Verified Answer :

Hippalectryon

We want a function \(g\) such that \(g(15x-1)=x\)

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

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

Due to the uniqueness of the inverse, this is the function we were searching for.

\(f(x)=15x-1\to y=15x-1\\\\15x-1=y\ \ \ \ |add\ 1\ to\ both\ sides\\\\15x=y+1\ \ \ \ \ |divide\ both\ sides\ by\ 15\\\\\boxed{x=\frac{y}{15}+\frac{1}{15}}\to\boxed{x=\frac{1}{15}y+\frac{1}{15}}\\\\Answer:\boxed{\boxed{[f(x)]^{-1}=\frac{1}{15}x+\frac{1}{15}}}\)

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