ksaimouli
  • ksaimouli
find global extrema in given intreval
Mathematics
jamiebookeater
  • jamiebookeater
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ksaimouli
  • ksaimouli
|dw:1353711362019:dw|
ksaimouli
  • ksaimouli
|dw:1353711391114:dw|
ksaimouli
  • ksaimouli
@zepdrix this one is easy for u :-)

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zepdrix
  • zepdrix
\[\huge y=x^3-9x^2-21x-11\] \[\huge y'=3x^2-18x-21\]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?
ksaimouli
  • ksaimouli
|dw:1353711927753:dw|
ksaimouli
  • ksaimouli
is that right
ksaimouli
  • ksaimouli
@zepdrix
ksaimouli
  • ksaimouli
but the answer is no global extrema how that is possible
zepdrix
  • zepdrix
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.
ksaimouli
  • ksaimouli
so what i found is local min and max
zepdrix
  • zepdrix
Yah that sounds right :D
ksaimouli
  • ksaimouli
ok thanks

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