Find the range of the function $f(x)=-1/|1-x|$ My procedure: Let $f(x)=y$.
$$|1-x|=-1/y$$
But,
$$|1-x|>0$$
so,
$$-1/y>0$$
Then, what will I do? And what is the answer?
 A: Range is $y < 0$ since $y$ can't be zero and $|1-x|$ is non negative.
A: $f(x) = -\frac{1}{|1-x|}$ for all x i.e domain being (R$-${1}) as it is discontinuous at $x=1$. 
Case 1 : $1-x>0$ $\implies$ $x<1$ i.e the domain is $(-∞,1)$. Now $f(x) = \frac{1}{x-1}$, here as $x \to 1^-$ , $f(x)\to-∞$ and as $x\to-∞$ , $f(x) \to 0$. Summarizing we can say that for domain $(-∞,1)$ the range corresponds to $(-∞,0)$.
Case 1 : $1-x<0$ $\implies$ $x>1$ i.e the domain is $(1,∞)$. Now $f(x) = \frac{1}{1-x}$, here as $x\to1^+$ , $f(x) \to -∞$ and as $x\to∞$ , $f(x) \to 0$. Summarizing we can say that for domain $(1,∞)$ the range corresponds to $(-∞,0)$.
Overall range is $(-∞,0)$ for the complete real line with a discontinuity at $x=1$.
A: Try to sketch graph you will get more clarity $f(x) = -\frac{1}{|1-x|}$  you will see range is $(-∞,0)$ for the complete real line with a discontinuity at $x=1$.
A: Begin with the following consider $f(x)$ as composition of $|1-x|$ then $1/x$ then $-x$ the first maps $R$ onto the positive $R^+\cup\{0\}$ the second maps the resulting set to itself finally the last one will flip so it maps it to $R^-\cup \{0\}$ I consider the extended real without considering that the result is $(-\infty,0)$
