what is the approximate value of the function at x = -4?
there is a graph.

- katherinesmith

what is the approximate value of the function at x = -4?
there is a graph.

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- katherinesmith

this is the graph

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- zepdrix

So when we look over at x=-4, we try to figure out at what height the red line is located.
Looks like it's between 1 and 2 yes?
Maybe 1.75ish?

- katherinesmith

is it closer to 1.5 or 1.8? i'm assuming 1.8

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## More answers

- zepdrix

Yah 1.8 sounds a little bit more accurate :)

- zepdrix

It's probably 1.5 at around x=-6

- katherinesmith

i have another one so stay here!

- katherinesmith

what is the approximate value of the function at x = 2?

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- zepdrix

Hmm, so our red line is below the x-axis at x=2, right?
Take a guess! :)

- katherinesmith

either 0.25 or 0.50... positive or negative

- zepdrix

Since it's below the x-axis, it will definitely be negative.
It's between -1 and 0.
Yah -0.25 seems like a good guess!

- katherinesmith

do you know how to do ones like :
solve x/(x-2) < 2

- zepdrix

\[\Large \frac{x}{x-2}\lt2\]Let's ummmm, multiply both sides by x-2,\[\Large x \lt 2(x-2)\]Understand that step? :o
The (x-2)'s cancel on the left side.

- katherinesmith

yes.

- katherinesmith

so that leaves you with x < 2x - 4

- katherinesmith

then you put x on one side and get -x < - 4, then it becomes x > 4. correct answer?

- zepdrix

Hmm one sec, yes that's part of our inequality.
I think we should be getting another part also though.
Lemme see what I messed up.

- katherinesmith

x > 4 could be a correct answer... i don't know why it wouldn't be. but let me know what you get.

- zepdrix

Oh you have some multiple choice answers or no? :

- katherinesmith

no i don't.

- zepdrix

oh :)

- zepdrix

The problem is this...
Remember what happens when we multiply/divide across an inequality?
The inequality sign `flips` directions.
Example:\[\Large -3x \lt 6\]Dividing both sides by -3 gives us,\[\Large x>-2\]See how the sign changed from `less than` to `greater than` ?

- katherinesmith

yes, which i did that.

- zepdrix

So in our problem, I think it's a bad idea to multiply both sides by (x-2) as I did.
We don't know whether or not that quantity is negative.
So let's try something else :O

- katherinesmith

okay

- zepdrix

\[\Large \frac{x}{x-2}\lt2\]Subtracting 2 from each side,\[\Large \frac{x}{x-2}-2<0\]Let's get a common denominator:\[\Large \frac{x}{x-2}-\frac{2(x-2)}{x-2}\lt0\]

- zepdrix

So what does that simplify down to? :)\[\Large \frac{x-2(x-2)}{x-2}<0\]

- zepdrix

Hmm I hope I'm doing this right, man I hate inequalities lol.
Lemme check one of the smarty pants guys :P
@SithsAndGiggles @satellite73

- katherinesmith

let me know. im still confused

- anonymous

I don't think putting the 2 on the other side is really necessary. I'd go with
\[x<2(x-2)\]
which gives
\[x<2x-4\\
-x<-4\]
Either way works, but I think this is simpler. ;)

- katherinesmith

so that leaves you with x > 4. because you divide by -x and flip the inequality.

- zepdrix

Well if you wolfram this, it gives 2 intervals, x<2, x>4.
So I was just getting a little confused :d

- katherinesmith

how did you get the two intervals? because thats' the answer i need

- zepdrix

Because from this point, where I combined the fractions:\[\Large \frac{x-2(x-2)}{x-2}<0\]After we simplify, we get,\[\Large \frac{-x+4}{x-2}\lt0\]

- zepdrix

If we look at the top and bottom separately (which I guess is what we're supposed to do),\[\Large -x+4\lt0 \qquad\qquad\qquad x-2\lt0\]

- zepdrix

\[\Large -x+4\lt0 \qquad\to\qquad x>4\]\[\Large x-2\lt0 \quad\qquad\to\qquad x<2\]

- zepdrix

That too confusing? :d
Or was it back when we combined the fractions the part you got stuck on?

- katherinesmith

this all just hurts my head. lord jesus help me

- zepdrix

XD

- katherinesmith

alright one more question

- zepdrix

k c:

- katherinesmith

|dw:1377555967193:dw|

- katherinesmith

that's an = sign and then a - sign

- zepdrix

Solve for x?

- katherinesmith

yes

- zepdrix

Note:\[\Large x^2+x \qquad=\qquad x(x+1)\]

- zepdrix

\[\Large \frac{x+5}{x(x+1)}=\frac{1}{x(x+1)}-\frac{x-6}{x+1}\]

- katherinesmith

then what the heck do you do

- zepdrix

If we multiply both sides by (x+1), can you see what will happen? :o

- katherinesmith

please write it out for me

- zepdrix

Lemme give you a simpler example really quick :)
Maybe it will help it to click.\[\Large \frac{x}{3}=\frac{7}{3}\]
If we multiply both sides by 3, what does it give us?

- katherinesmith

x = 7

- zepdrix

good, all of the denominator 3's cancel with the 3's that we multiplied through by, right?

- katherinesmith

yes

- katherinesmith

\[\frac{ x + 5 }{ x } = \frac{ 1 }{ x } - x - 6\]

- katherinesmith

then what

- zepdrix

One small thing to be careful of.
We don't usually write it, but there is a pair of parenthesis implied when we have a negative in front of a fraction.\[\Large \frac{x+5}{x(x+1)}=\frac{1}{x(x+1)}-\frac{(x-6)}{x+1}\]

- zepdrix

\[\Large \frac{ x + 5 }{ x } = \frac{ 1 }{ x } -(x - 6)\]

- katherinesmith

I got that. what next.

- zepdrix

To finish getting rid of the fractions, let's multiply both sides by x.

- zepdrix

Gives us something like this, yes? :)\[\Large (x+5)=1-x(x-6)\]

- katherinesmith

yes

- zepdrix

From here, distribute your -x to each term in the brackets.
And then try to get everything to one side, so the other side =0

- zepdrix

crap i gotta go :U sorryyyy :C gotta go take care of something

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