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Math Help - % Percent error

  1. #1
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    % Percent error

    Suppose f(x) = x^(1/4). If x is accurate to within 6%, within what percent is f(x) accurate ?

    i know that %percent error = delta F/ f(x) = f'(x)*deltaX / f(x)
    what i dont know is how to get delta f out this this information
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  2. #2
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    Let me guess,

    You said,
    %percent error = df / f = f' * dx / f

    f' = df/dx, thus:

    df / f = df/dx * dx / f

    Your x can change 6% so dx = 0.06

    df/dx = 1/(4*x^(3/4))

    error = 1/(4*x^(3/4)) * 0.06 / x^(1/4) = 3/(200*x)
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  3. #3
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    Plug it in

    f(x) = x^(1/4). If x is accurate to within 6%, within what percent is f(x) accurate

    x=1
    f(x) = 1

    x=1.06
    f(x) = 1.01467385

    Therefore it is accurate to within 1.467% when x=1.

    Why is this not exactly 3/(200*1) = 1.5% ?
    Last edited by MathGuru; April 12th 2005 at 06:29 PM.
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