Question

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

Answer 1

Hint: In order to find solution to this problem, we will see if the function $f\left( x \right)$ is differentiable or non-differentiable for any point by checking the limit $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}$ exists that is, it is a finite number, which is the slope of this tangent line.

Complete step by step solution:
The function $f$ is differentiable at $a$ if it has a non-vertical tangent at the corresponding point on the graph, that is, at \[\left( a,f\left( a \right) \right)\].
That means that the limit $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}$ exists that is, it is a finite number, which is the slope of this tangent line.
And when this limit exists, it is called derivative of $f$ at $a$ and denoted \[f'\left( a \right)\] or $\dfrac{df}{dx}\left( a \right)$ .
So, a point where the function is not differentiable is a point where this limit does not exist, that is, it is either infinite (which is a case of a vertical tangent), where the function is discontinuous, or where there are two different one-sided limits like for \[f\left( x \right)=\left| x \right|~\] at $0$.
Now we have our function as:
$f\left( x \right)=\left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right|$
Now we will use piecewise linear definition with different slopes at $x=1,2,3...$
Therefore, we have our equation as:
\[f=(1-x)+(2-x)+(3-x)=6-3x,x\in \left[ -\infty ,1 \right]\]
By differentiating, we get:
\[{f}'=-3,x\in (-\infty ,1)\]
We get our second equation as:
\[f=(x-1)+(2-x)+(3-x)=4-x,x\in [1,2]\]
By differentiating, we get:
\[{f}'=-1,x\in (1,2)\]
We get our third equation as:
\[f=(x-1)+(x-2)+(3-x)=x,x\in [2,3]\]
By differentiating, we get:
\[{f}'=1,x\in (2,3)\]
We get our fourth equation as:
\[f=(x-1)+(x-2)+(x-3)=3x-6,x\in \left[ 3,\infty \right]\]
By differentiating, we get:
\[f'=3,x\in (3,\infty )\]

Therefore, the function is differentiable at $x=-3,-1,1,3$.

Note: A function is non-differentiable at any point at which it is discontinuous, it has a corner point
 This happens at $a$ if $\displaystyle \lim_{h \to {{0}^{-}}}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}\ne \displaystyle \lim_{h \to {{0}^{+}}}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$ and it has a vertical tangent line.
This happens at $a$ if $\displaystyle \lim_{x \to {{a}^{-}}}\left| {f}'\left( x \right) \right|=\infty $ or $\displaystyle \lim_{x \to {{a}^{+}}}\left| {f}'\left( x \right) \right|=\infty $
If the above condition satisfies, then a function is non-differentiable.