It is known that the domain of function f (x) is R. for any real number a and B, f (a + b) = f (a) - f (b), find f (x) parity It is known that the domain of function f (x) is R. for any real number a and B, f (a + b) = f (a) - f (b), find f (x) parity

It is known that the domain of function f (x) is R. for any real number a and B, f (a + b) = f (a) - f (b), find f (x) parity It is known that the domain of function f (x) is R. for any real number a and B, f (a + b) = f (a) - f (b), find f (x) parity

The solution consists of F (a + b) = f (a) - f (b)
Let a = b = 0
That is, f (0 + 0) = f (0) - f (0)
That is, f (0) = 0
Then go to a = x, B = - X
Then f (a + b) = f (a) - f (b)
Becomes f (x + (- x)) = f (x) - f (- x)
That is, f (x) - f (- x) = f (0)
That is, f (- x) = f (x)
That is, f (x) is an even function

The known function f (x) is a monotonically increasing function defined on (0, + ∞), which satisfies f (XY) = f (x) + F (y), f (3) = 1 (1) Find f (1), f (1) 3) Value of; (2) If f (x) + F (X-8) ≤ 2, find the value range of X

(1) Let x = y = 1, then: F (1 • 1) = f (1) + F (1), ‡ f (1) = 0; Let y = 1X, then f (x • 1x) = f (x) + F (1x) = f (1) = 0, ∵ f (3) = 1, ∵ f (13) = - f (3) = - 1; （2）∵f（9）=f（3）+f（3）=2，∴f（x）+f（x-8）≤2⇔f[x（x-8）]≤f...

Let f (x) be a monotonically increasing function on the domain (0, positive infinity), and f (XY) = f (x) + F for any x, y in the domain All have f (XY) = f (x) + F (y), f (2) = 1, find the making inequality f (x) + 2

Remember to adopt it first^^
f(3-x)
≥f(x)+2
=f(x)+1+1
=f(x)+f(2)+f(2)
=f(2x)+f(2)
=f(4x)
That is, f (3-x) ≥ f (4x)
Because monotone increasing function
‡ 3-x ≥ 4x, i.e. x ≤ 3 / 5
∵ 3-x > 0, x > 0
∴0＜x＜3
In conclusion, 0 ＜ x ≤ 3 / 5

Let f (x) be a monotonically increasing function defined in (0, + ∞), and for any x, y in the definition domain, f (XY) = f (x) f (y), f (2) = 1, find the value range of X that makes the inequality f (x) + F (x-3) ≤ 2

It should be this. F (XY) = f (x) + F (y)
f(xy)=f(x)+f(y)
f(4)=f(2 × 2)=f(2)+f(2)=1+1=2
f(x)+f(x-3)≤2
f(x (x-3))≤2＝f(4)
And f (x) is a monotonically increasing function on the definition (0, + ∞)
x>0
And x-3 > 0
And 0

Let the domain of function f (x) be a monotone function on a positive real number, and satisfy f (XY) = f (x) + F (y), if f (1 / 3) = 1, f (1) = 0 If f (x) = - 1, find the value of X

Let x = 3, y = 1 / 3
Then f (XY) = f (x) + F (y), f (1) = f (3) + F (1 / 3) = 0
And f (1 / 3) = 1
So f (3) = - 1
Then the value of X is 3

It is known that the function f (x) has f (XY) = f (x) f (y) for any real number x, y, and f (- 1) = 1, f (27) = 9. When 0 ≤ x < 1, 0 ≤ f (x) < 1. 1. Find the values of F (0) and f (3). 2. Judge the parity of F (x). 3. Judge the monotonicity of F (x) on [0, + ∞), and give a proof. 4. If a ≥ 0 and f (a + 1) ≤ ³ √ 9, find the value range of A

(1) Let x = 0 y = 27 f (0) = 9F (0) f (0) = 0f (9) = f (3 * 3) = f (3) ^ 2F (27) = f (3 * 9) = f (3) ^ 3 = 9 F (3) = 9 ^ 1 / 3 (2) let y = - 1 f (- x) = f (x) f (- 1) = f (x) is an even function (3) f (1) = f (- 1) = 1 Let y = 1 / X f (1) = f (x) f (1 / x) = 1 f (1 / x) = 1 f (1 / x) = 1 / F (x) let x2 > X1 > 0