已知函数f(x)=2x*3-6x*2+3,试判断函数零点的个数和存在区间

f(x)=2x^3-6x^2+ 3f'(x) = 6x^2 - 12x =0x(x-2) =0x = 0 or 2f''(x) = 12x-12f''(0) = -12 ( max)f''(2) = 12 ( min)f(0) = 3 > 0f(2) = 2(8)-6(4) + 3 = -5 < 0f(x) is increasing on （-∞,0]from -ve -> +vef(x) ...