Need 1 Problem help and 1 answer check

Discussion in 'General Discussions' started by rarehunternick, Jan 29, 2009.

  1. rarehunternick

    rarehunternick Level II

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    Ok, i'm trying to find a domain of a function. Its been like a year since i did the domains of a function, except for my last question. Anyway,

    find the domain of:

    F(X)= sqrt((-17x+3)/(-19x-17))

    I think i did something wrong, but i can't figure it out. For the domain, i got (-infinity,-17/19)U(-17/19,3/17]. Maybe i did something wrong. So, help again please.
     
  2. mjnskivt

    mjnskivt Level III

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    Re: Need help again...math

    Are you allowed to deal with nonreal #s? If not, then by looking at the denominator, x<(-17/19), which would eliminate the second part of your domain.
     
  3. rarehunternick

    rarehunternick Level II

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    Re: Need help again...math

    no nonreal #s.
     
  4. Ice Nine

    Ice Nine Level IV

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    Re: Need help again...math

    You are almost correct...

    the answer is (-infinity,-17/19)U[3/17,infinity)

    Just see where the numerator is + or - and where the denominator is + or -
    numerator: + for x > 3/17, - for x < 3/17, numerator = 0 @ x = 3/17
    denominator: + for x < -17/19, - for x > -17/19, denominator = 0 @ x = -17/19

    f(x) is defined for when both numerator and denominator have the same sign (or when the numerator is 0)
     
  5. DeNo

    DeNo Level III

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    Re: Need help again...math

    Too easy.
    Gee sif u dont get it.

    LIke, solve this

    2+1x2=x

    I bet you got 6 right? xD
     
  6. Ice Nine

    Ice Nine Level IV

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    Re: Need help again...math

    >:[ no need to be mean...and his problem wasn't that easy...the negative signs made it easy to get tripped up, especially if you are doing the problem under time pressure.
     
  7. rarehunternick

    rarehunternick Level II

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    Re: Need help again...math (NEW PROBLEMS...HELP PLEASE)

    [​IMG]

    second problem:

    [​IMG]
    can anyone solve?

    ---

    edit: Images of problems...
     
  8. mjnskivt

    mjnskivt Level III

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    Re: Need help again...math (NEW PROBLEMS...HELP PLEASE)

    First question I think answer is C=6. For the function to be continuous, the values of both pieces must be equal at 2, so you plug 2 in for x to get 2C+6 and 4C-6. Then you set them equal, 2C+6=4C-6. From there you solve for C.

    The second question is the same idea, you plug in -2 for x to get -2m-10 and 4-20-8=-24. So -2m-10=-24, solve for m, giving you m+7.

    Hope I'm right :)
     
  9. Ice Nine

    Ice Nine Level IV

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    Re: Need help again...math (NEW PROBLEMS...HELP PLEASE)

    This is correct. C=6, m=7
     
  10. rarehunternick

    rarehunternick Level II

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    Re: Math Help again please (See new post and info)

    ok... i got some new problems that I'm stuck on. Oh, their are 2 that I just need to see if my answers are correct and 4 that I need help with. I'll give more details after i post them.

    Answers needed to be checked:
    [​IMG]

    ** Note: I'm just one answer off and can't figure it out
    ** Note: These are the problems that I don't understand. You don't have to do all of them because the last 3 are related so, you only have to explain one of them to me so I can attempt them.

    ------------------------------------------------------------------------------------------------------

    Thanks for all the help guys. Usually the thing that I post are just for checking and if i haven't covered them in the class. This topic will be constantly updated when i have trouble.
     
  11. rarehunternick

    rarehunternick Level II

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    Re: Math Help again please (See new post and info)

    Additional Problem that I need help with:

    [​IMG]
     
  12. rarehunternick

    rarehunternick Level II

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    Re: Math Help again please (See new post and info)

    Update: I only need help on 1 problem now and 1 answer check
     
  13. Will

    Will Level IV

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    Re: Math Help again please (See new post and info)





    If f(x)=root 2x
    f'(x)=1/(root2x)


    root 2x = 2x^(1/2)
    f'(x)=2x^(-1/2)
    When the power is negative, it implies that the base number is the denominator to 1 (the numerator).







    EDIT:

    Dude, stop spamming.
    If you don't have something worth while to say, don't spam for cash.
     
    Commy likes this.