Need math help!! Simple problem!

Discussion in 'After Hours Lounge (Off Topic)' started by Van Patton, Feb 3, 2004.

  1. Van Patton

    Van Patton Second Unit

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    OK I'm a freshman in college and it's been a while since I've done this so I'm rusty! The problem is: x^2-1/2x-4. Thanks for the help!
     
  2. Greg_R

    Greg_R Screenwriter

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    Ummm... I think we need the other side to the equation! (i.e. = 0 or something).
     
  3. Van Patton

    Van Patton Second Unit

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    Um........Well it stemmed from this. Here is the original equation.


    Given that f(x)=x^2-1 and g(x)=2x-4, find each of the following, if it exists.

    (f/g)(x)=
     
  4. Christ Reynolds

    Christ Reynolds Producer

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    i think it may be asking you to plug in, for every value of x in f(x), g(x). im not sure though, confused by how it is written. but if that is the case, you have (f/g)(x)=[(2x-4)^2]-1. just pretend that g(x), also 2x-4, is x in the f(x) equation, and plug it in, as i did above. you may be asking for something else entirely though [​IMG]

    CJ
     
  5. Van Patton

    Van Patton Second Unit

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    nah man thats good just solve that out and hoook me up with an answer. Thanks!!!
     
  6. Holadem

    Holadem Lead Actor

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    (f/g)(x)=

    This is f divided by g. There is nothing else to do other than what you already wrote, x^2-1/2x-4, unless you want to attempt long division. I don't see why though.

    Also, DYOFHW. I'll let you figure that one out.

    --
    H
     
  7. Edwin-S

    Edwin-S Producer
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    Just curious, do you think you will learn to do calculus by having someone else figure out the answers for you?
     
  8. Bryan X

    Bryan X Producer

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    [​IMG]
     
  9. Edwin-S

    Edwin-S Producer
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    Sorry, but this is a calculus question. It is not as simple as dividing the top by the bottom. It could be a case of substituting g(x) into f(x), but it could also be question that uses f' and g'.

    like f/g(x)= f(g'(x)) + f'(g(x)/g(x)^2

    I'm not sure if that equation is totally correct. I don't have my calculus text available, and even though I took the course I didn't do too well. I would actually have to look it up and see if it correlates.
     
  10. NickSo

    NickSo Producer

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    The best i could do dividng F by G is reducing it by factoring the numerator...

    I think its more complicated than that...
     
  11. Holadem

    Holadem Lead Actor

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    What makes you infer such things from the simple, if perhaps erronous statement f/g(x)?

    That's what I don't understand. How does someone start thinking of derivatives, with something like f/g(x)?!

    --
    H
     
  12. Morgan Jolley

    Morgan Jolley Lead Actor

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    What would really be helpful is the entire text of the equation, as well as an idea of what topic this deals with.
     
  13. Edwin-S

    Edwin-S Producer
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    Ah, yes, I vaguely remember now. You might be correct. It could be a substitution problem too, though. Damn, I wish I had my text book and notes here. Now it is going to bother me. I agree with your other comment though.......DYOFHW. [​IMG]

    Edit: I should add, wouldn't that require a limit to be implied? God....I can't believe I only took this stuff a little while ago and I'm already having trouble remembering how it went.
     
  14. Van Patton

    Van Patton Second Unit

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    Its not hard.......you just divide what I gave you the first time and that is our answer. It is SO easy and all you all are just making it much more harder than it needs to be.
     
  15. Haggai

    Haggai Producer

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    The college math lecturer (me) should probably step in here. Edwin, the rule you were thinking of is the quotient rule for derivatives. It allows you find the derivative of a function that has the form f(x)/g(x), where the / stands for division (as it usually does). That rule can be expressed this way:

    [f(x)/g(x)]' = [f'(x)g(x) - f(x)g'(x)]/g(x)^2

    However, the original question:



    does not appear to involve taking any derivatives. It's most likely just looking for f(x) divided by g(x). A similar notation can be used for other ways of combining functions:

    (f+g)(x) = f(x) + g(x)
    (f-g)(x) = f(x) - g(x)
    (fg)(x) = f(x)g(x)

    Lastly, I second (or third) Holadem's advice, DYOFHW. Ask for help, OK, but don't expect someone else to just do the problem.
     
  16. Van Patton

    Van Patton Second Unit

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    Look I did four out of five of the same problems just like this! So far NO ONE has helped me. I could've reiterated the babble that has come out of everyones mouths so far. Please help me divide that or dont post anything.
     
  17. Edwin-S

    Edwin-S Producer
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    Yes. Thank you, Haggai. I finally decided to use this wonderful thing called the internet and actually look up the notation and see if I was making a complete hash of it. Sure enough I was.

    I have always wanted to pass calculus and have yet to do so. I've taken the bloody 1st year course twice, but I just can't seem to get my head around it. With your explanation some of it started to come back a bit.
     
  18. Will Pomeroy

    Will Pomeroy Stunt Coordinator

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    If you're going to ask people to DYFHW, you could at least be polite...

    Its already in its simplest form. You might want to double check the question.
     
  19. Edwin-S

    Edwin-S Producer
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    Yes. I factored it, and it doesn't seem to be able to be simplified beyond the original form.
     
  20. Haggai

    Haggai Producer

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    As it was written here, it would factor as:

    [(x+1)(x-1)] / [2(x-2)]

    No other simplifying possible from that.
     

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