Newton-Raphson iteration formula for finding ∛c, where c > 0, is:

1996

Answer: C. xₙ₊₁ = (2xₙ³ + c) / (3xₙ²) — To find ∛c, solve f(x) = x³ - c = 0. Newton-Raphson uses xₙ₊₁ = xₙ - f(xₙ)/f′(xₙ). Here, f′(x) = 3x². So, xₙ₊₁ = xₙ - (xₙ³ - c)/(3xₙ²) = (3xₙ³ - xₙ³ +…

1. A.

   xₙ₊₁ = (2xₙ³ + ∛c) / (3xₙ²)

2. B.

   xₙ₊₁ = (2xₙ³ - ∛c) / (3xₙ²)

3. C.

   xₙ₊₁ = (2xₙ³ + c) / (3xₙ²)

4. D.

   xₙ₊₁ = (2xₙ³ - c) / (3xₙ²)

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