Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Review - Exercises - Page 367: 62

Answer

$x$ $=$ $3.732051$ $or$ $0.267949$

Work Step by Step

$x$ $=$ $x^{2}$$-$$3x$ $+$$1$ $x_{n+1}$ $=$ $x_{n}$ $-$ $\frac{f(x)}{f^{'}(x)}$ $f(x)$ $=$ $x$ $-$ $x^{2}$ $+$ $3x$ $-$ $1$ ${f^{'}(x)}$ $=$ $-2x$ $+$ $4$ $Let$ $x_{1}$ $=$ $3$ $x_{2}$ $=$ $3$ $-$ $\frac{f(1)}{f'(1)}$ $=$ $4$ $x_{3}$ $=$ $4$ $-$ $\frac{f(2)}{f'(2)}$ $=$ $3.75$ $x_{4}$ $=$ $3.75$ $-$ $\frac{f(3)}{f'(3)}$ $=$ $3.73214$ $x_{5}$ $=$ $3.73214$ $-$ $\frac{f(4)}{f'(4)}$ $=$ $3.7320508$ $x_{6}$ $=$ $3.7320508$ $-$ $\frac{f(5)}{f'(5)}$ $=$ $3.7320508$ $=$ $x_{5}$ $Let$ $x_{1}$ $=$ $0.3$ $x_{2}$ $=$ $0.3$ $-$ $\frac{f(1)}{f'(1)}$ $=$ $0.267647$ $x_{3}$ $=$ $0.267647$ $-$ $\frac{f(2)}{f'(2)}$ $=$ $0.26794917$ $x_{4}$ $=$ $0.26794917$ $-$ $\frac{f(3)}{f'(3)}$ $=$ $0.26794919$ $x_{5}$ $=$ $0.26794919$ $-$ $\frac{f(4)}{f'(4)}$ $=$ $0.26794919$ $=$ $x_{4}$ $ Keep$ $six$ $decimal$ $places$ $So,$ $x$ $=$ $3.732051$ $or$ $0.267949$
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