The function f (x) whose domain is R is known to satisfy f [f (x) - x 2 + x] = f (x) - x 2 + X If there is only one real number a such that f (a) = a, the analytic expression of real number f (x) is obtained   I did this: Because f (a) = a, f [f (x) - x? + x] = f (x) - x? + X When x = a, f (2x-x? 2) = 2x-x Because there is and only one real number a, f (a) = a   And I don't know what to do

  The function f (x) whose domain is R is known to satisfy f [f (x) - x 2 + x] = f (x) - x 2 + X If there is only one real number a such that f (a) = a, the analytic expression of real number f (x) is obtained   I did this: Because f (a) = a, f [f (x) - x? + x] = f (x) - x? + X When x = a, f (2x-x? 2) = 2x-x Because there is and only one real number a, f (a) = a   And I don't know what to do

When x = a, f (a) = a, then f [f (x) - x? + x] = f [A-A * a + a] = f (2a-a * a) = 2a-a * a; and because there is and only one real number a, f (a) = A and f (2a-a * a) = 2a-a * a; then 2a-a * a = a, then a * a = a, a = 0 or 1 because: F [f (x) - x? + x] = f (x) - x? + X, and because there is and only one real

I'm poor at math. I've got a good answer to the same question. But I still can't understand it. I hope someone will analyze it for me. We know that f (x) is a quadratic function, and f (x + 1) + F (- 1) = 2x ^ 2-4x + 4, Let f (x) = ax ^ 2 + BX + C, then get a (x + 1) ^ 2 + B (x + 1) + C + a (x-1) ^ 2 + B (x-1) + C = 2x ^ 2-4x + 4, and then get 2aX ^ 2 + 2bx + 2A + 2C = 2x ^ 2-4x + 4,

F (x) is a quadratic function, f (x) = ax ^ 2 + BX + CF (x + 1) = a (x + 1) ^ 2 + B (x + 1) + C = ax ^ 2 + 2aX + A + BX + B + B + CF (x-1) = a (x-1) ^ 2 + 2 + B (x-1) + C = ax ^ 2-2ax + A + bx-b + C, the above two formulas add, 2aX, and - 2A s offset, B and - B offset: there are: F (x + 1) + F (x + 1) + F (x-1) = 2A x ^ 2 + 2A + 2bx + 2C x + 2C + 2C (x-1) = 2aX ^ 2 + 2A + 2bx + 2C + 2C + 2C 2x x + 2C + F (x + 1) + F (x + 1) + andby the known condition: F

According to the condition, find the expressions of the following functions respectively 1. F (x + 1 / x) = the square of X + 1 / X 2. F (1 + 1 / x) = 1 / X squared - 1

1. F (x + 1 / x) = the square of X + 1 / X
t=x+1/x
t^2-2=x^2+1/x^2
f(t)=t^2-2,f(x)=x^2-2
2. F (1 + 1 / x) = 1 / X squared - 1
t=1+1/x,1/x=t-1
f(t)=(t-1)^2-1=t^2-2t
f(x)=x^2-2x

A problem of function expression in the winter vacation assignment of senior one mathematics The definition domain of the function f (x) = ax + B / 1 + X2 is (- 1,1), and for any x belonging to (- 1,1), there is f (- x) = - f (x), and f (1 / 2) = 2 / 5 (1) The analytic expression of the function f (x) is determined; (2) It is proved by definition that f (x) is an increasing function on (- 1,1); (3) Solving inequality f (t-1) + F (T) < 0 I have finished all the reward points recently, so I haven't got them. I'm sorry. But please give me the process and result,

I don't know what you mean by B / 1 + X2, so we can say that the following method is: F (- x) = - f (x) is an odd function, then f (0) = 0, the original formula can find B, and with F (1 / 2) = 2 / 5, we can find a, so two or three questions are simple. There are examples in the textbook

The function f (x) = asin (3x + φ), (a > 0,0 < φ < π, when x = π / 12, the maximum value is 4 ① Find f (x), minimum positive period (2) find the analytic formula of F (x)

Y = SiNx the minimum positive period is 2 π
Therefore, the minimum positive period of F (x) = asin (3x + a) is 2 π / 3
F (x) = SiNx the minimum positive period is 2 π
Therefore, the minimum positive period of F (x) = asin (3x + a) is 2 π / 3
0Asin（π/4+a）=4
A=4
π/4+a=π/2+π*n
0a=π/4
f（x）=4sin（3x+π/4）

The function f (x) = x ^ 2 + 2aX + 2, X ∈ [- 5,5]. Find the maximum and minimum of this function

The opening of this function is upward, and the axis of symmetry is x = - A, so we need to discuss the range of A
If - a ＜ - 5, that is, a ＞ 5. The minimum value is f (- 5) = 27-10a
The maximum value is f (5) = 27 + 10A
If - 5 ≤ - a ≤ 5, that is - 5 ≤ a ≤ 5. The minimum value is f (- a) = 2-A ^ 2
The maximum value is Max {f (5), f (- 5)}
[it can be further subdivided here. When - 5 ≤ a ≤ 0, the maximum value is f (- 5); when 0 < a ≤ 5, the maximum value is f (5)]
If - a > 5, that is, a ＜ - 5. The maximum value is f (- 5) = 27-10a
The minimum value is f (5) = 27 + 10A
If you don't understand, ask, if you understand, please accept!

1. When x > 0, the minimum value of the function y = x + 1 / X is____ 2. When x0, the minimum value of the function y = x + 1 / X is____ . 2. When the maximum value of x0) is____ . 5. The solution set of inequality | 1-3x > 7 is_____ 6. When x=____ The function y = x (1-2x) (0

1. The minimum value of the function y = x + 1 / X is 2
2. The maximum value of the function y = x + 1 / X is - 2
3. When x = 1, the minimum value of the function is 3
4. When x = √ 6 / 2, the maximum value of the function is 1-2 √ 6
5. The solution set of inequality | 1-3x > 7 is x > 8 / 3 or X

Find the maximum value and minimum value of the following function, and obtain the maximum value by using the function. The minimum value x y = 3-2cosx, X belongs to R

Because y = 3-2cosx, cosx ∈ [- 1,1],
So the maximum value of y = 3 is - 1,
The minimum value of the function is y = 3-2 = 1,
Because x belongs to R, cosx = - 1, x = π + 2K π (K ∈ z)
When the minimum value of the function is 1, cosx = 1, x = 2K π (K ∈ z)

Symmetric axis equation of function y = asin (Wx + φ)

Sin axis of symmetry is the best place to take
That is sin (Wx + φ) = ± 1
wx+φ=kπ+π/2
So the axis of symmetry x = (K π + π / 2 - φ) / W

The function y = asin (Wx + φ) + B has the highest point (π / 11,3) and the lowest point (7 π / 12, - 5) in the same period

0