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

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

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

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

âˆ´fâ€²(x)=15x2-6x

x0=-0.81

1stÂ iteration :

f(x0)=f(-0.81)=5â‹…(-0.81)3-3â‹…(-0.81)2+8=3.3745

fâ€²(x0)=fâ€²(-0.81)=15â‹…(-0.81)2-6â‹…(-0.81)=14.7015

x1=x0-f(x0)fâ€²(x0)

x1=-0.81-3.374514.7015

x1=-1.0395

2ndÂ iteration :

f(x1)=f(-1.0395)=5â‹…(-1.0395)3-3â‹…(-1.0395)2+8=-0.8587

fâ€²(x1)=fâ€²(-1.0395)=15â‹…(-1.0395)2-6â‹…(-1.0395)=22.4467

x2=x1-f(x1)fâ€²(x1)

x2=-1.0395--0.858722.4467

x2=-1.0013

3rdÂ iteration :

f(x2)=f(-1.0013)=5â‹…(-1.0013)3-3â‹…(-1.0013)2+8=-0.0269

fâ€²(x2)=fâ€²(-1.0013)=15â‹…(-1.0013)2-6â‹…(-1.0013)=21.0461

x3=x2-f(x2)fâ€²(x2)

x3=-1.0013--0.026921.0461

x3=-1

4thÂ iteration :

Â f(x3)=f(-1)=5â‹…(-1)3-3â‹…(-1)2+8=0

fâ€²(x3)=fâ€²(-1)=15â‹…(-1)2-6â‹…(-1)=21.0001

x4=x3-f(x3)fâ€²(x3)

x4=-1-021.0001

x4=-1

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

root is-1

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

Free

90.2K

Verified Solution

Question

Advance Math

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

Answer & Explanation Solved by verified expert

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