ITERACION TAREA 2 MOSHEH DAVID BENITEZ MENDOZA

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ITERACIONES 
Segunda raíz del ejercicio de clase 
DESCRIPCIÓN BREVE 
Aplicar paso por paso el método de 
bisección a las siguientes funciones 
hasta alcanzar el criterio de paro para 
aproximarse a la raíz 
Mosheh David Benitez Mendoza 
10050549 
 
PROBLEMA 1 
Aproximar a la raíz: 
𝐹(𝑥) = −0.874𝑥2 + 1.75𝑥 + 2.627 
Datos de entrada: 
𝑥1 = 2 
 𝑥2 = 4 
Criterio de Paro 
𝜀 = 1.1 
 
 
ITERACIÓN 1 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
2 + 4
2
= 3 
 
𝑓(𝑥1) = −0.874(2)
2 + 1.75(2) + 2.627 = 2.631 
𝐹(𝑥𝑟) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (2.631) ∗ (0.011) = 0.028941 > 0 𝑛𝑜 ℎ𝑎𝑦 𝑟𝑎𝑖𝑧 
 
ITERACIÓN 2 
𝑥1 = 3 
𝑥2 = 4 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
3 + 4
2
= 3.5 
𝐸𝑎 = |
3.5 − 3
3.5
| ⋅ 100 = 14.2857% 
 
𝑓(𝑥1) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝐹(𝑥𝑟) = −0.874(3.5)
2 + 1.75(3.5) + 2.627 = −1.9545 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (0.011) ∗ (−1.9545) = −0.021499 
 
 
ITERACIÓN 3 
𝑥1 = 3 
𝑥2 = 3.5 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
3 + 3.5
2
= 3.25 
 
𝐸𝑎 = |
3.25 − 3.5
3.25
| ⋅ 100 = 7.6923% 
 
𝑓(𝑥1) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝐹(𝑥𝑟) = −0.874(3.25)
2 + 1.75(3.25) + 2.627 = −0.917125 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (0.011) ∗ (−1.9545) = −0.0100 
 
ITERACIÓN 4 
𝑥1 = 3 
𝑥2 = 3.25 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
3 + 3.25
2
= 3.125 
 
𝐸𝑎 = |
3.125 − 3.25
3.125
| ⋅ 100 = 4% 
 
𝑓(𝑥1) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝐹(𝑥𝑟) = −0.874(3.125)
2 + 1.75(3.125) + 2.627 = −0.439406 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (0.011) ∗ (= −0.439406) = −4.83 ∗ 10
−3 
 
 
ITERACIÓN 5 
𝑥1 = 3 
𝑥2 = 3.125 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
3 + 3.125
2
= 3.0625 
 
𝐸𝑎 = |
3.0625 − 3.125
3.0625
| ⋅ 100 = 2.04% 
 
𝑓(𝑥1) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝐹(𝑥𝑟) = −0.874(3.0625)
2 + 1.75(3.0625) + 2.627 = −0.2107 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (0.011) ∗ (−0.2107) = −4.3186 ∗ 10
−3 
 
ITERACIÓN 6 
𝑥1 = 3 
𝑥2 = 3.0625 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
3 + 3.0625
2
= 3.03125 
𝐸𝑎 = |
3.03125 − 3.0625
3.03125
| ⋅ 100 = 1.0309% 
 
𝑓(𝑥1) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝐹(𝑥𝑟) = −0.874(3.03125)
2 + 1.75(3.03125) + 2.627 = −0.0990 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (0.011) ∗ (−0.0990) = −1.08945 ∗ 10
−3 
 
ITERACIÓN 7 
𝑥1 = 3 
𝑥2 = 3.03125 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
3 + 3.0625
2
= 3.015625 
𝐸𝑎 = |
3.015625 − 3.03125
3.015625
| ⋅ 100 = 0.518134% 
 
𝑓(𝑥1) = −0.874(3)
2 + 1.75(3) + 2.627 = 0.011 
𝐹(𝑥𝑟) = −0.874(3.03125)
2 + 1.75(3.03125) + 2.627 = −0.043807 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (0.011) ∗ (−0.0990) = −4.8187 ∗ 10
−4 
𝑟𝑎𝑖𝑧 𝑥𝑟 = 3.015625 
 
 
 
 
 
