<s>[INST] I don't understand functional equations. A complicated question. 
We are given the following:
$f(x)=5x+3$
$g(2x−1)=x−3$.
We have to find:
$f^{-1}∘g(x) = ?$

I have the solution to the problem, but I don't understand two things:

*

*In order to solve for $g(x)$, we need to have substitution where $t = 2x-1$, and then plug that $x$ onto $x-3$. Why? Given my understanding of functions, it's more logical to me that we should just plug $2x-1$ into $x-3$, so we get $g(x) = (2x-1)-3$.


*Later on in the exercise, after we find $g(x)$ and $f^{-1}(x)$, we are faced with what, for me, seems like the same problem from above. Yet we solve it differently. Here is what I mean:
$$f^{-1}(g(x)) = f^{-1}(\frac{x-5}{2}) = \frac{ \frac{x-5}{2}-3 }{5} $$
Why do we in case (2) simply plug $\frac{x-5}{2}$ into $x$, while in case (1) we need to substitute with $t$ and then plug into $x$? Apologies for the complexity.
 [/INST] Let's find the inverse of $f$ first. Solving for $x$, we find that $x = \frac{f(x) - 3}{5}$. "Renaming" variables gives you $f^{-1}(x) = \frac{x - 3}{5}$. Now, to find the expression for $g(x)$, we need to perform a change of variables. Let $x^* = 2x - 1$. Then $x = \frac{x^{*} + 1}{2}$. Substituting this into the expression for $g$, you get $g(x^{*}) = \frac{x^{*} + 1}{2} - 3 = \frac{x^{*} - 5}{2}$. Now, substitute this expression into $f^{-1}(x)$, and you get $f^{-1}(g(x)) = \frac{1}{5} \times (\frac{x - 5}{2} - 3) = \frac{x - 11}{10}$.
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