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Perform the bisection method to provide a root for f (x) = x² + 3x – 2 given ɛ = 0.001. The first three (3) steps were already done in class. Let x1 = 0, x2 = 1, and x3 = 0.5. Then (0.5) = –0.25 . Redefine x1 = 0.5, x3 = 0.75. Then f (0.75) = 0.8125. Redefine x2 = 0.75, x3 = 0.625. Then f(0.625) = 0.2656. = 0.5625. Then f (0.5625) = 0.0039. Redefine x2 = 0.625, x3 %3D Redefine X3 = Then f(x3) = Redefine X3 = Then f (x3) Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Since |f (x3)| < take x =

Perform the bisection method to provide a root for f (x) = x² + 3x – 2 given ɛ = 0.001. The first three (3) steps were already done in class. Let x1 = 0, x2 = 1, and x3 = 0.5. Then (0.5) = –0.25 . Redefine x1 = 0.5, x3 = 0.75. Then f (0.75) = 0.8125. Redefine x2 = 0.75, x3 = 0.625. Then f(0.625) = 0.2656. = 0.5625. Then f (0.5625) = 0.0039. Redefine x2 = 0.625, x3 %3D Redefine X3 = Then f(x3) = Redefine X3 = Then f (x3) Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Since |f (x3)| < take x =

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter3: Functions
Section3.4: Composition Of Functions
Problem 97SE: The number of bacteria in a refrigerated food product is given by N(T)=23T256T+1,3T33 ,where T is...
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Perform the bisection method to provide a root for f (x) = x² + 3x – 2 given ɛ = 0.001. The first three (3) steps were already done in class. Let x1 = 0, x2 = 1, and x3 = 0.5. Then (0.5) = –0.25 . Redefine x1 = 0.5, x3 = 0.75. Then f (0.75) = 0.8125. Redefine x2 = 0.75, хз = 0.625. Then f(0.625) = 0.2656. = 0.5625. Then f (0.5625) = 0.0039. Redefine x2 = 0.625, x3 Redefine ,X3 = Then f (x3) = Redefine X3 = Then f (x3) Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine , X3 = Then f (x3) = Since |f (x3)| < take x =
Transcribed Image Text:Perform the bisection method to provide a root for f (x) = x² + 3x – 2 given ɛ = 0.001. The first three (3) steps were already done in class. Let x1 = 0, x2 = 1, and x3 = 0.5. Then (0.5) = –0.25 . Redefine x1 = 0.5, x3 = 0.75. Then f (0.75) = 0.8125. Redefine x2 = 0.75, хз = 0.625. Then f(0.625) = 0.2656. = 0.5625. Then f (0.5625) = 0.0039. Redefine x2 = 0.625, x3 Redefine ,X3 = Then f (x3) = Redefine X3 = Then f (x3) Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine X3 = Then f (x3) = Redefine , X3 = Then f (x3) = Since |f (x3)| < take x =
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