Answer: Choice D)
Work Shown:
[tex](f+g)(x) = f(x)+g(x)\\\\(f+g)(x) = \frac{x-23}{x^2+9x-36}+\frac{1}{x+12}\\\\(f+g)(x) = \frac{x-23}{(x+12)(x-3)}+\frac{1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+x-3}{x^2+9x-36}\\\\(f+g)(x) = \frac{2x-26}{x^2+9x-36}\\\\[/tex]
We must require that [tex]x \ne -12[/tex] and [tex]x \ne 3[/tex] to avoid having 0 in the denominator. This is why choice D is the answer.
