Homework Answers
NEWTON'S METHOD:
Formula:
Xn+1= Xn - (F(Xn) / F'(Xn))
find the approximate root of 2X3-6X-10 upto three iteration
Let F(X)=2X3-6X-10
F'(X)=2(3X2 )-6
=6X2-6
F(0) =2(0)-6(0)-10
=-10
F(1)=2(1)-6(1)-10
= -14
F(2)=2(23)-6(2)-10
= -6
X0 = -4 [ F2 is closer to -4 because 2 to -6 I am taking middle value]
First iteration:
X1= -4 -(-6/90) (substitute x0 value in above formula)
= -4+9.0667
= 5.0667
Second iteration:
X2= 5.0667 - ( (2(5.0667)3-6(5.0667) -10) / (6(5.0667)2-6))
= 5.0667 - ((20.2668-30.4002-10) /(35.4669-6))
= 5.0667 - (20.1334 / 35.4669)
= 4.499
Third iteration:
X3 =4.499 - ((2(4.499)3-6(4.499)-10) / (6(4.499)2-6))
=4.499 - ((182.129-26.994-10) / (121.446-6))
= 4.499 -(145.135 /115.446)
= 4.499 - 1.258
= 3.241
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