Question

Let \( f(x) = x^3 - x - 3 \). In finding a positive real root of \( f(x) = 0 \) using Newton-Raphson method, if the starting guess \( x_0 = 2 \), then the numerical value of the root \( x_1 \) after the first iteration is .........

• A. 1.5
• B. 1.6
• C. 1.7
• D. 1.8

Hint:

Use the Newton-Raphson iteration formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) and calculate both \( f(x_n) \) and \( f'(x_n) \) precisely for better convergence.

Answer 1

The Correct Option is C

The Newton-Raphson formula is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Given: \[ f(x) = x^3 - x - 3 ⇒ f'(x) = 3x^2 - 1 \] Starting with \( x_0 = 2 \): \[ f(2) = 2^3 - 2 - 3 = 8 - 2 - 3 = 3 \] \[ f'(2) = 3 \cdot 2^2 - 1 = 3 \cdot 4 - 1 = 12 - 1 = 11 \] \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{3}{11} = \frac{22 - 3}{11} = \frac{19}{11} \approx 1.727 \approx 1.7 \]