Is a function that is both odd and even odd?

Yes, this function can only be a function that coincides with the x-axis and is equal in length from the origin.

Like what. -infinity, +infinity) (c,c)(0).

That is, it must be the origin of the x-axis or a line segment that is symmetrical with respect to y-axis and coincides with the x-axis.

The teacher said: It is said that it is both odd and even, non-odd and non-even, and there is no intersection between odd and even. That is, a function that is both odd and even is not an odd function. Mistake.

Some odd functions can be even functions correct.

Aren't your teachers good?

Your teacher is wrong It is both an odd function and an even function It must be an odd function The answer is wrong You always talk about the question according to the answer, which is the most infuriating!

For example, the function x=0 is symmetrical with respect to the origin and symmetry with respect to the y-axis, so it is both an odd function and an even function.

However, for some complex functions, it is possible that one defined interval is symmetric with respect to the origin, and another defined interval is symmetric with respect to the y-axis, so we can say that this function is odd in one interval and even in another, but we cannot simply say that this function is odd or even (this term refers to the entire custom interval).

Good luck.

If there is f(-x) = - f(x) for any x in the domain of the function f(x), then the function f(x) is called an odd function.

As long as it meets the definition, it is an odd function, so both an odd and even function is an odd function.

logical relationship" or, and, non" learned!

The relationship between arise and even is and so so it must be an odd function

A function that is both odd and even is an odd function with respect to the symmetry of the domain with respect to the origin f(x)=0.

Of course, it's an odd function.

Because it satisfies f(-x)=-f(x)=0.

is an odd function. The teacher said no, said that it was both odd and even, and that it was not odd and not even, and that it was definitely not true.

Take your teacher off!

I really thought about it carefully, there should only be a constant function of y=0, "The teacher said no, said that it is both odd and even, non-odd and non-even" is definitely not right. After 300,000 years of experience, it's impossible to test, this thing doesn't make any sense.

Of course there is an intersection.

For example, y=0 is both odd and even.

It seems that there are no odd and even functions.

Functions are divided into non-odd and non-even, and even odd are both odd and even.

Functions that are both odd and even functions are f(x)=f(-x) and f(-x)=-f(x), and functions that satisfy f(x)=0 and define the symmetry of the domain with respect to the origin are called odd and even functions.

This function defines the domain to be 1,1, because for every x that defines the domain, there is f(x) 0, so f(-x)=f(x)=-f(x)=0. In general, if for the function f(x) is defined in any one of the x's.

Both have f(x)=f(-x), then the function f(x) is called an even function. If there is f(-x) = - f(x) for any x in the domain of the function f(x), then the function f(x) is called an odd function.

Algorithms. The sum of two even functions is an even function. The sum of two odd functions is an odd function. The sum of an even function and an odd function is a non-odd function and a non-even function.

The product of two even functions multiplied is an even function. The product of two odd functions multiplied is an even function. The product of multiplying an even function by an odd function is the odd function definition.

As long as f(-x)=-f(x) (odd function) and f(-x)=f(x) (even function) can be true for any x in the function definition domain, then the function f(x) is both odd and even, and is called both odd and even.

Proof method: Since f(x) is both an odd and an even function, the domain is defined to be symmetric with respect to the origin.

When x=0, if f(x) is defined, because f(x) is an odd function, i.e., f(0)=-f(-0) holds, i.e., f(0)=-f(0) holds, and f(0)=0 is obtained.

When x≠0, since f(x) is an odd function, f(x)=-f(-x) holds; Because f(x) is also an even function, f(x) = f(-x).

So f(x)=-f(-x) and f(x)=f(-x) are both true, so we get f(x)=-f(x), so f(x)=0.

So f(x) is the constant equals 0 and defines the domain as a function of the origin symmetry.

Odd and even function properties:

1. Properties of odd functions.

1.The difference between the sum or subtraction of two odd functions is the odd function.

2.The difference between the sum or subtraction of an even function and an odd function is a non-odd and non-even function.

3.The product of two odd functions multiplied or the quotient obtained by division is an even function.

4.The product of an even function multiplied by an odd function or the quotient obtained by division is an odd function.

5.The integral of the odd function on the symmetry interval is zero.

2. Properties of odd functions.

1. If you know the function expression, for any x in the definition domain of the function f(x), it satisfies f(x)=f(-x) such as y=x*x.

2. If you know the image, the even function image is symmetrical with respect to the y-axis (straight line x=0).

3. Defining domain d with respect to origin symmetry is a necessary and insufficient condition for this function to become an even function.

Odd function definition: f( x) f(x).

Even function definition: f( x) f(x).

is both an odd function and an even function, then.

f(－x)＝－f(x)＝f(x)

Get f(x) 0

Therefore, only this one function satisfies the condition.

What is a function that is both odd and even?

The function y = x 2 is a function that is both odd and even.

y = 0*x

It is both an odd function.

It's an even function again.

Because; f(x) =0*x

Then: f(-x) =0*(-x) =0*x = f(x)f(-x) =0*(-x) =0*x =f(x) So it is both an odd function and an even function.

Yes, the constant function f(x)=0 defines the domain r

Defined according to odd functions:

f(x)+f(-x)=0

According to the even function definition:

f（x）=f（-x）

then f(x)=0

The only condition that is met is the constant function y=0, and nothing else works.