The function f (XY) = f (x) + F (y) of non-zero real number set, and f (x) is an increasing function on (0, + 00), the solution of inequality f (2) + F (x-1 / 2) is less than or equal to 0

The function f (XY) = f (x) + F (y) of non-zero real number set, and f (x) is an increasing function on (0, + 00), the solution of inequality f (2) + F (x-1 / 2) is less than or equal to 0

F (1) = f (1 * 1) = f (1) + F (1)
So f (1) = 0
Because f (x) is an increasing function on (0, + 00)
So when x ∈ (0,1], f (x) ≤ 0
When x ∈ (1, + ∞), f (x) > 0
f(2)+f(x-1/2) = f(2x - 1) ≤ 0
That is, 0 ＜ 2x - 1 ≤ 1
The solution is 1 / 2 < x ≤ 1
Known inequality about X (AX-5) (x ^ 2-A)
∵3∈M
Substituting x = 3, the inequality holds
∴(3a-5)/(9-a)0
==>a9 ①
∵ 5 does not belong to M
Substituting x = 5, the inequality does not hold
(5a-5) / (25-A) ≥ 0 or A-25 = 0
(A-1) / (A-25) ≤ 0 or A-25 = 0
==> 1≤a≤25 ②
The range of intersection number a
1≤a
From 3 to m, (3a-5) (9-A) 0 is obtained,
∴a9.①
5 does not belong to M
（5a-5)(25-a)>=0,
(a-1)(a-25)
Given that the two zeros of function f (x) = x2-ax-b are 2 and 3, then the zeros of function g (x) = bx2-ax-1 are ()
A. - 1 and - 2b. 1 and 2C. − 12 and − 13D. 12 and 13
The two zeros of ∵ function f (x) = x2-ax-b are 2 and 3 ∵ the two real roots of equation x2-ax-b = 0 are 2 and 3. According to Weida's theorem, the two zeros of 2 + 3 = a, 2 × 3 = - B, ∵ a = 5, B = - 6 ∵ G (x) = - 6x2-5x-1 ∵ - 6x2-5x-1 = 0 ⇔ (2x + 1) (3x + 1) = 0 ∵ g (x) = 0 are − 12 and − 13, that is to say, the zeros of function g (x) = bx2-ax-1 are − 12 and − 13
If f (x-1) = 3x ^ 2-2x + 1, then f (x + 1)=
RT
Consider X-1 as a term
f(X-1)=3x^2-2x+1=3(x-1)^2+4(x-1)+2
f(X+1)=3(x+1)^2+4(x+1)+2=3x^2+10x+9
f(X-1)=3x^2-2x+1
=3x^2-6x+3+4x+1
=3(x-1)^2+4x-4+4+1
=3(x-1)^2+4(x-1)+5
f(x)=3x^2+4x+5
f(x+1)=3(x+1)^2+4(x+1)+5
=3x^2+6x+3+4x+4+5
=3x^2+10x+12
Given the function f (x) = ax ^ 2 + BX + C, a + B + C = 0, a > b > C, it is proved that the function f (x) has two different zeros
Δ=b²-4ac=（-a-c）²-4ac=（a-c）²
Because a > C means a-c > 0
So Δ 0
So there are two different zeros
Questions are welcome
Obviously, a + B + C ＜ 3a, so a > 0, and f (1) = a + B + C = 0, so X1 = 1, X2 = C / A
If f (2x + 1) = 3x-2 and f (a) = 4, then the value of a is______ .
Let t = 2x + 1, x = t − 12, substitute f (2x + 1) = 3x-2, f (T) = 32t − 72, then f (x) = 32x − 72, then f (a) = 32a − 72 = 4, the solution is a = 5, so the answer is: 5
Given the quadratic function f (x) = ax & # 178; + BX (a ≠ 0, a, B are constants), f (2) = 0, and the function g (x) = f (x) - X has only one zero point, find the analytic expression of F (x)
There is only one origin in (B-1) x, + (B-1) x, so (B-1) &\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\chenchen 178;
If f (2x + 1) = 3x-2 and f (a) = 4, then the value of a is?
f(2x+1)=3x-2
have to
f(2x+1)=3/2(2x+1)-7/2
therefore
f(a)=3/2a-7/2
When f (a) = 3 / 2a-7 / 2 = 4
A = 5
Five
If f (2x + 1) = 3x-2 and f (a) = 4, then the value of a is?
3x-2=4 => x=2
2x+1=2 => x=1/2
If f (2x + 1) = 3x-2, then f (2x + 1) = 3x-2
f(x)=(3/2)x-7/2
∵f(a)=4
∴a=5
Let y = 2x + 1, then x = Y / 2 - 1
The original formula F (2x + 1) = 3x-2 = 2x + 1 + x-3
Substituting y, we get f (y) = y + (Y / 2 - 1) - 3 = 3Y / 2 - 4
So f (a) = 3A / 2 - 4 = 4,
The solution is a = 16 / 3
Let a = 2x + 1, then x = (A-1) / 2;
f(a)=3[(a-1)/2]-2=4;
3[(a-1)/2]=6;
(a-1)/2=2;
A=5
If the two zeros of the function FX = x & # 178; - ax-b are 2 and 3, then the zeros of the function GX = BX & # 178; - AX-1 are 2 and 3
It is known from the title that when x = 2 and 3, FX = x2-ax-b = 0, the values of a and B can be obtained by substituting 2 and 3 into the original formula, and then the values of a and B can be substituted into GX to find the zero point
Given that f (2x + 1) = 3x + 2, and f (a) = 2, find the value of A
f(2x+1)=3/2(2X+1)+1/2
So f (a) = 3 / 2A + 1 / 2 = 2
3/2a=3/2
A=1
I hope I can help you,