已知函数f(x)=2x3 -6x2+m(m为常数)在[-2,2]上有最大值3,则该函数在 [-2,2]上的最小值是: A. 3 B. -5 C. -40...

作者: tihaiku 人气: - 评论: 0
问题 已知函数f(x)=2x3 -6x2+m(m为常数)在[-2,2]上有最大值3,则该函数在 [-2,2]上的最小值是: A. 3 B. -5 C. -40 D. -37
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答案 D
解析 提示:已知最大值为3,经以下计算得m=3。 f'(x) = 6x2 -12x = 6x(x-2),令 f'(x) = 0,得x1= 0,x2 = 2 f''(x) = 12x-12,f''(0) =-120,所以在x= 0取得极大值 代入f(x),f极大(0) = 0-0 + m = 3,m= 3 端点 x = 2,x= -2 比较f(0)、f(2)、f(-2)函数值大小,得: fmin(-2) =-37

相关内容:函数,-6x2+m,常数,最大值,-40...

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