X^3-3x^2-9x+2>=m,在[-2,2]恒成立求m求职范围

令f(x)=x³-3x²-9x+2则f '(x)=3x²-6x-9=3(x-3)(x+1)令f '(x)=0,得x1=3（舍去）,x2=-1①当x∈[-2,-1)时,f '(x)＞0,为增函数②当x∈(-1,2]时,f '(x)＜0,为减函数又f(-2)=0 f(2)=-20故f(2)＜f(-2)所以f(x...