Newton-Raphson method is used to compute a root of the equation x2-13=0 with 3.5 as the initial value. The approximation after one iteration is
(A) 3.575
(B) 3.676
(C) 3.667
(D) 3.607
Answer: (D)
Explanation: In Newton-Raphson’s method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x).
f(x) = x2-13 f'(x) = 2x Applying the above formula, we get Next x = 3.5 - (3.5*3.5 - 13)/2*3.5 Next x = 3.607
Recommended Posts:
- GATE | GATE CS 2010 | Question 1
- GATE | GATE CS 2010 | Question 3
- GATE | GATE CS 2010 | Question 4
- GATE | GATE CS 2010 | Question 5
- GATE | GATE CS 2010 | Question 65
- GATE | GATE CS 2010 | Question 8
- GATE | GATE CS 2010 | Question 65
- GATE | GATE CS 2010 | Question 10
- GATE | GATE CS 2010 | Question 11
- GATE | GATE CS 2010 | Question 65
- GATE | GATE CS 2010 | Question 13
- GATE | GATE CS 2010 | Question 14
- GATE | GATE CS 2010 | Question 15
- GATE | GATE CS 2010 | Question 16
- GATE | GATE CS 2010 | Question 65
- GATE | GATE CS 2010 | Question 18
- GATE | GATE CS 2010 | Question 19
- GATE | GATE CS 2010 | Question 20
- GATE | GATE CS 2010 | Question 21
- GATE | GATE CS 2010 | Question 22
