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: 24



Work Step by Step

We have $f(x)=2x^3+x^2+3x-8 \implies f(1)=-2$; $f'(x)=6x^2+2x+3 \implies f'(1)=11$; $f''(x)=12x+2 \implies f''(1)=14\\ f'''(x)=12 \implies f'''(2)=12$ Therefore, the Taylor's series at $x=1$ is as follows: $f(x)=f(1)+f'(1) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+\dfrac{f'''(1)(x-1)^3 }{3!}$ or, $=-2+11(x-1)+\dfrac{(14) (x-1)^2 }{2 }+\dfrac{(12)(x-1)^3 }{6}$ or, $f(x)=-2+11(x-1)+7(x-1)^2+2(x-1)^3$
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