Let FX be an increasing function defined on R +, f (x) = f (x / y) + F (y). If f (3) = 1, f (x) - f (1 / X-8) > 2, find the value of X Where > 2 has an equal sign, R + means 0 to positive infinity,

Let FX be an increasing function defined on R +, f (x) = f (x / y) + F (y). If f (3) = 1, f (x) - f (1 / X-8) > 2, find the value of X Where > 2 has an equal sign, R + means 0 to positive infinity,

When y = 1, f (x) = f (x) + F (1) f (1) = 0x = 1, f (1) = f (1 / y) + F (y) f (1 / y) + F (y) = 0f (3) = 1, we can get f (x) = log3 (x) f (1 / X-8) = - f (X-8) f (x) - f (1 / X-8) = f (x) + F (X-8) = log3 (x) + log3 (X-8) log3 (x ^ 2-8x) > 2x ^ 2-8x > 9 (X-9) (x + 1) > 0, because x > 0, so x > 9

Let f (x) be an increasing function defined on R +, and for any x > 0, Y > 0, fxy = FX + FY always holds
(1) When x > 1, f (x) > 0
(2) If f (3) = 1, solve the inequality f (x) > F (x-1) + 2

(1) It is proved that: let x = y = 1, ∧ f (1) = f (1 * 1) = f (1) + F (1) ∧ f (1) = 0 and f (x) be an increasing function defined on R +. If x > 1, f (x) > 0 (2). F (3) = 1 ∧ let x = y = 3, f (3) + F (3) = f (9) = 2. The inequality f (x) > F (x-1) + 2 is equivalent to f (x) > F (x-1) + F (...)

If the function FX is an increasing function on R, then FX = fx-f (- x) + 1 is monotonic on R?

Let x 1, x 2 ∈ R, and X 1-x 2,
Because the function FX is an increasing function on R, so
f(x1)f(-x2),f(-x2)

Given that the domain r satisfies f (x + 5) = - f (x) + 2 in function f (x) and f (x) = x when x ∈ (0,5), then the value of F (2008) is
How to simplify the step f (x + 5) = - f (x) + 2? And the answer should be - 1, the process will not!! = =|||

F (x + 5N) is discussed by F (x + 5) = - f (x) + 2, which can be divided into the following two cases
(1) N is odd, f (x + 5N) = - f (x) + 2
(2) N is even, f (x + 5N) = f (x)
Then f (2008) = f (3 + 5 * 401) = - f (3) + 2 = - 3 + 2 = - 1

The function f (x) defined on R satisfies f (x) = 0, f (x) + F (1-x) = 1, f (x / 5) = f (x) / 2, and when 0 ≤ a < B ≤ 1, f (a) ≤ f (b), then f (1 / 2008)=
Sorry, wrong number
The function f (x) defined on R satisfies f (0) = 0, f (x) + F (1-x) = 1, f (x / 5) = f (x) / 2, and when 0 ≤ a < B ≤ 1, f (a) ≤ f (b), then f (1 / 2008) =?

Let x = 0f (0) + F (1) = 1, so f (1) = 1F (1 / 5) = f (1) / 2 = 1 / 2F (1 / 25) = f (1 / 5) / 2 = 1 / 4, and so on, we can get f (1 / 3125) = 1 / 32, let x = 1 / 2F (1 / 2) + F (1 / 2) = 1F (1 / 2) = 1 / 2F (1 / 10) = f (1 / 2) / 2 = 1 / 4

It is known that f '(x) is the derivative of F (x) and is defined on R. for any x, there is 2F (x) + XF' (x) > x ^ 2. Try to prove that f (x) > 0

1 when x = 0, the original formula is; 2F (0) + 0 > 0 = = > F (0) > 02 when x > 0, [x & # 178; f (x)] '= 2xf (x) + X & # 178; f' (x) = x [2F (x) + XF '(x)] > X & # 179; > 0

Let the derivative of function f (x) on R be f '(x), and 2F (x) + XF' (x) > x2
A. f（x）＞0B. f（x）＜0C. f（x）＞xD. f（x）＜x

∵ 2F (x) + XF ′ (x) > X2, let x = 0, then f (x) > 0, so B and D can be excluded. If f (x) = x2 + 0.1, the known condition 2F (x) + XF ′ (x) > x2 holds, but f (x) > x may not hold, so C is also wrong, so a is chosen, so a is chosen

It is known that f (x) is a differentiable function defined on (0, + ∞) and satisfies XF ′ (x) - f (x) ≥ 0. For any positive number a, B, if a > b, there must be ()
A. af（a）≤bf（b）B. bf（b）≤af（a）C. af（b）≤bf（a）D. bf（a）≤af（b）

F (x) = f (x) x, we can get f '(x) = 1x2 [XF' (x) - f (x)], and from XF '(x) - f (x) ≥ 0, we discuss it in two cases: ① XF' (x) - F (x) > 0, so F '(x) > 0, that is, f (x) is an increasing function, that is, when a > b > 0, f (a) > F (b), (f (b) B < f (a) a, thus AF (b) < BF (a); ② XF' (x) - f (x) = 0, so f (x) is a constant function, where f (b) B = f (a) a, that is, AF (b) = BF (a); in general, AF (b) ≤ BF (a); therefore, C is selected;

It is known that f (x) is a differentiable function defined on (0, + ∞) and satisfies XF ′ (x) - f (x) ≥ 0. For any positive number a, B, if a > b, there must be ()
A. af（a）≤bf（b）B. bf（b）≤af（a）C. af（b）≤bf（a）D. bf（a）≤af（b）

F (x) = f (x) x, we can get f '(x) = 1x2 [XF' (x) - f (x)], and from XF '(x) - f (x) ≥ 0, we discuss it in two cases: ① XF' (x) - F (x) > 0, so F '(x) > 0, that is, f (x) is an increasing function, that is, when a > b > 0, f (a) > F (b), (f (b) B < f (a) a, thus AF (b) < BF (a); ② XF' (x) - f (x) = 0, so f (x) is a constant function, where f (b) B = f (a) a, that is, AF (b) = BF (a); in general, AF (b) ≤ BF (a); therefore, C is selected;

F & nbsp; (x) a nonnegative differentiable function defined on (0, + ∞) satisfying XF '(x) + F (x) ≤ 0. For any positive number a, B, if a < B, then there must be ()
A. af（b）≤bf（a）B. bf（a）≤af （b）C. af（a）≤bf （b）D. bf（b）≤af （a）

Let g (x) = XF (x), X ∈ (0, + ∞), then G ′ (x) = XF ′ (x) + F (x) ≤ 0, ∵ g (x) monotonically decreases in the interval x ∈ (0, + ∞). ∵ a < B, ∵ g (a) ≥ g (b), that is, AF (a) ≥ BF (b)