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# A Non-Inversible Function

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

Note by Daniel Liu
3 years, 3 months ago

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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 $$f(0)=1,f(2)=1$$. · 3 years, 3 months ago

Excellent Solution. Did not think that way! · 3 years, 3 months ago

Yep, that's correct. Quite an interesting problem, until you realize that the solution is very quick and easy. · 3 years, 3 months ago

Comment deleted Jun 03, 2014

nevermind I see it now! anyway I'm sorry for your loss...I wonder who downvoted your comment. · 3 years, 3 months ago

Comment deleted Jun 03, 2014

Oh so once you know $$f(4)=1$$, plug in $$x=-2$$ and say $$f(-4)=a$$. The problem becomes

$2a-1=1$

and thus $$a=f(-4)=1$$. · 3 years, 3 months ago

For what domain? · 3 years, 3 months ago

I think all reals works, but if you see any contradiction feel free to tell me and I can change it. · 3 years, 3 months ago

That's a good question to ask. Staff · 3 years, 3 months ago

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 .. ! · 3 years, 3 months ago