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$$ \frac{\sqrt{4 + \arccos\left|\frac{2-x}{x+3}\right|}}{\sqrt{x^2 - 4x + 5} - 3} $$ I'm trying to find the natural domain of the function above. I set up this conditions:
$$ \begin{cases}\sqrt{x^2 - 4x + 5} - 3\neq0&(denominator)\\x^2 - 4x + 5\ge0&(root)\\4 + \arccos\left|\frac{2-x}{x+3}\right|\ge0&(root)\\\left|\frac{2-x}{x+3}\right|\ge-1\cup\left|\frac{2-x}{x+3}\right|\le1&(arccos)\end{cases} $$
Now I know that is necessary to set up two more conditions: the existence condition of the absolute value, and the existence condition of the fraction denominator. But I don't know how to deal with absolute values. Could someone explain it to me maybe using this example?

Note: These are not homework, but an exercise chose, because has an absolute value inside, to understand the theory.

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    The last condition should be really a intersection.2012-01-26
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    I am not sure if calculus tag is right.2012-01-26
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    I think Calculus tag is alright because, this is the first time, functions are learnt for their own sake. Domain and range problems are one of the first ones that a Calculus student faces!2012-01-26
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    @KannappanSampath: Domain and range is pre-calculus, isn't it? Anyway, I won't argue too much about it :-)2012-01-26

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