Given that x satisfies 8 ≤ x ≤ 32, find the maximum and minimum values of the function f (x) = log2x / 8 · (log2x-1)

Given that x satisfies 8 ≤ x ≤ 32, find the maximum and minimum values of the function f (x) = log2x / 8 · (log2x-1)

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8> = x > = root 2 find the maximum and minimum values of the function log2 (2 / x) times log2 (4 / x) I'm sorry. It's log2 (2 / x) times log2 (4 / x)

Let t = ㏒ 2 (x) 2 (x) 2 (x / 2) = ㏒ 2 (x / 4) = ㏒ 2 (x) - 2. Let t = ㏒ 2 (x) 2 (x) - 2 (x / 2) = (2) = = > 1 / 2 ≤㏒ 2 (x) ≤ 3. = = > 1 / 2 ≤ t ≤ 3. Therefore, the problem can be solved to find the maximum value of y = (t-1) (T-2) on [1 / 2,3], 3]. The combination of number and shape is easy to know, Ymin = y (3 / 2) = - 1 / 1 / 2 / 2) = - 1 / 1 / 2,3.it is easy to know the combination of number and shape, Ymin = y (3 / 2) = - 1 / 1 / 2) 4, ymax = y (3) = 2

Find the function y = log2 (x / 2) * logx (x / 4), where x belongs to the maximum and minimum values of [1,8]

First, change the bottom formula into a unified formula with the base of 10 ~ and then simplify it to the simplest form: y = - lgx + 2lg2 ~ and then bring in the definition field ~ the answer is [- lg2,2lg2]

The function f (x) = 2 (log2x) 2 + A * log2x + B has a minimum value of 1 when x = 1 / 2. Try to determine the values of a and B

Let t = log2x, then f (x) can be expressed as
2t²+at+b=2（t+a/4）²+b-a²/8
The value range of log2x is r, so
The minimum value of F (x) is
When t = - A / 4, i.e. log (1 / 2) = - A / 4, there is B-A / 8 = 1
A = 4log2, B = 1 + 2log? 2 is obtained

The known function f (x) = (log2x-2) (log4x-1) 2） (1) When x ∈ [2,4], find the range of the function; (2) If f (x) > mlog2x holds for X ∈ [4,16], find the value range of M

（1）f（x）=（log2x-2）（log4x-1
2）
=1
2（log2x）2-3
2log2x+1，2≤x≤4
Let t = log2x, then y = 1
2t2-3
2t+1=1
2（t-3
2）2-1
8，
∵2≤x≤4，
∴1≤t≤2．
When t = 3
At 2, Ymin = - 1
When t = 1, or T = 2, ymax = 0
The value range of the function is [- 1
8，0]．
(2) Let t = log2x, get 1
2t2-3
2T + 1 > MT holds for 2 ≤ t ≤ 4
∴m＜1
2t+1
T-3
2 holds for t ∈ [2,4],
Let g (T) = 1
2t+1
T-3
2，t∈[2，4]，
∴g（t）=1
2t+1
T-3
2=1
2（t+2
t）-3
2，
∵g（t）=1
2t+1
T-3
2 is an increasing function on [2,4],
When t = 2, G (T) min = g (2) = 0,
∴m＜0．

The known function f (x) = log2x + 3, X ∈ [1,4] (1) Find the value range of function f (x); (2) If G (x) = f (x 2) - [f (x)] 2, find the minimum value of G (x) and the corresponding value of X

(1) ∵ f (x) = log2x + 3 is an increasing function on X ∈ [1,4],

Given that f (x) = (1-log2x) / (1 + log2x) (x > 1 / 2), what is the inverse function of F (3 / 5)? (2 is the base)

Let f (x) = (1-log2x) / (1 + log2x) = 3 / 5
5-5log2x=3+3log2x
log2x=1/4
x=2^(1/4)
That is, f [2 ^ (1 / 4)] = 3 / 5
So F-1 (3 / 5) = 2 ^ (1 / 4)

Given the piecewise function f (x) = log2x (x > 0) 3 ^ x (x ≤ 0), then the value of F [f (1 / 4)] is

1/9

Find the minimum value of the function f (x) = 2log2 ^ (X-2) - log2 ^ (x-3) Thank you

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The function f (x) = (log2x-1) / (log2x + 1), if f (x1) + F (x2) = 1 (where X1 and X2 are greater than 2), then the minimum value of F (x1x2) is obtained

Let X1 = a, X2 = B, where a and B are greater than 2. Let f (x) = (log2x-1) / (log2x + 1), if f (a) + F (2b) = 1, where a, b > 2. Find the minimum value of F (AB)