OHH, its because i did the graph by hand so it looks bad and sloppy. i would take a picture of my calculator's screen but i wouldve thought "close enough" was good enough. i still dont quite understand how to get the points though(referring to #6). the book has examples with intervals and this one doesnt, so basically i plot points into the original(draw it, etc), then plot points into the second derivative? am i following right?

Ecael

United States6703 Posts

June 05 2008 05:02 GMT

#22

For points of inflection, one can define it as points where the second derivative of the function = 0. So you simply set that equation to 0 and solve for x. Intervals doesn't matter in terms of point of inflections, as you are not going to see a point of inflection at infinity. I am not sure what do you mean by ploting points into the original though. If I take it that you are trying to solve it with a calculator, then yeah, you'd simply plot a graph of the second derivative and solve for where that function is 0.

In terms of the original function, the point of inflection changes how the function...um, flows, someone help me with the right term. You'll see a visible change of how the function's slope changes though starting at the point of inflection.

Raithed

China7078 Posts

June 05 2008 05:14 GMT

#23

if its not that big of trouble can you write out how you got 2(x^2-6)(x^-5) ? so basically the second derivative is set to 0.

2(x^2-6)/(x^-5) = 0 and just solve for x, so would it become x^2-6 = 0, x = 3? as one of the points?

Ecael

United States6703 Posts

June 05 2008 05:41 GMT

#24

No problem, let me do it out with quotient rule I guess, I used chain since I've always preferred the extra steps over me accidentally forgetting order for quotient rule (bad memory, lol). Oh, for inflection point, look at how you solved it again, x^2 - 6 = 0, x^2 = 6, root(6) =/= 3.

So first derivative = (3-x^2)/(x^4). Quotient rule states that f(x) = u/v, f'(x) = (du)(v) - (dv)(u) / v^2, so (-2x)(x^4) - (3-x^2)(4x^3) / (x^8) = (-2x^5) - (12x^3 - 4x^5) / x^8, factor out a x^3. -2x^2 - 12 + 4x^2 / x^5. or 2x^2 - 12 / x^5, 2((x^2)-6) / x^5.

Raithed

China7078 Posts

June 05 2008 05:49 GMT

#25

ah, okay yeah i did dumb mistakes.

x^2-6 = 0, x^2=6, x = sqrt. 6 <-- is this one of the inflection lines?

Ecael

United States6703 Posts

June 05 2008 05:52 GMT

#26

+/- sqrt(6) are the two points of inflection.

Raithed

China7078 Posts

June 05 2008 05:54 GMT

#27

ah.. so am i done or is there more?

Ecael

United States6703 Posts

June 05 2008 06:04 GMT

#28

lol, don't have us do all the work. Check yourself as to see what is missing

(And no, those are the only two, looking at the graphs on the previous page should make that pretty clear)

One really good point that jgad brought up is the usage of graph and simple mathematical rules for you to verify what you know. Knowing the graphical end of it will make checking a lot easier for you, and it is also a quick and reliable double check for you as far as simple mistakes are concerned. A quick look at the graph could've told you that your original derivative had something wrong since you were getting imaginary values for critical points (well, simple logic dictates that it is wrong, but that's beside the point). It'll also give you an idea of about where the critical and inflection points lie in regard to each other.

Raithed

China7078 Posts

June 05 2008 06:22 GMT

#29

thanks a lot Ecael for the help and many others, its 2am, going to bed. <3

jtan

Sweden5891 Posts

June 05 2008 06:45 GMT

#30

also you can see in the graph that the inflexion points are correct. sqrt(6)=~2.5 and if you look at x=2.5 you see that that's about where the curve changes from convex to concave.

so question about number 5, if its not defined does it even have a relative extrema?

This is a weird question. Your function is not defined IN the point x=0 because you get a division by zero in the function so you get no well defined value there. The function still has local extremas, but it can't have one in x=0 since the function is not defined IN that point.

jtan

Sweden5891 Posts

June 05 2008 06:46 GMT

#31

This small program is really useful for making plots on pc

http://www.padowan.dk/graph/Download.php

gwho

United States632 Posts

June 05 2008 08:58 GMT

#32

5) there are no local max/mins but an absolute max min. and that would be negative infinity and infinity.

6) points of inflection u just take double derivative and see if it's positive or negative.

Rambling.

Canada314 Posts

June 05 2008 09:35 GMT

#33

Hope this Helps

+ Show Spoiler +

Ecael

United States6703 Posts

June 05 2008 09:36 GMT

#34

There are local extremas, take a look at the function posted instead of the graph, which multiple people have noted as deficient.

That's a cool program, jtan, the powertoy calculator is just not up to par :p

jtan

Sweden5891 Posts

June 05 2008 10:42 GMT

#35

On June 05 2008 17:58 gwho wrote:
5) there are no local max/mins but an absolute max min. and that would be negative infinity and infinity.

6) points of inflection u just take double derivative and see if it's positive or negative.

wrong

+-sqrt(3) is local max/min

raitheads graph is wrong, he just drew that as an approximate to the graph in his calculator lol

Chef

10810 Posts

June 05 2008 11:10 GMT

#36

Sometimes in university, I think to myself, "boy, writing essays sure is fucking boring." Then I remember I haven't done any complicated math in 3 years, and I feel a little better.

jgad

Canada899 Posts

June 05 2008 11:49 GMT

#37

When I was in university, I sometimes would look around at my ~93% male Engineering Physics class and thinking to myself "boy, I wonder where all the girls end up?". Then I took an elective course in neuropsychology and it was like walking into the girls' locker room - wtf?! It was like 93% chicks - and then, going back to my soul-crushing complicated math courses, I felt a little worse