设函数f(t)=36x^2 + 12x + 1,函数在定义域内可导,则f(t)二阶导数为72>0,所以f(t)一阶导函数72x + 12 = 0时,函数f(t)=36x^2 + 12x + 1取得最小值,解方程72x + 12 = 0得x=-1/6,带入f(t)=36x^2 + 12x + 1,得最小值为0；即36x^2 + 12x + 1>=0,将36x^2 + 12x + 1>=0整理得-3x^2-x+1<=13/12