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Show that the equation;x+3sinx=2 has a root between x=0.4 and x=0.6 using newton raphson's formular only

  • calculus -

    f(x) = x + 3sinx - 2
    f'(x) = 1 + 3cosx

    1: f(0.6000000) = 0.2939274
    2: f(0.5154411) = -0.0058029
    3: f(0.5170484) = -0.0000019
    4: f(0.5170490) = -0.0000000

  • calculus -

    Farouk, It would be a good idea to check back the previous posts you have made before reposting. It may save you a long wait.

    See answer to your previous post:

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