y=(3x2+2x+1)/(x-1)求值域

Question

y=(3x2+2x+1)/(x-1)求值域

Answer 1

令t=x-1
x=t+1
y=[3(t^2+2t+1)+2t+2+1]/t=(3t^2+8t+6)/t=8+3(t+2/t)
由均值不等式：|t+2/t|>=2 √2
所以y>=8+3*2 √2=8+6 √2
或y