biseccion metodos numericos

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Hallar el error cometido en la cuarta interaci x x+1 -3 18 -2 -2 8 -1 -1 2 0 0 0 1 1 2 2 2 8 3 3 18 4 1ra iteración a -1 b 0 f(a)= = 0 f(b)= = -1 f(c)= -1 c= a+b)/2 -0.5 f(a)xf(c)= error |x-c|= 0.5 2da iteración a -0.5 b 0.0 f(a)= = -1 f(b)= = -1 f(c)= -1.125 c= a+b)/2 -0.25 f(a)xf(c)= error |x-c|= -0.25 3ra iteración a -0.25 b 0.0 f(a)= = -1.125 f(b)= = -1 f(c)= -1.09375 c= a+b)/2 -0.125 f(a)xf(c)= error |x-c|= 0.125 f(x)=2x 2 -(x+1) 2x 2 2(-1) 2 +(-1-1) 2(0) 2 +(0-1) 2(-0,5) 2 +(-0,5-1) 2(0) 2 +(0-1) 2(-0,5) 2 +(-0,5-1) 2(0) 2 +(0-1)

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metodos numericos

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Hoja1Hallar el error cometido en la cuarta interacionf(x)=2x2-(x+1)x2x2x+1-318-2-28-1-12000112228331841ra iteracina-1b0f(a)=2(-1)2+(-1-1)=0f(b)=2(0)2+(0-1)=-1f(c)=-1c=a+b)/2-0.5f(a)xf(c)=0Si f(a)xf(c)>=0 entonces c=aerror|x-c|=0.52da iteracina-0.5b0.0f(a)=2(-0,5)2+(-0,5-1)=-1f(b)=2(0)2+(0-1)=-1f(c)=-1.125c=a+b)/2-0.25f(a)xf(c)=1.125Si f(a)xf(c)>0 entonces a=cerror|x-c|=-0.253ra iteracina-0.25b0.0f(a)=2(-0,5)2+(-0,5-1)=-1.125f(b)=2(0)2+(0-1)=-1f(c)=-1.09375c=a+b)/2-0.125f(a)xf(c)=1.23046875Si f(a)xf(c)>0 entonces a=cerror|x-c|=0.1254ta iteracina-0.125b0.0f(a)=2(-0,5)2+(-0,5-1)=-1.09375f(b)=2(0)2+(0-1)=-1f(c)=-1.0546875c=a+b)/2-0.0625f(a)xf(c)=1.1535644531Si f(a)xf(c)>0 entonces a=cerror|x-c|=0.06252x2-3-2-10123188202818x+1-3-2-10123-2-101234