How do I find the domain and (if any) discontinuities for #g(x)=(x^2+6x+9)/(x+3)#? I don’t entirely understand what is being asked by domain and discontinuity.

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Answer 1 · 2017-12-23T20:36:22

The domain of #g(x)# is #(-oo, -3) uu (-3, oo)#.

It has a hole (removable singularity) at #x=-3#.

Given:

#g(x) = (x^2+6x+9)/(x+3)#

Note that when #x=-3# the denominator is #0#. Since division by #0# is undefined, #x=-3# is not part of the domain.

#g(x)# is well defined for any other real value of #x#, so its domain is #(-oo, -3) uu (-3, oo)#.

If we factor the numerator, we find:

#x^2+6x+9 = (x+3)(x+3)#

So:

#g(x) = (x^2+6x+9)/(x+3) = (color(red)(cancel(color(black)((x+3))))(x+3))/color(red)(cancel(color(black)((x+3)))) = x+3#

with exclusion #x != -3#

Note that the simplified expression for #g(x)# is continuous and well defined for all real values of #x#, including #x=-3#.

So #g(x)# has a hole at #x=-3#, that is a removable singularity - removable by redefining #g(-3) = (color(blue)(-3))+3 = 0#