The iteration formula of Newton-Raphson method for finding roots of the equation f(x) = 0 is

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TuteeHUB · Accepted Answer

${x_{n + 1}} = {x_n} - \frac{{f'\left( {{x_n}} \right)}}{{f\left( {{x_n}} \right)}}$

TuteeHUB · Answer

${x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}$

TuteeHUB · Answer

${x_{n + 1}} = {x_n} + \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}$

TuteeHUB · Answer

${x_{n + 1}} = {x_n}+ \frac{{f'\left( {{x_n}} \right)}}{{f\left( {{x_n}} \right)}}$

1.	The iteration formula of Newton-Raphson method for finding roots of the equation f(x) = 0 is
A.	\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)
B.	\({x_{n + 1}} = {x_n} - \frac{{f'\left( {{x_n}} \right)}}{{f\left( {{x_n}} \right)}}\)
C.	\({x_{n + 1}} = {x_n} + \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)
D.	\({x_{n + 1}} = {x_n}+ \frac{{f'\left( {{x_n}} \right)}}{{f\left( {{x_n}} \right)}}\)
Answer» B. \({x_{n + 1}} = {x_n} - \frac{{f'\left( {{x_n}} \right)}}{{f\left( {{x_n}} \right)}}\)

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