Stability of autonomous system $x'=f(x)$ In the book I'm using I found the following statement
Consider the autonomous system
$$x'=f(x)$$ with the the equilibrium point $\bar x$.
If there exists a $\delta >0$ such that $f(x)<0$ for $x \in (\bar x, \bar x+\delta)$ and $f(x)>0$ for $f \in (\bar x - \delta, \bar x)$, then $x\equiv \bar x$ asymptotically stable.
If there exists a $\delta >0$ such that $f(x)>0$ for $x \in (\bar x, \bar x+\delta)$ and $f(x)<0$ for $x \in (\bar x - \delta, \bar x)$, then $x \equiv \bar x$ is unstable.
This follows directly from the monotonicity of the solutions $x(t)$ in the specified intervals.
Unfortunately I can't see it. Is this really a short proof? If so, I would appreciate it if someone could elaborate a bit more. Thanks!

It is easier if you try to visualize what is going on here.  If you start near $\bar{x}$ but not quite at it, either


*

*$x<\bar{x}$: you are to the left of $\bar{x}$, so for (asymptotic) stability you want $\dot{x}=f$ to push you to the right, closer to $\bar{x}$, i.e. $f>0$; or

*$x>\bar{x}$: you are to the right of $\bar{x}$, so for (asymptotic) stability you want $\dot{x}=f$ to push you to the left, closer to $\bar{x}$, i.e. $f<0$.


Turn this around for unstable.
