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Use the Bisection method to find solutions accurate to within 10-2 for x³ – 7x2 + 14x – 6 = 0 on [0,1]. Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = –6 < 0 and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p E (0, 1) such that f(p) = 0. Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0. Let po = ao + sign as f(ao), therefore a1 = Po = 0.5, bị = bọ = 1 and repeat: pi = 0.75, .. This yields the following results for p, and f(Pn): - %3D bo-an 2 0.5, and we have f(po) = –0.6250 < 0 (the same 0.75, ... %3D %3D f (Pn) n Pn 0 0.5 -0.6250000

Use the Bisection method to find solutions accurate to within 10-2 for x³ – 7x2 + 14x – 6 = 0 on [0,1]. Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = –6 < 0 and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p E (0, 1) such that f(p) = 0. Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0. Let po = ao + sign as f(ao), therefore a1 = Po = 0.5, bị = bọ = 1 and repeat: pi = 0.75, .. This yields the following results for p, and f(Pn): - %3D bo-an 2 0.5, and we have f(po) = –0.6250 < 0 (the same 0.75, ... %3D %3D f (Pn) n Pn 0 0.5 -0.6250000

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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3a • Use the Bisection method to find solutions accurate to within 10-2 for x³ – 7x2 + 14x – 6 = 0 on [0,1]. Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = -6 < 0 and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p e (0, 1) such that f(p) = 0. Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0. Let po = ao + bn-an sign as f(ao), therefore a1 = Po = 0.5, b1 = This yields the following results for p, and f(pn): 0.5, and we have f(po) = -0.6250 < 0 (the same bo = 1 and repeat: P1 || 2 = 0.75, ... Pn f(Pn) 0.5 -0.6250000 1 0.75000000 +0.9843750 2 0.62500000 +0.2597656 3 0.56250000 -0.1618652 4 0.59375000 +0.0540466 5 0.57812500 -0.0526237 6 0.58593750 +0.0010313
Transcribed Image Text:3a • Use the Bisection method to find solutions accurate to within 10-2 for x³ – 7x2 + 14x – 6 = 0 on [0,1]. Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = -6 < 0 and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p e (0, 1) such that f(p) = 0. Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0. Let po = ao + bn-an sign as f(ao), therefore a1 = Po = 0.5, b1 = This yields the following results for p, and f(pn): 0.5, and we have f(po) = -0.6250 < 0 (the same bo = 1 and repeat: P1 || 2 = 0.75, ... Pn f(Pn) 0.5 -0.6250000 1 0.75000000 +0.9843750 2 0.62500000 +0.2597656 3 0.56250000 -0.1618652 4 0.59375000 +0.0540466 5 0.57812500 -0.0526237 6 0.58593750 +0.0010313
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