Find the absolute maximum and minimum values of a function $f$ given by
$f(x)=12 x^{4 / 3}-6 x^{1 / 3}, x \in[-1,1]$
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 $\frac{-9}{4}$.
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