Ask a question about function parity!! Worry!!!

g(x)= (a-1)·f(x)· [1 (a x-1)+1 2] = (a-1)·f(x)· (a x+1) [2(a x-1)] is obtained by passing the score).

f(x) is an odd function, f(-x)=-f(x)g(-x)= (a-1)·f(-x)· [1 (a (-x) -1) +1 2]

a-1)·f(x)· [1 (1 a x -1) +1 2]-(a-1)·f(x)· It is obtained through the score).

a-1)·f(x)· [a x (1-a x) +1 2]-(a-1)·f(x)· (a x+1) [2(1-a x)] is obtained by general fraction).

a-1)·f(x)· (a x+1) [2(a x-1)]g(x)g(x) is an even function.

My method has its own number f(x) is an odd function, just find the simplest odd function, f(x)=x

g（1）=？

g（-1）=？They're equal and even functions, and that's it.

Isn't that the topic of the third year of high school?

Categorize A.

Then follow the parity step of the function.

f(x+2)=-f(x) gives f(x 4)=f(x 2 2)= f(x 2)=f(x) then the period of the secondary function is 4, then f( can be obtained according to the odd function

According to f(x+2)=-f(x), there is f(

f(x) is an odd function.

f( =

If it's a fill-in-the-blank question, I have a way to get an answer quickly.

Since f(x) is an even function, let's assume that x2 is a model, and x+2 is equivalent to a negative from the y-axis along the x-axis.

2 units in direction, and because (-x+2) 2=(x-2) 2, and x-2 is equivalent to positive from the y-axis along the x-axis.

The direction moves by 2 units, even function, the y-axis is symmetrical on both sides, so it is equal.

Fill-in-the-blank questions are fastest with the special value method. Give me points.

f(x+2) even function, the axis of symmetry is x=0

Move it 2 units to the right.

is f[(x-2)+2]=f(x).

Then the axis of symmetry is also shifted 2 units to the right.

So the f(x) axis of symmetry is x=2

So there is f(2+x)=f(2-x).

i.e. f(-x+2) = f(x+2).

In fact, this is very obvious, there is no need for a process, it is that you do not understand it properly.

The so-called even function is for x, it has nothing to do with its addition or subtraction...

So f(x+2) is an even function, then it is clear that f(-x+2) = f(x+2).

The even function is f(-x)=f(x).

The odd function is f(-x)=-f(x).

1) The answer given is incorrect! should be discussed.

When a=0, even; When a is non-zero, it is not odd or even. Methods such as the second floor.

2) Segmented Discussion:

When x>=a, f(x)=x +x-a+1=(x+, because -1 2 a, f(x) is incremented.

So, the minimum value of f(x) = f(a) = a +1;

When x<=a, f(x)=x -x+a+1=(, because a 1 2, f(x) decreases, so the minimum value of f(x) = f(a) = a +1;

So, the minimum value of f(x) = a +1

Solution: Substitute -x into f(x) to see if it is equal to f(x), or if it is the opposite of each other, if it is equal it is an even function, if it is the opposite of each other, it is an odd function. If it is neither, it is a non-odd and non-even function. From this, this function is a non-odd and non-even function.

Problem 1: Let x=1 and substitute f(x)=x +|x-a|The +1 result is: Because f(x) is not equal to f(-x) nor is it equal to -f(-x). So non-odd non-even functions.

Question 2: Drawing comes out as soon as you draw it, and drawing is the easiest.

f(x)≠f(-x)≠—f(x) are non-odd and non-even functions with quadratic function images.

Even functions. When x!=0（!= is not equal to), f(0+x) = f(0-x), symmetrical with respect to the line x=0.

When x=0, f(x)=0, on a straight line x=0.

So it's an even function.

The original function f(x) = y=e to the power of -x, to the power of -x, to the power of -1 of e+1=1 (to the power of e x)-e x+1

f(-x)=e^x+1/(e^x)+1

f(x)≠f(-x)≠-f(-x)

So it's not odd or even (not so sure).

To be parity, the first condition is to define domain symmetry.

This function is defined in the following domains.

|a|,|a|]

Because a≠0, the definition domain is asymmetric, so it does not have parity.

Score a>0

a=0a<0.

The final answer is a<0

The odd function is about the origin symmetry.

The even function is symmetrical with respect to the y-axis.

f(x)=-f(-x)

Such as (1,4) and (-1,-4).

The one below is correct. Look at it as a whole.