I stumbled upon a question and I can't seem to find the answer. Here it is:
Suppose $f$ a continuous function (from $\Bbb R$ to $\Bbb R$) with 2 attractive fixed points. Let's call them $a$ and $b$. The basin of attraction of $a$ is $(-\infty,p)$ and the basin of attraction of $b$ is $(p, \infty)$ for $p$ in $\Bbb R$. What can you say about the point $p$? My intuition would be to say that the function $f$ is not derivable in $p$, but I don't really know why.
Also, what would be an example of a function that has these properties?
Thank you in advance for your answers!
If $f(p)>p$, then there is a neighborhood of $p$ on which $f(x)>p$. Therefore, that neighborhood would belong to the attractive basin of $b$. This is a contradiction, since all points $<p$ belong to the attractive basin of $a$.
Similarly, if $f(p)<p$ then a neighborhood of $p$ would belong to the attractive basin of $a$, which also cannot be since all points $>p$ must belong to the basin of $b$.
Therefore, $f(p)=p$, i.e. $p$ is also a fixed point.
The information given doesn't tell you anything about the differentiability at $p$. You were given the example of $\frac{4}{\pi}\arctan(x)$ with $p=0$, which is differentiable at $p$. You can bend it a bit to produce an example that is not differentiable at $p$:
$$f(x)=\begin{cases}\frac{4}{\pi}\arctan(x),&x\geq0\\\frac{4(1+\epsilon)}{\pi}\arctan\left(x/(1+\epsilon)\right),&x>0\end{cases}$$