A few high school math problems about parity

1.f(x) is an odd function, f(-x)=-f(x)g(x) is an even function, and g(-x)=g(x).

The above formula is established, so substitute -x into it, and you get.

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

According to the above properties, g(x)-f(x)=x 2+x+2 is a simultaneous g(x)+f(x)=x 2-x+2

The addition of the two formulas yields g(x)=x 2+2

Substituting g(x)+f(x)=x 2-x+2.

f(x)=-x g(x)=x²+2

2.From f(0)=0, m=0 and n belong to r

is an increasing function defined on 0 to positive infinity, and for everything x,y>0, satisfying f(x y) = f(x)-f(y).

Let x=y=1

f(1)=f(1)-f(1)=0

f(1)=0

f(6)=1

Let x=36, y=6

f(36/6)=f(36)-f(6)

f(6)=f(36)-f(6)

2f(6)=f(36)

f(36)=2

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

For all x,y>0, f(x y)=f(x)-f(y)f[(x+3) (1 3)]<2=f(36)f(3x+9)0

So the solution of the inequality is.

At 0, f(x) ax (x 2-1) a (x 1 x) functions x, 1 x are increments in the defined domain, so x 1 x is increments in (1,0) and (0,1). again a 0, so that f(x) is a subtraction function in (-1,1).

5.Is there a mistake in the title? The inference should be.

Is the image symmetrical about x= -2? If so, then.

x [ 6, 2], the expression f(x) is .

f(x) =－(x+4)² 1

Uh: Are you asking someone else to do your homework?

Even functions.

f(-x)=-x fraction numerator and denominator multiply by 2 x

x[2 x (1-2 x)+1 2] coefficient -1 with parentheses =x[2 x (2 x-1)-1 2] molecule-1 and then +1, and then split the fraction.

x[1+1 (2 x-1)-1 2] simplify=x[1 (2 x-1)+1 2].

f(x)

Man, you're messing up the brackets and the meaning isn't clear.

That's right. The odd function f(x) that defines (positive infinity, negative infinity) as an increasing function, must have f(x) 0 in the interval (0, positive infinity) and g(x) 0 in the interval (0, positive infinity).

And because the image of the even function g(x) in the interval (0, positive infinity) coincides with the image of f(x), and a>b>0, f(a) = g(a) f(b) = g(b) 0

1、f(b)-f(-a)=f(b)+f(a)g(a)-g(-b)=g(a)-g(b)

f(b)-f(-a) g(a)-g(-b) is correct.

3、f(a)-f(-b)=f(b)+f(a)g(b)-g(-a)=g(b)-g(a)

f(a)-f(-b)＞g(b)-g(-a)

Let x [-6,-2], then x+4 [-2,2] thus x+4 satisfy.

f(x+4)=-x+4)^2+1

f(x) is symmetrical to x=2, and f(2+x)=f(2-x) can obtain f(4-x)=f(x).

The ruler is f(x+4)=f(-x)=f(x), so f(x)=-x 2-8x-7,x [-6,-2].

f(x)=loga(x+1)

x+1>0

x>-1 (i)

g(x)=loga(1-x)

1-x>0

x<1 (ii)

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

Synthesis (i)(ii).

The domain of -1f(x) is (-1,1).

f(-x)=loga(-x+1)-loga(1+x)=loga(1-x)-log(x+1)

f(x) and therefore f(x) is an odd function.

The defined domain is (1,1).

As for parity, if you change x to x, you will find that the subtraction is swapped with the subtracted number, so it becomes the original inverse, i.e., f(-x)=-f(x).

Therefore, the original function is an odd function.

(1)x+1>0 1-x>0, then -1(2)f(-x)=f(-x)-g(-x).

Because f(-x) = g(x).

g(-x)=f(x)

So f(-x)=g(x)-f(x)=-f(x) is an odd function.

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

cf(x) is an even function, then f(x)=f(-x).

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

The odd function is because: when x>0, f(x)=x2+2 f(-x)=-x2-2 f(x)=f(-x).

Similarly; When x<0f(x)=-x2-2 f(-x)=x22 f(x)=f(-x).

x=0, needless to say.

By definition, it should be an odd function.

It is observed that f(x) is an odd function because it agrees.

f(-x)=-f(x)

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

Also, when x=0, f(x)=0 Unfortunately, the problem omits the odd function marked as the domain r, in addition to f(-x)=-f(x), it must also satisfy f(0)=0

The domain of this function is rSo don't miss it.

Drawing an image is about origin symmetry.

So, it's definitely an odd function.

Have fun! I hope it can help you, if you don't understand, please hi me, I wish you progress in your studies!

It is an odd function, which can be intuitively known through the image. Of course, it can also be proven!

Odd function, you might as well set x 0, -x 0, f-x=-x 2-2=-fx, the same reason can be proved when x 0 is true, because f0=0, is an odd function.

It is an odd function, according to the definition of odd function and even function to prove, first define the symmetry of the domain with respect to the origin, secondly, assume that x is greater than 0, then find f(-x), substitute negative x into the second expression, and then sort it out. is f(-x) = —f(x). So it's an odd function.

2.It is known that the even function f(x) increases monotonically at [0, positive infinity).

The value range of f(2x-1)> f(1 3) x is satisfied.

2x-1＞1/3

x＞2/32x-1＜-1/3

x＞-1/3

In where r is the odd function f(x) satisfies f(x+3)=f(x) and f(1)>0f(2)=(2m-3) m+1, we find the range of m f(1) = f(-2) >1

f(x) is an odd function.

So f(2) = -f(-2) <1 both (2m-3) m+1 < 1

Just solve the inequality (I don't know if your 1 is under the denominator).

1. According to the property of the even function, f(x)=f(-x), so (m-1)+2m+3=(m-1)-2m+3, so m=0, then f(x)=f(-x)=x)=x)=x=x, 2+3

Let x 2+3=t, then f(x)=f(t)=x 2+3=-3 4 f(a 2-a+1) is always greater than 0, and f f(a 2-a+1) when equal to 0

x-1>1 3 gives x>2 3

3. Let 2=-x then f(t)=f(x) according to the property of the odd function f(-x)=-f(x).

1.Because f(x) is an even function, m=0

So f(x)=-x 2+3 a 2-a+1=(a-1 2) 2+3 4 3 4

When a=1 2, a 2-a+1=3 4 f(-3 4) = f(3 4) because f(x) is an even function

When a≠1 2 f(a 2-a+1) f(-3 4).

2x-1f(a^2-a+1)≥f(-3/4)

2.Because the even function f(x) increases monotonically at [0, positive infinity). So f(x) decreases monotonically at (negative infinity, 0).

So 1When 2x-1 0, i.e. x 1 2, then 2x-1 1 3, so x 2 3

2.When 2x-1 0 is x 1 2 then 2x-1 1 3 so x 1 2

In summary, x 2 3 or x 1 2

3.Since f(x) is an odd function in r, f(0)=0 so f(3)=f(0)=0

And because f(1)>0, so f(2)=(2m-3) m+1 0, so m 1 or m 1 2