Given that the image of the function f (x) = ax & # 178; + BX + C passes through the origin, for any x ∈ R, f (x-1) = f (x + 1) holds, and the equation f (x) = x has two equal real roots. Find the analytic expression of F (x) Let f (1-x) = f (1 + x) hold,

Given that the image of the function f (x) = ax & # 178; + BX + C passes through the origin, for any x ∈ R, f (x-1) = f (x + 1) holds, and the equation f (x) = x has two equal real roots. Find the analytic expression of F (x) Let f (1-x) = f (1 + x) hold,

F (x-1) = f (x + 1) is a typical characteristic of periodic function. The quadratic function can not be a periodic function, so there may be some mistakes here. It is very likely that f (1-x) = f (x + 1). Under this condition, we can see that x = 1 is the symmetric axis of the function image, so - B / 2A = 1.3
Is the title wrong? It should be f (x + 1) = f (1-x)
If f (x-1) = f (x + 1) holds, it should be f (1-x) = f (x + 1). Otherwise, it is a periodic function.
The image of function f (x) = ax & # 178; + BX + C passes through the origin and gets C = 0. For any x ∈ R, f (1-x) = f (x + 1) holds,
So the axis of symmetry is x = 1, B = - 2A, and ax ^ 2 + (B-1) x + C = 0 has two equal real roots
(B-1) ^ 2-4ac = 0, the solution is b = 1, a = - 1 / 2, so f (x) = - 1 / 2... Expansion
If f (x-1) = f (x + 1) holds, it should be f (1-x) = f (x + 1). Otherwise, it is a periodic function.
The image of function f (x) = ax & # 178; + BX + C passes through the origin and gets C = 0. For any x ∈ R, f (1-x) = f (x + 1) holds,
So the axis of symmetry is x = 1, B = - 2A, and ax ^ 2 + (B-1) x + C = 0 has two equal real roots
(B-1) ^ 2-4ac = 0, the solution is b = 1, a = - 1 / 2, so f (x) = - 1 / 2x ^ 2 + X
The sum ellipse 9x2 + 4y2 = 36 has the same focus and passes through points (2, - 3)
∵ the standard equation of the ellipse 9x2 + 4y2 = 36 is x24 + & nbsp; Y29 = 1 ∵ its focal coordinates are (0, ± 5) ∵ the ellipse and the ellipse 9x2 + 4y2 = 36 have the same focal points, ∵ let the elliptic equation be x2b + y2b + 5 = 1 ∵ the ellipse passes through the point (2, - 3) ∵ 22b + (− 3) 2B + 5 = 1 ∵ B = 10 ∵ and the ellipse 9x2 + 4y2 = 36 have the same focal points, and the equation of the ellipse passing through the point (2, - 3) is x210 + y215 = 1
The product of all real numbers with absolute value less than 3 is? Detailed
A detailed explanation of the question is helpful for the respondent to give an accurate answer
Answer: 0
Real numbers with absolute value less than 3 refer to all real numbers in (- 3, 3) interval, including 0
The product of any number and 0 is still 0
A detailed explanation of the question is helpful for the respondent to give an accurate answer
Is 0
Because real numbers with absolute values less than 3 include 0, and 0 times any number is 0
Although there are many real numbers with absolute values less than 3... But a 0 can guarantee that the final answer is 0
All products of real numbers with absolute value less than 3 are zero
According to the owner's idea, it should not include zero
There are two cases excluding zero
When the domain of definition is from zero to positive infinity, all products can be transformed into the form of reciprocal corresponding to opportunity, that is, (0,1] corresponding to {1, positive infinity) can find a combination of principal and reciprocal, and the products are all 1
The result is the same when defined from negative infinity to zero, and the product is also 1
Obviously, in the domain of definition, the real number of [1,3] can not reach the corresponding (0,1] interval full... Expansion
All products of real numbers with absolute value less than 3 are zero
According to the owner's idea, it should not include zero
There are two cases excluding zero
When the domain of definition is from zero to positive infinity, all products can be transformed into the form of reciprocal corresponding to opportunity, that is, (0,1] corresponding to {1, positive infinity) can find a combination of principal and reciprocal, and the products are all 1
The result is the same when defined from negative infinity to zero, and the product is also 1
Obviously, in the domain of definition in the title, the real number of [1,3] can't reach all the real numbers in the corresponding (0,1] interval
So the product approaches zero. The case of (- 3, - 1) is the same as that of [- 1,0]
The product should be infinitely approaching zero
I wish you success in your studies
The sum ellipse 9x2 + 4y2 = 36 has the same focus and passes through points (2, - 3)
∵ the standard equation of ellipse 9x2 + 4y2 = 36 is x24 + & nbsp; Y29 = 1 ∵ its focus coordinate is (0, ± 5) ∵ the ellipse and ellipse 9x2 + 4y2 = 36 have the same focus, ∵ let the ellipse equation be x2b + y2b + 5 = 1 ∵ the ellipse passes through the point (2, - 3) ∵ 22b + (− 3) 2B + 5 = 1 ∵ B = 10 ∵ and ellipse 9x2 + 4y2 = 3
Given that the function f (x) = | x2-4x | - M has four zeros, the value range of real number m is ()
A. M ＞ - 4B. M ＞ 0C. 0 ＜ m ＜ 4D. M ＜ 0 or m ＞ 4
The function y = | x2-4x | has four intersections with the function y = m, as shown in the figure: combining with the image, we can get 0 < m < 4, so we choose C
Find the standard equation of the ellipse which passes through the point P (- 2,3) and has a common focus with the ellipse 9x + 4Y = 36
The elliptic equation is written as
x²/4+y²/9=1
The focus is on the y-axis, c-178; = a-178; - b-178; = 9-4 = 5
C of the second ellipse is the same as it, so let the equation be
x²/(a²-5)+y²/a²=1
Take (- 2,3) in there
4/(a²-5)+9/a²=1
To solve this problem, we can get that if a & # 178; = 3 or 15 and a & # 178; > 5, then 3 is discarded
So a & # 178; = 15
The elliptic equation is
x²/10+y²/15=1
The original equation is x / 4 + Y / 9 = 1
Focus of upper ellipse (positive and negative root sign 5,0)
Let the ellipse be the square of X / the square of M + the square of Y / (the square of M + 5) = 1
If x = - 2, y = 3, then M = 10 or - 2 (rounding off)
Therefore, X / 10 + Y / 15 = 1
If a function image is a line segment AB, endpoint a (- 2,1), endpoint B (2,3), then the value range of independent variable x is
If a function image is a line segment AB, endpoint a (- 2,1), endpoint B (2,3), then the value range of independent variable x is [- 2,3]
That is to say, the value range of abscissa corresponding to the line segment is the calculated value
After reading the answer and your answer just now, I don't think you understand what independent variable is. First, you can see what independent variable is, and the answer is [- 2,2], please give me points... You are right
Given the equation of ellipse is 25X ^ 2 + 36Y ^ 2 = 900, find the vertex coordinates of ellipse
25x²+36y²=900
x²/(900/25)+y²/(900/36)=1
x²/36+ y²/25=1
Vertex coordinates (6,0), (- 6,0), (0,5), (0, - 5)
The value range of the independent variable in the function y = √ (5-4x) / x + 3 is
The value range of the independent variable in the function y = {√ (5-4x)} / x + 3 is
Is the whole score under the root sign or only the numerator under the root sign
It is known that the distance between the two focuses of an ellipse is 8, and the coordinates of the two vertices are (- 6,0), (6,0), respectively
Focus distance = 8
So C = 4
The two vertices are (- 6,0) (6,0)
So a = 6
c²=16 a²=36 b²=a²-c²=20
So the equation is X & # 178 / 36 + Y & # 178 / 20 = 1
The distance between two focal points is 8, C = 4
When a = 6, B ^ 2 = 36 - 16 = 20
The equation is X & # 178 / 36 + Y & # 178 / 20 = 1
When B = 6, a ^ 2 = 36 + 16 = 52
The equation is y ^ 2 / 52 + x ^ 2 / 36 = 1
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