Consider the fixed point iteration formula: *x_(n+1) = (2/3)[(x_n)^3 - 1] - 3(x_n)^2 + 4x_n = g(x) *Note: "_" precedes a subscript and "^" precedes a superscript (a) Find an interval in which every starting point x_0 will definitely converge to alpha = 1. (b) Show that the order of the above fixed point iteration formula is 2 (quadratic convergence). ======================================= For (a), I took the derivative of g(x) and set it equal to zero. I found that when g'(x) = 2x^2 - 6x +4 = (2x - 2)(x - 2)= 0, x = 1, 2. But g'(alpha) = g'(1) = 2 - 6 + 4 = 0...? I want to say that the interval is (1,2]... For (b), I tried |alpha - x_(n + 1)| <= c|1 - x_n|^p, where p is the order and c is some constant >= 0. And Newton's method usually converges quadratically... I ended up with: |-(2/3)(x_n)^3 + 3(x_n)^2 - 4x_n + (5/3)| <= c|1 - x_n|^p I don't know how to conclude that p must be 2... or if this is even right...