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 229: 81



Work Step by Step

$f(x)=\frac{1}{8}x-3$ $y=\frac{1}{8}x-3~~$ (Interchange $x$ and $y$): $x=\frac{1}{8}y-3$ $x+3=\frac{1}{8}y$ $y=8x+24$ $f^{-1}(x)=8x+24$ $(f^{-1}~o~f^{-1})(4)=f^{-1}(f^{-1}(4))=f^{-1}(8(4)+24)=f^{-1}(56)=8(56)+24=472$
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