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!
 A: 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.
A: Note that $x=x(t)$ is a solution to your differential equation and  at the equilibrium point we have $f(x)=0$ which means $x'=0$. 
If there exists a $δ>0$ such that $f(x)<0 $for $x∈(x¯,x¯+δ)$ and $f(x)>0$ for $f∈(x¯−δ,x¯)$, then $x≡x¯$ asymptotically stable because if your initial point is on right side of $\bar x$ the solution starts decreasing and approaches to $\bar x$ and if your initial point is on the left side of $\bar x$ then the solution increases and again approaches $\bar x$.
Similar argument works for the case of instability of the equilibrium points where, solutions with initial conditions close to the equilibrium point escape from that point.
Make sure to keep in mind that  $x'=f(x)$ means with positive values of $f(x)$, $x(t)$ will increase and with negative values of $f(x)$, $x(t)$ decreases.
