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

1 Answer
Jan 13, 2016

Answer:

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)#