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anonymous
 3 years ago
First and second derivative analysis... Is this right so far?
anonymous
 3 years ago
First and second derivative analysis... Is this right so far?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@zepdrix can you help?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, sorry... Max, min, inflections..

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1We 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 \[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1x} }\cdot \frac{1}{(1+x)^2}\]\[f'(x)=\frac{1}{ \sqrt{1x} \sqrt{1+x}}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Did 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Didn't know you can do that.. :/

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm 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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm, interesting. So we know a critical point is x=1 right?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, x=1, 1... But what about domain issues?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmmmmm... i'm a little bit confused... what IS our domain?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Hmmm 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. \[1 < x \le 1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That looks right, automatically the end points of our domain are critical values right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because we can't attach a tangent line there?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Automatically? :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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just found this: http://screencast.com/t/b4Jx1qLkQlo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1yah, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 :(

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oohhh.. okay... now to move onto the second derivative tests... :)

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

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm 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.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so... When f''(x) = 0 x = 0

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha, yeah.. been there before.. I prefer the TI
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