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?
Wait so what would the relatives be?
How do you do the leftover?
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
Great. I have never done one like this before. So I have to make a table of values first.
Okay I have the table of values
So what do I do now?
It is not that table, the table of f"(x) to consider concave up or down.
oh okay so the table of the derivative
So the relative max is where it changes from concave up to concave down?
Of the derivative?
Know it?
the concept how to consider the inflection points are relative max or min?
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?
http://www.youtube.com/watch?v=2tmRPytHBuk very helpful
x= +1 and -1 are the critical numbers. So I put them into the second derivative.
read and watch the video tap and consider
if f''(c) > 0 then f(c) is a relative minimum if f'(c) < 0 then f(c) is a relative maximum
yes, you got it.
Thank you so much!
That video is very helpful
np
I did it!!!! I figured it out all my myself!
verygoodstudent!!!
Youareaverygoodteacher. :) If I could be a double fan I would be. Haha!
hahahaha.... be friend is good enough, mate
Do you know a video on finding asymptotes?
I am so exited! I can actually do math now!!!
Wait so is there a relative minimum or is there just a relative max at -sqrt 3
@OOOPS
a bunch of them http://www.youtube.com/results?search_query=finding+asymptotes&sm=1
Does there have to be a relative minimum?
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