Solve for f(1) when f(x) is always positive and x>0, satisfying $f(x)f^{-1}(x)=1$

Question

I've encountered this problem when I tried to solve for the value of f(1).

$f(x)f^{-1}(x)=1, x>0, f(x) > 0$

Answer 1

This can be derived from the definition of an inverse, namely,

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

To start, from your restriction,

$$f^{-1}(x) = \frac{1}{f(x)}$$

$$f^{-1}(f(x)) = \frac{1}{f(f(x))}=x$$

$$f(f(x))=\frac{1}{x}$$

Using a nearly identical argument where we merely swap $f$ and $f^{-1}$, we get

$$f^{-1}(f^{-1}(x)) = \frac{1}{x}$$

Additionally, since $$f(f(x))^{-1} = f^{-1}(f^{-1}(x))$$ By your restriction

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

Therefore, no such function exists, as if one did, given a number in its domain that is not equal to 1 called $\alpha$,

$1=\frac{1}{\alpha}^2 \neq 1$, which is a contradiction.