Studying non-differentiable points of an irrational function

[greg_rack]

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.

 

[Office_Shredder]

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

 

[greg_rack]

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

 

[PeroK]

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?

 

[greg_rack]

Do you know L'Hopital's rule?

No, not yet

 

[PeroK]

No, not yet

Okay, you don't need it actually - there is a simple factorisation you can do.

 

[Mark44]

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.

 

[Stephen Tashi]

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.

 

[Mayhem]

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.

 

[Mark44]

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.

 

[Mayhem]

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.

 

[greg_rack]

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

 

[greg_rack]

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?

 

[Mark44]

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.

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.

 

[Stephen Tashi]

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 $lim_{x \rightarrow a} g(x) = lim_{h \rightarrow 0} \frac{ f(a+h) - f(a)}{h}$.