Given the iteration $x_{n+1}=\sqrt[3]{3x_n+1}$ show that $p\in(1,2)$ is a fixed point
$$x=\sqrt[3]{3x+1}\iff f(x)=x^3-3x-1$$
Plugging in $1$ result in $-3$ whereas $2$ gives $1$ so there is a root in this interval, but how can I be sure that there is a fixed point and not just a root?
The fixed point is the real root of $x^3-3x-1=0$, we can show that this Eq. has one real root in $(1,2)$. It follows by Intermediate theorem as $f(1)f(2)=-3<0$, So the given cubic has one real root in (1,2) and it can be found numerically as 1.8793...