How to determine the stability of a fixed point if the derivative at the point is equal to one? ($\,\left\lvert\, f'(x^{\ast})\right\rvert = 1$)

Question

Context:

I am learning about 1-dimensional maps: For instance the logistic model of population growth.

Suppose I have the map $x_{n+1} = f(x_n)$. The point $x^{\ast}$ is called a fixed point of the map if $x^{\ast}=f(x^{\ast})$.

The fixed point is stable if $\,\left\lvert\, f'(x^{\ast})\right\rvert < 1$ and unstable if $\,\left\lvert\, f'(x^{\ast})\right\rvert > 1$.

Question:

I am trying to determine the stability of the fixed points for a particular instance of the map (given above):

$x_{n+1} = 2(x_n)^2+11(x_n)+12$

This map has two fixed points:

$x=-3,x=-2$

Unfortunately the derivative of $f(x_n)=2 (x_n)^2+11 (x_n)+12$ at the fixed point $x=-3$ is exactly equal to 1. (Not greater or lesser than 1.)

$\,\left\lvert\, f'(-3)\right\rvert = 1$

How do I determine the stability of it?

Answer 1

If $x_n = a_n/2 - 3$, then we get the logistic map $a_{n + 1} = a_n (a_n - 1)$. The cobweb diagram shows that $|a_{n + 2}| < |a_n|$ for small enough $a_0$, thus $a = 0$ is stable (in fact, asymptotically stable):