Critical number, sign analysis, and classification at a cusp

Differentiating, $f'(x) = \frac{2}{3}(x-2)^{-1/3} = \dfrac{2}{3(x-2)^{1/3}}$. This is never equal to zero, but it is undefined at $x=2$, so $x = 2$ is the only critical number.

For $x < 2$, $(x-2)^{1/3} < 0$, so $f'(x) < 0$, meaning $f$ is decreasing on $(-\infty, 2)$. For $x > 2$, $(x-2)^{1/3} > 0$, so $f'(x) > 0$, meaning $f$ is increasing on $(2, \infty)$.

Since $f'$ changes from negative to positive at $x = 2$, the First Derivative Test tells us $f$ has a local minimum at $x = 2$, with value $f(2) = 0$. This is consistent with the cusp shape of the graph, which points downward to a sharp corner at $(2,0)$.