A question about the parity of functions, the parity of functions

f(x)=|x-1|+|x+1|

f(-x)=|-x-1|+|x+1|=|-(x+1)|+x-1)|Because the absolute value of a negative number is its opposite, f(-x)=|x+1|+|x-1|=f(x)

Let me tell you: If you want to judge the parity of a function, first define the domain to be symmetrical, that is, the two sides of the domain must be the same and take the same or merge Second, f(x)=-f(-x) is an odd function If f(x)=f(-x) is an even function Method 2: If the function image is symmetric about the origin, it is an odd function If it is symmetric with respect to the y-axis, it is an even function.

f（x）=|x+1|+|x-1|

Replace x with -x:

f（-x）=|-x+1|+|x-1|

(x-1)|+x+1)|

x-1|+|x+1|

f(x) so f(x) is an even function.

If it is f(x)=|x+1|-|x-1|

Replace x with -x:

f（-x）=|-x+1|-|x-1|

(x-1)|-x+1)|

x-1|-|x+1|

|x+1|-|x-1|)

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

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

f(x+4)=f(x) period is 4, and f(6)=f(2) is an odd function. f(x+2)=-f(x)=f(-x) The axis of symmetry is x=1

Therefore f(2) = f(0).

Define the domain as r

f(0)=0

i.e. f(6)=0

f(x)=-(x +a) (bx +1) when x=0, f(x)=0 so bring x=0 into f(x)=-(x +a) (bx +1 ) to get a=0

Again, it is an odd function.

f(-x)=-f(x)

x -bx+1= x (bx+1) gives b=0 f(x)=-x

In the interval [- 1,1], take x1,x2, and x10 is f(x1)-f(x2)>0

f(x1)>f(x2)

It is a subtractive function.

When x=-1 there is a maximum value, and the maximum value is 1

First of all, you have to understand what an odd function is, on [-1,1] it is an odd function, then at x=0, its value should be 0, but from the expression of the function you gave, x=0 seems to be a singularity, you first check whether your question has been copied wrong.

The function you are giving is a piecewise function, and the exact way to write it should be:

g(x)=(1/2)x^2+1 (x<0) g(x)=(-1/2)x^2-1 (x>0)。This indicates:

When the independent variable x<0 the function correspondence is g(x)=(1 2)x 2+1, when the independent variable x>0 the function correspondence is g(x)=(-1 2)x 2-1.

So when x>0, then -x<0, thus.

g(-x)=-1 2(-x) 2-1 =-1 2x 2-1 =-(1 2x 2+1)= -g(x) is not true, the correct one should be: since the independent variable -x<0, the function correspondence is g(x)=(1 2)x 2+1, so g(-x)=(1 2)(-x) 2+1 =(1 2)x 2+1 = -g(x).

No, when x>0, then -x<0

g(-x)=(-x)^2+1=x^2+1

g(x)=-1/2x^2-1

The two have nothing to do with each other.

g(x) is neither an odd nor an even function.

It is known that f(x) is a function defined on the set of real numbers r, and satisfies f(x+2)=-1 f(x).

f(x+4)=-1 f(x+2)=1 f(x) is a function of 4 periods, f(1)=-1 8

f(2007)=f(3+4*501)= f(3)f(1+2)

1/f(1)

Periodic functions? Looks like it did.

f(x+4)=-1 f(x+2)=f(x), so the period of the function is 4f(2007)=f(3).

f(3)=-1/f(1)=8

f(2007)=8

f(x+1) is an even function, f(x+1)=f(-x+1), i.e. the axis of symmetry is x=1, f(x)=f(2-x).

Let x>1, then -x<-1, i.e., 2-x<1

At x<1, f(x)=x +1, so there is f(2-x)=(2-x) 2+1, and when x>1, f(x)=f(2-x)=(2-x) 2+1=x 2-4x+5

1 let a=b=0 f(0)=2f(0) f(0)=0

Let a=x b=-x f(0)=f(x)+f(-x)=0 function be a strange function.

2 f(12)=4f(3)=-4f(-3)=-4a