Question

若函数f(x)=x^3+ax^2+1在（0,2）内单调递减，则实数a的取值范围

Answer 1

f(x)=x^3+ax^2+1
f'(x) = 3x^2 + 2ax =3x(x+2a/3)=3(x+0){x-(-2a/3)}
当x∈(0，2a/3）时，f'(x)＜0，f(x)单调减
在（0,2）内单调递减，则 -2a/3 ≥ 2，解得：a≤-3

Answer 2

f'=3x^2+2ax<0
x(x+2a/3)<0
有题意得递减区间为0<x<-2a/3
所以要满足2<-2a/3
a<-3

Answer 3

f'(x)=3x^2+2ax
f(x)=x^3+ax^2+1在（0,2）内单调递减，
在（0,2）内f'(x)<=0
-a/3>0 a<0
f'(2)=12+4a<=0 a<=-3
实数a的取值范围 a<=-3

Answer 4

先求导然后按二次函数处理就可以了

Answer 5

求导