Given f (x) + 2F (1 / x) = x (x is not equal to 0), how to find f (x)?

Given f (x) + 2F (1 / x) = x (x is not equal to 0), how to find f (x)?

F (x) + 2F (1 / x) = x another equation obtained by replacing x with 1 / X is f (1 / x) + 2F (x) = 1 / X. solve the above equations to obtain f (x) = (2-x ^ 2) / 3x

Solving function analytic formula by solving equations
It is known that f (x) + 2F (- x) = 2x + 1
Find f (x)
According to the meaning of the title
f(-x)+2f(x)=-2x+1
Why is there such a step according to the known? Why can it become this form?
I don't understand how f (- x) and 2F (x) wait to return to the original formula!
The original formula is not f (x) and 2F (- x)
How do these two formulas work out

You can get that form by replacing x with - x, which is a common method to solve abstract functions. Pay attention to complete replacement when replacing. For example, if you give this question, you need to replace x with - x, then every x must be replaced, including the one representing the function and the one outside, so as to ensure the function is tenable

Let the inequality 2x-1 ＞ m (x-1) hold for all the value ranges of real numbers m satisfying | m | ≤ 2, and find the value range of X
Give me an idea

2x-1＞m（x-1）
(2-m)x>1-m
M = 2, X can be any value
m(1-m)/(2-m)=
This inequality holds for | m | ≤ 2, so x is greater than the maximum value of 1-1 / (2-m) on | m | ≤ 2, which is 1 / 4, so x > 1 / 4

Senior one mathematics [function & equation]
It is known that the equation (1 / 2) ^ 2 = 1 / (1-lg true a) about X has a positive root, and the value range of real number a is obtained
Give me an idea
The original problem is [it is known that the equation (1 / 2) ^ x = 1 / (1-lg true a) about X has a positive root, so we can find the value range of real number a. 】

(1 / 2) ^ x = 1 / (1-lga) has positive root
First, 1-lga ≠ 0, that is, a ≠ 10
[(1/2)^x]^(-1)=[1/(1-lga)]^(-1)
2 ^ x = 1-lga has positive roots
LGA = 1-2 ^ X has positive roots
When x > 0, 1-2 ^ x < 0
As long as LGA ＜ 0, we can satisfy the existence of X ＞ 0 so that the equation holds, that is, there is a positive root
∴a∈（0,1)
This is what we need

Senior one mathematics [equation & function]
Let f (x) = 2 ^| x-a |, for any real number T, there is always f (2 + T) = f (2-T), then the value of real number a is________ .
Give me an idea

For any real number T, there is always f (2 + T) = t (2-T), then f (x) is symmetric with respect to x = 2, that is, 2 ^| 2 + T-A | 2 ^| 2-t-a | 2 + T-A | = | 2-t-a | 2 + T-A = 2-t-a or 2 + T-A = - (2-t-a) | t = 0 or 4-2a = 0, because it is required that the first solution of T = 0 is invalid, that is, a = 2 satisfies the meaning of the problem

A detailed explanation of the method of solving function analytic solution equations in senior one mathematics
2f(x)+f(1/x)=x
To find the analytic expression of F (x), please explain in detail why 1 / X can be replaced by X

Let u = 1 / x, x = 1 / u
2f(1/u)+f(u)=1/u;
U and X are symbols of independent variables, which can be replaced by each other;
2f(1/x)+f(x)=1/x;

Senior one mathematics: how to use parameter separation method to solve function with parameters
How to use the parameter separation method to solve the interval and the range on the interval, and find the parameter range
For example:
The function f (x) = x ^ 2 + MX + 3, when x ∈ [- 2,2], f (x) ≥ m is constant, find the range of real number m?
Tell me the idea of parameter separation method, and show it in the process of example

Please, they want the parameter separation method!
F (x) = x ^ 2 + MX + 3 > = m holds
So (1-x) M

How to use elimination method to solve function relation in senior one

For example: given that f (x) satisfies 2F (x) - f (1 / x) = 2x + 1, find f (x)
Replace x with 1 / X first
2f(1/x)-f(x)=2/x+1
The binary equations of F (x) and f (1 / x) are obtained simultaneously. By the method of addition, subtraction and elimination, f (x) is solved

Function model and its application
A store has a kind of goods. If it is sold at the beginning of the month, it can make a profit of 100 yuan, and then deposit the principal and interest into the bank. It is known that the monthly interest of the bank is 2.4%. If it is sold at the end of the month, it can make a profit of 120 yuan, but it has to pay a storage fee of 5 yuan. Is it better to sell this kind of goods at the beginning of the month or at the end of the month?

The principal is set at X Yuan at the beginning of the month and sold at Y1 yuan at the end of the month
Sold at the beginning of the month and Y2 yuan at the end of the month
Then Y1 = (x + 100) (1 + 2.4%)
Y2=X+120-5
When Y1 > Y2, the solution is x > 525
The same goes for X

[Yang Yang, excuse me] senior one math problem, application of function model]
A farm needs to buy feed regularly. It is known that the factory needs 200kg feed per day, and the price is 1.8 yuan per kg. The average cost of feed storage and other expenses is 0.03 yuan per kg per day, and the daily freight is 300 yuan
(1) How many days can the factory buy feed to minimize the total cost of daily payment?
(2) If the company that provides feed stipulates that the price of feed purchased at a time is not less than 5 tons, it can enjoy 85% discount. Q: does the factory consider taking advantage of this discount? Please explain the reason
(1) seems to be 10; and (2) should be used
It's better to list the expression or explain the main idea

You have typed a wrong word
A farm needs to buy feed regularly. It is known that the factory needs 200kg feed per day, and the price per kg is 1.8 yuan. The average cost of feed storage and other expenses is 0.03 yuan per kilogram per day. If the freight is 300 yuan per time, then:
(1) How many days can the factory buy feed to minimize the total cost of daily payment?
Buy feed once every x days (6 yuan for each feed saved for one day)
The first feed should be kept for 0 days (if you buy one, you don't need to keep it after eating one day, otherwise 0 becomes 1), the second feed should be kept for 1 day, and the third feed should be kept for 2 days The last feed was kept for X-1 days
300+200*X*1.8+6*0+6*1…… 6 * (x-1) (6 yuan per feed for one day) = 300 + 360x + 6 * (0 + 1 + 2 + 3 +X-1）
1+2+3…… +X-1=（1+X-1）*（X-1-1+1）/2=（X*X-X）/2
So the original formula = 3x square + 357x + 300
Average daily = 3x + 357 + 300 / X
Because x > 0, the average daily > = (6x * 300 * 2 / x under the root sign twice) + 357 = 417
If and only if x = 300 / x, the equal sign holds, x = 10 days
(I did this question for one and a half hours because I made a small mistake n times
(2) If the company that provides feed stipulates that the price of feed purchased at a time is not less than 5 tons, it can enjoy 85% discount. Q: does the factory consider taking advantage of this discount? Please explain the reason
Assuming x > = 25, the feed is not less than 5 tons
300+200*X*1.8*0.85+3X*X-3*X=3X*X+303X+300
Average daily = 3x + 303 + 300 / X
When x = 25, y = 390