efghssgfsghssass


1)
f(x) = exp(6*x)+3*(log(2))^2*exp(2*x)-log(8)*exp(4*x)-(log(2))^3
df(x) = 6*exp(6*x) - 4*exp(4*x)*log(8) + 6*exp(2*x)*log(2)^2
si graficamos f(x) y df(x)c=newtonraphson(f,df,0,5e-15,25)
K= 1 c=-0.05114214 eabs=0.05114214 |f(c)|=0.00921157
K= 2 c=-0.08984248 eabs=0.03870034 |f(c)|=0.00288671
K= 3 c=-0.11824474 eabs=0.02840226 |f(c)|=0.00089158
K= 4 c=-0.13856560eabs=0.02032086 |f(c)|=0.00027220
K= 5 c=-0.15281619 eabs=0.01425060 |f(c)|=0.00008237
K= 6 c=-0.16266025 eabs=0.00984406 |f(c)|=0.00002477
K= 7 c=-0.16938618 eabs=0.00672592 |f(c)|=0.00000741
K= 8c=-0.17394606 eabs=0.00455989 |f(c)|=0.00000221
K= 9 c=-0.17702081 eabs=0.00307475 |f(c)|=0.00000066
K= 10 c=-0.17908646 eabs=0.00206564 |f(c)|=0.00000020
K= 11 c=-0.18047068eabs=0.00138422 |f(c)|=0.00000006
K= 12 c=-0.18139669 eabs=0.00092601 |f(c)|=0.00000002
K= 13 c=-0.18201546 eabs=0.00061877 |f(c)|=0.00000001
K= 14 c=-0.18242861 eabs=0.00041315 |f(c)|=0.00000000
K= 15c=-0.18270433 eabs=0.00027572 |f(c)|=0.00000000
K= 16 c=-0.18288828 eabs=0.00018394 |f(c)|=0.00000000
K= 17 c=-0.18301096 eabs=0.00012268 |f(c)|=0.00000000
K= 18 c=-0.18309277 eabs=0.00008181|f(c)|=0.00000000
K= 19 c=-0.18314733 eabs=0.00005455 |f(c)|=0.00000000
K= 20 c=-0.18318370 eabs=0.00003637 |f(c)|=0.00000000
K= 21 c=-0.18320795 eabs=0.00002425 |f(c)|=0.00000000
K= 22 c=-0.18322412eabs=0.00001617 |f(c)|=0.00000000
K= 23 c=-0.18323487 eabs=0.00001075 |f(c)|=0.00000000
K= 24 c=-0.18324204 eabs=0.00000717 |f(c)|=0.00000000
K= 25 c=-0.18324682 eabs=0.00000478 |f(c)|=0.00000000
laaproximacion a la raiz es
c =-0.1832
b) como el error absoluto tiende a la tol lentamente podemos decir que es de convergencia lineal.











2.
f(x) = (x^2)-2*x*exp(-x)+exp(-2*x)df(x) = 2*x - 2*exp(-x) - 2*exp(-2*x) + 2*x*exp(-x)
la grafica de f(x) y df(x) es

c=newtonraphson(f,df,0.5,5e-5,4)
K= 1 c=0.53315550 eabs=0.03315550 |f(c)|=0.00287243
K= 2 c=0.55004381...