Here x3+5x−3=0
Let f(x)=x3+5x−3
dxd(x3+5x−3)=3x2+5∴f(x)=3x2+5
Here
Here f(0)=−3<0 and f(1)=3>0
∴ Root lies between 0 and 1
x0=20+1=0.5x0=0.5
1st iteration :
f(x0)=f(0.5)=0.53+5⋅0.5−3=−0.375f′(x0)=f(0.5)=3⋅0.52+5=5.75
x1=x0−f(x0)f(x0)x1=0.5−5.75−0.375x1=0.5652
2nd iteration :
f(x1)=f(0.5652)=0.56523+5⋅0.5652−3=0.0067f′(x1)=f(0.5652)=3⋅0.56522+5=5.9584
x2=x1−f(x1)f(x1)x2=0.5652−5.95840.0067x2=0.5641
3rd iteration :
f(x2)=f(0.5641)=0.56413+5⋅0.5641−3=0
f′(x2)=f(0.5641)=3⋅0.56412+5=5.9546x3=x2−f(x2)f(x2)x3=0.5641−5.95460x3=0.5641 Approximate root of the equation x3+5x−3=0 using Newton Raphson mehtod is 0.5641 (After 3 iterations)
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