Question

What is the domain and range of `y =1/(2x-4)`?

Answer 1

The domain of `y` is `=RR-{2}`
The range of `y`, `=RR-{0}`

Explanation:

As you cannot divide by `0`,

`2x-4!=0`

`x!=2`

Therefore, the domain of `y` is `D_y=RR-{2}`

To determine the range, we calculate `y^-1`

`y=1/(2x-4)`

`(2x-4)=1/y`

`2x=1/y+4=(1+4y)/y`

`x=(1+4y)/(2y)`

So,

`y^-1=(1+4x)/(2x)`

The domain of `y^-1` is `D_(y^-1)=RR-{0}`

This is the range of `y`, `R_y=RR-{0}`
graph{1/(2x-4) [-11.25, 11.25, -5.625, 5.625]}

Answer 2

`"domain " x inRR,x!=2`

`"range " y inRR,y!=0`

Explanation:

The denominator of y cannot be zero as this would make y `color(blue)"undefined".`Equating the denominator to zero and solving gives the value that x cannot be.

`"solve " 2x-4=0rArrx=2larrcolor(red)" excluded value"`

`"domain " x inRR,x!=2`

`"to find excluded value/s in the range"`

`"Rearrange the function making x the subject"`

`rArry(2x-4)=1`

`rArr2xy-4y=1`

`rArr2xy=1+4y`

`rArrx=(1+4y)/(2y)`

`"the denominator cannot be zero"`

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

`"range " y inRR,y!=0`
graph{1/(2x-4) [-10, 10, -5, 5]}