What is the inverse function of y = 3x? Is the process x = Y / 3 and then y = x / 3? If so How to find the inverse function of y = x + LNX?

What is the inverse function of y = 3x? Is the process x = Y / 3 and then y = x / 3? If so How to find the inverse function of y = x + LNX?

A:
y=3x,x=y/3
So: the inverse function of y = 3x is y = x / 3
As you do, find X and swap X and y
y=x+lnx=ln(e^x)+lnx=ln(xe^x)
xe^x=e^y
It seems that there is no way to deal with the conventional practice

The inverse function of y = 2Sin 3x (- π / 6 ≤ x ≤ π / 6)

-π/2≤3x≤π/2
y/2=sin 3x ,-1≤y/2≤1,-2≤y≤2
3x=arc sin(y/2)
x=1/3*arc sin(y/2)
y=1/3*arc sin(x/2),-2≤x≤2

What is the inverse function of y = - 3x + 4?
What is the inverse function of y = - 3x + 4

Y = - 3x + 4, the range is r
-3x=y-4
∴ x=(-1/3)y+(4/3)
The inverse function of y = - 3x + 4 is y = (- 1 / 3) x + (4 / 3) x ∈ R

Let the inverse function of F (x) = log2x be y = g (x). If G (1a − 1) = 14, then a is equal to ()
A. -2B. −12C. 12D. 2

The inverse function of ∵ f (x) = log2x is y = g (x), G (1a − 1) = 14, ∵ f (14) = 1a − 1, that is log142 = 1a − 1-2 = 1a − 1, ∵ a = 12, so C

Given the function f (x) = x ^ 2-4x-5, X ∈ [1,3], judge whether there is an inverse function, if there is, find out the inverse function; if not, explain the reason
Today!

f(x)=(x-2)²+9
The axis of symmetry x = 2 is in the interval
So f (x0) is not a monotone function
So there is no inverse function

What is the domain of the inverse function of the function y = 2arccos (X-2) and how to find it?
What I did was 0=

Your answer is right
The definition of inverse function is the range of original function
Because 0

It is proved that the intersection of the original function image and its inverse function image is on the straight line y = X

The images of the functions y = f (x) and y = f - &# (x) are symmetric with respect to the line y = X
This is a true proposition, correct

Is the intersection of the original function and the inverse function necessarily on the line y = x?
For example!

Not necessarily. For example, y = (1 / 16) ^ X
And the focus of inverse function is (1 / 4,1 / 2), (1 / 2,1 / 4)

Is the intersection of the image of a function and its inverse function necessarily the intersection of the function and y = x

If there is an intersection point, it must be on y = X

It is proved that the intersection of a function and its inverse function is not necessarily on y = X
prove!
Of course, the intersection of a function and its inverse function is not necessarily on y = X. for example, the inverse proportion function y = 1 / X is clumsy

This is a false proposition if the inverse function exists
Let (x0, Y0) be a point on y = f (x), then for the inverse function, x = g (y), there must be (Y0, x0). If it intersects, then (x0, Y0) = (Y0, x0), since the intersection is on y = x,