Is this correct?

The function fails to be differentiable at x=0. Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity.

y=3-cuberoot(x)

Question

Is this correct?

The function fails to be differentiable at x=0. Tell whether the problem is a corner, a cusp, a vertical tangent, or a discontinuity.

y=3-cuberoot(x)

anonymous · Answer

$$3-\sqrt[3]{x}$$
Vertical tangent?

anonymous · Answer

No, it's not a vertical tangent because your equation exists at x=0. what you have here is a cusp.

this is basically because as you get infinitesimally closer to x=0 your equation is getting closer to 3 quicker and quicker. 

that explanation didn't make a lot of sense because I'm not sure how to exactly describe how a cusp is formed, I just know one when I see one. I suggest you graph this out, it might help.

anonymous · Answer

Ohh okay.. According to http://www.wolframalpha.com/input/?i=3-%28x%29^%281%2F3%29 I can see the cusp. But when I graph it on my graphing calculator (TI-89) I see a vertical tangent...

anonymous · Answer

For it to be a vertical tangent, Your equation Cannot exist at the point. since this function exists at 0 you should assume it's a cusp