Intuition for dealing with absolute-value functions

Question

I have always struggled with the intuition behind absolute-value functions. It's hard for me to know what they would exactly look like (I know that there would be cusps and so on), and how they behave. For example, for this function, $f(x) = \frac{1}{|x-2|}-x$, I have no idea how I would take its derivatives and so on, and how to deal with it in a piece-wise nature, overall. My past question (see Finding the intervals where $f(x)=\frac{1}{|x-2|}-x$ is monotonous) illustrates this problem. I usually have difficultly in breaking these absolute-value functions into parts, and then working with the parts.

So is there an intuitive approach for dealing with absolute-value functions?

Answer 1

I prefer to consider the function case by case and use calculus. This traditional approach gives the clearest picture. But if I want to avoid calculus, I will sketch the graph step by step.

(1) The graph of $\displaystyle y=\frac{1}{x-2}$ (a hyperbola)

(2) The graph of $\displaystyle y=\frac{1}{|x-2|}$ can be obtained by reflecting the part of the curve $\displaystyle y=\frac{1}{x-2}$ below the $x$-axis, in the $x$-axis.

(3) Draw also the line $y=x$. $\displaystyle \frac{1}{|x-2|}-x$ is the vertical distance between the curve $\displaystyle y=\frac{1}{|x-2|}$ and the line $y=x$. The distances are represented by the vertical line segments. Green line segments show positive distances and purple line segments show negative distances.

(4) Plot the vertical distances to obtain the graph of $\displaystyle y=\frac{1}{|x-2|}-x$

Answer 2

It sometime varies on the case, but usually the best option is trying to break the function into parts, where the part that is in absolute value is positive or negative. For example, with the function you mentioned:

$$f(x) = \frac{1}{|x-2|} -x = \begin{cases} \frac{1}{x-2} -x \text{, if $x>2$}\\ \frac{1}{-(x-2)} - x \text{, if $x<2$} \end{cases}$$

and trying to evaluate each part separately. The key here is to recognize the intervals where the absolute value is positive, usually solving an inequality function. In that case it is a simple inequalit which is $x-2<0$ where you instantly get $x<2$. Sometimes, it can be a bit harder.

You could also take $f(x) = \frac{1}{\sqrt{(x-2)^2}}-x$ where this is comfortable, since it operates the same as taking the absolute value.

Another option is as nourert suggested above is using the $sgn(f(x))$ function, but I usually find it less helpful and less intuitive.

($sgn(f(x)) = \begin{cases} -1 \text{, if $f(x)<0$}\\0 \text{, if $f(x)=0$} \\ 1, \text{if $f(x)>0$}\end{cases}$)

Hope this helps.