Number of Equilibrium Points in a system Suppose we know that the number of asymptotically stable equilibrium points is two for the system
\begin{equation}
\dot{x} = f(x) 
\end{equation}
where $x \in R^1$ and $f$ is continuous. Is the statement "the number of all equilibria must be at least three" true? If yes, why?
 A: This is true for scalar systems because if $x_e$ is asymptotically stable equilibrium point then around $x_e$ the system $\dot{x} = f(x)$ behaves like $\dot{x} = a x$ where $a = \frac{d}{dx}f(x)|_{x=x_e} < 0$. Therefore $f(x) > 0$ when "close to $x_e$ from the left" and $f(x) < 0$ when "close to $x_e$ from the right". So if you have two asymptically stable equilibrium points in a scalar system it means that $f(x)$ has to cross the x-axis "from positive to negative direction" twice. Because $f$ is continuous this is only possible if there is at least one more equilibrium.
If $f$ is not continuous, this is not true anymore. For example:
$$
\dot{x}=f(x)=\begin{cases}
-x-1 & x\leq 0\\
-x+1 & x > 0
\end{cases}
$$
Of course the system has only two equilibrium points at $x=-1$ and $x=1$ as you can see in the plot. Both are locally asymptotically stable because locally they are governed by linear stable dynamics. But this is only possible because $f$ has a discontinuity. When $f$ is continuous it has to "properly" cross the x-axis at some point.

