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Math Help - Numerical Analysis - Fixed Point Iteration

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    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.
    Last edited by dac1234; October 10th 2011 at 12:30 PM.
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