Answer:
[tex]\large\boxed{f^{-1}(x)=\dfrac{x+4}{3}=\dfrac{1}{3}x+\dfrac{4}{3}}[/tex]
Step-by-step explanation:
[tex]f(x)=3x-4\to y=3x-4\\\\\text{exchange x to y and vice versa}\\\\x=3y-4\\\\\text{solve for y}\\\\3y-4=x\qquad\text{add 4 to both sides}\\\\3y=x+4\qquad\text{divide both sides by 3}\\\\y=\dfrac{x+4}{3}[/tex]
