2

What is the domain of a rational function when the polynomial on the numerator and the denominator have a common factor? For example, consider the rational function $\frac{(x-1)(x+2)}{(x-1)(x+3)}$, now this function is the same as the function $\frac{x+2}{x+3}$ for all values of $x$, except for the $x = 1$, so my guess would be that the domain of the original function we were given is $\mathbb{R} \setminus \{ 1, -3 \}$, and the domain of the simplified version would be: $\mathbb{R} \setminus \{ -3 \} $. Am I correct about that? Also if I am correct, why doesn't Desmos show this? Demos, the graphing calculator, shows the function $\frac{(x-1)(x+2)}{(x-1)(x+3)}$ with no holes at $x = 1$, essentially it shows the same as $\frac{x+2}{x+3}$, so does that mean that Desmos is wrong? I zoomed in for a while and didn't see any holes.

zlaaemi
  • 1,037

1 Answers1

2

Your guess and the domains you specify for both of the functions you mention are correct.

Also, Desmos does in fact get this right even though it's not so good at explicitly displaying it. Try clicking and dragging your mouse along the function to the point where x=1, Desmos will display the co-ordinates "(1, undefined)" and a little open dot will appear along the function. While I also personally prefer using Desmos, GeoGebra does a better job of displaying discontinuities, try graphing it there to see what I mean.

Also see the linked answer for more on why: https://math.stackexchange.com/a/2799485/871321

BigBaller
  • 68
  • 1
    Thanks! You're right, it does show that it is undefined at $x = 1$, I guess it is just really hard to see unless you actually click on it. – zlaaemi Jan 05 '24 at 22:37