Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 2 - Functions - Strengthen Your Understanding - Page 114: 32



Work Step by Step

Add $6$ to both sides to get $$f(x)+6=\frac{3}{4}x$$ Multiply both sides by $\frac{4}{3}$ to get $$x=\frac{4}{3}f(x)+6(\frac{4}{3})=\frac{4}{3}f(x)+8$$ Thus, $f^{-1}(y)=\frac{4}{3}y+8$. $f^{-1}(8)$ can be found using $y=8$ to get $$f^{-1}(8)=\frac{4}{3}(8)+8=\frac{32}{3}+8=\frac{32}{3}+\frac{24}{3}=\frac{56}{3}$$ This isn't equal to $0$, so the statement is false.
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