I'm trying to figure out this question, or rather, part a of this question from a practice final. I know that if you consider the fixed points of $g(x) = (α+1)x-x^2$ (which are $ x = 0, x= α $), but I don't know how to find the interval for which this series converges to the fixed point, or why it is $α-\frac{1}{5} < x < α+\frac{1}{5}$ and don't really know how to start in general for these types of problems which ask for a convergence interval.
You need to find out
a) for which $x$ you get $|g_α'(x)|<1$, and that $α$ is inside that interval and
b) for which $α$ you get $g_α'(α)=0$.
b) Sanity check
The purpose of this answer is use parabola geometry in an intuitive way.
The iteration given by $f(x) = x^2$ converges quadratically to 0 for absolute value of $x$ less than 1.
The function $g(x)$ has roots at $0$ and $\alpha + 1$ with a fixed point at $\alpha$. But what if you choose $\alpha$ so that it is the midpoint the roots? Then you get that parabola axis of symmetry and all. Both $g$ and $f$ have the same shape and geometry.
You can check that $\alpha = 1$ sets this up.
But not only do you have the same shape, the axis of symmetry for both functions is at the fixed point. It sure seems like a good bet that the iterative sequence for $g$ would converge for any starting number in the open interval $(0, 2)$. Notice that $g(2) = 0$ in the same way that $f(-1) = 1$.
This is how you get quadratic convergence to the number $\alpha = 1$.
Intuitively, as you perturb $\alpha$ away from $1$ the convergence rate degrades.
