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

Explanation

`f(x)=12 x^{4 / 3}-6 x^{1 / 3} x in[-1,1]`

`f^{prime}(x)=16 x^{1 / 3}-2 x^{-2 / 3}`

`=frac{2(8 x-1)}{x^{2 / 3}}`

For critical points, `f^{prime}(x)=0`

`therefore f^{prime}(x)=0`

`Rightarrow frac{2(8 x-1)}{x^{2 / 3}}=0`

`Rightarrow x=frac{1}{8}`

`f(-1)=12+6=18`

...