Question

The Newton-Raphson method is used to find the root of the equation f(x) \(\equiv\) x\(^2\) \(-\) x \(-\) 1 \(=\) 0. If the initial guess for the root is 1, then the estimate of the root after two iteration is

• A. 2
• B. 1.80
• C. 1.67
• D. 1.82

Hint:

Newton-Raphson tips:
- Double check your derivative: \( \frac{d}{dx}(x^2 - x - 1) = 2x - 1 \).
- Write out intermediate steps clearly: \( x_1 = 2 \rightarrow x_2 = 2 - 1/3 = 1.67 \).
This prevents algebraic or arithmetic errors during multi-step numerical calculations.

Answer 1

The Correct Option is C

Step 1: Simplify the iteration formula algebraically before plugging in any numbers.
For \( f(x) = x^2 - x - 1 \) and \( f'(x) = 2x - 1 \), the Newton-Raphson update combines into a single explicit expression:
\[ x_{n+1} = x_n - \frac{x_n^2 - x_n - 1}{2x_n - 1} = \frac{x_n(2x_n-1) - (x_n^2 - x_n - 1)}{2x_n - 1} = \frac{x_n^2 + 1}{2x_n - 1} \]

Step 2: Apply this simplified formula for the first iteration.
With \( x_0 = 1 \):
\[ x_1 = \frac{1^2+1}{2(1)-1} = \frac{2}{1} = 2 \]

Step 3: Apply it again for the second iteration.
With \( x_1 = 2 \):
\[ x_2 = \frac{2^2+1}{2(2)-1} = \frac{5}{3} \approx 1.6667 \]

Step 4: Final Answer.
Rounded to two decimals, \( x_2 \approx \boxed{1.67} \).
Therefore, option (C) is correct.