已知函数f(x)是定义域在R上的偶函数,且在区间(-∞,0)上单调递减,求满足f(x2+2x+3)>f(-x2-4x-5)的x的集合.
人气:270 ℃ 时间:2019-08-18 12:11:21
解答
因为f(x)为R上的偶函数,所以f(x2+2x+3)=f(-x2-2x-3),则f(x2+2x+3)>f(-x2-4x-5)即为f(-x2-2x-3)>f(-x2-4x-5).又-x2-2x-3<0,-x2-4x-5<0,且f(x)在区间(-∞,0)上单调递减,所以-x2-2x-3<-x...
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