Question

100 баллов!!!!!!!!!!!

Answer 1

$y=2x^2-lnx+frac{1}{2}; ; ,; ; xin [, frac{1}{4}, ;, 1, ]\\y'=4x-frac{1}{x}=frac{4x^2-1}{x}=frac{(2x-1)(2x+1)}{x} =0\\Kriticheskie; tochki:; ; x_1=-frac{1}{2}; ,; x_2=frac{1}{2}; ,; x_3=0\\znaki; y':; ; ---(-frac{1}{2})+++(0)---(frac{1}{2})+++\\x=frac{1}{2}in [, frac{1}{4}, ;, 1, ]\\y(frac{1}{2})=2cdot frac{1}{4}-lnfrac{1}{2}+frac{1}{2}=1+ln2approx 1,69\\y(frac{1}{4})=2cdot frac{1}{16}-lnfrac{1}{4}+frac{1}{2}=frac{5}{8}+ln4approx 2,01\\y(1)=2-ln1+frac{1}{2}=2,5$

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