find domain and range of inverse functions?

Question

equation:
f(x) = 1/x-2
D(-oo, oo)
how do i figure out the range?
and finding the inverse function?
I know I have to switch the "y" and "x".
I end up getting 1/x + 2, which doesn't look correct.
So once I have the D and R for the f(x) equation the D and R for the F^-1 (x) is the opposite right?
Thanks for your help, really appreciate it.

Answer 1

Let f(x) = y

1) y = 1/(x-2)

2) y(x-2) = 1

3) xy-2y = 1

4) xy = 1+2y

5) x = (1+2y)/y "switch y and x"

6) y= (1+2x)/x "switch y to f^-1(x)"

7) f^-1(x) = (1+2x)/x "f^-1 is the proper way to symbolize inverse"

* to check inverse for correctness, plug f(x) into f^-1(x) or vice versa if done correctly, the solution is always x.

* so from y = 1/(x-2) Domain is (-infinity,2)u(2,infinity) because denominator can't equal 0.

* so from y^-1 = (1+2x)/x Range is (-infinity,0)u(0,infinity) because denominator can't equal 0.

Answer 2

f(x)=1/(x-2)
the domain is all real numbers excluding x=2, and the range of f(x) is all real numbers except for y=0.
now to find the inverse:
y=1/(x-2)
so x-2=1/y
and x=(1/y) + 2
f^{-1}(x)= (1/x) +2,
to verify
we do the following
f(f^{-1}(x))= f((1/x) +2)
=1/[( 1/x +2)-2]
=1/[1/x]=x

the domain of f^{-1} is all real numbers except x=0,
and the range is all real numbers with the exception of x=2.

Answer 3

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

as you can see

Domain
(-∞, 2) U (2, +∞)

Range
(-∞,0) U (0,+∞)

For a graph, go to http://www.calculator.com/calcs/GCalc.html

Answer 4

f^-1(x) = (1/x) +2
but for f^-1(x) the domain would be the range of the original and the range of f^-1(x) would be the domain of f(x)