y=(x-1)/2x²
2yx²-x+1=0
當(dāng)y≠0,△=(-1)²-8y≥0, y<=1/8所以y最大值1/8

y=(x-1)/2x²
y'=(1/2)[1/x²+(x-1)(1/x²)']
=(1/2)[1/x²-2(x-1)(1/x³)']
=x/2x³ - 2(x-1)/2x³
=(2-x)/2x³
x=2時,y'=0,極值點(diǎn)
再求一下y'',看一下x=2左右兩側(cè)y''的大小,基本上y'極大值點(diǎn).所以y最大值1/8