A Non-Inversible Function

If a function f(x)f(x) from reals to reals satisfies that 2f(2x)f(x2)2=12f(2x)-f(x^2)^2=1 then prove that f1(x)f^{-1}(x) does not exist.

source: me

Note by Daniel Liu
5 years, 4 months ago

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We basically have to prove that there exists distinctx1,x2x_1,x_2 such that f(x1)=f(x2)f(x_1)=f(x_2) because if f1(x)f^{-1}(x) does exist under this, then f1(f(x1))=x1,f1(f(x2))=x2f^{-1}(f(x_1))=x_1, f^{-1}(f(x_2))=x_2, which violates the definition of a function that f(x)f(x) corresponds to at most one value.

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

To elaberate, we first find a value xx such that 2x=x2=y2x=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)f(y) since (f(y)1)2=0f(y)=1(f(y)-1)^2=0\Rightarrow f(y)=1. Since x2=2xx^2=2x gives x=0,2x=0,2, we have f(0)=1,f(2)=1f(0)=1,f(2)=1.

Xuming Liang - 5 years, 4 months ago

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Excellent Solution. Did not think that way!

Avineil Jain - 5 years, 4 months ago

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

Daniel Liu - 5 years, 4 months ago

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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 .. !

Ramesh Goenka - 5 years, 4 months ago

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For what domain?

Kanthi Deep - 5 years, 4 months ago

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That's a good question to ask.

Calvin Lin Staff - 5 years, 4 months ago

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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.

Daniel Liu - 5 years, 4 months ago

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Put x=2, it gives f(4)=1 . Now differentiating the given equality, 4f'(2x)-2f(x^2)f'(x^2)*2x = 0. put x=2, it gives ,f(4)=1/2 ..clearly, the function is non-invertible..

Aamir Rafiq - 5 years, 4 months ago

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