Why isn't $]-1, 1]$ the domain of $f(x) = \arcsin\left(\frac{|x|}{x+1} \right)$? I have to find the domain of this function: $$f(x) = \arcsin\left(\frac{|x|}{x+1} \right)$$
My attempt:

*

*because arcsine is the inverse of sine function, then its domain is the image of sine: $[-1, 1]$.

*the absolute value is defined in $]-\infty, +\infty[$,

*the denominator is defined in $]-1, +\infty[$.

The domain of $f(x)$ is the intersection of these domain. Therefore, it's $]-1, 1]$.
But this is wrong, because the correct domain is $[-1/2, +\infty[$, where does this $-1/2$ come from?
 A: Lying in bed I thought of probably the clearest answer:
The domain of $\arcsin$ is $[-1,1]$ so the range of $\frac {|x|}{x+1}$ must be restricted to $-1 \le \frac {|x|}{x+1} \le 1$.
That requires that $0\le |\frac {|x|}{x+1}|=\frac {|x|}{|x+1|} \le 1$ and that $|x| \le |x+1|$.
If we consider the four possible positive/negative values of $|x|, |x+1|$ we can have

*

*$x+1< 0$
so $x< -1$ and $|x|=-x$ and $|x+1|= -(x+1)=-x-1$.  We need $-x < -x-1$ or $0 < -1$.  This never happens.  None of $(-\infty, -1)$ is in the domain of $\arcsin\frac{|x|}{x+1}$.


*$-1 \le x < 0$
(Note: $x=-1$ means $\frac {|x|}{x+1}$ will have a forbidden $0$ in the denominator so we can't have $x=-1$ no matter what we conclude below)
If $-1\le x < 0$ then $|x| = -x$ and $|x+1|=x+1$ and we must have $-x \le x+1$ or $-1 \le 2x$ or $x \ge -\frac 12$.
So, of $[-1,0)$ w have $[-\frac 12,0)$ is in the domain of $\arcsin\frac{|x|}{x+1}$.
(We can note that if $x \in (-1,-\frac 12)$ then $|x|\in(\frac 12, 1)$ and $x+1 \in (0,\frac 12)$ so $|x| > x+1> 0$ and $\frac {|x|}{x+1} > 1$.)
(We can further not if $x \in (-\frac 12,0)$ then $|x|\in (0, \frac 12)$ and $x+1\in (\frac 12, 1)$ so $0 < |x| < x+1$ so $0\le \frac {|x|}{x+1}< 1$.)


*$x \ge 0$
so $|x| = x$ and $|x+1| = x+1$ so we need $x \le x+1$ which is always the case.
So of $[0,\infty)$, all of $[0,\infty)$ is in the domain of $\arcsin\frac{|x|}{x+1}$.
Putting 1),2),3) together  the domain of $\arcsin\frac{|x|}{x+1}$ is $[-\frac 12,\infty).
=====old answer below====
One thing you can notice is if $-1 < x < -\frac 12$ then $|x|=-x \in (\frac 12, 1)$.  So $|x| > \frac 12$.
And $0 < x + 1 < \frac 12$.  So $\frac 1{x+1} > 2$.
So $\frac {|x|}{x+1} = |x|\cdot \frac 1{x+1} > \frac 12\cdot 2 = 1$. and is out of the domain for $\arctan$.
.....
But how were you supposed to figure that on your own?
Read on.
=======
Note we need $-1 \le \frac {|x|}{x+ 1} \le 1$.
As the denominator can not be equal to $0$ we must have $x \ne -1$.
If we multiply all terms by $x+1$ we will get $-(x+1) ?? |x| ?? (x+1)$... but we must take into account if $x+1$ is negative of positive.
If $x +1 > 0$ we get $-(x+1) \le |x| \le x+1$.  But if $x+1 < 0$ we will flip the inequality sign to get $-(x+1) \ge |x| \ge x+1$.
Okay... let start with $x+1 > 0$ or in other words $x > -1$.  Then we have $-(x+1)\le |x| \le x+1$ but as absolute values are never negative that is the same as $0\le |x| \le x+1$ which is the same as saying $-(x+1)=-x-1\le x \le x+ 1$.
As $x \le x+1$ always that condition is always satisfied.  But we need $-x-1\le x$ or in other words we need $-1 \le 2x$ or $x \ge -\frac 12$.
So if $x > -1$ the acceptable domain is $[-\frac 12, \infty)$.
But what if $x < -1$?
Then we get $-(x+1) \ge |x| \ge x+1$ but as $x+1$ is negative and absolute values are non-negative we have:  $0 \le |x| \le -(x+1)$ or $(x+1) \le x \le -(x+1)$.  But note $x+1 \le x$ is never possible.  So we can not have $x < -1$.
So the only acceptable domain is $[-\frac 12,\infty)$.
A: For the argument of $\arcsin$ we have
$-1\le \dfrac{|x|} {x+1}\le 1;$
1)Let $x \ge 0:$ Then $|x|=x;$
$-1\le \dfrac{x}{x+1}\le 1;$
We have $0\le \dfrac{x}{x+1} <1$ for $x \ge 0$.
2)Let $x<0:$ $|x|=-x;$
A bit trickier.
$-1 \le \dfrac{-x}{x+1} \le 1;$
a)Let $x >-1:$
$-(x+1)\le -x \le x+1;$
$-1\le 0 \le 2x+1;$
$-1/2 \le x;$
b)Let $x<-1:$
$-(x+1) \ge - x \ge (x+1);$
$-1 \ge 0 \ge 1;$
Hence rule out $x< -1.$
Putting together $1)$ and $2)a$
We get for the domain$_f$ : $[-1/2,\infty).$
