Math/ Calculus

Given an initial guess of x1 = 1, use the Newton-Raphson Method to fi�nd the second approximation x = x2 of the solution of the equation:

sqrt(x+3)=x

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asked by Angela
  1. let f(x) = √(x+3)-x
    f'(x) = 1/(2√(x+3)) - 1

    so, we get

    f(1.00000) = 1.00000 -- next x: 2.33333
    f(2.33333) = -0.02393 -- next x: 2.30279
    f(2.30279) = -0.00001 -- next x: 2.30278

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    posted by Steve

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