Question Number 77872 by berket last updated on 11/Jan/20
![show that f(x)=2r^3 +5x−1 has a zero in the interval [0.1].](Q77872.png)
$${show}\:{that}\:{f}\left({x}\right)=\mathrm{2}{r}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{1}\:{has}\:{a}\:{zero}\:{in}\:{the}\:{interval}\:\left[\mathrm{0}.\mathrm{1}\right]. \\ $$
Commented by key of knowledge last updated on 11/Jan/20
$$\mathrm{r}^{\mathrm{3}} \:\mathrm{or}\:\mathrm{x}^{\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 11/Jan/20
![f(x)=2x^3 +5x−1 ⇒f^′ (x)=6x^2 +5 >0 ⇒f is increasing on R f(0)=−1<0 and f(1)=2+5−1 =6>0 ⇒∃! α_0 ∈]0,1[ /f(α_0 )=0 we can use newton method by taking x_0 =(1/2) and x_(n+1) =x_n −((f(x_n ))/(f^′ (x_n ))) ⇒x_1 =x_0 −((f(x_0 ))/(f^′ (x_0 )))=(1/2)−((f((1/2)))/(f^′ ((1/2)))) f((1/2))=(1/4)+(5/2)−1 =((1+10−4)/4) =(7/4) f^′ ((1/2))=(6/4)+5 =(3/2)+5 =((13)/2) ⇒x_1 =(1/2)−((7/4)/((13)/2)) =(1/2)−(7/4)×(2/(13)) =(1/2)−(7/(26)) x_1 =((13−7)/(26)) =(6/(26)) =(3/(13)) ⇒x_1 ∼0,23 we fllow the same way to calculate x_2 ,x_3 ,x_4 ,...](Q77888.png)
$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{1}\:\Rightarrow{f}^{'} \left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} \:+\mathrm{5}\:>\mathrm{0}\:\Rightarrow{f}\:{is}\:{increasing}\:{on}\:{R} \\ $$$$\left.{f}\left(\mathrm{0}\right)=−\mathrm{1}<\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)=\mathrm{2}+\mathrm{5}−\mathrm{1}\:=\mathrm{6}>\mathrm{0}\:\Rightarrow\exists!\:\alpha_{\mathrm{0}} \:\in\right]\mathrm{0},\mathrm{1}\left[\:/{f}\left(\alpha_{\mathrm{0}} \right)=\mathrm{0}\right. \\ $$$${we}\:{can}\:{use}\:{newton}\:{method}\:\:{by}\:{taking}\:{x}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\:{and} \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} −\frac{{f}\left({x}_{{n}} \right)}{{f}^{'} \left({x}_{{n}} \right)}\:\Rightarrow{x}_{\mathrm{1}} ={x}_{\mathrm{0}} −\frac{{f}\left({x}_{\mathrm{0}} \right)}{{f}^{'} \left({x}_{\mathrm{0}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{1}\:=\frac{\mathrm{1}+\mathrm{10}−\mathrm{4}}{\mathrm{4}}\:=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$${f}^{'} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{6}}{\mathrm{4}}+\mathrm{5}\:=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{5}\:=\frac{\mathrm{13}}{\mathrm{2}}\:\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\frac{\mathrm{7}}{\mathrm{4}}}{\frac{\mathrm{13}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{4}}×\frac{\mathrm{2}}{\mathrm{13}}\:=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{26}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{13}−\mathrm{7}}{\mathrm{26}}\:=\frac{\mathrm{6}}{\mathrm{26}}\:=\frac{\mathrm{3}}{\mathrm{13}}\:\Rightarrow{x}_{\mathrm{1}} \sim\mathrm{0},\mathrm{23}\:\:\:{we}\:{fllow}\:{the}\:{same}\:{way}\:{to}\:{calculate} \\ $$$${x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,{x}_{\mathrm{4}} ,... \\ $$
Commented by jagoll last updated on 12/Jan/20
$${oo}\:{yes}.\:{it}\:{my}\:{typo}.\:{i}\:{correct} \\ $$
Commented by jagoll last updated on 12/Jan/20
$${haha}\:\:{same}\:{sir}\:\frac{\mathrm{277}×\mathrm{8}}{\mathrm{432}×\mathrm{8}}=\frac{\mathrm{2216}}{\mathrm{3456}} \\ $$
Commented by jagoll last updated on 12/Jan/20
$${ohh}\:{yes}\:{sir}\:\:{i}'{m}\:{typo}\:{againt}\:.\:{hahaha} \\ $$
Commented by jagoll last updated on 12/Jan/20
$${thank}\:{you}\:{for}\:{cheking},\:{sir}.\:{i}\:{tried} \\ $$$${to}\:{work}\:{with}\:{the}\:{Cardano}\: \\ $$$${process}.\:{it}\:{turns}\:{out}\:{there}\:{was}\: \\ $$$${a}\:{calculation}\:{error}.\:{thank}\:{you}\:{sir} \\ $$
Answered by Rio Michael last updated on 11/Jan/20
![say f(x) = 2x^3 + 5x − 1 f(0) = 2(0)^3 +5(0) − 1= −1 f(1) = 2 + 5 − 1 = 6 a change of sign from − to + in the interval [0,1] shows that f(x) has a root or zero in this interval.](Q77880.png)
$${say}\:{f}\left({x}\right)\:=\:\mathrm{2}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}\:−\:\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\mathrm{2}\left(\mathrm{0}\right)^{\mathrm{3}} +\mathrm{5}\left(\mathrm{0}\right)\:−\:\mathrm{1}=\:−\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{2}\:+\:\mathrm{5}\:−\:\mathrm{1}\:=\:\mathrm{6} \\ $$$${a}\:{change}\:{of}\:{sign}\:{from}\:−\:{to}\:+\:{in}\:{the} \\ $$$${interval}\:\left[\mathrm{0},\mathrm{1}\right]\:{shows}\:{that}\:{f}\left({x}\right)\:{has}\: \\ $$$${a}\:{root}\:{or}\:{zero}\:{in}\:{this}\:{interval}. \\ $$
Answered by mr W last updated on 11/Jan/20
$$\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{831}}}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{4}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{831}}}{\mathrm{36}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\approx\mathrm{0}.\mathrm{1969} \\ $$
Commented by john santu last updated on 13/Jan/20
$${how}\:{get}\:\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }+\:\frac{\mathrm{1}}{\mathrm{4}}}...? \\ $$$${i}\:{got}\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 13/Jan/20
