- #1

greg_rack

Gold Member

- 359

- 78

- 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.

$$\frac{6x^3-4x}{3\sqrt[3]{(x^3-x)^2}}$$

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.