We must check cases where $\textbar x-\textbar{} x-1\textbar\textbar$ is positive and cases where it is negative. $x\textgreater1$ and $x\textless1$ will cause different signs for $x-\textbar x-1\textbar$. Hence we must check both intervals. However, when $x\textless1$, we see that when $x=\frac{1}{2}$, the expression is equal to 0 , so we must also check the intervals $x\textless\frac{1}{2}$ and $\frac{1}{2}\textless x\textless1$. Solving on the interval $x\textless\frac{1}{2}$, we get $\textbar x+1+x\textbar-x=1 \Longleftrightarrow x=0$. Solving on the interval $\frac{1}{2}\textless x\textless1$, we get $\textbar x+1+x\textbar-x=1 \Longleftrightarrow x=0$. Solving on the interval $x\textgreater1$, we get $\textbar x-x+1\textbar-x=1 \Longleftrightarrow x=0$. Checking for extraneous solutions, we find that $x=0$ 
Find the sum of all possible values of $x$ such that $\textbar x-\textbar{} x-1\textbar\textbar-x=1$ .~\uline{~~~~~~~~~~}~ 
1. $$0$$ 
2. $$1$$ 
3. $$2$$ 
4. $$3$$ 