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This might sound like a really obvious question, but I wanted to make sure.

Is the absolute value of the expression (-x-1)/(x²+1) = (x+1)/(x²+1) because you can take the minus out and make the expression -(x+1)/(x²+1) and then the absolute value removes the minus sign? Or would it simply be (x-1)/(x²+1)?

Craig
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  • If $x^2+1$ is positive, it depends on the sign of $x+1$ – Vincent Nov 17 '16 at 15:03
  • x is positive in this expression. – Craig Nov 17 '16 at 15:04
  • Then you know the answer – Vincent Nov 17 '16 at 15:05
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    Neither: $$\left|\frac{-x-1}{x^2+1}\right|=\frac{|x+1|}{x^2+1};.$$ The quantity $x^2+1$ is always positive, but we don’t know the sign of $-x-1$; we do know, as you said, that $-x-1=-(x+1)$, so that $$|-x-1|=|-(x+1)|=|x+1|;,$$ and we might as well use the simpler form $|x+1|$, but we still need the absolute value signs: we don’t know whether $x+1$ is positive, negative, or zero. If you know that $x\ge-1$, then of course you can remove the absolute value signs around $x+1$. – Brian M. Scott Nov 17 '16 at 15:05

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