Let f (x) = ax & # 178; + BX + C (a > 0), the two roots X1 and X2 of the equation f (x) - x = 0 be full Connect 0

Let f (x) = ax & # 178; + BX + C (a > 0), the two roots X1 and X2 of the equation f (x) - x = 0 be full Connect 0

Let f (x) - x = a (x-x1) (x-x2), when x ∈ (0, x1), because 0 ＜ x1 ＜ x2 ＜ 1 / A, a ＞ 0, so a (x-x1) (x-x2) ＞ 0, that is, f (x) ＞ x holds, and x1-f (x) = x1-x-a (x-x1) (x-x2) = (x1-x) [1 + a (x-x2)] because 0 ＜ x ＜ x1 ＜ x2 ＜ 1 / A, so

When f (x) = x & # 178; - BX + C satisfies f (0) = 3, and the image is symmetrical about the straight line x = 1, x > 0, find the size relationship of F (b ^ x) and f (C ^ x)

If f (0) = 3 and the image is symmetric with respect to the line x = 1, then C = 3 and B = 2
When x is greater than or equal to 0, C ^ x > b ^ X and greater than or equal to 1
So f (b ^ x)

For example 4, it is known that the function y = f (x) is a periodic function defined on R, the period T = 5, and the function y = f (x) (- 1 ≤ x ≤ 1) is an odd function. It is also known that y = f (x) is a linear function on [0,1], a quadratic function on [1,4], and the minimum value of the function is - 5 when x = 2. ① prove that f (1) + F (4) = 0; ② find the analytic expression of y = f (x), X ∈ [1,4]; ③ find the minimum value of y = f (x) on [4,9] Analytical expression

① ∵ f (x) is a periodic function with period of 5 ∵ f (4) = f (4-5) = f (- 1) ∵ y = f (x) (- 1 ≤ x ≤ 1) is an odd function ∵ f (1) = - f (- 1) = - f (4) ∵ f (1) + F (4) = 0; From F (1) + (f (1) + (f (1) + F (4) = 0, we get the following (1-2) 2-5 + a (4-2) 2-5 + a (4-2) (2-2) 2 (X-2) 2-5 (1 ≤ x ≤ 4) (1 ≤ 5 (1 ≤ x ≤ 4) (1 ≤ 4) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; the following (1-2 (1) 2 (1 (1-2 (1-2 (1-2 (1-2) (1-2-2) 2 (1-2 (1-2-2-2-2) 2 (2 (2-2-2) (2) (2) (2) (2 (2) (2 (1) (2) (2 (1) (2) (2) (2 (1) (2 (1) (2) (2) (2) (2) (2) (2 (1) (2) (2) (2) (2) (2) (2) (2& nbsp; & nbsp; As a result, when the - 1 ≤ 1 ≤ x ≤ 1, f (x) = -f (x) = -f (- x) = -3x-3x & nbsp; & nbsp & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp & nbsp; & nbsp & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; so, when the - 1 ≤ x ≤ 1 ≤ x ≤ 1 ≤ 1, f (x (x) is the case of-1 ≤ 1 ≤ 1 ≤ x ≤ 1, f (f (x) when 4 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ x ≤ 1, f (1 ≤ 1 ≤ 1 ≤ 1 ≤ 1, f (1) when 4 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ 1 ≤ 1, f (1-1-1 ≤ 1 ≤ 1-1-1 ≤ 1, f (X-1-1 ≤ 1, f (X-5 ≤ 1, f (X-5 ≤ 1, f (X-5 ≤ 1, ≤ 62 (x − 7) 2 − 5 & nbsp;  6＜x≤9

Given the function f (x) = x2-2x, G (x) = ax + 2 (a > 0), if ∀ x1 ∈ [- 1,2], ∃ x2 ∈ [- 1,2], such that f (x1) = g (x2), then the value range of real number a is ()
A. (0，12]B. [12，3]C. （0，3]D. [3，+∞）

The image of ∵ function f (x) = x2-2x is a parabola with the opening upward, and the minimum value of F (x) is f (1) = - 1 and the maximum value is f (- 1) = 3 when ∵ x 1 ∈ [- 1,2] is symmetric with respect to the straight line x = 1. The range of F (x 1) is [- 1,3] and ∵ g (x) = ax + 2 (a > 0), x 2 ∈ [- 1,2], and ∵ g (x 2) is a monotone increasing function, and the range of G (x 2) is [g (- 1), G (2)], that is, G (x 2) ∈ [2-A, 2A + 2] ∀ x1 ∈ [- 1, 2], ∃ x2 ∈ [- 1, 2], such that f (x1) = g (x2), ∀ 2 − a ≤ − 12a + 2 ≥ 3 {a ≥ 3, so D is selected

It is known that f (x) is a function defined on R, and f (x + 2) = 1 + F (x) / 1-f (x). If f (1) = 2 + radical 3, then f (2010) =?

Because f (x + 2) = 1 + F (x) / 1-f (x)
F (x) is a periodic function, t = 4
Then f (1) = f (5) = =F (2010) = 2 + radical 3

Given that the function f (x) defined on R is full of F (3) = 2-radical 3, and f (x + 3) = 1 / - f (x) for any x, then f (2010)=
What I want to ask is, why is f (2010) = f (0)? And if so, why is f (2010) = - (2-radical 3) = - 2 + radical 3 - 2-radical 3?

f(x+6)
=f(x+3+3)
=1/-f(x+3)
=1/-[1/-f(x)]
=f(x)
f(2010)
=f(6*335)
=f(6)
=f(3+3)
=1/-f(3)
=1 / (radical 3-2)
=-Root 3-2

Let f (x) = x ^ 2 + 2x-3 (1). If the solution set of inequality f (x) > a about X is {x | x ≠ - 1}, try to find the value of real number a

F (x) = x ^ 2 + 2x-3, the axis of symmetry is x = - 1, so the minimum value of F (x) at the vertex is - 4
So if f (x) ≥ - 4, the equal sign holds, x = - 1
So when f (x) > A, X ≠ - 1
So the value of a is - 4

Let f (x) = | x + a | - | x-4 |, X belong to R. (1) when a = 1, solve the inequality f (x)

When a = 1, f (x) = | x + 1 | - | - x-4 | then f (x) < 2, that is, | x + 1 | - | - x-4 | < 2, is divided into three parts: ① when x ≤ - 1, the original inequality is - (x + 1) - [- (x-4)] < 2, and it is sorted out that: - 5 < 2 is tenable, so x ≤ - 1; ② when - 1 ＜ x ≤ 4, the original inequality is (x + 1) - [- (x-4)] < 2, and the whole principle is: 2x < 5

Inequality! The inequality | 2x + T | - 1 about X is known

|2x+t|-1

Solving inequality 4 ^ x > 2 ^ (3-2x)

4 ^ x = 2 ^ 2x y = 2 ^ x monotonically increasing 2x > 3 on R_ 2x x>3/4