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How do you find $f(x)$ if you know that: $$f(2x) = 2f(x) - f(x)^2$$

The result is: $f(x)=1-e^{cx}$ (where $c$ is an arbitrary constant).

What would be the steps to get to the result?

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    $\begingroup$ You want to assume $f$ is continuous, otherwise there are more exotic solutions. $\endgroup$ – Robert Israel Apr 10 '18 at 19:51
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    $\begingroup$ @ChristianF: The case $f(x)=0$ is already covered by the given function, by selecting $c=0$. $\endgroup$ – celtschk Apr 10 '18 at 20:06
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Hint

With $g(x)=1-f (x) $ you obtain: $$g (2x) =g (x)^2$$ You can then look at this

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