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Question Number 122162 by naka3546 last updated on 14/Nov/20

F i n d t h e r e a l s o l u t i o n o f e q u a l i t y :

x^{3} - 2 x^{2} + 4 x - 1 = 0

P l e a s e s h o w y o u r w o r k i n g s!

Answered by mathmax by abdo last updated on 14/Nov/20

f (x) = x^{3} - {2 x}^{2} + 4 x - 1 \Rightarrow f^{^{'}} (x) = {3 x}^{2} - 4 x + 4

f^{^{'}} (x) = 0 \Rightarrow {3 x}^{2} - 4 x + 4 = 0 \to Δ^{^{'}} = 4 - 12 = - 8 < 0

a = 3 > 0 \Rightarrow f^{^{'}} (x) > 0 \Rightarrow f is strictly increasing on R

f (0) = - 1 and f (1) = 1 - 2 + 4 - 1 = 2 > 0

f (0) . f (- 1) < 0 \Rightarrow \exists! α \in] 0, 1 [/ f (α) = 0 let begin by x_{0} = \frac{1}{2}

and x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{^{'}} (x_{n})} (newton method)

x_{1} = x_{0} - \frac{f (x_{0})}{f^{^{'}} (x_{0})} = \frac{1}{2} - \frac{f (\frac{1}{2})}{f^{^{'}} (\frac{1}{2})} we have f (\frac{1}{2}) = \frac{1}{8} - \frac{1}{2} + 2 - 1

= \frac{1 - 4}{8} + 1 = \frac{- 3}{8} + 1 = \frac{5}{8}

f^{^{'}} (\frac{1}{2}) = \frac{3}{4} - 2 + 4 = \frac{3}{4} + 2 = \frac{11}{4} \Rightarrow x_{1} = \frac{1}{2} - \frac{5}{8} \times \frac{4}{11}

= \frac{1}{2} - \frac{5}{22} = \frac{11 - 5}{22} = \frac{6}{22} = \frac{3}{11}

x_{2} = x_{1} - \frac{f (x_{1})}{f^{^{'}} (x_{1})} = \frac{3}{11} - \frac{f (\frac{3}{11})}{f^{^{'}} (\frac{3}{11})} = \dots

x_{4} give a better approximation to this root \dots