Question

已知函数f(x)=log3(x/27)*log3(ax).(1/27<=x<=9)求f(x)的最小值

已知函数f(x)=log3(x/27)*log3(ax).(1/27<=x<=9)求f(x)的最小值

Answer 1

f(x)=(log3x-log327)(log3a+log3x)
f(x)=(log3x-3)(log3a+log3x)
t=log3x
f(t)=(t-3)(log3a+t)
f(t)=log3at+t^2-3log3a-3t
f(t)=t^2+(log3a-3)t-3log3a
f(t)=(t-(log3a-3)/2)^2-(log3a-3)^2/4-3log3a
1/27<=x<=9
-3<=t<=2
t=(log3a-3)/2
1.-3<=(log3a-3)/2<=2,-6<=log3a-3<=4,-3<=log3a<=7,1/27<=a<=3^7
fmin=-(log3a-3)^2/4-3log3a
2.(log3a-3)/2<-3
t=-3,fmin=9-3(log3a-3)-3log3a=9-3log3a+9-3log3a=18-6log3a,(0<a<1/27)
3.log3a-3>2,t=2,fmin=4+2(loga3-3)-3log3a=4+2loga3-6-3log3a=-2-log3a(a>3^7)
所以有三种情况

3.

Answer 2

f(x)=(log3x-3)log3x
log3x单调函数故只需要代入边界值
f(1/27)=18
f(9)=-2