Question 143601
Is the function 2/(3-x) or (2/3)-x? This is important, as the answer will most likely be different.

I will assume it is the former, namely y=2/(3-x). 

To compute the inverse function, we just switch the y's and x's. 
x=2/(3-y)
Then solve for y
(3-y)x=2
3-y=2/x
-y=(2/x)-3
y=-(2/x)+3
That is f inverse.

Now, D&R
For f--> What can x not be? It cannot be 3 because that would cause division by zero. Thus D: ({{{-infinity}}},3)U(3,{{{infinity}}}).
What is the range? Note that the degree of the numerator is less than the degree of the denominator. This means the function will tend toward, but never reach, 0. Thus y=0 is a horizontal asymptote and the range is R:({{{-infinity}}},0)U(0,{{{infinity}}})
Plot:
{{{graph( 300, 200, -5, 5, -5, 5, 2/(3-x) )}}}

For f inverse--->What can x not be? It cannot be 0 because that would cause division by zero. Thus D: ({{{-infinity}}},0)U(0,{{{infinity}}}).
What is the range? Note that the degree of the numerator will equal the denominator (upon using a common denominator x). This means the function will tend toward, but never reach, the coefficient of the the numerator's highest degree x divided by the coefficient of the denominator's highest degree x. This is 3/1=3. Thus y=3 is a horizontal asymptote and the range is R:({{{-infinity}}},3)U(3,{{{infinity}}})
Plot:
{{{graph( 300, 200, -5, 5, -5, 5, -(2/x)+3 )}}}


Now if we plot both, we can see why they are inverses:
{{{graph( 300, 200, -10, 10, -10, 10, -(2/x)+3, 2/(3-x) )}}}

If you have further difficulty, or are unsure of anything I did, please E-mail me at enabla@gmail.com