College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 192: 20


$f(-2)=-12$ $f(-1)=-3$ $f(0)=0$ $f( \frac{1}{2})=\frac{9}{8}$

Work Step by Step

We are given: $f(x)=x^{3}+2x$ We evaluate: $f(-2)=(-2)^{3}+2(-2)=-8+-4=-12$ $f(-1)=(-1)^{3}+2(-1)=-1+-2=-3$ $f(0)=0^{3}+2(0)=0+0=0$ $f( \frac{1}{2})=(\frac{1}{2})^{3}+2(\frac{1}{2})=\frac{1}{8}+1=\frac{9}{8}$
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