## BloomLocke367 one year ago sketch a graph of function f that satisfies these conditions

1. mathmate

.

2. BloomLocke367

$\lim_{x \rightarrow 2} f(x)=-3$ $\lim_{x \rightarrow -3^-} f(x)=f(2)=5$ 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

it has to satisfy all of those D:

4. mathmate

Can you check the following? f is constant on (x, +inf)

5. BloomLocke367

Yes I got that. ^.^

6. BloomLocke367

it's an open circle on (2,2) right?

7. mathmate

is it "f(x) is constant on (2,+inf)?

8. BloomLocke367

no, it just says f. I'm pretty sure f and f(x) mean the same thing, right?

9. mathmate

but is "f is constant on (2,+inf)"? (i.e. not (x,+inf) )

10. BloomLocke367

yes

11. BloomLocke367

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

or is it "f is constant on (-3,+inf)"?

13. BloomLocke367

f is constant on (2, +inf)

14. mathmate

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

15. BloomLocke367

wait... why -3?

16. BloomLocke367

wait....nvm. I see now

17. mathmate

From the first condition, $$Lim_{x->2} ~~f(x)=-3$$

18. mathmate

so far so good?

19. mathmate

from f(2)=5 (in the second line), we have |dw:1441334065831:dw|

20. mathmate

ok there?

21. BloomLocke367

is that vertical line supposed to be there?

22. mathmate

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

sorry, third line (increasing up to x=-3) |dw:1441334303013:dw|

24. BloomLocke367

I'm confused now. :/

25. mathmate

fourth line says f(3+) < f(3-)

26. mathmate

Sorry, I went on since you did not respond. Starting from where did you get confused.

27. mathmate

The vertical line is just an indicator, not part of the graph. Functions cannot have vertical lines.

28. mathmate

|dw:1441334492160:dw|

29. mathmate

are you still following?

30. mathmate

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

31. BloomLocke367

ohhh okay. so (2,5) is just a point, there isn't a line connected to it

32. mathmate

no, the line just shows it's x-coordinate.

33. mathmate

*its

34. BloomLocke367

oh okay. what would the y-coordinate be for the line with the curve?

35. mathmate

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

36. mathmate

|dw:1441334766038:dw|

37. BloomLocke367

wait... how do you know the limit is 5?

38. mathmate

Line 2.

39. BloomLocke367

oh right *facepalm*

40. mathmate

The last part is more complicated.

41. mathmate

This is the part between x=-3 and +2

42. BloomLocke367

okay. I'm staying calm so far XD helping someone else helped me xp

43. mathmate

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

and we know Lim x->2 = -3.

45. mathmate

So all we can do is to join the two points with any line we want.

46. BloomLocke367

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

47. mathmate

|dw:1441335204853:dw|

48. mathmate

we don't even know what Lim x->-3+ is so it could be anything < 5.

49. mathmate

So that's the best we could do. Any comments?

50. BloomLocke367

would this work?|dw:1441335313929:dw|

51. BloomLocke367

man that is sloppy XD but idc

52. mathmate

Almost, it does not satisfy Lim x->2 = -3 (i.e. both sides, 2- and 2+)

53. mathmate

|dw:1441335463481:dw|

54. BloomLocke367

I put that the second closed dot at (1,0)

55. BloomLocke367

so would that work? please..before you go