How do you find the domain of #1/(x+2)#?

1 Answer
May 13, 2017

Answer:

You look for the values of #x# that are outside the domain.

Explanation:

If #x=-2# then #x+2=0# so the denominator of the fraction will become indefinite.
Other values of #x# are acceptable, so:

Domain #x!=-2#
Or in 'the language':
#lim_(x->-2^-) y=-oo# and #lim_(x->-2^+) y=+oo#

graph{1/(x+2) [-10, 10, -5, 5]}
As for range : as #x# gets larger, the function will approach to #0#.

#lim_(x->-oo) y=lim_(x->+oo) y=0#

So range: #y!=0#