f(x)=log1/2[(1/2)x^2-x+1],求值域
问题描述：
f(x)=log1/2[(1/2)x^2-x+1],求值域
答
1/2x²-x+1
=1/2(x-1)²+1/2≥1/2
因为log1/2(x)递减
所以f(x)≤log1/2(1/2)=1
所以值域是(-∞,1]