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: 80



Work Step by Step

$f(x)=\frac{1}{8}x-3$ $y=\frac{1}{8}x-3~~$ (Interchange $x$ and $y$): $x=\frac{1}{8}y-3$ $x+3=\frac{1}{8}y$ $y=8x+24$ $f^{-1}(x)=8x+24$ $g(x)=x^3$ $y=x^3~~$ (Interchange $x$ and $y$): $x=y^3$ $y=\sqrt[3] x$ $g^{-1}(x)=\sqrt[3] x$ $(g^{-1}~o~f^{-1})(-3)=g^{-1}(f^{-1}(-3))=g^{-1}(8(-3)+24)=g^{-1}(0)=\sqrt[3] 0=0$
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