Stephanie M. answered • 04/24/15
Tutor
5.0
(956)
Private Tutor - English, Mathematics, and Study Skills
(a)
Simply add the functions to find (f+g)(x):
(f+g)(x) = f(x) + g(x) = 6/(x-12) + 9/(x+3)
The domain for any function is just a set of numbers you can plug into the function for x. Our main concern here is that you're never allowed to divide by zero. So, x-12 can't be equal to zero and x+3 can't be equal to zero:
x - 12 ≠ 0
x ≠ 12
x + 3 ≠ 0
x ≠ -3
The domain is all real numbers except -3 and 12. In integral notation, that's:
(-infinity, -3)∪(-3, 12)∪(12, infinity)
(b)
Subtract the functions to get (f-g)(x):
(f-g)(x) = f(x) - g(x) = 6/(x-12) - 9/(x+3)
Just like in (a), the domain is all real numbers except -3 and 12:
(-infinity, -3)∪(-3, 12)∪(12, infinity)
(c)
Multiply the functions to get (fg)(x):
(fg)(x) = f(x)g(x) = (6/(x-12))(9/(x+3)) = 54/((x-12)(x+3))
Just like in (a) and (b), the domain is all real numbers except -3 and 12:
(-infinity, -3)∪(-3, 12)∪(12, infinity)
(d)
Divide the functions to get (f/g)(x):
(f/g)(x) = f(x)/g(x) = (6/(x-12)) ÷ (9/(x+3))
You can simplify further by replacing division with multiplication by the reciprocal:
(6/(x-12)) × ((x+3)/9) = 6(x+3) / 9(x-12) = 2(x+3) / 3(x-12)
This one actually has a different domain. It's fine to plug in -3, but you'd still wind up dividing by zero if you plugged in 12. So, the domain is all real numbers except 12:
(-infinity, 12)∪(12, infinity)
