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How do we solve for a domain of a function, when it involves absolute values? For example (I created the example myself, so it might be a bit weird):

$$f(x) = \frac{1}{\sqrt{|2x+1| - |x-3|}}$$

Thank you!

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3 Answers 3

up vote 3 down vote accepted

Not at all weird, very nice example!

What I would suggest as a general approach is this. Imagine you have a value of $x$ in a calculator and you want to compute your expression one step at a time. What, if anything, could go wrong? This takes a lot of time to write out but once you get used to doing it in your head it's not so bad. So for your example we have $x$. Now

  • calculate $2x$ - this will always work and there will be no problems

  • calculate $2x+1$ - still no problems

  • $|2x+1|$ - no problems

  • similarly, $|x-3|$ will give no problems

  • and now $|2x+1|-|x-3|$ will give no problems

  • now for $\sqrt{|2x+1|-|x-3|}$: this will fail if $|2x+1|-|x-3|<0$

  • and for $1/\sqrt{|2x+1|-|x-3|}$: this will now fail if $\sqrt{|2x+1|-|x-3|}=0$, that is, $|2x+1|-|x-3|=0$.

So what it all comes down to: the calculation will fail if $|2x+1|-|x-3|\le0$, and it will be OK if $$|2x+1|-|x-3|>0\ .$$ The domain consists of all $x$ satisfying this inequality, and if you know how to solve this inequality you can find specific values of $x$.

If you don't know how to solve it here is a hint: write as $$|2x+1|>|x-3|$$ and then square both sides.

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Very good explanation! I understand it now :) Thank you! –  Phantom Feb 21 at 0:38

The insides of the absolute values change signs at $\frac {-1}2$ and $3$, so you can work over each of three regions and resolve the signs. Then within those you need the expression under the square root sign to be strictly positive. So for $x \lt \frac {-1}2$, you have $f(x)=\frac 1{\sqrt {(1-2x)-(3-x)}}=\frac 1{\sqrt{-2-x}}$ so you need $x \lt -2$. You do the other two regions the same way.

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Hints:

  1. You need the denominator to be nonzero.

  2. You need the expression in the square root to be nonnegative.

Hence, you need $$|2x+1|-|x-3|>0$$

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