Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.1 Properties of Functions - 2.1 Exercises - Page 54: 10

Answer

Please see image range = $\{0,3,6,9,12,15\}$

Work Step by Step

$\left[\begin{array}{lll} x & -3x+9 & (x,y)\\ \hline-2 & 6+9=15 & (-2,15)\\ -1 & 3+9=12 & (1,12)\\ 0 & 0+9=9 & (0,9)\\ 1 & -3+9=6 & (1,6)\\ 2 & -6+9=3 & (2,3)\\ 3 & -9+9=0 & (3,9) \end{array}\right]$ The range (second column values) is $\{0,3,6,9,12,15\}$
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