Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Review - Exercises - Page 70: 23



Work Step by Step

let $f^{-1}(2)=x, f(x)=2$ $2=2x+\ln x$ try $x=1$ RHS= $2\times1+\ln1=2+0=2$ =LHS $\therefore f^{-1}(2)=1$
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