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