High school mathematics about odd function and even function It is known that f (x) is even function, G (x) is odd function, and f (x) + G (x) = 1 / (x-1) So how can f (- x) + G (- x) = 1 / (- x-1) be obtained?

Function parity is not used, just change x to - X
This should be the intermediate process of a problem. The parity property is obtained f (x) - G (x) = 1 / (- x-1), and then f (x) and G (x) can be calculated respectively

Content and concept of odd function and even function in high school mathematics

Odd function f (- x) = - f (x), even function f (- x) = f (x), odd function is symmetric about origin, even function is symmetric about X axis

The properties of odd function and even function?

Odd function: 1. In the odd function f (x), the sign of F (x) and f (- x) are opposite and the absolute values are equal, that is, f (- x) = - f (x), on the contrary, the function y = f (x) satisfying f (- x) = - f (x) must be odd function. For example: F (x) = x ^ (2n-1), n ∈ Z; (f (x) is equal to the 2N-1 power of X, n belongs to integer) 2

The properties of odd function and even function?

The properties of odd function: 1. The image is symmetric about the origin. 2. Satisfy f (- x) = - f (x) 3. Monotonicity is consistent in the interval of symmetry about origin. 4. If the odd function is defined on x = 0, then there is f (0) = 0 5. The definition domain is symmetric about the origin (shared by odd and even functions). Even function properties: 1. Image is symmetric about y-axis

Properties of odd function and even function

The image of odd function is symmetric about the origin
Even function images are symmetric about y values
So you don't have to worry about the properties. Every time you think about an odd function, you want to be symmetrical about the origin. You can think of its properties. F (x) = - f (- x) for example, y = x, the odd function
Even functions are y = x squared
f(x)=f(-x)

It is known that the proposition p: equation 4x ^ 2 + 4 (m-2) x + 1 = 0 has no real root; proposition q: function y = radical (MX ^ 2 + MX + 1) has no real root If "P or Q" is true, "P and Q" is false, the value range of real number m is obtained

If "P or Q" is true, "P and Q" is false
Then p is true, q is false, or P is false and Q is true
(I)
When p is true and Q is false
Δ1=16(m-2)^2-16＜0
Δ2=m^2-4m＞0
M has no solution
(ii)
When p is false and Q is true
Δ1=16(m-2)^2-16≥0
Δ2=m^2-4m≤0
So 0 ≤ m ≤ 1 or 3 ≤ m ≤ 4
Therefore, the value range of real number m is {m | 0 ≤ m ≤ 1 or 3 ≤ m ≤ 4}
If you do not understand, please hi me, I wish you a happy study!

It is known that the proposition p: equation x2 + MX + 1 = 0 has two unequal negative real roots. Proposition q: equation 4x2 + 4 (m-2) x + 1 = 0 has no real roots. If P or q are true, P and Q are false, then the range of real number m is () A. （1，2]∪[3，+∞） B. （1，2）∪（3，+∞） C. （1，2] D. [3，+∞）

If P is true, then
m2−4＞0
- M < 0, M > 2;
If q is true, then △ = [4 (m-2)] 2-16 < 0, the solution is: 1 < m < 3;
∵ P or q is true, P and Q are false,
/ / P and Q are true and false,
When p is true and Q is false, m ≥ 3 is obtained; when p is false and Q is true, the solution is 1 < m ≤ 2
In conclusion, 1 ＜ m ≤ 2 or m ≥ 3;
Therefore, a

It is known that proposition p: equation x ^ 2 + MX + 1 = 0 has two unequal negative roots; proposition q: equation 4x ^ 2 + 4 (M + 2) x + 1 = 0 has no real root,

P or q is a false proposition, indicating that "P is a false proposition" and "q is a false proposition" proposition p: x ^ 2 + MX + 1 = 0 has two unequal negative roots

It is known that the proposition p: equation x2 + MX + 1 = 0 has two unequal negative real roots. Proposition q: equation 4x2 + 4 (m-2) x + 1 = 0 has no real roots. If P or q are true, P and Q are false, then the range of real number m is () A. （1，2]∪[3，+∞） B. （1，2）∪（3，+∞） C. （1，2] D. [3，+∞）

Proposition p: equation x ^ 2 + MX + 1 = 0 has two unequal positive real roots, proposition q: equation 4x ^ 2 + 4 (M + 2) x + 1 = 0 has no real root Proposition p: equation x ^ 2 + MX + 1 = 0 has two unequal positive real roots, proposition q: equation 4x ^ 2 + 4 (M + 2) x + 1 = 0 has no real root. If P or q are true propositions, find the value range of M

P: If the discriminant is m 2-4 > 0, then M > 2 or m < - 2
Q: If the discriminant is 16 (M + 2) 2 - 16 < 0, then: - 1 because P or q is true, then at least one of P and Q is true
If P and Q are all false, then: - 2 ≤ m ≤ - 1
The result is: m < - 2 or M > - 1

High school mathematics about odd function and even function It is known that f (x) is even function, G (x) is odd function, and f (x) + G (x) = 1 / (x-1) So how can f (- x) + G (- x) = 1 / (- x-1) be obtained?