I recently got acquainted to the max min functions. They seem quite confusing to understand and plot.

For example: $min (|x|,|y|)=1$
What does this mean?

I sense these mean that first we plot the equality; And then cancel those portions of the graph that lie above another portion of graph to get the minimum graph.

I reduce this function into:

$ |x|=1 $ or $ |y|=1 $
$x=1,-1 $ or $ y=1,-1 $

Plotting these I get this.

On cancelling the portions of the graph which have some of the graph below them we get this.

But desmos tells me the graph is this.

Obviously my understanding of the functions is incorrect.
Can somebody walk me through how to make and understand the graph in the example I mentioned. Or you could use any other similar example as you wish.

I would be happy to further clarify the question if anyone asks.

  • $\begingroup$ The equation is invariant by the changes of variable $x\to\pm x$, $y\to\pm y$, and $x\to y$, $y\to x$. Each of these changes of variables corresponds to a symmetry of the graph with respect to the axes, origin, or with respect to $x=y$. Therefore, it is enough to assume $y\geq x\geq 0$. In that case, $\min(|x|,|y|)=x=1$. Therefore, you get the ray $\{(x,y):\ x=1, y\geq x\}$. Now, you just need to take symmetries with respect to the axes, origin, and to the diagonal line $x=y$. $\endgroup$ Mar 31 '18 at 13:39

you have forgotten that $$|x|\geq |y|$$ or $$|x|<|y|$$ must hold. You will get two cases: $$\min(|x|,|y|)=|y|$$ if $$|x|\geq |y|$$ and $$\min(|x|,|y|)=|x|$$ if $$|x|<|y|$$

  • $\begingroup$ Could you elaborate a bit please? $\endgroup$ Mar 31 '18 at 13:29
  • $\begingroup$ I have added some lines for you! $\endgroup$ Mar 31 '18 at 13:37
  • $\begingroup$ I understood, Thanks and nice answer! $\endgroup$ Mar 31 '18 at 13:46

$\min (|x|,|y|)=1$ means: take $y$ and $x$ value, if one of them has absolute value of $1$ and the other has absolute value of $1$ or more plot this point.

So we get the lines:$$\min(|x|,|y|)=|x|=\begin{cases}|x|=1,y\ge 1\\|x|=1,y\le -1\end{cases}\\\min(|x|,|y|)=|y|=\begin{cases}x\ge1,|y|=1\\x\le-1,|y|=1\end{cases}$$

  • $\begingroup$ Good Answer!, That really helped. $\endgroup$ Mar 31 '18 at 13:47

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