Find the absolute maximum and minimum values of a function $f$ given by

Given : $f(x)=12 x^{\frac{4}{3}}-6 x^{\frac{1}{3}}$

$\Rightarrow f^{\prime}(x)=16 x^{\frac{1}{3}}-2 x^{\frac{-2}{3}}=\frac{2(8 x-1)}{x^{\frac{2}{3}}}$

For a local maximum or a local minimum, we must have

$f^{\prime}(x)=0$

$\Rightarrow \frac{2(8 x-1)}{x^{\frac{2}{3}}}=0$

$\Rightarrow 8 x-1=0$

$\Rightarrow x=\frac{1}{8}$

Thus, the critical points of $f$ are $-1, \frac{1}{8}$ and 1 .

Now,

$f(-1)=12(-1)^{\frac{4}{3}}-6(-1)^{\frac{1}{3}}=18$

$f\left(\frac{1}{8}\right)=12\left(\frac{1}{8}\right)^{\frac{4}{3}}-6\left(\frac{1}{8}\right)^{\frac{1}{3}}=\frac{-9}{4}$

$f(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}}=6$

Hence, the absolute maximum value when $x=-1$ is 18 and the absolute minimum value when $x=\frac{1}{8}$ is $\frac{-9}{4}$.