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
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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