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.
 A: Think about how $g$ works:


*

*$g(2\cdot 7 - 1) = 7-3$. This means $g(13)=4$

*$g(2\cdot 8 - 1) = 8-3$. This means $g(15)=5$

*$g(2\cdot 9 - 1) = 9-3$. This means $g(17)=6$

*$g(2\cdot 10 - 1) = 10-3$. This means $g(19)=7$
The rule is that $g(2\cdot\square - 1) = \square - 3$.
Now focus on the list of values on the right. We have shown how to get $g(13)$, $g(15)$, $g(17$), and $g(19)$.
Suppose I want to determine the value of $g(31)$, but I don't want to continue the pattern and compute everything in between.
If I knew what to put into "$\square$" so that "$2\cdot\square-1$" was the number $31$, then the answer would just be "$\square-4$".
What I need to know, in other words, is how to solve "$2\cdot\square-1 = 31$" for "$\square$". But you can do that: just add $1$ to both sides and divide by $2$. That just means "$\square = \tfrac{31+1}{2}$".
There's nothing special about $31$. To find $g(N)$, you get $\square = \tfrac{N+1}{2}$ so that $g(N)=\square-3= \tfrac{N+1}{2} -3$.
$N$ here is just a placeholder. You can call it $x$ or $y$ or whatever. In particular, $\boxed{g(x)=\tfrac{x+1}{2} -3}$.
Now can you apply similar reasoning for the second part of the question?
A: You need to be consistent in your substitutions. If you replace $x$ in
$$
g(2x-1)=x-3
$$
with $t=2x-1$, you need to do it on both sides simultaneously,
$$
g(2(2x-1)-1)=(2x-1)-3,
$$
which does not simplify the situation.
What you want is to replace $2x-1$ with $t$ or $x$ with $\frac{t+1}2$. This leads then directly to the given solution
$$
g(t)=\frac{t-5}2,
$$
where you now can, again simultaneously on both sides, replace the variable name $t$ with $x$.
A: 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 $f^{-1}(g(x)) = \frac{1}{5} \times (\frac{x - 5}{2} - 3) = \frac{x - 11}{10}$.
