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

Question

I was teaching absolute function properties in a K-12 class. I made this question in my mind.

Suppose $f(x)$ is a one-to-one function, and its definition is $f(x)=max\left \{ x,3x\right \}=ax+b|x|+c$ find $a,b,c$

As I was teaching, I solved the question by two methods:

First, put three arbitrary numbers for $x$, for example, $x=0,1,2$ and solve the system of equations:

$$\left\{\begin{matrix} x=0 &max \left \{ 0,0 \right \}=0 &a(0)+b(0)+c=0 \rightarrow c=0\\ x=1&max \left \{ 1,3 \right \}=3 &a+b+c=3\\ x=-1& max \left \{ -1,-3 \right \}=-1 & -a+b+c=-1 \end{matrix}\right.\\b=1 ,a=2\\ max\left \{ x,3x\right \}=2x+1|x|+0$$

Second, plot max{x,3x} and determine $a,b,c$ with respect to a figure of the function:

Obviously, $c=0$ $$\rightarrow a=\left\{\begin{matrix} x>0 & max\left \{ x,3x \right \}=3x&=ax+bx+0 \\ x<0 & max\left \{ x,3x \right \}=x& =ax-bx+0 \end{matrix}\right.\\ \forall x \in \mathbb{R}\begin{Bmatrix} (a+b)x=3x\\ (a-b)x=x \end{Bmatrix}\rightarrow a=2,b=1 \\\rightarrow max\left \{ x,3x \right \}=2x+|x|+0 $$

...and then, one of the students asked for a third method. I told her, "I will think, and you must think too!"

Now I have two questions:

1. What do you think about my handmade question?
2. Can you give me suggestion for more methods to solve it?

Answer 1

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

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

Answer 2

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!

Answer 3

Note that $f(-x) = -ax + b|x|+ c$, thus breaking into symmetrical and anti symmetrical combinations one obtains $$ f(x) - f(-x) = 2ax\\ f(x) + f(-x) = 2b|x| + 2c $$ Thus $$ a = \frac{f(1) - f(-1)}{2} = \frac{3 + 1}{2} = 2\\ c = f(0) = 0\\ b = \frac{f(1) + f(-1)}{2} - c = \frac{3 - 1}{2} = 1 $$

Second variation of the same idea: $$ f(-x) = \max(-x, -3x) = -\min(x, 3x) $$ so $$ f(x) \pm f(-x) = \max(x, 3x) \mp \min(x, 3x)\\ f(x) + f(-x) = \max(x, 3x) - \min(x, 3x) = |x - 3x| = 2|x|\\ f(x) - f(-x) = \max(x, 3x) + \min(x, 3x) = x + 3x = 4x $$ Comparing with $$ f(x) - f(-x) = 2ax\\ f(x) + f(-x) = 2b|x| + 2c $$ we get $$ a = 2\quad b = 1\quad c = 0 $$

Answer 4

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

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

Answer 5

 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|