Matematica

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| | UNIVERSIDAD NACIONAL EXPERIMENTAL | | |
| | “SIMÓN RODRÍGUEZ” | | |
| | INGENIERÍA DE ALIMENTOS | | |
| | NÚCLEO CANOABO | | |
| | CALCULO III | | |

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Facilitador: Gilberto Toro Participante:
Rojas Mariolis
23.795.746

Ejercicio 1
Aplique elmétodo de bisección para obtener P3, para F(X)= √X –Cosx en [0.1]
A0= 0 b0=1 [xa xb]
0 , 1

2
2
P0= a0+ab = 0+1 = ½ = 0.5

F(p1) = √0.5 – cos0.5 =-0.29
F(a0) = √0 – cos0 = -1
F(bo) = √1 – cos1 = 0.00015
F(a0)* F(p1)= (-1)*(-0.29) = 0.29>0

2
2
P1 = a1+b1 = 0.5 +1 = 0.75

F(p2) = √0.75 – cos0.75 = -0.13
F(b1) = √1 – cos1 = 0.00015F(a1) = √1/2 – cos1/2 = -0.29
F(a1)* F(p2)= (-0.29)*(-0.13) = 0.038>0

2
P2= 0.75 +1 = 0.875

F(p2) = √0.875 – cos0.875 = -0.064
F(a3) = √0.75 – cos0.75 = -0.13
F(a3)* F(p2)=(-0.13)*(-0.064) = 8.32e-03 >0

2
P3= 0.875 +1 = 15/16= 0.9375

P3=0.9375

Ejercicio 2
Encuentre una aproximación de √3 correcta con una exactitud 10-4 usando el algoritmo de bisección [sugerenciaconsidere f(x)= x2-3]

x | y |
0 | -3 |
1 | -2 |
2 | 1 |

f(x)= x2-3
[ ] *

Ln(2)
n= E ln (b –a) - ln (8)

[ ] *

Ln(2)
n= E ln 1 - ln (10-4) = 13.28 ~ 13

A0= 0 b0=1

2
2
P1= a0+ab = 1+2 = 3/2 = 1.5

F(p1) = x2- 3= (1.5)2 – 3 = -0.75F(a0) = x2- 3= (2)2 – 3 = 1
F(a)* f(p1) = 1*(-0.75)= -0.75 <0 b0=-0.75

2
2
P2= a0+b1 = 1+1.5 = 1.25 f(b) *f(p2)

F(b) = x2- 3= (1.5)2 – 3 = -0.75
F(a0) = x2- 3= (1.125)2 – 3...