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NathanJHW

  • one year ago

The figure at below is the graph of f ′′ (x), the second derivative of a function f (x). The domain of the function f (x) is all real numbers, and the graph shows f ′′ (x) for −2.6≤ x ≤3.6.

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  1. NathanJHW
    • one year ago
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  2. NathanJHW
    • one year ago
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    A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent. B. Find all values of x in the interval (−2.6, 3.6) where f (x) is concave upwards. Explain your answer. C. Suppose it is known that in the interval (−3.6, 3.6), f (x) has critical points at x =1.37, and x = −0 .97. Classify these points as relative maxima or minima of f (x). Explain your answer.

  3. NathanJHW
    • one year ago
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    @jim_thompson5910

  4. jim_thompson5910
    • one year ago
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    what do you have so far?

  5. NathanJHW
    • one year ago
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    Don't know where to start.

  6. jim_thompson5910
    • one year ago
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    A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent.

  7. jim_thompson5910
    • one year ago
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    f ' (x) has a horizontal tangent whenever the tangent slope on f ' (x) is 0 that only happens when f '' (x) = 0

  8. jim_thompson5910
    • one year ago
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    It's very similar to saying "f(x) has a horizontal tangent at the point where f ' (x) = 0"

  9. NathanJHW
    • one year ago
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    So would occur at the maximums and minimums?

  10. jim_thompson5910
    • one year ago
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    no I'm saying that the horizontal tangent on f ' (x) happens at the roots of f '' (x)

  11. NathanJHW
    • one year ago
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    So how do we find those?

  12. jim_thompson5910
    • one year ago
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    what are the roots of f '' (x)

  13. jim_thompson5910
    • one year ago
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    you're given the graph

  14. NathanJHW
    • one year ago
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    Oh duh. -2,1,3

  15. jim_thompson5910
    • one year ago
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    yep

  16. NathanJHW
    • one year ago
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    So the answer would be -2, 1 , 3?

  17. NathanJHW
    • one year ago
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    @jim_thompson5910

  18. jim_thompson5910
    • one year ago
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    for part A, yes

  19. NathanJHW
    • one year ago
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    That was easier than I thought.

  20. jim_thompson5910
    • one year ago
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    at those x values, the value of f '' is 0, so that's where the horizontal tangent on f ' will be

  21. NathanJHW
    • one year ago
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    Now how do we do part B?

  22. jim_thompson5910
    • one year ago
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    f(x) is concave up whenever f '' (x) > 0

  23. jim_thompson5910
    • one year ago
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    again you're given the graph of f '' so just list the interval(s) when f '' is above the x axis to report the interval(s) when f is concave up

  24. NathanJHW
    • one year ago
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    Well I don't think I can from the graph the exact values of f''

  25. NathanJHW
    • one year ago
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    @jim_thompson5910

  26. jim_thompson5910
    • one year ago
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    again they want the interval along the x axis

  27. jim_thompson5910
    • one year ago
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    look at the graph. when is f '' (x) > 0 ? what x values?

  28. NathanJHW
    • one year ago
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    -1 and 0

  29. jim_thompson5910
    • one year ago
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    list it as an interval

  30. jim_thompson5910
    • one year ago
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    from x = ??? to x = ???, the value of f '' (x) is positive

  31. NathanJHW
    • one year ago
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    x=-2 to x=1?

  32. jim_thompson5910
    • one year ago
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    good, where else?

  33. NathanJHW
    • one year ago
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    x=1 to x=3

  34. jim_thompson5910
    • one year ago
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    you sure?

  35. NathanJHW
    • one year ago
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    Well its either that or x=3 to x = we don't know

  36. jim_thompson5910
    • one year ago
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    but it gives you the bounds on which f '' is restricted

  37. NathanJHW
    • one year ago
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    oh so x=3 to x=3.6

  38. jim_thompson5910
    • one year ago
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    yep

  39. jim_thompson5910
    • one year ago
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    "x = -2 to x = 1" can be written in interval notation as (-2,1) (-2,1) is NOT a point, it's an interval. yeah the notation is confusing sometimes

  40. jim_thompson5910
    • one year ago
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    3 to 3.6 can be written as (3,3.6)

  41. jim_thompson5910
    • one year ago
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    put the two together with a union symbol and you get \[\Large (-2,1) \cup (3,3.6)\]

  42. jim_thompson5910
    • one year ago
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    that represents where f '' (x) > 0, where f is concave up

  43. NathanJHW
    • one year ago
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    Wait, isn't concave up when the curve looks like a U not an upside down U? So doesn't that mean it is concave up between x=1 and x=3 only? Is that what the above expression means?

  44. jim_thompson5910
    • one year ago
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    you are correct, but you're thinking of f having those qualities (of having a U or upside down U)

  45. NathanJHW
    • one year ago
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    I'm confused.

  46. jim_thompson5910
    • one year ago
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    you're given the graph of f '' and they want to know info about f

  47. jim_thompson5910
    • one year ago
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    we don't know what f looks like but we can use f '' to figure out when f is concave up or down

  48. NathanJHW
    • one year ago
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    Oh I see. I forgot the graph shows f''(x).

  49. jim_thompson5910
    • one year ago
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    That's a common trick calc teachers use, so watch out

  50. NathanJHW
    • one year ago
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    Alright, know how do we do part C?

  51. jim_thompson5910
    • one year ago
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    the critical point x = 1.37 where is this point located on f '' ? is it in a concave up region? or concave down?

  52. NathanJHW
    • one year ago
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    concave up

  53. jim_thompson5910
    • one year ago
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    or put another way, if x = 1.37, is f '' positive or negative?

  54. NathanJHW
    • one year ago
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    negative

  55. jim_thompson5910
    • one year ago
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    since f '' is negative when x = 1.37, this means that x = 1.37 lies in a concave down region on f

  56. NathanJHW
    • one year ago
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    so it's a maxima?

  57. jim_thompson5910
    • one year ago
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    so this is what the small piece of f looks like |dw:1433375975175:dw|

  58. jim_thompson5910
    • one year ago
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    oops my bad

  59. jim_thompson5910
    • one year ago
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    swapped them lol

  60. jim_thompson5910
    • one year ago
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    |dw:1433376035069:dw|

  61. jim_thompson5910
    • one year ago
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    |dw:1433376059680:dw|

  62. NathanJHW
    • one year ago
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    so it's the maxima?

  63. jim_thompson5910
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    relative max, yeah

  64. NathanJHW
    • one year ago
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    and 0.97 is the minima for being positive

  65. NathanJHW
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    oops -0.97

  66. NathanJHW
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    actually it stays the same, it is the relative minima

  67. jim_thompson5910
    • one year ago
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    if x = -0.97, then f '' (x) is positive making f concave up here so if x = -0.97, then there is a relative min on f at x = -0.97 you are correct

  68. NathanJHW
    • one year ago
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    Alright, thank you for your help. Do you mind helping me with another problem if I do @jim_thompson5910 on the question?

  69. jim_thompson5910
    • one year ago
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    sure, one more

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