Let $f$ be function defined by
$$f(x)=\frac{x^3+1}{3}$$ has 3 fixed points say $\alpha ,\beta ,\gamma$ where $-2 < \alpha < -1$,$0 < \beta < 1$,$1 <\gamma <2$
I want to show the following :
for arbitray choice of $x_1$, define $x_n$ as $x_{n+1}=f(x_n)$
a) If $x_1$ < $\alpha$ then $x_n \to -\infty$ as $n\to \infty$
b) If $\alpha < x_1 < \gamma$ then $x_n \to \beta $ as $n\to \infty$
c) If $x_1$ > $\gamma$ then $x_n \to \infty$ as $n\to \infty$
My Attempt:
$$g(x)=f(x)-x$$
I can show using Intermediate value property there exist 3 fixed as per given condition. And $f'(x)>0$ therefore function is increasing on $\mathbb{R}$. I am not able to proceed further. Any help will be appreciated