Is it 2 cycle attracting?

Question

Let $c$ be a constant and let $f(x) = x^3-3x+c$, $c>0$. Determine the values of c for which ${0,c}$ is a 2-cycle. Is the 2-cycle attracting for the value of $c$? Explain.

I am having trouble seeing why the solution is $c = ±√2$, and that in fact the {0,c} is 2-cycle repelling.

Answer 1

You can read here that

for a function $f$ where $f^n(z_0)=z_0$ that $z_0$ is

repelling when $|(f^n)'(z_0)|>1$ and

attracting when $|(f^n)'(z_0)|<1$.

$$f(x)=x^3-3x+c \implies f'(x)=3x^2-3$$

Take $c =\pm\sqrt2$; this means $f(c)=0, f'(c)=3$

$$\bigg(f(f(x)) \bigg)' =f'(f(x))f'(x)\implies f'(f(c))f'(c)=f'(0)f'(c)=-3\times 3=-9. $$

So then we have $f^2(c)=c \,$ and $| (f^2)'(c)|>1$ and therefore $\{0, c \}$ is a repelling cycle.