GATE 2017 EC – Question 40
Starting with $x = 1$, the solution of the equation $x^3 + x = 1$, after two iterations of Newton-Raphson's method (up to two decimal places) is ________.
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Correct answer: 0.68 to 0.70
Explanation
Let $f(x) = x^3 + x - 1$ with $f'(x) = 3x^2 + 1$. From $x_0 = 1$: $f = 1$ and $f' = 4$, so $x_1 = 1 - \frac{1}{4} = 0.75$. Then $f(0.75) = 0.1719$ and $f'(0.75) = 2.6875$, so $x_2 = 0.75 - 0.0640 = 0.686$, which is 0.69 to two decimal places.