Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.2 New Functions From Old - Exercises Set 0.2 - Page 27: 75

Answer

It is not true, in general, that for any nonconstant functions $f$ and $g$ we have $f \circ g=g \circ f$; in some cases this condition is satisfied and in some cases it is not.

Work Step by Step

It is not true, in general, that for any nonconstant functions $f$ and $g$ we have $f \circ g=g \circ f$; in some cases this condition is satisfied and in some cases it is not. For example, for the functions $f(x)=x^2$ and $g(x)=x^3$ we have$$(f \circ g)(x)=f(g(x))= (x^3)^2=x^6=(x^2)^3=g(f(x))=(g \circ f)(x);$$in general, for the functions $f(x)=x^m$ and $g(x)=x^n$ we have$$(f \circ g)(x)=f(g(x))= (x^n)^m=x^{mn}=(x^m)^n=g(f(x))=(g \circ f)(x).$$ Now, consider the functions $f(x)=x^2$ and $g(x)=x+1$, so we have$$(f \circ g)(x)=f(g(x))= (x+1)^2=x^2+2x+1 \neq x^2+1=g(f(x))=(g \circ f)(x).$$
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