Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 156: 17

Answer

f(-1) = 0 f(0) = 2 f(1) = 8 f(2) = 18 f(3) = 32

Work Step by Step

$f(x) = 2(x+1)^2$ when x = -1 $f(-1) = 2(-1 + 1)^2$ $f(-1) = 2(0)^2$ $f(-1) = 2(0)$ $f(-1) = 0$ when x = 0 $f(0) = 2(0 + 1)^2$ $f(0) = 2(1)^2$ $f(0) = 2(1)$ $f(0) = 2$ when x = 1 $f(1) = 2(1 + 1)^2$ $f(1) = 2(2)^2$ $f(1) = 2(4)$ $f(1) = 8$ when x = 2 $f(2) = 2(2 + 1)^2$ $f(2) = 2(3)^2$ $f(2) = 2(9)$ $f(2) = 18$ when x = 3 $f(3) = 2(3 + 1)^2$ $f(3) = 2(4)^2$ $f(3) = 2(16)$ $f(3) = 36$
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