GATE 2010 CS (CS1) – Question 12
Newton–Raphson solves x²−13=0 starting at 3.5. What is the first iterate?
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Correct answer: (D) 3.607
Explanation
For f(x)=x²−13, Newton’s update is $x_1=x_0-(x_0^2-13)/(2x_0)$. At x₀=3.5, f=−0.75 and f′=7, so x₁=3.5+0.75/7=3.607142857. Rounded to the offered precision this is 3.607, D.