I need to find an equation $\dot x = f(x)$ (where $\dot x$ denotes $\frac{dx}{dt}$) where there are precisely three fixed points, and all of them are stable, or explain why such a situation is not possible.
I have seen a solution to this problem where it says that the situation is not possible because "the mean value theorem guarantees that between any two fixed points of the same type (stable, unstable), there must be a fixed point of the other type". What I don't understand is why that should be true, or how you could go about proving that.
My attempt thus far is: suppose $x=a$, $x=b$ are fixed points (I.e., $f(a)=f(b)=0$), then by the mean value theorem, there exists a point $x=c$ such that $c\in (a,b)$ and $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{0}{b-a}=0$, but all this tells us is that the function goes from increasing to decreasing (or vice versa) on this interval. How does it tell us that if, say, $a$ and $b$ are both stable, that there cannot be another fixed point between them that is also stable?
Thank you.
For an equilibrium point $ x=a$ to be stable, we have to have the graph of $f(x)$ changes sign from positive to negative.
If you also have another equilibrium point, $x=b$ which is also stable, $f(x)$ needs to change sign from positive to negative at $x=b.$
Therefore the graph of $f(x)$ is going from negative to positive somewhere between $a$ and $b.$
Thus there exists an unstable equilibrium point $x=c$ between $ a$ and $ b.$
Hence if there are only three equilibrium points, all three of them can not be stable.