Given that f (2x + 1) is an even function, find the axis of symmetry of F (x) I just don't understand why f (- 2x + 1) = f (2x + 1) is defined by even function Let 2x + 1 = t ∵ f (T) is an even function ∴f(-t)=f(t) Is f (- 2x-1) = f (2x + 1)

Given that f (2x + 1) is an even function, find the axis of symmetry of F (x) I just don't understand why f (- 2x + 1) = f (2x + 1) is defined by even function Let 2x + 1 = t ∵ f (T) is an even function ∴f(-t)=f(t) Is f (- 2x-1) = f (2x + 1)

F (2x + 1) is even function instead of F (x)
So it is wrong that f (T) is an even function

Find the analytic expression of the function corresponding to the image obtained by translating the line y = 3x along the positive direction of X axis for 2 unit lengths
Note that it's the x-axis, not the y-axis

Y=3(X-2)
Y=3X-6

The image of the function y = 2 / X is translated up three unit lengths along the y-axis to get the image of the function, and then one unit length is translated right along the x-axis to get the function
Thank you

Y up and down, X left and right
1,y=2/x+3
2,y=2/(x-1)+3

Let f (x) be defined as R, with period π, and f (x) = SiNx, - π 2 ≤ x < 0cosx, 0 ≤ x < π 2, then f (- 5 π 3) = -___ ．

It is known that the domain of function f (x) is r, the period is π, f (- 5 π 3) = f (- 2 π + π 3) = f (π 3). Since f (x) = SiNx, - π 2 ≤ x < 0cosx, 0 ≤ x < π 2F (π 3) = cos π 3 = 12, the answer is: 12

If f (x) = cosx, (− π 2 ≤ x < 0) SiNx, (0 ≤ x < π), then f (− 15 π 4) is equal to ()
A. 22B. 1C. 0D. −22

∵ f (x) = cosx, − π 2 ≤ X & lt; 0sinx, 0 ≤ X & lt; π, the minimum positive period is 3 π 2F (− 15 π 4) = f (3 π 4 − 3 × 3 π 2) = f (3 π 4) = sin 3 π 4 = 22, so a

Let f (x) be a periodic function whose domain of definition is R and the minimum positive period is 3 π / 2

f(-15π/4)=f(-18π/4 +3π/4)=f[-3•(3π/2)+3π/4]=f(3π/4)=sin3π/4=sin(π-π/4)=sin(π/4)=√2/2

Let f (x) = 1 / (xlnx), and for any x belonging to (0,1), a > LN2 * f (x) holds, and find the value range of real number a

If a > LN2 * f (x) is constant, then a is greater than g (x) = LN2 * f (x) (0

The function f (x) = - ax + xlnx is not monotone in the interval [1, the square of e]

f'(x)=-a+1*lnx+x*1/x=-a+lnx+1
If f '(x) is not monotone, then f' (x) has positive and negative in the interval
Because f '(x) = lnx-a + 1 is an increasing function
There are positive and negative
Then the minimum value is less than 0 and the maximum value is greater than 0
That is, f '(E & # 178;) = 2-A + 1 > 0
f'(1)=0-a+1

The graph of function f (x) and G (x) = (1 / 4) ^ x is symmetric with respect to the line y = X. then the monotone increasing interval of function H (x) = f (6x-x2) is

The image of function f (x) is symmetric to the image of G (x) = (1 / 4) ＾ x with respect to the line y = x, then f (x) = log (1 / 4) x is a decreasing function in its domain of definition
The increasing interval of 2a-x ^ 2 = - (x-1) ^ 2 + 1 is (0,1)
[the same increases, the different decreases]
So the decreasing interval of F (2x-x ＾) is (0,1)

It is known that the image of the quadratic function y = AX2 + 4x + C intersects with the X axis at point a. the analytic expression of the image intersecting with the Y axis at (0,4) is obtained

I don't quite understand the topic
Maybe there is only one intersection point between point a and x-axis
So the analytic formula is y = x ^ 2 + 4x + 4