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find global extrema in given intreval

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@zepdrix this one is easy for u :-)

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Other answers:

\[\huge y=x^3-9x^2-21x-11\] \[\huge y'=3x^2-18x-21\]Setting the first derivative equal to zero gives us,\[\huge 0=3(x^2-6x-7)\]And this one should factor pretty nicely from here. :) Getting stuck on any part? Or just looking to check your work?
is that right
but the answer is no global extrema how that is possible
Ummmm, do you remember what the graph of a cubed function looks like? It will go towards negative infinity to the left, and positive infinity to the right. So our graph will look something like this: |dw:1353712343847:dw| Clearly we can see a couple of critical points. But it makes sense that they are neither a max nor a min due to the shape of the graph. Sorry I'm a little rusty on the terminology, I guess when they say "global extrema" they only care about max and min, they didn't want the actual critical points.
so what i found is local min and max
Yah that sounds right :D
ok thanks

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