# Studying non-differentiable points of an irrational function

• greg_rack
In summary: The proof of this is a little bit involved, but essentially what you need to do is show that the function ##g## has a limit at ##x = a## and that this limit is equal to the function ##f## evaluated at ##x = a##. I think that's what you do. It would be interesting to try to prove that!

#### greg_rack

Gold Member
Homework Statement
$$y=x\sqrt[3]{x^3-x}$$
Relevant Equations
none
I calculated the derivative of this function as:
$$\frac{6x^3-4x}{3\sqrt[3]{(x^3-x)^2}}$$
Now, in order to find and later study non-differentiable points, I must find the values which make the argument of the root equal to zero:
$$x^3-x=0 \rightarrow x=0 \vee x=\pm 1$$
and then find the left and right limits of the derivative tending to those points.

The deal is that, for ##x=0##, the derivative doesn't exist(and I cannot even "unlock" the indeterminate form ##\frac{0}{0}## to calculate left and right limit ##x\to 0^{\pm}##)... and since that point is in the domain of the function but not on that of the derivative, I don't know how to behave.

What do you mean when you say you can't unlock the indeterminate form?

Office_Shredder said:
What do you mean when you say you can't unlock the indeterminate form?
I end up having an indeterminate form ##\frac{0}{0}## with the limit
$$\lim_{x\to 0^{\pm}}[\frac{6x^3-4x}{3\sqrt[3]{(x^3-x)^2}}]$$
that I don't know how to solve

greg_rack said:
I end up having an indeterminate form ##\frac{0}{0}## with the limit
$$\lim_{x\to 0^{\pm}}[\frac{6x^3-4x}{3\sqrt[3]{(x^3-x)^2}}]$$
that I don't know how to solve
Do you know L'Hopital's rule?

PeroK said:
Do you know L'Hopital's rule?
No, not yet

greg_rack said:
No, not yet
Okay, you don't need it actually - there is a simple factorisation you can do.

PeroK said:
Okay, you don't need it actually - there is a simple factorisation you can do.
Particularly by factoring whatever you can out of the radical in the denominator as well as factoring the numerator.

greg_rack said:
The deal is that, for ##x=0##, the derivative doesn't exist

Technically, you should not conclude the derivative does not exist based only on the fact that the algebraic expression for the derivative is not valid at x = 0.

It's possible to have cases where ##lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}## exists but where the algebraic expression given by the rules for computing derivatives is undefined at ##x = a##. Textbooks often contain problems whose purpose is to illustrate such situations.

greg_rack said:
No, not yet
Anyway, here is L'Hopital's rule: $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$. Super useful tool.

Mayhem said:
Anyway, here is L'Hopital's rule: $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$. Super useful tool.
I agree, but it seems that the OP hasn't come across it yet in his study.
In any case, if a limit can be found by other means, as alluded to in this thread, they should be exhausted before wheeling out L'Hopital's Rule, which can occasionally be of no use at all.

Mark44 said:
I agree, but it seems that the OP hasn't come across it yet in his study.
In any case, if a limit can be found by other means, as alluded to in this thread, they should be exhausted before wheeling out L'Hopital's Rule, which can occasionally be of no use at all.
Yes, and it's necessary to make sure that the limit you are trying to evaluate doesn't come up in the derivation of f'(x) and g'(x) either, which in itself can be tricky to prove at times.

Mark44 said:
Particularly by factoring whatever you can out of the radical in the denominator as well as factoring the numerator.
By playing a little bit with the function, I managed to get:
$$\frac{2\sqrt[3]{x}(3x^2-2)}{3\sqrt[3]{(x^2-1)^2}}$$
which results in ##\frac{0}{3}## for ##x\to 0##... or am I wrong?
That doesn't look like the best factorisation ever, but seems to work anyway

Stephen Tashi said:
Technically, you should not conclude the derivative does not exist based only on the fact that the algebraic expression for the derivative is not valid at x = 0.

It's possible to have cases where ##lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}## exists but where the algebraic expression given by the rules for computing derivatives is undefined at ##x = a##. Textbooks often contain problems whose purpose is to illustrate such situations.
And in those cases, how do you prevent making mistakes? By calculating the limits of both the left and right calculated derivative?

greg_rack said:
By playing a little bit with the function, I managed to get:
$$\frac{2\sqrt[3]{x}(3x^2-2)}{3\sqrt[3]{(x^2-1)^2}}$$
which results in ##\frac{0}{3}## for ##x\to 0##... or am I wrong?
That's what I get, as well.
greg_rack said:
That doesn't look like the best factorisation ever, but seems to work anyway
If it allows you to get the limit, it's good enough.

greg_rack
greg_rack said:
And in those cases, how do you prevent making mistakes? By calculating the limits of both the left and right calculated derivative?

I think that's what you do. It would be interesting to try to prove that!

In a typical textbook problem of this type, you are given ##f'(x) = lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = g(x)## where ##g## is a function that has a defined value in an interval around ##x = a## but is undefined at ## x =a ##. We want to prove that ##lim_{x \rightarrow a} g(x) = lim_{h \rightarrow 0} \frac{ f(a+h) - f(a)}{h} ##.

greg_rack

## 1. What is a non-differentiable point?

A non-differentiable point is a point on a function where the derivative does not exist. This means that the function is not smooth at that point and has a sharp corner or a vertical tangent.

## 2. Why is it important to study non-differentiable points of an irrational function?

Studying non-differentiable points of an irrational function can help us understand the behavior of the function and how it changes. It also allows us to identify critical points, which are important in optimization problems.

## 3. How do you determine if a point is non-differentiable on an irrational function?

A point is non-differentiable on an irrational function if the function is not continuous or if the derivative does not exist at that point. This can be determined by taking the limit of the function as it approaches the point in question.

## 4. What are some common techniques for studying non-differentiable points?

Some common techniques for studying non-differentiable points include using the definition of the derivative, analyzing the graph of the function, and using the first and second derivative tests to identify critical points.

## 5. How can studying non-differentiable points help in real-world applications?

In real-world applications, studying non-differentiable points can help us understand the behavior of a function and make predictions about its behavior in different scenarios. This is especially useful in fields such as economics, physics, and engineering.