Respuesta :

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]

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