Let $f\left(x\right)=6x^{4/3}-3x^{1/3},x\in [-1,1]$. Then

 

$f^{\mathrm{\prime }}\left(x\right)=\frac{8x-1}{x^{2/3}}$. Thus $f^{\mathrm{\prime }}\left(x\right)=0$ when $x=1/8$

and $f\text{'} \left(x\right)$ does not exist when $x = 0.$ Now $f (– 1) = 9, f \left(0\right)$

$=0,f(1/8)=-9/8$and $f\left(1\right)=3.$

The maximum value of $f \left(x\right)$ is 9 and the minimum value is $– 9/8 on [–1, 1].$