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Newton–Raphson method is used to compute a root of the equation $x^3 - 13 = 0$ with $3.5$ as the initial value. The approximation after one iteration is

1. 3.575
2. 3.677
3. 3.667
4. 3.607

Correct answer: 3.667

Solution

Newton-Raphson iteration is $x_{1}=x_0-\frac{f(x_0)}{f'(x_0)}$. With $x_0=3.5$, $f(3.5)=3.5^3-13=29.875$ and $f'(3.5)=3(3.5)^2=36.75$. Thus $x_1=3.5-\frac{29.875}{36.75}\approx 3.667$.

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