Given that the function f (x) is equal to {x2-a, X ≥ 0, 2x + 3, x < 0}, if the image of the function f (x) passes through the point (1, - 1) to find The second question about the value of F (f (0)) if the equation f (x) = 4 has a solution, find the value range of A

Given that the function f (x) is equal to {x2-a, X ≥ 0, 2x + 3, x < 0}, if the image of the function f (x) passes through the point (1, - 1) to find The second question about the value of F (f (0)) if the equation f (x) = 4 has a solution, find the value range of A

F (1) = 1 ^ 2-A = 1-A = - 1, so a = 2; & nbsp; FF (0) = f (- 2) = 2 * (- 2) + 3 = - 1;
When X & lt; 0, f (x) & lt; 3, so the solution of F (x) = 4 is on X & gt; = 0. X ^ 2-A = 4 has a solution on X & gt; = 0, that is, the minimum value of x ^ 2-A & lt; = 4, that is: - A & lt; = 4, so a & gt; = - 4

Let's give some examples of functions that satisfy "for any real number a and B in the domain of definition, f (a * b) = f (a) + F (b)", and show the common properties of these functions
Please give me a look, remember to write the nature~
Can you write more?

The common property of these functions is "transforming multiplication operation into addition operation". The slide rule used in the past was invented and designed based on this principle. There are many examples, but they are all related to logarithmic function. Because the logarithmic function f (x) = log a X has this property:
f(xy) = log a （xy） = log a x + log a y = f(x) + f(y)
Examples are as follows:
1. y(X) = log a X
2. y(X) = lg X
3. y(X) = ln X
4. f(X) = k log a X f(xy) = k log a (xy) = k log a (x) + k log a (y) = f(x) + f(y)
The above x, y, a are all greater than 0, K ≠ 0
If we take different a > 0 and different K ≠ 0, we can give many examples

Try to give some examples of functions satisfying "for any real number a and B in the domain, f (a + b) = f (a) times f (b)"
I want examples! Examples! Lots of examples. Sweat. Just two or three
Ask for advice from experts

f(x)=0
f(x)=1
f(x)=a^x a>0,a≠1.

The basic properties of the function f (x) = ax ^ 2 + B / x ^ 2 (a, B are normal numbers) are pointed out and explained
If the title is the first sub topic
(2) When a = 1 / 4, B = 4, draw the function diagram
The second question doesn't need to be written right away

F (x) = ax ^ 2 + B / x ^ 2 (a, B are normal numbers) 1, the definition field of function is x is not equal to 0.2, when x is not equal to 0, f (x) > 0.3, when x is not equal to 0, f (x) is even function. 4, the minimum value of F (x) is 2 (AB) ^ (1 / 2). (AX ^ 2 + B / x ^ 2) ^ 2 > = 4 (AX ^ 2) (B / x ^ 2) = 4AB, f (x) > = 2 (AB)

For example, in the same function, what is the relationship between y = f (x) and y = F-1 (x) images
Such as the title

The image of primitive function and inverse function is symmetric with respect to the line y = X
For example, logarithmic function and exponential function with the same base

The maximum value of function f (x) = XX + 1 is______ ．

According to the meaning of the question, if x ≥ 0, then f (x) = XX + 1 = 1 x + 1 x and X + 1 x ≥ & nbsp; 2, then f (x) ≤ 12, so the answer is 12

Give an example to illustrate the meaning of the function y = f (x), f (2), f (a), f (a square)

For example, y = f (x) = 3x + 2;
f(2)=3*2+2=8;
f(a)=3*a+2
F (a square) = 3 * a * a + 2

For any AB belonging to R, the function f (x) has f (a + b) = f (a) + F (b) - 1 and f (x) > 1 when x > 0
1. Prove that f (x) is an increasing function on R
2, if f (4) = 5, solve the inequality f (3m & # 178; - 7)

1 ∵ f (a + b) = f (a) + F (b) - 1 Let X10 ∵ f (x2) = f (x2-x1 + x1) = f (x2-x1) + F (x1) - 1 ∵ f (x2) - f (x1) = f (x2-x1) - 1 ∵ x > 0, f (x) > 1 ∵ f (x2-x1) > 1 ∵ f (x2-x1) - 1 > 0 ∵ f (x2) - f (x1) > 0 ∵ f (x2) > F (x1) ∵ f (x) is an increasing function 2 ∵ f (4) = 5F (4) = f (2) + F (2

For any AB belonging to R, the function f (x) has f (a + b) = f (a) * f (b) and if x1
(1) F (x) > 0
(2) F (x) minus function

1. Substituting a = - 1 / 2 and B = 0, we get the following results
f(－1/2)=f(0)f(－1/2)
Because x1, then:
f(－1/2)>0
So f (0) = 1
2. Substituting a = x, B = - x, where x > 0, then:
f(0)=f(x)f(－x)
That is: F (x) f (- x) = 1
Because x1, then:
F (- x) > 1, thus: 00
Thus, for all x ∈ R, f (x) > 0
3. Let: x 10, f (x 1-x 2) > 1
Then:
f(x1)－f(x2)>0
f(x1)>f(x2)
So the function f (x) is a decreasing function of R

FX is an increasing function on R, FX = fx-f [2-x]. It is proved that FX is an increasing function on R

gsd