Senior 1 Mathematics Compulsory 1 Parity .

When x 0, -x 0

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

So f(x) is the piecewise function, and the analytic formula is.

At x 0, f(x)=x +x-1

When x=0, f(x)=0

At x 0, f(x)=x +x-1

It is implied that f(x) is an even function and f(x) and g(x) is an odd function.

f(-x)g（-x）=-f（x）g（x）

i.e. g(-x)=-g(x).

f（-x）+g（-x）=-1/(x+1)

i.e. f(x)-g(x)=-1 (x+1)--f(x)+g(x)=1 (x-1)--

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

f(x)=1/(x²-1)

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

g(x)=x/(x²-1)

1) Because x>0, -x<0, f(-x)=-x 3-x+1=-f(x)

So f(x)=x3+x-1

2) Because f(x) is an even function and f(x) g(x) is an odd function, g(x) is an odd function

Since f(x)+g(x)=1 (x-1)one, then f(-x)+g(-x)=1 (-x-1), i.e., f(x)-g(x)=1 (-x-1)two, one + two gives 2f(x)=1 (-x-1)+1 (x-1)=2 (x-1)*(x+1), then f(x)=1 (x-1)*(x+1).

Substituting one yields g(x)=x(x+1)*(x-1).

13、f(8)=f(5+3)=f(5)=f(2+3)=f(2)=f(-1+3)=f(-1)=-f(1)=-1

14. f(x)=bx +a(2+b)x+2a is an even function, then there is no odd term, that is: a(2+b)=0, (1)a=0, then f(x)=bx, regardless of whether the opening is up or down, the value range is not (- 4], rounded.

2) 2+b=0, i.e. b=-2, then f(x)=-2x +2a, to make the range (- 4], then 2a =4;

So: f(x)=-2x +4

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

13.Since f(x+3)=f(x) so f(x) is a function with a period of 3, so f(8)=f(5)=f(2)=f(-1).

And f(x) is an odd function, so f(x)=-f(-x), that is, with respect to origin symmetry, it is known that f(x)=x 2 when 0<=x<=1

So when -1<=x<=0, f(x)=-x 2, so f(-1)=-1, i.e., f(8)=-1

Since it is an even function, f(x) = f(-x), i.e. f(x) cannot have a term for x.

Therefore, 2a+ab=0......(1)

Because the value range is <=4, b<0, the parabolic opening is upward, and the maximum value is obtained when x=0.

f(0)=2a^2=4……（2）

Solving equations (1) and (2) yields a = root number 2 and b = -2

So f(x)=-2x 2+4

13. -1 When -1 x 0, 0 -x 1, f(-x)=(-x) 2=x 2, f(x) is an odd function, f(-x)= -f(x).

f(x)= -x 2, according to f(x+3)=f(x), f(8)=f(5)=f(2)=f(-1), f(-1)= -(1) 2= -1

x 2 is the second power of x.

14. f(x)=(x+a)(bx+2a)=bx 2+(ab+2a)x+2a 2, the range of f(x) is (negative infinity,4), b 0,2a 2=4

a= root number2, f(x) is an even function, ab+2a=0, b= -2. ∴f(x)= -2x^2+4

It's been so hard for me to fight, can you answer it?

It should be right.

f(x+3)=f(x)

So f(8)=f(5+3)=f(5)=f(2+3)=f(2)=f(-1+3)=f(-1).

Since it is an odd function over the entire defined domain, there is f(-x)=-f(x)f(-1)=-f(1)=-1

f(8)=-f(-8)=-f(-5)=-f(-2)=-f(1)=-1 makes f(1)=f(-1) simplified to 2a+ab=0a=0, then it is not true, a≠0, b=-2, and substituting a=3 2 is like this.

Question 13: Because f(x+3)=f(x) so f(8)=f(5+3)=f(5)=f(3+2)=f(2)=f(-1+3)=f(-1) and because f(x) is an odd function, f(-1)=-f(1) =-1

It is known that f(x) and g(x)= are odd and even functions on (-a, a), respectively, so f(x)=-f(-x), g(x)=g(-x)

m(x)=f（x）·g(x)=-f(-x)*g(-x)=m(-x)

Let t(x)=f(x)·g(x) because f(x) and g(x)= are odd and even functions on (-a, a), respectively.

So f(-x)=-f(x) g(x)=g(x)so t(x)=f(x)·g(x).

t(-x)=f(-x)·g(-x)=-f(x)·g(x)so t(x)=-t(-x).

So it's an odd function.

Let f(x) be an odd function, g(x) be an even function, h(x)=f(x)·g(x), then f(x)=-f(-x), g(x)=g(-x), h(x)=f(x)·g(x)=-f(-x)·g(-x)=-h(-x).

h(x)=f(x)·g(x) is an odd function over (-a,a).

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.

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)

Originally, after (-a, f(-a)), and f(-a)=-f(a), the odd function that makes sense at the origin must pass the (0,0) point.

The high-return even function cannot have an odd power of x.

Others can also be y=|x|

7.Odd and even.

The odd Qi Tong function is added and subtracted, and the result is an odd function.

Even functions are added and subtracted, and the result is an even function.

Or we can verify the relationship between f(x) and f(-x) according to the meaning of the fixed world.

It's important to note that it's easy to overlook that the domain of an odd or even function must be symmetrical with respect to the origin.

At the same time, you can stop thinking about those problems, and quickly.

Because f(-x)=-f(x), let x<=0, that is, modulo -x>=0, then f(-x)=(x) (1-3 times the root number tells x);

i.e. -f(x)=(x)(1-3 times root number x), f(x)=x(1-3 times Sun Kaigen number x), x<=0.

1 Let x=y=1, get f(1)=0, and in the same way let x=y=—1, get f(-1)=0

2 Let y=-1 be brought in, we know that f(-x)=-f(x), so it is an odd function.