Given that the function f (x) is an increasing function on R, a, B ∈ R, it is proved that if f (a) + F (b) > F (- a) + F (- b), then a + b > 0

Given that the function f (x) is an increasing function on R, a, B ∈ R, it is proved that if f (a) + F (b) > F (- a) + F (- b), then a + b > 0

Prove the inverse no proposition of the original proposition:
"If a + B ≤ 0, then f (a) + F (b) ≤ f (- a) + F (- b)" is true
Certificate: a + B ≤ 0 ⇒ a ≤ - B, B ≤ - A
⇒f（a）≤f（-b），f（b）≤f（-a）
⇒f（a）+f（b）≤f（-b）+f（-a）．
Therefore, the original proposition: if f (a) + F (b) > F (- a) + F (- b), then a + b > 0 is also true

When f (x) has f (a + b) = f (a) + F (b) for any AB and X is greater than 0, f (x) > 0 is proved to be an increasing function on R If the function f (x) defined as a non constant zero function on R holds f (a + b) = f (a) + (b) for any AB, and f (x) > 0 holds when x is greater than 0. 1 prove that f (x) is an increasing function on R. 2 if f (4) = 1 / 4, solve the inequality f (x-3) + F (5-3x) ≤ 1 / 2 about X. in a hurry, just solve the first question, and the second question is arbitrary

If X1 > X2, then x1-x2 > 0 and f (x1-x2) > 0
f(x1)=f(x1-x2+x2)=f(x1-x2)+f(x2)>f(x2)
So f (x) is an increasing function on R

Let FX be an increasing function on R, FX = fx-f (2-x), and prove that FX is an increasing function on R

F (x) increasing function
2-x is a subtractive function
therefore
F (2-x) is a subtractive function
Namely
-F (2-x) is an increasing function
therefore
FX = fx-f (2-x) is an increasing function

It is known that f (x) has f (AB) = f (a) + F (b) for any real number a and B 1) Find the values of F (1) and f (0) (2) If f (2) = P, f (3) = q (both P and Q are constants), find the value of F (36)

（1）
Let a = b = 1
f(1 × 1)=f(1)+f(1)
f(1)=f(1)+f(1)
So f (1) = 0
Let a = b = 0
f(0 × 0)=f(0)+f(0)
f(0)=f(0)+f(0)
So f (0) = 0
（2）
f(36)
=f(2 × 18)
=f(2)+f(18)
=p+f(2 × 9)
=p+f(2)+f(9)
=p+p+f(3 × 3)
=p+p+f(3)+f(3)
=p+p+q+q
=2(p+q)

It is known that the function f (x) holds for any real number a and B. f (AB) = f (a) + F (b). Proof: F (1 / x) = - f (x)

x=y=1
f(1)=2f(1),f(1)=0
f(x)+f(1/x)=f(1)=0
f(1/x)=-f(x)

It is known that the function y = f (x) has for any real number AB: F (a + b) = f (a) + F (b) - 1, and when x > 0, f (x) > 1, It is known that the function y = f (x) has for any real number AB: F (a + b) = f (a) + F (b) - 1, and when x > 0, f (x) > 1, (1) Verification: F (x) is an increasing function on R; (2) If f (4) = 5, find the value of F (2) and solve the inequality f (3m2-m-2) < 3. Find the detailed analysis of F (x2) = f [(x2-x1) + X1] = f (x2-x1) + F (x1) - 1 in the first problem and the detailed analysis of the second problem!

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