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Question: How do you determine the differentiability of \(f\left( x \right)\) , where \(f\left( x \right)=\lef...

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

Explanation

Solution

In order to find solution to this problem, we will see if the function f(x)f\left( x \right) is differentiable or non-differentiable for any point by checking the limit lim⁡x→af(x)−f(a)x−a\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 ff is differentiable at aa if it has a non-vertical tangent at the corresponding point on the graph, that is, at (a,f(a))\left( a,f\left( a \right) \right).
That means that the limit lim⁡x→af(x)−f(a)x−a\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 ff at aa and denoted f′(a)f'\left( a \right) or dfdx(a)\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(x)=∣x∣ f\left( x \right)=\left| x \right|~ at 00.
Now we have our function as:
f(x)=∣x−1∣+∣x−2∣+∣x−3∣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...x=1,2,3...
Therefore, we have our equation as:
f=(1−x)+(2−x)+(3−x)=6−3x,x∈[−∞,1]f=(1-x)+(2-x)+(3-x)=6-3x,x\in \left[ -\infty ,1 \right]
By differentiating, we get:
f′=−3,x∈(−∞,1){f}'=-3,x\in (-\infty ,1)
We get our second equation as:
f=(x−1)+(2−x)+(3−x)=4−x,x∈[1,2]f=(x-1)+(2-x)+(3-x)=4-x,x\in [1,2]
By differentiating, we get:
f′=−1,x∈(1,2){f}'=-1,x\in (1,2)
We get our third equation as:
f=(x−1)+(x−2)+(3−x)=x,x∈[2,3]f=(x-1)+(x-2)+(3-x)=x,x\in [2,3]
By differentiating, we get:
f′=1,x∈(2,3){f}'=1,x\in (2,3)
We get our fourth equation as:
f=(x−1)+(x−2)+(x−3)=3x−6,x∈[3,∞]f=(x-1)+(x-2)+(x-3)=3x-6,x\in \left[ 3,\infty \right]
By differentiating, we get:
f′=3,x∈(3,∞)f'=3,x\in (3,\infty )

Therefore, the function is differentiable at x=−3,−1,1,3x=-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 aa if lim⁡h→0−f(a+h)−f(a)h≠lim⁡h→0+f(a+h)−f(a)h\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 aa if lim⁡x→a−∣f′(x)∣=∞\displaystyle \lim_{x \to {{a}^{-}}}\left| {f}'\left( x \right) \right|=\infty or lim⁡x→a+∣f′(x)∣=∞\displaystyle \lim_{x \to {{a}^{+}}}\left| {f}'\left( x \right) \right|=\infty
If the above condition satisfies, then a function is non-differentiable.