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Ephilo Group Title

HELP PLEASE?! use the Second Derivative Test to find all relative extrema.

  • one year ago
  • one year ago

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  1. Ephilo Group Title
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    |dw:1364048045697:dw|

    • one year ago
  2. satellite73 Group Title
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    \[h(t)=t-4\sqrt{t+1}\]?

    • one year ago
  3. satellite73 Group Title
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    i guess you need the first two derivatives what did you get for the first one?

    • one year ago
  4. Ephilo Group Title
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    @satellite73 give me a second im gonna do it right now

    • one year ago
  5. satellite73 Group Title
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    ok

    • one year ago
  6. Ephilo Group Title
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    the first derivate is

    • one year ago
  7. Ephilo Group Title
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    |dw:1364048369021:dw|

    • one year ago
  8. satellite73 Group Title
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    yes it is i would write it as \[1-\frac{2}{\sqrt{t+1}}\] but that is correct.

    • one year ago
  9. Ephilo Group Title
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    okay

    • one year ago
  10. satellite73 Group Title
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    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 \(t\)

    • one year ago
  11. Ephilo Group Title
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    a bit confused here?

    • one year ago
  12. satellite73 Group Title
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    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

    • one year ago
  13. satellite73 Group Title
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    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

    • one year ago
  14. satellite73 Group Title
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    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 \(t\)

    • one year ago
  15. satellite73 Group Title
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    is that okay?

    • one year ago
  16. Ephilo Group Title
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    so you put zero in place of t right

    • one year ago
  17. satellite73 Group Title
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    no, you set the derivative equal to zero and solve for \(t\)

    • one year ago
  18. satellite73 Group Title
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    i.e. solve \[1-\frac{2}{\sqrt{t+1}}+0\]

    • one year ago
  19. satellite73 Group Title
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    damn typo!!

    • one year ago
  20. satellite73 Group Title
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    \[1-\frac{2}{\sqrt{t+1}}=0\]

    • one year ago
  21. S.RaviTeja Group Title
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    t=+sqrt(2)-1,-sqrt(2)-1

    • one year ago
  22. satellite73 Group Title
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    thats better

    • one year ago
  23. Ephilo Group Title
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    okay ill give it a try

    • one year ago
  24. satellite73 Group Title
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    @S.RaviTeja no, that is not the solution

    • one year ago
  25. S.RaviTeja Group Title
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    now substitute these values in second derivative of the function and if value is greater than zero its minima and less than zero maxima.

    • one year ago
  26. satellite73 Group Title
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    @Ephilo don't stress over this one \[1-\frac{2}{\sqrt{t+1}}=0\iff\sqrt{t+1}=2\]

    • one year ago
  27. Ephilo Group Title
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    so t=3 ?

    • one year ago
  28. satellite73 Group Title
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    yes

    • one year ago
  29. Ephilo Group Title
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    yay! lol

    • one year ago
  30. satellite73 Group Title
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    now you have two critical points, \(t=0\) and \(t=3\)

    • one year ago
  31. S.RaviTeja Group Title
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    oh sorry in a hurry id did a mistake t=3 and -6

    • one year ago
  32. satellite73 Group Title
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    now it is time for the second derivative take the second derivative and evaluate at the critical points

    • one year ago
  33. satellite73 Group Title
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    @S.RaviTeja -6 is not a solution

    • one year ago
  34. satellite73 Group Title
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    second derivative is ... ?

    • one year ago
  35. Ephilo Group Title
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    okay im working on it @satellite73

    • one year ago
  36. S.RaviTeja Group Title
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    ok gotcha:) if we take -6 we get imaginary values if we substitute in the original eqn or any other eqn..Thanks for referring:)

    • one year ago
  37. satellite73 Group Title
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    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

    • one year ago
  38. S.RaviTeja Group Title
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    second derivative is (t+1)^(-3/2)

    • one year ago
  39. Ephilo Group Title
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    would it be

    • one year ago
  40. Ephilo Group Title
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    |dw:1364049467577:dw|

    • one year ago
  41. satellite73 Group Title
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    \[\frac{1}{\sqrt{t+1}^3}\]

    • one year ago
  42. Ephilo Group Title
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    close enough lol

    • one year ago
  43. S.RaviTeja Group Title
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    yep thats what i was referring to as derivative that is of satellite73

    • one year ago
  44. satellite73 Group Title
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    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 \(3\)

    • one year ago
  45. satellite73 Group Title
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    since the number you get is clearly positive (it is \(\frac{1}{8}\)) this means the function is concave up |dw:1364049678258:dw|

    • one year ago
  46. Ephilo Group Title
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    so im evaluating right ?

    • one year ago
  47. satellite73 Group Title
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    yes

    • one year ago
  48. Ephilo Group Title
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    1/16

    • one year ago
  49. satellite73 Group Title
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    no i get \(\frac{1}{8}\) but in any case it is positive which is all you really care about

    • one year ago
  50. S.RaviTeja Group Title
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    its 1/8

    • one year ago
  51. Ephilo Group Title
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    i forgot to simplify

    • one year ago
  52. Ephilo Group Title
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    would that be all

    • one year ago
  53. S.RaviTeja Group Title
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    yep

    • one year ago
  54. Ephilo Group Title
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    oh okay lol

    • one year ago
  55. Ephilo Group Title
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    thanks guys (:

    • one year ago
  56. satellite73 Group Title
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    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

    • one year ago
  57. Ephilo Group Title
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    oh okay @satellite73

    • one year ago
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