composite function problem If I have the following expression:
$$g(f(x))-g(x)=1,$$
it is possible to express $f(x)$ in terms of the $g(x)$:
$$f(x)=g^{-1}(1+g(x)).$$
Is it possible to express $g(x)$ in terms of $f(x)$?
 A: In general, no. $g$ may not even be invertible.
With a more refined class of functions, you might have better luck. E.g. restricting $g$ to an exponential function.
Let $g(x) := e^{ax+b}$, then
$$e^{af(x)+b} = e^{ax+b}+1 \implies af(x)+b = \ln(e^{ax+b}+1) \implies f(x) = \frac{\ln(e^{ax+b}+1)-b}{a}$$
Or, we could let $g$, $f$ be polynomials. Then
$$g(f(x)) = g(x)+1 \implies \deg(g) = \deg(g)\deg(f) \implies \deg(f) = 1$$
(noting that $\deg(g) \not = 0$)
So $f$ is some line $ax+b$. Let $g(x) := \sum\limits_{k=0}^n a_k \ x^k$
Then $g(ax+b) = g(x)+1 \implies a_n a^n = a_n \implies a = \pm 1$
If $a = -1$, then $g(-b+b) = g(b)+1 = g(0)$
But $g(b) = g(0)+1 \implies g(0) = g(b)-1$
So $a = 1$.
Then $g(2b) = g(b)+1 = g(0)+2$, etc...
Hopefully the pattern is clear, but (formally, by induction) we will have
$$g(mb) = g(0)+m$$
Note that $b$ cannot be $0$, therefore define $h(x) := g(x)-g(0)-\frac{1}{b} x$
$g(0) \not = g(b)$ says $n \geq 1$, so $\deg(h) = 0$ or $1$.
$$h(mb) = g(mb)-g(0)-\frac{1}{b} mb = g(mb) -g(0) -m = 0$$
$h$ cannot have more than one root, so it must be the zero function.
$\therefore 0 = g(x)-g(0)-\frac{1}{b}x \implies g(x) = \frac{1}{b}x +g(0)$
