定義在[-4,4]上的偶函數(shù)f(x)在[0,4]上是單調(diào)遞增函數(shù),f(xy)=f(x)+f(y),且f(3)=1,求解不等式f(x)+f(x-2)>1
定義在[-4,4]上的偶函數(shù)f(x)在[0,4]上是單調(diào)遞增函數(shù),f(xy)=f(x)+f(y),且f(3)=1,求解不等式f(x)+f(x-2)>1
優(yōu)質(zhì)解答
答:
f(x)定義在[-4,4]上的偶函數(shù):f(-x)=f(x)
在[0,4]上單調(diào)遞增,在[-4,0]上單調(diào)遞減
f(xy)=f(x)+f(y),f(3)=1=f(-3)
f(x)+f(x-2)>1
f[x(x-2)]>f(3)=f(-3)
所以:
3<|x(x-2)|<=4
所以:
-4<=x(x-2)<-3(無解)或者3所以:x^2-2x-3>0并且x^2-2x-4<=0
所以:1-√5<=x<-1或者3因為:-4<=x<=4并且-4<=x-2<=4,即-2<=x<=4
綜上所述:1-√5<=x<-1或者3