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: 24



Work Step by Step

Differentiate the given function $f(x)$ as follows: $f'(x)=6x^2+2x+3 \implies f'(1)=11 \\ f''(x)=12x+2 \implies f''(1)=14\\ f'''(x)=12 \implies f'''(2)=12$ The Taylor's series at $x=1$ can be written as: $f(x)=f(1)+f'(1) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+\dfrac{f'''(1)(x-1)^3 }{3!} =-2+11(x-1)+\dfrac{14}{2} (x-1)^2+\dfrac{12}{6} (x-1)^3$ So, $f(x)=-2+11(x-1)+7(x-1)^2+2(x-1)^3$
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