If f (2x-1) is even function, find the symmetry axis of y = f (2x)

Are you doing something wrong?
It should be x = - 1 / 2 symmetry
We can think that since f (2x-1) is an even function, it is symmetrical about the Y axis. In brackets, we can extract 2 = 2 (x-1 / 2), so the image must be shifted 1 / 2 units to the left to be f (2x). So x = - 1 / 2

If y = f (2x) is an even function, then the axis of symmetry of y = f (2x + 1) is

The symmetry axis of even function is y = 0,
So take 2x + 1 = 0, x = - 1 / 2 is the symmetry axis of y = f (2x + 1)

Let f (x) = MX2 - (1-m) x + m, where m is a real number
Let f (x) = MX ^ 2 - (1-m) x + m, where m is a real number. (1) if f (x) has zeros, find the value range of M; (2) let the solution set of inequality f (x) < MX + m be a, when m is a positive number, the set a & # 8838; (- ∞, 3)?

(1) F (x) has a zero point, that is, the function curve has an intersection with the X axis,
Δ=(1-M)^2-4M^2=-3M^2-2M+1=-(3M-1)(M+1)>=0
So: - 1

Given that the domain of the function f (x) = − MX2 + 6mx + m + 8 is r, then the real number m has the same value______ ．

The definition domain of ∵ function f (x) = − MX2 + 6mx + m + 8 is r, ∵ - MX2 + 6mx + m + 8 ≥ 0 is constant, when m = 0, 8 ≥ 0 is constant, when m ≠ 0, there is − m ＞ 036m2 + 4m (M + 8) ≤ 0 to solve the inequality, − 45 ≤ m < 0, so the answer is: − 45 ≤ m ≤ 0

Given the function f (x) = (K-2) x + (4-3k), when x ∈ [- 1,1], the image of function f (x) is always above the x-axis, the value range of real number k is obtained

When k = 2, the function f (x) = - 2, in which the image is always below the X axis, does not meet the requirements. When k ≠ 2, the function f (x) is a linear function, and its image is a straight line. If x ∈ [- 1,1], the image of function f (x) is always above the X axis, then f (− 1) > 0f (1) > 0, that is, 6 − 4K > 02 − 2K > 0, the solution is: K < 1

The image of a linear function passes through the point P (- 2,3) and is parallel to the line y = (- 1 / 2) X

Because the function is parallel to the line y = (- 1 / 2) x, the slope of the function is - 1 / 2
Let the function be y = (- 1 / 2) x + B
We also know that the function passes through point P and is substituted into b = 4
So the linear function is y = (- 1 / / 2) x + 4

How to draw the image of y = X-2 and y = 1 / 2x + 1, the key is how to confirm the two points, detailed points, online and so on!

Two points are intercept, let x, y = 0 respectively
Y = X-2, let x = 0. Get y = - 2, let y = 0, x = 2. The two points of the line are (0, - 2) (2,0), which are connected directly by a straight line
Y = 1 / 2x + 1, similarly (- 2,0) (0,1)

Image of the second function of elementary school: given that the point P is on the image of the first function y = - X-1, and the distance to the X axis is 2, the coordinates of the point P are obtained

P(a,b)
On y = - X-1
b=-a-1
a=-b-1
The distance to the x-axis is the absolute value of the ordinate
So | B | = 2
b=±2
a=-b-1
So B = 2, a = - 3
b=-2,a=1
So p (- 3,2) or (1, - 2)

If there is a point P on the image and the distance from the x-axis is 4, the coordinates of the point P are obtained

If the distance between P and X axis is 4, the ordinate is 4
So substituting y = 4 into y = x + 2 gives x = 2
That is, the coordinates (2,4) of point P

Given the function f (x) = f '(π 4) cosx + SiNx, then the value of F (π 4) is______ ．

Because f ′ (x) = - F ′ (π 4) · SiNx + cosx, the solution of F ′ (π 4) = - F ′ (π 4) · sin π 4 + cos π 4 is f ′ (π 4) = 2-1, so f (π 4) = f ′ (π 4) cos π 4 + sin π 4 = 22 (2-1) + 22 = 1, so the answer is 1

If f (2x-1) is even function, find the symmetry axis of y = f (2x)