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BloomLocke367

  • one year ago

sketch a graph of function f that satisfies these conditions

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  1. mathmate
    • one year ago
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    .

  2. BloomLocke367
    • one year ago
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    \[\lim_{x \rightarrow 2} f(x)=-3\] \[\lim_{x \rightarrow -3^-} f(x)=f(2)=5\] f is increasing on (-inf, -3) \[\lim_{x \rightarrow -3^-}f(x)>\lim_{x \rightarrow -3^+}f(x)\] f is constant on (x, +inf)

  3. BloomLocke367
    • one year ago
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    it has to satisfy all of those D:

  4. mathmate
    • one year ago
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    Can you check the following? f is constant on (x, +inf)

  5. BloomLocke367
    • one year ago
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    Yes I got that. ^.^

  6. BloomLocke367
    • one year ago
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    it's an open circle on (2,2) right?

  7. mathmate
    • one year ago
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    is it "f(x) is constant on (2,+inf)?

  8. BloomLocke367
    • one year ago
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    no, it just says f. I'm pretty sure f and f(x) mean the same thing, right?

  9. mathmate
    • one year ago
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    but is "f is constant on (2,+inf)"? (i.e. not (x,+inf) )

  10. BloomLocke367
    • one year ago
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    yes

  11. BloomLocke367
    • one year ago
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    I have, right now, an open circle on point (2,2) with a horizontal line going to the right c: that's right, right?

  12. mathmate
    • one year ago
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    or is it "f is constant on (-3,+inf)"?

  13. BloomLocke367
    • one year ago
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    f is constant on (2, +inf)

  14. mathmate
    • one year ago
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    Well, we have Lim f(x) x->2 = -3, so the horizontal line should be at y=-3. |dw:1441333851902:dw| agree?

  15. BloomLocke367
    • one year ago
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    wait... why -3?

  16. BloomLocke367
    • one year ago
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    wait....nvm. I see now

  17. mathmate
    • one year ago
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    From the first condition, \(Lim_{x->2} ~~f(x)=-3\)

  18. mathmate
    • one year ago
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    so far so good?

  19. mathmate
    • one year ago
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    from f(2)=5 (in the second line), we have |dw:1441334065831:dw|

  20. mathmate
    • one year ago
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    ok there?

  21. BloomLocke367
    • one year ago
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    is that vertical line supposed to be there?

  22. mathmate
    • one year ago
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    The rest of line 2 (lim x->3- = 5 ) gives: |dw:1441334208871:dw| then from the fourth line f is increasing up to x=-3

  23. mathmate
    • one year ago
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    sorry, third line (increasing up to x=-3) |dw:1441334303013:dw|

  24. BloomLocke367
    • one year ago
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    I'm confused now. :/

  25. mathmate
    • one year ago
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    fourth line says f(3+) < f(3-)

  26. mathmate
    • one year ago
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    Sorry, I went on since you did not respond. Starting from where did you get confused.

  27. mathmate
    • one year ago
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    The vertical line is just an indicator, not part of the graph. Functions cannot have vertical lines.

  28. mathmate
    • one year ago
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    |dw:1441334492160:dw|

  29. mathmate
    • one year ago
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    are you still following?

  30. mathmate
    • one year ago
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    fourth line says f(3+) < f(3-) but does not define anywhere between x=-3 and x=2, so it is arbitrary.

  31. BloomLocke367
    • one year ago
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    ohhh okay. so (2,5) is just a point, there isn't a line connected to it

  32. mathmate
    • one year ago
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    no, the line just shows it's x-coordinate.

  33. mathmate
    • one year ago
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    *its

  34. BloomLocke367
    • one year ago
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    oh okay. what would the y-coordinate be for the line with the curve?

  35. mathmate
    • one year ago
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    It's not defined. We only know that it is increasing, with a limit of 5 at x=-3-

  36. mathmate
    • one year ago
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    |dw:1441334766038:dw|

  37. BloomLocke367
    • one year ago
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    wait... how do you know the limit is 5?

  38. mathmate
    • one year ago
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    Line 2.

  39. BloomLocke367
    • one year ago
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    oh right *facepalm*

  40. mathmate
    • one year ago
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    The last part is more complicated.

  41. mathmate
    • one year ago
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    This is the part between x=-3 and +2

  42. BloomLocke367
    • one year ago
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    okay. I'm staying calm so far XD helping someone else helped me xp

  43. mathmate
    • one year ago
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    We don't know much about it. We only know that - lim x->3- f(x) is greater than lim x->3+ f(x), so there is a jump discontinuity.

  44. mathmate
    • one year ago
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    and we know Lim x->2 = -3.

  45. mathmate
    • one year ago
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    So all we can do is to join the two points with any line we want.

  46. BloomLocke367
    • one year ago
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    a jump. so this one will have to have a closed circle somewhere under the open circle at (-3,5)

  47. mathmate
    • one year ago
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    |dw:1441335204853:dw|

  48. mathmate
    • one year ago
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    we don't even know what Lim x->-3+ is so it could be anything < 5.

  49. mathmate
    • one year ago
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    So that's the best we could do. Any comments?

  50. BloomLocke367
    • one year ago
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    would this work?|dw:1441335313929:dw|

  51. BloomLocke367
    • one year ago
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    man that is sloppy XD but idc

  52. mathmate
    • one year ago
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    Almost, it does not satisfy Lim x->2 = -3 (i.e. both sides, 2- and 2+)

  53. mathmate
    • one year ago
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    |dw:1441335463481:dw|

  54. BloomLocke367
    • one year ago
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    I put that the second closed dot at (1,0)

  55. BloomLocke367
    • one year ago
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    so would that work? please..before you go

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