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Question Number 40407 by prof Abdo imad last updated on 21/Jul/18
![let f(x)= x^3 −x−1 1) prove that ∃ α ∈ ]1,2[ /f(α)=0 2) use the newton method with x_0 =(3/2) to find a better value for α (take onlly 5 terms)](Q40407.png)
$${let}\:{f}\left({x}\right)=\:{x}^{\mathrm{3}} −{x}−\mathrm{1} \\ $$$$\left.\mathrm{1}\left.\right)\:{prove}\:{that}\:\exists\:\alpha\:\in\:\right]\mathrm{1},\mathrm{2}\left[\:/{f}\left(\alpha\right)=\mathrm{0}\right. \\ $$$$\left.\mathrm{2}\right)\:{use}\:{the}\:{newton}\:{method}\:\:{with}\:{x}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${to}\:{find}\:{a}\:{better}\:{value}\:{for}\:\alpha\:\left({take}\:{onlly}\:\mathrm{5}\:{terms}\right) \\ $$
Answered by maxmathsup by imad last updated on 25/Jul/18
![1) we have f^′ (x)=3x^2 −1 >0 on interval ]1,2[ so is increasing on ]1,2[ f(1)=1−1−1=−1 and f(2)=8−3=5 ⇒f(1).f(2)<0 ⇒∃ ! α ∈]1,2[ /f(α)=0 2) we have x_0 =(3/2) and x_(n+1) =x_n −((f(x_n ))/(f^′ (x_n ))) ⇒ x_1 =x_0 −((f(x_o ))/(f^′ (x_0 ))) =(3/2) −((f((3/2)))/(f^′ ((3/2)))) but f((3/2))=((3/2))^3 −(3/2) −1 =((27)/8) −(5/2) =((27−20)/8)=(7/8) f^′ ((3/2)) =3((3/2))^2 −1 =((27)/4) −1 = ((23)/4) ⇒ x_1 =(3/2) −((7/8)/((23)/4)) =(3/2) −(7/8) .(4/(23)) =(3/2) −(7/(46)) =((138−14)/(92)) =((124)/(92)) =((62)/(46)) =((31)/(23)) x_2 =x_1 −((f(x_1 ))/(f^′ (x_1 ))) =((31)/(23)) −((f(((31)/(23))))/(f^′ (((31)/(23))))) .....be continued...](Q40651.png)
$$\left.\mathrm{1}\left.\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\:>\mathrm{0}\:{on}\:{interval}\:\right]\mathrm{1},\mathrm{2}\left[\:{so}\:{is}\:{increasing}\:{on}\:\right]\mathrm{1},\mathrm{2}\left[\right. \\ $$$$\left.{f}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{1}−\mathrm{1}=−\mathrm{1}\:{and}\:{f}\left(\mathrm{2}\right)=\mathrm{8}−\mathrm{3}=\mathrm{5}\:\Rightarrow{f}\left(\mathrm{1}\right).{f}\left(\mathrm{2}\right)<\mathrm{0}\:\:\Rightarrow\exists\:!\:\alpha\:\in\right]\mathrm{1},\mathrm{2}\left[\:/{f}\left(\alpha\right)=\mathrm{0}\right. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{x}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{2}}\:{and}\:{x}_{{n}+\mathrm{1}} \:={x}_{{n}} \:−\frac{{f}\left({x}_{{n}} \right)}{{f}^{'} \left({x}_{{n}} \right)}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} ={x}_{\mathrm{0}} \:−\frac{{f}\left({x}_{{o}} \right)}{{f}^{'} \left({x}_{\mathrm{0}} \right)}\:=\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{{f}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{{f}^{'} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\:{but}\:{f}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} \:−\frac{\mathrm{3}}{\mathrm{2}}\:−\mathrm{1}\:=\frac{\mathrm{27}}{\mathrm{8}}\:−\frac{\mathrm{5}}{\mathrm{2}}\:=\frac{\mathrm{27}−\mathrm{20}}{\mathrm{8}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${f}^{'} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \:−\mathrm{1}\:=\frac{\mathrm{27}}{\mathrm{4}}\:−\mathrm{1}\:=\:\frac{\mathrm{23}}{\mathrm{4}}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\frac{\mathrm{7}}{\mathrm{8}}}{\frac{\mathrm{23}}{\mathrm{4}}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{7}}{\mathrm{8}}\:.\frac{\mathrm{4}}{\mathrm{23}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{7}}{\mathrm{46}}\:=\frac{\mathrm{138}−\mathrm{14}}{\mathrm{92}}\:=\frac{\mathrm{124}}{\mathrm{92}}\:\:=\frac{\mathrm{62}}{\mathrm{46}}\:=\frac{\mathrm{31}}{\mathrm{23}} \\ $$$${x}_{\mathrm{2}} ={x}_{\mathrm{1}} \:−\frac{{f}\left({x}_{\mathrm{1}} \right)}{{f}^{'} \left({x}_{\mathrm{1}} \right)}\:=\frac{\mathrm{31}}{\mathrm{23}}\:−\frac{{f}\left(\frac{\mathrm{31}}{\mathrm{23}}\right)}{{f}^{'} \left(\frac{\mathrm{31}}{\mathrm{23}}\right)}\:\:\:.....{be}\:{continued}... \\ $$