[tex]f(x)=3x+1\to y=3x+1\\\\\text{change x to y and y to x}\\\\x=3y+1\ \ \ \ \text{solve for y}\\\\3y+1=x\ \ \ \ |-1\\\\3y=x-1\ \ \ \ |:3\\\\y=\dfrac{x-1}{3}\\\\\boxed{f^{-1}(x)=\dfrac{x-1}{3}}\\\\f^{-1}(6)\to\text{put x=6 to the equation of the function}\ f^{-1}(x)\\\\f^{-1}(6)=\dfrac{6-1}{3}=\dfrac{5}{3}=1\dfrac{2}{3}\\\\\boxed{f^{-1}(x)=1\dfrac{2}{3}}[/tex]
