Using Newton's method with first approximate x1 = 0 , find the root of

Question

Using Newton's method with first approximate x₁ = 0 , find the root of

x³ + 6x² + 9x + 1= 0

accurate to 7 decimal places.

Answer 1

x³ + 6x² + 9x + 1= 0

Newton's method:

x_n+1 = x_n - f(x_n)/f'(x_n) = x_n - (x_n³ + 6x_n² + 9x_n + 1) / (3x_n² + 12x_n + 9)

The successive approximations are best done in Excel.

x₁ = 0

x₂ = x₁ - (x₁³ + 6x₁² + 9x₁ + 1) / (3x₁² + 12x₁ + 9) = -0.111111111

x₃ = x₂ - (x₂³ + 6x₂² + 9x₂ + 1) / (3x₂² + 12x₂ + 9) = -0.120548433

x₄ = x₃ - (x₃³ + 6x₃² + 9x₃ + 1) / (3x₃² + 12x₃ + 9) = -0.120614755

x₅ = x₄ - (x₄³ + 6x₄² + 9x₄ + 1) / (3x₄² + 12x₄ + 9) = -0.120614758

We stop at x₅ because the first 7 decimal places stopped changing.

Rounded to 7 decimal places: x ≈ -0.1206148