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: 23



Work Step by Step

We have $f(x)=x^3+2x+4 \implies f(2)=8$; $f'(x)=3x^2-2 \implies f'(2)=10$; $f''(x)=6x \implies f''(2)=12\\ f'''(x)=6 \implies f'''(2)=6$ Therefore, the Taylor's series at $x=2$ is as follows: $f(x)=f(2)+f'(2) (x-2)+\dfrac{f''(2)(x-2)^2 }{2!}+\dfrac{f'''(2)(x-2)^3 }{3!}$ or, $=8+10(x-2)+\dfrac{(12) (x-2)^2 }{2 \cdot 1}+\dfrac{(6)(x-2)^3 }{3 cdot 2 \cdot 1}$ or, $f(x)=8+10(x-2)+6(x-2)^2+(x-2)^3$
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.