## baldymcgee6 Group Title First and second derivative analysis... Is this right so far? one year ago one year ago

1. baldymcgee6 Group Title
2. baldymcgee6 Group Title

@zepdrix can you help?

3. zepdrix Group Title

analysis? what does that mean? like you're trying to find critical points and such? :o

4. baldymcgee6 Group Title

Yeah, sorry... Max, min, inflections..

5. zepdrix Group Title

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 Group Title

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 Group Title

Didn't know you can do that.. :/

8. zepdrix Group Title

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 Group Title

Maybe not the most relevant if we just want max and min values :) but still kinda neat.

10. zepdrix Group Title

err asymptote at -1 at least.. 1 something else is going on maybe.. hmm i can't tell by that graph hehe

11. zepdrix Group Title

blah i dunno XD

12. baldymcgee6 Group Title

hmm, interesting. So we know a critical point is x=1 right?

13. zepdrix Group Title

yah it looks like x=1 and x=-1 are critical points. :D

14. zepdrix Group Title

If i did my math correctly.. :O i hope.

15. baldymcgee6 Group Title

yeah, x=1, -1... But what about domain issues?

16. zepdrix Group Title

Oh good call, -1 isn't in our domain :3 so we don't care about that value.

17. baldymcgee6 Group Title

hmmmmm... i'm a little bit confused... what IS our domain?

18. zepdrix Group Title

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 Group Title

That looks right, automatically the end points of our domain are critical values right?

20. baldymcgee6 Group Title

Because we can't attach a tangent line there?

21. zepdrix Group Title

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 Group Title

Maybe that's what you were saying though :)

23. baldymcgee6 Group Title

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 Group Title

Hmm interesting :D

25. baldymcgee6 Group Title

I just found this: http://screencast.com/t/b4Jx1qLkQlo

26. zepdrix Group Title

cool :O

27. baldymcgee6 Group Title

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 Group Title

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 Group Title

30. baldymcgee6 Group Title

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 Group Title

|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 Group Title

oohhh.. okay... now to move onto the second derivative tests... :)

33. zepdrix Group Title

I would probably use the simplified form that I had set up, I think it'll be easier to differentiate again :)

34. zepdrix Group Title

$\huge f'(x)=-(x-1)^{-1/2}(x+1)^{-1/2}$

35. baldymcgee6 Group Title

right :D

36. zepdrix Group Title

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 Group Title

I wanted to simplify it into a form that we can easily find inflection points from. So I combined the fractions :O

38. zepdrix Group Title

If you need to see my steps, I can post them :D

39. baldymcgee6 Group Title

Yep got that too!

40. baldymcgee6 Group Title

so... When f''(x) = 0 x = 0

41. zepdrix Group Title

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 Group Title

barely.. lol

43. zepdrix Group Title

XD

44. zepdrix Group Title

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

45. baldymcgee6 Group Title

haha, yeah.. been there before.. I prefer the TI

46. zepdrix Group Title

oh fair enough c: