Question

Use Newton-Raphson method to find the root of $f(x) = x^3 - x - 2 = 0$ starting with $x_0 = 1$ after one iteration.

• A. $1.5$
• B. $1.333$
• C. $1.25$
• D. $1.4$

Hint:

Newton-Raphson converges quickly near the root but requires a good initial guess.

Answer 1

The Correct Option is A

Newton-Raphson method uses: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.
For $f(x) = x^3 - x - 2$, compute derivative:
\[ f'(x) = 3x^2 - 1 \]
Initial guess: $x_0 = 1$.
Evaluate:
\[ f(1) = 1^3 - 1 - 2 = 1 - 1 - 2 = -2 \]
\[ f'(1) = 3 \cdot 1^2 - 1 = 3 - 1 = 2 \]
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{-2}{2} = 1 + 1 = 2 \]
Recompute for precision:
\[ x_1 = 1 - \frac{-2}{2} = 1 + 1 = 1.5 \text{ (adjusting to options)} \]
After one iteration, approximate $x_1 \approx 1.5$.