Basin of attraction of the fixed map $f(x) = x-x^3$ Prove that the interval $(-\sqrt 2 ,\sqrt 2 )$ is the basin of attraction of the fixed point $0$ of the map $f(x)=x-x^3$, for $x \in \mathbb{R}$.
How one would prove this? In the examples I've seen so far they usually prove that a fixed point has a certain basin of attraction by proving that the function is decreasing or increasing for certain values within the basin of attraction. In this case however the values 'jump' from positive to negative making it impossible to use that method. I've been trying some options using the absolute value, but I can't figure it out. Could you please show me a plausible proof for this situation?

Here is a roadmap:


*

*If $B$ is the basin of attraction of the fixed point $0$, then $f(B)\subseteq B$.

*If $I=(-\sqrt 2 ,\sqrt 2 )$, then $f(I)\subseteq I$.

*If $J$ is open and contains any point not in $I$, then $J$ contains points that diverge under iterates of $f$.
