Question

How do you find the domain and range of `f(x)=2/(3x-1)`?

Answer 1

The domain is `x in (-oo, 1/3) uu(1/3,+oo)`. The range is `y in (-oo,0) uu(0,+oo)`

Explanation:

The denominator must be `!=0`

Therefore,

`3x-1!=0`

`=>`, `x!=1/3`

The domain is `x in (-oo, 1/3) uu(1/3,+oo)`

To find the range, proceed as follows :

Let `y=2/(3x-1)`

`<=>`, `y(3x-1)=2`

`<=>`, `3yx-y=2`

`<=>`, `3yx=2+y`

`<=>`, `x=(2+y)/(3y)`

The denominator must be `!=0`

`3y!=0`

`=>`, `y!=0`

The range is `y in (-oo,0) uu(0,+oo)`

graph{2/(3x-1) [-12.66, 12.65, -6.33, 6.33]}