Can the definition domain of y = (1-x) / (1 + x) take 0? The value range of its inverse function cannot be - 1, so the definition field should not take 0, right?

Can the definition domain of y = (1-x) / (1 + x) take 0? The value range of its inverse function cannot be - 1, so the definition field should not take 0, right?

For invertible function, the definition domain of inverse function is the range of original function
y=(1-x)/(1+x)=(2-1-x)/(1+x)=-1+2/(1+x)
Then the value range of the original function is x, which is not equal to - 1
That is to say, the domain of inverse function is (- infinite, - 1) and (- 1, + infinite)
Notice that when x = 1, y = 0
So the inverse function can take a value at 0, and the value is 1

What is the definition domain of the inverse function of y = 1 / 2 (a ^ x-a ^ (- x))?

The definition domain of inverse function is the original function range R

The domain of the inverse function of y = (2 ^ x-1) / (2 ^ x + 1) is___ ?

We can set the x power of 2 as t, (T > 0), y = (t-1) / (T + 1), and change the denominator into a quadratic equation of one variable about t

The definition domain of inverse function problem y = 1 / 2 [e ^ X-1 / (e ^ x + 1)] is? The definition domain of the inverse function of y = 1 / 2 [e ^ X-1 / (e ^ x + 1)] is? p. S the first time I did this kind of topic, let me know how to do it Is that all right? thank you`~ But the answer our teacher said was [- 1 / 2,1 / 2]

Let t = e ^ x, because the value range of e ^ x is (0, + ∞), so t ∈ (0, + ∞) so y = (1 / 2) [(t-1) / (T + 1)] and then use the method of separating variables. Because the denominator has t + 1, we try to make a T + 1 in the molecule so that the molecule does not contain T. it is not difficult to think of: y = (1 / 2) [(T + 1-2) / (T + 1)] = (1 / 2)

Find the function y = 2x + 1 The inverse function of 2x − 1 (x ≥ 1 and X ≠ 0) and its definition domain

By y = 2x + 1
2x − 1 gives 2x = y + 1
y−1，
∴x=log2y+1
Y − 1 and Y ＞ - 1
That is, the function y = 2x + 1
The inverse function of 2x − 1 (x ≥ 1 and X ≠ 0): y = log2x + 1
x−1．
∵ x ≥ 1 and X ≠ 0, ᙽ 2x ≥ 2, ᙽ y + 1
y−1≥2，∴1＜y≤3，
The definition domain of inverse function is (1, 3]

Does FX have an inverse function and is FX monotone in the domain of definition

If not monotone, there will be no inverse function
You can think from the graph that if a non monotonic function is symmetric about y = x, then there must be one X corresponding to many y in the new function, which is not a function

Monotone function must have inverse function, but why is it not necessarily monotone function that has inverse function

This should be analyzed from the mapping
For the function with inverse function, the definition field to the range is 1-1 correspondence or bijection. The definition field and range are D and B respectively. If for x1, X2 ∈ D, x1 ≠ X2, we deduce that f (x1) ≠ f (x2), f (x1), f (x2) ∈ B. then it is called 1-1 correspondence or bijection
If the function is monotone, whether increasing or decreasing, we can guarantee x1, X2 ∈ D, x1 ≠ x2. We deduce that f (x1) ≠ f (x2), f (x1), f (x2) ∈ B, so there is inverse function in monotone function
But conversely: x1, X2 ∈ D, x1 ≠ X2, we can deduce f (x1) ≠ f (x2), f (x1), f (x2) ∈ B, can we deduce that for all x ∈ D, there is one of X1 > X2, f (x1) > F (x2), or F (x1) < f (x2)? No
The monotonicity of functions is a sufficient and unnecessary condition for the existence of inverse functions

Let the inverse function of the function y = f (x) exist and the image of y = f (2x-1) pass through the point (3,2), and find the function y = f ^ - 1 (4x + 1) Let the inverse function of function y = f (x) exist and the image of y = f (2x-1) pass through the point (3,2), find the coordinates of the point that the image of function y = f ^ - 1 (4x + 1) must pass through

Analysis,
The image of F (2x-1) passes (3,2)
That is, f (2 * 3-1) = 2, that is, f (5) = 2
The inverse function of F (x) exists,
Therefore, f ^ (- 1) (2) = 5 must hold,
4x+1=2,x=1/4,
Therefore, the image of f ^ (- 1) (4x + 1) must be over (1 / 4,5)

Let the inverse function of the function y = f (x) (the definition field is D and the range is a) is y = f ^ - 1 (x) Let y = f ^ - 1 (x) be the inverse function of the function y = f (x) (the definition field is D, the range of value is a) is y = f ^ - 1 (x), and the function y = f (x) is monotonically increasing on D. It is proved that the function y = f ^ - 1 (x) is also an increasing function on a (2) Let y = f (x) be an odd function on D. It is proved that the function y = f ^ - 1 (x) is also an odd function on a

Starting from the definition. (1) in F (x), as x increases, f (x) also increases,
When f (x) increases, X also increases
(2) In the same way. F (- x) = - f (x)... Or one-to-one correspondence

If the domain of the inverse function F-1 (x) of F (x) is [1 / A, 4 / a], then the domain of F (x) is

When f (AX + 3) = x, f (x) = (x-3) / A, the definition domain of inverse function F-1 (x) is [1 / A, 4 / a], that is, when the value range of function f (x) is [1 / A, 4 / A] 1 / a ≤ (x-3) / a ≤ 4 / AA > 0, the definition domain of 1 ≤ x-3 ≤ 44 ≤ x ≤ 7F (x) is: [4,7] a > 0, x-3 ≤ 4x ≤ 4, or, X ≥ 7F (x) is: (- ∞