PROBLEMA 2 
Aproximar a la raíz 𝑓(𝑥) = −2.1 + 6.21𝑥 − 3.9𝑥2 + 0.667𝑥3 
Datos de entrada: 
Intervalo donde se encuentra la raíz: 
𝑋1 = 0.3 
𝑋2 = 0.5 
Criterio de paro 
Épsilon = 0.01 
 
ITERACION 1 
 
𝑥1 = 0.3 
𝑥2 = 0.5 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.3 + 0.5
2
= 0.4 
 
𝑓(𝑥1) = −2.1 + 6.21(0.3) − 3.9(0.3)2 + 0.667(0.3)3 = −0.569991 
𝑓(𝑥1) = −2.1 + 6.21(0.4) − 3.9(0.4)2 + 0.667(0.4)3 = −0.197312 
 
(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.569991) ∗ (−0.197312) = 0.112466 
 
ITERACION 2 
 
𝑥1 = 0.4 
𝑥2 = 0.5 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4 + 0.5
2
= 0.45 
 
𝐸𝑎 = |
0.45 − 0.4
0.45
| ⋅ 100 = 11.11% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.4) − 3.9(0.4)2 + 0.667(0.4)3 = −0.197312 
𝑓(𝑥1) = −2.1 + 6.21(0.45) − 3.9(0.45)2 + 0.667(0.45)3 = −0.034469 | 
 
(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.569991) ∗ (−0.197312) = 0.112466 
 
ITERACION 3 
 
𝑥1 = 0.45 
𝑥2 = 0.5 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.45 + 0.5
2
= 0.475 
𝐸𝑎 = |
0.45 − 0.4
0.45
| ⋅ 100 = 5.26% 
𝑓(𝑥1) = −2.1 + 6.21(0.45) − 3.9(0.45)
2 + 0.667(0.45)3 = −0.034469 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.475) − 3.9(0.475)
2 + 0.667(0.475)3 = 0.041229 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.034469) ∗ (0.041229) = −1.4234 ∗ 10
−3 
 
ITERACION 4 
 
𝑥1 = 0.45 
𝑥2 = 0.475 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.45 + 0.475
2
= 0.4625 
𝐸𝑎 = |
0.4625 − 0.475
0.4625
| ⋅ 100 = 2.70% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.45) − 3.9(0.45)
2 + 0.667(0.45)3 = −0.034469 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.4625) − 3.9(0.4625)
2 + 0.667(0.4625)3 = 3.8780 ∗ 103 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.034469) ∗ (3.8780 ∗ 10
3) = −1.336 ∗ 10−4 
 
ITERACION 5 
 
𝑥1 = 0.45 
𝑥2 = 0.4625 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
 
𝑥𝑟 =
0.45 + 0.4625
2
= 0.45625 
 
𝐸𝑎 = |
0.45625 − 0.4625
0.45625
| ⋅ 100 = 1.36% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.45) − 3.9(0.45)
2 + 0.667(0.45)3 = −0.034469 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.45625) − 3.9(0.45625)
2 + 0.667(0.45625)3 = −0.141875 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.034469) ∗ (−0.128096) = 4.89 ∗ 10
−3 
 
ITERACION 6 
 
𝑥1 = 0.45625 
𝑥2 = 0.4625 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.45625 + 0.4625
2
= 0.459375 
 
𝐸𝑎 = |
0.459375 − 0.45625
0.459375
| ⋅ 100 = 0.68% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.45625) − 3.9(0.45625)
2 + 0.667(0.45625)3 = −0.141875 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.459375) − 3.9(0.459375)
2 + 0.667(0.459375)3 = −5.6214 ∗ 10−3 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.141875) ∗ (−5.6214 ∗ 10
−3) = 7.975 ∗ 10−4 
 
ITERACION 7 
 
𝑥1 = 0.459375 
𝑥2 = 0.4625 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.459375 + 0.4625
2
= 0.4609375 
 
𝐸𝑎 = |
0.4609375 − 0.459375
0.4609375
| ⋅ 100 = 0.338% 
 
(𝑥1) = −2.1 + 6.21(0.459375) − 3.9(0.459375)
2 + 0.667(0.459375)3 = −5.6214 ∗ 10−3 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.4609375) − 3.9(0.4609375)
2 + 0.667(0.4609375)3 = −8.64 ∗ 10−4 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−0.128096) ∗ (8.60 ∗ 10
−3) = 4.8593 ∗ 10−6 
 
