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ksaimouli
 4 years ago
find global extrema in given intreval
ksaimouli
 4 years ago
find global extrema in given intreval

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ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353711362019:dw

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353711391114:dw

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0@zepdrix this one is easy for u :)

zepdrix
 4 years ago
Best 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?

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353711927753:dw

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0but the answer is no global extrema how that is possible

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Ummmm, 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.

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.0so what i found is local min and max

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Yah that sounds right :D
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