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:

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