Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 26



Work Step by Step

Differentiate the given function $f(x)$ as follows: $f'(x)=15x^4-4x^3+6x^2+2x \implies f'(-1)=23 \\ f''(x)=60x^3-12x^2+12x+2 \implies f''(-1)=-82\\ f'''(x)=180x^2-24x+12 \implies f'''(-1)=216$ The Taylor series at $x=-1$ is as follows: $f(x)=f(-1)+f'(-1) (x+1)+\dfrac{f''(-1)}{2!}(x+1)^2 +\dfrac{f''(-1)}{3!} (x+1)^3 +\dfrac{f''(-1)}{4!} (x+1)^4 \\=-7+23(x+1)-41(x+2)^2+36(x+1)^3-16(x+1)^4+3(x+1)^5$
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