Question

f<x>=1/3ax2+1/2(a-1)X2-X+4求单调区间

Answer 1

问：f=1/3ax2+1/2(a-1)X2-X+4求单调区间

答：f'(x)=x^2-(a+1)+a=(x-a)(x-1)(1)a>1时有f'(x)>0时有x>a,x0时有x>1,x=0,故在R上是单调增函数.

问：题目打错了 前面是三次方

答：f(x)=(a/3)x^3-(1/2)*(a^2+1)x^2+ax+1求导:f'(x)=ax^2-(a^2+1)x+a f'(x)=(ax-1)(x-a)解f(x)=0得:ax=1或者x=a1)a=0时:f'(x)=-x,f(x)=-(1/2)x^2+1单调递增区间(-00]单调递减区间[0+)2)a=-1时f(x)=(-x-1)(x+1)=-(x+1)^2