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数学
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函数f(x)=(m-1)x²+mx+3是偶函数,则f(x)的递减区间__ 求详解
人气:487 ℃ 时间:2019-10-24 10:41:41
解答
函数f(x)=(m-1)x²+mx+3是偶函数,所以它关于y轴对称,可得:
m=0
所以函数可化为:
f(x)=-x²+3
因此其递减区间为[0,+∞)
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