If a function \(f(x)\) from reals to reals satisfies that \[2f(2x)-f(x^2)^2=1\] then prove that \(f^{-1}(x)\) does not exist.

*source: me*

If a function \(f(x)\) from reals to reals satisfies that \[2f(2x)-f(x^2)^2=1\] then prove that \(f^{-1}(x)\) does not exist.

*source: me*

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x=2, it givesf(4)=1. Now differentiating the given equality,4f'(2x)-2f(x^2)f'(x^2)*2x = 0. putx=2, it gives ,f(4)=1/2..clearly, the function is non-invertible..Log in to reply

For what domain?

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I think all reals works, but if you see any contradiction feel free to tell me and I can change it.

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That's a good question to ask.

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for x(2-x) has two solutions 0 and 2 .. !! the values of the function is 1 for two elements in its domain i.e x=0 and 2 .. its not onto .. !

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We basically have to prove that there exists distinct\(x_1,x_2\) such that \(f(x_1)=f(x_2)\) because if \(f^{-1}(x)\) does exist under this, then \(f^{-1}(f(x_1))=x_1, f^{-1}(f(x_2))=x_2\), which violates the definition of a function that \(f(x)\) corresponds to at most one value.

For this problem simply take \(x=0,2\) and we have it.

To elaberate, we first find a value \(x\) such that \(2x=x^2=y\) so that we get an equation with one variable, it turns out that this would lead to only one solution for \(f(y)\) since \((f(y)-1)^2=0\Rightarrow f(y)=1\). Since \(x^2=2x\) gives \(x=0,2\), we have \(f(0)=1,f(2)=1\).

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Excellent Solution. Did not think that way!

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Yep, that's correct. Quite an interesting problem, until you realize that the solution is very quick and easy.

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Comment deleted Jun 03, 2014

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nevermind I see it now! anyway I'm sorry for your loss...I wonder who downvoted your comment.

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Comment deleted Jun 03, 2014

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\[2a-1=1\]

and thus \(a=f(-4)=1\).

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