It is known that the function FX is meaningful for x > 0 and satisfies F2 = 1, fxy = FX + FY, and FX is an increasing function. If FX + F (X-2) = 2 is true, the value range of X is

It is known that the function FX is meaningful for x > 0 and satisfies F2 = 1, fxy = FX + FY, and FX is an increasing function. If FX + F (X-2) = 2 is true, the value range of X is

f(2)=1
f(4)=f(2 × 2)=f(2)+f(2)=2
f(x)+f(x-2)=f[x(x-2)]
F (x) is an increasing function, so the original inequality is equivalent to
x>0
x-2>0
x(x-2)≥4
The solution of the inequality system is obtained
x≥1+√5

An increasing function defined on (0, + infinity) satisfies f (x / y) = f (x) - f (y). If f (3) = 1, solve the inequality f (x + 5) < 2

f(x/y)=f(x)-f(y)
f(3)=f(9/3)=f(9)-f(3)
f(9)=2f(3)=2
Increasing function defined on (0, + infinity)
f(x+5)<2=f(9)
x+5<9
x<4

F (x) is an increasing function defined on (0, + ∞), then the solution set of inequality f (x) > F [8 (X-2)] is () A. （0，+∞） B. （0，2） C. （2，+∞） D. （2，16 7）

From the fact that f (x) is an increasing function defined on (0, + ∞),
x＞0
8(x−2)＞0
x＞8(x−2) ⇒2＜x＜16
7，
Therefore, D

Let f (x) be an increasing function defined on (0, positive infinity), and f (x / y) = f (x) - f (y), solve the inequality f (X-5) - f (1 / x + 1) < = f (7)

∵f(x/y)=f(x)-f(y)
∴f(x-5)-f(1/x+1)=f [(x-5)(x+1)]≤f(7)
∵ f (x) is an increasing function defined on (0, positive infinity)
∴(x-5)(x+1)≤7
Get (X-6) (x + 2) ≤ 0, X belongs to 0 to positive infinity
0＜x≤6
x-5≥0 x≥5
1/x+1≥0 x＞-1
To sum up, 5 ≤ x ≤ 6

If f (x + y) = f (x) + F (y) holds for any real number x and y, and f (x) is not equal to zero, judge the parity of function f (x) Come on! I only have two hours. Speed! What about the process?

If x = 0 and y = 0, there is
f(0)=f(0)+f(0)
f(0)=2f(0)
f(0)=0
Let x = x, y = - X
be
f(x-x)=f(x)+f(-x)
f(0)=f(x)+f(-x)=0
f(-x)=-f(x)
Because f (x) is not equal to zero
therefore
F (x) is an odd function

To judge whether a function has parity, do you have to judge the definition field of the function first. What if the definition field is all real numbers r? Isn't odd function symmetric about origin? Is even function symmetric about y axis? Do you

Firstly, we should consider whether the definition field of a function is symmetrical about the origin, that is, the definition field is either R or (- K, + k), where k is greater than 0. If the definition field is not symmetrical about the origin, the basis for discussing the parity of a function will be lost
In addition, odd functions are symmetric about the origin and even functions are symmetric about the y-axis, which is only the image of the function, not the domain
I hope you can read more carefully