Finding critical numbers including a cusp

Differentiating, $f'(x) = \frac{2}{3}(x-1)^{-1/3} = \frac{2}{3(x-1)^{1/3}}$. This is never equal to zero since the numerator is a nonzero constant, but it is undefined at $x=1$ because of division by zero. Since $f(1) = 0$ is defined, $x=1$ is a critical number (it corresponds to the cusp visible in the graph).

Now apply the Closed Interval Method: evaluate $f$ at the critical number $x=1$ and at the endpoints $x=-1$ and $x=3$. We get $f(-1) = (-2)^{2/3} = \sqrt[3]{4} \approx 1.587$, $f(1) = 0$, and $f(3) = 2^{2/3} = \sqrt[3]{4} \approx 1.587$. Comparing these three values, the smallest is $0$ at $x=1$ and the largest is $\sqrt[3]{4}$, attained at both $x=-1$ and $x=3$.

Therefore the critical number is $x=1$, the absolute minimum value is $f(1)=0$, and the absolute maximum value is $f(-1)=f(3)=\sqrt[3]{4}$.