# Thread: Numerical Analysis - Fixed Point Iteration

1. ## Numerical Analysis - Fixed Point Iteration

Assume that g is a continuously differentiable function and that the Fixed-Point Iteration g(x) = x has exactly three ﬁxed points, −3, 1, and 2. Assume that g'(−3) = 2.4 and that FPI started sufﬁciently near the ﬁxed point 2 converges to 2. Find g'(1).

Guess
F(x) = x^3 - 7x + 6 where -3 , 1 , and 2 are roots

Using FPI equation F(x). --- g(x) = x -f(x) ----
f(x) = -x^3 + 8x - 6
By pluging in the values
f(-3) = - 3 ,
f(1) = 1
f(2) = 2

Given g'(-3) = 2.4
Use general form of fixed point iteration to get
f'(x) = c * (3x^2 - 7) , where c is a constant
f'(-3) = c * (27 - 7) = -1.4
solve for c
c= - .07

f(x) = (x^3) - 7x + 6 = 0
f(x) = -.07x^3 + .49x - .42
USE FPI form to get
g(x) = .07x^3 + .51x +.42
g'(x) = .21x^2 +.51
g'(-3) = .21 * 9 + .51 = 1.89 +.51 =2.4

Hence g'(x) = .21x^2 +.51.
g'(1) = .21 +.51 =.72
g'(2) = 1.35

The equation I found satisfy that there are three fixed points and F'(-3) = 2.4, but it does not satisfy that when FPI is near 2 it converges to 2. My answer converges to 1. This is because g'(1) < g'(2).

my other attempt [no computation].
Since g'(2) converges to 2. The slope or derivative g'(2) < 1.
My guess is that g'(1) should be greater than 1 or greater than g'(2).
Given g'-(3) = 2.4......I don't know how this helps me in finding g'(1).
hmm maybe can I assume g(3) = 2.4????
g'(1) = 1.2???

Right now I don't see any approach where I can find g'(1). Some help would be nice.