Let y = f (x), y = g (x) have inverse functions, and the function images of F (x-1) and g-1 (X-2) are symmetric with respect to the straight line y = X. if G (5) = 1999, then f (4)=______ .

From the meaning of the title, f (x-1) and g-1 (X-2) are inverse functions,
The inverse function of y = g-1 (X-2) is y = g (x) + 2,
∴f（x-1）=g（x）+2，
∴f（4）=g（5）+2=1999+2=2001，
So the answer is: 2001

It is known that y = f (x-1) is an odd function and y = g (x) is an inverse function of y = f (x). If x 1 + x 2 = 0, then G (x1) + G (x2)=______ .

The function y = f (x-1) is a singular function defined on R, and its image is symmetric about the origin of the function y = f (x). The image of the function y = f (x-1) is shifted one unit to the left. The image of the function y = f (x) is symmetric with respect to (- 1,0) and ∵ y = g (x) is the inverse function of y = f (x)

It is known that the function f (x) has an inverse function on the domain R, and f (9) = 18. If the inverse function of F (x + 1) is y = f ^ (- 1) (x + 1), then f (2008) =?

The inverse function of y = f (x + 1) y = f ^ (- 1) (x) - 1
∴ f^(-1)(x+1)=f^(-1)(x) -1
f(9)=18
That is, f ^ (- 1) (18) = 9
∴ f^(-1) (18+1999)=f^(-1)(18)+1999=2008
That is, f ^ (- 1) (2017) = 2008
∴ f(2008)=2017

If y = f (x + 1) inverse function is F-1 (x + 1), what information can we get? The inverse function is y = F-1 (x + 1)

Just push it down
F-1 (y) = x + 1, y = F-1 (x) - 1 = F-1 (x + 1), so F-1 (x) - F-1 (x + 1) = 1

The inverse function of the function y = 2 ^ (x + 1) (x ∈ R) is - -- the definition domain of the inverse function is ---

x+1=log2(y)
x=-1+log2(y)
So the inverse function is y = - 1 + log2 (x)
The definition domain is x > 0

It is known that the function y = f (x) is the inverse function of the function g (x) = logax (a > 0 and a ≠ 1), and f (1) = 2 to find the analytic expression of F (x); If b > 0, f (x) ≥ b-X is always true on X ∈ [1, + ∞), and find the value range of B

The function y = f (x) is the inverse function of the function g (x) = logax (a > 0 and a ≠ 1)
∴ f(x)=a^x
∵ f(1)=2
That is, a = 2
∴ f(x)=2^x
If 2 ^ x ≥ b-X is known to be constant on X ∈ [1, + ∞),
If + X ∈∞, then + X ∈∞ holds
Let H (x) = 2 ^ x + X, that is, the minimum value of H (x) ≥ B
∵ y = 2 ^ x, y = x are all increasing functions,
Then H (x) is an increasing function,
The minimum value of H (x) is h (1) = 2 + 1 = 3
The value range of B is B ≤ 3

It is known that the inverse function of the function f (x) = 2x is f − 1 (x). If f − 1 (a) + F − 1 (b) = 4, then 1 A+1 The minimum value of B is () A. 1 B. 1 Two C. 1 Three D. 1 Four

The inverse function of the function y = 2x is y = F-1 (x) = log2x,
So F-1 (a) + F-1 (b) = 4 is log2a + log2b = 4,
AB = 16 (a, b > 0)
One
A+1
b≥2
One
a×1
B=1
2, (if and only if a = b)
Therefore, B

The inverse function F-1 (x) of the function f (x) = 2 ^ x satisfies the minimum value of F-1 (a) + F-1 (b) = 4,1 / A + 1 / b

f-1(x)=log2 x
f-1(a)+f-1(b)=log2 ab=4
So AB = 16
1 / A + 1 / b > = 2 roots 1 / AB = 1 / 2
So the minimum is 1 / 2

Given the function f (x) = log2 (x + 1 / x-1), find the inverse function f ^ - 1 (x) of F (x), and find the function g (x) = f ^ - 1 (x) - log2 (k) Known function f (x) = log2 (x + 1 / x-1) Find the inverse function f ^ - 1 (x) of F (x), and find that the real number k whose function g (x) = f ^ - 1 (x) - log2 (k) has zero point is the value range

Known function f (x) = log2 (x + 1 / x-1)
Find the inverse function f ^ - 1 (x) of F (x), and find that the real number k whose function g (x) = f ^ - 1 (x) - log2 (k) has zero point is the value range
Analysis: ∵ function f (x) = log [2, (x + 1) / (x-1)]
y=log[2,(x+1)/(x-1)]
2^y=(x+1)/(x-1)
x=(2^y+1)/(2^y-1)
y=(2^x+1)/(2^x-1)
Inverse function of F (x) = (2 ^ x + 1) / (2 ^ x-1)
When x → - ∞, f ^ - 1 (x) → - 1; when x → + ∞, f ^ - 1 (x) → 1;
Let g (x) = f ^ - 1 (x) - log2 (k) have zeros
Make | log (2, K) | 1 = = > log (2, K) 0k > 2
∴0

Given that the function f (x) = log2 (x), X belongs to [2,8], the minimum value of function g (x) = f (x) ^ 2-2af (x) + 3 is h (a) (1) Find H (a) (2) Whether there is a real number m and the following conditions are satisfied: 1. M ＞ n ＞ 3; 2. When the definition field of H (a) is [n, M], the value field is [n ^ 2, m ^ 2]. If it exists, find the value of M, N; if not, please explain the reason

Let t = f (x), G (x) = g (T) = t? - 2at + 3. Define the domain [1,3]
Consider the parabola corresponding to G (T), and the axis of symmetry t = a
When 1 ≤ a ≤ 3, the minimum value is h (a) = g (a) = 3-A 2
When a > 3, the minimum value is h (a) = g (3) = 12-6a
When a 3, H (a) = 12-6a decreases monotonically. If M > n > 3, then there must be h (n) > H (m)
H (n) = m 2, H (m) = N 2, i.e. 12-6 n = m 2, 12-6 M = N 2
That is (m-n) (M + n-6) = 0. M = n (omitting the contradiction with the question design), M + n = 6 (rejecting the contradiction with the question design)
The equation has no solution

Let y = f (x), y = g (x) have inverse functions, and the function images of F (x-1) and g-1 (X-2) are symmetric with respect to the straight line y = X. if G (5) = 1999, then f (4)=______ .