Are even functions even? If x ^ 2 + B, x + 3 is even, B should be equal to 0 because it is odd Are even functions even? If x ^ 2 + B, x + 3 is even, B must be equal to 0 because it is odd. Then is 3 not odd? Does it become a contradiction between non odd and non even? Does the often said odd and even term mean that the exponent of X is odd or even?

Are even functions even? If x ^ 2 + B, x + 3 is even, B should be equal to 0 because it is odd Are even functions even? If x ^ 2 + B, x + 3 is even, B must be equal to 0 because it is odd. Then is 3 not odd? Does it become a contradiction between non odd and non even? Does the often said odd and even term mean that the exponent of X is odd or even?

Even function is an independent variable, and the degree of X is even. For example, the reason why you want B = 0 is to ensure that there is no item of degree 1 of X. but mathematically, the definition of dual function is that f (x) = f (- x) under the condition that the domain of definition is symmetric with respect to X. The even term of odd term means that the degree of X is odd or even

If the power of X is even (y = n power of x), then this function is even?

It's best to draw a picture if you can't tell the difference, as long as the domain is symmetrical
Image definition of even function: About Y-axis symmetry
Image definition of odd function: on origin symmetry
On the other hand, we can prove that even functions are symmetric about the Y-axis and odd functions are symmetric about the origin

Why is the product of an odd function and an even function odd

Let f (x) and G (x) be odd and even functions respectively
Let y (x) = f (x) g (x)
Then y (- x) = f (- x) g (- x) = - f (x) g (x) = - Y (x)
So y (x) is an odd function

Given that f (x) is an odd function defined on R and satisfies f (x + 2) = - f (x), when 0 ≤ x ≤ 1, f (x) = 12x, then the value of X of F (x) = 12 is ()
A. 2n（n∈Z）B. 2n-1（n∈Z）C. 4n+1（n∈Z）D. 4n-1（n∈Z）

∵ f (x) is an odd function and f (x + 2) = - f (x), ∵ f (x + 4) = - f (x + 2) = f (x) ∵ the period of F (x) is t = 4. ∵ when 0 ≤ x ≤ 1, f (x) = 12x, and f (x) is an odd function, ∵ when - 1 ≤ x ≤ 0, f (x) = 12x, let 12x = - 12, the solution is: x = - 1, and the function f (x) takes 4 as a cycle

The odd function f (x) = [(- 2 ^ x) + b] / {[2 ^ (x + 1)] + a} defined on R
1, find the value of a and B
The function defined on R is an odd function f (x) = [(- 2 ^ x) + b] / {[2 ^ (x + 1)] + a}
Finding the values of a and B

You should understand the concept of odd function before doing the problem
1. In the odd function f (x), f (x) and f (- x) have opposite signs and equal absolute values, that is, f (- x) = - f (x). On the contrary, the function y = f (x) satisfying f (- x) = - f (x) must be odd. For example, f (x) = x ^ (2n-1), n ∈ Z; (f (x) is equal to the 2N-1 power of X, n is an integer)
2. The image of odd function is centrosymmetric with respect to the origin (0,0)
3. The domain of definition of odd function must be symmetric about the center of origin (0,0), otherwise it cannot be an odd function
4. If f (x) is an odd function and X belongs to R, then f (0) = 0
According to the fourth article, f (0) = 0, x = 0 is taken into the solution, 0 = (- 1 + b) / (2 + a), - 1 + B = 0, B = 1;
According to the first, f (- 1) = - f (1), then (- 1 / 2 + b) / (1 + a) = - (- 2 + b) / (4 + a), bring in B, (1 / 2) / (1 + a) = 1 / (4 + a), get a = 2
You have mastered the concept and characteristics of odd function. It's very easy to do this kind of problem. Have you learned it?

Given the function & nbsp; f (x) = x2-2 | x | - 1, try to judge the parity of function f (x) and draw the image of function

The domain of definition is R. for any x ∈ R, f (- x) = (- x) 2-2 | - x | - 1 = x2-2 | - 1 = f (x). Therefore, y = f (x) is an even function. When x > 0, f (x) = x2-2x-1, so the image of the function is as follows:

To determine whether the function is symmetrical about the origin, we only need to determine if f (x) is equal to 0 when x = 0. If it is symmetrical about the origin, then take - x into f (x) to find out the parity? What's the role of the domain? Can senior people answer? Is f (x) = cosx symmetrical about the origin?

The definition of even function is f (x) = f (- x), that is to say, a function in which - x is substituted into a formula and the formula remains unchanged. In the coordinate system, its image is symmetrical about the y-axis. The definition of odd function is f (x) = - f (x), that is, a function in which - x is substituted into a formula and all terms of the formula change sign

The function y = (9-x ^ 2 under the root sign) divided by (| x + 4 | + | x-3 |), that is, the numerator is 9-x ^ 2 under the root sign, and the denominator is | x + 4 | + | x-3 |

Because 9-x & sup2; ≥ 0, so x & sup2; ≤ 9, - 3 ≤ x ≤ 3, so x + 4 > 0, | x + 4 | = x + 4; x-3 ≤ 0, | x-3 | = 3-x, so y = f (x) = [√ (9-x & sup2;)] / [(x + 4) + (3-x)] = [√ (9-x & sup2;)] / 7, so f (- x) = [√ (9 - (- x) & sup2;)] / 7 = f (x) and

Given that f (x + 1) is odd, f (x-1) is even, and f (0) = 2, then f (4) =?
The answer is as follows:
Because f (x-1) = f (- x-1)
So f (0) = f (- 2) = 2
From F (x + 1) = - f (- x-1)
We get f (4) = - f (- 2) = - 2

f(X-1)=f(-X-1)
Let x = 1
Then f (0) = f (- 2) = 2
f(X+1)=-f(-X-1)
Let x = 3
Then f (4) = - f (- 2) = - 2

（1）f（x）=1/x+2
(2) F (x) = the third power of X + 2
(3) F (x) = root X of degree 3

F (- x) = 1 / (- x) + 2 = - 1 / x + 2 = - (1 / X-2), non parity
F (- x) = (- x) ^ 3 + 2 = - x ^ 3 + 2 = - (x ^ 3-2), non parity
F (- x) = cubic root (- x) = - cubic root x = - f (x), odd