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: 33

Answer

$f(-4) = 8$ $f(-\frac{3}{2}) = -\frac{3}{4}$ $f(-1) = -1$ $f(0) = 0$ $f(25) = -1$

Work Step by Step

Because $-4$ is less than $-1$ we have $f(-4) = (-4)^2 + 2(-4) = 16-8 = 8$ For $-\frac{3}{2}$ it less than $-1$, therefore $f(-\frac{3}{2})$ = $(-\frac{3}{2})^2$ + $2(-\frac{3}{2})$ = $(\frac{9}{4})$ - $2(-\frac{3}{2})$ = $(\frac{9}{4}) - 3 = -\frac{3}{4}$ Because $-1$ is less or equal $-1$, we have $f(-1) = (-1)^2 + 2(-1) = 1-2 = -1 $ For $0$ is greater than $-1$ and less than $1$, therefore $f(0) = 0$ Because $25$ is greater than $1$, therefore $f(25) = -1$
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