Respuesta :

naǫ
[tex]f(x)=4(x+3)-5 \\ f(x)=4x+12-5 \\ f(x)=4x+7 \\ \\ \hbox{solve for x:} \\ y=4x+7 \\ y-7=4x \\ x=\frac{y-7}{4} \\ \\ \hbox{swap x and y:} \\ y=\frac{x-7}{4} \\ \\ f^{-1}(x)=\frac{x-7}{4} \\ f^{-1}(3)=\frac{3-7}{4}=\frac{-4}{4}=-1 \\ \boxed{f^{-1}(3)=-1}[/tex]