Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 32

Answer

$$ \approx 8.06007$$

Work Step by Step

$$\eqalign{ & {4.02^{3/2}} \cr & {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr & {\text{in the exercise 24 above }}\left( {page{\text{ 662}}} \right){\text{. for }}f\left( x \right) = {\left( {4 + x} \right)^{3/2}} \cr & {\text{we found that}} \cr & {P_4}\left( x \right) = 8 + 3x + \frac{3}{{16}}{x^2} - \frac{1}{{128}}{x^3} + \frac{3}{{4096}}{x^4} \cr & {\text{To approximate }}{4.02^{3/2}},{\text{ we must evaluate }}{\left( {4 + 0.02} \right)^{3/2}},{\text{ }} \cr & {\text{then taking }}x = 0.02 \cr & {P_4}\left( {0.02} \right) = 8 + 3\left( {0.02} \right) + \frac{3}{{16}}{\left( {0.02} \right)^2} - \frac{1}{{128}}{\left( {0.02} \right)^3} + \frac{3}{{4096}}{\left( {0.02} \right)^4} \cr & {P_4}\left( {0.02} \right) = 8.06007 \cr & {\text{Thus}}{\text{, }}{4.02^{3/2}} \approx 8.06007 \cr} $$
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