Answer :

Hippalectryon
We want a function [tex]g[/tex] such that [tex]g(15x-1)=x[/tex]

One easy way to do to that is to choose [tex]g(x)=\dfrac{x+1}{15}[/tex]

That function works : [tex]g(15x-1)=\dfrac{15x-1+1}{15}=x[/tex]

Due to the uniqueness of the inverse, this is the function we were searching for.
[tex]f(x)=15x-1\to y=15x-1\\\\15x-1=y\ \ \ \ |add\ 1\ to\ both\ sides\\\\15x=y+1\ \ \ \ \ |divide\ both\ sides\ by\ 15\\\\\boxed{x=\frac{y}{15}+\frac{1}{15}}\to\boxed{x=\frac{1}{15}y+\frac{1}{15}}\\\\Answer:\boxed{\boxed{[f(x)]^{-1}=\frac{1}{15}x+\frac{1}{15}}}[/tex]

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