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October 12th, 2013, 06:34 AM   #1
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What does this mean: f' does not exist (1e derivative test)

Hi all,

Can someone explain to me what the following means? A derivative test where f'(x) = 0 or does not exist.

I am confused about what they mean about does not exist. I have only learned to make a derivative of the function and then make it equal to zero. I have never heard of does not exist.

Do they mean by undefined in the meaning as a fraction or log where zero is undefined? Is that then not the same as make the derivative equal to zero? Also, does this only apply to a first derivative test or also the second derivative test?

I also have two other questions.

Why is it that a corner also can be a min. or max.? I thought it was only applied to concave and convex, because in the interval of c it is decreasing and increasing. I also read somewhere that if it is a corner it is not differentiable. What do they mean by that?

What is the purpose of a second, third, fourth, etc., derivative? Because you can endlessly make a derivative until it becomes a constant and then the derivative is zero.

I am really curious about this, because in the books it does not tell a lot about it and is also vague. Thanks in advance.

Last edited by skipjack; January 19th, 2016 at 04:49 PM.
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October 12th, 2013, 08:02 AM   #2
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Re: What does this mean: f' does not exist (1e derivative te

Don't forget there are very ugly functions, not just the nice linear function, parabola and sinus. A function in most general sense just a map from real number to real number. So you can give the values f(x) for every x pointwisely without any connection between x1 and x2. This f(x) is even not necessary continous, the function doesn't look locally linear at all, the limit in the definition of derivative doesn't exist.

Other example the abs(x) function at 0. Although It is a continuos function, but the limit doesn't exist at 0, because

does not exist.

If the derivative exists and f'(x0) = 0, then the function looks locally as a horizontal line (special case of linear function where the slope is 0).

So we can think about the first derivative as a best linear approximation of the function at x0. More precisely the linear function L(x) = f(x0) + f'(x0)*(x-x0) is the best fitting linear function for f(x) around x0.

The higher rank derivates have similar play. The second derivative plays a great part in approximating the funciton locally with a quadratic function. So you can analyse a (differentiable or analitic) function only with its derivatives, that is why this area of math sometimes called Analysis.
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October 12th, 2013, 11:07 AM   #3
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Re: What does this mean: f' does not exist (1e derivative te

Thanks for the reply Csak.

I don't understand your explanation. I still don't know what the mean by f'(x) = zero or undefined. What do they mean by undefined? I understand what they mean by zero it is simply make the derivative equal to zero. The x coordinate that comes out of the calculation is the x coordinate of that critical point. Do they perhaps mean by undefined as fractions, log, etc., where zero is undefined? But is that the same as set the derivative equal to zero?

I also don't understand your explanation about the purpose about higher derivatives.

I hope you can explain it to me in more non-mathematical terms. I am not super familiar with math. I am new in math and only have math courses for 3 months.

Last edited by skipjack; January 19th, 2016 at 04:53 PM.
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October 13th, 2013, 03:15 AM   #4
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Re: What does this mean: f' does not exist (1e derivative te

The fact that derivative is undefined and the fact the derivative is well-defined and it's equal to zero are as different as night and day. Are you sure you know what a derivative means? Do you know its definition?

Quote:
I understand what they mean by zero it is simply make the derivative equal to zero.
It seems to me you want to start from the fact the derivative is given. No, the function f(x) is given. And the definition of the derivative is given. So you can not make the derivative to zero, instead you can apply the definition and you can calculate it. If a function is given, then you can decide (with the definition of derivative) whether the derivative exists or not at a particular point. So you get two sets, one for where the derivative is undefined and one where the derivative is defined. The latter set splits up into two sets again, where the derivative is zero and where it is not zero.

Yes, it is similar to square root, log etc. Square root x by definition is a real number -- if it exists -- what you multiply itself you get x. This definition splits up the real numbers into two mutual exclusive sets. For negative numbers the square root is undefined, for non-negative numbers the square root is defined. And of course the subset of the second set where square root x = 0 is different from the first set.

Quote:
I hope you can explain it to me in more non-mathematical terms.
You misunderstood the point of this forum.

Last edited by skipjack; January 19th, 2016 at 05:00 PM.
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October 13th, 2013, 12:14 PM   #5
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Re: What does this mean: f' does not exist (1e derivative te

I do understand what a derivative is, but I don't understand why f'(x) must be equal to zero or undefined. I don't understand what they mean by undefined.

