Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 67: 22



Work Step by Step

$f(x)=\dfrac{4x-1}{2x+3}$ Substitute $f(x)$ by $y$: $y=\dfrac{4x-1}{2x+3}$. Take $2x+3$ to multiply both sides: $y(2x+3)=4x-1$. Evaluate the product on the left: $2xy+3y=4x-1$. Take $4x$ to the left side and $3y$ to the right side: $2xy-4x=-1-3y$. Take out common factor $x$ on the left side: $x(2y-4)=-1-3y$. Take $2y-4$ to divide both sides: $x=\dfrac{-1-3y}{2y-4}$. Interchange $x$ and $y$: $y=\dfrac{-1-3x}{2x-4}$. Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=\dfrac{-1-3x}{2x-4}$.
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.