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mathcalculus Group Title

PLEASE HELP :( Please help: sketch the graph of a function. label the intercepts, relative extrema, points of inflection, and asymptotes! then state the domain of the function! y= (5-3x)/(x-2)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    so i found the vertical asymptote: x=2 and horizontal asymptote: y=3 and the intercepts i found that it is (0, -2.5) and (1.67,0) so the domain is: (-inf,2)(2, inf) now I'm stuck on how to do the relative extrema and the points of inflections. :( please help.

    • one year ago
  2. mathcalculus Group Title
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    i tried to also find the derivative and i got 1/ (x-2)^2

    • one year ago
  3. mathcalculus Group Title
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    @zepdrix please help

    • one year ago
  4. zepdrix Group Title
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    So did you apply the Quotient Rule? One sec, I'll check it :O

    • one year ago
  5. mathcalculus Group Title
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    yeah i used the quotient rule

    • one year ago
  6. zepdrix Group Title
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    Mmm yah I think your first answer was correct: 1/ (x-2)^2

    • one year ago
  7. mathcalculus Group Title
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    yeah the first answer was correct.

    • one year ago
  8. mathcalculus Group Title
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    i mean I'm trying to find the relative extrema but i forgot how to. i know i have to find the derivative first. then I'm not sure where to go . :(

    • one year ago
  9. zepdrix Group Title
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    So we have our first derivative:\[\Large y'\quad=\quad \frac{1}{(x-2)^2}\] To find extrema/critical points, we set the first derivative equal to zero and solve for x.

    • one year ago
  10. mathcalculus Group Title
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    right but wouldn't it be false?

    • one year ago
  11. mathcalculus Group Title
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    1=0? doesn't work.

    • one year ago
  12. zepdrix Group Title
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    Oh there is another little tid bit that I often forget. critical points also exist when the first derivative is `undefined`. Do you see any points where that might be true?

    • one year ago
  13. mathcalculus Group Title
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    by undefined what do you mean?

    • one year ago
  14. zepdrix Group Title
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    Numbers that would cause a problem :D Like dividing by zero.

    • one year ago
  15. zepdrix Group Title
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    Any x values that would cause a problem for our derivative?

    • one year ago
  16. mathcalculus Group Title
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    like this one?

    • one year ago
  17. mathcalculus Group Title
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    oh ya 2!

    • one year ago
  18. zepdrix Group Title
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    Yes! Good. x=2 is something we might consider to be a critical point. Before we can say that for certain though, is x=2 in the domain of our function?

    • one year ago
  19. zepdrix Group Title
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    You stated the domain earlier, so go back and look at it! :)

    • one year ago
  20. mathcalculus Group Title
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    no.

    • one year ago
  21. mathcalculus Group Title
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    x cannot equal to 2.

    • one year ago
  22. zepdrix Group Title
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    Good good. Since it's not in the domain of our function, it can't be a critical point.

    • one year ago
  23. mathcalculus Group Title
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    all real numbers except 2.

    • one year ago
  24. zepdrix Group Title
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    So there are no critical points. Should we look for inflection points next? :o

    • one year ago
  25. mathcalculus Group Title
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    so theres no relative extrema?

    • one year ago
  26. mathcalculus Group Title
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    yes last but not least, inflection points. :)

    • one year ago
  27. zepdrix Group Title
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    So we'll need to find the second derivative for that. Have you tried that step yet?

    • one year ago
  28. mathcalculus Group Title
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    no but let me try real quick

    • one year ago
  29. mathcalculus Group Title
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    so inflection points is always SECOND DERIVATIVE right?

    • one year ago
  30. zepdrix Group Title
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    y=0 gives us roots of the function. y'=0 gives us critical points of the function. y''=0 gives us inflection points of the function. yes.

    • one year ago
  31. mathcalculus Group Title
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    i got (-3x^3+17x^2-32x+20)/(x^2-4x+4)

    • one year ago
  32. mathcalculus Group Title
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    @zepdrix

    • one year ago
  33. zepdrix Group Title
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    Dude dude dude -_- what..? Don't expand out squares, especially if they're in the denominator.

    • one year ago
  34. zepdrix Group Title
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    Did you apply the quotient rule? We don't want to do that.

    • one year ago
  35. mathcalculus Group Title
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    yeah i did:/

    • one year ago
  36. mathcalculus Group Title
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    i used the rule book.

    • one year ago
  37. zepdrix Group Title
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    \[\Large y'\quad=\quad \frac{1}{(x-2)^2}\quad=\quad (x-2)^{-2}\]Power rule from there :O

    • one year ago
  38. mathcalculus Group Title
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    ohhhhh yes

    • one year ago
  39. mathcalculus Group Title
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    crap.

    • one year ago
  40. zepdrix Group Title
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    XD

    • one year ago
  41. mathcalculus Group Title
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    -2/(x-2)^3

    • one year ago
  42. mathcalculus Group Title
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    ok :D

    • one year ago
  43. zepdrix Group Title
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    Ok cool. Setting our second derivative equal to zero:\[\Large 0\quad=\quad \frac{-2}{(x-2)^3}\]

    • one year ago
  44. zepdrix Group Title
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    Hmm I think inflection points work the same way as critical points ~ they exist where the second derivative is undefined.. mmmmm

    • one year ago
  45. mathcalculus Group Title
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    huh?

    • one year ago
  46. mathcalculus Group Title
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    don't we plug in severals x's between the intervals or something?

    • one year ago
  47. mathcalculus Group Title
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    so threres no inflection point?

    • one year ago
  48. zepdrix Group Title
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    between what intervals though? We would need to find inflection points to determine which intervals to check. I guess we just check where the function has a discontinuity since we have no inflection points. So ya, check on the left side of 2 and also on the right. We should get different concavity on each side for this problem.

    • one year ago
  49. zepdrix Group Title
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    Oh the problem didn't ask for concavity did it? -_- oh.. hmm

    • one year ago
  50. mathcalculus Group Title
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    no it didn't...

    • one year ago
  51. mathcalculus Group Title
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    so help me with inflection point, do we have to include other x's between the intervals?im not sure

    • one year ago
  52. zepdrix Group Title
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    include other x's... whu? +_+

    • one year ago
  53. mathcalculus Group Title
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    so this is the answer: it doesn't say which is for which but i got the idea of what is what. my question is how'd they get the other curve on the right side

    • one year ago
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  54. zepdrix Group Title
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    So they labeled the intercepts (as you found), the asymptotes x=2, y=-3, and nothing else... So I think we did everything correctly, seeing as how we found no extrema for this one and they didn't label any on their graph.. :\

    • one year ago
  55. zepdrix Group Title
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    Oh the other curve, umm

    • one year ago
  56. zepdrix Group Title
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    bah i dunno >:O my head hurts.. i need a maf break!

    • one year ago
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