Question

What is the range of $f\left(x\right)=\frac{|x-1|}{x-1}.$

Answer 1

We have,

$f\left(x\right)=\frac{|x-1|}{x-1}$

Now for,

$|x-1|=(x-1)when,x>1$

$|x-1|=-(x-1)when,x<1$

$\therefore f\left(x\right)=\frac{x-1}{x-1}=1x\in (1,\infty )$

$f\left(x\right)=-\frac{(x-1)}{x-1}=-1x\in (-\infty ,1)$

Now, for $x=1$

Then,

$R.H.L.$

${lim}_{x\to 1+}f\left(x\right)={lim}_{h\to 0}f(1+h)={lim}_{h\to 0}\frac{|1+h-1|}{1+h-1}=\frac{h}{h}=1$

$L.H.L.$

${lim}_{x\to 1-}f\left(x\right)={lim}_{h\to 0}f(1-h)={lim}_{h\to 0}\frac{|1-h-1|}{1-h-1}=-\frac{h}{h}=-1$

Hence, It does not exits.

Then, Range is $[-1,1]$

Domain is $[-\infty ,\infty ]$

Range is $R-\left\{1\right\}$

Domain is $(-\infty ,\infty )-\left\{1\right\}$

Hence, this is the answer.