If f (x) = 2x + 5, then f (2x)= f(X+3)= f(2x+1)= f（-x）= F (x + 1) f（x-4）=

If f (x) = 2x + 5, then f (2x)= f(X+3)= f(2x+1)= f（-x）= F (x + 1) f（x-4）=

Hello: F (2x) = 4x + 5, f (x + 3) = 2 (x + 3) + 5 = 2x + 11F (2x + 1) = 2 (2x + 1) + 5 = 4x + 7F (- x) = - 2x + 5F (x + 1) = 2 (X & # 178; + 1) + 5 = 2x & # 178; + 7F (x-4) = 2 (x-4) + 5 = 2x-3. If you don't understand this question, you can ask. If you are satisfied, please click "adopt as satisfied" in the lower right corner

Given the function f (x) = (x ^ 2 + 2x + a) / x, X ∈ [1, + ∞). (1) when a = 0.5, find the minimum value of function f (x) (2) if
The known function f (x) = (x ^ 2 + 2x + a) / x, X ∈ [1, + ∞)
(1) When a = 0.5, find the minimum value of function f (x)
(2) If f (x) > 0 holds for any x ∈ [1, + ∞), try to find the value range of real number a

Let y = f (x) = (x ^ 2 + 2x + a) / x, X ∈ [1, + ∞) so there is YX = x ^ 2 + 2x + A, sort out x ^ 2 + (2-y) x + a = 0, and treat x ^ 2 + (2-y) x + a = 0 as a quadratic equation of one variable with respect to X. to make the equation solvable, the discriminant (2-y) ^ 2-4a is greater than or equal to 0, and the solution X1 = [(Y-2) + radical [(2-y) ^ 2-4a] / 2, X2 = [(Y-2

Given the function f (x) = loga (2x-1) + m ^ 2 + 1 (a > 0, a ≠ 1), then n = (), M = ()

n=1 m=±2
Because a is uncertain, it must be 2x-1 = 1 to pass the fixed point, so n = 1
In this case, M = ± 2

It is known that the quadratic function f (x) satisfies f (0) = 1, f (x + 1) - f (x) = 2x. In the interval [- 1,1], the image of y = f (x) is always above y = MX + m, and the value range of m is obtained

F (x) = ax ^ 2 + BX + CF (0) = C = 1F (x) = ax ^ 2 + BX + 1F (x + 1) = a (x + 1) ^ 2 + B (x + 1) + 1 = a (x ^ 2 + 2x + 1) + BX + B + 1 = ax ^ 2 + (2a + b) x + A + B + 1F (x + 1) - f (x) = ax ^ 2 + (2a + b) x + A + B + 1-ax ^ 2-bx-1 = 2aX + A + B = 2x, then a = 1 and a + B = 0b = - 1F (x) = x ^ 2-x + 1, the symmetry axis is 1 / 2 straight line y = m (x + 1)

If f (x) = m + loga (x-3), then the maximum value of G (x) = m ^ (x + 2) / m ^ (2x) + 4 is

A:
If f (x) = m + loga (x-3), the image constant crossing point (4,2): F (x) = m + loga (x-3)
2=m+loga(4-3)
m=2
g(x)= m^(x+2) / [m^(2x)+4]
=2^(x+2)/[2^(2x)+4]
=4/(2^x+4/2^x)
=

On the image of function f (x) = (x2 + 2x-1) / X-1
On what symmetry is the image of function f (x) = (x2 + 2x-1) / X-1

The image of F (x) = (x ^ 2 + 2x-1) / X-1 = X-1 / x + 1 is symmetric with respect to point (0,1),
This is because f (- x) = - x + 1 / x + 1 = 2-F (x)