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        riddles >> general problem-solving / chatting / whatever >> Real valued functions (f(x))^2=x^2
        
(Message started by: knightfischer on Mar 17^th, 2008, 9:09am)

Title: Real valued functions (f(x))^2=x^2
Post by knightfischer on Mar 17^th, 2008, 9:09am

How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2=x^2 for each x in [-1,1]?

The answer is four. I find two: f(x)=x, f(x)=-x.

Would the other two be f(x)=|x|, f(x)=|-x|?

Can anyone help with this?

Title: Re: Real valued functions (f(x))^2=x^2
Post by pex on Mar 17^th, 2008, 9:23am

on 03/17/08 at 09:09:39, knightfischer wrote:

This seems correct, except that the last one should, of course, be -|x| instead of |-x|.

Consider that [f(x)]² = x² implies that, for each x, f(x) is either x or -x (otherwise, its square wouldn't be x²). By continuity, the only place where the function can "move" from being x to being -x (or vice versa) is where x = -x; thus, at x=0.

Therefore, we have indeed four possibilities:
1. f(x) = x everywhere,
2. f(x) = -x everywhere,
3. f(x) = -x for x<0, x for x>0; that is, f(x) = |x|,
4. f(x) = x for x<0, -x for x>0; that is, f(x) = -|x|.

Title: Re: Real valued functions (f(x))^2=x^2
Post by knightfischer on Mar 17^th, 2008, 11:10am

Thanks, again!