Let f (x) = (AX + 3) / (2-x), if f (x) is equal to its inverse function, find the real number a

y=f(x)=(ax+3)/(2-x)
y(2-x)=ax+3
2y-yx=ax+3
(a+y)x=2y-3
x=(2y-3)/(a+y)
Inverse function of F (x): G (x) = (2x-3) / (a + x)
f(x)=g(x)
a=-2

If the point (2,4) is on the image of the function f (x) = 2 ^ (AX + b), and on the image of its inverse function, find the values of real numbers a and B

The inverse function of F (x) = 2 ^ (AX + b) is
F（X）=[log2x-b]/a
Put (2,4) into the above two formulas
4=2^(2a+b),2a+b=2
4=(1-b)/a,4a=1-b
A = - 1 / 2, B = 3

Given that f (x) and G (x) are inverse functions of each other, find the inverse function of F (AX + b) This is the original problem. How to write the specific steps of solving the problem

Let g (f (AX + b)) = ax + b let P (x + b)) = x be the inverse function of F (x), and let P (x) be the inverse function of F (AX + b)) = x, then p (f (AX + b)) = x have a * P (f (AX + b)) + B = ax + B = g (f (AX + b)) Let f (AX + b) = t have a * P (T) + B = g (T) to get P (T) =

Let P (3,1) be a point of intersection between the image of quadratic function f (x) = ax square-2ax + B (x is greater than or equal to 1) and its inverse function, and find the values of a and B

Let P (3,1) be a point of intersection between the image of quadratic function f (x) = ax square-2ax + B (x ≥ 1) and its inverse function
Then f (3) = 9a-6a + B = 1 3A + B = 1 (1)
f(1)=a-2a+b=3 -a+b=3 (2)
(1)-(2) 4a=-2 a=-1/2
Substituting (2) B = 5 / 2
That's what you want

If the point (1,7) is not only on the image of AX + B under the function y = radical, but also on the image of its inverse function, the values of a and B are obtained

Y = √ (AX + b), inverse function, x = √ (ay + b), they are symmetric about the line y = X,
All in the first quadrant,
Put x = 1, y = 7 into the above two formulas,
a+b=49,7a+b=1,
a=-8,
b=57.

It is known that the point (1,2) is on the image of the function y = radical ax + B and on the image of its inverse function, then a =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B =? B? The root sign is ax + B

f(x)=√(ax+b)
Point on two images
Then f (1) = 2
f(2)=1
√(a+b)=2
√(2a+b)=1
a+b=4
2a+b=1
So a = - 3, B = 7

If the point (1,2) is y = On the graph of AX + B and its inverse function, then a= ，b= .

Method 1: from the known:
A + B = 2, that is, a + B = 4,
By y =
Ax + B solution X: x = 1
a(y2−b)，
Then y =
The inverse function of AX + B is y = 1
a(x2−b)，
∵ the point (1,2) is on the image of the inverse function
∴2＝1
a(1 −b)
The simultaneous solution of a + B = 4 leads to a = - 3, B = 7,
Method 2: from the known point (1,2) at y =
On the image of AX + B
be
A + B = 2, that is, a + B = 4,
And ∵ the graph of functions which are inverse functions of each other is symmetric with respect to y = X
The point (2,1) is also in the function y =
On the image of AX + B
It is concluded that:
2A + B = 1, that is, 2A + B = 1,
By combining this with a + B = 4, a = - 3, B = 7,
answer:
a=-3，b=7，

If f (x) = the radical (AX + b) and its inverse function both cross the point (1,2), then the number of intersection points between F (x) and its inverse function image is___ Wei .

First of all, the inverse function of a function has passed (1,2), indicating that this function has passed (2,1)
Then f (x) passes through (1,2), (2,1) two points at the same time, and takes it in and solves a = - 3, B = 7
So f (x) = radical (- 3x + 7)
This is understandable

Why is the image of the function and its inverse function symmetrical about the line y = x? Can this be proved mathematically? The process of proof

According to the definition of inverse function, if the original function y = f (x) passes (a, b), then the inverse function F-1 (x) passes (B, a). Let P (x, y) be any point on y = f (x), then f (x) = y, then the symmetric point of P (x, y) with respect to y = x is (y, x) ∵ y = f (x), so F-1 (y) = F-1 (f (x)) = x, that is (y, x) on the image of y = F-1 (x), it can be proved that y = F-1 (x)

Why are the images of two functions inverse to each other symmetric with respect to y = x? Exploration and discovery of P76 page in mathematics compulsory course of PEP

The inverse function, to put it bluntly, is to exchange the independent variable and dependent variable in the function. In the graph, it is to exchange X and Y. therefore, the simplest deformation is to make the original image symmetrical about y = X

Let f (x) = (AX + 3) / (2-x), if f (x) is equal to its inverse function, find the real number a