Ex2-_graficos_e_bisseccao_waylla[1]

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UEMASUL UNIVERSIDADE ESTADUAL DA REGIÃO TOCANTINA DO MARANHÃO CCHSTL - CAMPUS AÇAILÂNDIA LISTA DE EXERCÍCIOS - MATEMÁTICA COMPUTACIONAL WAYLLA ANDRADE DOS SANTOS [05] us gráficos y = y = tg(x). Em seguda utilize da com precisão de 10-5 do valor w de entre as duas curvas. 2 = no tg(x) no 2 [ 2 tal que X = = - = f(4,2) = - 2,422220225 (4,7) = no = 76,01276297 = - (tem raiz) 1. 4,2+4,7 = 4,45 = 2 f(x) = 0,726731427 X E 2. = 4,45+4,7 = 4,575 = 4,575 2 E 3. X = 4,45+ 4,575 = 4,5125 = (4,5125) 2 = 4,5125] 4. X = 4,45 + 4,5125 = 4,48125 2 f(x)= - 0,232170781 = E [ 4,5125]= # 7 X to = (X) - = C 10. X X= = 4,49296875+ 4,493945313 = 4,493457032 E 4,493945313] - = 2 = + = * = 7 X - = 2 8. X = 4,49296875+ + 4,496875 = 4,4949 21875 - E- = [4,49296875; 4,496875] B7 = C = = * - = ] X = (X) 4 = + = = = 7 - = 5. 4,48125 + 4,5125 = 4,496875f(x)= (4,493212891) - 4,493212891 = - 3,965153304.1 X E 4,493457032] = *12. X = 4,493212891 + 4,493457032 = 4,493334962 2 = (4,493334962) - 4,49 3334 962 in - 10⁻³ E [4,493334962; 4,493,457032] X 13. X= = 4,493457032 = 4,493395997 2 = tg 4,493395997 f(x)= = - 2,71768313 XE [ 4,493395997; 4,493457032] X= 4,493395997 + 4,493457032 = 4,4934 26515 2 f(x) = tg (4,493426515) - 4,493426515 = 3,44412197. 15. X= 4,493395997 + 4,493426515 = 4,493411256 2 = tg - (4,493411256 f(x) = 3,6305067 10-5 [ 4,493395997; 4,493411256] # 16. 4,493395997 + 4,493411256 = 2 f(x) = to (4,493403627) - 4,493403627 = - [4,493403627;UEMASUL TABELA RESUMO ITERAÇÃO X F(x) |b-a| 1 4,45 -01726731427 0,2 2 4,575 2,65775007 0,125 3 4,5125 0,0625 4 4,48125 0,03125 5 4,496875 0,071133786 0,015625 6 4,4890625 7 4,49296875 8 4,494921875 1,953125.10⁻³ 9 4,493945313 0,010846712 9,76563.10⁻⁴ 10 4,493457032 11 4,4932 12891 -3,965153304.10³ 2,44141.10⁻⁴ 12 4,493334962 1,2207.10-4 13 4,493395997 6,1035.10-5 14 4,4934 26515 15 4,493411256 16 4,49 3403627 4 7,629.10⁻⁶5 (4.49341, 4.49341) -10 -5 5 10 (0,0) (-4.49341, -4.49341) -5[06] us de = 2 e Em utilize metodo da com ole 10-5 clo valor positive de as duas no [0,5; 1,5]. no ex-2 = no = no ex-2 = Cos = cos f(a). f(b) f (0,5) = - 1,290212201 (a) (1,5) = 3,271740413 (b) = - 4,221239399 raiz) 1. = 0,5+1,5 = 1 2 f(x) = e-2- 1 - 2 (x) = X E 1,5] 2. = 1,25 f(x) = e 1,25 Cos 2) 2 f(x) = 1,409976352 E 3. = 1,125 f(x) = e 1,125 - - 2) 2 = 4. X = 1 + = 1,0625 f(x) = e 1,0625 - - ( 1,0625 - 2) 2 f (x) = 0,2669822876 X = 1 + 1,0625 = 1,03125 f (x) = e 1,03125 - - Cos 1,03125 - 2) =1+ 1,03125 = 1,015625 2 X E ] f(x) = e 1,015625 1015625 - 2) = 0,0370028746 " *7. X= 1+ 1,015625 = 1,0078125 2 2 f(x) = e - - cos - *8. = 1+ = 1,00390625 2 = 1,00390625 - 2) X E 1,0078125] = - 0,01697 27169 1,00390625+ 1,0078125 = 1,005859375 2 f(x) = e 1,005859375 1,005859375 -2 - - cos - -2) f (x) = 7" E [ * 10. X = = 1,0068359375 2 f(x) = e - cos ( 1,0068359375 - 2) = - 0,0036 09334 11 X E [ 1,0068359375; 11. X = + = 2 = e 1,00732421875 1,00732421875 - f(x)= -] - - = 2 = + = X ] X = = - - a = [ I Э = - - C- =(x)f C = + T = * = - - - C - a = C = =16. 1,0075988769531+ = 2 f(x)= e 1,0076141357422 - 2 Coo (t -2) - = E [ 1,0076141357422; X 17. X = 1,0076293945313 = 2 = - - 1,0076217651367 - 2) f(x) = - 7" 18. X = + 1,0076293945313 = 2 f(x)= = e 1,0076255798340 - - cos - 2) = 1,0076255798340]UEMASUL TABELA RESUMO ITERAÇÃO X F(x) |b-a| 1 1 05 2 1,25 1,409976 352 0,25 3 1,125 0,609079747 0,125 4 1,0625 0,0625 5 1,03125 0,111147 7643 0,03125 6 1,015625 0,0370028746 0,0515625 7 1,0078125 8 1,00390625 9 1,005859375 10 4,0068 359375 -0,0036093348 11 1,00732421875 -0,0013736620 12 1,0075 68359375 2,44141.10⁻⁴ 13 1,0076904296875 0,0003046770 1,2207.10-4 14 1,0076293945313 15 16 1,0076 141357422 17 1,0076217651367 7,629.10-6 18 1,00 76255798340 0,00000 73722 3,814.10⁻⁶1 (1.00762, 0.73909) -2 0 2 3