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Given the function : f(x) = x³ + 5x² + 7x − 20; solve using Bisection Method with initial values of (1.000, 2.000) 1st iteration xn = 1.000 xp = 2.000 xnew = 1.500 2nd iteration xn = 1.000 xp = 1.500 xnew = 1.250 3rd iteration xn = 1.250 xp = 1.500 xnew = 1.375 4th iteration xn = 1.250 xp = 1.375 xnew = 1.312 5th iteration xn = 1.250 xp = 1.312 xnew = 1.281 f(xn) = -7.000 f(xp) = 22.000 f(xnew) = 5.125 f(x) = -7.000 f(xp) = 5.125 f(xnew) = -1.484 f(xn) = -1.484 f(xp) = 5.125 f(xnew) = 1.678 f(xn) = -1.484 f(xp) = 1.676 f(xnew) = 0.049 f(x) = -1.484 f(xp) = 0.049 f(xnew) = -0.726

Given the function : f(x) = x³ + 5x² + 7x − 20; solve using Bisection Method with initial values of (1.000, 2.000) 1st iteration xn = 1.000 xp = 2.000 xnew = 1.500 2nd iteration xn = 1.000 xp = 1.500 xnew = 1.250 3rd iteration xn = 1.250 xp = 1.500 xnew = 1.375 4th iteration xn = 1.250 xp = 1.375 xnew = 1.312 5th iteration xn = 1.250 xp = 1.312 xnew = 1.281 f(xn) = -7.000 f(xp) = 22.000 f(xnew) = 5.125 f(x) = -7.000 f(xp) = 5.125 f(xnew) = -1.484 f(xn) = -1.484 f(xp) = 5.125 f(xnew) = 1.678 f(xn) = -1.484 f(xp) = 1.676 f(xnew) = 0.049 f(x) = -1.484 f(xp) = 0.049 f(xnew) = -0.726

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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Given the function : f(x) = x³ + 5x² + 7x − 20; solve using Bisection Method with initial values of (1.000, 2.000) 1st iteration xn = 1.000 xp = 2.000 xnew = 1.500 2nd iteration xn = 1.000 xp = 1.500 xnew = 1.250 3rd iteration xn = 1.250 xp = 1.500 xnew = 1.375 4th iteration xn = 1.250 xp = 1.375 xnew = 1.312 5th iteration xn = 1.250 xp = 1.312 xnew = 1.281 f(xn) = -7.000 f(xp) = 22.000 f(xnew) = 5.125 f(x) = -7.000 f(xp) = 5.125 f(xnew) = -1.484 f(xn) = -1.484 f(xp) = 5.125 f(xnew) = 1.678 f(xn) = -1.484 f(xp) = 1.676 f(xnew) = 0.049 f(x) = -1.484 f(xp) = 0.049 f(xnew) = -0.726
Transcribed Image Text:Given the function : f(x) = x³ + 5x² + 7x − 20; solve using Bisection Method with initial values of (1.000, 2.000) 1st iteration xn = 1.000 xp = 2.000 xnew = 1.500 2nd iteration xn = 1.000 xp = 1.500 xnew = 1.250 3rd iteration xn = 1.250 xp = 1.500 xnew = 1.375 4th iteration xn = 1.250 xp = 1.375 xnew = 1.312 5th iteration xn = 1.250 xp = 1.312 xnew = 1.281 f(xn) = -7.000 f(xp) = 22.000 f(xnew) = 5.125 f(x) = -7.000 f(xp) = 5.125 f(xnew) = -1.484 f(xn) = -1.484 f(xp) = 5.125 f(xnew) = 1.678 f(xn) = -1.484 f(xp) = 1.676 f(xnew) = 0.049 f(x) = -1.484 f(xp) = 0.049 f(xnew) = -0.726
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