# How I can prove this? (fixed points of $f(x)=(x^3+1)/3$)

the function f defined by $f(x)=(x^3+1)/3$ has three fixed points say α,β,γ where $-2<α<-1$, $0<β<1$, $1<γ<2$. For arbitrarily chosen $x_{1}$, define ${x_{n}}$ by setting $x_{n+1}=f(x_{n})$ If $α<x_{1}<γ$, prove that $x_{n}\rightarrow β$ as $n \rightarrow \infty$

I think I must prove three things, but not sure:

1: if $α<x_{1}<γ$, then $α<f(x)<γ$

2: if $α<x_{1}<β$, then $x_{1}<f(x)<β$

3: if $β<x<γ$, then $β<f(x)<x_{1}$

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Hints: Since there are only those three fixed points, we either have $f(x) < x$ for all $x$ with $\alpha < x < \beta$ or $f(x) > x$ for all those $x$. Check one point $x$ in that interval to see which it is. Similarly for $\beta < x < \gamma$.
Since $f$ is an increasing function, if $x > \alpha$ then $f(x) > f(\alpha) = \alpha$, and similarly ....