Finding critical numbers of a cube-root function

Rewrite $f(x) = x^{5/3} - 5x^{2/3}$, so $f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{-1/3}(x - 2)$. This can also be written as $f'(x) = \dfrac{5(x-2)}{3x^{1/3}}$. Setting the numerator equal to zero gives $x = 2$, and the derivative is undefined at $x = 0$ (division by zero), so the critical numbers are $x = 0$ and $x = 2$.

Now test the sign of $f'(x) = \dfrac{5(x-2)}{3x^{1/3}}$ on each interval determined by $0$ and $2$. For $x<0$: the numerator $(x-2)$ is negative, and the denominator $3x^{1/3}$ is negative (since the real cube root of a negative number is negative), so $f' = \frac{\text{negative}}{\text{negative}} > 0$ (increasing). For $0<x<2$: numerator negative, denominator positive, so $f'<0$ (decreasing). For $x>2$: numerator positive, denominator positive, so $f'>0$ (increasing).

Since $f'$ changes from positive to negative at $x=0$, $f$ has a local maximum there. Since $f'$ changes from negative to positive at $x=2$, $f$ has a local minimum there. Thus $f$ is increasing on $(-\infty, 0)$ and $(2,\infty)$, and decreasing on $(0,2)$. The critical point $x=0$ is a local maximum (where $f'$ is undefined), and the critical point $x=2$ is a local minimum (where $f'=0$).