2.5 "piecewise functions" and absolute value

What do you get if you put a positive number into an absolute value? Answer: you get that same number back. $\mid 5\mid =5$ . $\mid \pi \mid =\pi$ . And so on. We can say, as a generalization, that $\mid x\mid =x$ ; but only if $x$ is positive .

OK, so, what happens if you put a negative number into an absolute value? Answer: you get that same number back, but made positive. OK, how do you make a negative number positive? Mathematically, you multiply it by –1 . $\mid -5\mid =-(-5)=5$ . $\mid -\pi \mid =-(-\pi )=\pi$ . We can say, as a generalization, that $\mid x\mid =-x$ ; but only if $x$ is negative .

So the absolute value function can be defined like this.

If you’ve never seen this before, it looks extremely odd. If you try to pin that feeling down, I think you’ll find this looks odd for some combination of these three reasons.

1. The whole idea of a “piecewise function”—that is, a function which is defined differently on different domains—may be unfamiliar. Think about it in terms of the function game. Imagine getting a card that says “If you are given a positive number or 0, respond with the same number you were given. If you are given a negative number, multiply it by –1 and give that back.” This is one of those “can a function do that ?” moments. Yes, it can—and, in fact, functions defined in this “piecewise manner” are more common than you might think.
2. The $-x$ looks suspicious. “I thought an absolute value could never be negative!” Well, that’s right. But if $x$ is negative, then $-x$ is positive. Instead of thinking of the $-x$ as “negative $x$ ” it may help to think of it as “change the sign of $x$ .”
3. Even if you get past those objections, you may feel that we have taken a perfectly ordinary, easy to understand function, and redefined it in a terribly complicated way. Why bother?

Surprisingly, the piecewise definition makes many problems easier . Let’s consider a few graphing problems.

You already know how to graph $y=\mid x\mid$ . But you can explain the V shape very easily with the piecewise definition. On the right side of the graph (where $x\ge 0$ ), it is the graph of $y=x$ . On the left side of the graph (where ), it is the graph of $y=-x$ .

Still, that’s just a new way of graphing something that we already knew how to graph, right? But now consider this problem: graph $y=x+\mid x\mid$ . How do we approach that? With the piecewise definition, it becomes a snap.

So we graph $y=\mathrm{2x}$ on the right, and $y=0$ on the left. (You may want to try doing this in three separate drawings, as I did above.)

Our final example requires us to use the piecewise definition of the absolute value for both $x$ and $y$ .

Graph |x|+|y|=4

We saw that in order to graph $\mid x\mid$ we had to view the left and right sides separately. Similarly, $\mid y\mid$ divides the graph vertically .

• On top, where $y\ge 0$ , $\mid y\mid =y$ .
• Where , on the bottom, $\mid y\mid =-y$ .

Since this equation has both variables under absolute values, we have to divide the graph both horizontally and vertically, which means we look at each quadrant separately . $\mid x\mid +\mid y\mid =4$

Second Quadrant	First Quadrant
$x\le 0$ , so $\mid x\mid =-x$	$x\ge 0$ , so $\mid x\mid =x$
$y\ge 0$ , so $\mid y\mid =y$	$y\ge 0$ , so $\mid y\mid =y$
$(-x)+y=4$	$x+y=4$
$y=x+4$	$y=-x+4$
Third Quadrant	Fourth Quadrant
, so $\mid x\mid =-x$	, so $\mid x\mid =x$
, so $\mid y\mid =-y$	, so $\mid y\mid =-y$
$(-x)+(-y)=4$	$x+(-y)=4$
$y=-x-4$	$y=x-4$

Now we graph each line, but only in its respective quadrant. For instance, in the fourth quadrant, we are graphing the line $y=x-4$ . So we draw the line, but use only the part of it that is in the fourth quadrant.

Repeating this process in all four quadrants, we arrive at the proper graph.

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