Question

7)f(x)= 10 -x^3-3cosx, estimate therootusing secant method up to 3 significant

Answer 1

given fra)= 10-13-30054 7 0 2 3 7.3751 3.24841-14-03 Root Is lies between 2013 1st Iteration 2-188 let ao=2:21=3 fero)=f(2)= 3.2484 : $ $(2,7=463)=-14.03 22 = 20 462,0–2,6620) 26-14:03)-3(3.2484) $(7,7--f(20) -14.03 — 3.2484 fc2a) = f(2-16€) =(-2-18532- 3605(2-188)+10 = 1.2615 and Iteralios: let 7,= 3 : 23-2-188 f(20) = f(3) = -14-03 : f(na) = f(2-186) – 1.2615 23= a fras)- 22 fca,)3(1-261S) – 21856–1401) fras)-f(21) 2.255 1-2615-(-14.03) f(+3) = f(2-255) = 0.4295

and Iteration 4 = 2-688 : 213=2255 f(n) = f(2-18€) = 1.2615; $(33) = $12-255)= 0.4295 A4 = 22 f(na)– 13 frag) 2-168(0.4295)-2.255(1-315) = 2.2896 0.4295-1-2615 f(12) -f(%) 415 Iteration 22=2255 & ny=2.2856 $(29) - f(a.ass) = 0.4295 8 fay) = f28:9890) = -0.0269 Das = 72 fcay)- auf(ns) 3.2551-0,0269)-2389660.485) flow-fr29) -0.0269-0.4295 =) 95= 2.2875 v 515 Iteration 24 = 2.2896: 45=2-2875 frau) = f(7-1896)=-0.0165 $ $(25)=f63.9875)= 0.0005 As = xu f'(as) – as f(x4)_Que36.60.0005)-&#81-0.0245) fras) - frey) 0.0005-(-0.0269) = 2.2876 → Approximate 25=26=27876 root is 2-2876 fhankeyn.