When I was teaching absolute function properties, I suddenly made this question ...

Solution 1:

You can use the fact that, $\max\{x,y\}=\frac{x+y+|x-y|}{2}$.

Therefore, $\max\{x,3x\}=\frac{x+3x+|x-3x|}{2} = \frac{4x+|-2x|}{2} =2x+|x|$.

Solution 2:

I will probably put more emphasis on those transformation properties of the $\max(\cdot)$ function which is reasonably "obvious" and easy to remember/use. For example,

1. $\max(a+b,a+c) = a + \max(b,c)$.
2. $\max(ab,ac) = a \max(b,c)$ when $a > 0$.
3. $\max(a,-a) = |a|$.

This may help a student to get familiar with such "tools" for attacking similar problems.
In any event, once a student knows about above properties, we have

$$\max(x,3x) = \max(2x - x,2x + x) \stackrel{(*1)}{=} 2x + \max(-x,x) \stackrel{(*3)}{=} 2x + |x|$$

Solution 3:

 max{x,3x} + min{x,3x} = 4x
 max{x,3x} - min{x,3x} = |2x|
 ----------------------------
 2max{x,3x} = 4x + 2|x|

 max{x,3x} = 2x + |x|


As a bonus, min{x,3x} = 2x - |x|

Solution 4:

I like the question! It is not immediately obvious to me, however, that it suffices to match your form to the given function at three points. Perhaps that is the case, but if so it requires a separate argument. Alternatively, having discovered the final form you could verify it directly.

I would address the original problem this way: Your function, $f$, is $3x$ if $x ≥ 0$ and $x$ if $x < 0$. We know functions that look like that! $g(x) = x + |x|$, for example, is given by $2x$ if $x≥0$ and $0$ if $x<0$. Ah, but that's just almost your function already. All we have to do is add x.

Less formal than your method, clearly, but perhaps it has the advantage of drawing from a list of functions your students may already have looked at. Anyway, it's an alternative!