Question

已知函数f（x）=-1/2x^2+4x-3lnx在[t,t+1]上不单调，则t的范围

Answer 1

f'(x)=-x+4-3/x=(-x^2+4x-3)/2
f(x在[t,t+1]不单调，即f'(x)在[t,t+1]内有零点
而f'(x)的零点为1,3
故t<1<t+1或t<3<t+1
所以t得范围为(0,1)∪(2,3)

Answer 2

求导f'(x)=-x+4-3/x
令f'(x)>0 得x(x-1)(x-3)<0 得x<0或1<x<3
令f'(x)<0 得x(x-1)(x-3)>0 得0<x<1或x>3
函数在0,1,3处取得极值
f(x)在[t,t+1]上不单调 即为在[t,t+1]内函数有极值
t<0<t+1 t<1<t+1 t<3<t+1
得-t属于{-1,0}并上{0,1}并上{2,3}