## Calculus: Early Transcendentals 8th Edition

$g^{-1}(4)=0$
Let $g^{-1}(4)=a$ $g(a)=3+a+e^a=4$ $a+e^a=1$ try $a=0$ LHS=$0+e^0=1$=RHS $\therefore g^{-1}(4)=0$