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I used Newton-Raphson's method to find the approximate root of the equation 0 = 1+ x3 - 5x and it was 0.5 x, so the approximate root at the second step is (a) X2 = 0.221568 (b) x2 = 0.201568 %3D X2 = 0.241568 None of the above is correct (d)

I used Newton-Raphson's method to find the approximate root of the equation 0 = 1+ x3 - 5x and it was 0.5 x, so the approximate root at the second step is (a) X2 = 0.221568 (b) x2 = 0.201568 %3D X2 = 0.241568 None of the above is correct (d)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I used Newton-Raphson's method to find the approximate root of the equation 0 = 1+ x3 - 5x and it was 0.5 = x, so the approximate root at the second step is (a) x2 = 0.221568 (b) x2 = 0.201568 (c) x2 = 0.241568 None of the above is correct (d)
Transcribed Image Text:I used Newton-Raphson's method to find the approximate root of the equation 0 = 1+ x3 - 5x and it was 0.5 = x, so the approximate root at the second step is (a) x2 = 0.221568 (b) x2 = 0.201568 (c) x2 = 0.241568 None of the above is correct (d)
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