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Let $f,g :R \to R $ one to one functions such that $f(x)< g(x), \forall x \in R $

Is it true that $f^{-1}(x)>g^{-1}(x), \forall x \in R$??

I'd say yes, thinking at their graphs: if the graph of f is below the graph of g, when we take the symmetry with respect to the line $y=x$ the graph of $f^{-1}(x)$ will be above the graph of $g^{-1}$. But i am interested in a rigorously formulated proof . If it is true...

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No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^{-1}(x)$ is to the right of the graph of $f^{-1}(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^{-1}(x)=-x$ and $g^{-1}(x)=-x+1$, so $g^{-1}(x)>f^{-1}(x)$ for all $x$.

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Right! I'd say like this: Let $f^{-1}(x)=a, g^{-1}(x)=b$ So we have $x=f(a)=g(b)$

Then from the hypothesis $g(b)>f(b)$ so $f(a)>f(b)$

And if we assume that f is strictly increasing it follows that $a>b$ So it's sufficient to add the condition that $f$ is increasing, $g$ doesn't need to be.

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  • $\begingroup$ kccu's answer gives a counterexample. I also don't see how at all your argument is meant to hold in the first place. $\endgroup$ – Eevee Trainer Apr 21 at 23:31
  • $\begingroup$ What exactly you don't see? I showed that the initial assertion is true if we add a condition. What exactly is your problem? Why you give -1? A lot of weird reactions here unfortunately... But if you feel better doing this... $\endgroup$ – amarius8312 Apr 22 at 20:00

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