Is there a continuous function f(x) defined on [tex](-\infty,+\infty)[/tex] such that f(f(x))=[tex]e^{-x}[/tex]?(adsbygoogle = window.adsbygoogle || []).push({});

My opinion is "no", and here is how i think:

first of all if such a function exists, it should be a "one-to-one" function, that is for every y>0, there should be exaclty one x such that f(x)=y.

Thus by the "one-to-one" property of f(x), for every [tex]x_{1}>x_{2}[/tex], either [tex]f(x_{1})>f(x_{2}) or f(x_{1})<f(x_{2})[/tex], but not [tex]f(x_{1})=f(x_{2})[/tex].

However we notice that in both case, [tex]f(f(x_{1}))>f(f(x_{2}))[/tex], and this is contradicting to the fact that [tex]e^{-x_{1}}<e^{-x_{2}}[/tex].

So I conclude that no such function f(x) exists, let along any "continuous" function.

Am I right so far?

And why emphasize "continuous" since I haven't find anything to do with continuity in my prove?

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# Is there such a continuous function?

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