Question

The Newton-Raphson method is used to find the root of \[ f(x)=x^2-x-1=0 \] If the initial guess is \(x_0=1\), then the estimate after two iterations is

• A. \(2\)
• B. \(1.80\)
• C. \(1.67\)
• D. \(1.82\)

Hint:

Newton-Raphson converges quickly: \[ \boxed{x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}} \] Always recompute derivative at each step.

Answer 1

The Correct Option is D

Concept: Newton-Raphson formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Given: \[ f(x)=x^2-x-1,\quad f'(x)=2x-1 \]

Step 1: First iteration.
Start with \(x_0=1\) \[ f(1)=1-1-1=-1 \] \[ f'(1)=2(1)-1=1 \] \[ x_1 = 1 - \frac{-1}{1} = 2 \]

Step 2: Second iteration.
\[ f(2)=4-2-1=1 \] \[ f'(2)=4-1=3 \] \[ x_2 = 2 - \frac{1}{3} = \frac{5}{3} \approx 1.67 \]

Step 3: Interpretation.
After two iterations: \[ x_2 \approx 1.67 \] Final Answer: \[ \boxed{1.67} \Rightarrow Option (C) \]