Suppose $f$ is a map from $\mathbb{R} \rightarrow \mathbb{R}$, $f'$ exists everywhere, and $(0,0)$ is an attracting fixed point. Does the basin of attraction of $(0,0)$ have to be one interval? Either provide a proof or counterexample.
I'm not sure about this. I think it may have to be just one interval because I've drawn a few cobweb plots and found no case where the basin consists of multiple intervals.
In principle there is no reason to expect that the basin of attraction of $(0,0)$ is just one interval. For example, suppose there exists a value $a \in \mathbb{R}$ such that $f(a)=0$, which forces some small open interval around $x=a$ to be in the basin of attraction of $(0,0)$. But it might well happen that there exists a point $b$ between $0$ and $a$ which is not attracted to zero, hence the basin of attraction cannot be one interval.
It looks to me like $f(x)=16x^4 - 16x^2$ is a counterexample. Note that if $a=1$ then $f(a)=0$, so the basin of attraction of $0$ contains a small interval around $a=1$. But starting with $b=\frac{1}{2}$ and iterating we get $$f(\frac{1}{2}) = -3, \qquad f(-3) = 1152, \ldots $$ which is going to shoot off to infinity and so $b$ is not in the basin of attraction.
I got this example by starting with $x^2$ to get an attractor at $0$, then multiplying by $x^2-1$ to get points at $-1$ and $+1$ mapping to $0$, then multiplying by a big constant to make $\frac{1}{2}$ iterate off to infinity.
