Question

Let f(x) = ||x|-1|, then what are the points at which f is not differentiable?

Answer 1

If you think typing here is not easy, you may go where we did yesterday

Answer 2

it's okay here

Answer 3

Please provide a little details, as I am new to this thing

Answer 4

one second, i'm just finishing up elsewhere

I will

Answer 5

Its ok

Answer 6

iam, can i get back to you soon on this? i need to go do a few things. i shouldn't be too long.

Answer 7

Sure, :)

Answer 8

@amit7808, can you help?

Answer 9

okay

Answer 10

Thanks for coming back

Answer 11

Well, should I say, what I understand about the problem?

Answer 12

Sorry, had to clear some programs.

Answer 13

Its ok

Answer 14

No, it's okay. You want to find those points where it's not differentiable.

Answer 15

Yes

Answer 16

You have a composition of two functions that lose differentiability at certain points. You have\[g(x)=|x|\]and\[f(x)=|x-1|\]then\[(f )o (g)(x)=f(g(x))=||x|-1|\]

Answer 17

Firstly, do you know why the absolute value function loses differentiability?

Answer 18

If you do, I can move on.

Answer 19

No, I don't have any idea

Answer 20

Okay. A function is differentiable at a point x on a domain if its limit from all directions exists, which means, among other things, it must have the same value whether you approach from the left or the right (on an interval, which is what you have here).

Answer 21

If you look at the absolute value function, |x|, it's a line of gradient 1 from x=0 to infinity and a line of gradient -1 from x=0 to -infinity.

Answer 22

You have then the following:\[y=x\]for \[x \in [0,\infty)\]and \[y=-x \]\[x \in (-\infty,0]\]

Answer 23

Yes, I understand that

Answer 24

(i need to eat while i do this)

Answer 25

So do you want to come back later?

Answer 26

no it's ok. may be a little slower

Answer 27

NP

Answer 28

So, the slope of the function |x| as you approach 0 from the right is 1, while the slope of the function |x| as you approach from the left is -1. That is, the limit of the definition of the derivative is different depending on your direction. We say that the function is not differentiable at that point. Here, that point is zero.

Answer 29

Is this okay so far?

Answer 30

The derivative of |x| IS defined at any OTHER point because we can ALWAYS find a small enough interval in which to approach x so that the limit is the same from both directions.

It's only at zero here that the function is not differentiable.

Answer 31

Yeah. So you can see what I'm saying?

Answer 32

Ok, so the point where it is differentiable is that where it has the same slope while approaching from both side?

Answer 33

Yes

Answer 34

Ok then its clear. Please proceed

Answer 35

At x=0, this falls down.

Answer 36

So, to the crux of this, the function g(x) defined above is not differentiable at the point x=0. It means then, that f(g(x)) will not be continuous there also.

Answer 37

i.e.\[||x|-1|\]won't be differentiable at x=0.

Answer 38

Now, that's half of it.

Answer 39

Just let me eat some more.

Answer 40

Ok let me speak then, while you eat

Answer 41

So from what you said, I can see that the other point would be 1 from the function f(x)

Answer 42

I mean 1 would be the point where the curve will not be differentiable

Answer 43

Nearly

Answer 44

When you look at ||x|-1|, you have to now consider where the composite is not differentiable. We have another absolute function, and so it won't be differentiable at the points where\[|x|-1=0\]That is, where\[|x|=1\]i.e. when \[x= \pm 1\]since either value can satisfy |x|=1.

Answer 45

So your points are 0, +1, -1.

Answer 46

I didn't understand the line "so it won't be differentiable at the points where |x|−1=0"

Answer 47

Why do you say that?

Answer 48

Okay, let u=|x|-1. Then \[||x|-1|=|u|\]If I ask you where |u| is not differentiable, where would you say that would take place?

Answer 49

i.e. where is |u| not differentiable?

Answer 50

Aah I see it now.

Answer 51

Good.

Answer 52

In general, what are things to consider while solving this type of problems

Answer 53

You have a function of a function...as in 'chain rule', but it's just composition of functions.

Answer 54

I mean problems dealing with differentiability

Answer 55

Start from the 'inside-out'. I look for functions within functions and start with the one that is nested the most. You then move out, but keep note of what's going on (e.g. here you might need to think of ||x|-1| = |u|, say, so don't slip up, until you get experience).

Answer 56

You also need to look for places where a function ceases to exist, as in the step function. Just look it up and you'll see the function moves up in steps, but the end-point on each step isn't included and the function jumps.

Answer 57

Be familiar with common functions, like the absolute value function.

Answer 58

Yes, I understand

Answer 59

okay

Answer 60

But what are the things to consider with problems involving differentiability ?

Answer 61

Also, nothing beats a plot!

Answer 62

Just what I said: anywhere where the function goes through a sharp change (e.g. absolute) and anywhere where a function ceases to exist or be continuous.

Answer 63

So thats all

Answer 64

Pretty much.

Answer 65

Well, have you heard of the exam STEP

Answer 66

No

Answer 67

Actually I am preparing for it

Answer 68

anyhow I have another question

Answer 69

http://openstudy.com/updates/4d9abb520b248b0b471ce9e2?source=email#/updates/4d9ac0cf0b248b0bdc1fe9e2

Answer 70

STEP for cambridge

Answer 71

Yes, its not just for cambridge

Answer 72

Many other accepts it

Answer 73

Well, here is my next question
http://openstudy.com/updates/4d9abb520b248b0b471ce9e2?source=email#/updates/4d9ac0cf0b248b0bdc1fe9e2