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# Inequalities and Absolute Value Concepts -- "Piecewise Functions" and Absolute Value

Module by: Kenny M. Felder. E-mail the author

Summary: This module introduces piecewise functions for the purpose of understanding absolute value equations.

What do you get if you put a positive number into an absolute value? Answer: you get that same number back. 5=55=5 size 12{ lline 5 rline =5} {}. π=ππ=π size 12{ lline π rline =π} {}. And so on. We can say, as a generalization, that x=xx=x size 12{ lline x rline =x} {}; but only if xx size 12{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. 5=(5)=55=(5)=5 size 12{ lline - 5 rline = - $$- 5$$ =5} {}. π=(π)=ππ=(π)=π size 12{ lline - π rline = - $$- π$$ =π} {}. We can say, as a generalization, that x=xx=x size 12{ lline x rline = - x} {}; but only if xx size 12{x} {} is negative.

So the absolute value function can be defined like this.

## The “Piecewise” Definition of Absolute Value

x = { x , x 0 x , x < 0 x = { x , x 0 x , x < 0
(1)

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 xx size 12{ - x} {} looks suspicious. “I thought an absolute value could never be negative!” Well, that’s right. But if xx size 12{x} {} is negative, then xx size 12{ - x} {} is positive. Instead of thinking of the xx size 12{ - x} {} as “negative xx size 12{x} {}” it may help to think of it as “change the sign of xx size 12{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=xy=x size 12{y= lline x rline } {}. But you can explain the V shape very easily with the piecewise definition. On the right side of the graph (where x0x0 size 12{x >= 0} {}), it is the graph of y=xy=x size 12{y=x} {}. On the left side of the graph (where x<0x<0 size 12{x<0} {}), it is the graph of y=xy=x size 12{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+xy=x+x size 12{y=x+ lline x rline } {}. How do we approach that? With the piecewise definition, it becomes a snap.

x + | x | = { x + x = 2 x x 0 x + ( - x ) = 0 x < 0 x + | x | = { x + x = 2 x x 0 x + ( - x ) = 0 x < 0
(2)

So we graph y=2xy=2x size 12{y=2x} {} on the right, and y=0y=0 size 12{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 xx and yy.

## Example 1: Graph |x|+|y|=4

We saw that in order to graph xx size 12{ lline x rline } {} we had to view the left and right sides separately. Similarly, yy size 12{ lline y rline } {} divides the graph vertically.

• On top, where y0y0 size 12{y >= 0} {}, y=yy=y size 12{ lline y rline =y} {}.
• Where y<0y<0 size 12{y<0} {}, on the bottom, y=yy=y size 12{ lline y rline = - 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. x+y=4x+y=4 size 12{ lline x rline + lline y rline =4} {}
 Second Quadrant First Quadrant x≤0x≤0 size 12{x >= 0} {}, so ∣x∣=−x∣x∣=−x size 12{ lline x rline = - x} {} x≥0x≥0 size 12{x >= 0} {}, so ∣x∣=x∣x∣=x size 12{ lline x rline =x} {} y≥0y≥0 size 12{y >= 0} {}, so ∣y∣=y∣y∣=y size 12{ lline y rline =y} {} y≥0y≥0 size 12{y >= 0} {}, so ∣y∣=y∣y∣=y size 12{ lline y rline =y} {} (−x)+y=4(−x)+y=4 size 12{ $$- x$$ +y=4} {} x+y=4x+y=4 size 12{x+y=4} {} y=x+4y=x+4 size 12{y=x+4} {} y=−x+4y=−x+4 size 12{y= - x+4} {} Third Quadrant Fourth Quadrant x≤0x≤0 size 12{x <= 0} {}, so ∣x∣=−x∣x∣=−x size 12{ lline x rline = - x} {} x≥0x≥0 size 12{x <= 0} {}, so ∣x∣=x∣x∣=x size 12{ lline x rline =x} {} y≤0y≤0 size 12{y <= 0} {}, so ∣y∣=−y∣y∣=−y size 12{ lline y rline = - y} {} y≤0y≤0 size 12{y <= 0} {}, so ∣y∣=−y∣y∣=−y size 12{ lline y rline = - y} {} (−x)+(−y)=4(−x)+(−y)=4 size 12{ $$- x$$ + $$- y$$ =4} {} x+(−y)=4x+(−y)=4 size 12{x+ $$- y$$ =4} {} y=−x−4y=−x−4 size 12{y= - x - 4} {} y=x−4y=x−4 size 12{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=x4y=x4 size 12{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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