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: 23

Answer

$f(0)=0$ $f(3)=15$ $f(-3)=3$ $f(a)=a^{2}+2a$ $f(-x)=x^{2}-2x$ $f( \frac{1}{a})=\frac{1}{a^{2}}+\frac{2}{a}$

Work Step by Step

We are given: $f(x)=x^{2}+2x$ We evaluate: $f(0)=0^{2}+2(0)=0+0=0$ $f(3)=3^{2}+2(3)=9+6=15$ $f(-3)=(-3)^{2}+2(-3)=9-6=3$ $f(a)=a^{2}+2(a)=a^{2}+2a$ $f(-x)=(-x)^{2}+2(-x)=x^{2}-2x$ $f( \frac{1}{a})=(\frac{1}{a})^{2}+2(\frac{1}{a})=\frac{1}{a^{2}}+\frac{2}{a}$
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