What is the approximate value of X4 using Newton's method for f(x) = x^3 - 2x + 2 starting from X1 = -1.0000?

What is the approximate value of X4 using Newton's method for f(x) = x^3 - 2x + 2 starting from X1 = -1.0000?

• A. -2.0000
• B. -2.1467
• C. -1.5000
• D. -2.3000

Answer: (B)

To determine the approximate value of X4 using Newton's method, we start with the given function \( f(x) = x^3 - 2x + 2 \). Newton's method is based on the formula: \[ X_{n+1} = X_n - \frac{f(X_n)}{f'(X_n)} \] To apply this method, we first need to calculate the derivative of \( f(x) \): \[ f'(x) = 3x^2 - 2 \] We start with the initial guess \( X_1 = -1.0000 \). 1. **Calculate \( f(X_1) \) and \( f'(X_1) \):** \[ f(-1) = (-1)^3 - 2(-1) + 2 = -1 + 2 + 2 = 3 \] \[ f'(-1) = 3(-1)^2 - 2 = 3(1) - 2 = 1 \] 2. **Update to find \( X_2 \):** \[ X_2 = X_1 - \frac{f(X_1)}{f

To determine the approximate value of X4 using Newton's method, we start with the given function ( f(x) = x^3 - 2x + 2 ). Newton's method is based on the formula:

[

X_{n+1} = X_n - \frac{f(X_n)}{f'(X_n)}

]

To apply this method, we first need to calculate the derivative of ( f(x) ):

[

f'(x) = 3x^2 - 2

]

We start with the initial guess ( X_1 = -1.0000 ).

1. Calculate ( f(X_1) ) and ( f'(X_1) ):

[

f(-1) = (-1)^3 - 2(-1) + 2 = -1 + 2 + 2 = 3

]

[

f'(-1) = 3(-1)^2 - 2 = 3(1) - 2 = 1

]

2. Update to find ( X_2 ):

[

X_2 = X_1 - \frac{f(X_1)}{f