## Calculus: Early Transcendentals 8th Edition

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