The inverse function f ^ - 1 (x) of function y + 2 ^ x (x > 0) and its definition domain are given

The inverse function f ^ - 1 (x) of function y + 2 ^ x (x > 0) and its definition domain are given

y=2^x
x> 0 then Y > 2 ^ 0 = 1
That is, the definition domain (0, + ∞) and the range of value (1, + ∞)
So the inverse function
Definition domain (1, + ∞), range (0, + ∞)

Find the inverse function of y = (e ^ x-e ^ - x) / (e ^ x + e ^ - x), and point out its definition domain

y=(e^x-e^(-x))/(e^x+e^(-x))…… The numerator and denominator are multiplied by e ^ X
= [e^(2x)-1]/ [e^(2x)+1]= [e^(2x)+1-2]/ [e^(2x)+1]
=1-2/[e^(2x)+1],
∵e^(2x)>0,e^(2x)+1>1,
∴0

What is the inverse function of (y) = 1? RTRT

Definition domain of inverse function of function y = [(1 / 2) ^ x] - 1
namely
Value range of function y = [(1 / 2) ^ x] - 1!
Because the value range of the function y = [(1 / 2) ^ x] - 1 is Y > - 1
Therefore, the definition domain of the inverse function of the function y = [(1 / 2) ^ x] - 1 is (- 1, + infinity)

The definition domain of the inverse function of the function y = 2 ^ X-1 / 2 ^ x is

R

Definition domain of inverse function of function y = 2 ^ x + 1

Inverse function definition domain = original function range = (1, + ∞)
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Is there any inverse function for monotone function in the domain of definition? As the title

If it is a monotone function, there must be an inverse function

There is an inverse function in a function, which is not necessarily monotonic? As the title Is the inverse of the piecewise function a function?

For example, the piecewise function f (x) = {1 + X (x > = 0)
-x (x<0)
}

Monotone function must have monotone inverse function. Does not monotone function have monotone inverse function? If so, please give an example

Monotone function must have single value inverse function;
Non monotonic continuous functions have no single valued inverse functions;
If the function is not monotonic and discontinuous, it may still have inverse functions
The definition domain of F (x) is {0,1,2} and f (0) = 2, f (1) = 0, f (2) = 1 is not monotonic, but it has inverse function

If the image of function y = FX passes through point (0,1), then the image of function y = f (x + 4) crosses the point

If the image with y=f (x) passes through a point (0,1), the function y=f (x+4) passes through a point (-4,1), and its inverse function must pass through a point (1, -4)

Given the function f (x) = (A-X) / (x-a-1), the image symmetry center of the inverse function f ^ 1 (x) is (- 1,3), then the real number a is equal to many

It's difficult to write,
f=(a-x)/(x-a-1)
x=(a*y+y+a)/(y+1)=a+1-1/(y+1).
The center of this function is (- 1, a + 1) = (- 1,3)
a+1=3
a=2.