Critical numbers and classification via first derivative test

Rewrite $f(x) = x^{4/3} - 4x^{1/3}$ and differentiate: $f'(x) = \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3} = \frac{4}{3}x^{-2/3}(x-1) = \frac{4(x-1)}{3x^{2/3}}$. This is zero when $x = 1$, and undefined when $x = 0$ (since $x^{2/3}$ appears in the denominator). So the critical numbers are $x = 0$ and $x = 1$.

Since $x^{2/3} \ge 0$ always, the sign of $f'(x)$ matches the sign of $(x-1)$ where $f'$ is defined. For $x<0$: $(x-1)<0$ so $f'<0$ (decreasing on $(-\infty, 0)$). For $0<x<1$: $(x-1)<0$ so $f'<0$ (decreasing on $(0, 1)$). For $x>1$: $(x-1)>0$ so $f'>0$ (increasing on $(1, \infty)$). Note that $f'$ is undefined at $x=0$ where the graph has a vertical tangent (cusp).

$\text{python-graph}$

Since $f'$ is undefined at $x=0$ and does not change sign across this point (negative on both sides where defined), $x=0$ is neither a local max nor min. Since $f'$ changes from negative to positive at $x=1$, $f$ has a local minimum there, with value $f(1) = 1^{1/3}(1-4) = -3$. Thus $f$ is decreasing on $(-\infty, 0)$ and on $(0, 1)$, and increasing on $(1, \infty)$, with a local minimum at $x=1$.