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baldymcgee6

  • 2 years ago

First and second derivative analysis... Is this right so far?

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  1. baldymcgee6
    • 2 years ago
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    http://screencast.com/t/5vrFs8Ggn

  2. baldymcgee6
    • 2 years ago
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    @zepdrix can you help?

  3. zepdrix
    • 2 years ago
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    analysis? what does that mean? like you're trying to find critical points and such? :o

  4. baldymcgee6
    • 2 years ago
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    Yeah, sorry... Max, min, inflections..

  5. zepdrix
    • 2 years ago
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    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 \[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1-x} }\cdot \frac{-1}{(1+x)^2}\]\[f'(x)=-\frac{1}{ \sqrt{1-x} \sqrt{1+x}}\]

  6. zepdrix
    • 2 years ago
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    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

  7. baldymcgee6
    • 2 years ago
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    Didn't know you can do that.. :/

  8. zepdrix
    • 2 years ago
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    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

  9. zepdrix
    • 2 years ago
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    Maybe not the most relevant if we just want max and min values :) but still kinda neat.

  10. zepdrix
    • 2 years ago
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    err asymptote at -1 at least.. 1 something else is going on maybe.. hmm i can't tell by that graph hehe

  11. zepdrix
    • 2 years ago
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    blah i dunno XD

  12. baldymcgee6
    • 2 years ago
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    hmm, interesting. So we know a critical point is x=1 right?

  13. zepdrix
    • 2 years ago
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    yah it looks like x=1 and x=-1 are critical points. :D

  14. zepdrix
    • 2 years ago
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    If i did my math correctly.. :O i hope.

  15. baldymcgee6
    • 2 years ago
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    yeah, x=1, -1... But what about domain issues?

  16. zepdrix
    • 2 years ago
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    Oh good call, -1 isn't in our domain :3 so we don't care about that value.

  17. baldymcgee6
    • 2 years ago
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    hmmmmm... i'm a little bit confused... what IS our domain?

  18. zepdrix
    • 2 years ago
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    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. \[-1 < x \le 1\]

  19. baldymcgee6
    • 2 years ago
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    That looks right, automatically the end points of our domain are critical values right?

  20. baldymcgee6
    • 2 years ago
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    Because we can't attach a tangent line there?

  21. zepdrix
    • 2 years ago
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    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

  22. zepdrix
    • 2 years ago
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    Maybe that's what you were saying though :)

  23. baldymcgee6
    • 2 years ago
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    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.

  24. zepdrix
    • 2 years ago
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    Hmm interesting :D

  25. baldymcgee6
    • 2 years ago
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    I just found this: http://screencast.com/t/b4Jx1qLkQlo

  26. zepdrix
    • 2 years ago
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    cool :O

  27. baldymcgee6
    • 2 years ago
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    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?

  28. zepdrix
    • 2 years ago
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    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.

  29. baldymcgee6
    • 2 years ago
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  30. baldymcgee6
    • 2 years ago
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    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 :(

  31. zepdrix
    • 2 years ago
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    |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.

  32. baldymcgee6
    • 2 years ago
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    oohhh.. okay... now to move onto the second derivative tests... :)

  33. zepdrix
    • 2 years ago
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    I would probably use the simplified form that I had set up, I think it'll be easier to differentiate again :)

  34. zepdrix
    • 2 years ago
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    \[\huge f'(x)=-(x-1)^{-1/2}(x+1)^{-1/2}\]

  35. baldymcgee6
    • 2 years ago
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    right :D

  36. zepdrix
    • 2 years ago
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    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.

  37. zepdrix
    • 2 years ago
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    I wanted to simplify it into a form that we can easily find inflection points from. So I combined the fractions :O

  38. zepdrix
    • 2 years ago
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    If you need to see my steps, I can post them :D

  39. baldymcgee6
    • 2 years ago
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    Yep got that too!

  40. baldymcgee6
    • 2 years ago
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    so... When f''(x) = 0 x = 0

  41. zepdrix
    • 2 years ago
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    If you refer back to our graph on wolfram, you'll see that makes sense! It changes from CCU to CCD around x=0!

  42. baldymcgee6
    • 2 years ago
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    barely.. lol

  43. zepdrix
    • 2 years ago
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    XD

  44. zepdrix
    • 2 years ago
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    If you ever use Wolfram, make sure you're looking at the "Real-valued plot" from the graph drop down box.

  45. baldymcgee6
    • 2 years ago
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    haha, yeah.. been there before.. I prefer the TI

  46. zepdrix
    • 2 years ago
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    oh fair enough c:

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