Question

若x＜0,求函数y=2x/x^2+x+1的最小值

Answer 1

x=-t,t>0

y=-2t/[t^2-t+1]=-2/[t+1/t-1]

t+1/t>=2,

y>=-2/(2-1)=-2

x=-1时最小,为-2