高一数学帮帮忙俊急急!

2024-11-23 22:26:14
推荐回答(2个)
回答1:

f(x)=log(2){x/2}*log(2) {x/4}= {log(2)x-1}*{ log(2)x -2}
根据-3≤log(1/2)x≤-3/2
可得2√2≤x≤8
设log(2)x=t,则3/2≤t≤3
所以所求式子=t²-3t+2(3/2≤t≤3)
此式子的取值范围为[-1/4,2]
所以f(x)值域为[-1/4,2]

回答2:

-3<=log1/2(x)<=-3/2
=> (1/2)^(-3/2) <=x <=(1/2)^(-3)
=> genhao8<=x<=8
=> 3/2<=log2x<=3

f(x) = (log2x-1)(log2x-2) = log2x*log2x-3log2x+2 = (log2x-3/2)^2 -1/4
值域 [-1/4,2]