Finding critical numbers

Rewrite $f(x) = x^{4/3} + 4x^{1/3}$, so $f'(x) = \frac{4}{3}x^{1/3} + \frac{4}{3}x^{-2/3}$. Factor out $\frac{4}{3}x^{-2/3}$ to get $f'(x) = \frac{4}{3}x^{-2/3}(x+1) = \frac{4(x+1)}{3x^{2/3}}$.

This derivative equals zero when $x + 1 = 0$, i.e. $x = -1$, and is undefined when $x^{2/3} = 0$, i.e. $x = 0$. Both values are in the domain of $f$, so both are critical numbers.

Differentiating $f'(x) = \frac{4}{3}x^{1/3} + \frac{4}{3}x^{-2/3}$ gives $f''(x) = \frac{4}{9}x^{-2/3} - \frac{8}{9}x^{-5/3}$.

At $x=-1$, we use the convention $x^{p/q} = (x^p)^{1/q}$ with odd denominators: $(-1)^{-2/3} = \frac{1}{(-1)^{2/3}} = \frac{1}{((-1)^2)^{1/3}} = \frac{1}{1} = 1$ and $(-1)^{-5/3} = \frac{1}{(-1)^{5/3}} = \frac{1}{((-1)^5)^{1/3}} = \frac{1}{(-1)^{1/3}} = \frac{1}{-1} = -1$. Therefore, $f''(-1) = \frac{4}{9}(1) - \frac{8}{9}(-1) = \frac{4}{9} + \frac{8}{9} = \frac{12}{9} = \frac{4}{3} > 0$. Since $f'(-1)=0$ and $f''(-1) > 0$, the Second Derivative Test says $x=-1$ is a local minimum.

Write $f''(x) = \frac{4}{9}x^{-5/3}\left(x - 2\right)$ by factoring out $\frac{4}{9}x^{-5/3}$ from both terms (check: $\frac{4}{9}x^{-5/3}\cdot x = \frac{4}{9}x^{-2/3}$ and $\frac{4}{9}x^{-5/3}\cdot(-2) = -\frac{8}{9}x^{-5/3}$, matching the original expression).

For $x$ in $(0,\infty)$, $x^{-5/3} > 0$ always, so the sign of $f''(x)$ matches the sign of $(x-2)$. On the interval $(0,2)$, we have $x - 2 < 0$, so $f''(x) < 0$ and $f$ is concave down. On the interval $(2,\infty)$, we have $x - 2 > 0$, so $f''(x) > 0$ and $f$ is concave up. The concavity changes at $x=2$ where $f''(2) = 0$.

The second derivative test requires plugging the critical number into the second derivative to check its sign, but this only works if the second derivative is defined there. Since $f''(x) = \frac{4}{9}x^{-2/3} - \frac{8}{9}x^{-5/3}$ involves $x$ raised to negative powers, $f''(x)$ is undefined at $x=0$. Therefore the test cannot be applied at that critical number, and a different method would be needed to classify it.