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Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Do three iterations (by hand) of the bisection method, applied to f(x) = x³ – 2, using a = 0 and b = 2.

Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Do three iterations (by hand) of the bisection method, applied to f(x) = x³ – 2, using a = 0 and b = 2.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.6: Quadratic Functions
Problem 32E
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answer this and I provide some examples and the table of bisection method  (pls pls follow the attachment that i gave and pls complete the table )

Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Do three iterations (by hand) of the bisection method, applied to f(x) = x³ – 2, using a = 0 and b = 2.
Transcribed Image Text:Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Do three iterations (by hand) of the bisection method, applied to f(x) = x³ – 2, using a = 0 and b = 2.
BISECTION METHOD TABLE!!!!! 0.625 0.562 0.593 -0.040 0.593 0.562 0.577 -0.015 0.577 0.562 0.569 -0.002 k br f(Ck+1) dk Remarks 0.569 0.562 0.565 0.003 0.007 Continue ak Ck+1 8 0.569 0.565 0.567 0.0002 0.003 Continue 9 0.569 0.567 0.568 0.001 Stop ΕΧΑMPLES therefore, X = 0. 568 = x10 Example 1.) If f (x) = 2 – e*, and take the original interval to be [a, b] = [0,1]. Example 3.) f(x) = 6x-5x2 + 7x - 2 Solution: Solution: to -10 2 40 f(x) -20 -2 f(a) = 1, f(b) = -0.7183 c = 10+ 1] - 1 f(c) = 0.3513 > 0. Therefore, [a, b] = ;,1| k S(Ck+1) Remarks Ck+1 0.5 1 1.000 Continue 1 0.5 0.25 -0.040 2 .5 0.375 0.375 2 0.5 0.25 0.375 0.238 3. 0.25 0.312 0.343 0.327 -0.12 %3D 4 0.312 0.054 - 0.343 0.312 -0.035 0.343 0.327 0.335 0.009 0.023 Continue f(a) = 0.3513, f(b) = -0.7183 %3D %3D Continue solution until X = 0.333 3 c = 2 4 Example 4.) Solve 2x3 – 2.5x – 5 = O for the root in the interval [1, 2]. f(c) = -0.1170 < 0. Therefore, [a, b] = Solution: k-0; a, = 1; bo = 2 s (do* bo) = s (2) - r(1.5) = -2 < 0 = F(1.5) = -2 < o Therefore, f(ao)s (be) = 11 > 0 and If(1.5)| > E (ao + f(a) = 0.3513, f (b) = –0.1170 C = 8. k = 1; a, = 1.5; b, = 2 f(c) = 0.1318 > 0. Therefore, [a, b] = r() - r (15* 2) s(aD:) = s (15*2) = f(1.75) = 1.344 %3D Therefore, f(a)f (b) = -2.688 < 0 and IS(1.75)| >e k = 2; az = 1.5; b, = 1.75 Details of the remaining steps are provided in the table below: Example 2.) f(x) = e¬*-x k (Ck+1) Remarks Solution: ak 1 Ck+1 1.5 -2.000 Continue 1 15 175 1 344 x2 = 0; f(x?) = 1 xº = 1; f(xº) = -0.632 1.5 1.75 1.625 -0.481 3 1.625 1.75 1.688 0.392 4 1.625 1.688 1.656 -0.054 5 1.656 1.688 1.672 0.167 k f(ck+1) dk Remarks 6. 1.656 1.672 1.664 0.056 Ck+1 0.5 7 1.656 1.664 1.660 0.001 1 0.106 Continue 8 1.656 1.660 1.658 -0.027 1 0.5 0.75 -0.277 9 1.658 1.660 1.659 -0.013 10 1.660 1.660 1.660 -0.006 2 0.75 0.5 0.625 -0.089 11 1.660 1.660 1.660 -0.003 3 0.625 0.5 0.562 0.008 12 1.660 1.660 1.660 -0.001 --
Transcribed Image Text:BISECTION METHOD TABLE!!!!! 0.625 0.562 0.593 -0.040 0.593 0.562 0.577 -0.015 0.577 0.562 0.569 -0.002 k br f(Ck+1) dk Remarks 0.569 0.562 0.565 0.003 0.007 Continue ak Ck+1 8 0.569 0.565 0.567 0.0002 0.003 Continue 9 0.569 0.567 0.568 0.001 Stop ΕΧΑMPLES therefore, X = 0. 568 = x10 Example 1.) If f (x) = 2 – e*, and take the original interval to be [a, b] = [0,1]. Example 3.) f(x) = 6x-5x2 + 7x - 2 Solution: Solution: to -10 2 40 f(x) -20 -2 f(a) = 1, f(b) = -0.7183 c = 10+ 1] - 1 f(c) = 0.3513 > 0. Therefore, [a, b] = ;,1| k S(Ck+1) Remarks Ck+1 0.5 1 1.000 Continue 1 0.5 0.25 -0.040 2 .5 0.375 0.375 2 0.5 0.25 0.375 0.238 3. 0.25 0.312 0.343 0.327 -0.12 %3D 4 0.312 0.054 - 0.343 0.312 -0.035 0.343 0.327 0.335 0.009 0.023 Continue f(a) = 0.3513, f(b) = -0.7183 %3D %3D Continue solution until X = 0.333 3 c = 2 4 Example 4.) Solve 2x3 – 2.5x – 5 = O for the root in the interval [1, 2]. f(c) = -0.1170 < 0. Therefore, [a, b] = Solution: k-0; a, = 1; bo = 2 s (do* bo) = s (2) - r(1.5) = -2 < 0 = F(1.5) = -2 < o Therefore, f(ao)s (be) = 11 > 0 and If(1.5)| > E (ao + f(a) = 0.3513, f (b) = –0.1170 C = 8. k = 1; a, = 1.5; b, = 2 f(c) = 0.1318 > 0. Therefore, [a, b] = r() - r (15* 2) s(aD:) = s (15*2) = f(1.75) = 1.344 %3D Therefore, f(a)f (b) = -2.688 < 0 and IS(1.75)| >e k = 2; az = 1.5; b, = 1.75 Details of the remaining steps are provided in the table below: Example 2.) f(x) = e¬*-x k (Ck+1) Remarks Solution: ak 1 Ck+1 1.5 -2.000 Continue 1 15 175 1 344 x2 = 0; f(x?) = 1 xº = 1; f(xº) = -0.632 1.5 1.75 1.625 -0.481 3 1.625 1.75 1.688 0.392 4 1.625 1.688 1.656 -0.054 5 1.656 1.688 1.672 0.167 k f(ck+1) dk Remarks 6. 1.656 1.672 1.664 0.056 Ck+1 0.5 7 1.656 1.664 1.660 0.001 1 0.106 Continue 8 1.656 1.660 1.658 -0.027 1 0.5 0.75 -0.277 9 1.658 1.660 1.659 -0.013 10 1.660 1.660 1.660 -0.006 2 0.75 0.5 0.625 -0.089 11 1.660 1.660 1.660 -0.003 3 0.625 0.5 0.562 0.008 12 1.660 1.660 1.660 -0.001 --
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