Question

How do you determine the differentiability of f(x), where `f(x)=|x-1|+|x-2|+|x-3|`?

Answer 1

Piecewise linear, with different slopes. See the illustrative graph revealing nodes, with double slopes, at x = 1, 2, 3..

Explanation:

This is the combined equation for four piecewise definitions.

`f=(1-x)+(2-x)+(3-x)=6-3x, x in (-oo,1] and f'=-3, x in (-oo, 1)`

`f=(x-1)+(2-x)+(3-x)=4-x, x in [1, 2] and f'=-1, x in (1, 2)`

`f=(x-1)+(x-2)+(3-x)=x, x in [2, 3] and f'=1, x in (2, 3)`

`f=(x-1)+(x-2)+(x-3)=3x-6, x in [3, oo} and f'-2, x in (3, oo)`
graph{y-|x-1|-|x-2|-|x-3|=0x^2 [-8.69, 8.685, -4.345, 4.345]}