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Is there a decreasing bijective function $f: \Bbb R \rightarrow \Bbb R$ such that $f(f(x)) -f(x^3)=2^{f(x)}$ is true $\forall x \in \Bbb R$? If it exists, find an example for such function, if not, explain why.

Ok so, if it's decreasing, that means $(\forall x_1,x_2)(x_1<x_2)\Rightarrow (f(x_1)>f(x_2))$. If it's bijective, that means it maps onto the entire $\Bbb R$.

Let's assume the opposite of $f(f(x)) -f(x^3)=2^{f(x)}$, i.e.:

  1. $f(f(x)) -f(x^3)>2^{f(x)}$

If it's decreasing, that means $f(f(x))=x$. That means $x -f(x^3)>2^{f(x)}$.

(This also means that $x-f(x^3)$ has to be greater than $0$ because $2^{f(x)}>0$. I don't know if this is important but I'm out of ideas what else to do with this problem.)

  1. $f(f(x)) -f(x^3)<2^{f(x)}$

Mostly the same reasoning as the first case.

What should I do from here? Is this a good approach to this problem?

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No, say $a=f(-1)$, $b=f(0)$ and $c=f(1)$. Then $a>b>c$ since it is decreasing. Thus $f(a)<f(b)<f(c)$ and $2^a>2^b>2^b$ since $x\mapsto 2^x$ is increasing. So $$f(a)-a<f(b)-b<f(c)-c$$ and now we get $$2^a<2^b<2^c$$ A contradiction. Must be mistake here since I never use that it is bijective.

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