5 Determine las raíces reales de
+,
= + , +,-
usando el Método de la Regla Falsa aproximar en el
intervalo [0.5 , 1] con ξa = 0,1% Solución
= 26 + 82,3 88 +45,4 -9 +0,65 , [0.5,1] = 0.5 = 26+82,30.5 880.5 +45,40.5-90.5 +0,650.5 = 1.7171875 = 1 = 26+82,3 26+82,31 881 +45,41-91 +0,651 =5.35 = =1 0.51 10.5 =0.621490 = 0.5 1 0.5 =2 10.621490 0.621490 0.6214901 =0.557419 = 1 0.621490 0.621490 1 =3 0.6214900.557419 0.557419 0.5574190.621490 =0.580567 = 0.621490 0.557419 0.557419 0.621490 =4 0.5574190.580567 0.580567 0.5805670.557419 =0.57936 = 0.557419 0.580567 0.580567 0.557419
| | 0.5793270.57936 % ∈= 0.579327 ∗100=5.69∗10−% 6 Demostrar que f(x) = x3 + 2x2 – 6 tiene una raíz en [1, 2] y utilizando el Método de bisección determine una aproximación a la raíz con una precisión de al menos 10-4. Solución
= + 2 6, =1, =2 = 1 =1 +21 6=3 = 2 =2 +22 6=10
[1,2]
Como las respuestas son de signos opuestos se demuestra que hay una solución en el intervalo dado.
= 3/2 =3/2 +23/2 6 = 158 =
= + 2 = 1 + 2 2 1 = 32
=1, = 32
5 = 1 + 3/21 = 2 4 5/4 =5/4 +25/4 6 = 59 64 =
=5/4, = 32 11 = 54 + 3/25/4 = 2 8 11/8 =11/8 +211/8 6=195/512 =
=5/4, = 118 21 = 54 + 11/85/4 = 2 16
21/16 =21/16 +221/16 6=0.293701 =
=21/16, = 118 11/821/16 = 43 =1.34375 = 21 + 16 2 32
43/32 =43/32 +243/32 6=0.0377 =6
=21/16, = 43 32 43/3221/16 = 85 =1.328125 = 21 + 16 2 64
85 85 (64) = (64) + 2(85 64)6=0.1295
=7 =85/64, = 43 32 43/3285/64 = 171 =1.3359 = 85 + 64 2 128 =8 (343 256)=0.00473 =9 (687 512)=0.0166 =10
=171/128, = 43 32 43/32171/128 = 343 =1.3398 = 171 + 128 2 256 =343/256, = 43 32 43/32343/256 = 687 =1.3417 = 343 + 256 2 512
=343/256, = 687 512 687/512343/256 = 1373 =1.3408 = 343 + 256 2 1024