baldymcgee6
  • baldymcgee6
First and second derivative analysis... Is this right so far?
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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baldymcgee6
  • baldymcgee6
http://screencast.com/t/5vrFs8Ggn
baldymcgee6
  • baldymcgee6
@zepdrix can you help?
zepdrix
  • zepdrix
analysis? what does that mean? like you're trying to find critical points and such? :o

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baldymcgee6
  • baldymcgee6
Yeah, sorry... Max, min, inflections..
zepdrix
  • zepdrix
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 \[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1-x} }\cdot \frac{-1}{(1+x)^2}\]\[f'(x)=-\frac{1}{ \sqrt{1-x} \sqrt{1+x}}\]
zepdrix
  • zepdrix
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
baldymcgee6
  • baldymcgee6
Didn't know you can do that.. :/
zepdrix
  • zepdrix
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%281-x%29%2F%281%2Bx%29%29%5E%281%2F2%29
zepdrix
  • zepdrix
Maybe not the most relevant if we just want max and min values :) but still kinda neat.
zepdrix
  • zepdrix
err asymptote at -1 at least.. 1 something else is going on maybe.. hmm i can't tell by that graph hehe
zepdrix
  • zepdrix
blah i dunno XD
baldymcgee6
  • baldymcgee6
hmm, interesting. So we know a critical point is x=1 right?
zepdrix
  • zepdrix
yah it looks like x=1 and x=-1 are critical points. :D
zepdrix
  • zepdrix
If i did my math correctly.. :O i hope.
baldymcgee6
  • baldymcgee6
yeah, x=1, -1... But what about domain issues?
zepdrix
  • zepdrix
Oh good call, -1 isn't in our domain :3 so we don't care about that value.
baldymcgee6
  • baldymcgee6
hmmmmm... i'm a little bit confused... what IS our domain?
zepdrix
  • zepdrix
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. \[-1 < x \le 1\]
baldymcgee6
  • baldymcgee6
That looks right, automatically the end points of our domain are critical values right?
baldymcgee6
  • baldymcgee6
Because we can't attach a tangent line there?
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
Maybe that's what you were saying though :)
baldymcgee6
  • baldymcgee6
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.
zepdrix
  • zepdrix
Hmm interesting :D
baldymcgee6
  • baldymcgee6
I just found this: http://screencast.com/t/b4Jx1qLkQlo
zepdrix
  • zepdrix
cool :O
baldymcgee6
  • baldymcgee6
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?
zepdrix
  • zepdrix
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.
baldymcgee6
  • baldymcgee6
1 Attachment
baldymcgee6
  • baldymcgee6
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 :(
zepdrix
  • zepdrix
|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.
baldymcgee6
  • baldymcgee6
oohhh.. okay... now to move onto the second derivative tests... :)
zepdrix
  • zepdrix
I would probably use the simplified form that I had set up, I think it'll be easier to differentiate again :)
zepdrix
  • zepdrix
\[\huge f'(x)=-(x-1)^{-1/2}(x+1)^{-1/2}\]
baldymcgee6
  • baldymcgee6
right :D
zepdrix
  • zepdrix
Hmm this one is a pain in the butt, I'm coming up with: \[\huge f''(x)=\frac{x}{(x-1)^{3/2}(x+1)^{3/2}}\] It's quite possible that I made a mistake though :D So make sure you check your work.
zepdrix
  • zepdrix
I wanted to simplify it into a form that we can easily find inflection points from. So I combined the fractions :O
zepdrix
  • zepdrix
If you need to see my steps, I can post them :D
baldymcgee6
  • baldymcgee6
Yep got that too!
baldymcgee6
  • baldymcgee6
so... When f''(x) = 0 x = 0
zepdrix
  • zepdrix
If you refer back to our graph on wolfram, you'll see that makes sense! It changes from CCU to CCD around x=0!
baldymcgee6
  • baldymcgee6
barely.. lol
zepdrix
  • zepdrix
XD
zepdrix
  • zepdrix
If you ever use Wolfram, make sure you're looking at the "Real-valued plot" from the graph drop down box.
baldymcgee6
  • baldymcgee6
haha, yeah.. been there before.. I prefer the TI
zepdrix
  • zepdrix
oh fair enough c:

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