The derivative at x = 3 does not exist.
To see why, recall the definition of absolute value:
[tex]|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x<0\end{cases}[/tex]
If x - 3 ≥ 0, then
2 + |x - 3| = 2 + (x - 3) = x - 1
and the derivative of this piece is 1 for all x > 3.
If x - 3 < 0, then
2 + |x - 3| = 2 - (x - 3) = 5 - x
and its derivative is -1 for all x < 3.
The derivatives to either side of x = 3 are not the same, so f '(x) is not continuous and indeed does not take on any value at this point.
