Using Newton-Raphson method, a root correct to 3 decimal places of the…
2017
Using Newton-Raphson method, a root correct to 3 decimal places of the equation x3 - 3x - 5 = 0
- A.
2.222
- B.
2.275
- C.
2.279
- D.
None of the above
Attempted by 13 students.
Show answer & explanation
Correct answer: C
To find the root of x^3 - 3x - 5 = 0 using Newton-Raphson, let f(x) = x^3 - 3x - 5 and f'(x) = 3x^2 - 3. The iteration formula is x_{n+1} = x_n - f(x_n)/f'(x_n). Starting with initial guess x_0 = 2, we calculate f(2) = -3 and f'(2) = 9. Thus x_1 = 2 - (-3/9) ≈ 2.333. Next, f(2.333) ≈ 0.70 and f'(2.333) ≈ 13.33, giving x_2 ≈ 2.280. Further iterations converge to the root x ≈ 2.279 correct to three decimal places.