Mathematical Parity Problem 5, Mathematical Parity Problem

Personally, I don't think it's very appropriate to use the image method.

The title is about an abstract function f(x), and the meaning of the title is that for any odd function f(x), the following options must be odd functions, so it is possible to use the image method to obtain some special cases, so that the original correct options are still correct, but some incorrect options are regarded as correct.

I think the safest way to do this is to use the formula method. Then, for the correct option, it is made with strict derivation; For incorrect options, use the image method to give a counterexample. (For example, take the function f(x)=x and draw an image of 1 4.) ）

However, the answer to this question is an even function, so if you use the image method, as long as it is not constant 0, the solution is correct. If there are some non-odd and non-even functions mixed in with the results, the answer may not be accurate.

Here's my advice

Feel free to assume an odd function, e.g. f(x)=x

Turn the line below the x-axis, with the x-axis as the axis of symmetry, and turn it up into a V-shape, even function.

Turn the line on the right side of the y-axis under the x-axis, and do the same as the side up, odd function.

y=x*x=x2, the image is a parabola with a fixed point of (0,0) opening upward, symmetrical on the y-axis, even function.

y=x+x=2x, odd function, you should understand...

This method has particularity, multiple-choice questions can be used to try, proof questions can only be formulas...

Let x=0, y=-1, then f(0)=-f(0), so f(0)=0...1)

Let x=y=1, then f(1)=2f(1), so f(1)=0.

Let x=y=1, then f(1)=2f(1), so f(1)=0.

Let y=-1, so f(-x)=x*f(-1)-f(x)=-f(x)....2)

From (1) and (2), we know that f(x) is an odd function.

-sign, x=y=0

f(0)=0

f(-y)=0*f(-y)+y*f(0)=0,f(y)=0 odd and even functions.

Indicates a multiplication sign. x=y=0

f(0)=0

f(1*1)=2f(1),f1=0,f(1=-1*-1)=-1*f[-1]-1*f[-1}=0,f-1=0

f[-x]=-1*f[x]-x*f(-1)=-f(x) fx odd function.

Odd and even numbers are only for integers.

A number divisible by 2 is an even number, such as 2, 4, 6, ......

Not divisible by 2 are odd numbers such as 1, 3, 5, ......

Because g(x) is an even function, g(-x)=g(x)g(x)=x 2+ (x-a)*f(x) =x 2+ a-x-1 g(-x)=(x) 2+ (x-a)*f(-x) =x 2+ a+x-1

So, a-x-1=a+x-1

x=0, not true.

or a-x-1=-(a+x-1).

a=1 process I don't know if there is a miscalculation.,It's been too long for the wood to touch the math.。。。

1) a≠0, then the domain is not symmetric with respect to 0, non-odd and non-even functions.

a=0 f(x)=1/(-x)-1

f(-x)=1 x-1 , non-odd and non-even functions.

2) a≠0, then the definition domain is not symmetric with respect to 0, non-odd and non-even functions.

So consider a=0.

In this case, g(x)=x +|1-1|=x +2 is an even function, so a=0

Regardless of whether a is any constant.

None of them are even functions.

There is a problem with the question.

Before you do the problem, you should understand the concept: odd functions.

1. In the odd function f(x), the signs of f(x) and f(-x) are opposite and the absolute values are equal, that is, f(-x)=-f(x), and conversely, the function y=f(x) that satisfies f(-x)=-f(x) must be an odd function. For example: f(x)=x (2n-1),n z; (f(x) is equal to x to the power of 2n-1, n is an integer).

2. The odd function image is symmetrical with respect to the center of the origin (0,0).

3. The domain of the odd function must be symmetrical with respect to the center of the origin (0,0), otherwise it cannot be an odd function.

4. If f(x) is an odd function and x belongs to r, then f(0)=0

According to the fourth clause, f(0)=0, bring x=0 into the solution 0=(-1+b) (2+a), -1+b=0, b=1;

According to the first clause, f(-1)=-f(1), then (-1 2+b) (1+a)=-(-2+b) (4+a), bring in b, (1 2) (1+a) = 1 (4+a), and get a=2

You have mastered the concepts and characteristics of odd functions, and it is very easy to do this kind of problem, have you learned it?

I don't know what your education is.

In elementary school, it refers to whether a natural number can be divisible by 2, if it can it be an odd number, and if it can't, it will be an even number. Primary schools are also called singular and even.

If it is junior high school [second and third year of junior high school], parity refers to a function.

For any x r, there is f(-x)=(-x) 2=x 2=f(x)In this case, we call the function f(x)=x 2 an even function.

For any x in the defined domain r of the function f(x)=x, there is f(-x)=-f(x), and we call the function f(x)=x an odd function.

In addition to functions, there are also those who are studying.

If a number satisfies xmod2=1, then it is an odd number;

If a number satisfies xmod2=0, then it is even.

When the domain of the function is defined symmetrically with respect to the coordinate origin. If there is f(x)=f(-x) for any x, then the function f(x) is an even function. If f(x)=-f(-x), then the function f(x) is odd.

The function image of the even function is symmetrical with respect to the y-axis, and the function image of the odd function is symmetrical with respect to the coordinate origin.

If you don't understand, you can continue to ask.

Seeing that you are asking about primary school mathematics, it is estimated that if it is good, it is odd + odd = even + even = even odd + even = odd In other words, the sum of two numbers with the same oddity and parity is equal to even, and the addition of different numbers is equal to odd.

Parity in functions: The image of the odd function in the coordinates is symmetrical with respect to the origin, and the image of the even function is symmetrical with respect to the y-axis.

Odd numbers are not divisible by two, and even numbers are divisible by two.

f(x)=x 7+ax 5+bx 3+cx+2, f(-x)=-(x 7+ax 5+bx 3+cx)+2, f(x)+f(-x)=4, i.e., f(-3)+f(3)=4, and f(-3)=-3, then f(3)=7

Remember f(x) = odd function + constant type m

f(x)+f(-x)=2m, 2 times constant, know one to find one, oral arithmetic is enough, odd functions are common three positive (proportional, sine, tangent), odd subtype, exponential reduction a x-a (-x), etc.

It's not Tanano Chie! The chain shouted.

Definition of odd and even functions: For the function f(x), 1. The domain of the function f(x) is required to be symmetric with respect to the origin.

2. If f(-x)=-f(x), then the function f(x) is called an odd function.

If f(-x) = f(x), then the function f(x) is called an even function.

From the graph, you can see that the odd function is symmetrical with respect to the origin, and the even function is symmetric with respect to the y-axis.

The following judgment is judged according to the definition, not odd and even functions, you can find a special point so that it does not meet the definition of odd function or even function.

The four domains of a, b, c, and d are all real numbers r. Define the domain symmetry with respect to the origin.

For a, y=-8 due to x=-1;When x=1, y=2. Therefore, it is neither an odd nor an even function.

For b, since (-x)sin(-x)=(x)(-sinx)=xsinx.

Therefore, it is an even function.

For c, due to (-x) = x. Therefore, it is an odd function.

For d, e(-1)≠e, or e(-1)≠-e.

Therefore it is neither an odd nor an even function.