## anonymous one year ago Find the extreme values of the function and where they occur

1. anonymous

$1/(x^2-1)$

2. anonymous

@Loser66

3. Loser66

What is the definition of extreme value?

4. anonymous

The relative (or local) max/min points.

5. Loser66

yes, so that just take derivative of f

6. anonymous

Only 1 derivative?

7. Loser66

let it =0, find x replace that x into the original function, and you are done.

8. Loser66

first derivative only.

9. anonymous

Oh! So when i write the answer i just say y=??

10. Loser66

When you get the result (x =0) , your conclusion is f(x) has extreme value and x =0 and f(0) = -1

11. Loser66

*at, not and f(x) has extreme value AT x =0 ,....

12. anonymous

ohh, okay, let me solve it.

13. anonymous

$f'(x)=\left(\begin{matrix}-2x \\ (x^2-1)^2\end{matrix}\right)$

14. anonymous

correct?

15. Loser66

It is NOT a matrix, it is a fraction. but it is correct.

16. anonymous

yeah sorry, meant to be a fraction

17. anonymous

so how do i solve for x from here?

18. Loser66

you solve for f' =0 iff the numerator =0

19. Loser66

and the numerator is -2x =0 iff x =0, right?

20. anonymous

correct

21. anonymous

so x=0?

22. Loser66

plug back to original one to find value of f(0)=??

23. anonymous

-1?

24. Loser66

yup

25. anonymous

So the answer doesn't have to be like, absolute max = ?? absolute min= ??

26. anonymous

sorry just making sure haha

27. amoodarya

|dw:1437521757173:dw|

28. Loser66

If you have 2 solutions, plug back and compare which one is larger, the larger one is max, the lesser one is min In this case, you have only 1 solution. You have no comparison, how to say whether it is max or min?

29. amoodarya

|dw:1437521827659:dw| this is sketch of f(X) so x=0 is not abs max

30. anonymous

So no absolute max or min?

31. Loser66

yup, just local

32. anonymous

so is 0 the local max or local min? I'm confused sorry

33. amoodarya

|dw:1437523053515:dw|local max