Question

已知函数f(x)=(1/3)ax^3+(1/2)bx^2+cx+d在点(1f(1))处切线的斜率为

已知函数f(x)=(1/3)ax^3+(1/2)bx^2+cx+d在点(1f(1))处切线的斜率为0，a<b<c,若函数f(x)在区间(m,n)单调递增，求n-m的取值范围

Answer 1

依题意得;
f'(x)=ax^2+bx+c
f'(1)=c=0
f'(x)=x(ax+b)
又因为a<b<c=0
f'(x)>=0,则x(ax+b)>=0
x(-ax-b)<=0
又因为a<b<c=0
-1<=-b/a<=0
则-b/a<=x<=0
即函数的增区间为[-b/a,0]
若函数f(x)在区间(m,n)单调递增

则-b/a<=m,n<=0
0<n-m<=0-(-b/a)=b/a<1
即0<n-m<1

追问

不对

f'(1)=a+b+c=0

追答

不好意思，看错了，看成是0
依题意得;
f'(x)=ax^2+bx+c
f'(1)
=a+b+c=0
c=-(a+b)
f'(x)=ax^2+bx-(a+b)=(x-1)(ax+a+b)
又因为a0，a=0,则(x-1)(ax+a+b)>=0
(x-1)(x+1+b/a)<=0
又因为a<b<-(a+b),a<0
-1/2<b/a<1
即函数的增区间为[-1-b/a,1]
若函数f(x)在区间(m,n)单调递增
则-1-b/a<=m,n<=1
0<n-m<=1-[-1(-b/a)]=2+b/a
3/2<2+b/a<3
即0<n-m<3/2