## Ephilo 2 years ago HELP PLEASE?! use the Second Derivative Test to find all relative extrema.

1. Ephilo

|dw:1364048045697:dw|

2. satellite73

$h(t)=t-4\sqrt{t+1}$?

3. satellite73

i guess you need the first two derivatives what did you get for the first one?

4. Ephilo

@satellite73 give me a second im gonna do it right now

5. satellite73

ok

6. Ephilo

the first derivate is

7. Ephilo

|dw:1364048369021:dw|

8. satellite73

yes it is i would write it as $1-\frac{2}{\sqrt{t+1}}$ but that is correct.

9. Ephilo

okay

10. satellite73

so before finding the second derivative, we need the critical points one critical point is where the derivative is undefined, at $$t=-1$$ which is also the left hand endpoint of the domain of $$h$$ since the domain of $$h$$ is $$[-1,\infty)$$ to find the other set $1-\frac{2}{\sqrt{t+1}}=0$ and solve for $$t$$

11. Ephilo

a bit confused here?

12. satellite73

you are asked to use the second derivative to find the relative max and min of your function that means you are asked to use the second derivative to test the critical points to see if they are max or min or neither in order to do that, you need to find the critical points first

13. satellite73

critical points where the derivative is zero or where the derivative is undefined it is pretty clearly undefined at $$t=-1$$ because you have $$\sqrt{t+1}$$ in the denominator, and you cannot divide by zero

14. satellite73

so your next job is to see where the derivative is equal to zero that is, set $1-\frac{2}{\sqrt{t+1}}+0$and solve for $$t$$

15. satellite73

is that okay?

16. Ephilo

so you put zero in place of t right

17. satellite73

no, you set the derivative equal to zero and solve for $$t$$

18. satellite73

i.e. solve $1-\frac{2}{\sqrt{t+1}}+0$

19. satellite73

damn typo!!

20. satellite73

$1-\frac{2}{\sqrt{t+1}}=0$

21. S.RaviTeja

t=+sqrt(2)-1,-sqrt(2)-1

22. satellite73

thats better

23. Ephilo

okay ill give it a try

24. satellite73

@S.RaviTeja no, that is not the solution

25. S.RaviTeja

now substitute these values in second derivative of the function and if value is greater than zero its minima and less than zero maxima.

26. satellite73

@Ephilo don't stress over this one $1-\frac{2}{\sqrt{t+1}}=0\iff\sqrt{t+1}=2$

27. Ephilo

so t=3 ?

28. satellite73

yes

29. Ephilo

yay! lol

30. satellite73

now you have two critical points, $$t=0$$ and $$t=3$$

31. S.RaviTeja

oh sorry in a hurry id did a mistake t=3 and -6

32. satellite73

now it is time for the second derivative take the second derivative and evaluate at the critical points

33. satellite73

@S.RaviTeja -6 is not a solution

34. satellite73

second derivative is ... ?

35. Ephilo

okay im working on it @satellite73

36. S.RaviTeja

ok gotcha:) if we take -6 we get imaginary values if we substitute in the original eqn or any other eqn..Thanks for referring:)

37. satellite73

ok. use the first answer you wrote for the derivative and use the power rule. then rewrite in radical notation so you can actually compute the number

38. S.RaviTeja

second derivative is (t+1)^(-3/2)

39. Ephilo

would it be

40. Ephilo

|dw:1364049467577:dw|

41. satellite73

$\frac{1}{\sqrt{t+1}^3}$

42. Ephilo

close enough lol

43. S.RaviTeja

yep thats what i was referring to as derivative that is of satellite73

44. satellite73

replace $$t$$ by $$-1$$ and see that it is undefined so forget that, it is the left point of the domain then replace $$t$$ by $$3$$

45. satellite73

since the number you get is clearly positive (it is $$\frac{1}{8}$$) this means the function is concave up |dw:1364049678258:dw|

46. Ephilo

so im evaluating right ?

47. satellite73

yes

48. Ephilo

1/16

49. satellite73

no i get $$\frac{1}{8}$$ but in any case it is positive which is all you really care about

50. S.RaviTeja

its 1/8

51. Ephilo

i forgot to simplify

52. Ephilo

would that be all

53. S.RaviTeja

yep

54. Ephilo

oh okay lol

55. Ephilo

thanks guys (:

56. satellite73

that is not all you have to say that since the second derivative evaluated at the critical point is a positive number, the function is concave up, meaning the critical point is a local minimum

57. Ephilo

oh okay @satellite73