f'(x)=3x²-6x+3
f'(x)=3x²-6x+3=3(x-1)²=0,x=1已知三次函數(shù)f(x)=xˇ3-3xˇ2+3x+c,若對(duì)任意x∈[-1,4]都有f(x)〉f(x)的導(dǎo)數(shù)成立,求c的取值范圍?f(x)=x³-3x²+3x+cf'(x)=3x²-6x+3取g(x)=f(x)-f'(x)=(x³-3x²+3x+c)-(3x²-6x+3)=x³-6x²+9x+c-3g'(x)=3x²-12x+9=3(x-3)(x-1)考察g'(x),x=1或x=3時(shí),g'(x)=0x<1時(shí),g'(x)>0,單調(diào)遞減1