University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 26



Work Step by Step

We have $f'(x)=15x^4-4x^3+6x^2+2x \implies f'(-1)=23$; $f''(x)=60x^3-12x^2+12x+2 \implies f''(-1)=-82\\ f'''(x)=180x^2-24x+12 \implies f'''(2)=216$ and $f^{4}(-1)=-384$ Therefore, the Taylor's series at $x=-1$ is as follows: $f(x)=f(-1)+f'(-1) (x+1)+\dfrac{f''(-1)(x+1)^2 }{2!}+\dfrac{f''(-1)(x+1)^3 }{3!}+\dfrac{f''(-1)(x+1)^4 }{4!}$ or, $f(x)=-7+23(x+1)-41(x+2)^2+36(x+1)^3-16(x+1)^4+3(x+1)^5$
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