G(x) and F(X) of following (I have work done)

In summary: So:[2(x-2) - 4(x-2)(x+2)^2]/(x+2)would become:2(x-2) - 4(x-2)(x+2)and then I would just do like so:subtracting like terms to leave:[2(x-2)]-(4x-4)
  • #1
woox
17
0

Homework Statement


f(x)=(2x-4)/(x+2)
g(x) = 4x2-16

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


Homework Equations





The Attempt at a Solution



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

I got (-16x^2+4)/(x^2), but I can't 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
 
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  • #2
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
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
woox said:
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

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
AssyriaQ said:
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).

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

Is that correct?
 
  • #6
AssyriaQ said:
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).

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

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

Is that correct?

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:
  • #7
NoMoreExams said:
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)

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
No, if

[tex] g(x) = 4x^2 - 16 [/tex]

Then

[tex] g(blah) = 4(blah)^2 - 16 [/tex]
 
  • #9
NoMoreExams said:
No, if

[tex] g(x) = 4x^2 - 16 [/tex]

Then

[tex] g(blah) = 4(blah)^2 - 16 [/tex]

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

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

?
 
  • #10
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
NoMoreExams said:
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?

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
By simplify I mean find common denominator
 
  • #13
ah so the common denominator being x^2 resulting in:
[(-16x^2)+4]/x^2

correct? (multiplied by x^2/x^2)
 
  • #14
Yes, so what's 1/that?
 
  • #15
x^2/[(-16x^2)+4] ?
 
  • #16
Yes, although I would write it as

[tex] \frac{x^2}{4-16x^2}[/tex]
 
  • #17
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
No... you need to find a common denominator again!
 
  • #19
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
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.
 

FAQ: G(x) and F(X) of following (I have work done)

1. What is the difference between G(x) and F(x)?

G(x) and F(x) are both mathematical functions that relate an input variable (x) to an output variable. However, G(x) and F(x) may differ in the way they manipulate the input variable and the resulting output.

2. How are G(x) and F(x) used in scientific research?

G(x) and F(x) are commonly used in scientific research to model and analyze various phenomena. They can help to identify patterns, make predictions, and understand complex systems.

3. Can G(x) and F(x) be used interchangeably?

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4. What is the domain and range of G(x) and F(x)?

The domain of a function refers to all the possible input values, while the range refers to all the possible output values. The domain and range of G(x) and F(x) may vary depending on the specific function and its limitations.

5. How can I determine the value of G(x) and F(x) for a given input?

To determine the value of G(x) and F(x) for a specific input, simply substitute the input value for x in the function and solve for the resulting output variable. This process can be repeated for different input values to generate a set of corresponding output values.

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