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b)
\The function is 2x7 + 7x4+ 2 = 0
\Let f(x) = 2x7 + 7x4+ 2
\Given x1 = 1
\Find x2 = ? and x3 = ?
\f \\'(x) = 2*7x7-1 + 7*4x4-1
\f \\'(x) = 14x6 + 28x3
\According to Newton Method
x2 = x1 – [ f(x1) / f’(x1) ]
\f(x1) = 2(x1)7 + 7(x1)4+ 2
\f(x1) = 2(1)7 + 7(1)4+ 2 = 2 + 7 + 2 = 11
\f \\'(x1) = 14(x1)6 + 28(x1)3 = 14(1)6 + 28(1)3 = 14 + 28 = 42
\x2 = x1 – [ f(x1) / f’(x1) ]
\x2 = 1 – [ 11 / 42 ]
\x2 = 1 – 0.2619
\x2 = 0.738
\x3 = x2 – [ f(x2) / f’(x2) ]
\f(x2) = 2(x2)7 + 7(x2)4+ 2
\f(x2) = 2(0.738)7 + 7(0.738)4+ 2
\= 0.238 + 2.076 + 2
\= 4.314
\f \\'(x2) = 14(x2)6 + 28(x21)3
\= 14(0.738)6 + 28(0.738)3
\= 2.26 + 11.25
\= 13.51
\x3 = x3 – [ f(x3) / f’(x2) ]
\x3 = 1 – [ 4.314 / 13.51 ]
\x3 = 1 – 0.319
\x3 = 0.681
\Solution :
\x3 = 0.681
\