Known: F (x) + F (13 / 42) = f (1 / 6) + F (1 / 7); Verification: F (x) is a periodic function

Known: F (x) + F (13 / 42) = f (1 / 6) + F (1 / 7); Verification: F (x) is a periodic function

Because for any x ∈ R, there is
f(x+13/42)+f(x)=f(x+1/6)+f(x+1/7),
Therefore, f (x + 7 / 42) - f (x) = f (x + 13 / 42) - f (x + 6 / 42)
````````````````=f(x+19/43)-f(x+12/42)
````````````````=……=f(x+49/42)-f(x+42/42).
That is, f (x + 42 / 42) - f (x) = f (x + 49 / 42) - f (x + 7 / 42)... (1)
Similarly, yes
f(x+7/42)-f(x+1/42)=f(x+14/42)-f(x+8/42)
```````````````````=f(x+21/42)-f(x+15/42)
```````````````````=……=f(x+49/42)-f(x+43/42)-f(x)
That is, f (x + 49 / 42) - f (x) = f (x + 43 / 42) - f (x + 1 / 42)... (2)
From (1) (2), the
f(x+42/42)-f(x)=f(x+43/42)-f(x+1/42)
```````````````=f(x+44/42)-f(x+2/42)
```````````````=……=f(x+84/42)-f(x+42/42),
That is, f (x + 1) - f (x) = f (x + 2) - f (x + 1)
Therefore, f (x + n) = f (x) + n [f (x + 1) - f (x)] holds for all n ∈ n
And because for all x ∈ R, | f (x) | ≤ 1, that is, f (x) is bounded, only f (x + 1) - f (x) | 0
Therefore, for all x ∈ R, f (x + 1) = f (x), that is, f (x) is a periodic function

If the periodic function f (x) is continuous in (- ∞, + ∞), then f (x) is bounded in (- ∞, + ∞)? How to prove it?

Let it exist Chous T and have f (x + T) = f (x), then the function exists on [0, t], and the continuous function on the closed interval has M = max (ABS (f (x)), x = [0, t]), that is, the function is bounded

How to prove whether a function is an increasing function or a decreasing function? Give examples

Let x1 ＜ x2
F (x2) - f (x1) ＞ 0 increase
F (x2) - f (x1) ＜ 0 subtraction function
Example: F (x) = x + 1
Let x1 ＜ x2
f(x2)-f(x1)=x2+1-x1-1=x2-x1
∵x1＜x2
∴f(x2)-f(x1)=x2+1-x1-1=x2-x1＞0
So FX is an increasing function

Prove that the following functions are homogeneous functions, and explain that they are several homogeneous functions (1) f(x,y)=x^3+xy^2(2) f(x.y)=(x^5)[e^(-y/x)]

No, I'm wrong,
Multiply the independent variable of the function by a factor. If the dependent variable is equivalent to the power of the original function multiplied by this factor, the function is called homogeneous function
So f (TX, ty) = t ³ F (x, y), cubic homogeneous
F (TX, ty) = T ^ 5F (x, y), quintic homogeneous
Come on, this shouldn't be something in class. I don't remember middle school textbooks. It's good to read more extracurricular books

Why do continuous functions on closed intervals [a, b] have maximum and minimum values on [a, b]

You should pay attention to understanding these two properties: 1. Continuous 2. The closed interval [a, b] shows that the function is uninterrupted on the closed interval [a, b], and has definite and finite values at points a and B. of course, all values on the interval [a, b] are definite and finite, so there must be maximum and minimum values. If it is an open interval (a, b), semi

I feel dizzy If the domain of function y = f (x + 1) is [- 2,3], it is - 2

Your classmate is right. For the function y = f (x + 1) = f (T) t = x + 1, the function definition field is [- 2,3], which means that the definition field of T is [- 2,3]. Naturally, the definition field of X + 1 is also [- 2,3], and the definition field of X can be known. If in the function y = f (x + 1), the definition field of X is [- 2,3], then - 2 = 0, x + 1 > = 0. You need to pay attention to the function