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Find a rational function $f: \mathbb R \rightarrow \mathbb R$ with range $f(\mathbb R)=[-1,1]$

(Thus $f(x)=\frac{P(x)}{Q(x)}$ for all $x \in \mathbb R$ for suitable polynomials P and Q, where Q has no real root).

I'm not entirely sure where on this question to start? Any suggestions would be appreciated.

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2 Answers 2

How about the following function? $$\dfrac{2x}{x^2+1}$$

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EDIT: Oops, my bad –  Hagen von Eitzen Nov 21 '12 at 17:24
    
$f(1)=2/2=1$ and $f(-1)=-2/2=1$, so it's $[-1,1]$. –  Max Morin Nov 21 '12 at 17:29
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One way to significantly restrict the range is by using functions like $x \mapsto x^2$ : its range is half of the real numbers. By composing this with a rational function sending $0,1,\infty$ to $-1,0,1$, you get what you want.

For example, $x \mapsto \frac{x-1}{x+1}$ does this, so after composing we get : $x \mapsto \frac{x^2-1}{x^2+1}$

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Not quite. There is no $x \in \mathbb{R}$ such that $f(x) = 1$. –  user17762 Nov 21 '12 at 17:38
    
right. I was too hasty –  mercio Nov 21 '12 at 17:52
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