Derivatives: Newtons Method, Help!?

Question

1)Use Newtos Method to approximate the root of the equation: 
X^3=100x+220 
Start with x0=12 and preform 3 iterations, ie, find x1-x2 and x3 
Calculate Ix0-x1I, Ix1-x2I, and Ix3-x2I 
Used Newtons Method x_n+1=x_n-(x_n)/f1(x_n) 

Answer 1

The function f(x) = x³ - 100x  - 220

f'(x) = 3x² - 100

Newtons method iterative formula :

x₀ = 12

f(x₀) = x₀³ - 100x₀ - 220

= (12)³ - 100(12) - 220 = 1728 - 1200 - 220

f(x₀) = 308

f'(x₀) = 3x₀² - 100

= 3(12)² - 100 = 3(144) - 100

f'(x₀) = 332

x₁  = 12 - (308/332) = 12 - 0.9277

x₁  = 11.07

 

f(x₁) = x₁³ - 100x₁ - 220

= (11.07)³ - 100(11.07) - 220 = 1356.57 - 1107 - 220

f(x₁) = 29.57

f'(x₁) = 3x₁² - 100

= 3(11.07)² - 100 = 367.63  - 100

f'(x₁) = 267.63

x₂  = 11.07 - (29.57/267.63) = 11.07 - 0.1104

x₂  = 10.95

 

f(x₂) = x₂³ - 100x₂ - 220

= (10.95)³ - 100(10.95) - 220 = 1312.93 - 1095 - 220

f(x₂) = - 2.07

f'(x₂) = 3x₂² - 100

= 3(10.95)² - 100 = 359.70 - 100

f'(x₂) = 259.70

x₃ = 10.95 - (-2.07/259.70)

x₃ = 10.9579

|x₀ - x₁| = |12 - 11.07| = 0.93

|x₁ - x₂| = |11.07 - 10.95| = 0.12

|x₃ - x₂| = |10.9579 - 10.95| = 0.0079.