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Topic: Real valued functions (f(x))^2=x^2 (Read 638 times) |
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knightfischer
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Real valued functions (f(x))^2=x^2
« on: Mar 17th, 2008, 9:09am » |
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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?
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pex
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Re: Real valued functions (f(x))^2=x^2
« Reply #1 on: Mar 17th, 2008, 9:23am » |
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on Mar 17th, 2008, 9:09am, knightfischer wrote: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? |
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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|.
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knightfischer
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Re: Real valued functions (f(x))^2=x^2
« Reply #2 on: Mar 17th, 2008, 11:10am » |
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Thanks, again!
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