[tex]f(x)=3x-\dfrac{1}{2}\\\\y=3x-\dfrac{1}{2}\ \ \ \ |\cdot2\\\\2y=6x-1\ \ \ |+1\\\\2y+1=6x\ \ \ \ |:6\\\\x=\dfrac{2y+1}{6}\\\\therefore\ f^{-1}(x)=\dfrac{2x+1}{6}[/tex]
[tex]\text{substitute x=3 to the equation:}\\\\f^{-1}(3)=\dfrac{2\cdot3+1}{6}=\dfrac{7}{6}[/tex]
