The specific way to find f x in high school, thank you!

Guo Dunyun: You "want to ask what method you --- and when, and the idea of solving the problem (mainly), because I just learned," which is very good. What you mean is that you want to elevate the perceptual understanding to the rational understanding, from the special to the general.

This is a bit philosophical, in fact, to learn mathematics well, you need to know some philosophical knowledge.

You don't have to think of this as too mysterious, but in fact, all kinds of basic concepts, theorems, and inferences in textbooks belong to rational cognition, and each example problem belongs to perceptual cognition. Practice makes perfect, and if you are exposed to a variety of questions, you will naturally deepen your rational understanding. The following is a detailed answer to the questions raised

1 It is known that f(x) is a one-time function, and it satisfies 3f(x 1) f(x) 2x 9 to find f(x).

Let f(x)=kx+ b, then 3f(x 1)=3[k(x+1)+b]=3kx+3+3b

3f(x＋1)－f（x）＝3kx+3+3b－kx－b=2kx+2b+3

3f(x 1) f(x) 2x 9, the corresponding terms are equal.

2kx=2x，k=1；2b+3=9，∴b=3，f（x）= x+3

2 It is known that f(x) satisfies 2f(x) f(1 x) 3x and finds f(x).

Let f(x)=kx, then.

2f(x) f(1 x) 2kx+k x= 3x, k x= 3x 2kx, k=x (3 2k), k (3 2k) = x, k 0, and (3 2k) 0, 3 2k, k 3 2

0≤k＜3/2；

f（x）=kx，0≤k＜3/2。

3 It is known that the equation f(x y) f(x) y(2x y 1) holds for a real number x, y, and f(0) 1, find f(x)

According to the given conditions, it is clear that f(x)=x +1, for f(0) 1 is true;

f(x－y)＝（x－y）²+1=x²－2xy+y²

f(x) y(2x y 1) = x +1 y(2x y 1) = x 2xy+y, for the equation f(x y) f(x) y(2x y 1) for a solid number x, y is true, f(x) = x +1.

These three questions are quite difficult, the first question is easier, it is easier to stop, the key is "the corresponding terms are equal" and get k=1, b=3;Problem 2 is more difficult, it is difficult from beginning to end, and it will not be expected in advance to involve inequalities and the range of values of k; The difficulty of the third problem is that there is no way to start the problem, so it is intuitive (there is intuitionism in mathematics), starting from the easiest place, judging from "f(0) 1", there must be "f(x)=x+1" or "f(x)=x +1"; Judging from "f(x y) f(x) y(2x y 1)", it must be "f(x)=x +1". Judgment does not mean proof, the proof here only needs to be verified, followed by two verifications, and the given conditions achieve the purpose of proof.

Can you give me the value of my wealth?

[If you can't see clearly, you can click to enlarge.] 】

1. Treat the x+1 x in parentheses as a whole, and the cube on the right and the formula are factored (x+1 x) (x 2+1 x 2 -1) to continue the formula:

x+1 x)[(x+1 x) 2 -3] makes x+1 x=t

f(t)=t(t^2-3)

f(x)=x(x^2-3)

2. This kind of problem is to constantly iterate (iteration is understood as continuous incoming), and always pay attention to defining the domain during the iteration process!

f(5), 5=10, f(10)=10-2=8, f(5)=f[f(10)]=f(8)=f[f(13)]=f[11]=9 That's it, f(11)=9 is a definite number, there is no function symbol outside, so there is no need to count.

f(0)=f[f(5)]=f[9]=f[f(14)]=f[12]=10

f'(-x)+xf'(x)=x 1)

Substituting -x into the equation gives :f'(x)-xf'(-x)=-x2 )1)*x+2), eliminate f'(-x), get: f'(x)(x^2+1)=x^2-x

i.e. f'(x)=(x^2-x)/(x^2+1)=1-(x+1)/(x^2+1)

Integral: so f(x)=x- xdx (x 2+1)- dx (x 2+1)==

Turn the inside into 2-cos2x, and then it's good to draw.

For example, an interval of algorithms, in fact, this kind of questions are only a test of computing ability, to have an overall concept, proficient in the application of formulas, and solve into a simple algorithm or an algorithm suitable for yourself. It is necessary to distinguish the relationship and difference between f(x) and x. f(x) is an arithmetic relation, and x is an unknown.

In f(f(x)), the whole of f(x) becomes an unknown, and f(f(x)) is the arithmetic relation. If you understand this relationship, f(f(f(x))) and so on are just a matter of calculation skills.