The problem of function parity, the problem of function parity

f(x)=x|sinx+a|+b is an odd function, then f(-x)=-f(x).

x|-sinx+a|+b=-x|sinx+a|-b is true for any x so that x=0 gets: b=-b, b=0

x|-sinx+a|=-x|sinx+a|-sinx+a|=|sinx+a|

Let x= 2, get |a-1|=|a+1|, a=0, so a=0 and b=0 are a=b=0, and the necessary condition is ab=0

is an odd function by f(x).

know f(0)=0

Both 0|sin0+a|+b=0

Therefore b=0 therefore ab=0

Since f(x) is an odd function, ab=0 can be deduced

Therefore, ab=0 is a necessary condition.

When a=2 b=0.

f(x)=x|sin(x+π/2)|+0=x|cosx|It is also an odd function.

So there are no sufficient conditions for these four conditions.

1.Necessity:

f(x)=x|sinx+a|+b is an odd function, then f(-x)=-f(x).

x|-sinx+a|+b=-x|sinx+a|-b is true for any x so that x=0 gets: b=-b, b=0

x|-sinx+a|=-x|sinx+a|-sinx+a|=|sinx+a|

Let x= 2, get |a-1|=|a+1|, a=0 so there is ab=0 a+b=0 a=b a 2+b 2=02Sufficiency:

If a=b, a 2+b 2=0;Then 2*a 2=0, i.e. a=0, then b=0.

Then f(x)= x|sinx+a|+b = x|sinx|f(0)=0;

f(-x)=-x|-sinx|=-x|sinx|=-f(x);

then f(x) is an odd function.

Odd function f(0) = 0, so b = 0 f (-x) = -f (x) only a also = 0 ah look at it, you also know, choose d

f(0)=0, the function crosses the origin.

Monotonically increasing on [0,1), on [0,1) there is a minimum value f(0)=0 and on (-1,0] there is a monotonically increasing value, and on (-1,0] there is a maximum value f(0)=0 on [0,1) that the y value cannot be smaller than the y value on (-1,0].

Since it is an odd function, since f(x) is an increasing function at x [0,1), then f(x) is also an increasing function at x [-1,0).

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)=2x-1 at [0,2], so f(x)=f(-x)=-2x-1f(2+x)=f(2-x) at [-2,0].

f(2+x+2)=f[2-(x+2)]=f(-x)=f(x) The period is 4

In [-4, -2].

f(x)=2x+3

is a piecewise function.

So f(x)=2x+3 x [-4,-2]f(x)=-2x-1 x belongs to [-2,0].

F(2+x)=f(2-x) so the period is 4 and f(x) is an even function.

Therefore, when x belongs to [-2,0], f(x)=-2x-1, because the period is 4, so when x belongs to [-4,-2], f(x+4)=2(x+4)-1=2x+7

Piecewise functions are good to write as soon as they are written.

f(2+x)=f(2-x), that is, f(2+x)=f[4-(2+x)] can be transformed into f(x)=f(4-x), and f(x) is an even function, that is, there is f(x)=f(-x), so f(4-x)=f(x-4), so f(x)=f(x-4) can also be transformed into f(x+4)=f(x)

x belongs to [-4,0] divided into two segments, [-4,-2] and (-2,0], when x belongs to [-4,-2], x+4 belongs to [0,2], and on the [0,2] interval, f(x)=2x-1, so when x belongs to [-4,-2], f(x)=f(x+4)=2(x+4)-1=2x+7

And when x belongs to (-2,0], -x belongs to [0,2), so f(x)=f(-x)=2(-x)-1=-2x-1

The symmetry of the function with respect to x=2 can be found, and since it is an even function, i.e., symmetry with respect to x=0, then the function can be graphed according to the known conditions.

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.

First of all, we can determine that the definition domain is symmetrical with respect to the origin, so that g(x)=f(x)+f(-x), so g(-x)=f(-x)+f(x)=g(x), which is an even function;

Let h(x)=f(x)-f(-x), so h(-x)=f(-x)-f(x)=-h(x), which is an odd function.

If for functions.

Define any x in the domain.

Or then functions.

It's called an even function. With respect to y-axis symmetry, if for functions.

Define any x in the domain.

Or then functions.

It's called an odd function. Regarding origin symmetry,. ⑶

If there is any x in the function definition field, there is.

and , (x r, and r is symmetrical with respect to the origin. Then the function.

It is both an odd and even function and is called both an odd and even function.

If there is an a a within the domain for the function definition, such that.

There is a b, so that.

Then the function. Functions that are neither odd nor even are called non-odd and non-even functions.

The defined domains are opposites of each other, and the defining domains must be symmetrical with respect to the origin.

The special one is both an odd and an even function.

Note: Odd and evenness are integral properties of a function, for the entire defined domain.

The domain of an odd and even function must be symmetrical with respect to the origin, and if the domain of a function is not symmetrical with respect to the origin, then the function must not be parity.

The comparison draws conclusions)

The basis for judging or proving whether a function is parity is by definition.

If an odd function.

If it makes sense at x=0, then the function must have a value of 0 at x=0. And about the origin symmetry.

If the function definition domain is not symmetric about the origin or does not meet the conditions of odd or even functions, it is called a non-odd and non-even function. For example.

or [(the definition domain is not symmetric with respect to the origin).

If a function conforms to both odd and even functions, it is called both odd and even. For example.

Note: Any constant function (which defines the domain with respect to the origin symmetry) is an even function, only.

is both odd and even functions.

According to the definition of parity functions.

g(x) = f(x) +f(-x)

g(-x) = f(-x) +f(x)

Thus g(x) is an even function.

h(x) = f(x) -f(-x)

h(-x) = f(-x) -f(x) = -[f(x) -f(-x)]

H(x) is therefore an odd function.

f(x) is the annihilation even function defined on r, and its gram is symmetrical at x=1, so the function is a periodic function of t=2.

Then an=f(1 2n).

a=f(1)=f(1/2+1/2)=f(1/4+1/4+1/2)……

Because f(x) is an odd function, f(-x) = -f(x) so when the god hall x<0, f(x) =f(-x) =lg(2-x) so the expression for f(x) is:

lg(x+2) (x>0).

LG(2-x) (at x<0).

0 (x=0)

This is a split-pure splitting segment function.

f(x)=lg(x+2) when x 0 and f(x)=-lg(-x+2) when x 0

When x=0, f(x)=0

Note that this function must be equal to 0 at x=0, because it is a function of the source macro on r.

The second equation is obtained according to the definition of the odd function.