first off, just to remind you, to get the "inverse relation" of any expression, you first do a quick switcharoo on the variables, and then solve for "y", or whatever the dependent is. So let's do so.
[tex]\bf \stackrel{f(x)}{y}=3x-\cfrac{1}{2}\qquad inverse\implies \boxed{x}=3\boxed{y}-\cfrac{1}{2}
\\\\\\
x+\cfrac{1}{2}=3y\implies \cfrac{x+\frac{1}{2}}{3}=y\implies \cfrac{x+\frac{1}{2}}{\frac{3}{1}}=y\implies \left( x+\cfrac{1}{2} \right)\cfrac{1}{3}=y
\\\\\\
\cfrac{x}{3}+\cfrac{1}{6}=y\impliedby f^{-1}(x)\\\\
-------------------------------\\\\\cfrac{(4)}{3}+\cfrac{1}{6}=f^{-1}(4)\implies \cfrac{8+1}{6}=f^{-1}(4)\implies \cfrac{9}{6}=f^{-1}(4)
\\\\\\
\cfrac{3}{2}=f^{-1}(4)[/tex]
