Question

The iterative formula to find the root of \( \sqrt[3]{x^2} = 2 \) using Newton-Raphson method is \( x_{n+1} = \)

• A. \( \frac{3x_n - \frac{2x_n^2}{3}}{3} \)
• B. \( \frac{1}{3} \left[ 2x_n + \frac{32}{x_n^2} \right] \)
• C. \( \frac{1}{3} \left[ 3x_n + \frac{32}{x_n^2} \right] \)
• D. \( \frac{1}{3} \left[ 3x_n - \frac{32}{x_n^2} \right] \)

Hint:

Remember: In Newton-Raphson method, iterative formula is \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Always carefully compute both \( f(x) \) and \( f'(x) \) before substitution.

Answer 1

The Correct Option is B

We use Newton-Raphson formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] For \( f(x) = x^{\frac{2}{3}} - 2 \) Derivative: \[ f'(x) = \frac{2}{3} x^{-\frac{1}{3}} \] Substituting in the formula: \[ x_{n+1} = x_n - \frac{x_n^{\frac{2}{3}} - 2}{\frac{2}{3} x_n^{-\frac{1}{3}}} \] Simplify and rearrange terms to get: \[ x_{n+1} = \frac{1}{3} \left[ 2x_n + \frac{32}{x_n^2} \right] \]