## Ephilo Group Title HELP PLEASE?! use the Second Derivative Test to find all relative extrema. one year ago one year ago

1. Ephilo Group Title

|dw:1364048045697:dw|

2. satellite73 Group Title

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

3. satellite73 Group Title

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

4. Ephilo Group Title

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

5. satellite73 Group Title

ok

6. Ephilo Group Title

the first derivate is

7. Ephilo Group Title

|dw:1364048369021:dw|

8. satellite73 Group Title

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

9. Ephilo Group Title

okay

10. satellite73 Group Title

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 Group Title

a bit confused here?

12. satellite73 Group Title

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 Group Title

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 Group Title

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 Group Title

is that okay?

16. Ephilo Group Title

so you put zero in place of t right

17. satellite73 Group Title

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

18. satellite73 Group Title

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

19. satellite73 Group Title

damn typo!!

20. satellite73 Group Title

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

21. S.RaviTeja Group Title

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

22. satellite73 Group Title

thats better

23. Ephilo Group Title

okay ill give it a try

24. satellite73 Group Title

@S.RaviTeja no, that is not the solution

25. S.RaviTeja Group Title

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 Group Title

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

27. Ephilo Group Title

so t=3 ?

28. satellite73 Group Title

yes

29. Ephilo Group Title

yay! lol

30. satellite73 Group Title

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

31. S.RaviTeja Group Title

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

32. satellite73 Group Title

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

33. satellite73 Group Title

@S.RaviTeja -6 is not a solution

34. satellite73 Group Title

second derivative is ... ?

35. Ephilo Group Title

okay im working on it @satellite73

36. S.RaviTeja Group Title

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 Group Title

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 Group Title

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

39. Ephilo Group Title

would it be

40. Ephilo Group Title

|dw:1364049467577:dw|

41. satellite73 Group Title

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

42. Ephilo Group Title

close enough lol

43. S.RaviTeja Group Title

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

44. satellite73 Group Title

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 Group Title

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 Group Title

so im evaluating right ?

47. satellite73 Group Title

yes

48. Ephilo Group Title

1/16

49. satellite73 Group Title

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

50. S.RaviTeja Group Title

its 1/8

51. Ephilo Group Title

i forgot to simplify

52. Ephilo Group Title

would that be all

53. S.RaviTeja Group Title

yep

54. Ephilo Group Title

oh okay lol

55. Ephilo Group Title

thanks guys (:

56. satellite73 Group Title

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 Group Title

oh okay @satellite73