Consider the function f(x) = x² - 2x - 1. Suppose an execution of the…
2008
Consider the function f(x) = x² - 2x - 1. Suppose an execution of the Newton-Raphson method to find a zero of f(x) starts with an approximation x₀ = 2 of x. What is the value of x₂, the approximation of x that the algorithm produces after two iterations, rounded to three decimal places?
- A.
2.417
- B.
2.419
- C.
2.423
- D.
2.425
Show answer & explanation
Correct answer: A
For f(x) = x² - 2x - 1, the derivative is f'(x) = 2x - 2.
Newton-Raphson iteration is
xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ).
Starting with x₀ = 2:
f(2) = 2² - 2(2) - 1 = -1, and f'(2) = 2.
So, x₁ = 2 - (-1/2) = 2.5.
Now,
f(2.5) = 2.5² - 2(2.5) - 1 = 0.25, and f'(2.5) = 3.
So, x₂ = 2.5 - 0.25/3 = 2.416666... .
Rounded to three decimal places, x₂ = 2.417.