Given $f(x) = x^2-1$. Prove that the basin of attraction of the 2-cycle $\{-1,0\}$ consists of all numbers in the interval $\left(\frac{1-\sqrt5}2, \frac{1+\sqrt5}2\right)$, except for the points where iterates are eventually the fixed point $\frac{1-\sqrt5}2$
What I first did was set \begin{align}f(x) &= x \\ x^2-x-1 &= 0 \\x &= \frac{1 \pm \sqrt5}2 \end{align}
To find the periodic orbit: I took the second iterate ie. \begin{align}f^2(x) &= f(f(x)) \\ (x^2-1)^2 - 1 &= x \\ x^4-2x^2 -x &= 0 \\ x (x+1)(x^2-x-1) &= 0 \end{align}
which gives four points at $x = 0 , -1, \frac{1 + \sqrt5}2$ and $\frac{1 - \sqrt5}2 $
I also found the maximum and minimum to be $ (0,0) \to \text{Max} $ and $ (1,-1) \to \text{Min} $, and $(-1,-1)\to \text{Min}$.
I showed graphically of $f^2(x) = x^4-2x^2$ with points (above), and I traced how some points do not reach the basin attraction 2-cycle, and will eventually hit the periodic point $(-0.618, -0.618)$,
I found a point to the left of $(-1,-1)$ it's eventually fixed point, there was also $\frac{1- \sqrt5}{-2}$ and another value to the left of $(\frac{1 + \sqrt5}2,\frac{1 + \sqrt5}2))$
But how can I write this as a proof format that there are multiple points? What I explained above, is it enough to show that there exist multiple points that are eventually fixed to $\left(-\frac{1−\sqrt5}2,-\frac{1−\sqrt5}2\right)$ ?