Question

The Newton-Raphson method is used to solve \( f(x) = 0 \), where \( f(x) = e^x - 5x \). If the initial guess \( x^{(0) = 1.0 \), the value of the next iterate, \( x^{(1)} \), rounded off to 2 decimal places, is:}

Hint:

The Newton-Raphson method converges faster if the initial guess is close to the actual root. Always evaluate both \( f(x) \) and \( f'(x) \) accurately.

Answer 1

Given: \[ f(x) = e^x - 5x. \] Initial guess: \[ x^{(0)} = x_0 = 1. \] Step 1: Differentiating \( f(x) \) with respect to \( x \): \[ f'(x) = e^x - 5. \] Step 2: Newton-Raphson's Iterative Method: The first iteration is given by: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. \] Substitute \( f(x) \) and \( f'(x) \): \[ x_1 = x_0 - \frac{e^{x_0} - 5x_0}{e^{x_0} - 5}. \] Step 3: Substituting \( x_0 = 1 \): \[ f(x_0) = e^1 - 5 \cdot 1 = e - 5, \] \[ f'(x_0) = e^1 - 5 = e - 5. \] Substituting into the formula for \( x_1 \): \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{e - 5}{e - 5}. \] Simplify: \[ x_1 = 1 - 1 = 0. \] Conclusion: The value of the next iterate \( x^{(1)} \) is: \[ x^{(1)} = 0. \]