Menu Close

let-f-x-x-3-x-1-1-prove-that-1-2-f-0-2-use-the-newton-method-with-x-0-3-2-to-find-a-better-value-for-take-onlly-5-terms-




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)
letf(x)=x3x11)provethatα]1,2[/f(α)=02)usethenewtonmethodwithx0=32tofindabettervalueforα(takeonlly5terms)
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...
1)wehavef(x)=3x21>0oninterval]1,2[soisincreasingon]1,2[f(1)=111=1andf(2)=83=5f(1).f(2)<0!α]1,2[/f(α)=02)wehavex0=32andxn+1=xnf(xn)f(xn)x1=x0f(xo)f(x0)=32f(32)f(32)butf(32)=(32)3321=27852=27208=78f(32)=3(32)21=2741=234x1=3278234=3278.423=32746=1381492=12492=6246=3123x2=x1f(x1)f(x1)=3123f(3123)f(3123)..becontinued

Leave a Reply

Your email address will not be published. Required fields are marked *