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baldymcgee6Best ResponseYou've already chosen the best response.0
http://screencast.com/t/5vrFs8Ggn
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
@zepdrix can you help?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
analysis? what does that mean? like you're trying to find critical points and such? :o
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Yeah, sorry... Max, min, inflections..
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
We have critical points anywhere the first derivative is 0 or UNDEFINED. They won't necessarily give us max or min values, but they are critical points, giving us some information :O \[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1x} }\cdot \frac{1}{(1+x)^2}\]\[f'(x)=\frac{1}{ \sqrt{1x} \sqrt{1+x}}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Did i simplify that correctly? The first square root term you had, the 1/2 power, i just flipped the fraction so it wasn't negative anymore :O
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Didn't know you can do that.. :/
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmm if you graph this on wolfram, you'll see that the critical points are actually telling us where the asymptotes are located :D http://www.wolframalpha.com/input/?i=y%3D%28%281x%29%2F%281%2Bx%29%29%5E%281%2F2%29
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Maybe not the most relevant if we just want max and min values :) but still kinda neat.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
err asymptote at 1 at least.. 1 something else is going on maybe.. hmm i can't tell by that graph hehe
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
hmm, interesting. So we know a critical point is x=1 right?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
yah it looks like x=1 and x=1 are critical points. :D
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
If i did my math correctly.. :O i hope.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
yeah, x=1, 1... But what about domain issues?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Oh good call, 1 isn't in our domain :3 so we don't care about that value.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
hmmmmm... i'm a little bit confused... what IS our domain?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmmm so we can't divide by 0, so 1 is out. Anddd we can't take the square root of a negative number, meaning, Anything smaller than 1 is out (it will make the bottom negative). Anything larger than 1 is out (it will make the top negative). I think our domain looks like this. \[1 < x \le 1\]
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
That looks right, automatically the end points of our domain are critical values right?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Because we can't attach a tangent line there?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Automatically? :o hmm i dunno. I just know that they're critical points because the first derivative told us so. It turns out that this is probably one of those weird functions where we'll end up having a MINIMUM at x=1 even though it doesn't have a slope of 0 there :D
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Maybe that's what you were saying though :)
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Well, our prof said that the end points of the domain are critical values, because we cannot attach a tangent line there... I'm not sure I really understand that, I had never heard that before.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
I just found this: http://screencast.com/t/b4Jx1qLkQlo
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
I'm still unsure of our critical points though, if x=1 is outside of our domain, then we can't use that,.. so I guess just x=1 is critical?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
yah, f'(x)=0 will give us critical points (which there were none). and also a and b are considered critical points. But b wasn't in our domain, so we don't even consider that point. Yah sounds good :D So just the lonely little x=1.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
I don't understand how there is a minimum at x=1.... We can't use a test value x>1 because it is out of our domain.. again :(
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
dw:1353277622511:dw True, we can't test both sides, all we can do is this.. see that it's decreasing on the left, and nonexistant on the right. So it's a minimum.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
oohhh.. okay... now to move onto the second derivative tests... :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
I would probably use the simplified form that I had set up, I think it'll be easier to differentiate again :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\huge f'(x)=(x1)^{1/2}(x+1)^{1/2}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmm this one is a pain in the butt, I'm coming up with: \[\huge f''(x)=\frac{x}{(x1)^{3/2}(x+1)^{3/2}}\] It's quite possible that I made a mistake though :D So make sure you check your work.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
I wanted to simplify it into a form that we can easily find inflection points from. So I combined the fractions :O
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
If you need to see my steps, I can post them :D
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
so... When f''(x) = 0 x = 0
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
If you refer back to our graph on wolfram, you'll see that makes sense! It changes from CCU to CCD around x=0!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
If you ever use Wolfram, make sure you're looking at the "Realvalued plot" from the graph drop down box.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
haha, yeah.. been there before.. I prefer the TI
 one year ago
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