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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    If you think typing here is not easy, you may go where we did yesterday

  2. anonymous
    • 5 years ago
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    it's okay here

  3. anonymous
    • 5 years ago
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    Please provide a little details, as I am new to this thing

  4. anonymous
    • 5 years ago
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    one second, i'm just finishing up elsewhere I will

  5. anonymous
    • 5 years ago
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    Its ok

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Sure, :)

  8. anonymous
    • 5 years ago
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    @amit7808, can you help?

  9. anonymous
    • 5 years ago
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    okay

  10. anonymous
    • 5 years ago
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    Thanks for coming back

  11. anonymous
    • 5 years ago
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    Well, should I say, what I understand about the problem?

  12. anonymous
    • 5 years ago
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    Sorry, had to clear some programs.

  13. anonymous
    • 5 years ago
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    Its ok

  14. anonymous
    • 5 years ago
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    No, it's okay. You want to find those points where it's not differentiable.

  15. anonymous
    • 5 years ago
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    Yes

  16. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Firstly, do you know why the absolute value function loses differentiability?

  18. anonymous
    • 5 years ago
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    If you do, I can move on.

  19. anonymous
    • 5 years ago
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    No, I don't have any idea

  20. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    You have then the following:\[y=x\]for \[x \in [0,\infty)\]and \[y=-x \]\[x \in (-\infty,0]\]

  23. anonymous
    • 5 years ago
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    Yes, I understand that

  24. anonymous
    • 5 years ago
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    (i need to eat while i do this)

  25. anonymous
    • 5 years ago
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    So do you want to come back later?

  26. anonymous
    • 5 years ago
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    no it's ok. may be a little slower

  27. anonymous
    • 5 years ago
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    NP

  28. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Is this okay so far?

  30. anonymous
    • 5 years ago
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  31. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Yeah. So you can see what I'm saying?

  33. anonymous
    • 5 years ago
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    Ok, so the point where it is differentiable is that where it has the same slope while approaching from both side?

  34. anonymous
    • 5 years ago
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    Yes

  35. anonymous
    • 5 years ago
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    Ok then its clear. Please proceed

  36. anonymous
    • 5 years ago
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    At x=0, this falls down.

  37. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    i.e.\[||x|-1|\]won't be differentiable at x=0.

  39. anonymous
    • 5 years ago
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    Now, that's half of it.

  40. anonymous
    • 5 years ago
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    Just let me eat some more.

  41. anonymous
    • 5 years ago
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    Ok let me speak then, while you eat

  42. anonymous
    • 5 years ago
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    So from what you said, I can see that the other point would be 1 from the function f(x)

  43. anonymous
    • 5 years ago
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    I mean 1 would be the point where the curve will not be differentiable

  44. anonymous
    • 5 years ago
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    Nearly

  45. anonymous
    • 5 years ago
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  46. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    So your points are 0, +1, -1.

  48. anonymous
    • 5 years ago
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    I didn't understand the line "so it won't be differentiable at the points where |x|−1=0"

  49. anonymous
    • 5 years ago
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    Why do you say that?

  50. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    i.e. where is |u| not differentiable?

  52. anonymous
    • 5 years ago
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    Aah I see it now.

  53. anonymous
    • 5 years ago
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    Good.

  54. anonymous
    • 5 years ago
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    In general, what are things to consider while solving this type of problems

  55. anonymous
    • 5 years ago
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    You have a function of a function...as in 'chain rule', but it's just composition of functions.

  56. anonymous
    • 5 years ago
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    I mean problems dealing with differentiability

  57. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Be familiar with common functions, like the absolute value function.

  60. anonymous
    • 5 years ago
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    Yes, I understand

  61. anonymous
    • 5 years ago
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    okay

  62. anonymous
    • 5 years ago
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    But what are the things to consider with problems involving differentiability ?

  63. anonymous
    • 5 years ago
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    Also, nothing beats a plot!

  64. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    So thats all

  66. anonymous
    • 5 years ago
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    Pretty much.

  67. anonymous
    • 5 years ago
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    Well, have you heard of the exam STEP

  68. anonymous
    • 5 years ago
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    No

  69. anonymous
    • 5 years ago
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    Actually I am preparing for it

  70. anonymous
    • 5 years ago
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    anyhow I have another question

  71. anonymous
    • 5 years ago
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    STEP for cambridge

  72. anonymous
    • 5 years ago
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    Yes, its not just for cambridge

  73. anonymous
    • 5 years ago
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    Many other accepts it

  74. anonymous
    • 5 years ago
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    Well, here is my next question http://openstudy.com/updates/4d9abb520b248b0b471ce9e2?source=email#/updates/4d9ac0cf0b248b0bdc1fe9e2

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