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ksaimouliBest ResponseYou've already chosen the best response.0
dw:1353711362019:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
dw:1353711391114:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
@zepdrix this one is easy for u :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\huge y=x^39x^221x11\] \[\huge y'=3x^218x21\]Setting the first derivative equal to zero gives us,\[\huge 0=3(x^26x7)\]And this one should factor pretty nicely from here. :) Getting stuck on any part? Or just looking to check your work?
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
dw:1353711927753:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
but the answer is no global extrema how that is possible
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
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.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.0
so what i found is local min and max
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Yah that sounds right :D
 one year ago
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