已知函数f(x)=－2x3－3x2＋12x+1在〔m,1〕上的最小值为－17,求m

f(x) = -2x - 3x + 12x + 1 f'(x) = -6x - 6x + 12 令f'(x) = 0,即 -6x - 6x + 12 = 0 x + x - 2 = 0 (x + 1/2) = 9/4 x = -1/2±3/2 x = 1 x = -2 即f(x)在(1,8)处有极大值,在(-2,-19)处有极小值.因为m < 1,所以在...