分析:根據(jù)對數(shù)的運算性質(zhì)得出
y=log(x2+2)=-log
2(x
2+2),判斷出真數(shù)大于等于2恒成立,再由以2為底對數(shù)函數(shù)是增函數(shù),求出原函數(shù)的值域.
解答:解:∵x
2+2≥2恒成立,∴函數(shù)的定義域是R,
∵函數(shù)y=log
2x在定義域上是增函數(shù),
∴y≥log
22=1,
又∵
y=log(x2+2)=-log
2(x
2+2),
∴函數(shù)的值域是(-∞,-1].
故答案為:(-∞,-1].
點評:本題的考點是復合函數(shù)的值域,對于對數(shù)型的復合函數(shù)應先求定義域,再根據(jù)對數(shù)函數(shù)的單調(diào)性求出值域.