I went down the wrong path in trying to show that; if $f(x)=ax+b$ then there is a $g(x)=dx+e$ such that $f(g(x)) = x$. This wrong path led me to a baffling result of x being a constant. Can anyone explain either my error or what it means? Note that on my 2nd attempt I did derive $g(x)$, so that is not my question. My question is what does the following mean? How can x be a constant?
I did very simple substitution, just to see where it led me:
- if $f(g(x)) = x$ then
- $f(dx+e) = x$ and substituting;
- $a(dx+e) +b = x$ and expanding
- $adx +ae + b = x$ and grouping
- $(ad-1)x = -ae - b$
- $x = \frac{-ae-b}{ad-1}$ which is a constant
I know this is not how to get the inverse function. I was able to figure that out. But why do the above steps lead to an obviously impossible value for x?
The problem with the reasoning is the same old division by zero. But this is very disguised.
Now, you start off with $f(g(x)) = x$, as an equality of functions : so what this actually means, is that for every real number $r$ we have $f(g(r)) = r$.
Your manipulations led you to $(ad-1)x = -ae-b$, which continued to be an equality of functions.
At this point, you have divided both sides by $ad-1$. This has led to the functional equation $x = c$ for some constant $c$, which is a contradiction, and you feel you have done something wrong.
The truth is, you have obtained by contradiction that $ad-1 = 0$, because if indeed $ad-1 \neq0$ then the functional inequality $(6)$ would have held, which is not possible.
So no , $x$ cannot be a constant. Instead, the assumption that $ad-1\neq 0$ is false and has led you down this path, and once you tread it you know a contradiction awaits you. So that tells you $ad-1=0$.
Now, knowing that $ad-1 = 0$, from $(5)$ we get $ae+b = 0$. Thus, your objective of obtaining $d$ and $e$ in terms of $a$ and $b$ is achieved. Note that if $a= 0$ these constants can't be found.
Just want to add that if you want to derive $d$ and $e$ there's the easier method of substitution which you can use. As I mentioned earlier, a functional equality, upon substitution of appropriate numbers, becomes a numerical equality. So substitute numbers.
For example, suppose that $f(g(x)) = x$. Then substitute $x = 0$, to get $ae+b = 0$. Substitute $x = 1$ to get $a(d+e)+b = 1$, and solve these.