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: 25



Work Step by Step

We have $f(x)=x^4+x^2+1 \implies f(-2)=21$; $f'(x)=4x^3+2x \implies f'(-2)=-36$; $f''(x)=12x^2+2 \implies f''(-2)=50\\ f'''(x)=24x \implies f'''(2)=24$ Therefore, the Taylor's series at $x=-2$ is as follows: $f(x)=f(-2)+f'(-2) (x+2)+\dfrac{f''(-2)(x+2)^2 }{2!}+\dfrac{f''(-2)(x+2)^3 }{3!}+\dfrac{f''(-2)(x+2)^4 }{4!}$ or, $f(x)=21-36(x+2)+25(x+2)^2-8(x+2)^3+(x+2)^4$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.