Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.8 Newton's Method - 4.8 Exercises - Page 316: 6

Answer

$$\eqalign{ & {x_{n + 1}} = \frac{{x_n^2 + 3}}{{2{x_n} - 2}} \cr & {x_1} = 3.5 \cr & {x_2} = 3.05 \cr} $$

Work Step by Step

$$\eqalign{ & {\text{Let }}f\left( x \right) = {x^2} - 2x - 3,{\text{ and }}{x_0} = 2 \cr & {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr & f'\left( x \right) = 2x - 2 \cr & {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr & {x_{n + 1}} = {x_n} - \frac{{x_n^2 - 2{x_n} - 3}}{{2{x_n} - 2}} \cr & {x_{n + 1}} = \frac{{2x_n^2 - 2{x_n} - x_n^2 + 2{x_n} + 3}}{{2{x_n} - 2}} \cr & {x_{n + 1}} = \frac{{x_n^2 + 3}}{{2{x_n} - 2}} \cr & \cr & {\text{Taking }}{x_0} = 2 \cr & {x_0} = 2 \cr & {x_{0 + 1}} = {x_1} = \frac{{{{\left( 2 \right)}^2} + 3}}{{2{{\left( 2 \right)}^2} - 2}} = 3.5 \cr & {x_{1 + 1}} = {x_2} = \frac{{{{\left( {3.5} \right)}^2} + 3}}{{2{{\left( {3.5} \right)}^2} - 2}} = 3.05 \cr & \cr & {\text{Thus}}{\text{, we obtain}} \cr & {x_{n + 1}} = \frac{{x_n^2 + 3}}{{2{x_n} - 2}} \cr & {x_1} = 3.5 \cr & {x_2} = 3.05 \cr} $$
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