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Math Help - Finding root of equation

  1. #1
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    Finding root of equation

    Show that the equation x^3 + x = 3 has exactly one root, ∝. Find an integer n such that ∝ lies between n and n + 1. Starting with x0 = n as a first approximation, use Newton's Method to find 2 further approximations to ∝.

    f(x) = x^3 + x - 3, [n, n + 1]

    f'(x) = 3x^2 + 1

    x2 = ( x1 + x1^3 + x1 - 3 ) / ( 3x1^2 + 1 )
    x2 = ( n + n^3 + n - 3 ) / ( 3n^2 + 1 )
    x2 = ( n^3 + 2n - 3 ) / ( 3n^2 + 1 )
    x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 )

    I dont you what to do from here.
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    Show that the equation x^3 + x = 3 has exactly one root, ∝. Find an integer n such that ∝ lies between n and n + 1. Starting with x0 = n as a first approximation, use Newton's Method to find 2 further approximations to ∝.

    f(x) = x^3 + x - 3, [n, n + 1]

    f'(x) = 3x^2 + 1

    x2 = ( x1 + x1^3 + x1 - 3 ) / ( 3x1^2 + 1 )
    x2 = ( n + n^3 + n - 3 ) / ( 3n^2 + 1 )
    x2 = ( n^3 + 2n - 3 ) / ( 3n^2 + 1 )
    x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 )

    I dont you what to do from here.
    If there is only one root then the function does not turn anywhere. So you need to show that you can't solve the derivative = 0

    So if f(x) = x^3 + x - 3

    f'(x) = 3x^2 + 1

    3x^2 + 1 = 0

    3x^2 = -1

    x^2 = -\frac{1}{3}.

    There does not exist any x that solves this equation, so the function does not turn anywhere. Therefore there's only one root.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    If there is only one root then the function does not turn anywhere. So you need to show that you can't solve the derivative = 0

    So if f(x) = x^3 + x - 3

    f'(x) = 3x^2 + 1

    3x^2 + 1 = 0

    3x^2 = -1

    x^2 = -\frac{1}{3}.

    There does not exist any x that solves this equation, so the function does not turn anywhere. Therefore there's only one root.
    But there is a root. The answer in my memo is x2 = 17/14 = 1.2142857
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    Quote Originally Posted by TsAmE View Post
    But there is a root. The answer in my memo is x2 = 17/14 = 1.2142857

    It is close , the solution is  x = 1.213411663...
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    Quote Originally Posted by TsAmE View Post
    But there is a root. The answer in my memo is x2 = 17/14 = 1.2142857
    The point is, if there's more than one root, then the function has to turn somewhere. But we have shown that it doesn't turn, so there is at most one root.

    To prove that there is a root, substitute some values in and see if you get both positive and negative values. If so then it must have crossed the x axis somewhere.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    The point is, if there's more than one root, then the function has to turn somewhere. But we have shown that it doesn't turn, so there is at most one root.

    To prove that there is a root, substitute some values in and see if you get both positive and negative values. If so then it must have crossed the x axis somewhere.
    Yeah I recognize what you saying: use the intermediate value theorem. I subbed in n = 1 into x2 = ( n( n^2 + 2 ) - 3 ) / ( 3n^2 + 1 ), and got x2 = 0, which means this is the root right? How in the answer did they get x2 = 17/14?
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