If the inverse function f ^ - 1 (x) = (x + a) / (x + b) of function f (x) = (x + a) / (x + b), find a, B

If the inverse function f ^ - 1 (x) = (x + a) / (x + b) of function f (x) = (x + a) / (x + b), find a, B

y=(x+a)/(x+b)
xy+by=x+a
x(y-1)=a-by
x=(a-by)/(y-1)
f^-1(x)=(x+a)/(x+b)
=(a-bx)/(x-1)
A = any real number b = - 1

The inverse function of y = f (x) is y = F-1 (x). What is the inverse function of y = f (x + 2) It is very urgent to give the specific process of pushing down, and give the general conclusion

In fact, the inverse function is to solve x = F-1 (y) from y = f (x), and then exchange the positions of X and Y. therefore, the inverse function of y = f (x + 2) is x + 2 = F-1 (y), that is, x = F-1 (Y) - 2, and the inverse function is y = F-1 (x) - 2

Are y = f (x + 1) and y = f - (x + 1) inverse functions

Non reciprocal functions
Just draw a picture
F (x) and F-1 (x) are symmetric about y = X
Y = f (x + 1) and y = f - (x + 1) are the images after they are both shifted one unit to the right
In this case, the axis of symmetry also shifts to y = x + 1, and is no longer y = X
So they are not inverse functions of each other

Y = - 2 / x, (x ∈ R, X ≠ 0) find its inverse function y = f ^ - 1 (x) Seeking process

The inverse function of the inverse proportional function is itself
I hope it will help you
Welcome to inquire

Yes no question: given that y = f (x) has an inverse function, then the function must pass (0,0)

For example, y = x + 1 has an inverse function x = y + 1, but it is not (0,0) point

If the inverse function of function y = f (x) is F-1 (x), and f (0) = 1, then f (12) = () The answer is 13,

0

0

If there is an inverse function, it is a monotone function in the interval
y=(x-a)²-a²
Axis of symmetry x = a
If x = a is monotone, then x = a is not in the interval
So a = 2
y=(x-a)²-a²
a=1,x-a>=0
So x-a = √ (y 2 + a 2)
x=a+√(y²+a²)
Similarly, if a > = 2, then x-a

The function f (x) = x ^ 2-2tx + 1 (t ∈ R), the definition domain is x ∈ [0,1] ∪ [7,8], f (x) has inverse function, the value of T

X ∈ [0,1] ∪ [7,8], f (x) has inverse function
Then the function value of F (x) on X ∈ [0,1] ∪ [7,8] cannot be repeated
In other words, there is no intersection between the symmetric interval [2t-1,2t] and [7,8] of [0,1] with respect to the symmetric axis X = t
So:
（1）2t9/2；
To sum up, the value range of T is: T9 / 2;

If y = - (1-2x) ^ (1 / 2) x ∈ [a, b], the inverse function is itself The definition domain of whether the condition is satisfied is only [- 1,0]. If not, all the qualified [a, b] are obtained If so, please explain why

A:
It should be the inverse function of y = - (1-x ^ 2) ^ (1 / 2), and the inverse function of y = - (1-x ^ 2) / 2
y=-(1-x^2)^(1/2)=-√(1-x^2)
√(1-x^2)=-y>=0,y=0,-1

Inverse function of function y = 2x-1______ .

From y = 2x-1, x = 1 + log2y and Y > 0
That is, y = 1 + log2x, x > 0
So the inverse function of the function y = 2x-1 is y = 1 + log2x (x > 0)
So the answer is: y = 1 + log2x (x > 0)