Discussion
The first approximation of the root lying between 0 and 1 of the equation
\[ x^{3} + 3x - 1 = 0 \]
by Newton-Raphson method with initial approximation \(x_{0} = 0\), is :
Answer: B
The first approximation is 1/3.
Newton-Raphson: \(x_{1} = x_{0} - f(x_{0})/f'(x_{0})\), with \(f(x) = x^{3} + 3x - 1\) and \(f'(x) = 3x^{2} + 3\).
\[ \begin{aligned} x_{1} &= 0 - \frac{f(0)}{f'(0)} \\ &= 0 - \frac{-1}{3} = \frac{1}{3} \end{aligned} \]
As a check, \(f(1/3) = 1/27 \gt 0\) and \(f(0) = -1\), so the root lies just below 1/3, and the next iterate moves only slightly. 1/2 would be the bisection midpoint, and 1/4 and 1/5 do not come from one Newton step.
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