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Math Help - function question

  1. #1
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    function question

    I don't understand why I'm not able to manipulate this quite basic function into another form

    the function is:
    \\x=\sqrt[3]{9}\\x^{3}=9\\f(x)=x^{3}-9

    But then this function is exactly the same:
    \\x=\sqrt[3]{9}\\x-\sqrt[3]{9}=0\\\text{If i try to produce a 9 and a}\;x^3\;\text{by cubing both sides}\\(x^3-\sqrt[3]{9})^3=x^3-3x^2\times\sqrt[3]{9}+3x\times(\sqrt[3]{9})^2-9\\\neq x^{3}-9

    Am I misunderstanding something basic here?
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  2. #2
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    Re: function question

    Consider a^3-b^3 = (a-b)(a^2+ab+b^2) so

    x^3-9 = x^3-(\sqrt[3]{9})^3 = (x-\sqrt[3]{9})(x^2 +\sqrt[3]{9}x+(\sqrt[3]{9})^2)
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  3. #3
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    Re: Using Newtons method to solve

    the actual question was to solve x=\sqrt[3]{9} using Newtons method with the first approximation x=2, if you cube both sides and do Newtons method:
    \\x^3=9\\f(x)=x^3-9\\f'(x)=3x^2\\x_2=2\frac{1}{12}=2.0833333333

    but if you continue with the original equation without cubing both sides
    \\f(x)=x-\sqrt[3]{9}\\f'(x)=1\\x_2=\sqrt[3]{9}=2.080083823

    Why are the solutions similar but not exactly the same?
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  4. #4
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    Re: Using Newtons method to solve

    f(x)=x^3-9


    \\f(x)=x-\sqrt[3]{9}

    These are different equations, that is why you have different solutions. One is a cubic and one is a straight line.

    Newton's will only give an approximation of the root.
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