当x=_时,y=(x-2)(x-4)(x-6)(x-8)+12有最小值.
问题描述:
当x=______时,y=(x-2)(x-4)(x-6)(x-8)+12有最小值.
答
y=(x-2)(x-4)(x-6)(x-8)+12=[(x-2)(x-8)][(x-4)(x-6)]+12=(x2-10x+16)(x2-10x+24)+12=(x2-10x)2+40(x2-10x)+396=[(x2-10x)2+40(x2-10x)+400]-4=(x2-10x+20)2-4 当x2-10x+20=0时,即x=...