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EXAMPLE 1 Starting with x, = 2, find the third approximation x3 to the root of the equation x3 - 2x - 2 = 0. SOLUTION We apply Newton's method with f(x) = x³ - 2x – 2 and f'(x) = So the iterative equation becomes x,3 - 2x, – 2 Xn+1 = X, - With n = 1, we have x3 - 2x, - 2 ]xx² - [ 23 - 2(2) – 2 X2 = X1 - = 2 - %3D Then with n = 2, we obtain X23 — 2х2 — 2 X3 = x2 - |x,² – (Round to four decimal places.) It turns out that this third approximation x3 is accurate to three decimal places. 22

EXAMPLE 1 Starting with x, = 2, find the third approximation x3 to the root of the equation x3 - 2x - 2 = 0. SOLUTION We apply Newton's method with f(x) = x³ - 2x – 2 and f'(x) = So the iterative equation becomes x,3 - 2x, – 2 Xn+1 = X, - With n = 1, we have x3 - 2x, - 2 ]xx² - [ 23 - 2(2) – 2 X2 = X1 - = 2 - %3D Then with n = 2, we obtain X23 — 2х2 — 2 X3 = x2 - |x,² – (Round to four decimal places.) It turns out that this third approximation x3 is accurate to three decimal places. 22

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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EXAMPLE 1 Starting with x1 = 2, find the third approximation x3 to the root of the equation x3 - 2x - 2 = 0. SOLUTION We apply Newton's method with f(x) = x³ - 2x – 2 and f'(x) = So the iterative equation becomes x,3 – 2x, - 2 Xn+1 = X, - With n = 1, we have x,3 - 2x, - 2 |xx² - [ 23 - 2(2) – 2 X2 = X1 = 2 - Then with n = 2, we obtain x23 - 2x2 - 2 X3 = x2 - 2 - 2 2 (Round to four decimal places.) It turns out that this third approximation x3 2 is accurate to three decimal places. 22
Transcribed Image Text:EXAMPLE 1 Starting with x1 = 2, find the third approximation x3 to the root of the equation x3 - 2x - 2 = 0. SOLUTION We apply Newton's method with f(x) = x³ - 2x – 2 and f'(x) = So the iterative equation becomes x,3 – 2x, - 2 Xn+1 = X, - With n = 1, we have x,3 - 2x, - 2 |xx² - [ 23 - 2(2) – 2 X2 = X1 = 2 - Then with n = 2, we obtain x23 - 2x2 - 2 X3 = x2 - 2 - 2 2 (Round to four decimal places.) It turns out that this third approximation x3 2 is accurate to three decimal places. 22
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