Question
Find the positive real root of x3 - x - 3 = 0 using Newton-Raphson method. If the starting guess (x0) is 2, the numerical value of the root after two iterations (x2) is __________ (round off to two decimal places).
Answer :
Correct answer is : 1.67
f(x) = x3 - x - 3
Starting guess (x0 = 2)
Now first iterations
Second iterations
x1 = 1.7273
x2 = 1.67
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