Does y=x^3+(x-3)^(1/3) Have a Vertical Tangent, Cusp, Corner, or None at x=3?

[crazco]

Homework Statement

At x=3 does the function y=x^3 + (x-3)^(1/3) have a vertical tangent, cusp, corner or none?

The Attempt at a Solution

I took the derivative

y'=3x^2 + 1 / 3x^(2/3) then i replaced 3 in which gave 27 + 1 / 0

I don't understand how to come to the correct conclusion following this step.

Vertical tangent comes to mind since 1 / 0 is a vertical line, but I don't know how to prove it using limits.

 

[Dick]

You mean y'=3x^2 + 1 / 3(x-3)^(2/3), right? If you approach x=3 from below then y' goes to +infinity. I you approach it from above it also goes to +infinity. y is also continuous at x=3. Looks like a vertical tangent to me.

 

[LCKurtz]

Homework Statement

At x=3 does the function y=x^3 + (x-3)^(1/3) have a vertical tangent, cusp, corner or none?

The Attempt at a Solution

I took the derivative

y'=3x^2 + 1 / 3x^(2/3)

Is that really what you got?? It's either a typo or a mis-calculation.

then i replaced 3 in which gave 27 + 1 / 0

I don't understand how to come to the correct conclusion following this step.

Vertical tangent comes to mind since 1 / 0 is a vertical line, but I don't know how to prove it using limits.

If you have a positive infinite limit from both the left right that suggests a vertical line alright. You can tell whether it is vertical tangent line or cusp by looking at concavity on each side of x = 3.