For Polynomial Lovers 2

Suppose $|f(x)|-|g(x)|+|h(x)|=F(x)$

And since $x=-1$ and $x=0$ are critical points of $F(x)$ we can write it as

$\large F(x)=a|x+1|+b|x|+cx+d$ , where $a$ , $b$ , $c$ , $d$ are constants.

$\large F(x) =\begin{cases} (c-a-b)x+d-a, \quad x<-1\\ (a+c-b)x+a+d,\quad -1 \leq x \leq 0\\ (a+b+c)x+a+d,\quad x\geq 0\end{cases}$

On comparing with the given we get $a=\frac{3}{2},b=\frac{-5}{2},c=-1,d=\frac{1}{2}$

This gives $f(x)= \frac{3x+3}{2}, g(x)= \frac{5x}{2}, h(x)= -x+ \frac{1}{2}$

Note that by symmetry $f(x)$ and $h(x)$ may be interchanged

And finally $g(2)= \boxed{5}$