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

1 Answer
Lucio Falabella
Jan 13, 2016

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)

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