Answer
There are no absolute maximum and absolute minimum values.
Critical points $x=1,\frac{11}{3}$
Work Step by Step
Differentiate and equate it 0 then obtain critical points.
$f'(x)=\frac{-4x+4}{(11-3x)^\frac{2}{3}}$
$\begin{aligned}-4x+4&=0\\-4x&=-4\\x&=1\end{aligned}$
$\begin{aligned} \sqrt[3]{\left(11-3x\right)^2}\neq 0\\\left(\sqrt[3]{\left(11-3x\right)^2}\right)^3&>0^3\\121-66x+9x^2&>0\\\left(3x\right)^2-66x+11^2&>0\\\left(3x\right)^2-2\cdot \:3x\cdot \:11+11^2&>0\\\left(3x-11\right)^2&>0\\3x-11&>0\\x&>\frac{11}{3}\\x&=\frac{11}{3}\end{aligned}$
There is no global extrema since the function increases without bound as $x\rightarrow\infty$ and it decreases as $x\rightarrow- \infty$.
