For any real number x, there is always fxy = FX + FY, where XY is not equal to 0, the value of F1 and F4 is proved, and the value range of FX plus FX-3 is less than or equal to 2 need

For any real number x, there is always fxy = FX + FY, where XY is not equal to 0, the value of F1 and F4 is proved, and the value range of FX plus FX-3 is less than or equal to 2 need

Your problem is simple enough, f (1 × 1) = f (1) + F (1), i.e. f (1) = 0. I can't solve the following problem, because there is less value of F (2) and less condition of F (4) = 2F (2). I can't work out f (4)? I can get f (4) = 2 directly. The second problem f (x) + F (x-3) ≤ 2 can be changed according to the formula of the original problem
Let f (x) = LG (2 / (1 + x) + a) be an odd function, then the value range of X with F (x) ＜ 0?
F (- x) = - f (x), f (0) = 0, we can get a = - 1, and then we can find the range
The function f (x) defined on the set of non-zero real numbers satisfies f (XY) = f (x) + F (y), and f (x) is an increasing function on the interval (0, positive infinity)
1. Find the value of F (1), f (- 1) 2. Prove f (x) = f (x) 3. Solve the inequality f (2) + F (x-0.5)
1. Let x = 1, f (y) = f (1) + F (y), then f (1) = 0
Let x = y = - 1, f (1) = f (- 1) + F (- 1), f (- 1) = 0
2. We know that f (- 1) = 0, let y = - 1, f (- x) = f (x) + F (- 1), so f (x) = f (- x)
3.f(2)+f(x-0.5)=f(x+1.5)
F (x) = LG (2 / 1-x + a) is an odd function f (x)
Because of the odd function, f (0) = 0 LG (2 + a) = 0 2 + a = 1 a = - 1
f(x)=lg[2/(1-x)-1]
-1
The function f (x) defined on the set of non-zero real numbers satisfies f (XY) = f (x) + F (y), and f (x) is an increasing function on the interval (0, + ∞)
The inequality of solution about X: F (2) + F (x-1 / 2) ≤ 0
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Taking x = y = 1, we can get f (1) = f (1) + F (1), and the solution is f (1) = 0,
Taking x = y = - 1, we can get f (1) = 2F (- 1), so f (- 1) = 0,
Take y = - 1 to get f (- x) = f (x) + F (- 1) = f (x), so the function is even,
Therefore, from the fact that f (x) is an increasing function on (0, + ∞), f (x) is a decreasing function on (- ∞, 0),
So, from F (2) + F (x-1 / 2)
The function f (x) defined on R, for any x, y ∈ R, f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 1, it is proved that f (x) is an odd function
Let f (x) = f (TaNx), we prove that the equation f (x) = 0 has at least one real root; if the equation f (x) = 0 has n real roots on (- π / 2, π / 2), then n must be odd
Let y = 0
f(x)+f(x)=2f(x)f(0)
So f (x) = f (x) f (0)
f(x)[f(0)-1]=0
f(0)≠1
So only f (x) = 0
So f (- x) = 0 = - f (x)
The domain R is symmetric about the origin
So it's an odd function
Let x = y = 0, then f (0) + F (0) = 2F (0) f (0)
It is deduced that f (0) = 1 or F (0) = 0
∵ f (0) is not equal to 1
∴f(0)=0
Let x = 0, then f (0 + y) + F (0-y) = 2F (0) f (y)
It is deduced that f (y) + F (- y) = 0
If you replace y with X, you get
f(x)+f(-x)=0
∴f(x)=-f(-x)
ψ f (x) is an odd function
This is easy to do
Let X be equal to 0, y be equal to 0, then the original formula is 2F (0) = the square of 2F (0), and because f (0) is not equal to 1, then f (0) = 0
Let x = 0, the original formula becomes f (y) + F (- y) = 0, which shows that the function is odd
Let x = y = 0
Then f (0) = 0 or 1 (rounding off)
Let x = 0
Then f (y) + F (- y) = 0
That is, f (- y) = - f (y)
Given the function f (x) = cos ^ 2 (x + π / 12), G (x) = 1 + 1 / 2sin2x (1), let x = x0 be an axis of symmetry of the image of the function y = f (x), and find the value of G (x0);
Why is the axis of symmetry 2x0 + π / 6 = π + 2K π instead of 2x0 + π / 6 = π + K π
Find the result
F (x) = cos ^ 2 (x + π / 12) = 1 / 2 * cos (2x + π / 6) + 1 / 2x = x0 is a symmetry axis of the function y = f (x) image. The symmetry axis of y = cosx is: x = k π, (K ∈ z), so: 2x + π / 6 = k π, (K ∈ z). The solution is: x = k π / 2 - π / 12, (K ∈ z), substituting: G (x) = 1 + 1 / 2sin2x = 1 ± 1 / 4, so the result is: 5 / 4 (k is odd)
The axis of symmetry of cosine function is π, 3 π,. 2K π + π. The axis of symmetry of sine function is π / 2, 5 π / 2,. 2K π + π / 2 F (x) = cos ^ 2 (x + π / 12) = (1 / 2) cos (2x + π / 6). Question: is x = 2K π not the axis of symmetry of cosine function
Given that the function f (x), G (x) is defined on R, for any x, y belongs to R, f (X-Y) = f (x) g (y) - G (x) f (y) and f (1) is not equal to 0, finding f (x) is an odd function
If f (1) = f (2), find the value of G (1) + G (- 1)
2. Let f (x) = - | X-1 | + | X-2 |, if the inequality | a + B | + | A-B | > = | a | f (x) (a is not equal to 0, AB belongs to R) find the range of real number X
If f (x) = 1 / (2 ^ x-1) + A is an odd function, then a=
3. The minimum value of y = ((sin3x * sin ^ 3x + cos3x * cos ^ 3x) / cos ^ 2x) + sin2x
Thank you very much.)
You don't have to answer the first question
Let x = 0, y = 0. Then f (0) = f (0) g (0) - G (0) f (0) = 0 get f (0) = 0, let y = 0, x = 1. Then f (1) = f (1) g (0) - G (1) f (0) and f (1)! = 0 get g (0) = 1. Let x = 0 then f (- y) = f (0) g (y) - G (0) f (y). Substitute f (0) = 0, G (0) = 1 to get f (- y) = - f (y), and f (x) is an odd function
Given the function f (x) = Cos2 (x2 − π 12), G (x) = sin2x. Let x = x0 be an axis of symmetry of the image of the function y = f (x), then the value of G (x0) is equal to______ .
Let f (x) = 12 [1 + cos (x - π 6)]. Because x = x0 is a symmetry axis of the graph of function y = f (x), so x0 − π 6 = k π, that is, 2x0 = 2K π + π 3 (K ∈ z). So g (x0) = sin2x0 = sin (2k π + π 3) = 32
Given that the function y = f (x) (x is not equal to 0) is an odd function, f (3) = 1, and x > 0, the function f (x) = loga (x + 1), if x
3>0
f(3)=loga(3+1)=1=loga(4)=1
A=4
x> 0, f (x) = log4 (x + 1)
X0, satisfying the known expression
f(-x)=loga(-x+1)=loga(1-x)
Function is odd, f (x) = - f (- x)
f(x)=-f(-x)=-loga(1-x)
X
Because when f (3) = 1, x > 0, f (x) = loga (x + 1)
So a = 4
So when x > 0, f (x) = log4 (x + 1)
Because y = f (x) (x is not equal to 0) is an odd function
So f (x) = - f (- x)
So x