# Homework Help: G(x) and F(X) of following (I have work done!)

1. Feb 3, 2009

### woox

1. The problem statement, all variables and given/known data
f(x)=(2x-4)/(x+2)
g(x) = 4x2-16

find and simplyify:
1/[g(1/x)]
f(x)-g(x)

2. Relevant equations

3. The attempt at a solution

For 1/[g(1/x)]:

I got (-16x^2+4)/(x^2), but I cant figure out how to simplify this down further, this assignment had other problems and they all simplified down to a whole number so I assume this one would too.

For f(x)-g(x)
Plugged in variables, found common denominator, and ended with this mess:
[(4x^3)+(8x^2)-14x+36]/(x+2)

I am sure that can get simplified further, and tried and was really geting no where, any help on these two? Am I doing something wrong or is this the right answer ? Thanks

2. Feb 3, 2009

### NoMoreExams

How did you get that for 1/g(1/x)

To work with f - g, first note that f = 2(x-2)/(x+2) and g = 4(x^2 - 4) = 4(x-2)(x+2)

3. Feb 3, 2009

### woox

For 1/[g(1/x)] I did was plug in (1/x) in 1/g(x) which gave me -> 1/[4*((1/x)^2)] and then I just flipped it adding the ^-1

For the fx-gx, I did see that but I didnt know what to do from there either, i still needed a common denominator and basically it sort of lead back to what I got.

4. Feb 3, 2009

### AssyriaQ

That sounds like you flipped twice, which brought you back to g(1/x). Do this stepwise: first substitute x by 1/x in g(x) to get g(1/x), and then take the inverse, i.e., 1/g(1/x).

5. Feb 3, 2009

### woox

Did that and got :
[(x^2)-1] / 16

Is that correct?

6. Feb 3, 2009

### NoMoreExams

Correct - first plug in 1/x then flip it but I would be careful to use "inverse".

Incorrect. If you want people to catch your mistake show your work! First show how you got g(1/x), SIMPLIFY FIRST. Then show what you do to get 1/g(1/x)

Last edited: Feb 3, 2009
7. Feb 3, 2009

### woox

ok so I start with:
gx=4x^2-16

inverse of that is:
1/[4x^2]-16

Then I plugin (1/x)
and get:
1/[4(1/x)^2]-16

Is that correct so far? Then do I flip it back? making it:
([4(1/x)^2]-16)^-1

8. Feb 3, 2009

### NoMoreExams

No, if

$$g(x) = 4x^2 - 16$$

Then

$$g(blah) = 4(blah)^2 - 16$$

9. Feb 3, 2009

### woox

So :
4(1/1/x)^2 - 16

taking inverse:
4(1/x)^2 - 16

?

10. Feb 3, 2009

### NoMoreExams

How are you getting 1/1/x? that notation doesn't even make sense since a/(b/c) <> (a/b)/c.

Your 2nd line is correct, g(1/x) = 4(1/x)^2 - 16, what does simplify to?

11. Feb 3, 2009

### woox

it should be:
g(1/x) = 4(1/x)^2 - 16

g(1/x) = 4(1^2/x^2) - 16

g(1/x) = 4(1/x^2) - 16

g(1/x) = (4/x^2) - 16

g(1/x) = 4[(1/x^2) - 4]

correct?

12. Feb 3, 2009

### NoMoreExams

By simplify I mean find common denominator

13. Feb 3, 2009

### woox

ah so the common denominator being x^2 resulting in:
[(-16x^2)+4]/x^2

correct? (multiplied by x^2/x^2)

14. Feb 3, 2009

### NoMoreExams

Yes, so what's 1/that?

15. Feb 3, 2009

### woox

x^2/[(-16x^2)+4] ?

16. Feb 3, 2009

### NoMoreExams

Yes, although I would write it as

$$\frac{x^2}{4-16x^2}$$

17. Feb 3, 2009

### woox

Cool, and for the second one I think i figured it out:

f = 2(x-2)/(x+2) and g = 4(x^2 - 4) = 4(x-2)(x+2)

We can factor out the x-2 and x+2, which gives us:
(x+2)(x-2)(-2)

I am not sure if I should multiple this out or leave it as is?

18. Feb 3, 2009

### NoMoreExams

No... you need to find a common denominator again!

19. Feb 3, 2009

### woox

f = 2(x-2)/(x+2) and g = 4(x-2)(x+2)

the common denominator would be (x+2) so:
2(x-2)-4(x-2)(x+2)(x+2)/(x+2)

now do I multiple variables together, and subtract likes leaving it over denominator? (That gave me this mess:

[(4x^3)+(8x^2)-14x+36]/(x+2)

)

20. Feb 3, 2009

### NoMoreExams

You need to use parantheses better, you mean

[2(x-2) - 4(x-2)(x+2)^2]/(x+2)

At this point I would factor out the terms that are common to both parts in the numerator.

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