What goes up comes down

The maximum value of $f(|x-2|)$ is identical to the maximum value of $f(|x|)$ as it is just a horizontal translation.

$f(|x|)=-11|x|^3-12|x|+6$ . Since $|x| \ge 0$ for all $x$ , and $f(|x|)$ will decrease as $|x|$ increases, the maximum value of the function is found when $|x|=0$ .

$f(0)=\large \color{#20A900}{\boxed{6}}$

we have $f(x)=-11x^3-12x+6 , \text{ if } x\rightarrow|x-2| \\ f(|x-2|)=-11|x-2|^3-12|x-2|+6$

To Find $f_{Max}(|x-2|)$ , i will Differentiate the function and equate it to 0 .

\color{#333333}{f^{'}(|x-2|)=-33|x-2|^2\dfrac{x-2}{|x-2|}-12\dfrac{x-2}{|x-2|}+0 = 0 \\ \implies \dfrac{x-2}{|x-2|}\bigg{(}33|x-2|^2+12\bigg{)}=0\\ \implies \dfrac{x-2}{|x-2|}=0 \\ \text{Now observe that , } \lim_{x\to 2^{+}}\dfrac{x-2}{|x-2|}=1 \text{ and } \lim_{x\to 2^{-}}\dfrac{x-2}{|x-2|}=-1 }

which means that at $x=2$ function is maximum , $\color{#3D99F6}{\boxed{f(0) =6}}$ is our answer