sketch a graph of function f that satisfies these conditions

- BloomLocke367

sketch a graph of function f that satisfies these conditions

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- mathmate

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- 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

it has to satisfy all of those D:

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## More answers

- mathmate

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

- BloomLocke367

Yes I got that. ^.^

- BloomLocke367

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

- mathmate

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

- BloomLocke367

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

- mathmate

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

- BloomLocke367

yes

- 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

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

- BloomLocke367

f is constant on (2, +inf)

- mathmate

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

- BloomLocke367

wait... why -3?

- BloomLocke367

wait....nvm. I see now

- mathmate

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

- mathmate

so far so good?

- mathmate

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

- mathmate

ok there?

- BloomLocke367

is that vertical line supposed to be there?

- 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

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

- BloomLocke367

I'm confused now. :/

- mathmate

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

- mathmate

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

- mathmate

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

- mathmate

|dw:1441334492160:dw|

- mathmate

are you still following?

- mathmate

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

- BloomLocke367

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

- mathmate

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

- mathmate

*its

- BloomLocke367

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

- mathmate

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

- mathmate

|dw:1441334766038:dw|

- BloomLocke367

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

- mathmate

Line 2.

- BloomLocke367

oh right *facepalm*

- mathmate

The last part is more complicated.

- mathmate

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

- BloomLocke367

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

- 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

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

- mathmate

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

- BloomLocke367

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

- mathmate

|dw:1441335204853:dw|

- mathmate

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

- mathmate

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

- BloomLocke367

would this work?|dw:1441335313929:dw|

- BloomLocke367

man that is sloppy XD but idc

- mathmate

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

- mathmate

|dw:1441335463481:dw|

- BloomLocke367

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

- BloomLocke367

so would that work? please..before you go

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