BloomLocke367
  • BloomLocke367
sketch a graph of function f that satisfies these conditions
Mathematics
chestercat
  • chestercat
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mathmate
  • mathmate
.
BloomLocke367
  • 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)
BloomLocke367
  • BloomLocke367
it has to satisfy all of those D:

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mathmate
  • mathmate
Can you check the following? f is constant on (x, +inf)
BloomLocke367
  • BloomLocke367
Yes I got that. ^.^
BloomLocke367
  • BloomLocke367
it's an open circle on (2,2) right?
mathmate
  • mathmate
is it "f(x) is constant on (2,+inf)?
BloomLocke367
  • BloomLocke367
no, it just says f. I'm pretty sure f and f(x) mean the same thing, right?
mathmate
  • mathmate
but is "f is constant on (2,+inf)"? (i.e. not (x,+inf) )
BloomLocke367
  • BloomLocke367
yes
BloomLocke367
  • 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?
mathmate
  • mathmate
or is it "f is constant on (-3,+inf)"?
BloomLocke367
  • BloomLocke367
f is constant on (2, +inf)
mathmate
  • mathmate
Well, we have Lim f(x) x->2 = -3, so the horizontal line should be at y=-3. |dw:1441333851902:dw| agree?
BloomLocke367
  • BloomLocke367
wait... why -3?
BloomLocke367
  • BloomLocke367
wait....nvm. I see now
mathmate
  • mathmate
From the first condition, \(Lim_{x->2} ~~f(x)=-3\)
mathmate
  • mathmate
so far so good?
mathmate
  • mathmate
from f(2)=5 (in the second line), we have |dw:1441334065831:dw|
mathmate
  • mathmate
ok there?
BloomLocke367
  • BloomLocke367
is that vertical line supposed to be there?
mathmate
  • 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
mathmate
  • mathmate
sorry, third line (increasing up to x=-3) |dw:1441334303013:dw|
BloomLocke367
  • BloomLocke367
I'm confused now. :/
mathmate
  • mathmate
fourth line says f(3+) < f(3-)
mathmate
  • mathmate
Sorry, I went on since you did not respond. Starting from where did you get confused.
mathmate
  • mathmate
The vertical line is just an indicator, not part of the graph. Functions cannot have vertical lines.
mathmate
  • mathmate
|dw:1441334492160:dw|
mathmate
  • mathmate
are you still following?
mathmate
  • mathmate
fourth line says f(3+) < f(3-) but does not define anywhere between x=-3 and x=2, so it is arbitrary.
BloomLocke367
  • BloomLocke367
ohhh okay. so (2,5) is just a point, there isn't a line connected to it
mathmate
  • mathmate
no, the line just shows it's x-coordinate.
mathmate
  • mathmate
*its
BloomLocke367
  • BloomLocke367
oh okay. what would the y-coordinate be for the line with the curve?
mathmate
  • mathmate
It's not defined. We only know that it is increasing, with a limit of 5 at x=-3-
mathmate
  • mathmate
|dw:1441334766038:dw|
BloomLocke367
  • BloomLocke367
wait... how do you know the limit is 5?
mathmate
  • mathmate
Line 2.
BloomLocke367
  • BloomLocke367
oh right *facepalm*
mathmate
  • mathmate
The last part is more complicated.
mathmate
  • mathmate
This is the part between x=-3 and +2
BloomLocke367
  • BloomLocke367
okay. I'm staying calm so far XD helping someone else helped me xp
mathmate
  • 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.
mathmate
  • mathmate
and we know Lim x->2 = -3.
mathmate
  • mathmate
So all we can do is to join the two points with any line we want.
BloomLocke367
  • BloomLocke367
a jump. so this one will have to have a closed circle somewhere under the open circle at (-3,5)
mathmate
  • mathmate
|dw:1441335204853:dw|
mathmate
  • mathmate
we don't even know what Lim x->-3+ is so it could be anything < 5.
mathmate
  • mathmate
So that's the best we could do. Any comments?
BloomLocke367
  • BloomLocke367
would this work?|dw:1441335313929:dw|
BloomLocke367
  • BloomLocke367
man that is sloppy XD but idc
mathmate
  • mathmate
Almost, it does not satisfy Lim x->2 = -3 (i.e. both sides, 2- and 2+)
mathmate
  • mathmate
|dw:1441335463481:dw|
BloomLocke367
  • BloomLocke367
I put that the second closed dot at (1,0)
BloomLocke367
  • BloomLocke367
so would that work? please..before you go

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