定义在R上的函数f(x)满足①对任意x,y∈R有f(x+y)=f(x)+f(y)②当x>0时,f(x)<0,f(1)=-2(

2024-12-01 17:36:28
推荐回答(1个)
回答1:

∵对任意x,y∈R有f(x+y)=f(x)+f(y)
(1)取x=y=0,可得f(0)=0,
(2)取y=-x,可得f(x)+f(-x)=f(0)=0,
所以f(-x)=-f(x),f(x)是奇函数
(3)任取x1<x2
则 x2-x1>0
∴f(x2)-f(x1)=f(x2)+f(-x1)=f(x2-x1
又∵当x>0时,f(x)<0,
f(x2)-f(x1)<0,
可得 f(x1)>f(x2),
所以f(x) 在R上是减函数 
(4)∵f(1)=-2
∴f(2)=f(1)+f(1)=-4,
f(4)=f(2)+f(2)=-8
∴不等式f(x2-2x)-f(x)≥-8
可化为f(x2-2x)-f(x)≥f(4)
即f(x2-2x)≥f(x)+f(4)
即x2-2x≤x+4
即x2-3x-4≤0
解得-1≤x≤4
故不等式f(x2-2x)-f(x)≥-8的解集为[-1,4]

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