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 7.3 - Page 427: 9

Answer

$(F^{-1} \circ F)(x)=F^{-1}(F(x))=F^{-1}(3x+2)$ $\Rightarrow \frac{(3x+2)-2}{3}=\frac{3x}{3}=x=I_{\mathbb{R}}$ as was to be shown. Hence $F^{-1}\circ F=I_{\mathbb{R}}$ by definition of equality of functions. $(F\circ F^{-1})(x)=F(F^{-1}(x))=F\left( \frac{y-2}{3} \right)$ $\Rightarrow 3\left( \frac{y-2}{3} \right)+2=y-2+2=y=I_{\mathbb{R}}$ as was to be shown. Hence $F\circ F^{-1}=I_{\mathbb{R}}$ by definition of equality of functions.

Work Step by Step

$(F^{-1} \circ F)(x)=F^{-1}(F(x))=F^{-1}(3x+2)$ $\Rightarrow \frac{(3x+2)-2}{3}=\frac{3x}{3}=x=I_{\mathbb{R}}$ as was to be shown. Hence $F^{-1}\circ F=I_{\mathbb{R}}$ by definition of equality of functions. $(F\circ F^{-1})(x)=F(F^{-1}(x))=F\left( \frac{y-2}{3} \right)$ $\Rightarrow 3\left( \frac{y-2}{3} \right)+2=y-2+2=y=I_{\mathbb{R}}$ as was to be shown. Hence $F\circ F^{-1}=I_{\mathbb{R}}$ by definition of equality of functions.
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