ITERACION 8 
𝑥1 = 0.4609375 
𝑥2 = 0.4625 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4609375 + 0.4625
2
= 0.46171875 
 
𝐸𝑎 = |
0.46171875 − 0.4609375
0.46171875
| ⋅ 100 = 0.169% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.4609375) − 3.9(0.4609375)
2 + 0.667(0.4609375)3 = −8.64 ∗ 10−4 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.46171875) − 3.9(0.46171875)
2 + 0.667(0.46171875)3 = 1.58 ∗ 10−3 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−8.64 ∗ 10
−4) ∗ (1.58 ∗ 10−3) = −1.303 ∗ 10−6 
 
ITERACION 9 
𝑥1 = 0.4609375 
𝑥2 = 0.46171875 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4609375 + 0.46171875
2
= 0.461328125 
 
𝐸𝑎 = |
0.462890625 − 0.46328125
0.462890625
| ⋅ 100 = 0.084% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.4609375) − 3.9(0.4609375)
2 + 0.667(0.4609375)3 = −8.64 ∗ 10−4 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.461328125) − 3.9(0.461328125)
2 + 0.667(0.461328125)3
= 3.225 ∗ 10−4 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−8.64 ∗ 10
−4) ∗ (3.225 ∗ 10−4) = −2.786 ∗ 10−7 
 
ITERACION 10 
𝑥1 = 0.4609375 
𝑥2 = 0.461328125 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4609375 + 0.461328125
2
= 0.4611328125 
 
𝐸𝑎 = |
0.462890625 − 0.46328125
0.462890625
| ⋅ 100 = 0.042% 
 
𝑓(𝑥1) = −2.1 + 6.21(0.4609375) − 3.9(0.4609375)
2 + 0.667(0.4609375)3 = −8.64 ∗ 10−4 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.4611328125) − 3.9(0.4611328125)
2 + 0.667(0.4611328125)3
= −2.7083 ∗ 10−4 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−8.64 ∗ 10
−4) ∗ (−2.7083 ∗ 10−4) = 2.339 ∗ 10−7 
 
ITERACION 11 
𝑥1 = 0.4611328125 
𝑥2 = 0.461328125 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4609375 + 0.461328125
2
= 0.4612304688 
 
𝐸𝑎 = |
0.4612304688 − 0.4611328125
0.4612304688
| ⋅ 100 = 0.021% 
 
𝑓(𝑥1) == −2.1 + 6.21(0.4611328125) − 3.9(0.4611328125)
2 + 0.667(0.4611328125)3
= −2.7083 ∗ 10−4 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.4612304688) − 3.9(0.4612304688)
2 + 0.667(0.4612304688)3
= 2.588 ∗ 10−5 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−2.7083 ∗ 10
−4) ∗ (2.588 ∗ 10−5) = −7.010 ∗ 10−9 
 
ITERACION 12 
 
𝑥1 = 0.4611328125 
𝑥2 = 0.4612304688 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4611328125 + 0.4612304688
2
= 0.4611816406 
 
𝐸𝑎 = |
0.4612304688 − 0.4612304688
0.4612304688
| ⋅ 100 = 0.0105% 
 
𝑓(𝑥1) == −2.1 + 6.21(0.4611328125) − 3.9(0.4611328125)
2 + 0.667(0.4611328125)3
= −2.7083 ∗ 10−4 
𝑓(𝑥𝑟) = −2.1 + 6.21(0.4611816406) − 3.9(0.4611816406)
2 + 0.667(0.4611816406)3
= −1.224 ∗ 10−4 
𝑓(𝑥1) ⋅ 𝑓(𝑥𝑟) = (−2.7083 ∗ 10
−4) ∗ (−1.224 ∗ 10−4) = 3.3149 ∗ 10−8 
 
ITERACION 13 
 
𝑥1 = 0.4611816406 
𝑥2 = 0.4612304688 
 
𝑥𝑟 =
𝑥1 + 𝑥2
2
 
𝑥𝑟 =
0.4611816406 + 0.4612304688
2
= 0.4612060547 
 
𝐸𝑎 = |
0.4612060547 − 0.4612304688
0.4612060547
| ⋅ 100 = 0.00529%𝑟𝑎𝑖𝑧 𝑥𝑟 = 0.4612060547