The function f (x) is defined on the interval (0, + ∞), and for any x ∈ positive real number, y ∈ R, f (x ^ y) = YF (x) 1) Find the value of F (1); (2) if f (1 / 2) < 0, find that f (x) is an increasing function on (0, + ∞)

1) F (x ^ y) = YF (x) let x = 1F (1) = YF (1) hold for any y, so f (1) = 0, X 〉 02) let y = - 1, f (1 / x) = - f (x) take any X1 〉 x2 〉 0, let X1 = (1 / 2) ^ T1, X2 = (1 / 2) ^ T2, then T1 〈 T2, then f (x1) - f (x2) = f ((1 / 2) ^ T1) - f ((1 / 2) ^ T2) = (T1-T2) f (1 / 2) 〉 0f

The function f (x) defined on the interval (0, positive infinity) satisfies that for any real number x.y, f (x ^ y) = YF (x)
If a > b > C > 1, and a, B, C are equal difference sequence, prove f (a) f (c)

Prove it first
If a > b > C > 1, and a, B, C form an arithmetic sequence, and f (a) f (c) 0, then a = B + D, C = b-d
Let a = B ^ P, C = B ^ Q,
From a > b > C > 1, we know that P and Q are positive numbers, and P! = Q
f(a)f(c) = f(b^p)f(b^q) = pq f(b)^2
Now prove that PQ 0
therefore
F (x) is an increasing function

The function f (x) is defined on the interval (0, + ∞), and for any x ∈ positive real number, y ∈ R, f (x ^ y) = YF (x)
(1) Find the value of F (1); (2) if f (1 / 2) < 0, find that f (x) is an increasing function on (0, + ∞)

Another x = 1, y = 0
f（1）=0f（1）=0
And x ^ y = t,
For the convenience of writing, you take logx # t as the logarithm of base X and t. then y = logx # t
If f (T) = logx # t * f (x), x = 1 / 2, then f (T) = log (1 / 2) # t * f (1 / 2)
The logarithm function of 1 / 2 base is a decreasing function, f (1 / 2) is a negative number, and f (T) is an increasing function

The function y = FX defined on the interval (0, + infinity) satisfies that for any positive number x, f (x ^ y) = YF (x), and f (2) 0, where a > 0 and a is not equal to 1

1, x = 2 ^ (log2 x) the same as y = 2 ^ (log2 y) all f (x) = f (2 ^ log2 x) = f (2) log2 X. if y is also changed into this form, then the answer to question 1 is solved. 2, let x > y, then log2 x > log2 y, all log2 x-log2 Y > 0, and because f (2) y, if f (x) - f (y) 0, then x must be > 0, all only require loga

Can you help me to solve a problem: the function f (x) defined on the interval (0, positive infinity) satisfies that any real number x, y has f (x ^ y) = YF (x)
Question 1, if a > b > C > 1, and a, B, C are equal ratio sequence, prove f (a) f (b)

(1) Let the common ratio of a, B and C be K
From the meaning of the title: F (a) f (b) = f (a) f (KA) = f (a) KF (a) = KF (a) ^ 2
Similarly, f (b) ^ 2 = f (b) f (b) = f (KA) f (KA) = k ^ 2F (a) ^ 2
Because a > b > C > 1, k > 1 means K ^ 2 > K
So f (a) f (b)

Why are two formulas equal? ∫ [b, a] f (x) DX * ∫ [b, a] 1 / F (y) dy = ∫ [b, a] f (x) / F (y) DXDY
D:a

The integral field of double integral is rectangular, so double integral can be transformed into the product of two definite integrals

Let f (x) satisfy AF (x) + B (1 / x) = C / x, where a, B and C are constants, and | a | ≠ | B |
Detailed process, thank you!

(1) AF (x) + B (1 / x) = C / X (1) AF (- x) - B (1 / x) = - C / X (2) (1) + (2) a (f (x) + F (- x)) = 0f (x) = - f (- x) f (x) is an odd function (2) AF (x) + B (1 / x) = C / XAF '(x) - B (1 / x ^ 2) = - C / x ^ 2F' (x) = (B-C) / (AX ^ 2) f '' (x) = - 3 (B-C) / (AX ^ 3)

Let f (x) be suitable for AF (x) + BF (1 / x) = C / X (a, B, C are all constants), and | a | = | B |, we prove that f (- x) = - f (x)

Is there something wrong with the question? It should be and | a ≠ | B | right
af(x)+bf(1/x)=c/x
a²f(x)+abf(1/x)=ac/x .(1)
af(1/x)+bf(x)=cx
abf(1/x)+b²f(x)=bcx .(2)
(1) Formula - (2) formula: (A & # 178; - B & # 178;) f (x) = AC / x-bcx (| a ≠| B |)
F (x) = C (A / x-bx) / (A & # 178; - B & # 178;) is the solution
It is not difficult to prove that f (- x) = - f (x) is an odd function

AF (x) + BF (1 / x) = C / x, / A / is not equal to / B / x, which belongs to the interval except 0. Try to prove that f (x) is an odd function

Because of the fact that (x) + BF (1 / x) = C / x, because (x) + BF (1 / x) = C / x = C / x, so the BF (x) + AF (1 / x) = CX is added to get (a + b) + (f (x) + F (1 / x)) = C (x + 1 / x) (a + B = 0 and (a + b) + ((a + b) + (f (- x) + F (- (- 1 / x) + F (- (- 1 / x)) = C (- x-x-x-1 / x) to add to (a + b) add (a + b) to add (a + b) + (f (f (x) + (f (x) + (f (f (x) + (f (f (x) + (f (x) + (f (f (x) + (f (f (x) + (f (f (x) + (f (f (x) + (f (x) + (f (x) + (f (f (x) + (f (f (x) + (f (x let g (x) = f (x) + F (- x) (x does not = 0), then G (x) + G (1 / x) = 0. Suppose g (x) does not = 0, then G (1 / x) does not = 0, so g (x) = - G (1 / x) In addition, if AF (x) + BF (1 / x) = C / x, the hypothesis is not tenable, so g (x) = 0, that is to say, f (x) + F (- x) = 0, so f (x) is an odd function

F (x) is a nonnegative differentiable function defined on (0, + ∞), and satisfies XF '(x) + F (x) ≤ 0. For any positive numbers a and B, if a

F (x) = f (x) / x, then f '(x) = [XF' (x) - f (x)] / x ^ 2 = [XF '(x) + F (x)] / x ^ 2-2f (x) / x ^ 2F (b) / b, equivalent to
bf(a)>af(b).
Your conclusion is right and can be used to prove it
af(b)

The function f (x) is defined on the interval (0, + ∞), and for any x ∈ positive real number, y ∈ R, f (x ^ y) = YF (x) 1) Find the value of F (1); (2) if f (1 / 2) < 0, find that f (x) is an increasing function on (0, + ∞)