Práctica resuelta
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1.
xf(x)
-4-25.9816844
-3-16.9502129
-2-9.86466472
-1-4.63212056
0-1
11.71828183
25.3890561
315.0855369
444.59815
Del cambio de signos e toma a = 0 y b = 1
abxrf(xr)f(xa)f(xa)*f(xr)Error (%)
010.50.39872127-1-0.39872127
00.50.25-0.27847458-10.27847458100
0.250.50.3750.06436641-0.27847458-0.0179244133.3333333
0.250.3750.3125-0.10581831-0.278474580.0294677120
0.31250.3750.34375-0.02043803-0.105818310.002162729.09090909
0.343750.3750.3593750.02203347-0.02043803-0.000450324.34782609
0.343750.3593750.35156250.00081538-0.02043803-1.6665E-052.22222222
0.343750.35156250.34765625-0.00980686-0.020438030.000200431.12359551
0.347656250.35156250.34960938-0.00449463-0.009806864.4078E-050.55865922
Como el error es menor al 0.5% se toma como valor para la raz a xr = 0.34960938.
2.
De la grfica se tomara los puntos con un Para g1: x0 = g(x0)
x0g(x0)
g0-0.7-0.54663507
g1-0.54663507-0.63878371
g2-0.63878371-0.58554389
g3-0.58554389-0.61691811
g4-0.61691811-0.59866244
g5-0.59866244-0.6093599
g6-0.6093599-0.60311796
g7-0.60311796-0.606769
g8-0.606769-0.6046365
g9-0.6046365-0.60588309
g10-0.60588309-0.60515473
g11-0.60515473-0.60558042
g12-0.60558042-0.60533167
g13-0.60533167-0.60547704
g14-0.60547704-0.60539209
g15-0.60539209-0.60544173
Se toma la raz g15: x = -0.60544173