This is a past prelim question,

Let $F(x)=2x+qx^2$ with $\frac{1}{2}\leq q \leq 1$ and $x_{n+1}=F(x_{n})$.

## For which interval does this iteration converge ?

I tried these but did not work;

Fixed points are $0, -\frac{1}{q}$, Since $F'(0)>1$ if there is such an interval with initials from that interval converging to a fixed point this fixed point will be $-\frac{1}{q}$. I tried to use the idea; find closed interval C with $|F'|\leq\lambda <1$ on $C$ and $F(C)\subset C$. Suppose we just want to find an interval not the maximal one. Then $F' = 2+2px$, $|2+2qx|\leq \frac{1}{2}$ this will give smaller interval, then $-\frac{1}{2}\leq 2+2qx \leq \frac{1}{2}$, $-\frac{5}{4q} \leq x \leq -\frac{3}{4q} =: C$. But $F(C)$ is not a subset of $C$. I tried for smaller number $a$ for $|F'| < a$, but it did not work.

Question 1: Is there any way to find such $C$ ?

Question 2: How can we find maximal interval ?

any clues/ideas would be appreciated.