Помогите пожалуйста!!!!!!!!!!!!!!

$\displaystyle y= \frac{27x^4}{4}-x+2; [0;2]$
$\displaystyle y'= \frac{108x^3 *4}{16}-1=27x^3-1$
$27x^3-1=0$
$\displaystyle x^3= \frac{1}{27}$
$\displaystyle x= \frac{1}{3}$
$y(0)=2$
$y(2)=108-2+2=108$
$\displaystyle y( \frac{1}{3})= \frac{27}{81*4}- \frac{1}{3}+2= \frac{1}{12}- \frac{1}{3}+2=- \frac{1}{4}+ \frac{8}{4}= \frac{7}{4}$

Ответ: y(наиб.)=108, y(наим.)=7/4