Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - 2.7 - Inverse Functions - 2.7 Exercises - Page 228: 9

Answer

It has been verified that $f^{-1}[f (x)]=f[f^{-1} (x)]$, that is, the function is the inverse function. So, the inverse function is: $\dfrac{x-1}{3}$

Work Step by Step

We have the inverse function: $f^{-1} (x)=\dfrac{x-1}{3}$ Therefore, $f[f^{-1} (x)]=f(\dfrac{x-1}{3})=3(\dfrac{x-1}{3})+1=x$ and $f^{-1}[f (x)]=f^{-1}(3x+1)=\dfrac{(3x+1)-1}{3}=x$ Therefore, it has been verified that $f^{-1}[f (x)]=f[f^{-1} (x)]$, that is, the function is the inverse function. So, the inverse function is: $\dfrac{x-1}{3}$
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