let-f-x-x-3-x-1-1-prove-that-1-2-f-0-2-use-the-newton-method-with-x-0-3-2-to-find-a-better-value-for-take-onlly-5-terms-

Question Number 40407 by prof Abdo imad last updated on 21/Jul/18

l e t f (x) = x^{3} - x - 1

1) p r o v e t h a t \exists α \in] 1, 2 [/ f (α) = 0

2) u s e t h e n e w t o n m e t h o d w i t h x_{0} = \frac{3}{2}

t o f i n d a b e t t e r v a l u e f o r α (t a k e o n l l y 5 t e r m s)

Answered by maxmathsup by imad last updated on 25/Jul/18

1) w e h a v e f^{^{'}} (x) = 3 x^{2} - 1 > 0 o n i n t e r v a l] 1, 2 [s o i s i n c r e a s i n g o n] 1, 2 [

f (1) = 1 - 1 - 1 = - 1 a n d f (2) = 8 - 3 = 5 \Rightarrow f (1) . f (2) < 0 \Rightarrow \exists! α \in] 1, 2 [/ f (α) = 0

2) w e h a v e x_{0} = \frac{3}{2} a n d x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{^{'}} (x_{n})} \Rightarrow

x_{1} = x_{0} - \frac{f (x_{o})}{f^{^{'}} (x_{0})} = \frac{3}{2} - \frac{f (\frac{3}{2})}{f^{^{'}} (\frac{3}{2})} b u t f (\frac{3}{2}) = {(\frac{3}{2})}^{3} - \frac{3}{2} - 1 = \frac{27}{8} - \frac{5}{2} = \frac{27 - 20}{8} = \frac{7}{8}

f^{^{'}} (\frac{3}{2}) = 3 {(\frac{3}{2})}^{2} - 1 = \frac{27}{4} - 1 = \frac{23}{4} \Rightarrow

x_{1} = \frac{3}{2} - \frac{\frac{7}{8}}{\frac{23}{4}} = \frac{3}{2} - \frac{7}{8} . \frac{4}{23} = \frac{3}{2} - \frac{7}{46} = \frac{138 - 14}{92} = \frac{124}{92} = \frac{62}{46} = \frac{31}{23}

x_{2} = x_{1} - \frac{f (x_{1})}{f^{^{'}} (x_{1})} = \frac{31}{23} - \frac{f (\frac{31}{23})}{f^{^{'}} (\frac{31}{23})} \dots . . b e c o n t i n u e d \dots