If 0 ≤ x < 3, f (x) = x & sup2; + 2x, if f (3) = f (0), find the analytic expression of F (x)

Because even functions are symmetric about the Y axis, f (x) = f (- x)
When you think about X

The function f (x) is an even function on the domain R. when x belongs to (0, + 00) is, f (x) = x ^ 2 + 2x, then when x belongs to (- 00,0), f (x) is an analytic expression

When x belongs to (0, + ∞), f (x) = x & # 178; + 2x
So when x belongs to (- ∞, 0)
-X belongs to (0, + ∞)
So f (- x) = (- x) &# 178; + 2 (- x) = x & # 178; - 2x = f (x)
So when x belongs to (- ∞, 0), f (x) = x & # 178; - 2x

The functions y = f (x) and y = g (x) are both odd functions whose domain of definition is R. how to prove that y = f (x) * g (x) is even function
Don't jump. I don't understand.

Since f (x) and G (x) are all odd functions with the domain of R, so
The domain of F (x) g (x) is r, which is symmetric about the origin
And
f（-x）=-f（x）,g（-x）=-g（x）
Let H (x) = f (x) g (x)
Then H (- x) = f (- x) g (- x) = - f (x) * - G (x)) = f (x) g (x) = H (x)
So, H (x) = f (x) g (x) is even function

Let f (x) be defined as (- L, l). It is proved that there must exist even and odd functions H (x) on (- L, l) such that f (x) = g (x) + H (x)
It is proved in the book that if G (x) and H (x) exist, f (x) = g (x) + H (x), (1),
And G (- x) = g (x), H (- x) = - H (x)
So f (- x) = g (- x) + H (- x) = g (x) - H (x), (2)
Using (1) and (2), we can make g (x) and H (x)
g(x)=[f(x)+f(-x)]/2
h(x)=[f(x)-f(-x)]/2
Then G (x) + H (x) = f (x),
g(-x)=[f(-x)+f(x)]/2=g(x),
h(-x)=[f(-x)-f(x)]/2=h(x).
The proof is complete
I didn't understand the problem, and I didn't understand the process
What are the conditions and conclusions of this problem?
What's the process of proof above? Some people say it's counter evidence, but it seems it's not?
This is to prove that any function on (- L, l) can be represented by the sum of an odd function and an even function. How can it be so unclear? Thank you~~

It is proved that there exist even and odd functions H (x) on (- L, l) such that f (x) = g (x) + H (x)
If f (x) is defined as (- L, l)
Suppose g (x) and H (x) exist, such that f (x) = g (x) + H (x), (1),
And G (- x) = g (x), H (- x) = - H (x)
So f (- x) = g (- x) + H (- x) = g (x) - H (x), (2)
These sentences are inevitable, without proof or any conditions, and are purely constructions
It's just a foreshadowing to introduce g (x) and H (x)
It is mainly to prove that one of the two functions is odd and the other is even, which is the core of the proof,
As long as we find an odd function and an even function to represent f (x), the proof is complete
So here's the following statement
g(x)=[f(x)+f(-x)]/2
h(x)=[f(x)-f(-x)]/2
Then G (x) + H (x) = f (x),
g(-x)=[f(-x)+f(x)]/2=g(x),
h(-x)=[f(-x)-f(x)]/2=h(x).
It is to express g (x) and H (x) by F (x)
Then we prove that one of F (x) g (x) is an odd function and the other is an even function

The number a is 120% of the number B. how many percent more is the number a than the number B?

The number a is 120% of the number B. this sentence indicates that the unit B is "1"
120%-1=120%-100%=20%
A: number a is 20% more than number B

If | M-1 | + | n-3 | = 0, then the value of (m-n) 3 is ()
A. 6B. -6C. 8D. -8

According to the meaning of the question, M-1 = 0, n-3 = 0, the solution is m = 1, n = 3, so, (m-n) 3 = (1-3) 3 = - 8

The average number of the three numbers is 4, and their ratio is 23:56:12. The smallest number is 4______ ．

23:56:12 = (23 × 6): (56 × 6): (12 × 6) = 4:5:3, 4 + 5 + 3 = 12, 4 × 3 × 312 = 3; answer: the smallest number is 3; so the answer is: 3

In a right triangle, if the sum of two right sides is 8, what is the maximum area of the triangle?

If the sum of two numbers (integers) is fixed, the maximum value of their score is the closest value of the sum divided by 2
therefore
The right angle side is 4,4
The maximum area is 4 * 4 / 2 = 8

98 divided by [120 times (60% plus 1 / 3)]

98 divided by [120 times (60% plus 1 / 3)]
=98÷（120x60%+120x1/3）
=98÷（72+40）
=98÷112
=7/8
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Let a be mxn matrix and B be nxm matrix, then the system of linear equations ABX = 0
The answer is that when m > N, there must be a non-zero solution?

When m > N, R (a)

If 0 ≤ x < 3, f (x) = x & sup2; + 2x, if f (3) = f (0), find the analytic expression of F (x)