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abannavong

  • 2 years ago

help with AP calc hw plz!

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  1. Hero
    • 2 years ago
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    Simply posting your actual question would be great

  2. abannavong
    • 2 years ago
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    im going to take a pic :3

  3. abannavong
    • 2 years ago
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  4. abannavong
    • 2 years ago
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    the limit is x->-1/2+

  5. Hero
    • 2 years ago
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    You've got like 5 questions circled. Which one is it?

  6. abannavong
    • 2 years ago
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    #38

  7. campbell_st
    • 2 years ago
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    perhaps factorising the numerator and denominator will allow you to cancel a common factor

  8. abannavong
    • 2 years ago
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    kk

  9. satellite73
    • 2 years ago
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    limit as x goes to ???

  10. Hero
    • 2 years ago
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    -1/2

  11. campbell_st
    • 2 years ago
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    there is a common factor that is evident when you factorise numerator and denominator

  12. satellite73
    • 2 years ago
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    then replace \(x\) by \(-\frac{1}{2}\) if you get a number, that is your answer any rational function is continuous on its domain so the limit and the value are the same

  13. Hero
    • 2 years ago
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    I think @campbell_st is right. You should try factoring the numerator and denominator first before plugging in x = -1/2

  14. satellite73
    • 2 years ago
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    if you get \(\frac{2}{0}\) then you know you can factor both top and bottom as \((2x+1)(\text{something)}\)

  15. satellite73
    • 2 years ago
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    i wouldn't do that, i would plug in first

  16. campbell_st
    • 2 years ago
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    the other thing about eliminating a common factor is that it makes the substitution calculation easier

  17. satellite73
    • 2 years ago
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    if you get something other than \(\frac{0}{0}\) then that is the answer if you do get \(\frac{0}{0}\) then you know how to factor

  18. Hero
    • 2 years ago
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    If you plug in first, you're definitely going to get a denominator you won't like

  19. satellite73
    • 2 years ago
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    @campbell_st i suck at factoring, so i like to know if i can factor first if i know i have a zero of a polynomial, i know already how to factor it that is why i would check first and not waste my time factoring, in case for example it does not factor over the integers

  20. abannavong
    • 2 years ago
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    so plugin -1/2 im kinda confused

  21. Hero
    • 2 years ago
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    Why waste your time trying to evaluate before factoring? Seems silly

  22. campbell_st
    • 2 years ago
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    lol... people get lost with squaring and substituting negative fractions... its really what works best you each of us

  23. satellite73
    • 2 years ago
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    what i am trying to say is that if i replace \(x\) by \(-\frac{1}{2}\) i know it must factor as \((2x+1)\times (\text{whatever})\)

  24. Zarkon
    • 2 years ago
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    rational functions are continuous on their domain..if you plug in the value and you get a real number then that is the answer to the limit problem

  25. satellite73
    • 2 years ago
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    because it tells me exactly how to factor otherwise i am doing the dance of what two numbers blah blah blah

  26. campbell_st
    • 2 years ago
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    here is the factorised version I hope... \[\lim_{x \rightarrow -\frac{1}{2}} \frac{(2x + 1)(3x -1)}{(2x + 1)(2x - 3)}\]

  27. abannavong
    • 2 years ago
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    kk

  28. Hero
    • 2 years ago
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    @satellite lost me. You're telling me that plugging in x = -1/2 tells you which two numbers you use to factor it?

  29. satellite73
    • 2 years ago
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    @hero both @zarkon and i answered the question "why evaluate first" i think

  30. satellite73
    • 2 years ago
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    yes, by the "factor theorem" if \(r\) is a root of a polynomial \(p(x)\) then we know \(p(x)=(x-r)q(x)\)

  31. Hero
    • 2 years ago
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    No, @zarkon said something I already knew. The problem is, if you plug in x = -1/2 in this case, you clearly get a zero denominator

  32. Agent47
    • 2 years ago
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    This limit is begging for L'Hopital's Rule.....

  33. satellite73
    • 2 years ago
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    it is not some miracle that each time you get \(\frac{0}{0}\) you can factor and cancel. it is precisely because of the factor theorem. you HAVE to be able to factor out the zeros

  34. campbell_st
    • 2 years ago
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    it doesn't need L'Hopital ... just factor to find it.... don't over think it

  35. abannavong
    • 2 years ago
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    im kinda confuzzled a bit :3

  36. Agent47
    • 2 years ago
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    i can do lhopital in my head on this... but i guess it depends

  37. Hero
    • 2 years ago
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    That's because there's too many people trying to give their version of how to solve it.

  38. satellite73
    • 2 years ago
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    what i am trying to say is this: you want the limit of a rational function as x approaches some number first thing you should do is check what you get when you evaluate the function at that number if you get a number back, that is your answer and you are done

  39. campbell_st
    • 2 years ago
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    looking at the type of questions you need to factorise to find the limit..

  40. Agent47
    • 2 years ago
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    abanna, just listen to satellite, he's always right lol

  41. Hero
    • 2 years ago
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    I object to that statement^

  42. satellite73
    • 2 years ago
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    if you get \(\frac{a}{0}\) where \(a\) is some non zero number, then there is no limit

  43. campbell_st
    • 2 years ago
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    @abannavong have a look above at my factorised version.. eliminate the common factor then substitute for your limit.

  44. satellite73
    • 2 years ago
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    and if you get \(\frac{0}{0}\) then you know to factor and cancel AND you know how to factor

  45. Hero
    • 2 years ago
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    I see what you're saying now, @satellite. I usually do that part in my head first, but not on paper.

  46. abannavong
    • 2 years ago
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    oh ok

  47. satellite73
    • 2 years ago
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    because for example, in this case we see that \(-\frac{1}{2}\) gives 0 that tells us that it MUST factor as \((2x+1)(\text{whatever})\) because \(2x+1\) has a zero at \(-\frac{1}{2}\)

  48. campbell_st
    • 2 years ago
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    @abannavong what level of maths is this...

  49. satellite73
    • 2 years ago
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    and your job is then only to find the other factor, as @campbell_st wrote above

  50. Agent47
    • 2 years ago
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    AB calculus, aka cal 1

  51. abannavong
    • 2 years ago
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    AP Calc :3

  52. campbell_st
    • 2 years ago
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    ok... so still high school... just I'm in australia...

  53. abannavong
    • 2 years ago
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    yeah in grade 12

  54. campbell_st
    • 2 years ago
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    thanks... I think you get 5/8 as the limit... good luck

  55. abannavong
    • 2 years ago
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    thank you @campbell_st

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