## anonymous 5 years ago Let f(x) = ||x|-1|, then what are the points at which f is not differentiable?

1. anonymous

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

2. anonymous

it's okay here

3. anonymous

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

4. anonymous

one second, i'm just finishing up elsewhere I will

5. anonymous

Its ok

6. anonymous

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

7. anonymous

Sure, :)

8. anonymous

@amit7808, can you help?

9. anonymous

okay

10. anonymous

Thanks for coming back

11. anonymous

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

12. anonymous

Sorry, had to clear some programs.

13. anonymous

Its ok

14. anonymous

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

15. anonymous

Yes

16. anonymous

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

17. anonymous

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

18. anonymous

If you do, I can move on.

19. anonymous

No, I don't have any idea

20. anonymous

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).

21. anonymous

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.

22. anonymous

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

23. anonymous

Yes, I understand that

24. anonymous

(i need to eat while i do this)

25. anonymous

So do you want to come back later?

26. anonymous

no it's ok. may be a little slower

27. anonymous

NP

28. anonymous

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.

29. anonymous

Is this okay so far?

30. anonymous

31. anonymous

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.

32. anonymous

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

33. anonymous

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

34. anonymous

Yes

35. anonymous

Ok then its clear. Please proceed

36. anonymous

At x=0, this falls down.

37. anonymous

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.

38. anonymous

i.e.$||x|-1|$won't be differentiable at x=0.

39. anonymous

Now, that's half of it.

40. anonymous

Just let me eat some more.

41. anonymous

Ok let me speak then, while you eat

42. anonymous

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

43. anonymous

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

44. anonymous

Nearly

45. anonymous

46. anonymous

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.

47. anonymous

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

48. anonymous

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

49. anonymous

Why do you say that?

50. anonymous

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?

51. anonymous

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

52. anonymous

Aah I see it now.

53. anonymous

Good.

54. anonymous

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

55. anonymous

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

56. anonymous

I mean problems dealing with differentiability

57. anonymous

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).

58. anonymous

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.

59. anonymous

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

60. anonymous

Yes, I understand

61. anonymous

okay

62. anonymous

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

63. anonymous

Also, nothing beats a plot!

64. anonymous

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.

65. anonymous

So thats all

66. anonymous

Pretty much.

67. anonymous

Well, have you heard of the exam STEP

68. anonymous

No

69. anonymous

Actually I am preparing for it

70. anonymous

anyhow I have another question

71. anonymous
72. anonymous

STEP for cambridge

73. anonymous

Yes, its not just for cambridge

74. anonymous

Many other accepts it

75. anonymous