Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Exponents and Exponential Functions - 9.4 Taylor Polynomials - Preliminary Questions - Page 492: 1

Answer

$$T_{3}(x)=9+8(x-3)+2(x-3)^{2}+2(x-3)^{3}$$

Work Step by Step

In general, with $a=3$ $$ T_{3}(x)=f(3)+f^{\prime}(3)(x-3)+\frac{f^{\prime \prime}(3)}{2}(x-3)^{2}+\frac{f^{\prime \prime \prime}(3)}{6}(x-3)^{3} $$ Using the information provided, we find $$ T_{3}(x)=9+8(x-3)+2(x-3)^{2}+2(x-3)^{3} $$
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