It is known that f (x) is a function defined on R which is not always zero, and for any a, B belongs to R, f (AB) = AF (b) + BF (a) is satisfied Find the value of 1, f (0), f (1) 2. Judge the parity of F (x) and prove it

It is known that f (x) is a function defined on R which is not always zero, and for any a, B belongs to R, f (AB) = AF (b) + BF (a) is satisfied Find the value of 1, f (0), f (1) 2. Judge the parity of F (x) and prove it

1: Let a = 0, B = 0, f (0) = 0
Let a = 1, B = 1. F (1) = 0
2: Let a = b = - 1, f (- 1) = 0
Let B = - 1, then f (- a) = AF (- 1) - f (a)
F (- a) = - f (a) odd

It is known that y = f (x) is a function defined on R which is not always zero, and for any a, B ∈ R, it satisfies: F (a · b) = AF (b) + BF (a); (1) find the value of F (1); (2) judge the parity of y = f (x), and prove your conclusion

(1) Let a = b = 1, we can get f (1) = f (1) + F (1), let a = b = 1, y = f (x) is an odd function. Let a = b = - 1, we can get f (1) = - f (- 1) - f (- 1), so f (- 1) = 0; let a = x, B = - 1, so f (- x) = x & nbsp; f (- 1) - f (x) = - f (x); let y = f (x) be an odd function

It is known that f (x) is a function defined on R which is not always zero, and for any a, the value B ∈ r of F (0) satisfies: F (a * b) = AF (b) + BF (a)
It is known that f (x) is a function defined on R which is not always zero, and for any a, B ∈ R, f (a * b) = AF (b) + BF (a)
Finding the value of F (0)

f(a*a) = 2a * f(a) ,
f[(-a)*(-a)] = -2a * f(-a),
f(a^2) = 2a * f(a) = -2a * f(-a)
2A * [f (a) + F (- a)] = 0, a arbitrary
f(a) + f(-a) = 0,
It is shown that f (x) is an odd function, if defined in x = 0, then
f(0) = 0

It is known that f (x) is a function defined on R which is not always zero, and for any a, B ∈ R, f (a * b) = AF (b) + BF (a)
Find the value of F (0), f (1)

Idea: the abstract function can be assigned
Let a = 0, B = 0, then f (0) = 0
Let a = b = 1, then f (1) = f (1) + F (1), so f (1) = 0

It is known that f (x) is a function defined on R which is not always zero, and for any a, B belongs to R, it satisfies the following conditions: F (AB) = AF (b) + BF (a) to find the value of F (1);
Let a = b = 1 and substitute it to get the solution: (1) Let f (1) = 1 * f (1) + 1 * f (1), then f (1) = 0
Why is f (1) = 0 instead of F (1) = 2?
I don't know about this
Who can help me

f(1)=1*f(1)+1*f(1)
So f (1) = f (1) + F (1) = 2 * f (1)
So 2F (1) - f (1) = 0
So f (1) = 0

If f (x) satisfies AF (x) + BF (1 / x) = CX, if f (x), A.B.C is not equal to 0, and the square of A-B is not equal to 0

af(x)+bf(1/x)=cx (1)
Let x = 1 / X
af(1/x)+bf(x)=c/x (2)
(1)×a-(2)×b
(a²-b²)f(x)=acx-bc/x
f(x)=(acx-bc/x)/(a²-b²)

Let f (x) be differentiable at x = 0, f (0) not equal to 0, the derivative of F (x) not equal to 0, AF (H) + BF (H) - f (0) = O (H), (H tends to 0), and find ab

Let g (x) = AF (x) + BF (x) - f (0), then G (0) = 0, a + B = 1
LIM (x - > 0) g (x) / x = g '(x) = 0 also gives a + B = 1
If you want to continue, you have to change one of AF (H) and BF (H), such as BF (2H)
Such a new problem, G (0) = f (0) (a + B-1) = 0, a + B = 1
And G '(x) = a'f (x) + 2F' (2x), G '(0) = (a + 2b) f' (x) = 0, a + 2B = 0
b=-1 a=2 ab=-2

1. AF (x) + BF (- x) = CX (| a | is not equal to | B |) 2. F (0) = 1 and for any real number x, y has f (Y-X) = f (y) - x (2y-x + 1)

(1) Let x = u, then: AF (U) + BF (- U) = Cu, let x = - u, then: AF (- U) + BF (U) = - Cu simultaneous equations, have: (a ^ 2-B ^ 2) f (U) = C (a + b) u should be: | a | not equal to | B |, so: F (U) = Cu / (a-b) so: F (x) = CX / (a-b) (2) let x = y, have: F (0) = f (x) - x (x + 1), so: F (x) = x ^ 2 + 2

Let f (x) = SiNx + √ 3cosx + 1, if the real numbers a, B, C make AF (x) + BF (x-C) = 1 hold for any x ∈ R, find the values of a and B

a　=　b　=　1/2　?

For any x1, X2 ∈ [0, + ∞) (x1 ≠ x2), there is f (x2) - f (x1) / x2-x1

When it is greater than 0, the function is a decreasing function, so f (3)