f'(x) = x^3+3x^2 -9x+cf(x)=1/4x4+x3-9/2x2+cx有三个极值点 f'(x) = 0 必须有三个根,f'(x) 有两个极值点,f'‘(x) = 3x^2+6x-9 = 0 有两个根 x=-3, x=1,f'(x)的极值点为 f'(-3) = 27 +C, f' (1) = -5+C要是f'(x) = 0...