Métodos numericos parte 11

1. Indicaciones
1.- Lee el documentoMETODO DE LA SECANTE.pdf yresuelve losiguiente:
a) Resolverlaecuación (𝑥) = 42𝑥
− (8)4𝑥
𝑥 + 12 el métodode lasecante.Obtengael resultado
con cuatro decimalesexactos.
Valoresinicialesx00= 0.1 x0= 0.2
f(0.2)= 3.185
f(0.1)= 4.1299
x= 0.2 - ( 3.185/ ((3.185 - (4.1299))/ (0.2 - 0.1)))
x= 0.5371
Error =|0.5371-0.2|=0.3371
f(0.5371)= -0.4111
f(0.2)= 3.185
x= 0.5371 - ( -0.4111/ ((-0.4111 - (3.185))/ (0.5371 - 0.2)))
x= 0.4986
Error =|0.4986-0.5371|=0.0385
f(0.4986)= 0.0155
f(0.5371)= -0.4111
x= 0.4986 - ( 0.0155/ ((0.0155 - (-0.4111))/ (0.4986 - 0.5371)))
x= 0.5
Error =|0.5-0.4986|=0.0014
f(0.5)= 0
f(0.4986)= 0.0155
x= 0.5 - ( 0/ ((0 - (0.0155))/ (0.5 - 0.4986)))
x= 0.5
Error =|0.5-0.5|=0

2. X=0.5
En este caso el resultadofue exactocomolomuestralagráfica
b) Resolverlaecuación(𝑥) = 𝑒−𝑥 − 𝑥 por el métodode lasecante.
Valoresinicialesx00= 0.1 x0= 0.2
f(0.2)= 0.61873
f(0.1)= 0.80484
x= 0.2 - ( 0.61873/ ((0.61873 - (0.80484))/ (0.2 - 0.1)))
x= 0.53245
Error =|0.53245-0.2|=0.33245
f(0.53245)= 0.05471
f(0.2)= 0.61873
x= 0.53245 - ( 0.05471/ ((0.05471 - (0.61873))/ (0.53245 - 0.2)))
x= 0.5647
Error =|0.5647-0.53245|=0.03225
f(0.5647)= 0.00383
f(0.53245)= 0.05471

3. x= 0.5647 - ( 0.00383/ ((0.00383 - (0.05471))/ (0.5647 - 0.53245)))
x= 0.56713
Error =|0.56713-0.5647|=0.00243
f(0.56713)= 0.00002
f(0.5647)= 0.00383
x= 0.56713 - ( 0.00002/ ((0.00002- (0.00383))/ (0.56713 - 0.5647)))
x= 0.56714
Error =|0.56714-0.56713|=0.00001
x= 0.56714