Prove that the function f given by f(x)=|x-1|, x ∈ R is not differentiable at x = 1.

Question

anonymous · Answer

You plot the function and show ...

diyadiya · Answer

Plot?

anonymous · Answer

Graph

diyadiya · Answer

i dont think there's a graph hm and idk to plot it ;/

diyadiya · Answer

ok wait

turingtest · Answer

the formula$$|x|=\sqrt{x^2}$$ should do the trick, right?

anonymous · Answer

f'(1) should be undefined.

turingtest · Answer

right, if we use the formula I posted it the derivative will be undefined at 1

anonymous · Answer

To the right of x=1 line the graph of f(x) has slope +1. To the left of the x=1 it has slope −1.

anonymous · Answer

However, this description of the derivative leaves out the value of f'(0).

anonymous · Answer

Alternative analytical way is to show that right hand derivative is not equal to left hand and hence the proof.

anonymous · Answer

Turing, we should not need any formula to show this.

diyadiya · Answer

Alternative is better

mr.math · Answer

Redefine the function as:

$$\left| x-1 \right|= x-1 \text{ for } x\ge 1,  $$
$$\text{         }  =1-x \text{    for } x<1$$
$$f'(1^+)=1 \ne f'(1^-)=-1.$$

turingtest · Answer

Hey, I'm letting you run the show here...

anonymous · Answer

but I would like you to do it geometrically, since it clears intuition :)

anonymous · Answer

I am grateful to you, Turing :D

mr.math · Answer

Since $f'(1^+)\ne f'(1^-)$, then f(x) is not differentiable at x=1.

diyadiya · Answer

Ok Thanks MrMath!

and Thanks FoolForMath & TuringTest :)

mr.math · Answer

You're welcome!