<s>[INST] find $g^{-1}(x)$ in terms of $f^{-1}$ Consider $g(2x-3) = \frac{2f(x-2) + 3}{5 - f(x-2)}$ . Also $f$ and $g$ are invertible . Now find $g^{-1}(x)$ in terms of $f^{-1}$. 
My try : Because in the parentheses we have $x-2$ and $2x-3$ instead of $x$ , I don't know how we can solve this problem.
 [/INST] HInt:
first $$x \to x+2 \\\to 
g(2(x+2)-3)=\dfrac{2f(x+2-2) + 3}{5 - f(x+2-2)}\to \\
g(2x+1)=\dfrac{2f(x) + 3}{5 - f(x)}$$
now find $f(x)$ then apply $f^{-1}$
$$y=\dfrac{2f(x) + 3}{5 - f(x)} \to \\
5y-yf(x)=2f(x)+3 \to \\f(x)(y+2)=5y-3 \\ \to \\
f(x)=\dfrac{5y-3}{y+2}$$
apply $f^{-1}$to both sides 
$$f^{-1}f(x)=f^{-1}(\dfrac{5y-3}{y+2})\\
x=f^{-1}(\dfrac{5y-3}{y+2})\\
x=f^{-1}(\dfrac{5g(2x+1)-3}{g(2x+1)+2})$$
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