College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.2 - Page 240: 41


a. 8 b. 3 c. 6

Work Step by Step

a. $5\neq 3$, so we calculate $h(5)$ using the $\displaystyle \frac{x^{2}-9}{x-3}$ part $h(5)=\displaystyle \frac{5^{2}-9}{5-3}=\frac{25-9}{2}=\frac{16}{2}=8$ b. $0\neq 3$, so we calculate $h(0)$ using the $\displaystyle \frac{x^{2}-9}{x-3}$ part $h(5)=\displaystyle \frac{0^{2}-9}{0-3}=\frac{-9}{-3}=3$ c. for $x=3, h(3)=6$
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