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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.

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    I would start by putting $t=x-2$ then cross multiply and then group by $f(t)$ terms. – 2017-02-22

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Defining $h(x) = x-2$, $k(x) = \frac{2x+3}{5-x}$ and $\ell(x)=2x-3$, which are both invertible functions, we have $$g(\ell(x)) = k(f(h(x))$$ Inverting the functions yields $$\ell^{-1}(g^{-1}(x)) = h^{-1}(f^{-1}(k^{-1}(x)))$$ so that, applying $\ell$ to both sides, we get $$g^{-1}(x) = \ell(h^{-1}(f^{-1}(k^{-1}(x))))$$ [Note that the order of the functions in a composite is reversed when you invert them.]

Finding expressions for $h^{-1}$, $k^{-1}$ and $\ell^{-1}$ isn't too difficult; doing so will yield the desired expression for $g^{-1}$ in terms of $f^{-1}$.

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    I think the right structure is $g(2x-3) = k(f(h(x))$ – 2017-02-22
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    @S.H.W: Good catch, I've updated my answer. Thanks! – 2017-02-22
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    Very good answer , thank you very much! – 2017-02-22
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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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    Okay but we won't give $g^{-1}$ in terms of $f^{-1}$ – 2017-02-22
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    @S.H.W ,you can do it for $g^{-1}$ – 2017-02-22