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Given the function : f(x) = x³ + 5x² + 7x-20; solve using Secant Method with initial values of (1.000, 2.000) st

Given the function : f(x) = x³ + 5x² + 7x-20; solve using Secant Method with initial values of (1.000, 2.000) st

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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Question
Given the function : f(x) = x³ + 5x² + 7x-20; solve using Secant Method with initial values of (1.000, 2.000) 1st iteration X0 = 1.000 X1 2.000 xnew = 1.241 2nd iteration X0 = 2.000 X1 2.000 xnew 1.296 3rd iteration X0 = 1.241 X1 = 1.296 xnew= 1.310 4th iteration X0 = 1.296 X1 = 1.310 xnew= 1.310 5th iteration X0 = -0.001 X1 = -0.001 xnew= 1.310 f(x0) = -7.000 f(x1) = 22.000 f(xnew) = -1.701 f(x0) = 22.000 f(x1) = -1.701 f(xnew) = -0.353 f(x0) = -1.701 f(x1) = -0.353 f(xnew) -0.001 = f(x0) = -0.001 f(x1) = -0.001 f(xnew) = -0.001 f(x0) = 1.310 f(x1) = -0.001 f(xnew) = -0.001
Transcribed Image Text:Given the function : f(x) = x³ + 5x² + 7x-20; solve using Secant Method with initial values of (1.000, 2.000) 1st iteration X0 = 1.000 X1 2.000 xnew = 1.241 2nd iteration X0 = 2.000 X1 2.000 xnew 1.296 3rd iteration X0 = 1.241 X1 = 1.296 xnew= 1.310 4th iteration X0 = 1.296 X1 = 1.310 xnew= 1.310 5th iteration X0 = -0.001 X1 = -0.001 xnew= 1.310 f(x0) = -7.000 f(x1) = 22.000 f(xnew) = -1.701 f(x0) = 22.000 f(x1) = -1.701 f(xnew) = -0.353 f(x0) = -1.701 f(x1) = -0.353 f(xnew) -0.001 = f(x0) = -0.001 f(x1) = -0.001 f(xnew) = -0.001 f(x0) = 1.310 f(x1) = -0.001 f(xnew) = -0.001
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