Let $f$ be a function such that $f[f(x) - e^x] = e + 1$ for all $x \in \Bbb R$. Find $f(\ln 2)$.
I've considered two cases:
- $f(x) = e^x + c$, where $c$ is constant. Then $f(c) = e^c + c = e + 1$, which implies that $c = 1$, thus $f(x) = e^x + 1$ and $f(\ln 2) = 3$.
- $f(x) = e + 1$. This clearly satisfies the conditions and thus $f(\ln 2) = e + 1$.
Now I want to know how to analyse this equation further. And is it possible to find all $f$ that satisfy the equation above?
The general solution for $f$ is ugly and heavily relies on the axiom of choice. It goes like this:
Choose a non-empty set of real numbers $A$, no matter how big or small. It can be continuous, or maybe countable, or finite, or consist of just one point. Define $f(x)=e+1$ for any $x\in A$.
$\forall x\not\in A$ (if any), select some $C\in A$ and define $f(x)=e^x+C$.
Your first solution arises if we choose $A=\{1\}$ and the second is the result of $A=\mathbb R$.
That's what happens when the continuity is neither required explicitly nor enforced by the equation itself.
Back to the original question: $f(\ln2)$ can be anything except $2+\ln2$.