Given the ordinary differential equation
$$\frac{dx}{dt} = x^3 - 2x^2 + (1- \mu)x,$$
where $\mu$ is a parameter, I want to find an expression for the fixed point(s) $x^*$ as a function of $\mu$; and to determine the stability of fixed points.
In my working out I find that the fixed points are $x^* = 0$ and also $x^* = 1±\sqrt{\mu}$. So when I try to find the stability of the fixed points, my lecturer says to consider the three regions:
$$\mu < 0 ;$$ $$0 < \mu < 1;$$ $$\mu > 1.$$
My question is: I do not get why he considers these regions. Why not consider $\mu < 0$, $\mu = 0$ and $\mu >0$ for example?
I am fine evaluating the stability, but I do not get understand how to determine the intervals - any clarification on this would be welcome.
This is the sketch of the curve for reference: https://www.desmos.com/calculator/t1y2i9y7lr
Edit - I understand that I will need to calculate the eigenvalues - can anyone please tell me how to do this.
As you found, there will be equilibrium points at $x=0$ and $x = 1 \pm \sqrt{\mu}$.
If $\mu < 0$, then $\sqrt{\mu}$ is undefined (over the reals), so there is only one equilibrium point.
If $\mu = 0$, then $1 \pm \sqrt{\mu} = 1 \pm 0 = 1$, so there are only two equilibrium points.
If $ 0 < \mu < 1$, then $1 + \sqrt{\mu}>1 - \sqrt{\mu}>0$ are all distinct values and we have three equilibrium points.
Therefore, we have three regions with distinctly different dynamics: $\mu < 0$, $0 < \mu < 1$, and $\mu > 1$, with bifurcations occuring at $\mu =0$ and $\mu =1$. See the following plot which gives the equilibrium values as a function of $\mu$: