A function problem in the first year of senior high school: it is known that the odd function f (x) over R satisfies f (x-4) = - f (x), and is an increasing function in the interval [0,2] It is known that the odd function f (x) over R satisfies f (x-4) = - f (x) and is an increasing function in the interval [0,2]. If the equation f (x) = m (M > 0) has four different roots x1, X2, X3, X4 in the interval [- 8,8], find X1 + x2 + X3 + X4 Please give the process.. thank you..

A function problem in the first year of senior high school: it is known that the odd function f (x) over R satisfies f (x-4) = - f (x), and is an increasing function in the interval [0,2] It is known that the odd function f (x) over R satisfies f (x-4) = - f (x) and is an increasing function in the interval [0,2]. If the equation f (x) = m (M > 0) has four different roots x1, X2, X3, X4 in the interval [- 8,8], find X1 + x2 + X3 + X4 Please give the process.. thank you..

Let x = t + 2 be substituted into f (x-4) = - f (x) to get f (T + 2-4) = - f (T + 2), that is, f (T-2) = - f (T + 2) and f (x) is an odd function f (T-2) = - f (2-T), so - f (T + 2) = - f (2-T), that is, f (2 + T) = f (2-T) (1), that is, the straight line x = 2 is the axis of symmetry of F (x)
The known set a = {x belongs to R | ax2-3x + 2 = 0}
1. If a is a single element set, find the value of a and the set a
2. Find the set P = {a belongs to R | a such that a contains at least one element}
1. When {a = 2, a = 3};
When a ≠ 0, △ = 0, that is, when a = 9 / 8, a = {4 / 3}
2. There are two situations: when a = 0, it is obviously in line with the meaning of the question;
When a ≠ 0, only △ 0 is needed, that is, a ≤ 9 / 8 and a ≠ 0;
In conclusion: set P = = {a | a ≤ 9 / 8}
What is the Veda theorem of quartic equation
This is a problem in solving such equations
A (known) = cosacosb sinaasinb
B (known) = sinacosb + cosasinb
Sin square + cos square = 1 (2)
Then find the specific values of the four trigonometric functions, where angle AB is the angle in the triangle
This has nothing to do with Vader's theorem
cosAcosB-sinAsinB=cos(A+B)=a
sinAcosB+cosAsinB=sin(A+B)=b
If f (x), an odd function defined on R, satisfies f (x-4) = - f (x) and is an increasing function on interval [0,2], then ()
A. f（15）＜f（0）＜f（-5）B. f（0）＜f（15）＜f（-5）C. f（-5）＜f（15）＜f（0）D. f（-5）＜f（0）＜f（15）
∵ f (x) satisfies that f (x-4) = - f (x), ∵ f (X-8) = f (x), ∵ function is a periodic function with period 8, then f (- 5) = f (3) = - f (- 1) = f (1), f (15) = f (- 1), and ∵ f (x) is an odd function on R, f (0) = 0, then f (0) = 0, and ∵ f (x) is in the interval [0, 2]
The known set a = {X / ax-3x-4 = 0, X belongs to R}
(1) If there are two elements in a, find the value range of real number a;
(2) If there is at most one element in a, find the value range of real number a
(1) If there are two elements in a, then ax-3x-4 = 0, if there are two unequal real roots, then a ≠ 0 and 3 & # 178; + 4 * 4 * a > 0, then a ≠ 0, if there is at most one element in a ＞ - 9 / 16 (2) a, then ax-3x-4 = 0 has two equal real roots, or if there is no real root, then a = 0 or a ≠ 0 and 3 & # 178; + 4 * 4 * a ≤ 0, then a = 0 or a ≠ 0 and a ≤ - 9 / 16
Exercises related to quadratic equation of one variable (Weida theorem)
1. Given that ab ≠ 0, the coefficients of the equation AX & # 178; + BX + C = 0 satisfy (B / 2) &# 178; = AC, then the ratio of the two equations is
A：0:1 B：1:1 C:1:2 D:2:3
2. The bivariate quadratic equation with the reciprocal of two of the equations X & # 178; - 6x + 2 = 0 as the root (the coefficient of quadratic term is 1) is?
3. If n ＞ 0, the equation x & # 178; - (m-2n) x + 1 / 4nm = 0 about X has two equal positive real roots, find the value of M / N?
Quadratic function:
If the minimum value of quadratic function y = ax & # 178; + 4x + A-1 is 2, then a =?
Given X & # 178; + Y & # 178; + 4x-6y + 13 = 0, then the value of the Y power of X is?
1. (B / 2) &# 178; = AC B & # 178; = 4ac B & # 178; - 4ac = 0 △ = B & # 178; - 4ac = 0 the equation AX & # 178; + BX + C = 0 has only one real root, or two equal real roots. Therefore, the ratio of the two roots of the equation is 1:1, and B2 is selected. Let X & # 178; - 6x + 2 = 0 be X1 and X2, then X1 + x2 = 6, X1 * x2 = 2
I don't know
The odd function f (x) defined on R is an increasing function, and the image of even function g (x) on the left closed right open interval from zero to positive infinity coincides with the image of F (x). Let a > b > 0
f（b）-f（-a）>g(a)-g(-b)
f（b）-f（-a）g(b)-g(-a)
f（a）-f（-b）
f（b）-f（-a）>g(a)-g(-b)
f（a）-f（-b）>g(b)-g(-a)
The odd function f (x) on R is an increasing function,
The image of even function g (x) on the left closed right open interval from zero to positive infinity coincides with the image of F (x),
a>b>0
[f(b)-f(-a)]-[g(a)-g(-b)]=f(b)+f(a)-g(a)+g(b)>0
[f(b)-f(-a)]-[g(b)-g(-a)]=f(b)+f(a)-g(b)+g(a)>0;
So: F (b) - f (- a) > G (a) - G... expansion
The odd function f (x) on R is an increasing function,
The image of even function g (x) on the left closed right open interval from zero to positive infinity coincides with the image of F (x),
a>b>0
[f(b)-f(-a)]-[g(a)-g(-b)]=f(b)+f(a)-g(a)+g(b)>0
[f(b)-f(-a)]-[g(b)-g(-a)]=f(b)+f(a)-g(b)+g(a)>0;
So: F (b) - f (- a) > G (a) - G (- b)
F (b) - f (- a) > G (b) - G (- a)
Given u = {x | x ^ 2-3x + 2 > = 0}, a = {x | X-2 | > 1 |, B = {x | (x-1) / (X-2) > 0, find a ∪ B, a ∩ B, (CUA) ∪ B, a ∩ (cub), CUA, cub
From x ^ 2 to 3x + 2 ≥ 0
(X-2) (x-1) ≥ 0
The solution is x ≥ 2 or X ≤ 1
So u = {x | x ≤ 1 or X ≥ 2}
From | X-2 | > 1
We get (X-2) ^ 2 > 1
We get x ^ 2-4x + 3 > 0
(x-3) (x-1) > 0
The solution is x > 3 or x0
Get x > 2 or X
x²-3x+2>=0
(x-2)(x-1)>=0
x> = 2 or x = 2 or X1 x > 3 or X3 or X0
(x-1)>0 x-2>0 x>2
X-13 or X3 or X0
(x-1)>0 x-2>0 x>2
X-1
How to deduce the formula of quadratic equation of one variable
It's a derivation process, not a formula. That's how this formula came into being
Ax & # 178; + BX + C = 0 divide both sides by a X & # 178; + (BX / a) + C / a = 0, add formula item (B / 2a) &# 178; X & # 178; + (BX / a) + (B / 2a) &# 178; + C / a = (B / 2a) &# 178; the left side is the complete square formula, and move C / A to the right side (x + (B / 2a)) &# 178; = (B / 2a) &# 178; - (C / a) right
The odd function f (x) defined on R is an increasing function, and the image of even function g (x) in the interval [0, positive infinity] coincides with the image of F (x)
Let a > b > 0, the following inequality is correct
①f(b)-f(-a)＞g(a)-g(-b)
②f(b)-f(-a)＜g(a)-g(-b)
③f(a)-f(-b)＞g(b)-g(-a)
④f(a)-f(-b)＜g(b)-g(-a)
1. If f (x) = 0, a > b > 0, then f (a) > F (b) > 0, f (a) = - f (- a), f (b) = - f (- b), f (a) = g (a), f (b) = g (b), G (a) = g (- a), G (b) = g (- b) prove that f (b) - f (- a) > G (a) - G (- b), that is, f (b) + G (- b) > G (a) + F (- a) because f (b) + G (- b) = 2F b) > 0, G (a) + F (- a) = 0