设函数f(x)=x^3-3ax^2+3bx的图像与直线12x+y-1=0相切于点(1,-11).(1)求a,b的值(2)解不等式:f'(x)
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设函数f(x)=x^3-3ax^2+3bx的图像与直线12x+y-1=0相切于点(1,-11).(1)求a,b的值(2)解不等式:f'(x)
设函数f(x)=x^3-3ax^2+3bx的图像与直线12x+y-1=0相切于点(1,-11).
(1)求a,b的值
(2)解不等式:f'(x)
设函数f(x)=x^3-3ax^2+3bx的图像与直线12x+y-1=0相切于点(1,-11).(1)求a,b的值(2)解不等式:f'(x)
(Ⅰ)求导得f′(x)=3x2-6ax+3b.
由于f(x)的图象与直线12x+y-1=0相切于点(1,-11),
所以f(1)=-11,f′(1)=-12,即:
1-3a+3b=-11解得:a=1,b=-3.
3-6a+3b=-12
(Ⅱ)由a=1,b=-3得:f′(x)=3x2-6ax+3b=3(x2-2x-3)=3(x+1)(x-3)
令f′(x)>0,解得x<-1或x>3;
又令f′(x)<0,解得-1<x<3.
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