What is the inverse of function f(x)=(1/x-3)+2

can you give me detailed step by step because i dont understand this ( i have the answer but i dont understand how to get it)

y=(1/x-3)+2
y-2=1/x+3
x+3=1/y-2
x=(1/y-2)-3
f^-1(x) = (1/x-2) - 3

An inverse function undoes a function which means if put a number in the original function and get an answer, then put the answer in the inverse function, you will get the original number. Simply solve the original question for the other unknown.
f(x) = (1/x - 3) + 2
y = (1/x - 3) + 2 subtract 2 from both sides
y - 2 = 1/x - 3 add 3 to both sides
y + 1 = 1/x invert both sides
x = 1/(y+1)
g(y) = 1/(y+1) ANSWER

check it out by solving f(1)
f(1) = (1/1 - 3) + 2
f(1) = 1 - 3 + 2
f(1) = 0

g(0) = 1/(0+1)
g(0) = 1/1
g(0) = 1

Finding Inverses:
Step 1: replace f(x) with y. y = (1/x-3)+2
Step 2: swap y and x x = (1/y-3)+2
**Now here i'm confused. Is it (1/x)-3 or 1/(x-3)
If its the first one, stick to left column, if its the second one,
stick to right column**
Step 3: solve for y x-2 = (1/y)-3 x-2 = 1/(y-3)
x+1 = 1/y (x-2)(y-3) = 1
y(x+1) = 1 y-3 = 1/(x-2)
y = 1/(x+1) y = [1/(x-2)] + 3
Step 4: replace y with f^-1(x) f^-1(x) = 1/(x+1) f^-1(x) = [1/(x-2)] + 3
Step 5: you're done!

f(x)= y = (1/x-3)+2
1) interchange x and y --> x = (1/y -3) + 2
2) solve for y, so move the 2 --> x -2 = 1/y -3
3) move the -3 --> x-2+3 = 1/y
4) add #'s --> x+1 = 1/y
5) invert --> 1/(x+1) = y (inverse fct)

Rewrite:
y=(1/(x-3))+2

Switch x and y:
x=(1/(y-3))+2

Solve for y;
x-2 = 1/(y-3)
(y-3)(x-2) = 1
y-3 = 1/(x-2)
y = 1/(x-2) + 3

Substitute y = f(x):
f(x) = 1/(x-2) + 3