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Solve the following equations and compare the roots using the five methods. Tabulate the data following  5 decimal values and stop at  € ≤ 0.0001 1. f(x)= x³ + 3x² - 2x - 7   The Picture below is the example of solving the problem

Solve the following equations and compare the roots using the five methods. Tabulate the data following  5 decimal values and stop at  € ≤ 0.0001 1. f(x)= x³ + 3x² - 2x - 7   The Picture below is the example of solving the problem

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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Solve the following equations and compare the roots using the five methods. Tabulate the data following 
5 decimal values and stop at  € ≤ 0.0001

1. f(x)= x³ + 3x² - 2x - 7

 

The Picture below is the example of solving the problem 

Sample Problem: Estimate the roots of equation f(x) = x³ − 4x² + x - 10 = 0 Tabulate the possible x values close to zero and substitute to f(x): X -3 -2 -1 0 1 2 f(x) -76 -36 -16 -10 -12 -16 Assuming that initial values are not equal. 1st Iteration: Select initial value, say xo = 3 → f(x) = -16 Select initial value, say x₁ = 4 ⇒ f(x₁) = −6 Using the iterative formula to get x2, 4-3 x₂ = 3 − (−16) [−6 − (−16) -= 4.60 Now, we replace new xo with x₁ = 4 and new x₁ with x₂ = 4.60. 2nd Iteration: Use replaced value, x, = 4 ⇒ f(x) = -6 Use replaced value, x₁ = 4.60 ➜ f(x₁) = 7.296 Using the iterative formula to get x2, x₂ =4-1 - (-6)[ 4.60 4 ¹7.296-(-6) = 4.2708 Now, we replace new xo with x₁ = 4.60 and new x₁ with x₂ = 4.2708. 3rd Iteration: Use replaced value, xo = 4.60 ➜ f(x) = 7.296 Use replaced value, x₁ = 4.2708 → f(x₁) = −0.7899 Using the iterative formula to get x2, x₂ = 4.6 (7.296)[ 4.27084.60 -0.7899 (7.296) = 4.3029 Now, we replace new xo with x₁ = 4.2708 and new x₁ with x₂ = 4.3029. 4th Iteration: Use replaced value, xo = 4.2708 ⇒ f(x) = -0.7899 Use replaced value, x₁ = 4.3029 → f(x₁) = -0.0899 Using the iterative formula to get X2, x₂ = 4.2708- (-0.7899) 4.30294.2708 -0.0899- (-0.7899) Now, we replace new xo with x₂ = 4.3029 and new x₁with x₂ = 4.3070. 5th Iteration: Use replaced value, xo = 4.3029 → f(x) = -0.0899 Use replaced value, x₁ = 4.3070 ➜ f(x₁) = 0.001926 Using the iterative formula to get X2, x₂ = 4.3029-(-0.0899) [ 4.3070 .3029 ¹0.001926 - (-0.0899)- = 4.3069 Then we check for the absolute difference of x2: 14.3069-4.3070|= 0.0001. Note that this method is faster than bisection method (13 iterations) and Regula-Falsi (7 iterations) having same error criteria. x0 x1 x2 AbsError, € Iteration 1 3.0000 4.0000 4.6000 2 4.0000 4.6000 4.2708 0.3292 3 4.6000 4.2708 4.3030 0.0322 4 4.2708 4.3030 4.3070 0.0040 5 4.3030 4.3070 4.3069 0.0001 3 -16 = 4.3070 4 -6 5 20
Transcribed Image Text:Sample Problem: Estimate the roots of equation f(x) = x³ − 4x² + x - 10 = 0 Tabulate the possible x values close to zero and substitute to f(x): X -3 -2 -1 0 1 2 f(x) -76 -36 -16 -10 -12 -16 Assuming that initial values are not equal. 1st Iteration: Select initial value, say xo = 3 → f(x) = -16 Select initial value, say x₁ = 4 ⇒ f(x₁) = −6 Using the iterative formula to get x2, 4-3 x₂ = 3 − (−16) [−6 − (−16) -= 4.60 Now, we replace new xo with x₁ = 4 and new x₁ with x₂ = 4.60. 2nd Iteration: Use replaced value, x, = 4 ⇒ f(x) = -6 Use replaced value, x₁ = 4.60 ➜ f(x₁) = 7.296 Using the iterative formula to get x2, x₂ =4-1 - (-6)[ 4.60 4 ¹7.296-(-6) = 4.2708 Now, we replace new xo with x₁ = 4.60 and new x₁ with x₂ = 4.2708. 3rd Iteration: Use replaced value, xo = 4.60 ➜ f(x) = 7.296 Use replaced value, x₁ = 4.2708 → f(x₁) = −0.7899 Using the iterative formula to get x2, x₂ = 4.6 (7.296)[ 4.27084.60 -0.7899 (7.296) = 4.3029 Now, we replace new xo with x₁ = 4.2708 and new x₁ with x₂ = 4.3029. 4th Iteration: Use replaced value, xo = 4.2708 ⇒ f(x) = -0.7899 Use replaced value, x₁ = 4.3029 → f(x₁) = -0.0899 Using the iterative formula to get X2, x₂ = 4.2708- (-0.7899) 4.30294.2708 -0.0899- (-0.7899) Now, we replace new xo with x₂ = 4.3029 and new x₁with x₂ = 4.3070. 5th Iteration: Use replaced value, xo = 4.3029 → f(x) = -0.0899 Use replaced value, x₁ = 4.3070 ➜ f(x₁) = 0.001926 Using the iterative formula to get X2, x₂ = 4.3029-(-0.0899) [ 4.3070 .3029 ¹0.001926 - (-0.0899)- = 4.3069 Then we check for the absolute difference of x2: 14.3069-4.3070|= 0.0001. Note that this method is faster than bisection method (13 iterations) and Regula-Falsi (7 iterations) having same error criteria. x0 x1 x2 AbsError, € Iteration 1 3.0000 4.0000 4.6000 2 4.0000 4.6000 4.2708 0.3292 3 4.6000 4.2708 4.3030 0.0322 4 4.2708 4.3030 4.3070 0.0040 5 4.3030 4.3070 4.3069 0.0001 3 -16 = 4.3070 4 -6 5 20
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