Question

Newton Raphson method is used to solve the equation \(x^3 - 2x - 5 = 0\) with an initial guess \(x_0 = 3\). What is \(x_1\), the value after ONE iteration?

Hint:

In the Newton-Raphson method, use the derivative of the function to iteratively approach the root. The accuracy improves with each iteration.

Answer 1

The Newton-Raphson method is given by the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Where:
\( f(x) = x^3 - 2x - 5 \)
\( f'(x) = 3x^2 - 2 \)
Given \( x_0 = 3 \), let's calculate \( x_1 \): 1. Calculate \( f(x_0) \): \[ f(3) = 3^3 - 2(3) - 5 = 27 - 6 - 5 = 16 \] 2. Calculate \( f'(x_0) \): \[ f'(3) = 3(3)^2 - 2 = 3(9) - 2 = 27 - 2 = 25 \] 3. Apply the Newton-Raphson formula: \[ x_1 = 3 - \frac{16}{25} = 3 - 0.64 = 2.36 \] Thus, the value of \( x_1 \) after one iteration is \( 2.36 \).