Question

已知函数f(x)=(log3(x/3))(log3(9x)),x∈[1/27,3].

已知函数f(x)=(log3(x/3))(log3(9x)),x∈[1/27,3].求其函数的最大值和最小值，并求出相应的x值.

求帮忙....

Answer 1

解：f(x)=[log3(x）-1][log3(x)+2]
=[log3(x)]^2-log3(x）-2
=[log3(x) -1/2]^2 -9/4
又x∈[1/27,3]所以log3(1/27)<=log3（x)<=log3(3)即-3<=log3(x）<=1,所以log3(x）∈[-3,1/2]
上单调减，log3(x)∈[1/2,1]单调增，又log3(x)=1/2时x=√3，,所以x∈[1/27,√3],单调减，x∈[√3,3]上单调增，
所以f(x）min=(1/2-1/2)^2-9/4=-9/4,当log3(x)=1/2,即x=√3时取得；
f(1/27)=(-3-1/2)^2-9/4=10
f(3)=1*2=2
所以f(x)max=10,即x=1/27时取得。

Answer 2

把函数变形得f(x)=
=[log3(X)-log3(1/3)][log3(9)+log3(x)]
=[log3(x)+1][2+log3(x)]
=2log3(3)+[log3(x)][log3(x)]+2+log3(x)
因为log3(x)以3为底，所以是一个增函数
x=3时有最大值，代入函数解得f(X)max=6
x=1和1/27时有最小值 解得 f(x)min=2

Answer 3

f(x)=[log3(x/3)][log3(9x)]
=[log3(x)-1][log3(x)+2]
=[log3(x)]²+log3(x)-2
=[log3(x) +1/2]² -9/4
x∈[1/27，3]
-3≤log3(x)≤1
当log3(x)=-1/2时，即x=√3/3时，f(x)有最小值[f(x)]min=-9/4
当log3(x)=-3时，即x=1/27时，f(x)有最大值[f(x)]max=4