I just did a question: Given any function f (x), for any x, y, f (x + y) = f (x) + 2Y (x + y). F (1) = 1, find the analytic expression of F (x) The answer is this Let x = 0, y = 1 Then f (0) = - 1 Re order x=0,y=x Then f (x) = f (0) + 2x (0 + x) So f (x) = 2x ^ 2-1 Look at my assignment again Let x = 1, y = - 1, f(0)=f(1)+0. Why is f (0) = 1 now? Please point out my mistakes or assignment requirements

I just did a question: Given any function f (x), for any x, y, f (x + y) = f (x) + 2Y (x + y). F (1) = 1, find the analytic expression of F (x) The answer is this Let x = 0, y = 1 Then f (0) = - 1 Re order x=0,y=x Then f (x) = f (0) + 2x (0 + x) So f (x) = 2x ^ 2-1 Look at my assignment again Let x = 1, y = - 1, f(0)=f(1)+0. Why is f (0) = 1 now? Please point out my mistakes or assignment requirements

In fact, there is something wrong with the title
F (x + y) = f (x) + 2Y (x + y). Let y = 1-x, then f (1) = f (x) + 2Y, that is, f (x) = 1-2y = 1-2 (1-x) = 2x-1
Substituting into the original formula 2 (x + y) - 1 = 2x-1 + 2Y (x + y), we get 2Y (x + Y-1) = 0, that is to say, this formula holds only when x + y = 1 or y = 0, not for any x, y

Given the function f (x) = X2 (the square of x) + 1 and G (x) = f [f (x)], G (x) = g (x) - in (x), is there a real number in such that G (x) is a decreasing function on (- infinite, - 1] and (- 1,0) is an increasing function on (- 1,0)

F (x) = x & # 178; + 1, G (x) = f [f (x)] = f (X & # 178; + 1) = x ^ 4 + 2x & # 178; + 2, ‖ g (x) = x ^ 4 + 2x & # 178; + 2 - λ X & # 178; - λ = x ^ 4 + (2 - λ) x & # 178; + 2 - λ G '(x) = 4x & # 179; + (2 - λ) x from the meaning of the question, G' (- 1) = 0, that is - 4 - (2 - λ) = 0, the solution is λ = 6

The domain of F (2x-1) is to find the domain of F (x)
1.1≤X≤2
2.1≤2x-1≤3
3. Then the definition field is [1,3]
Help me write some ideas for each step

You are right to do this. Please remember that the domain of a function must be for X, just like f (2x – 1) whose domain is [1,2] means that its X is between 1 and 2. If 2x – 1 is regarded as an unknown quantity, then its domain is [1,3]. You can also understand it this way. F (2x – 1) = f (2 (x – 1,2) is equivalent to moving the original f (x) to the right by 1 or 2 units, After that, the length of the domain is expanded by two times. So see 0.5 on the left and add 0.5 on the right, so it is [1,3]

Abstract function domain,
Beginners = =, can't understand
Given that the domain of F (x) is [- 1,1], find the domain of F (1-3x) + F (2-2x)
Given that the domain of F (2x-1) is [0,1], find the domain of F (1-3x) + F (2-2x)

(1) That is, if * - 1 ≤ 1-3x ≤ 1 and - 1 ≤ 2-2x ≤ 1, 1 / 2 ≤ x ≤ 2 / 3
(2) 0≤2x-1x≤1 1/2≤x≤1
If 1 / 2 ≤ 1-3x ≤ 1 and 1 / 2 ≤ 2-2x ≤ 1, X has no solution, so the domain is empty
If you don't know, you can ask again