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Ephilo

  • 2 years ago

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

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  1. Ephilo
    • 2 years ago
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    |dw:1364048045697:dw|

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

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

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

  5. satellite73
    • 2 years ago
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    ok

  6. Ephilo
    • 2 years ago
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    the first derivate is

  7. Ephilo
    • 2 years ago
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    |dw:1364048369021:dw|

  8. satellite73
    • 2 years 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
    • 2 years ago
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    okay

  10. satellite73
    • 2 years 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
    • 2 years ago
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    a bit confused here?

  12. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    is that okay?

  16. Ephilo
    • 2 years ago
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    so you put zero in place of t right

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

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

  19. satellite73
    • 2 years ago
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    damn typo!!

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

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

  22. satellite73
    • 2 years ago
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    thats better

  23. Ephilo
    • 2 years ago
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    okay ill give it a try

  24. satellite73
    • 2 years ago
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    @S.RaviTeja no, that is not the solution

  25. S.RaviTeja
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    so t=3 ?

  28. satellite73
    • 2 years ago
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    yes

  29. Ephilo
    • 2 years ago
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    yay! lol

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

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

  32. satellite73
    • 2 years 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
    • 2 years ago
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    @S.RaviTeja -6 is not a solution

  34. satellite73
    • 2 years ago
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    second derivative is ... ?

  35. Ephilo
    • 2 years ago
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    okay im working on it @satellite73

  36. S.RaviTeja
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    second derivative is (t+1)^(-3/2)

  39. Ephilo
    • 2 years ago
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    would it be

  40. Ephilo
    • 2 years ago
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    |dw:1364049467577:dw|

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

  42. Ephilo
    • 2 years ago
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    close enough lol

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

  44. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    so im evaluating right ?

  47. satellite73
    • 2 years ago
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    yes

  48. Ephilo
    • 2 years ago
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    1/16

  49. satellite73
    • 2 years 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
    • 2 years ago
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    its 1/8

  51. Ephilo
    • 2 years ago
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    i forgot to simplify

  52. Ephilo
    • 2 years ago
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    would that be all

  53. S.RaviTeja
    • 2 years ago
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    yep

  54. Ephilo
    • 2 years ago
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    oh okay lol

  55. Ephilo
    • 2 years ago
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    thanks guys (:

  56. satellite73
    • 2 years 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
    • 2 years ago
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    oh okay @satellite73

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