How to show there's no differentiable f:R-->R such that f(f(x) = e^(-x) -1?

Question

I think this is interesting. I posted this before, but there was a typo and the definition of g was wrong. Now it's correct

Show there's no f:R-->R, differentiable on R, such that the composite g = f o f is given by g(x) = e^(-x) -1

Answer 1

Assume such an f exists. Then f'(0) must exist and be a real number.
First note that f(f(x)) has a single fixed point at x=0. Next, f(f(f(0)))=f(0), so f(0) is a fixed point of f(f(x)), hence f(0)=0. Taking derivatives we get f'(f(x))f'(x)=-e^(-x), hence at x=0 we get f'(f(0))f'(0)=-1. Recalling f(0)=0, this becomes f'(0)^2=-1 for a contradiction.