onegirl 2 years ago Find all critical numbers and use the First Derivative Test to classify each as the location of a location maximum, local minimum, or neither. y = x^4 + 4x^3 - 2

1. onegirl

@electrokid can u help?

2. electrokid

using our previous discussions, you can start the work and I guide.

3. electrokid

step 1) first derivative \(y'=?\)

4. onegirl

Sorry i replied late i had internet issues. but the first derivative is 4x^2(x + 3)

5. electrokid

good. step 2) solve for "x": \(4x^2(x+3)=0\)

6. onegirl

okay i got x = -3 and x = 0

7. onegirl

are you there?

8. electrokid

good. those are the critical points. step 3) classify as maxima, minima or neither step 3.1) find second derivative of "f" -> find f''(x)

9. onegirl

how do i do that?

10. onegirl

are u there @electrokid

11. electrokid

step 3.1) find the second derivative.

12. onegirl

okay

13. onegirl

I got 12x^2 + 24x

14. PeterPan

Remember the critical points that you got? If the second derivative is positive when evaluated at one of those critical points, that critical point would be a local minimum. On the flip-side, if the second derivative is negative when evaluated at a critical point, that point would be a local maximum.

15. onegirl

so i substitute those critical point into the second derivative

16. PeterPan

Yes.

17. electrokid

yes. \(f''(x)=12x^2+24x\) step 3.2) plug in x=-3 in this equation and check the sign..

18. onegirl

okay

19. onegirl

i got 36

20. onegirl

and when i plugged in 0 i got 0

21. PeterPan

Okay, so, f''(-3) = 36 This is positive, therefore...?

22. onegirl

so it is a minimum?

23. PeterPan

a local minimum ;) What about f''(0) ?

24. onegirl

I got 0

25. PeterPan

It's neither positive nor negative, right?

26. onegirl

yes

27. PeterPan

Well, what does that mean, then?

28. onegirl

so it will be neither

29. PeterPan

It will be neither :D

30. onegirl

Okay thanks

31. electrokid

@PeterPan and @onegirl both did a great job and effort and both deserve a medal :) so, @onegirl give one to P-Pan ☮

32. PeterPan