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mathcalculus

  • 2 years ago

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)

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  1. mathcalculus
    • 2 years ago
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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.

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

  3. mathcalculus
    • 2 years ago
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    @zepdrix please help

  4. zepdrix
    • 2 years ago
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    So did you apply the Quotient Rule? One sec, I'll check it :O

  5. mathcalculus
    • 2 years ago
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    yeah i used the quotient rule

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

  7. mathcalculus
    • 2 years ago
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    yeah the first answer was correct.

  8. mathcalculus
    • 2 years ago
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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 . :(

  9. zepdrix
    • 2 years ago
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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.

  10. mathcalculus
    • 2 years ago
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    right but wouldn't it be false?

  11. mathcalculus
    • 2 years ago
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    1=0? doesn't work.

  12. zepdrix
    • 2 years ago
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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?

  13. mathcalculus
    • 2 years ago
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    by undefined what do you mean?

  14. zepdrix
    • 2 years ago
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    Numbers that would cause a problem :D Like dividing by zero.

  15. zepdrix
    • 2 years ago
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    Any x values that would cause a problem for our derivative?

  16. mathcalculus
    • 2 years ago
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    like this one?

  17. mathcalculus
    • 2 years ago
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    oh ya 2!

  18. zepdrix
    • 2 years ago
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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?

  19. zepdrix
    • 2 years ago
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    You stated the domain earlier, so go back and look at it! :)

  20. mathcalculus
    • 2 years ago
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    no.

  21. mathcalculus
    • 2 years ago
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    x cannot equal to 2.

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

  23. mathcalculus
    • 2 years ago
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    all real numbers except 2.

  24. zepdrix
    • 2 years ago
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    So there are no critical points. Should we look for inflection points next? :o

  25. mathcalculus
    • 2 years ago
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    so theres no relative extrema?

  26. mathcalculus
    • 2 years ago
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    yes last but not least, inflection points. :)

  27. zepdrix
    • 2 years ago
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    So we'll need to find the second derivative for that. Have you tried that step yet?

  28. mathcalculus
    • 2 years ago
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    no but let me try real quick

  29. mathcalculus
    • 2 years ago
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    so inflection points is always SECOND DERIVATIVE right?

  30. zepdrix
    • 2 years ago
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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.

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

  32. mathcalculus
    • 2 years ago
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    @zepdrix

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

  34. zepdrix
    • 2 years ago
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    Did you apply the quotient rule? We don't want to do that.

  35. mathcalculus
    • 2 years ago
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    yeah i did:/

  36. mathcalculus
    • 2 years ago
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    i used the rule book.

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

  38. mathcalculus
    • 2 years ago
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    ohhhhh yes

  39. mathcalculus
    • 2 years ago
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    crap.

  40. zepdrix
    • 2 years ago
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    XD

  41. mathcalculus
    • 2 years ago
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    -2/(x-2)^3

  42. mathcalculus
    • 2 years ago
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    ok :D

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

  44. zepdrix
    • 2 years ago
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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

  45. mathcalculus
    • 2 years ago
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    huh?

  46. mathcalculus
    • 2 years ago
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    don't we plug in severals x's between the intervals or something?

  47. mathcalculus
    • 2 years ago
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    so threres no inflection point?

  48. zepdrix
    • 2 years ago
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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.

  49. zepdrix
    • 2 years ago
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    Oh the problem didn't ask for concavity did it? -_- oh.. hmm

  50. mathcalculus
    • 2 years ago
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    no it didn't...

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

  52. zepdrix
    • 2 years ago
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    include other x's... whu? +_+

  53. mathcalculus
    • 2 years ago
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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

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  54. zepdrix
    • 2 years ago
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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.. :\

  55. zepdrix
    • 2 years ago
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    Oh the other curve, umm

  56. zepdrix
    • 2 years ago
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    bah i dunno >:O my head hurts.. i need a maf break!

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