Question

HELPPPPPPP PLZZZZ ASPPPPPP

Answer 1

Answer: Choice B

`(f * g)(x) = (x^2+6x+8)/(x^2+2x-15), \ \text{ for } x \\e -5 \text{ and } x \\e 3\\\\`

=================================================

Work Shown:

`h(x) = (f * g)(x)\\\\h(x) = f(x) * g(x)\\\\h(x) = (x^2-16)/(x^2+3x-10)*(x^2-4)/(x^2-7x+12)\\\\h(x) = ((x-4)(x+4))/((x+5)(x-2))*((x-2)(x+2))/((x-3)(x-4))\\\\h(x) = (x+4)/((x+5)(x-2))*((x-2)(x+2))/(x-3) \ \ \text{ ... see note 1}\\\\h(x) = (x+4)/(x+5)*(x+2)/(x-3) \ \ \text{ ... see note 2}\\\\h(x) = ((x+4)(x+2))/((x+5)(x-3))\\\\h(x) = (x^2+6x+8)/(x^2+2x-15)\\\\`

note 1: A pair of (x-4) terms canceled

note 2: A pair of (x-2) terms canceled

------------------------------

Extra info (optional section):

The fact that
`x \\e -5 \text{ and } x \\e 3` is to avoid a division by zero error in the simplified version of h(x).

I would argue that
`x \\e 2 \text{ and } x \\e 4` should be thrown in as well simply so that the domains match up perfectly with the original f(x) and g(x) functions.

So I think the full domain should be that x is any real number but

`x \\e -5 \text{ and } x \\e 2\\x \\e 3 \text{ and } x \\e 4`

Put another way: if x = 2 is allowed in h(x), then that clashes with the fact that it's not allowed in f(x). The same idea happens with x = 4 but with g(x) this time. It's possible your teacher glossed this fact over, or ran out of room.