Given π = 3.1415926535... Let f (n) = K (n ∈ n +), K be the nth digit after the decimal point of π, denoted as f {f... [f (n)]} = FM (n) ｝｝｝｝｝｝｝｝｝ M f If f (0) = 1, is there a minimum positive integer m? If M ≥ m, FM (n) is a constant for all n ∈ n? If it exists, find m and constant K. if it does not exist, explain the reason

Given π = 3.1415926535... Let f (n) = K (n ∈ n +), K be the nth digit after the decimal point of π, denoted as f {f... [f (n)]} = FM (n) ｝｝｝｝｝｝｝｝｝ M f If f (0) = 1, is there a minimum positive integer m? If M ≥ m, FM (n) is a constant for all n ∈ n? If it exists, find m and constant K. if it does not exist, explain the reason

f（1）=1 f（2）=4 f（3）=1 f（4）=5 f（5）=9
f（6）=2 f（7）=6 f（8）=5 f（9）=3
Any f (n) is one of the 10 numbers 0-9, so the value of F 2 (n) is obtained from F (1) to f (9)
The number that F 2 (n) can be equal to is 1 2 3 4 5 6 9
Similarly,
The number F3 (n) can be equal to is 1 2 3 4 5 9
The number F4 (n) can be equal to is 1 3 4 5 9
The number F5 (n) can be equal to is 1 359
The number that F6 (n) can be equal to is 1 3 9
The number that F7 (n) can be equal to is 1 3
The number that F8 (n) can be equal to is 1
The number FM (n) can be equal to is 1 m > = 8
So m = 8, k = 1

Given that f (x) is a quadratic function and f (x + 1) + F (x-1) = the square of 2x + 6x-4.. then f (x) =?
PS: there should be a detailed analysis process

Let f (x) = ax ^ 2 + BX + C, where a is not equal to 0
Using the substitution method, we get f (x + 1) = a (x + 1) ^ 2 + B (x + 1) + C
f(x-1)=a(x-1)^2+b(x-1)+c
From the square of known f (x + 1) + F (x-1) = 2x + 6x-4
[a(x+1)^2+b(x+1)+c]+[a(x-1)^2+b(x-1)+c]=2x^2+6x-4
The left side of the above formula is reduced to 2aX ^ 2 + 2bx + 2 (a ^ 2 + C) = 2x ^ 2 + 6x-4
The analytic expressions of the two functions are equal, so the coefficients of x ^ 2 (x square), the coefficients of X first term and the constant terms are equal respectively
So we get 2A = 2,2b = 6,2 (a ^ 2 + C) = - 4, the solution is a = 1, B = 3, C = - 3
So we find that f (x) = x ^ 2 + 3x-3

Given the function f (x) = x2-2ax + 3 (1), if the monotone decreasing interval of function f (x) (- ∞, 2), find the maximum value of function f (x) in the interval [3, 5]. (2) if the function f (x) is monotone decreasing in the interval (- ∞, 2], find the maximum value of function f (1)

∵ function f (x) = x2-2ax + 3, so the monotone decreasing interval of function f (x) (- ∞, a)], (1) is determined by the monotone decreasing interval of F (x) (- ∞, 2), so if a = 2, then f (x) = x2-4x + 3, and ∵ function f (x) monotonically increases in the interval [3,5], so when x = 5, the maximum value of function f (x) is 8 (6 points) (2) if f (x) monotonically decreases in the interval (- ∞, 2], then f (1) = 4-2 if a ≥ 2 A ≤ 0 means that the maximum value of function f (1) is 0 - - (12 points)

The domain of F (x) is known
1. Find the domain of F [g (x)]
2. Given that the domain of F (x) is [0,3], find the domain of F (x-3)

1. In general method, the domain of F (x) is the range of G (x)
2. F (x) definition field is [0,3], then the value range of x ^ 2-3 is [0,3], so there is 0 "x ^ 2-3" 3, so the value range of X is [- radical 6, - radical 3] and above [radical 3, radical 6]
Note that the domain of function is the scope of X, especially the domain of compound function is the scope of X

The minimum positive period of the function f (x) = 1-2sin2x is______ ．

F (x) = 1-2sin2x = cos2x  function minimum positive period T = 2 π 2 = π, so the answer is: π

Given that the image of a quadratic function passes through the coordinate origin, its vertex coordinates are (1, - 2)

Let the relation of the quadratic function be y = a (x-1) 2-2, ∵ the image of the quadratic function passes through the origin of the coordinate, ∵ 0 = a (0-1) 2-2, the solution is a = 2, so the relation of the quadratic function is y = 2 (x-1) 2-2, that is, y = 2x2-4x

Given that the image of a quadratic function passes through the coordinate origin, its vertex coordinates are (1, - 2)

Let the relation of the quadratic function be y = a (x-1) 2-2, ∵ the image of the quadratic function passes through the origin of the coordinate, ∵ 0 = a (0-1) 2-2, the solution is a = 2, so the relation of the quadratic function is y = 2 (x-1) 2-2, that is, y = 2x2-4x

What does f mean in the function y = f (x)

F () represents the expression of some algorithm for the independent variable in brackets
For example, if f (x) = 3x + 2
Then f () is the expression of multiplying the number in brackets by 3 and then adding 2
This is a mathematical convention, F and () together express the meaning of the above description, do not explain separately

What does the mathematical function f (x0) mean?

Is this 0 a footmark? If so, this is the value of F (x) at x0

(solving process) (1) f (x) = the square of X, X ∈ (1,3)
(2) Y (x) = x under radical
(3) F (x) = x + the third power of X + the fifth power of X

1) Because the domain of definition is not symmetric about the origin, f (x) is not odd or even;
2) The domain of definition is x > = 0, which is not symmetric about the origin, so f (x) is not odd or even;
3) F (x) = x + x ^ 3 + x ^ 5, the domain is r
F (- x) = - f (x), so f (x) is an odd function