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After checking this answer, I did not quite get what I was trying to understand.

I have a routine exercise which states: Let $f: \Bbb R^2 \rightarrow \Bbb R$, $(x,y) \rightarrow \text{max} \{|x|,|y|\}$ Sketch $f^{-1}([-1,1] \cup [2,4])$

Ok, so here's how the answer goes for this:

By definition, $f ^{-1}([-1, 1] \cup [2, 4]) = \{(x, y) \in \Bbb R^2 : -1 \leq \text{max}\{|x|,|y|\} \leq 1 \text{ or } 2 \leq \text{max}\{|x|,|y|\} \leq 4\}$. Hence, an appropriate sketch is: enter image description here

But how do they sketch this? What are the steps for this?

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1 Answer 1

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Since change of $x$ to $-x$ and $y$ to $-y$ leaves the curves invariant, they symmetric in all quadrant. So consider only the first quadrant where $x,y \ge 0$.

(1) The first figure is bounded by four lines: $ x=0, y=0, x=1$ and $y=1$. Finally it is a square with corners $(1,1), (-1,1), (-1,-1),(1,-1)$. So, it is a region with all points inside it and hence it is full grey.

(2) Similarly, $max{|x|,|y|} \ge 2$ indicates all points outside the square with corners $(2,2),(-2,2),(-2,-2),(2,-2)$ and $max{|x|,|y|}\le 4$ means all points inside the square with corners $(4,4),{-4,4),(-4,-4),(4,-4)$. So the required area is the overlap of these two squares and it is greyed..

Finally, required region is the union of these two grey regions.

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  • $\begingroup$ so basically sketching a max(|x|,|y|) < a means the region inside the square which is centred at the origin with side of length 2a? $\endgroup$
    – user634512
    Mar 14, 2020 at 15:24
  • $\begingroup$ I have talked in terms of corners center as $(0,0)$ is understood.! Cheers. $\endgroup$
    – Z Ahmed
    Mar 14, 2020 at 15:27

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