f (x) = x2 + 1If xi is very close to the root then according to Newton Raphson iterative procedure, xi+1 is

f (x) = x2 + 1If xi is very close to the root then according to Newton

1.	f (x) = x2 + 1If xi is very close to the root then according to Newton Raphson iterative procedure, xi+1 is
A.	\(\dfrac{x^2_i - 1}{2x_i}\)
B.	\(\dfrac{2x_i}{x^2_i - 1}\)
C.	\(\dfrac{2x_i}{x^2_i + 1}\)
D.	\(\dfrac{x^2_i + 1}{2x_i}\)
Answer» B. \(\dfrac{2x_i}{x^2_i - 1}\)