## ksaimouli 2 years ago find global extrema in given intreval

1. ksaimouli

|dw:1353711362019:dw|

2. ksaimouli

|dw:1353711391114:dw|

3. ksaimouli

@zepdrix this one is easy for u :-)

4. 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?

5. ksaimouli

|dw:1353711927753:dw|

6. ksaimouli

is that right

7. ksaimouli

@zepdrix

8. ksaimouli

but the answer is no global extrema how that is possible

9. 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.

10. ksaimouli

so what i found is local min and max

11. zepdrix

Yah that sounds right :D

12. ksaimouli

ok thanks