Answer
$g^{-1}(4)=0$
Work Step by Step
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
$\therefore g^{-1}(4)=0$
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