2 months and 3 weeks ago the user Milly Choochoo asked in this thread a question about how to deal with expressions involving max and min functions. Unfortunately, he didn't receive much attention. But I wrote to him an answer, that I will share with you if it can help anyone.

Dear Milly;

What is $\operatorname{max}\{a,b\}$? It's just a function that gives us the biggest number from $a,b$. The same thing for $\operatorname{min}\{a,b\}$, except that this time, it gives us the smallest number from $a,b$. Let's begin with some few examples to wrap our heads around those two functions. From the definitions we got $\operatorname{max}\{-3,8\}$ is equal to $8$. Since out of $-3$ and $8$ it's the latter that is the biggest. And for $\operatorname{min}\{-7,1\}$ we can see that it is equal to $-7$ since this last one is the smallest number out of $-7$ and $1$.

Now that we have understood those functions, we need to determine a way that will reveal to us a formula to find the min and max of any two numbers. Let's take two arbitrary numbers as an example like $x,y$ such that $\displaystyle\underline{x\lt y}$, which is a very important assumption for what will come. If we graph those into the real number line then we will see that $x$ is in the left in regards of $y$ as shown in this image:

One of the numbers we can define based on $x$ and $y$ is the their average, $(x+y)/2$. What this average graphically means, is that it represent the number whose corresponding point on the number line is a midpoint of the segment $[x,y]$, as this image show:

We've made some neat progress till now, but we'll get to the next point only by seeing that this midpoint unlocks to us a new possibility. This midpoint makes an axis of symmetry, and so with the rightly chosen way we can go from the quantity $(x+y)/2$ to the smallest number, that is $x$, or to the biggest number $y$. To see what I mean look carefully at this image:

So by adding the distance from $(x+y)/2$ to $y$ to $(x+y)/2$ itself we will obtain $y$. And if we substract the distance from $(x+y)/2$ to $x$ to $(x+y)/2$ itself we will obtain $x$. So we actually found a way to get $x$ and $y$ out of some formula that will soon discover. Note however that those two distances are the same. And how is distance represented? I mean how to find the distance from $a$ to $b$? We use absolute values! To get: $\text{distance from }a\text{ to }b=|a-b|.$ So if we use that property we will get that:

\eqalign{ \operatorname{max}\{x,y\}&=\dfrac{x+y}{2}+\left|\dfrac{x+y}{2}-y\right|\\ &=\dfrac{x+y}{2}+\left|\dfrac{x+y-2y}{2}\right| \\ &=\dfrac{x+y}{2}+\left|\dfrac{x-y}{2}\right| \\ &=\dfrac{x+y}{2}+\left|\dfrac{1}2(x-y)\right| \\ &=\dfrac{x+y}{2}+\dfrac12\left|(x-y)\right| \quad\text{using properties of the }|\,\cdot\,| \\ &=\dfrac{x+y}{2}+\dfrac{\left|(x-y)\right|}2 \\ &=\dfrac{x+y+\left|x-y\right|}2\quad\blacksquare \\ }

Now I'm sure you can do the same yourself to prove that:

$\operatorname{min}\{x,y\}=\dfrac{x+y-\left|x-y\right|}2.$

So now whenever you see a problem involving $\operatorname{min}\{x,y\}$ or $\operatorname{max}\{x,y\}$ just replace it with the formulas we found above, and the rest would be easy.

1) Prove that $\operatorname{min}\{x,x\}=\operatorname{max}\{x,x\}=x.$

2) Derive a formula for $\operatorname{min}\{x,y,z\}$ and $\operatorname{max}\{x,y,z\}.$

I hope this helps. Best wishes, $\cal H$akim.

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TopNewestWell done Champ

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Nice approach .. :)

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Great note!!! Thanks.... :)

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Thank you!

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Out of topic, are you Hakim on MSE?

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Yeah, and you are Tunk-Fey on MSE.

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