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Ephilo

  • one year ago

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

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  1. Ephilo
    • one year ago
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    |dw:1364048045697:dw|

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

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

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

  5. satellite73
    • one year ago
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    ok

  6. Ephilo
    • one year ago
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    the first derivate is

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

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

  9. Ephilo
    • one year ago
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    okay

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

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

  12. satellite73
    • one year ago
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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

  13. satellite73
    • one year ago
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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

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

  15. satellite73
    • one year ago
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    is that okay?

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

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

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

  19. satellite73
    • one year ago
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    damn typo!!

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

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

  22. satellite73
    • one year ago
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    thats better

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

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

  25. S.RaviTeja
    • one year ago
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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.

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

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

  28. satellite73
    • one year ago
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    yes

  29. Ephilo
    • one year ago
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    yay! lol

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

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

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

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

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

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

  36. S.RaviTeja
    • one year ago
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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:)

  37. satellite73
    • one year ago
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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

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

  39. Ephilo
    • one year ago
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    would it be

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

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

  42. Ephilo
    • one year ago
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    close enough lol

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

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

  45. satellite73
    • one year ago
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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|

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

  47. satellite73
    • one year ago
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    yes

  48. Ephilo
    • one year ago
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    1/16

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

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

  51. Ephilo
    • one year ago
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    i forgot to simplify

  52. Ephilo
    • one year ago
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    would that be all

  53. S.RaviTeja
    • one year ago
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    yep

  54. Ephilo
    • one year ago
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    oh okay lol

  55. Ephilo
    • one year ago
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    thanks guys (:

  56. satellite73
    • one year ago
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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

  57. Ephilo
    • one year ago
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    oh okay @satellite73

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