a) Prove that there exists a differentiable function f:(0, ∞)->(0, ∞) such that f(f'(x))=x, for all x>0.
b) Prove that there is no differentiable function f:R->R such that f(f'(x))=x, for all x∈R.
I guess that the function is of the form f(x) = ax^b.
Then f'(x) = abx^(b-1).
f(f'(x)) = a(abx^(b-1))^b = a^(b+1) * b^b * x^(b^2 - b)
If f(f'(x)) = x, then (b^2 - b) = 1, so b = Phi, the golden ratio, which is a pleasant surprise.
also, a^(b+1) * b^b = 1, so a = b^(-b/(b+1)) which probably simplifies.
Hope I did not make a math mistake.
But this is an existence proof that an f(x) exists
Part a -- spoiler
