Finding critical numbers of f(x) = x^(2/3)

Differentiate using the power rule: $f'(x) = \frac{2}{3}x^{2/3 - 1} = \frac{2}{3}x^{-1/3}$. Rewriting with a positive exponent gives $f'(x) = \dfrac{2}{3x^{1/3}} = \dfrac{2}{3\sqrt[3]{x}}$.

This expression is never equal to zero, since the numerator is a nonzero constant. However, $f'(x)$ is undefined at $x = 0$ because the denominator $3\sqrt[3]{x}$ becomes $0$, and $0$ lies within the open interval $(-8,8)$. Therefore $x = 0$ is the only critical number.

The derivative is undefined at that point, not zero. The derivative formula has the variable in the denominator, so plugging in that critical number causes division by zero rather than giving a zero numerator. This means the function has a critical number purely because the derivative fails to exist there, even though the derivative is never actually equal to zero anywhere on the interval.