The minimum value of $ f(x) = | x - 1 | + | x - 2 | + | x - 3 | $ is? I don't get it why my solution is wrong : 
My solving : 
$ f(x) = | x - 1 | + | x - 2 | + | x - 3 | $ 
When $ x\leq 1 $ 
$ f(x) = | x - 1 | + | x - 2 | + | x - 3 | = 0 $ 
= $ 6 -3x $
since $ x\leq 1 $ 
$ f(x)\leq 3 $    also it clear $ f(x)\geq 0 $ 
=> The min value is 0 . 
I dont have to consider other cases, since 0 is the min value f(x) can take . 
Book has given the answer : $ 2 $ . 

We have
\begin{align}
f(x) & =
\begin{cases}
-(x-1)-(x-2)-(x-3) & = -3x+6 & \text{if } x \leq 1\\
(x-1) - (x-2) - (x-3) & = -x+4 & \text{if } x\in[1,2]\\
(x-1) + (x-2) - (x-3) & = x & \text{if } x\in[2,3]\\
(x-1) + (x-2) + (x-3) & = 3x-6 & \text{if }x \geq 3
\end{cases}
\end{align}
Hence, the minimum occurs at $x=2$ and take a value of $2$ as well. The plot is as shown below.

