Question

In: Advanced Math

Find a root of an equation f(x)=5x3-3x2+8 initial solution x0=-0.81, using Newton Raphson method

Expert Solution

Approximate root of the equation 5x3-3x2+8=0 Let f(x)=5x3-3x2+8

ddx(5x3-3x2+8)=15x2-6x

∴f′(x)=15x2-6x

x0=-0.81

1st iteration :

f(x0)=f(-0.81)=5⋅(-0.81)3-3⋅(-0.81)2+8=3.3745

f′(x0)=f′(-0.81)=15⋅(-0.81)2-6⋅(-0.81)=14.7015

x1=x0-f(x0)f′(x0)

x1=-0.81-3.374514.7015

x1=-1.0395

2nd iteration :

f(x1)=f(-1.0395)=5⋅(-1.0395)3-3⋅(-1.0395)2+8=-0.8587

f′(x1)=f′(-1.0395)=15⋅(-1.0395)2-6⋅(-1.0395)=22.4467

x2=x1-f(x1)f′(x1)

x2=-1.0395--0.858722.4467

x2=-1.0013

3rd iteration :

f(x2)=f(-1.0013)=5⋅(-1.0013)3-3⋅(-1.0013)2+8=-0.0269

f′(x2)=f′(-1.0013)=15⋅(-1.0013)2-6⋅(-1.0013)=21.0461

x3=x2-f(x2)f′(x2)

x3=-1.0013--0.026921.0461

x3=-1

4th iteration :

f(x3)=f(-1)=5⋅(-1)3-3⋅(-1)2+8=0

f′(x3)=f′(-1)=15⋅(-1)2-6⋅(-1)=21.0001

x4=x3-f(x3)f′(x3)

x4=-1-021.0001

x4=-1

Approximate root of the equation 5x3-3x2+8=0 is -1

root is-1

Sankar Maity answered 2 years ago

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