Question

What is the domain and range of `f(x) =(x-3)/(7x+4)`?

Answer 1

`dom f in RR \\\ {-4/7}` and `ran f in RR \\\ {1/7}`.

Explanation:

So this is a rational function, so to get `f(x)` into hyperbolic form (from which you can read off your asymptotes) perform polynomial division.

`x/(7x) = 1/7`

`1/7*7x = x` and `1/7*4=4/7`

`x - x =0` and `-3 - 4/7 = -3 4/7 = -25/7`.

So your quotient, `Q(x) = 1/7` and your remainder `R(x) = -25/7`.

Therefore, `f(x)=1/7 + -25/7 -: 7x +4`

Rearrange this to make your solution of the form `f(x) = a/(x-h) +k`

Therefore, `f(x) = -25/ (49(x+4/7)) + 1/7`

This means that you have asymptotes at `x=-4/7` and `y=1/7`. Since hyperbolas have infinite domains and ranges excluding any point on the asymptote, you have:

`dom f in RR \\\ {-4/7}` and `ran f in RR \\\ {1/7}`.

Answer 2

`x inRR,x!=-4/7`
`y inRR,y!=1/7`

Explanation:

`"f(x) is defined for all real values of x apart from values that"`
`"make the denominator equal to zero"`

`"equating the denominator to zero and solving gives the"`
`"value that x cannot be"`

`"solve " 7x+4=0rArrx=-4/7larrcolor(red)" excluded value"`

`rArr"domain is " x inRR,x!=-4/7`

`"to find any excluded value in the range, rearrange "`

`y=f(x)" making x the subject"`

`rArry(7x+4)=x-3larrcolor(blue)" cross-multiplying"`

`rArr7xy+4y=x-3`

`rArr7xy-x=-3-4y`

`rArrx(7y-1)=-(3+4y)`

`rArrx=-(3+4y)/(7y-1)`

`"the denominator cannot equal zero"`

`"solve " 7y-1=0rArry=1/7larrcolor(red)" excluded value"`

`rArr"range is " y inRR,y!=1/7`