The way I have learned is as follow:

First you make a derivative of f(x). Next to know what the extreme points are you set f'(x) equal to zero and then you know the x coordinates of the extreme points. If you put those x coordinates into f(x), you get the y coordinates. But now I read that f'(x) must be zero or undefined. And undefined I don't understand what they mean by that. It only says this:

interior points in interval where f'(x) = 0
End points of interval (if included in interval)
interior pints in interval where f' does not exist

If you want to find the local extremes.

And there is no explanation what they mean. I have searched on the internet and it says that f'(x) must be zero or undefined.

I also know the purpose of the forum, but I am new to all of these methods. The things you talk for instance linear approximation. I just started to read something about it and only about the formula and some math problems. My course only talks about the basic and not in depth, but when I search for more information I find info about it and want to understand it.

Sorry for the misunderstanding. I should had said it differently.

Last edited by skipjack; January 19th, 2016 at 05:03 PM.
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October 13th, 2013, 02:31 PM   #6
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Re: What does this mean: f' does not exist (1e derivative te

I really don't understand what you don't understand. By the way you keep using a word that doesn't make any sense in your context. This word is 'mean' as a noun. I think you want to type 'meaning', but please verify it.

Quote:
First you make a derivative of f(x).
No. First you decide by the definition whether the derivative exists or not.

I think this would be the best if you try to calculate the derivative of some very concrete functions, and you will understand why it doesn't exist always.

So please calculate the derivation of the following function at the following point.

1) let the domain of f(x) be the closed interval [100, 102], and let f(x) = 2, if 100<=x<101, and let f(x)=3, if 101<x<=102. The question is the derivative of f(x) at the point 1.

2) Same function as previous. The question is the derivative of f(x) at the point 101.

3) Let the domain of f(x) be the whole real numbers. Let f(x) = 0, if x is rational. Let f(x) = 1, if x is irrational. The question is the derivative of f(x) at the point 0.

4) f(x) = abs(x) = |x|. Derivative of f(x) at 0?

Please answer those four questions.

Last edited by skipjack; January 19th, 2016 at 05:05 PM.
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October 13th, 2013, 05:50 PM   #7
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Re: What does this mean: f' does not exist (1e derivative te

Ok I'll admit that I haven't read all your posts with great scrutiny but you appear to be butting heads or going around in circles.

A function must be continuous at any given x value within the defined range (by definition)
This means it must be curved or a straight line, it can't be a v shape.
Only functions can be differentiated so a 'corner' cannot be differentiated.

I don't think that the derivative of a function can be undefined.
y=abs(x) is not a function. Not unless you restrict the domain so that it is continuous and hence does not include x=0 which is the 'corner'
since it is not a function at x=0 it cannot have a derivative at x=0, defined or undefined. I suppose you could say that the derivative does not exist.

You use derivatives to measure rates
for example,
If x is a distance and t is a time
then dx/dt is literally the difference in distance / difference in time (as time unit gets infintesimally small) and that is velocity.
If you differentiate again you get difference in velocity / difference in time which is acceleration
if you differentiated a third time you will get change in acceleration with respect to time.
This is just one obvious example where higher derivatives are used.
When you learn curve sketching you will use the second derivative all the time.

When you are solving maximum and minimum questions maybe a corner is the answer but it is not a part of the calculus. It may be compared to something that is a calculus solution, but it won't come from any differentiation itself.

Have I helped ?
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October 13th, 2013, 05:55 PM   #8
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Re: What does this mean: f' does not exist (1e derivative te

y = |x| is a function, even though it is not differentiable at x = 0.
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October 13th, 2013, 06:12 PM   #9
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Re: What does this mean: f' does not exist (1e derivative te

Ok Greg I think you are right. Thanks.

Would it be correct if I said: (please comment, I want to be accurate)

A function must be continuous for it to be differentiable at any given point.
This means it must be curved or a straight line, it can't be a v shape.
Only continuous functions can be differentiated so a 'corner' cannot be differentiated.

y=abs(x) is not a continuous function. Not unless you restrict the domain so that does not include x=0 which is the 'corner'
since it is not a continuous function at x=0 it cannot have a defined derivative at x=0. Can it have an undefined one? maybe I don't know, I'd just say the derivative does not exist.
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October 13th, 2013, 06:25 PM   #10
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Re: What does this mean: f' does not exist (1e derivative te

Continuity alone does not imply differentiability. There are functions that are continuous everywhere and differentiable nowhere. The Weierstrass function is one such function.
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