Question

An approximate solution of the equation \( x^3 - 17 = 0 \) is to be obtained using the Newton-Raphson method. If the initial guess is \( x_0 = 2 \), the value at the end of the first iteration is \( x_1 = \) \_\_\_\_\_\ (rounded off to two decimal places).

Hint:

The Newton-Raphson method quickly converges to the root by iteratively applying the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Make sure to compute both \( f(x_n) \) and \( f'(x_n) \) at each step.

Answer 1

The Newton-Raphson method is given by the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] For the equation \( f(x) = x^3 - 17 \), we have: \[ f'(x) = 3x^2 \] Starting with the initial guess \( x_0 = 2 \), we can calculate the value of \( x_1 \). Step 1: Calculate \( f(x_0) \) and \( f'(x_0) \): \[ f(x_0) = 2^3 - 17 = 8 - 17 = -9 \] \[ f'(x_0) = 3 \times 2^2 = 3 \times 4 = 12 \] Step 2: Apply the Newton-Raphson formula: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-9}{12} = 2 + 0.75 = 2.75 \] Therefore, the value at the end of the first iteration is \( x_1 = 2.75 \).