Question

What is the domain and range of `f(x)=(2x-1)/(3-x)`?

Answer 1

`x inRR,x!=3`
`y inRR,y!=-2`

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

`"solve "3-x=0rArrx=3larrcolor(red)" excluded value"`

`"domain is "x inRR,x!=3`

To find any excluded values in the range rearrange f(x) making x the subject.

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

`rArry(3-x)=2x-1larrcolor(blue)" cross-multiplying"`

`rArr3y-xy=2x-1`

`rArr-xy-2x=-3y-1larrcolor(blue)" collecting terms in x together"`

`rArrx(-y-2)=-(3y+1)`

`rArrx=-(3y+1)/(-y-2)`

`"the denominator cannot equal zero"`

`"solve "-y-2=0rArry=-2larrcolor(red)" excluded value"`

`rArr"range is "y inRR,y!=-2`

Answer 2

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

Explanation:

The function is `f(x)=(2x-1)/(3-x)`

The denominator must be `!=0`

So,

`3-x!=0`, `=>`, `x!=3`

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

Let,

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

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

`3y-yx=2x-1`

`2x+yx=1+3y`

`x=(1+3y)/(2+y)`

`2+y!=0`

`y!=-1`

The range is `y in (-oo,-1) uu(-1,+oo)`

graph{(y-(2x-1)/(3-x))=0 [-58.53, 58.54, -29.26, 29.24]}