Also confused on this a bit, I tried graphing and could find an answer.
Consider the function f(x) = 6 (x − 4)^2/3 . For this function there are two important open intervals: ( − ∞ , A ) and ( A , ∞ ) where A is a critical number.
Find A:
use
www.desmos.com/calculator
to graph it, and you will see that there is a cusp at A
A is (4,0)
or, if you don't have desmos handy,
f'(x) = 4/(x-4)^(1/3)
so the slope is undefined at (4,0)