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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..

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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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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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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