
At $r=3$, the two roots coalesce with $x_*=2/3$, 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 $r=1+\sqrt5$, 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.
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