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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.
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 $x<0$ ), 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 $y$ .
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 .
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 |
$x\le 0$ , so $\mid x\mid =-x$ | $x\ge 0$ , so $\mid x\mid =x$ |
$y\le 0$ , so $\mid y\mid =-y$ | $y\le 0$ , so $\mid y\mid =-y$ |
$(-x)+(-y)=4$ | $x+(-y)=4$ |
$y=-x-4$ | $y=x-4$ |
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