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

#Algebra #Functions #InverseFunction #Existence

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $f(0)=1,f(2)=1$ .

Xuming Liang - 7 years ago

Excellent Solution. Did not think that way!

Avineil Jain - 7 years ago

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

Daniel Liu - 7 years 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 .. !

Ramesh Goenka - 7 years ago

For what domain?

Kanthi Deep - 7 years ago

That's a good question to ask.

Calvin Lin Staff - 7 years ago

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

Daniel Liu - 7 years ago

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 - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$