Q:

Domain of an absolute value

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!

A:

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.

A:

Hints:

You need the denominator to be nonzero.
You need the expression in the square root to be nonnegative.

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

A:

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.

