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).

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Answer :

Correct answer is : 1.67

f(x) = x3 - x - 3

f ( x ) = 3 x 2 1

Starting guess (x0 = 2)

Now first iterations

x 1 = x 0 ( x 0 3 x 0 3 ) 3 x 0 2 1

x 1 = x 0 f ( x 0 ) f ( x 0 ) = 2 ( 2 ) 3 2 3 3 ( 2 ) 2 1 = 1.7273

Second iterations

x 2 = x 1 ( x 1 3 x 1 3 ) 3 x 1 2 1

x1 = 1.7273

x 2 = x 1 f ( x 1 ) f ( x 1 ) = 1.7273 ( 1.7273 ) 3 1.7273 3 3 ( 1.7273 ) 2 1 = 1.67369

x2 = 1.67

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