Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 7 - Functions - Exercise Set 7.1 - Page 394: 15

Answer

$F.G:\mathbb{R}\rightarrow \mathbb{R}\\ G.F:\mathbb{R}\rightarrow \mathbb{R} for\,all\,x\in \mathbb{R} $ $(F.G)(x)=F(x).G(x)\,\,\,\,\,\,by\,definition\,of\,(F.G)(x)\\ (F.G)(x)=F(x).G(x)=G(x).F(x)\,\,\,\\ (by\,the\,commutative\,law\,for\,addition\,of\,real\,numbers)\\ but\,G(x).F(x)=(G.F)(x)\\ so\,\,F.G=G.F $

Work Step by Step

$F.G:\mathbb{R}\rightarrow \mathbb{R}\\ G.F:\mathbb{R}\rightarrow \mathbb{R} for\,all\,x\in \mathbb{R} $ $(F.G)(x)=F(x).G(x)\,\,\,\,\,\,by\,definition\,of\,(F.G)(x)\\ (F.G)(x)=F(x).G(x)=G(x).F(x)\,\,\,\\ (by\,the\,commutative\,law\,for\,addition\,of\,real\,numbers)\\ but\,G(x).F(x)=(G.F)(x)\\ so\,\,F.G=G.F $
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