Answer:
[tex]\large\boxed{f^{-1}(x)=\dfrac{1}{5}x+\dfrac{8}{5}}[/tex]
Step-by-step explanation:
[tex]f(x)=5x-8\to y=5x-8\\\\\text{change}\ x\ \text{to}\ y\ \text{and vice versa}\\\\x=5y-8\\\\\text{solve for}\ y:\\\\5y-8=x\qquad\text{add 8 to both sides}\\\\5y-8+8=x+8\\\\5y=x+8\qquad\text{divide both sides by 5}\\\\\dfrac{5y}{5}=\dfrac{x}{5}+\dfrac{8}{5}\\\\y=\dfrac{1}{5}x+\dfrac{8}{5}[/tex]
