Find the absolute maximum and minimum values of the function $f$ given by $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$ for $x \in [-1, 1]$.

(N/A) We have $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$.
First,we find the derivative $f'(x)$:
$f'(x) = 12 \cdot \frac{4}{3}x^{\frac{1}{3}} - 6 \cdot \frac{1}{3}x^{-\frac{2}{3}} = 16x^{\frac{1}{3}} - \frac{2}{x^{\frac{2}{3}}} = \frac{16x - 2}{x^{\frac{2}{3}}} = \frac{2(8x - 1)}{x^{\frac{2}{3}}}$.
Setting $f'(x) = 0$ gives $8x - 1 = 0$,so $x = \frac{1}{8}$.
Also,$f'(x)$ is undefined at $x = 0$. Thus,the critical points are $x = 0$ and $x = \frac{1}{8}$.
Now,we evaluate $f(x)$ at the critical points and the endpoints $x = -1$ and $x = 1$:
$f(-1) = 12(-1)^{\frac{4}{3}} - 6(-1)^{\frac{1}{3}} = 12(1) - 6(-1) = 12 + 6 = 18$.
$f(0) = 12(0)^{\frac{4}{3}} - 6(0)^{\frac{1}{3}} = 0$.
$f(\frac{1}{8}) = 12(\frac{1}{8})^{\frac{4}{3}} - 6(\frac{1}{8})^{\frac{1}{3}} = 12(\frac{1}{16}) - 6(\frac{1}{2}) = \frac{3}{4} - 3 = \frac{3 - 12}{4} = -\frac{9}{4}$.
$f(1) = 12(1)^{\frac{4}{3}} - 6(1)^{\frac{1}{3}} = 12 - 6 = 6$.
Comparing these values,the absolute maximum value is $18$ at $x = -1$ and the absolute minimum value is $-\frac{9}{4}$ at $x = \frac{1}{8}$.