Question

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

Answer 1

http://screencast.com/t/5vrFs8Ggn

Answer 2

@zepdrix can you help?

Answer 3

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

Answer 4

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

Answer 5

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
\[f'(x)=\frac{ \sqrt{1+x} }{ \sqrt{1-x} }\cdot \frac{-1}{(1+x)^2}\]\[f'(x)=-\frac{1}{ \sqrt{1-x} \sqrt{1+x}}\]

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

blah i dunno XD

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.
\[-1 < x \le 1\]

Answer 19

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

Answer 20

Because we can't attach a tangent line there?

Answer 21

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

Answer 22

Maybe that's what you were saying though :)

Answer 23

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.

Answer 24

Hmm interesting :D

Answer 25

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

Answer 26

cool :O

Answer 27

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?

Answer 28

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.

Answer 29

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 :(

Answer 30

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.