Question

What is the domain and range of `f(x) = 5/(x-9)`?

Answer 1

DOMAIN: `x in (-oo,9)uu(9,+oo)`

RANGE: `y in (-oo,0)uu(0,+oo)`

Explanation:

`y=f(x)=k/g(x)`

Existence Condition is:

`g(x)!=0`

`:.x-9!=0`

`:.x!=9`

Then:

`F.E.`= Field of Existence=Domain: `x in (-oo,9)uu(9,+oo)`

`x=9` could be a vertical asymptote

To find the range we have to study the behavior for:

• `x rarr +-oo`

`lim_(x rarr -oo) f(x)=lim_(x rarr -oo) 5/(x-9)=5/-oo=0^-`

`lim_(x rarr +oo) f(x)=lim_(x rarr +oo) 5/(x-9)=5/(+oo)=0^+`

Then

`y=0` is a horizontal asymptote.

Indeed,

`f(x)!=0 AAx in F.E.`

• `x rarr 9^(+-)`

`lim_(x rarr 9^-) f(x)=lim_(x rarr 9^-) 5/(x-9)=5/0^(-)=-oo`

`lim_(x rarr 9^+) f(x)=lim_(x rarr 9^+) 5/(x-9)=5/0^(+)=+oo`

Then

`x=9` it's a vertical asympote

`:.` Range of `f(x)`: `y in (-oo,0)uu(0,+oo)`