Question

高一数学 函数

Answer 1

令t=log½ x
不等式变为
2t²+7t+3≤0
(2t+1)(t+3)≤0
-3≤t≤-1/2
即
-3≤log½ x≤-1/2
log½ x≥-3且log½ x≤-1/2且x>0
解得
√2≤x≤8

追答

因为y=log½ x在定义域范围内是单调递减函数，所以log½ x≥-3=log½ 8 ,x≤8
log½ x≤-1/2=log½ √2,x≥√2,所以是√2≤x≤8

f(x)=log₂ (x*2^(-2))*log₂ (x*2^(-1))
=(log₂x-2)(log₂x-1)
=(log₂x)²-3*log₂x+2
=(log₂x-3/2)²-1/4
√2≤x≤8
log₂√2≤log₂x≤log₂8
1/2≤log₂x≤3
log₂x=3/2,f(x)有最小值-1/4
log₂x=3,f(x)有最大值2
所以f(x)的最小值为-1/4，此时x=2^1.5,f(x)的最大值为2，此时x=8

Answer 2

（1）(2log(1/2)x+1)(log(1/2)x+3)=<0
1/8=<x=<根号2
（2）=1/4×log(2)x
最大值=1/8

Answer 3

Answer 4

追答

望采纳

Answer 5

换元