Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises: 25

Answer

$f(2)=12$ $f(-2)=16$ $f(a)=3a^{2}-a+2$ $f(-a)=3(-a)^{2}-(-a)+2=3a^{2}+a+2$ $f(a+1)=3(a+1)^{2}-(a+1)+2=3(a^{2}+2a+1)-a-1+2=3a^{2}+5a+4$ $2f(a)=6a^{2}-2a+4$ $f(2a)=12a^{2}-2a+2$ $f(a^{2})=3a^{4}-a^{2}+2$ $[f(a)]^{2}=(3a^{2}-a+2)^{2}$ $f(a+h)=3(a+h)^{2}-(a+h)+2$

Work Step by Step

$f(x)=3x^{2}-x+2$ $f(2)=3\times2^{2}-2+2=3=3\times4=12$ $f(-2)=3\times(-2)^{2}+2+2=16$ $f(a)=3a^{2}-a+2$ $f(-a)=3(-a)^{2}-(-a)+2=3a^{2}+a+2$ $f(a+1)=3(a+1)^{2}-(a+1)+2=3(a^{2}+2a+1)-a-1+2=3a^{2}+5a+4$ $2f(a)=6a^{2}-2a+4$ $f(2a)=12a^{2}-2a+2$ $f(a^{2})=3a^{4}-a^{2}+2$ $[f(a)]^{2}=(3a^{2}-a+2)^{2}$ $f(a+h)=3(a+h)^{2}-(a+h)+2$
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