Piecewise from Rational Absolute Value Function

Question

How would one separate a function like the following into piecewise?

$$f(x)={\left|4-x\right|\over{\left|x-4\right|}}$$

I've been taught that with a rational function with an absolute value in the numerator only, one does the following:

$$g(x)={{\left|4-x\right|\over{x-4}} = \begin{cases}{4-x\over{x-4}} & x<4 \\ {-(4-x)\over{x-4}} & x>4 \end{cases}}$$

Eventually, of course, the pieces would be simplified, but I'll leave it like that for simplicity's sake.

Meanwhile, with an absolute value over an absolute value, I can't find the correct piecewise. When I take the limit of the function, I should get the answer $1$, but I can't do so without graphing the problem. Is there any way to create a correct piecewise version of this function?

Answer 1

You’re making it much too complicated. No matter what $x$ is, $4-x=-(x-4)$, so as long as $x\ne 4$,

$$\frac{4-x}{x-4}=-1\;,$$

and therefore

$$\left|\frac{4-x}{x-4}\right|=|-1|=1\;.$$

Of course the expression is undefined when $x=4$.

Answer 2

$f(x)=1$ for all $x\neq 4$:

$dom(f)=\{x\in\mathbb{R}:f(x)\mbox{ exists }\}=\{x\in\mathbb{R}:|x-4|\neq 0\}=\mathbb{R}\backslash\{4\}$

Now, if $x\in dom(f)$ (i.e. $x\neq 4$) we have $f(x)=\frac{|4-x|}{|x-4|}$. But all real $a$ we have $|a|=|-a|$. Hence $f(x)=\frac{|4-x|}{|x-4|}=\frac{|-(4-x)|}{|x-4|}=\frac{|x-4|}{|x-4|}=1$.