分析:先由函數(shù)求f′(x)=-x+4-
,再由“函數(shù)
f(x)=-x2+4x-3lnx在[t,t+1]上不單調”轉化為“f′(x)=-x+4-
=0在區(qū)間[t,t+1]上有解”從而有
=0在[t,t+1]上有解,進而轉化為:g(x)=x
2-4x+3=0在[t,t+1]上有解,用二次函數(shù)的性質研究.
解答:解:∵函數(shù)
f(x)=-x2+4x-3lnx∴f′(x)=-x+4-
∵函數(shù)
f(x)=-x2+4x-3lnx在[t,t+1]上不單調,
∴f′(x)=-x+4-
=0在[t,t+1]上有解
∴
=0在[t,t+1]上有解
∴g(x)=x
2-4x+3=0在[t,t+1]上有解
∴g(t)g(t+1)≤0或
| t<2<t+1 | g(t)≥0 | g(t+1)≥0 | △=4>0 |
| |
∴0<t≤1或2≤t<3.
故答案為:0<t≤1或2≤t<3
點評:本題主要考查導數(shù)法研究函數(shù)的單調性,基本思路:當函數(shù)是增函數(shù)時,導數(shù)大于等于零恒成立,當函數(shù)是減函數(shù)時,導數(shù)小于等于零恒成立,然后轉化為求相應函數(shù)的最值問題.注意判別式的應用.