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Solve one real root of e* – 2x – 5 = 0 with xo = -2 using the Fixed-Point Iteration Method until absolute error < 0.00001. -

Solve one real root of e* – 2x – 5 = 0 with xo = -2 using the Fixed-Point Iteration Method until absolute error < 0.00001. -

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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1. Solve one real root of e* – 2x – 5 = 0 with - xo = -2 using the Fixed-Point Iteration Method until absolute error < 0.00001. 2. Compute for a real root of sin Vx – x = Ousing three iterations of Fixed-Point Iteration Method with xo = 0.50 until absolute error < 0.00001. 3. Calculate for one real root of tan x = 4x with 1.2 up to four decimal places using the Newton-Raphson Method until absolute error < 0.00001. 4. Determine one real root of 8 (sin x) e¯* – 1 = 0 with xo = 0.30 using Newton-Raphson Method until absolute error < -х — 0.00001.
Transcribed Image Text:1. Solve one real root of e* – 2x – 5 = 0 with - xo = -2 using the Fixed-Point Iteration Method until absolute error < 0.00001. 2. Compute for a real root of sin Vx – x = Ousing three iterations of Fixed-Point Iteration Method with xo = 0.50 until absolute error < 0.00001. 3. Calculate for one real root of tan x = 4x with 1.2 up to four decimal places using the Newton-Raphson Method until absolute error < 0.00001. 4. Determine one real root of 8 (sin x) e¯* – 1 = 0 with xo = 0.30 using Newton-Raphson Method until absolute error < -х — 0.00001.
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