The expression of a function with a period of 2 does not necessarily satisfy f x 1 f x

No, it must be an odd function with a period of 2 to meet such a condition, if it is not an odd function, it is impossible to have this condition, in other words, the original function can be deduced from the condition given by the landlord is an odd function, but the function with a period of 2 is definitely not all odd functions, such as y=cos x period is 2, but it is not an odd function and does not meet that condition.

Landlord, there is a problem with the example you cited, since the odd function f(x) is lgx on (0,1), then it is no longer lgx on (-1,0), but -lg(-x)=-lgx+1, you will find that the conclusion is right, and you can't give x= as an example, this time beyond the (0,1) definition domain, to subtract the period and then calculate, it is very inconvenient to use the computer to say, the landlord you just need to calculate the function relationship by heart, it is not difficult at all, As long as you are careful, I hope mine can help you.

No, it should be f(x+2)=f(x) and your period is 1.

It can be seen that f(x) = f(x+1)Let x=x+1 give f(x+2) = f(x+1), so f(x+2) = f(x)The period is 2

Similarly, if the period is 2, then there must be f(x+2)=f(x)They are the sufficient and necessary conditions of each.

The details are as follows:f(x)=x^2 [0,2)

f(x)=(x-2)^2 [2,4)

f(x)=(x-4)^2 [4,6)

f(6)=f(0)=0

Monotonicity of functions:Let the domain of the function f(x) be defined.

is d, and interval i is contained in d. If for any two points x1 and x2 on the interval and when x1 is for any two points x1 and x2 on the interval i, when x1f(x2), then the function f(x) is said to be monotonically decreasing on the interval i. Monotonically increasing and monotonically decreasing functions are collectively referred to as monotonic functions.

Solution: f(x)=x when x [0,2) and f(x) is a function with a period of 2.

Let 2 x 4, then 0 x—2 2

f（x）=f（x－2）=（x－2）²

2 x 4 i.e.: f(x)=(x 2).

2 x 4 order 4 x 6, then 0 x—4 2

f（x）=f（x－4）=（x－4）²

4 x 6 i.e.: f(x) = (x 4) 4 x 6 ,.. in summary

x²0≤x＜2

f（x）={

x－2）²2≤x＜4

x－4）²4≤x＜6

The details are as follows: f(x)=x 2 [0,2)f(x)=(x-2) 2 [2,4)f(x)=(x-4) 2 [4,6)f(6)=f(0)=0 The monotonicity of the function: let the function f(x) be defined in the domain d, and the interval i is contained in d.

If for any two points on the interval x1 and x2, when x1

Your zone should be half-open and half-closed, otherwise there will be two at the endpoints.

Function value. f(x)=x^2

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

f(x)=(x-4)^2

f(6)=f(0)=0

When x is in the service [0,2), f(x)=x 3;

When x is in [2,4), then x-2 is in [0,2), f(x)=f(x-2)=(x-2) 3;

When x is in [4,6), then the old dead limb x-4 is in [0,2) in defeat, f(x)=f(x-2)=f(x-4)=(x-4) 3;

And f(6)=f(4)=f(2)=f(0)=0.

1) Treat 2x as a new variable u, then the minimum positive period of the sinu is 2, that is, when u increases to u 2 and must be increased to u 2, the value of the function sinu is repeated and u 2 2x 2 2 (x) so when the empty independent variable x is increased to x and must be increased to x, the value of the function is repeated, therefore, the period of y sin2x is

Demolition respects <>

Because the topic of your Yuandan is Qiaoyou's function of taking 2 as the period, that is, every time x is added or subtracted by 2, y is equal to the same number. So this function is again a periodic function. Because when x=2, y=3

So x=4,6,8,10,12, y is equal to 2.

For the function y=f(x), if there is a constant t that is not zero, such that when x takes every value in the defined domain, f(x+t)=f(x) holds, then the function y=f(x) is called the periodic function, and the non-zero constant grinding side number is called the period of this function. In fact, any constant kt (k z and k ≠0) is its period.

So 3=f(2)=f(4)=f(6)=f(8)=3

f(x+2)=-1/f(x)

f(x+4)=-1 f(x+2)=f(x), i.e., f(x+4)=f(x).

So f(x) is a periodic function, and one of its periods is 4

f(x+2)=-1/f(x)

So -1 f(x+2) = f(x).

So f(x+4).

f[(x+2)+2]

1/f(x+2)

f(x) so f(x) is a periodic function, and one of its periods is 4

The ordinate expands by 2 times for each additional unit x.

If the previous 2 times are removed, the period of 1 is used.