Answer to Question #177625 in Mechanical Engineering for Von Nikko

Question #177625

Determine the root of f(x)=x3-6x2+11x-6.1 using the Newton-Rapson Method(three iterations, xi=3.5)


1
Expert's answer
2021-04-12T01:53:12-0400

Given

f(x)=x36x2+11x6f(x)=x^3-6x^2+11x-6

f(x)=3x212x+11f'(x)=3x^2-12x+11

x0=3.5x_0=3.5

xn+1=xnf(xn)f(xn)x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}




Stepx0f(x0)x1f(x1)13.5000001.8750003.1739130.44382323.1739130.4438233.0323070.06777933.0323070.0677793.0014560.00291843.0014560.0029183.0000030.000006\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & 3.500000 & 1.875000 & 3.173913 & 0.443823 \\ \hdashline 2 & 3.173913 & 0.443823 & 3.032307 & 0.067779 \\ \hdashline 3 & 3.032307 & 0.067779 & 3.001456 & 0.002918 \\ \hdashline 4 & 3.001456 & 0.002918 & 3.000003 & 0.000006 \\ \end{array}


Root is : x=3.000003x=3.000003



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