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I need help figuring out how the derivative of the graph would look approximatially.

Mathematics
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|dw:1362964903781:dw|
This is the original graph how would the derivative look
rigth

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Other answers:

so
|dw:1362965404144:dw|These are the easy points to start with. Draw tangent lines. See how the slope of these tangent lines is 0. That means the VALUE of the derivative function will be zero.
|dw:1362965478556:dw|
so then the derivative just goes along the x line
No, the derivative passes through the x-axis at those 3 points :) now let's find some other points.
ok
|dw:1362965586761:dw|Understand how I drew the line tangent to the curve? See how it's pointing downward (read from left to right). It has an extremely negative slope. Meaning the derivative function will have an extremely negative VALUE at this x coordinate.
|dw:1362965677365:dw|
ok
well then there you ave it
|dw:1362965734778:dw|Now the function is starting to come together! c:
Have you learned about inflection points? :o Is this all too confusing farmergal? :c
no so then does it just go up to the x line then hit the dots on the x line.
Yes, but it will keep going past the x line, how far? Well until we reach the inflection point.
Now I'm confussed when you say it keeps going to the inflection point.
|dw:1362965901972:dw|Right around that point, it changes from `Concave Up` to `Concave Down`.
|dw:1362965979553:dw|SO it will look like this
|dw:1362965980976:dw|
Ummmm yah it looks like that's where we're headed :)
ok I got it so far.
|dw:1362966105601:dw|
ok
|dw:1362966148110:dw|So at that x coordinate, our derivative function reaches the bottom of the bowl :D
ok go it
|dw:1362966245521:dw|If we check one last point way on the right over here, it appears to be very positive slope, so the derivative function will have a very positive value.
|dw:1362966295539:dw|
Something like that, the original graph was a tad sloppy. So it's hard to say for certain :) heh
ok got it thanks

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