f(m-1)
(1)|m-1|=<2,解得-1<=m<=3
(2)|1-2m|<=2,-2<=1-2m<=2,解得：-1/2<=m<=3/2
(3)|m-1|<|1-2m|
平方得：m^2-2m+1<1-4m+4m^2
3m^2-2m>0
m(3m-2)>0
m>2/3,m<0
(1)(2)(3) 取交集得到：2/3
已知偶函數(shù)定義域為[-2,2]且在[0,2]上遞增.求不等式f(m-1)
已知偶函數(shù)定義域為[-2,2]且在[0,2]上遞增.求不等式f(m-1)
數(shù)學人氣：867 ℃時間：2020-09-23 05:57:29
優(yōu)質(zhì)解答
偶函數(shù),則有f(x)=f(|x|)
f(m-1)
(1)|m-1|=<2,解得-1<=m<=3
(2)|1-2m|<=2,-2<=1-2m<=2,解得：-1/2<=m<=3/2
(3)|m-1|<|1-2m|
平方得：m^2-2m+1<1-4m+4m^2
3m^2-2m>0
m(3m-2)>0
m>2/3,m<0
(1)(2)(3) 取交集得到：2/3
f(m-1)
(1)|m-1|=<2,解得-1<=m<=3
(2)|1-2m|<=2,-2<=1-2m<=2,解得：-1/2<=m<=3/2
(3)|m-1|<|1-2m|
平方得：m^2-2m+1<1-4m+4m^2
3m^2-2m>0
m(3m-2)>0
m>2/3,m<0
(1)(2)(3) 取交集得到：2/3
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