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is there any proof or theorem to say that if the inverse function $ y=f^{-1}(x) $ is POSITIVE in the sense $ f^{-1}(x) >0 $ for $x \ge 0 $ then the function $ f(x) \ge 0 $ will be also positive on the interval $ (0, \infty) $ ? in the sense that $ f(f^{-1}(x))=x $

so if a function IS POSITIVE its inverse will be also positive :D on the same interval

since taking the inverse function may be considered as reflection of the function $ y=f(x) $ across the line $ y=x $ i think that my aseveration is true.

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I think there is a bijective constraint, as you say nothing about $x<0$ behavior –  Bernhard Oct 3 '12 at 11:19
    
thanks Bernhard.. however if we assume $ f(x)=f(-x) $ i think i this case we could always take the positive branch of the function –  Jose Garcia Oct 3 '12 at 11:20
    
Your affirmation concerning the reflection is indeed sufficient (provided there is an inverse of course). –  user39572 Oct 3 '12 at 11:22
    
you can always DRAW an inverse for any function at least for piecewise continous function, so there is always an inverse –  Jose Garcia Oct 3 '12 at 11:24
    
My comment was aimed at the interval $(0,\infty)$ –  user39572 Oct 3 '12 at 11:30
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1 Answer 1

Assume $f,g\colon\mathbb R\to\mathbb R$ are functions with $f(g(x))=x$ for all $x\in \mathbb R$ and $g(x)>0$ for $x\ge 0$. Then it need not be the case that $f>0$ implies $f(x)\ge 0$.

As a simple counterexample consider $f(x)=x-1$ and $g(x)=x+1$.

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