1. Given f (x) + 2F (1 / x) = 2x + 1, find the analytic expression of F (x) 2. F (x) = 2x + 3, g = (x) = 1 / (x ^ 2-2) (1) Find f (x ^ 2) (2) Find g (1 / x) (3) Find f [g (x)] (4)g[f(x)+2] It's better to have a specific solution process

1. Given f (x) + 2F (1 / x) = 2x + 1, find the analytic expression of F (x) 2. F (x) = 2x + 3, g = (x) = 1 / (x ^ 2-2) (1) Find f (x ^ 2) (2) Find g (1 / x) (3) Find f [g (x)] (4)g[f(x)+2] It's better to have a specific solution process

one
Using (1 / x) to take x, you have to
f(1/x)+2f(x)=2/x+1
The former is, the former is
2*-
have to
f(x)=-2x/3+4/3x+1/3
two
(1)
f(x^2)=2(x^2)+3
(2)
G (1 / x) = 1 / [(1 / x) ^ 2-2]
(3)
F [g (x)] = 2 [1 / (x ^ 2-2)] + 3
(4)
G [f (x) + 2] = 1 / [(2x + 5) ^ 2-2]

If the function f (x) is an even function on R, and the function f (x) decreases monotonically when [0, + ∞), then the solution set of the inequality f (1) − f (1x) < 0 is ()
A. {x | x ≠ 0} B. {x | - 1 ＜ x ＜ 1} C. {x | - 1 or X ＞ 1} D. {x | - 1 ＜ x ＜ 1 and X ≠ 0}

∵ f (1) − f (1x) < 0 ∵ f (1x) > F (1) ∵ f (x) is an even function, and when [0, + ∞), the function f (x) monotonically decreases ∵ 1x | 1 | x | 1 | x ＞ 1 or X ＜ - 1, so C is selected

F (x) linear function problem
f（1）= -1,f（2）=1
I don't know how to write the detailed process

Standard expression of linear function
f(x)=ax+b
f(1)=a+b=-1
f(2)=2a+b=1
a=2
B = - 3 so
The linear function is:
f(x)=2x-3

1： It is known that f (x) is an odd function, G (x) is an even function, and f (x) + G (x) = 1 / (2x + 1)
Finding the analytic expressions of F (x) and G (x)
2： Let f (x) be any function, and the domain of definition is symmetric about the origin. Proof: F (x) must be expressed as the sum of an odd function and an even function
3： It is known that f (x) is an odd function, the domain of definition is D, G (x) is an even function, and the domain of definition is d,
Let f (x) = f (x) g (x), judge the parity of F (x)
② If f (x) = f (x) and G (x) are even functions, the sum of F (x) and G (x) is studied
The parity of G (x)

1. F (- x) = - f (x) g (- x) = g (x) let H (x) = f (x) + G (x) = 1 / (2x + 1) (1) H (- x) = f (- x) + G (- x) = - f (x) + G (x) = 1 / (- 2x + 1) (2) (1) + (2) 2g (x) = 1 / (2x + 1) + 1 / (- 2x + 1) = 2 / (1-4x & sup2;) g (x) = 1 / (1-4x & sup2;) f (x) = H (x) - G (x) = - 2x / (1-4x & sup2;) 2, let H (x) =

Given that the function f (x) = Xax + B (a, B are constants, and a ≠ 0) satisfies that f (2) = 1, f (x) = x has a unique solution, the analytic expression of function f (x) is obtained______ ，f[f（-3）]=______ ．

F (x) = Xax + B = x, we get AX2 + (B-1) x = 0, and have a unique solution △ = (B-1) 2 = 0, ① f (2) = 22a + B = 1, ② the simultaneous equations get a = 12, B = 1 ｜ f (x) = 2XX + 2F (- 3) = 6, ｜ f [f (- 3)] = f (6) = 32, so the answer is f (x) = 2XX + 2, 32

For any m, n belonging to R, the function F X has f (M + n) = f (m) + F (n) - 1, and when x > 0, f (x) > 1
(1) If f (3) = 4, solve the inequality f (a ^ 2 + a-5)

I don't think so. This can only be inferred according to the nature of function
Firstly, monotonicity is proved
Let X1 > X2, then: x1-x2 > 0, then f (x1-x2) > 1
So: F (x1) = f (x2 + x1-x2) = f (x2) + F (x1-x2) - 1 > F (x2)
So the function f (x) is monotonically increasing
Then find the value of the independent variable when the function value is 2
F (3) = f (2) + F (1) - 1 = [f (1) + F (1) - 1] + F (1) - 1 = 3f (1) - 2, so f (1) = 2
Then the original inequality is equivalent to: F (a ^ 2 + a-5)

Given a > 0 function FX = ax-bx ^ 2 (1) when b > 1, for any x belongs to [0,1], FX belongs to [- 1,1] if and only if? (2) when B belongs to (0,1), other invariants, find (1)

For any x ∈ [0,1], | f (x) | ≤ 1, for any x ∈ [0,1], | ax BX & sup2; | ≤ 1
For any x ∈ [0,1], - 1 ≤ ax BX & sup2; ≤ 1
For any x ∈ (0,1], BX-1 / X ≤ a ≤ BX + 1 / xb-1 ≤ a ≤ 2 √ B

Let f (1 / x + 1) = 1 / x ^ 2 - 1, then f (x) =?

The first step is: let 1 / x + 1 = t, then 1 / x = T-1, x = 1 / (t-1). The second step is to substitute it into the original function f (T) = (t-1) ^ 2-1 = T ^ 2-2t. The third step is: F (x) = x ^ 2-2x. The special method above should be expressed in this way: F (1 / x + 1) = 1 / X & sup2; - 1 = (1 / x-1) (1 / x + 1) = (1 / x + 1-2) (1 / x + 1) so that 1 / x + 1 = X

If the function y = x & # 178; + BX + C (x < 1) is not a monotone function, then the value range of real number B?
If the quadratic function y = x & # 178; + 2aX + B monotonically increases on [1, + ∞), then the value range of real number a

A:
(1) The quadratic function y = x & # 178; + BX + C is a monotone increasing function on x = 1
Explain the axis of symmetry of the corresponding parabola x = - a = - 1

Prove that the function f (x) = X-1 / X is an increasing function on (0, + 00)

X1 > x2 > 0f (x1) - f (x2) = x1-1 / x1-x2 + 1 / x2 denominator = x1x2 > 0, numerator = X1 & sup2; x2-x2-x1x2 & sup2; + X1 = x1x2 (x1-x2) + (x1-x2) = (x1x2 + 1) (x1-x2) X1 > X2, so x1-x2 > 0x1 > 0, X2 > 0, so x1x2 + 1 > 0, so numerator is greater than 0, so f (x1) - f (x2) > 0, so X1 > x2 > 0