Question

For anyone looking for a challenge ...
f(x)= (5x+4)/(x^2-36)
g(x)= (4x-2)/(x^2-36)

Solve: (fg)(x) and (f/g)(x)
Find: Domain

Answer 1

Have you done this part: -->Solve: (fg)(x) and (f/g)(x)

Answer 2

I'm having trouble with that part...
I know I'm not doing it right

Answer 3

Solve: (fg)(x). Now, by (fg)(x), you're saying the product of f and g and not the composition of f and g which would be f(g(x)). Right?

Answer 4

I'm not sure honestly... probably the product...

Answer 5

If this --> (fg)(x) is how the question appears in your book or problem source, then it is the product of the two functions.

Answer 6

So, (fg)(x) = f(x) times g(x).

Answer 7

oh well then that makes it so much easier...

Answer 8

(fg)(x) = (5x+4)/(x^2-36) times (4x-2)/(x^2-36)

Crank that out, and let's see what we get.

Answer 9

Okay !

Answer 10

Post what you cranked out as the product. The fun begins now in finding the domain and range.

Answer 11

so far I have 20x^2+6x-8/x^4-72x^2+1296

Answer 12

I think that is correct. Write it this way: (20x^2+6x-8) / (x^4-72x^2+1296) so that it is clear which terms are in the numerator and which are in the denominator.

Answer 13

Okay but now I have to simplify

Answer 14

To get the domain, a person always checks to see which values of x cause the denominator to be 0. Those values are not acceptable in the domain.

Answer 15

Nothing in the instructions about simplifying.

Answer 16

What is needed is to reverse all that multiplication and look at the factors of the product.

Answer 17

fg(x) = [((5x+4)*(4x-2))/ [(x^2 - 36)^2 ]

Answer 18

The domain is the set of all acceptable values of x.

Answer 19

What values of x make the denominator [(x^2 - 36)^2 ] = 0?

We need to know those because they have to be excluded from the domain.

Answer 20

okay that makes sense

Answer 21

so it would be [-6,infinity) U (-6,6) U (-infinity,6) ??

Answer 22

Which values of x are you excluding and why?

Answer 23

What values of x make the denominator [(x^2 - 36)^2 ] = 0?

Answer 24

(x^2 - 36) = (x + ? ) * ( x - ?)

Answer 25

x+6 and x-6

Answer 26

So, x = 6 and x = -6 are values that cause (x^2 - 36) to equal zero AND cause the denominator of (fg)(x) to be zero.

Answer 27

They (6 and -6) must be excluded from the domain.

Answer 28

I am not sure about this interval notation --> [-6,infinity) U (-6,6) U (-infinity,6).

Answer 29

Okay i thought that might have been it because you cant have -6 and 6

Answer 30

[-6,infinity) means -6 ≤ x < infinity.

That half open inverval includes -6 which cannot be part of the domain.

Answer 31

> thought that might have been it because you cant have -6 and 6
True but you wrote something different.

Answer 32

The domain of (fg)(x) is all the Reals except 6 and -6. Now, how do you write that in interval notation?

Answer 33

Ugh this sucks. Haha sorry I'm in Calc and haven't done these since Algebra

Answer 34

You know the concept. So, rewrite your interval notation.

I think the first section will begin ( - ∞ , -6 ) U ?

Answer 35

In this --> [-6,infinity) you have -6 is greater than infinity.

No big deal. Just think about the symbols.

Answer 36

Yes that's it !!

Answer 37

Hey, we are not finished with the domain.

( - ∞ , -6 ) U is the first of 3 sections of the domain answer to the first part of the problem.

Answer 38

It was multiple choice and there was only one with ( - ∞ , -6 ) U

Answer 39

( - ∞ , -6 ) U ( , ) U ( 6, )

Answer 40

(f/g)(x) = [(5x+4)/(x^2-36)] / [(4x-2)/(x^2-36)]

Answer 41

Haha woah watch it there mister.. leave the attitude at home. But I guess if we must continue..

Answer 42

(f/g)(x) = [(5x+4)/(x^2-36)] / [(4x-2)/(x^2-36)]

I'll leave this for you to work. If you post your result, I'll check it for you.