If $f\big(f(x)\big)=2^x-1$ for every $x\in\mathbb{R}$, then what is $f(0)+f(1)$?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $$f\big(f(x)\big)=2^x-1$$ for all $x\in\mathbb{R}$. Determine the value of $f(0)+f(1)$.

My approach: Let $a_1=x$ and $a_2=f(x)$. Then $a_3=f(f(x))=f(a_2)$, and continue the process $a_n=f(a_{n-1})$ hence $f(a_n)=f(f(a_{n-1}))=2^{a_{n-1}}-1=a_{n+1}$. Now how can I continue to get the result? Or is this recurrence approach is not good here?

Note : It is easy to verify that $f$ is injective.

Answer 1

Set $x\to f(x)$ to get:

$$f(2^x-1)=2^{f(x)}-1$$

Therefore (setting $x=0$ and $x=1$): $f(1)=2^{f(1)}-1$ and $f(0)=2^{f(0)}-1$. The equation $2^x=x+1$ has two roots $0$ and $1$, so, since $f$ is injective, we must have:

$$\{f(0),f(1)\}=\{0,1\}$$

The sum is therefore $1$.

Answer 2

We have $$\{f(0), f(1)\} \stackrel{f}\mapsto \{0,1\} \stackrel{f}\mapsto \{f(0),f(1)\}$$ so the function $g(x) = (f \circ f)(x) = 2^{x}-1$ maps the set $\{f(0),f(1)\}$ to itself.

If $f(0) < f(1)$ then also $g(f(0)) < g(f(1))$ so it must be $g(f(0)) = f(0)$ and $g(f(1)) = f(1)$. Therefore $f(0)$ and $f(1)$ satisfy $g(x) = x$ and the only two such points are $0,1$. We conclude $\{f(0),f(1)\} = \{0,1\}$.