show-that-f-x-2r-3-5x-1-has-a-zero-in-the-interval-0-1-

Question Number 77872 by berket last updated on 11/Jan/20

s h o w t h a t f (x) = 2 r^{3} + 5 x - 1 h a s a z e r o i n t h e i n t e r v a l [0.1] .

Commented by key of knowledge last updated on 11/Jan/20

r^{3} or x^{3}

Commented by mathmax by abdo last updated on 11/Jan/20

f (x) = 2 x^{3} + 5 x - 1 \Rightarrow f^{^{'}} (x) = 6 x^{2} + 5 > 0 \Rightarrow f i s i n c r e a s i n g o n R

f (0) = - 1 < 0 a n d f (1) = 2 + 5 - 1 = 6 > 0 \Rightarrow \exists! α_{0} \in] 0, 1 [/ f (α_{0}) = 0

w e c a n u s e n e w t o n m e t h o d b y t a k i n g x_{0} = \frac{1}{2} a n d

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{^{'}} (x_{n})} \Rightarrow x_{1} = x_{0} - \frac{f (x_{0})}{f^{^{'}} (x_{0})} = \frac{1}{2} - \frac{f (\frac{1}{2})}{f^{^{'}} (\frac{1}{2})}

f (\frac{1}{2}) = \frac{1}{4} + \frac{5}{2} - 1 = \frac{1 + 10 - 4}{4} = \frac{7}{4}

f^{^{'}} (\frac{1}{2}) = \frac{6}{4} + 5 = \frac{3}{2} + 5 = \frac{13}{2} \Rightarrow x_{1} = \frac{1}{2} - \frac{\frac{7}{4}}{\frac{13}{2}} = \frac{1}{2} - \frac{7}{4} \times \frac{2}{13} = \frac{1}{2} - \frac{7}{26}

x_{1} = \frac{13 - 7}{26} = \frac{6}{26} = \frac{3}{13} \Rightarrow x_{1} \sim 0, 23 w e f l l o w t h e s a m e w a y t o c a l c u l a t e

x_{2}, x_{3}, x_{4}, \dots

Commented by jagoll last updated on 12/Jan/20

o o y e s . i t m y t y p o . i c o r r e c t

Commented by jagoll last updated on 12/Jan/20

h a h a s a m e s i r \frac{277 \times 8}{432 \times 8} = \frac{2216}{3456}

Commented by jagoll last updated on 12/Jan/20

o h h y e s s i r i^{'} m t y p o a g a i n t . h a h a h a

Commented by jagoll last updated on 12/Jan/20

t h a n k y o u f o r c h e k i n g, s i r . i t r i e d

t o w o r k w i t h t h e C a r d a n o

p r o c e s s . i t t u r n s o u t t h e r e w a s

a c a l c u l a t i o n e r r o r . t h a n k y o u s i r

Answered by Rio Michael last updated on 11/Jan/20

s a y f (x) = 2 x^{3} + 5 x - 1

f (0) = 2 {(0)}^{3} + 5 (0) - 1 = - 1

f (1) = 2 + 5 - 1 = 6

a c h a n g e o f s i g n f r o m - t o + i n t h e

i n t e r v a l [0, 1] s h o w s t h a t f (x) h a s

a r o o t o r z e r o i n t h i s i n t e r v a l .

Answered by mr W last updated on 11/Jan/20

2 x^{3} + 5 x - 1 = 0

x = \sqrt[3]{\sqrt{{(\frac{5}{6})}^{3} + {(\frac{1}{4})}^{2}} + \frac{1}{4}} - \sqrt[3]{\sqrt{{(\frac{5}{6})}^{3} + {(\frac{1}{4})}^{2}} - \frac{1}{4}}

= \sqrt[3]{\frac{\sqrt{831}}{36} + \frac{1}{4}} - \sqrt[3]{\frac{\sqrt{831}}{36} - \frac{1}{4}}

\approx 0.1969

Commented by john santu last updated on 13/Jan/20

h o w g e t \sqrt[3]{\sqrt{{(\frac{5}{6})}^{3} + {(\frac{1}{4})}^{2}} + \frac{1}{4}} \dots ?

i g o t - \frac{1}{4}

Commented by mr W last updated on 13/Jan/20

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