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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:

1. 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 $g(x) = (2x-1)-3$.




2. 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:

$$f^-1(g(x)) = f^-1(fracx-52) = frac fracx-52-3 5 $$

Why do we in case (2) simply plug $fracx-52$ into $x$, while in case (1) we need to substitute with $t$ and then plug into $x$? Apologies for the complexity.

Think about how $g$ works:

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


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


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


• 
  $g(2cdot 10 - 1) = 10-3$. This means $g(19)=7$
  

The rule is that $g(2cdotsquare - 1) = square - 3$.

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 "$2cdotsquare-1$" was the number $31$, then the answer would just be "$square-4$".

What I need to know, in other words, is how to solve "$2cdotsquare-1 = 31$" for "$square$". But you can do that: just add $1$ to both sides and divide by $2$. That just means "$square = tfrac31+12$".

There's nothing special about $31$. To find $g(N)$, you get $square = tfracN+12$ so that $g(N)=square-3= tfracN+12 -3$.

$N$ here is just a placeholder. You can call it $x$ or $y$ or whatever. In particular, $boxedg(x)=tfracx+12 -3$.

Now can you apply similar reasoning for the second part of the question?

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 $fract+12$. This leads then directly to the given solution
$$
g(t)=fract-52,
$$
where you now can, again simultaneously on both sides, replace the variable name $t$ with $x$.

Let's find the inverse of $f$ first. Solving for $x$, we find that $x = fracf(x) - 35$. "Renaming" variables gives you $f^-1(x) = fracx - 35$. Now, to find the expression for $g(x)$, we need to perform a change of variables. Let $x^* = 2x - 1$. Then $x = fracx^* + 12$. Substituting this into the expression for $g$, you get $g(x^*) = fracx^* + 12 - 3 = fracx^* - 52$. Now, substitute this expression into $f^-1(x)$, and you get $f^-1(g(x)) = frac15 times (fracx - 52 - 3) = fracx - 1110$.