Investigacion

-->a=0 , b=90;

-->mifuncion(a)

ans = - 1.

-->mifuncion(b)

ans = 1.9965213

-->c=(a+b)/2 punto medio

c = 45.

-->mifuncion(c)

ans = 0.948486

Método de Newtonn. x_n f(x_n)

2, 5.556820 194.103158

3, 5.214460 50.979166

4, 5.043822 7.221652

5, 5.010768 0.212894

6, 5.009733 0.0002007, 5.009732 0.000000

Se satisface la tolerancia.

Resultado final: Raíz = 5.009732

x0=6

-->x1=x0-(mifuncion(x0)/dmifuncion(x0))

x1 =5.5568195-->x2=x1-(mifuncion(x1)/dmifuncion(x1))

x2 = 5.2144605

-->x3=x2-(mifuncion(x2)/dmifuncion(x2))

x3 = 5.0438222

-->x4=x3-(mifuncion(x3)/dmifuncion(x3))

x4 = 5.0107678-->x5=x4-(mifuncion(x4)/dmifuncion(x4))

x5 = 5.0097329

-->x6=x5-(mifuncion(x5)/dmifuncion(x5))

x6 = 5.009732

-->x7=x6-(mifuncion(x6)/dmifuncion(x6))

x7 = 5.009732

-->abs(x7-x6)a=0 , b=(%pi/2);

a= 0.

-->mifuncion(a)

ans = - 1.

-->mifuncion(b)

ans = 1.7320508

-->c=(a+b)/2

c = 0.7853982

-->mifuncion(c)

ans = 0.5176381

-->c1=(a+c)/2

c1 = 0.3926991-->mifuncion(c1)

ans = - 0.2610524

-->c2=(c+c1)/2

c2 = 0.5890486

-->mifuncion(c2)

ans = 0.1308063

-->c3=(c1+c2)/2

c3 = 0.4908739

-->mifuncion(c3)

ans =- 0.0654382

-->c4=(c2+c3)/2

c4 = 0.5399612

-->mifuncion(c4)

ans = 0.0327235

-->c5=(c3+c4)/2

c5 = 0.5154175

-->mifuncion(c5)

ans = - 0.0163623-->c6=(c4+c5)/2

c6 = 0.5276894

-->mifuncion(c6)

ans = 0.0081812

-->c7=(c5+c6)/2

c7 = 0.5215535

-->mifuncion(c7)

ans = - 0.0040906

-->c8=(c6+c7)/2

c8 = 0.5246214-->mifuncion(c8)

ans = 0.0020453

-->c9=(c7+c8)/2

c9 = 0.5230874

-->mifuncion(c9)

ans = - 0.0010227

-->c10=(c8+c9)/2

c10 = 0.5238544

-->mifuncion(c10)...