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While studying the concept of maxima and minima, I came across a question:

The function $f(x)=2|x| + |x+2| - \left| |x+2| - 2|x| \right|$ Then at which points does the given function has a local minimum or a local maximum

Other than using software can this type of question be solved by plotting graphs of these functions??

Malcolm
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user550174
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    I'd break it into cases. For instance, if $0≤x≤2$ we have $f(x)=2x+(x+2)-(2-x)=4x$. For $x≥2$ we have $f(x)=2x+x+2-(x-2)=2x+4$. And so on. – lulu Apr 08 '18 at 12:25
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    Welcome to stackexchange. This is the type of question where drawing the graph is the best way to understand what's going on. Then you can verify your conclusion with a little algebra. If I were your instructor I wouldn't even need that. I certainly wouldn't want you to use any software. – Ethan Bolker Apr 08 '18 at 12:33

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Perhaps this will help:

$$\min\{a,b\} ={a+b-|a-b|\over 2}$$

so in your case $$f(x) =2\min\{|2x|,|x+2|\}$$

If $4x^2\geq (x+2)^2$ or $x\in (-\infty,-{2\over 3})\cup (2,\infty)$ we have $$f(x) = 4|x|$$ and for $x\in (-{2\over 3},2)$ we have $$f(x)=2|x+2|$$

nonuser
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  • When I tried plotting it the graph in graph generating application the shape was kind of like "W".. But still the solution that you gave is kinda hard for me to understand... So I would be really grateful if you could break it down for me, especially the part where you redefined the function based on their interval, because I'm not sure if we really could do it.. also thanks for help.. – user550174 Apr 08 '18 at 16:45