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: 79



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 $y$): $x=y^3$ $y=\sqrt[3] x$ $g^{-1}(x)=\sqrt[3] x$ $(f^{-1}~o~g^{-1})(-1)=f^{-1}(g^{-1}(1))=f^{-1}(\sqrt[3] 1)=f^{-1}(1)=8(1)+24=32$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.