How to find the inverse function of y = x + 1 / X

How to find the inverse function of y = x + 1 / X

This is not a monotone function
So the inverse function doesn't exist

Find the inverse function of y = x + 1 / 2

The inverse function of y = 1 / (x + 2) is
y=(1-2x)/x

Finding the inverse function of y = 1-x / 1 + X

y(1+x)=1-x
y+yx=1-x
yx+x=1-y
x(y+1)=1-y
x=(1-y)/(1+y)
So the inverse function y = (1-x) / (1 + x), X ≠ - 1

Inverse function of y = (1-x) / (1 + x) I first solve x = and then call X and Y. here y = (1-x) / (1 + x) how to change it into x =,

y(1+x)=1-x
y+yx=1-x
yx+x=1-y
x(y+1)=1-y
x=(1-y)/(1+y)
So the inverse function y = (1-x) / (1 + x), X ≠ - 1

Inverse function y = 2x-3 / 5x + 1 (x ∈ R and X ≠ 3)

y(5x+1)=2x-3
5yx+y=2x-3
5yx-2x=-y-3
x(5y-2)=-y-3
x=-y-3/5y-2
X is not equal to 3, so y is not equal to 3 / 16,2 / 5

Is the inverse function of F (x / 3) = (2x + 3) / x Is to find f ^ - 1 (x / 3)

f(x/3)=(2x+3)/x (x≠0)
Let X / 3 = t, x = 3T
Then (2x + 3) / x = (6T + 3) / 3T = (2t + 1) / T
f(t)=(2t+1)/t
That is, f (x) = (2x + 1) / X
y=(2x+1)/x
xy=2x+1
x=1/(y-2)
So the inverse function y = 1 / (X-2) (x ≠ 2)

The process of inverse function of y = 1 + LG (2x-3)

The inverse function is to take y as an independent variable and find X
lg(2x-3)=y-1
10^(y-1)=2x-3
x=[10^(y-1)+3]/2
In general, we use X as the independent variable and y as the dependent variable, so we can replace it
y=[10^(x-1)+3]/2

How to calculate the inverse function of y = (2x + 3) / (x-1)

y=(2X+3)/(x-1)
(x-1)y=2x+3
xy-y=2x+3
xy-2x=3+y
x(y-2)=3+y
x=(3+y)/(y-2)
Change x to y
y=(3+x)/(x-2)

Finding the inverse function of y = 2x-3 / x + 1

y(x+1)=2x-3
yx+y=2x-3
2x-yx=y+3
x=(y+3)/(2-y)
So the inverse function is y = (x + 3) / (2-x)

The image of the inverse function of the function y = 1 / x + 2 (x ≠ - 2) passes through the point?

The y = 1 / (x + 2) image is obtained by moving the y = 1 / X image two units to the left
According to the image, there is an intersection point with the upper half axis of Y axis, and the intersection point is (0,1 / 2)