Answer:
[tex]\huge\boxed{A.\ f^{-1}(x)=\dfrac{x-3}{5}}[/tex]
Step-by-step explanation:
[tex]f(x)=5x+3\to y=5x+3\\\\\text{exchange}\ x\ \text{with}\ y\\\\5y+3=x\\\\\text{solve for}\ y\\\\5y+3=x\qquad|\text{subtract 3 from both sides}\\\\5y+3-3=x-3\\\\5y=x-3\qquad|\text{divide both sides by 5}\\\\\dfrac{5\!\!\!\!\diagup y}{5\!\!\!\!\diagup}=\dfrac{x-3}{5}\\\\y=\dfrac{x-3}{5}\to f^{-1}(x)=\dfrac{x-3}{5}[/tex]
