= Solution
The <fixed points> satisfy $x[rx(1-x)-1]=0$. There is always $x=0$, and the other branches are
$$
x_\pm(r)=\frac{1\pm\sqrt{1-4/r}}2,\qquad r\ge4.
$$
They coincide at $(r,x)=(4,1/2)$. For $0<x<1$, $F(x)>0$ and its maximum occurs at $x=2/3$, with value $4r/27$. Consequently $\boxed{F((0,1))\subset(0,1)\iff0<r<27/4}$; equality at the upper endpoint fails because the maximum is attained inside the interval.
The <fixed-point multiplier> at zero is zero, so zero is attracting for every $r>0$. On either nonzero branch,
$$
F'(x)=\frac{2-3x}{1-x}.
$$
The lower branch has multiplier greater than one for $r>4$ and is repelling. The upper branch is attracting for $4<r<16/3$, reaches multiplier $-1$ at $x=3/4$, and is repelling for $r>16/3$. At $r=4$ the multiplier is $+1$, and the two branches are created in a <saddle-node bifurcation>. The fixed point there is attracting from one side and repelling from the other: $F(x)-x=-4x(x-1/2)^2$.
The second change is a supercritical <period-doubling bifurcation> at $\boxed{(r,x)=(16/3,3/4)}$. To check its type, expand at that fixed point. The local map has quadratic coefficient $a=-20/3$ and cubic coefficient $b=-16/3$ when the multiplier is $-1$. Its second iterate has cubic coefficient $-2(a^2+b)=-704/9<0$. The multiplier decreases through $-1$ as $r$ increases, so a small attracting two-cycle is born on the $r>16/3$ side. These exhaust the fixed-point <bifurcations>, since their multipliers have no other crossings of $\pm1$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-3-map-bifurcation.png]
{title=Fixed-point branches of the cubic map, with attracting solid curves and repelling dashed curves}
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