# Complex inverse function

I've got a problem when solving an inverse function. Usually when I have a basic function and trying to find its inverse is not a problem. I just solve for X and find it. But now I've got a more complicated one and I don't know how to start.

If the inverse function is $$f^{-1}(x+2) = \frac{3x-2}{2x+3}$$

I have to find

$$f\left( \frac{1}{x-4} \right).$$

I really don't know where to start. I would know to solve it if it was simple as just f(x) but this is more complex. If someone could show me step by step it would really help me. Thank you ! :D

Hint: If $f^{-1}(x+2)=\frac{3x-2}{2x+3}$, then $f^{-1}(x)=\frac{3(x-2)-2}{2(x-2)+3}=\frac{3x-8}{2x-1}$ and you can solve for $f$ accordingly.
Brute force: $$f^{-1}(x) = \frac{3x-8}{2x-1} \Rightarrow y(2x-1) =3x -8 \Rightarrow x(2y-3) = y - 8 \Rightarrow f(x) = \frac{x-8}{2x-3}.$$
$$f\left(\frac{1}{x-4}\right) = \frac{\frac{1}{x-4} - 8 }{\frac{2}{x-4} - 3} = \frac{1 - 8(x-4)}{2-3(x-4)} = \frac{8x-33}{3x-14}$$
This is how you can do: first, write $z=x-2$. Then, solve the equation for $z$, which becomes: $$f^{-1}(z)=\frac{3(z-2)-2}{2(z-2)+3}$$ hence, solve the equation $f^{-1}(z)=y$ for $z$ and obtain $$z=\frac{8-y}{3-2y}$$ which therefore gives the explicit expression for $f$ to be $$f(y)=\frac{8-y}{3-2y}$$ Finally, substitute $y=\frac{1}{x-4}$ in this equation and obtain the desired result, which is $$f(\frac{1}{x-4})=\frac{8x-33}{3x-14}.$$