1 Answer

Chrona · Answer 1 · 2023-01-13T03:15:15+0000

Explanation:

So generally whenever you have a problem like this, you want to factor out the denominator and numerator, that way when dividing, you can easily see what the domain is by looking at factors.

So let's start by factoring the numerator and denominator of f(x)

$f(x) = (3x-6)/(2x+10)\implies (3(x-2))/(2(x+5))$

Now let's do this with the g(x)

$g(x) = (-2x+8)/(x+2)\implies(-2(x-4))/(x+2)$

Now when you divide f(x) by g(x), you keep f(x), change division to multiplication, and flip the g(x) so it's the reciprocal. This gives you the expression

(f(x))/(g(x))=(3(x-2))/(2(x+5))*(x+2)/(-2(x-4))

Now just multiply the numerator and denominator

(f(x))/(g(x))=(3(x-2)(x+2))/(2(x+5)*-2(x-4))

So the main point of factoring was to easily see when the denominator is zero, (since if one of the factors is zero, the entire thing will be zero, since zero * anything = zero), but it was also to find any removable discontinuities. Which are just when you have factor in the denominator and numerator, so you can simplify the fraction further by canceling it out. To give an example:
$f(x)=((x-2)(x+4))/((x-2)(x+3))\implies(x+4)/(x+3)$ . But the domain of f(x) is still restricted so that:
$x\\e 2, -3$ . But there will not be a vertical asymptote at x=2 like with traditional domain restrictions in a rational function. This is because you can technically define f(2), but only if you use the most simplified form. This will result in a hole being at that point to signify that it's a removable discontinuity.

Anyways a bit off topic, but something that may be necessary in other examples. Anyways, now to find the domain, we just need to find when f(x)/g(x) is not defined. Or more specifically we find when the denominator is going to be zero. So we just need to set both factors equal to zero.

$0 = 2(x+5)\\\text{divide both sides by 2 (a bit redundant)}\\0 = x+5\\\text{subtract 5 from both sides}\\-5 = x$