100 баллов!!!!!!!!!!!

Решите задачу:

$y=2x^2-lnx+\frac{1}{2}\; \; ,\; \; x\in [\, \frac{1}{4}\, ;\, 1\, ]\\\\y'=4x-\frac{1}{x}=\frac{4x^2-1}{x}=\frac{(2x-1)(2x+1)}{x} =0\\\\Kriticheskie\; tochki:\; \; x_1=-\frac{1}{2}\; ,\; x_2=\frac{1}{2}\; ,\; x_3=0\\\\znaki\; y':\; \; ---(-\frac{1}{2})+++(0)---(\frac{1}{2})+++\\\\x=\frac{1}{2}\in [\, \frac{1}{4}\, ;\, 1\, ]\\\\y(\frac{1}{2})=2\cdot \frac{1}{4}-ln\frac{1}{2}+\frac{1}{2}=1+ln2\approx 1,69\\\\y(\frac{1}{4})=2\cdot \frac{1}{16}-ln\frac{1}{4}+\frac{1}{2}=\frac{5}{8}+ln4\approx 2,01\\\\y(1)=2-ln1+\frac{1}{2}=2,5$

$y(naimen.)=y(\frac{1}{2})=1+ln2\\\\y(naibol.)=y(1)=2,5$