已知奇函数f(x)是定义在(-3,3)上的减函数,且满足不等式f(x-3)+f(x2-3)

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已知奇函数f(x)是定义在(-3,3)上的减函数,且满足不等式f(x-3)+f(x2-3)
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已知奇函数f(x)是定义在(-3,3)上的减函数,且满足不等式f(x-3)+f(x2-3)
已知奇函数f(x)是定义在(-3,3)上的减函数,且满足不等式f(x-3)+f(x2-3)

已知奇函数f(x)是定义在(-3,3)上的减函数,且满足不等式f(x-3)+f(x2-3)
既然是(-3,3)上的函数,那么x-3及x2-3不在定义域内,f(x-3)+f(x2-3)无意义,题目肯定是不对的,楼主应该确定题目正确

根据题意,可得-3<x-3<3-3<x2-3<3​,
解可得0<x<6-
6<x<
6​且x≠0,
故0<x<6,
又∵f(x)是奇函数,
∴f(x-3)<-f(x2-3)=f(3-x2),
又f(x)在(-3,3)上是减函数,
∴x-3>3-x2,即x2+x-6>0,
解得x>2或x<-3,综上得2<...

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根据题意,可得-3<x-3<3-3<x2-3<3​,
解可得0<x<6-
6<x<
6​且x≠0,
故0<x<6,
又∵f(x)是奇函数,
∴f(x-3)<-f(x2-3)=f(3-x2),
又f(x)在(-3,3)上是减函数,
∴x-3>3-x2,即x2+x-6>0,
解得x>2或x<-3,综上得2<x<6,即A={x|2<x<6},
∴B=A∪{x|1≤x≤5}={x|1≤x<6},
又g(x)=-3x2+3x-4=-3(x-12)2-知:g(x)在B上为减函数,
∴g(x)max=g(1)=-4.

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=9

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