anonymous
  • anonymous
Let f(x) = ||x|-1|, then what are the points at which f is not differentiable?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
If you think typing here is not easy, you may go where we did yesterday
anonymous
  • anonymous
it's okay here
anonymous
  • anonymous
Please provide a little details, as I am new to this thing

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anonymous
  • anonymous
one second, i'm just finishing up elsewhere I will
anonymous
  • anonymous
Its ok
anonymous
  • 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.
anonymous
  • anonymous
Sure, :)
anonymous
  • anonymous
@amit7808, can you help?
anonymous
  • anonymous
okay
anonymous
  • anonymous
Thanks for coming back
anonymous
  • anonymous
Well, should I say, what I understand about the problem?
anonymous
  • anonymous
Sorry, had to clear some programs.
anonymous
  • anonymous
Its ok
anonymous
  • anonymous
No, it's okay. You want to find those points where it's not differentiable.
anonymous
  • anonymous
Yes
anonymous
  • 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\[(f )o (g)(x)=f(g(x))=||x|-1|\]
anonymous
  • anonymous
Firstly, do you know why the absolute value function loses differentiability?
anonymous
  • anonymous
If you do, I can move on.
anonymous
  • anonymous
No, I don't have any idea
anonymous
  • 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).
anonymous
  • 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.
anonymous
  • anonymous
You have then the following:\[y=x\]for \[x \in [0,\infty)\]and \[y=-x \]\[x \in (-\infty,0]\]
anonymous
  • anonymous
Yes, I understand that
anonymous
  • anonymous
(i need to eat while i do this)
anonymous
  • anonymous
So do you want to come back later?
anonymous
  • anonymous
no it's ok. may be a little slower
anonymous
  • anonymous
NP
anonymous
  • 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.
anonymous
  • anonymous
Is this okay so far?
anonymous
  • anonymous
anonymous
  • 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.
anonymous
  • anonymous
Yeah. So you can see what I'm saying?
anonymous
  • anonymous
Ok, so the point where it is differentiable is that where it has the same slope while approaching from both side?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Ok then its clear. Please proceed
anonymous
  • anonymous
At x=0, this falls down.
anonymous
  • 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.
anonymous
  • anonymous
i.e.\[||x|-1|\]won't be differentiable at x=0.
anonymous
  • anonymous
Now, that's half of it.
anonymous
  • anonymous
Just let me eat some more.
anonymous
  • anonymous
Ok let me speak then, while you eat
anonymous
  • anonymous
So from what you said, I can see that the other point would be 1 from the function f(x)
anonymous
  • anonymous
I mean 1 would be the point where the curve will not be differentiable
anonymous
  • anonymous
Nearly
anonymous
  • anonymous
anonymous
  • 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.
anonymous
  • anonymous
So your points are 0, +1, -1.
anonymous
  • anonymous
I didn't understand the line "so it won't be differentiable at the points where |x|−1=0"
anonymous
  • anonymous
Why do you say that?
anonymous
  • 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?
anonymous
  • anonymous
i.e. where is |u| not differentiable?
anonymous
  • anonymous
Aah I see it now.
anonymous
  • anonymous
Good.
anonymous
  • anonymous
In general, what are things to consider while solving this type of problems
anonymous
  • anonymous
You have a function of a function...as in 'chain rule', but it's just composition of functions.
anonymous
  • anonymous
I mean problems dealing with differentiability
anonymous
  • 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).
anonymous
  • 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.
anonymous
  • anonymous
Be familiar with common functions, like the absolute value function.
anonymous
  • anonymous
Yes, I understand
anonymous
  • anonymous
okay
anonymous
  • anonymous
But what are the things to consider with problems involving differentiability ?
anonymous
  • anonymous
Also, nothing beats a plot!
anonymous
  • 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.
anonymous
  • anonymous
So thats all
anonymous
  • anonymous
Pretty much.
anonymous
  • anonymous
Well, have you heard of the exam STEP
anonymous
  • anonymous
No
anonymous
  • anonymous
Actually I am preparing for it
anonymous
  • anonymous
anyhow I have another question
anonymous
  • anonymous
STEP for cambridge
anonymous
  • anonymous
Yes, its not just for cambridge
anonymous
  • anonymous
Many other accepts it

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