(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

3. 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.

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# Homework Help: Determining if a function has a cusp, vertical tangent, corner or none at a point

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