高一数学;已知二次函数f(x)=(m+2)x^2-2mx+4是偶函数.1:试求m的值.2:求f(x)的单调区间.谢
人气:212 ℃ 时间:2019-11-04 16:22:39
解答
因为函数是偶函数,所以f(x)=f(-x)
即(m+2)x^2-2mx+4=(m+2)(-x)^2+2mx+4
所以m=o
所以f(x)=2x^2+4
画出其二次函数图象,可知单调增区间为[0,+∞),单调减区间为(-∞,0]
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