= Solution
The <Ricker population map> has equilibria $N=0$ and $N=K$. Since $f'(N)=e^{r(1-N/K)}(1-rN/K)$, their linearization multipliers are
$$
\boxed{f'(0)=e^r>1,\qquad f'(K)=1-r.}
$$
Thus extinction is unstable and the positive equilibrium is locally asymptotically stable for $0<r<2$, becoming unstable for $r>2$.
To verify an actual <period-doubling bifurcation>, rather than only a multiplier crossing, put $N/K=1+x$ and $r=2+\varepsilon$. The centered map is
$$
x\mapsto-(1+\varepsilon)x+O(\varepsilon x^2)+\frac23x^3+O(\varepsilon x^3+x^4).
$$
Its second iterate minus $x$ is $2\varepsilon x-\tfrac43x^3$ to leading order. After rescaling $x=\sqrt{\varepsilon}y$, the limiting nonzero roots are simple, so the <implicit function theorem> continues them. Besides $x=0$, two nearby points therefore appear for $\varepsilon>0$, with $x_\pm=\pm\sqrt{3\varepsilon/2}+O(\varepsilon)$. The map exchanges these points, and the derivative of its second iterate there is $1-4\varepsilon+O(\varepsilon^{3/2})$, so the emerging two-cycle is attracting. \b[The bifurcation at $r=2$ is supercritical period doubling.]
For nonnegative populations, $f$ increases up to $N=K/r$ and decreases thereafter to zero. Hence every update after the initial time satisfies
$$
\boxed{0\le N_t\le f(K/r)=\frac Kr e^{r-1},\qquad t\ge1.}
$$
The bound is attained by choosing $N_0=K/r$; it is not a bound on an arbitrary initial population.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-2-ricker.png]
{title=Unimodal Ricker update with its maximum at K divided by r and the positive fixed point at K}
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