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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 =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.2: Exponential Functions
Problem 58E
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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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