Let f ^ - 1 (x) be the inverse function of the function f (x) = log2 ^ (x + 1). If (1 + f ^ - 1 (a)) is multiplied by (1 + f ^ - 1 (b) = 8, then the value of F (a + B) is obtained

Let f ^ - 1 (x) be the inverse function of the function f (x) = log2 ^ (x + 1). If (1 + f ^ - 1 (a)) is multiplied by (1 + f ^ - 1 (b) = 8, then the value of F (a + B) is obtained

The inverse function is X-1 of 2. From the meaning of the question, the a power of 2 times the B power of 2 = 8, thus a + B = 3, f (3) = log24 = 2

Given the function f (x) = (log2x / 8) (log2x / 4) (2 "X" 8), find its maximum and minimum

f(x)=[log2(x)-log2(8)][log2(x)-log2(4)]
=[log2(x)-3][log2(x)-2]
Loga (x = 2)
Two

Given that the function f (x) = 2 (log2x) ^ 2-2alog2x + B has a minimum value of 1 when x = 1 / 2, try to find a, B

Let t = log2x, then f (x) = 2T ^ 2-2at + B (t is any real number)
When t = A / 2, f (x) takes the minimum value - A ^ 2 / 2 + B
When x = 1 / 2, t = - 1
So a = - 2, B = 3

Given that f (x) = {log2x (x > 0), 2 is the base number {3 ^ x (x ≤ 0), then f [f (1 / 4)]=

because
f(1/4)=log 2 (1/4)=-2
f(-2)=3^(-2(=1/9
So: F [f (1 / 4)] = 1 / 9

What is the minimum value of the function f (x) = 2log2 (x + 2) - log2x? Log after the 2 are the base, find the process! Thank you

f(x)=2log2(x+2)-log2x
=log2[(x+2)²/x]
=log2[x+(4/x)+4] x>0
≥log2[4+4]=log2[8]=3
The minimum value of 3 is obtained when x = 2

Let x satisfy the inequality - 3 ≤ Log1 2x≤-1 2. Find the function f (x) = (log2x) 4）•（log2x 2) Maximum and minimum values of

∵-3≤log1
2x≤-1
2，
Qi
2≤x≤8，
f（x）=（log2x
4）•(log2x
2)=(log2x)2−3log2x+2，
Let log2x = t, (1
2 ≤ t ≤ 3), then y = t2-3t + 2,
When t = 3
At 2, Ymin = - 1
4. When t = 3, ymax = 2
So y = f (x) has a maximum value of 2 and a minimum value of - 1
4．

Given that the definition domain of function f (x) = Log1 / 2 (8-2x) is (negative infinity, 2), find the range of (1) function and (2) the inverse function of function 1 / 2 is the base number, X is the exponent of 2

(1) ∵ u = 8-2 ^ x is a decreasing function, y = log (1 / 2, U) is also a decreasing function, ᙽ y = f (x) = log (1 / 2,8-2 ^ x) is an increasing function
It is known that the domain of its definition is (- ∞, 2], that is, X ≤ 2, ν y = f (x) ≤ f (2) = log (1 / 2,4) = - 2
The value range of F (x) is (- ∞, - 2]
(2) From the known function, X: is solved
8-2^x=(1/2)^y,2^x=8-2^(-y),x=log[2,8-2^(-y)]
So the inverse function of a given function is
y=log[2,8-2^(-x)],x∈(-∞,-2]
Note: log (a, b) denotes the logarithm of B with a as the base

Using inverse representation to find the value range of function y = X-1 / x + 2 (x > = - 1)

Using inverse representation method to find the value range of function y = (x-1) / (x + 2) (x ≥ - 1)
∵y=(x-1)/(x+2)
∴y(x+2)= x-1
Therefore, YX + 2Y = X-1
Therefore, yx-x = - 1-2y
Therefore, X (Y-1) = - 1-2y
Therefore, x = (- 1-2y) / (Y-1)
∵x≥-1
Therefore: (- 1-2y) / (Y-1) ≥ - 1
Therefore: (- 1-2y) / (Y-1) + 1 ≥ 0
Therefore: - (y + 2) / (Y-1) ≥ 0
Therefore: (y + 2) / (Y-1) ≤ 0
Therefore: - 2 ≤ y ≤ 1

If the value range of the limiting function FX = 2log1 / 2x is (- 1,1), then the value range of the inverse function is

-1/2(1/2)^-1/2>x>(1/2)^1/2
√2>x>√2/2
The range of inverse function is the domain of function definition
Therefore (√ 2 / 2, √ 2)

High school mathematics: given that the inverse function of the function f (x) = x ^ 2; + 1, X ∈ [0,1] is g (x), then what is the value range of the function y = [g (x)] ^ 2; + G (2x)? Given that the inverse function of the function f (x) = x ^ 2; + 1, X ∈ [0,1] is g (x), then what is the range of the function y = [g (x)] ^ 2; + G (2x)? The answer is {1}. Why?

It is found that G (x) = √ X-1 x ∈ [1,2]
y=[g(x)]²+g(2x)
Here we should pay attention to the definition domain of Y. from G (2x), 1 ≤ 2x ≤ 2,1 / 2 ≤ x ≤ 1
So the domain of Y is [1 / 2,1] ∩ [1,2] = 1
Therefore, the range of Y is {1}