Need Help with Newton's method?

Question

Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.)
1/3x^3 + 1/2x^2 + 12 = 0, x1 = −3

Answer 1

  

The function f(x) = (1/3)x³ + (1/2)x²  + 12

f'(x) = x² + x

Newtons method iterative formula : 

For n = 1,

x₁ = - 3

f(x₁) = (1/3)x₁³ + (1/2)x₁²  + 12

= (1/3)(- 3)³ + (1/2)(- 3)²  + 12

= (- 27/3) + (9/2) + 12 = - 9 + 4.5 + 12 = 7.5

f(x₁) = 7.5

f'(x₁) = x₁² + x₁

= (- 3)² + (- 3) = 9 - 3

f'(x₁) = 6

For n = 2,

 

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