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Given f(x)=x3+2x−4f ( x ) = x ^ { 3 } + 2 x - 4

Question 171

Multiple Choice

Given f(x) =x3+2x−4f ( x ) = x ^ { 3 } + 2 x - 4 , use Newton's method to find the iterative formula for xn+1x _ { n + 1 } .


A) xn+1=−xn3+2xn3xn2+2x _ { n + 1 } = - \frac { x _ { n } ^ { 3 } + 2 x _ { n } } { 3 x _ { n } ^ { 2 } + 2 }

B) xn+1=xn−xn3+2xn3xn2+2x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { 3 } + 2 x _ { n } } { 3 x _ { n } ^ { 2 } + 2 }
C) xn+1=xn−xn3+2xn−43xn2+2x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { 3 } + 2 x _ { n } - 4 } { 3 x _ { n } ^ { 2 } + 2 }
D) xn+1=xn+3xnx _ { n + 1 } = x _ { n } + \frac { 3 } { x _ { n } }
E) xn+1=xn3+2xn3xn2+2x _ { n + 1 } = \frac { x _ { n } ^ { 3 } + 2 x _ { n } } { 3 x _ { n } ^ { 2 } + 2 }
F) xn+1=xn+xn3+2xn3xn2+2x _ { n + 1 } = x _ { n } + \frac { x _ { n } ^ { 3 } + 2 x _ { n } } { 3 x _ { n } ^ { 2 } + 2 }

G) xn+1=xn+xn3+2xn−43xn2+2x _ { n + 1 } = x _ { n } + \frac { x _ { n } ^ { 3 } + 2 x _ { n } - 4 } { 3 x _ { n } ^ { 2 } + 2 }

H) None of these

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