The math problem in the first year of high school is to find parity, and I don t know how to solve i

Let x -x then bring -x into f(x).

If f(-x)=f(x) is an even function.

If f(-x)=-f(x) is an odd function.

None of them are non-odd and non-even functions.

f(-x)=|-x+1|-|x-1|=|-(x-1)|-x+1)|=|x-1|-|x+1|=f(x) is an even function.

f（-x）＝-x｜-x｜=- x|x|=-f(x) is an odd function.

Both are odd functions, and you will be introduced to use the geometric sketchpad and draw it yourself to know what it is.

But you can do it yourself, just find f(-x) and see how it relates to f(x).

a² -2a + 3

a - 1)² 2 ≥ 2

Since f(x) is an even function defined on r and an increasing function on (- 0), then f(-2) = f(2) and f(x) is a subtraction function on (0,+, so f(2) f(a -2a + 3).

When x and (x+3) (x+4) symbols are the same, the function values of the two are equal only when x=(x+3) (x+4) are simplified to obtain a quadratic equation x 2+3x-3=0The sum of the solutions is x1+x2=-3

When x is different from the (x+3) (x+4) symbol, due to f(x)=f(x+3) (x+4)=f(-x). x and (x+3) (x+4) symbols are the same, only when -x=(x+3) (x+4) the function values of the two are equal, and the equation x 2 + 5 x + 3 = 0, x3 + x4 = -5 is obtained

The sum of the two solutions is the answer sought. -8

Req -x=(x+3) (x+4).

x=1-1/(x+4)

1=(x+4)(x+1)

x^2+5x+3=0

Add the root formulas together.

x1+x2=-b/a

Then let x=(x+3) (x+4).

x=1-1/(x+4)

1=(x-1)(x+4)

x^2+3x-3=0

x1+x2=-b/a=-3

Solution: f(x) is an even function.

The domain [2a-3,1] is symmetrical with respect to the y-axis, i.e., [-1,1]2a-3=-1, i.e., a=1

and f(x) is an even function.

f(-x)=f(x)

i.e. a(-x) +b(-x)+c=ax +bx+cbx=0, i.e., b=0

In summary, a=1, b=0, c r

i.e. a, b, c r

According to the power (x) of 2 and the power of x, it is easy to get f(x) f(x), where f(x) is an odd function.

The definition method is used for the big questions, and the special value method is used for the small questions.

(1) f(x)+f(y)=f[(x+y) (1+xy)]f(1 2)+f(0)=f[(1 2+0) (1+1 2*0)]] is converted to f(0)=0

2) Proof: f(x)+f(y)=f[(x+y) (1+xy)]f(x)+f(-x)=f[(x-x) (1-x 2)]=f(0)=0

then f(x) is an odd function.

3)f(2x-1)<1=f(1/2)

Whereas f(x) is a monotonically increasing function on (-1,1).

So -1<2x-1<1 2

Solution: 0

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

2. Let x=-y f(x)+f(y)=f(x)+f(-x)=f(0)=0 So f(x) is an odd function.

3. Because f(x) increases monotonically and f(1 2)=1, the formula 2x-1<1 2 1 has -1<2x-1< 1 2 formulas in the defined domain.

is obtained by f(x)=a x-(1 a) x+1.

f（-x）=a^-x-(1/a)^-x+1=(1/a)^x-a^-x+1

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

f(x)-f(-x)=2a x-2(1 a) x, so f(x) is a non-odd and non-even function.

The question has been changed, it does not affect, and the answer is the same.

1.It is obtained from f(x) is an even function, g(x) is an odd function, and f(x)+g(x)=x +x-2.

f(-x)+g(-x)=f(x)-g(x)=x²-x-2.

The solution of the above two formulas gives f(x)=x -2, g(x)=x.

2.f(x) defines the domain as [a-1,2a] and is an even function, a-1=-2ai.e. a=1 3

f(-x)=f(x), b=0

then f(x)=1 3x +1+d.

f(x) has a maximum value f(2 3) = 27 31 + d

Therefore, the f(x) range is [27 31+d,1+d].

3.Don't know where the square is**? If it is 1+x: f(x) increments at (- 0) and decrements on (0, +). Proof by definition.

That is, within the interval range of x1< x2 to prove that f(x1) and f(x2) satisfy the above relationship.

f（x）=x，g（x）=2

It seems to be a question on the innovation assignment.

1. Assuming x=-x, bring in the equation to get: f(-x)+g(-x)=

According to the parity of the function, we know that f(-x) = f(x) and g(-x) = -g(x), so the equation is .

f(x)-g(x)=x-x-2 squared, and the equation is solved by combining it with the equation given in the problem.

f(x) = x squared - 2

g（x）=x

2. According to the definition of even function f(x)=f(-x), so.

The square of ax - bx+3a+d=the square of ax+bx+3a+d, we get b=0, and the domain of the even function is also symmetrical about the origin, so a-1=-2a, we get a=1 3, so the range is (d+1, d+31 27).

x 2 is a subtraction function at (- 0), and 1+x 2>0, so f(x) is an increasing function.

Similarly, at (0,+ is a subtraction function.

The odd function f(x) defined on r has f(0)=0f(x+2)=-f(x), so there is f[(x+2)+2]=-f(x+2) so, f(x+4)=-f(x+2)=f(x) then f(6)=f(4+2)=f(2)=f(0+2)=-f(0)=0

Because of the odd function, so -f(x)=f(-x) So this problem can be reduced to f(x+2)=f(-x) So the periodicity is 2 then f(6)=f(0) Since the odd function, f(0)=0 So this question is b