How do you find the domain and range of #y = (x^2 - 1) / (x+1)#?

1 Answer
Apr 4, 2017

Answer:

Domain: #x in {RR - {-1})#
Range: #y in {RR - {-2}}#

Explanation:

#(x^2-1)/(x+1)# is defined for all value of #x# except any value that would make the denominator equal to #0#.
That is #(x+1)!=0color(white)("XX")rarrcolor(white)("XX")x!=-1#
(This gives us the Domain).

Notice that if #x!=-1#
then #(x^2-1)/(x+1)=x-1#
Any real number #r in RR# can be generated from #x-1# by picking a value of #x=r+1# except for the already excluded #color(black)(x=-1)# That is, we can not generate #r=x-1# when #x=-1#. We can not generate #r=-2#.
So the Range must exclude #{-2}#