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Math Help - newton's method of approximation

  1. #1
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    newton's method of approximation

    A first approximation to the equation x^3 - 3x^2 + 1 = 0 is 0.5. Using one application of Newton's method, find a better approximation to two decimal places.

    is this correct??
    P(0) = 1 -pos
    P(1) = -1 -neg
    therefore there is a root between 0 and 1

    x^3 - 3x^2 + 1 = 0
    f(x) = x^3 - 3x^2 + 1
    f ' (x) = 3x^2 - 6x
    f(a) = a^3 - 3a^2 + 1
    f ' (a) = 3a^2 - 6a

    a_1 = a - f(a) / f ' (a)

    = 0.5 - [(0.5)^3 - 3x(0.5)2 + 1] / [3x(0.5)^2 - 6x(0.5)]

    =2/3

    =0.67
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  2. #2
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    Correct. If you use a calculator, you can do it very rapidly and get a better answer of about 0.652704
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  3. #3
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    ok thanks
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