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Given the function : f(x) = x3 + 39x2 – 2462x – 3444; solve using Regula-Falsi Method with initial values of (30.00, 40.00)

Given the function : f(x) = x3 + 39x2 – 2462x – 3444; solve using Regula-Falsi Method with initial values of (30.00, 40.00)

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 + 39x2 – 2462x – 3444; solve using Regula-Falsi Method with initial values of (30.00, 40.00)
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All answers must be rounded off up to 2 decimal places.
Given the function : f(x) = x³ + 39x² – 2462x - 3444; solve using Regula-Falsi Method with initial values of (30.00, 40.00) - 1st iteration xn = xp = xnew = 2nd iteration xn = xp = xnew = 3rd iteration xn= xp = xnew = 4th iteration xn= xp = xnew = 5th iteration xn = xp = xnew = f(xn) = f(xp) = f(xnew) = f(xn) = f(xp) = f(xnew): = f(xn) = f(xp) = f(xnew): f(xn) = f(xp) = f(xnew) = f(xn) = f(xp): f(xnew) =
Transcribed Image Text:Given the function : f(x) = x³ + 39x² – 2462x - 3444; solve using Regula-Falsi Method with initial values of (30.00, 40.00) - 1st iteration xn = xp = xnew = 2nd iteration xn = xp = xnew = 3rd iteration xn= xp = xnew = 4th iteration xn= xp = xnew = 5th iteration xn = xp = xnew = f(xn) = f(xp) = f(xnew) = f(xn) = f(xp) = f(xnew): = f(xn) = f(xp) = f(xnew): f(xn) = f(xp) = f(xnew) = f(xn) = f(xp): f(xnew) =
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