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Mathematics 30 Online
OpenStudy (anonymous):

Relative extrema of x/(x^2+1) is it 1/2 at x=1 and -1/2 at x=-1 or are those the absolute extrema?

OpenStudy (anonymous):

Wait so what would the relatives be?

OpenStudy (anonymous):

How do you do the leftover?

OpenStudy (anonymous):

to find it out, you must construct the table to consider in what interval, the concave up/down to consider whether the critical points give us relative max or min

OpenStudy (anonymous):

Great. I have never done one like this before. So I have to make a table of values first.

OpenStudy (anonymous):

Okay I have the table of values

OpenStudy (anonymous):

So what do I do now?

OpenStudy (anonymous):

It is not that table, the table of f"(x) to consider concave up or down.

OpenStudy (anonymous):

oh okay so the table of the derivative

OpenStudy (anonymous):

So the relative max is where it changes from concave up to concave down?

OpenStudy (anonymous):

Of the derivative?

OpenStudy (anonymous):

Know it?

OpenStudy (anonymous):

the concept how to consider the inflection points are relative max or min?

OpenStudy (anonymous):

No I didn't know that. My teacher does not explain things she just gives us homework. So the reflection points of my function are the relative max and min?

OpenStudy (anonymous):

http://www.youtube.com/watch?v=2tmRPytHBuk very helpful

OpenStudy (anonymous):

x= +1 and -1 are the critical numbers. So I put them into the second derivative.

OpenStudy (anonymous):

read and watch the video tap and consider

OpenStudy (anonymous):

if f''(c) > 0 then f(c) is a relative minimum if f'(c) < 0 then f(c) is a relative maximum

OpenStudy (anonymous):

yes, you got it.

OpenStudy (anonymous):

Thank you so much!

OpenStudy (anonymous):

That video is very helpful

OpenStudy (anonymous):

np

OpenStudy (anonymous):

I did it!!!! I figured it out all my myself!

OpenStudy (anonymous):

verygoodstudent!!!

OpenStudy (anonymous):

Youareaverygoodteacher. :) If I could be a double fan I would be. Haha!

OpenStudy (anonymous):

hahahaha.... be friend is good enough, mate

OpenStudy (anonymous):

Do you know a video on finding asymptotes?

OpenStudy (anonymous):

I am so exited! I can actually do math now!!!

OpenStudy (anonymous):

Wait so is there a relative minimum or is there just a relative max at -sqrt 3

OpenStudy (anonymous):

@OOOPS

OpenStudy (anonymous):

a bunch of them http://www.youtube.com/results?search_query=finding+asymptotes&sm=1

OpenStudy (anonymous):

Does there have to be a relative minimum?

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