Problem 1:
[tex]y=f\left( x \right) ,\\ \\ y=6-3x\\ \\ 3x=6-y\\ \\ x=\frac { 6 }{ 3 } -\frac { y }{ 3 } \\ \\ x=2-\frac { 1 }{ 3 } y\\ \\ \therefore \quad { f }^{ -1 }\left( x \right) =2-\frac { 1 }{ 3 } x[/tex]
Problem 2:
[tex]y=g\left( x \right) ,\\ \\ y=x+2\\ \\ x=y-2\\ \\ \therefore \quad { g }^{ -1 }\left( x \right) =x-2[/tex]
