How to Find Polynomials for a Piecewise Function?

[Ackbach]

Here is this week's POTW:

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Find polynomials $f(x), \: g(x),$ and $h(x),$ if they exist, such that for all $x,$
\[
|f(x)|-|g(x)|+h(x) = \begin{cases} -1 & \mbox{if $x<-1$} \\
3x+2 & \mbox{if $-1 \leq x \leq 0$} \\
-2x+2 & \mbox{if $x>0$.}
\end{cases}
\]

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Re: Problem Of The Week # 248 - Jan 11, 2017

This was Problem A-1 in the 1999 William Lowell Putnam Mathematical Competition.

Honorable mention to kiwi for a solution that was $90\%$ correct. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

Note that if $r(x)$ and $s(x)$ are any two functions, then
\[ \max(r,s) = (r+s + |r-s|)/2.\]
Therefore, if $F(x)$ is the given function, we have
\begin{align*}
F(x)\ &= \max\{-3x-3,0\}-\max\{5x,0\}+3x+2 \\
&= (-3x-3+|3x+3|)/2 \\
& \qquad - (5x + |5x|)/2 + 3x+2 \\
&= |(3x+3)/2| - |5x/2| -x + \frac{1}{2},
\end{align*}
so we may set $f(x)=(3x+3)/2$, $g(x) = 5x/2$, and $h(x)=-x+\frac{1}{2}$.