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