Question

已知x∈[1\27,1\9],函数f(x)=(log3 x\27)(log3 3x)求函数f(x)的最大值和最小值,

若方程f(x)=m=0有两根a,β，求aβ的值

Answer 1

设log3(x)=t
x∈[1/27,1/9]
则t∈[-3,-2]

f(x)=log3(x/27)*log3(3x)
=[log3(x)-log3(27)]*[log3(x)+log3(3)]
=(log3(x)-3)*(log3(x)+1)
即f(t)=(t-3)*(t+1)=t^2-2t-3=(t-1)^2-4
对称轴t=1,所以,当t=-3时,有最大值=12,当t=-2时有最小值=5

f(t)+m=0
t^2-2t-3+m=0
由根与系数的关系
t1+t2=2
即log3(a)+log3(b)=2
log3(ab)=2
ab=9

Answer 2

f(x)=(log3 x\27)(log3 3x)=(log3 x -log3 27)(log3 x +log3 3)=(log3 x -3)(log3 x +1)
=(log3 x)ˆ2-2log3 x -3=(log3 x -1)ˆ2-4
∵1\27≤x≤1\9 ∴－3≤x≤－2
∴f(x)的最小值和最大值分别为5和12

Answer 3

设log3(x)=t
x∈[1/27,1/9]
则t∈[-3,-2]

f(x)=log3(x/27)*log3(3x)
=[log3(x)-log3(27)]*[log3(x)+log3(3)]
=(log3(x)-3)*(log3(x)+1)
即f(t)=(t-3)*(t+1)=t^2-2t-3=(t-1)^2-4
对称轴t=1,所以,当t=-3时,有最大值=12,当t=-2时有最小值=5

f(t)+m=0
t^2-2t-3+m=0
由根与系数的关系
t1+t2=2
即log3(a)+log3(b)=2
log3(ab)=2
ab=9
谢谢。。

Answer 4

f(x)=(log3 (x/27))*(log3 (3x))
=(log3 x-log3 27)(log3 3+log3 x)
=(log3 x-3)(log3 x+1)
设t=log3 x
则f(x)=(t-3)(t+1)=t²-2t-3
剩下的自己做