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Can we find $f(x)$ given that $1-f(x) = f(-x)$ for all real $x$?

I start by rearranging to: $f(-x) + f(x) = 1$. I can find an example such as $f(x) = |x|$ that works for some values of $x$, but not all. Is there a method here? Is this possible?

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    Homework? . . .2010-07-30
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    Nope, just a problem posed to me - what is the line on homework, by the way? If I show I've sufficiently worked on a problem (i.e. not just posting the problem), then that's okay?2010-07-30
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    @Danish: That's correct.2010-07-30
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    @Danish, @Kaestur: As a rule, it's good to say why you care about getting the answer to your question.2010-07-30
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    @Danish Nice cricket name.2010-07-30

6 Answers 6