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.

1 Answer
Dec 23, 2017

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

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

Explanation:

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