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baldymcgee6
Group Title
First and second derivative analysis... Is this right so far?
 2 years ago
 2 years ago
baldymcgee6 Group Title
First and second derivative analysis... Is this right so far?
 2 years ago
 2 years ago

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baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
http://screencast.com/t/5vrFs8Ggn
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix can you help?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
analysis? what does that mean? like you're trying to find critical points and such? :o
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, sorry... Max, min, inflections..
 2 years ago

zepdrix Group TitleBest 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 \[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1x} }\cdot \frac{1}{(1+x)^2}\]\[f'(x)=\frac{1}{ \sqrt{1x} \sqrt{1+x}}\]
 2 years ago

zepdrix Group TitleBest 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
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
Didn't know you can do that.. :/
 2 years ago

zepdrix Group TitleBest 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
 2 years ago

zepdrix Group TitleBest 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.
 2 years ago

zepdrix Group TitleBest 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
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
blah i dunno XD
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
hmm, interesting. So we know a critical point is x=1 right?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
yah it looks like x=1 and x=1 are critical points. :D
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
If i did my math correctly.. :O i hope.
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
yeah, x=1, 1... But what about domain issues?
 2 years ago

zepdrix Group TitleBest 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.
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
hmmmmm... i'm a little bit confused... what IS our domain?
 2 years ago

zepdrix Group TitleBest 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. \[1 < x \le 1\]
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
That looks right, automatically the end points of our domain are critical values right?
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
Because we can't attach a tangent line there?
 2 years ago

zepdrix Group TitleBest 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
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Maybe that's what you were saying though :)
 2 years ago

baldymcgee6 Group TitleBest 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.
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Hmm interesting :D
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I just found this: http://screencast.com/t/b4Jx1qLkQlo
 2 years ago

baldymcgee6 Group TitleBest 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?
 2 years ago

zepdrix Group TitleBest 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.
 2 years ago

baldymcgee6 Group TitleBest 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 :(
 2 years ago

zepdrix Group TitleBest 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.
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
oohhh.. okay... now to move onto the second derivative tests... :)
 2 years ago

zepdrix Group TitleBest 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 :)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge f'(x)=(x1)^{1/2}(x+1)^{1/2}\]
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
right :D
 2 years ago

zepdrix Group TitleBest 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.
 2 years ago

zepdrix Group TitleBest 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
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
If you need to see my steps, I can post them :D
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
Yep got that too!
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
so... When f''(x) = 0 x = 0
 2 years ago

zepdrix Group TitleBest 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!
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
barely.. lol
 2 years ago

zepdrix Group TitleBest 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.
 2 years ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
haha, yeah.. been there before.. I prefer the TI
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
oh fair enough c:
 2 years ago
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