The function f (x) = x (square) + 1 / 2mx-m (square) + 1 (X &; R) (1) is known when m = 1 The known function f (x) = 2mx-m (square) + 1 / 2 x (square) + 1 (X &; R) (1) When m = 1, find the tangent equation of curve y = f (x) at point (2, f (2))! (2) When m > 0, find the monotone interval and extremum of function f (x)!

The function f (x) = x (square) + 1 / 2mx-m (square) + 1 (X &; R) (1) is known when m = 1 The known function f (x) = 2mx-m (square) + 1 / 2 x (square) + 1 (X &; R) (1) When m = 1, find the tangent equation of curve y = f (x) at point (2, f (2))! (2) When m > 0, find the monotone interval and extremum of function f (x)!

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Given the function f (x) = | x-a | + 4 / X (a belongs to R) (1) if a = 0, find the solution set of the inequality f (x) ≥ 0 (2). When the equation f (x) = 2 has exactly two real roots, find the value of A
(3) If x belongs to (0, positive infinity) and the inequality f (x) ≥ 1 holds, the value range of a is obtained

(1) F (x) ≥ 0, namely) | x | + 4 / X ≥ 0
When x > 0, it becomes x ^ 2 ≥ - 4, and the solution set is all positive real numbers;
When x

Given the square-x of the function f (x + 1) = 2x, find the analytic expression of F (x)

Let a = x + 1
x=a-1
So f (a) = 2 (A-1) & #178; - (A-1) = 2A & #178; - 5A + 3
So f (x) = 2x & # 178; - 5x + 3

Given the function f (2x) = the square of X + X, find the analytic expressions of F (x) and f (x + 1)

Let 2x = t
Then, x = 1 / 2T
F (T) = (1 / 2t) square + 1 / 2T = 1 / 4T ^ 2 + 1 / 2T
So, f (x) = (1 / 2x) ^ 2 + 1 / 2x = 1 / 4x ^ 2 + 1 / 2x
f(x+1)=1/4(x+1)^2+1/2(x+1)=1/4x^2+x+3/4

Given the function f (2x + 1) = x square + 1, find f (x)

We can use the substitution method here
Let 2x + 1 = t
Then x = (t-1) / 2
The original equation can be reduced to
f(t)=(t-1)/2+1=2t-1
Here, the values of X and t should be in the real number field
So f (x) = 2x-1

Given that the function f (x) is an odd function on R, if f (x + 4) = f (x), X belongs to the square of (0,2), f (x) = 2x, then f (7) =?

f(7)=f(3+4)=f(3)
f(3)=f(-1+4)=f(-1)
Because it's an odd function
f（x）=-f(-x)
f(-1)=-f(1)=-2

Given the function f (x) = - 2x square + (a + 3) x + 1-2a, G (x) = x (1-2x) + A, where a belongs to R
For any a belonging to [- 3, + ∞), the image of function f (x) is always in the value range of real number x above the image of function g (x).

f(x)-g(x)=(a+2)x+1-3a > 0
(a+2)x > 3a-1
When a = - 2, 0 > - 7, for any x, X ∈ (- ∞, + ∞)
When a > - 2, x > (3a-1) / (a + 2), X ∈ ((3a-1) / (a + 2), + ∞)
When a

Given the function f (x + 1) + F (x) = 2x square - 2x-3, find the analytic expression of FX

It is known that f (x) is a quadratic function. Let f (x) = ax ^ 2 + BX + C, then f (x + 1) + F (x) = a (x + 1) ^ 2 + B (x + 1) + C + ax ^ 2 + BX + C = 2aX ^ 2 + 2 (a + b) x + A + B + 2c, that is, 2aX ^ 2 + 2 (a + b) x + A + B + 2C = 2x ^ 2-2x-3. Thus, 2A = 2,2 (a + b) = - 2, a + B + 2C = - 3, and a = 1, B = - 2, C = - 1

Given that the minimum value of the function f (x) = the square of X + x = A-1 in the interval [0,1] is 0, then the value of a is 0

1
The axis of symmetry is x = - 1 / 2. So the function increases on [0,1], and the minimum value is x = 0, so the minimum value is 1-a.1-a = 0, a = 1

Find the minimum value of the square of the function f (x) = (x-1) in the interval [a, A-1]

F (x) = (x-1) ^ 2; the axis of symmetry is x = 1; [A-1, a]
1. When a = 1, i.e. a > = 2, the minimum value is f (A-1) = (A-2) ^ 2;
3. When A-1