How do you find the domain and range of #y= 1 / (x-3)#?

1 Answer
Jun 13, 2016

Answer:

Domain #= {x:x epsilon RR, x != 3}#
Range #y ={y:y epsilon RR, y!= 0}#

Explanation:

The domain is the set of all the x-values.

We notice that #x# is in the denominator, and the only restriction for a denominator is that it may not be equal to 0.
If #x - 3 = 0 rArr x =3#

So #x# can have any value except 3.

This equation can also be written as #(x - 3) = 1/y#
The range is the set of all the #y# values

In this case we can see that #y# may not be equal to 0

So #y# can have any value except 0.

Domain #= {x:x epsilon RR, x != 3}#
Range #y ={y:y epsilon RR, y!= 0}#