For $f(x)=12x^{4/3}-6x^{1/3}$ on $[-1,1]$, the absolute maximum value is

1. $f'(x)=16x^{1/3}-2x^{-2/3}=\dfrac{2(8x-1)}{x^{2/3}}$. 2. $f'(x)=0$ at $x=\tfrac{1}{8}$; undefined at $x=0$. Critical points $0,\tfrac{1}{8}$. 3. Values: $f(-1)=18$, $f(0)=0$, $f(\tfrac{1}{8})=-\tfrac{9}{4}$, $f(1)=6$. 4. The largest is $18$ at $x=-1$. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.27_