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