摘要：2．已知函数f (x)对任意x.y∈R.总有f (x) + f ( y) = f (x + y).且当x＞0时.f (x)＜0.f (1) =. (1)求证f (x)是R上的减函数, (2)求f (x)在[–3.3]上的最大值和最小值. 分析:抽象函数的性质要紧扣定义.并同时注意特殊值的应用. 证明:(1)令x = y =0.f (0) = 0.令x = – y可得: f (–x) = – f (x). 在R上任取x1＞x2.则f (x1) – f (x2) = f (x1) + f (– x2) = f (x1–x2). ∵x1＞x2.∴x1–x2＞0. 又∵x＞0时.f (x)＜0. ∴f (x1–x2)＜0. 即f (x1) – f (x2)＞0. 由定义可知f (x)在R上为单调递减函数. (2)∵f (x)在R上是减函数.∴f (x)在[–3.3]上也是减函数. ∴f (–3)最大.f (3)最小. f (3) = f (2) + f (1) = f (1) + f (1) + f (1) =3×() = –2. ∴f (–3) = – f (3) =2. 即f (–3)在[–3.3]上最大值为2.最小值为–2.

网址：http://m.1010jiajiao.com/timu_id_3815345[举报]