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ksaimouli

  • 2 years ago

find global extrema in given intreval

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  1. ksaimouli
    • 2 years ago
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    |dw:1353711362019:dw|

  2. ksaimouli
    • 2 years ago
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    |dw:1353711391114:dw|

  3. ksaimouli
    • 2 years ago
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    @zepdrix this one is easy for u :-)

  4. zepdrix
    • 2 years ago
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    \[\huge y=x^3-9x^2-21x-11\] \[\huge y'=3x^2-18x-21\]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?

  5. ksaimouli
    • 2 years ago
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    |dw:1353711927753:dw|

  6. ksaimouli
    • 2 years ago
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    is that right

  7. ksaimouli
    • 2 years ago
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    @zepdrix

  8. ksaimouli
    • 2 years ago
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    but the answer is no global extrema how that is possible

  9. zepdrix
    • 2 years ago
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    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.

  10. ksaimouli
    • 2 years ago
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    so what i found is local min and max

  11. zepdrix
    • 2 years ago
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    Yah that sounds right :D

  12. ksaimouli
    • 2 years ago
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    ok thanks

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