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Question Number 66104 by Rio Michael last updated on 09/Aug/19

f (x) = 2 x^{3} - x - 4

s h o w t h a t t h e e q u a t i o n f (x) = 0 h a s r o o t b e t w e e n 1 a n d 2

s h o w t h a t t h e e q u a t i o n f (x) = 0 c a n b e w r i t t e n a s

x = \sqrt{(\frac{2}{x} + \frac{1}{2})}

u s e t h e i t e r a t i o n

x_{n + 1} = \sqrt{(\frac{2}{x_{n}} + \frac{1}{2}),}

w i t h x_{0} = 1.385 t o f i n d t o 3 d e c i m a l p l a c e s t h e v a l u e o f x_{1} .

Commented by mathmax by abdo last updated on 09/Aug/19

w e h a v e f (x) = 2 x^{3} - x - 4 \Rightarrow f^{^{'}} (x) = 6 x^{2} - 1 = (\sqrt{6} x - 1) (\sqrt{6} x + 1)

f^{^{'}} (x) = 0 \Leftrightarrow x = \bar{+} \frac{1}{\sqrt{6}} v a r i a t i o n o f f (x)

x - \infty - \frac{1}{\sqrt{6}} \frac{1}{\sqrt{6}} + \infty

f^{^{'}} (x) + - +

f (x) - \infty i n c r . f (- \frac{1}{\sqrt{6}}) d e c r f (\frac{1}{\sqrt{6}}) i n c r + \infty

f i s i n c r e a s i n g o n [\frac{1}{\sqrt{6}}, + \infty [

f (1) = 2 - 1 - 4 = - 5

f (2) = 16 - 2 - 4 = 10 \Rightarrow f (1) f (2) < 0 \Rightarrow \exists α \in] 1, 2 [/ f (α) = 0

x = \sqrt{\frac{2}{x} + \frac{1}{2}} a n d x > 0 \Rightarrow x^{2} = \frac{4 + x}{2 x} \Rightarrow 2 x^{3} = 4 + x \Rightarrow 2 x^{3} - x - 4 = 0 \Rightarrow

f (x) = 0 s o f (x) = 0 a n d x > 0 \Leftrightarrow x = \sqrt{\frac{2}{x} + \frac{1}{2}}

i f w e c o n s i d e r t h e i t e r a t i o n x_{n + 1} = \sqrt{\frac{2}{x_{n}} + \frac{1}{2}}

w e g e t x_{1} = \sqrt{\frac{2}{x_{0}} + \frac{1}{2}} = \sqrt{\frac{4 + x_{0}}{2 x_{0}}} = \sqrt{\frac{4 + 1, 385}{2 \times 1, 385}}

r e s t t o f i n i s h t h e c a l c u l u s \dots .