Writing inverse without using piecewise function

Question

I am trying to find the inverse function for the given function f(x) = 3x - |x| + |x - 2|. I have already found that the inverse function can be expressed as the piecewise function f^{-1}(x) = \begin{cases} \frac{x - 2}{3} & \text{if } x < 2 \\ x - 2 & \text{if } 2 \leq x < 4 \\ \frac{x + 2}{3} & \text{if } x \geq 4 \end{cases} ] but I am struggling to see the method for writing it with absolute values instead. The answer key provides the inverse as f^{-1}(x) = 1/3(x + |x - 2| - |x - 4|). These are clearly the same, which can be seen by graphing, but I can't see a method for how one would reduce the piecewise answer to this simpler form. Is there a standard method, does it require some creativity, or do you just have to get lucky and see it?

Answer 1

If you already know a function $g:\mathbb R\to\mathbb R$ is continuous and (finite-pieces) piecewise linear. You can always have the formula $g(x)=a_0x+\sum_{k=1}^na_k|x-b_k|+a_{n+1}$ where $b_1<\dots<b_n$ are the points where $g $ is non-differentiable. By solving a linear system $a_0,\dots,a_{n+1}$ can be determined.

Update: Although it is duplicated with this post and mvw's answer there explained how to get $a_k$. Here is my reformulation.

Notice that $g$ is completely determined by $g(b_1),\dots,g(b_n)$ and $g'(b_1-)=\lim_{x\to-\infty}\frac{g(x)}x,g'(b_n+)=\lim_{x\to+\infty}\frac{g(x)}x$. By plugging their values we get the linear system.

Alternatively, we have $g'(b_j+)-g'(b_j-)=2a_j$ for $1\le j\le n$. Note that $g'(b_1-)=a_0-a_1-\dots-a_n$ and $g'(b_n+)=a_0+\dots+a_n$ which gives $2a_n=g'(b_1-)+g'(b_n+)$. Lastly $a_{n+1}=g(0)-\sum_{k=1}^na_k|b_k|$.

Back to your case $(b_1,b_2)=(2,4)$. We have $a_0=\frac12(\frac13+\frac13)=\frac13$ since the leftmost and rightmost slope are both $0$.

We have $a_1=\frac{g'(2+)-g'(2-)}2=\frac12(1-\frac13)=\frac13$ and $a_2=\frac{g'(4-)-g'(4+)}2=\frac12(\frac13-1)=-\frac13$.

Now $g(x)=\frac13x+\frac13|x-2|-\frac13|x-4|+a_3$. Taking $x=0$ we have $-\frac23=g(0)=\frac23-\frac43+a_3$, giving $a_3=0$.