Mathematical biology problem involving discrete single-species models The question is: Consider the nonlinear equation for population growth $$N_{t+1}=\frac{rN_t}{1+aN_t}$$ where $r>0$ is the basic reproduction rate, and $a>0$ is a constant.
Part (a) 
Does this equation exhibit over,under or exact-compensation?
My attempt is as follows: The map can be written as $$N_{t+1}=F(N_t)N_t=(\frac{r}{1+aN_t})N_t$$ so $F(N)=\frac{r}{1+aN}$ and when $N>>1$ this can be approximated by $F(N)\approx\frac{r}{aN}$ so $F(N)\to \frac{r}{aN}$ as $N\to \infty$ which is of the standard form for compensation with $b=1$ hence, this map represents exact-compensation.
Part (b) Use the change of variable $x_t=aN_t$ to eliminate a from the model.
My attempt is as follows: First note the change of variable also gives $x_{t+1}=aN_{t+1}$ so $$N_{t+1}=\frac{rN_t}{1+aN_t}$$ multiplying both sides by a and making the required substitution i get $$x_{t+1}=\frac{rx_t}{1+x_t}$$.
Part (c) Determine the steady-states of the re normalised equation,noting any restriction on physicality. My attempt is as follows: Steady-states occur when $$x^*=\frac{rx^*}{1+x^*}$$, clearly the trivial steady-state $x^*=0$ satisfies this, alternately, diving through by $x^*$ gives $$1=\frac{r}{1+x^*}$$ rearranging we get that $x^*=r-1$. This produces a positive (physical) steady state when $r>0$.
Part (d) Draw the cobweb plots for each case that you have identified. What does this tell you about the stability in each case. 
This is the part i'm stuck on, as i don't know what to do as i've only identified two steady states $x^*=0$ which is trivial and $x^*=r-1$ which produces a positive (physical) steady state when $r>1$, i also dont know how to draw the function i know i would draw the line $y=x$ which is $N_{t+1}=N_{t}$ but how would i draw the other curve?.
Part (e) using linear analysis, confirm the stability seen in the cobweb plots of each steady state. i think i would need to differentiate $F(N)$ and evaluate it at $x^*=r-1$ is this correct? and how would i go about doing this?
Part (f) is to comment on the implications of this model for the long-term survival of the population. ( is this refering to if the population will die-out or continue to grow)?.
Any help would be highly appreciated, Thanks for taking the time for reading through this post. 
 A: Note that
$$T(z):={rz\over1+az}$$
is a Moebius transformation with fixed points $z=0$ and $z={r-1\over a}$. In terms of the new complex coordinate
$$w:={z\over z-{r-1\over a}},\qquad{\rm resp.,}\qquad  z={r-1\over a}\>{w\over w-1}\>,$$  this transformation  appears as
$$\hat T(w)=r\>w\ ,$$
hence $\bigl(\hat T\bigr)^{\circ n}(w)=r^n w$. This allows to obtain an explicit representation of $T^{\circ n}(z)$. The computation gives
$$T^{\circ n}(z)={(r-1)r^n z\over(r-1)+az(r^n-1)}\qquad(n\geq0)\ .$$
In the notations of the question this means that
$$N_n={(r-1)r^n N_0\over(r-1)+a(r^n-1)N_0}\qquad(n\geq0)\ .\tag{1}$$
If you are not familiar with Moebius transformations you can easily prove $(1)$ by induction directly. But we needed the Moebius transformations in order to arrive at $(1)$.
A: I do not know if the following could help you.
Considering
$$N_{t+1}=\frac{r\,N_t}{1+a\,N_t}\implies \frac 1 {N_{t+1}}=\frac 1r \frac 1{N_t}+\frac ar$$ which makes $$\frac 1 {N_{t+1}}=a\,\frac{1-r^{-(t+1)}}{r-1}+c\, r^{-t}$$ $c$ being fixed by some condition.
Edit
Assuming that $N_0=A$ and computing the derivative, we find that $$\frac{dN}{dt}=-\frac{A (r-1)\, r^{-t}  (a
   A-r+1)\,\log \left({r}\right)}{\left(a A-r^{-t} (a A-r+1)\right)^2}$$
