Inverse function of y = ln + 1 Come on, I'm in a hurry

Inverse function of y = ln + 1 Come on, I'm in a hurry

y=1/2*lnx+1
lnx=2(y-1)
x=e^(2y-2)
So the inverse function y = e ^ (2x-2), X ∈ R

Is the inverse function of y = ln (x + 1)?

X = e ^ Y-1 domain: y belongs to negative infinity to positive infinity
Right

What is the inverse function of y = ln (1-x)?

Replace x = ln (Y-1) Y-1 = e ^ x y = e ^ x + 1. The original function value field is r, so the inverse function definition field is also r

What is the inverse function of the function y = 1 + ln (x - 1) (x > 1)?

It's OK to solve x inversely. First, the value range of Y is determined as y belongs to R
Y = 1 + ln (x-1), Y-1 = ln (x-1), X-1 = e ^ (Y-1), x = 1 + e ^ (Y-1); so the inverse function is
Y = 1 + e ^ (x-1), X belongs to R
What's the problem hi me

The inverse function of the function y = 1 + ln (x-1) (x > 1) is () A. y=ex-1-1（x＞0） B. y=ex-1+1（x＞0） C. y=ex-1-1（x∈R） D. y=ex-1+1（x∈R）

∵ function y = 1 + ln (x-1) (x > 1),
∴ln（x-1）=y-1，x-1=ey-1，
By exchanging X and y, y = EX-1 + 1 can be obtained
The inverse function of the function y = 1 + ln (x-1) (x > 1) is y = EX-1 + 1
Therefore, D

The inverse function of the power 2 of y = 1 + ln (x + 1) Is it nonexistent or uncertain?

non-existent;
The necessary and sufficient condition for the existence of inverse function is that the function is one-to-one corresponding
The functions in the problem are not one by one. Therefore, there is no inverse function

1. Y = 2sinx, X belongs to the closed interval - π / 6, π / 62, y = 1 + ln (x + 2) 3, y = 2 x power / 2 inverse function Can you give me the knowledge of inverse function in these three questions

1. For y = 2sinx, X belongs to the closed interval - π / 6, π / 6, the range of Y is [- 1 / 2,1 / 2]. By x = arcsin (Y / 2), X and y are interchanged into the form of inverse function, y = arcsin (x / 2), the definition domain is [- 1 / 2,1 / 2]. 2

Y = x? 2 + 6x (x > 0) y = A's 2x + 1 power inverse function!

Y = x? 2 + 6 x = (x + 3) 2? - 9 (x + 3) 2? = y + 9x = radical (y + 9) - 3Y = x? + 6x (x > 0) inverse function y = radical (x + 9) - 3 y = 2x + 1 power of lgY = LGA's 2x + 1 power = (2x + 1) lga2x + 1 = lgY / lgax = lgY / 2lga-1 / 2Y = 2x + 1 inverse function of a y = lgx / 2lga

Function problem: given f (x) = ln (x + 1), let the inverse function of F (x) be f '(x) 1. Find the monotone interval of G (x) = f (x) - f '(x). 2. If the inequality LNF' (x) - f (E's x power) < (4 / 3) x-a holds for any x > 0, many students in the class can't do it. Thank you, online, etc

Obviously x > - 1
g(x)=f(x)-f`(x)=ln(x+1)-1/(x+1)
g`(x)=1/(x+1)+1/(x+1)^2
Therefore, G '(x) > 0 g (x) is a monotone increasing function in the domain of definition
2 lnf`(x)-f(e^x)
=ln[1/(x+1)]-ln(e^x+1)<4/3x-a
Let H (x) = ln [1 / (x + 1)] - ln (e ^ x + 1) - 4 / 3x = - ln (x + 1) - ln (e ^ x + 1) - 4 / 3x
h`(x)=-1/(x+1)-e^x/(e^x+1)-4/3<0
I need to know a clear expression of the inequality

The inverse function of Y (x) = e ^ x is x (y) = LN

Generally speaking, it's not. It's the rudiment of the inverse function. If you exchange x with y, you get y (x) = LNX