Answer:
[tex]\large{\begin{array}{c|c}f^{-1}(x)&x\\\\-2&0\\0&4\\3&10\\4&12\end{array}}[/tex]
Step-by-step explanation:
[tex]f(x)=2x+4\\\\f^{-1}(x)=f(y)=2y+4\\\\f^{-1}(x)=-2\to x=f(-2)=2(-2)+4=-4+4=0\\\\f^{-1}(x)=0\to x=f(0)=2(0)+4=0+4=4\\\\f^{-1}(x)=3\to x=f(3)=2(3)+4=6+4=10\\\\f^{-1}(x)=4\to x=f(4)=2(4)+4=8+4=12[/tex]
