show that f(x)=2r^3 +5x−1 has a zero in the interval [0.1].


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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

$f (x) = 2 x^{3} + 5 x - 1 \Rightarrow f^{^{'}} (x) = 6 x^{2} + 5 > 0 \Rightarrow f i s i n c r e a s i n g o n R$ $f (0) = - 1 < 0 a n d f (1) = 2 + 5 - 1 = 6 > 0 \Rightarrow \exists! α_{0} \in] 0, 1 [/ f (α_{0}) = 0$ $w e c a n u s e n e w t o n m e t h o d b y t a k i n g x_{0} = \frac{1}{2} a n d$ $x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{^{'}} (x_{n})} \Rightarrow x_{1} = x_{0} - \frac{f (x_{0})}{f^{^{'}} (x_{0})} = \frac{1}{2} - \frac{f (\frac{1}{2})}{f^{^{'}} (\frac{1}{2})}$ $f (\frac{1}{2}) = \frac{1}{4} + \frac{5}{2} - 1 = \frac{1 + 10 - 4}{4} = \frac{7}{4}$ $f^{^{'}} (\frac{1}{2}) = \frac{6}{4} + 5 = \frac{3}{2} + 5 = \frac{13}{2} \Rightarrow x_{1} = \frac{1}{2} - \frac{\frac{7}{4}}{\frac{13}{2}} = \frac{1}{2} - \frac{7}{4} \times \frac{2}{13} = \frac{1}{2} - \frac{7}{26}$ $x_{1} = \frac{13 - 7}{26} = \frac{6}{26} = \frac{3}{13} \Rightarrow x_{1} \sim 0, 23 w e f l l o w t h e s a m e w a y t o c a l c u l a t e$ $x_{2}, x_{3}, x_{4}, . . .$
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

	$h o w g e t \sqrt[3]{\sqrt{{(\frac{5}{6})}^{3} + {(\frac{1}{4})}^{2}} + \frac{1}{4}} . . . ?$ $i g o t - \frac{1}{4}$
	Commented by mr W last updated on 13/Jan/20

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