若函数为f(x)=ax²+(a+1)x+2为偶函数,则f(x)在区间(-∞,1)上是先增后减.为什么啊.谁能给我讲明白啊,RT,周一就考试啦,·
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![若函数为f(x)=ax²+(a+1)x+2为偶函数,则f(x)在区间(-∞,1)上是先增后减.为什么啊.谁能给我讲明白啊,RT,周一就考试啦,·](/uploads/image/z/7238302-70-2.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0%E4%B8%BAf%28x%29%3Dax%26%23178%3B%2B%28a%2B1%29x%2B2%E4%B8%BA%E5%81%B6%E5%87%BD%E6%95%B0%2C%E5%88%99f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%28-%E2%88%9E%2C1%29%E4%B8%8A%E6%98%AF%E5%85%88%E5%A2%9E%E5%90%8E%E5%87%8F.%E4%B8%BA%E4%BB%80%E4%B9%88%E5%95%8A.%E8%B0%81%E8%83%BD%E7%BB%99%E6%88%91%E8%AE%B2%E6%98%8E%E7%99%BD%E5%95%8A%2CRT%2C%E5%91%A8%E4%B8%80%E5%B0%B1%E8%80%83%E8%AF%95%E5%95%A6%2C%C2%B7)
若函数为f(x)=ax²+(a+1)x+2为偶函数,则f(x)在区间(-∞,1)上是先增后减.为什么啊.谁能给我讲明白啊,RT,周一就考试啦,·
若函数为f(x)=ax²+(a+1)x+2为偶函数,则f(x)在区间(-∞,1)上是先增后减.为什么啊.
谁能给我讲明白啊,RT,周一就考试啦,·
若函数为f(x)=ax²+(a+1)x+2为偶函数,则f(x)在区间(-∞,1)上是先增后减.为什么啊.谁能给我讲明白啊,RT,周一就考试啦,·
因为f(x)=ax²+(a+1)x+2为偶函数
所以f(x)=f(-x)即f(-x)=a(-x)²+(a+1)(-x)+2=ax²-(a+1)x+2=f(x)=ax²+(a+1)x+2
可得a+1=0,即a=-1
则f(x)=-x²+2,对称轴为x=0
因为f(x)是开口向下的,故在(-∞,0)单调递增,在(0,1)单调递减