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Jgeurts Group Title

Below you are given a function f(x) and its first and second derivatives. Use this information to solve the following.

  • one year ago
  • one year ago

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  1. Abarnett Group Title
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    what is the function?

    • one year ago
  2. Abarnett Group Title
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    ok

    • one year ago
  3. Jgeurts Group Title
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    Determine the intervals where the function is concave up and concave down.

    • one year ago
  4. Jgeurts Group Title
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    \[f(x) = (x^2 - 4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(1-3x^2)/(x^2 +1)^3\]

    • one year ago
  5. Abarnett Group Title
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    for the first one the lowest point is at y=-4

    • one year ago
  6. Jgeurts Group Title
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    at f(0) i got that, isn't there a rule having to do with the second derivative i can use?

    • one year ago
  7. Abarnett Group Title
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    Are you thinking of Leibniz's notation?

    • one year ago
  8. agent0smith Group Title
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    Use the second derivative to find concavity. It's concave down where f'' is negative, ie f'' < 0 Concave up where f'' is positive, f'' > 0 Find where f'' = 0 firstly.

    • one year ago
  9. Abarnett Group Title
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    f' is equal to 0 when x is 2

    • one year ago
  10. Abarnett Group Title
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    @Jgeurts

    • one year ago
  11. Jgeurts Group Title
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    oh i see

    • one year ago
  12. Jgeurts Group Title
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    i got negative concavity for both sides of zero in [-2,3]

    • one year ago
  13. Jgeurts Group Title
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    so does that mean it looks something like this? |dw:1367705711666:dw|

    • one year ago
  14. Abarnett Group Title
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    for f'... yes

    • one year ago
  15. Jgeurts Group Title
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    what about f(x)?

    • one year ago
  16. Abarnett Group Title
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    the graph you drew is f(x)... sorry my bad!

    • one year ago
  17. Abarnett Group Title
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    f(x)' will look like this

    • one year ago
  18. Jgeurts Group Title
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    I also need to label the inflection point and asymptotes. To find the inflection point i need to find f''(0) right?

    • one year ago
  19. Abarnett Group Title
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    |dw:1367706115115:dw|

    • one year ago
  20. Jgeurts Group Title
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    what the.. how did you get that?

    • one year ago
  21. Abarnett Group Title
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    when it is down x is approx. -0.638 and y is approx. -3.222

    • one year ago
  22. Abarnett Group Title
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    Graphing calculator

    • one year ago
  23. Abarnett Group Title
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    and when it is up it is the same numbers... just positive

    • one year ago
  24. Abarnett Group Title
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    @Jgeurts

    • one year ago
  25. Jgeurts Group Title
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    ahhh I see

    • one year ago
  26. Jgeurts Group Title
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    so what does that tell us?

    • one year ago
  27. Jgeurts Group Title
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    oh the slope! duh

    • one year ago
  28. Abarnett Group Title
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    Just type in the functions. https://www.desmos.com/calculator this is a good online graphing calculator

    • one year ago
  29. Jgeurts Group Title
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    Thanks

    • one year ago
  30. Abarnett Group Title
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    np, tag me if you need me!

    • one year ago
  31. Jgeurts Group Title
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    Thumps up! Im working on my practice final all afternoon so i might need you! @abarnett

    • one year ago
  32. Abarnett Group Title
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    ok I will be on as long as i can! what class are you taking?

    • one year ago
  33. Jgeurts Group Title
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    Calc 1

    • one year ago
  34. Abarnett Group Title
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    cool

    • one year ago
  35. agent0smith Group Title
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    Did you find f''? once you do, you can find where it's concave up and down. Just from looking at the graph, I'd guess concave down between about -inf and -1, and +1 and +infinity Concave up between about -1 and 1.

    • one year ago
  36. Jgeurts Group Title
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    Yea I needed to completely work the problem beginning to end

    • one year ago
  37. agent0smith Group Title
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    Once you get the x values where f'' = 0, pick points to the left and right of those x values, and check if f'' is positive (concave down) or negative (concave up) to help find the intervals.

    • one year ago
  38. Jgeurts Group Title
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    @agent0smith ok, and that is really the only thing i need with the second derivative right?

    • one year ago
  39. Abarnett Group Title
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    |dw:1367707501528:dw| very top y=10 and x=0

    • one year ago
  40. agent0smith Group Title
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    Yep = it's just used to determine inflection points and concavity. Wherever f'' is neg, it's concave down, and vice versa.

    • one year ago
  41. Abarnett Group Title
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    that is f''

    • one year ago
  42. Jgeurts Group Title
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    f'' just gives me inflection points at f'' = 0 and concavity.. cool! :)

    • one year ago
  43. agent0smith Group Title
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    @Abarnett probably easier to just link to one than hand draw a graph :P f'': https://www.google.com/search?q=10(1-3x%5E2)%2F(x%5E2+%2B1)%5E3&aq=f&oq=10(1-3x%5E2)%2F(x%5E2+%2B1)%5E3&aqs=chrome.0.57j60l3j62l2.174j0&sourceid=chrome&ie=UTF-8

    • one year ago
  44. Abarnett Group Title
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    @agent0smith you are right my bad! :P

    • one year ago
  45. agent0smith Group Title
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    Oh and @Jgeurts f'' can also be used to find if a critical point (where f' = 0) is a max or min - at a maximum, f'' is negative. At a minimum, f'' is positive. But that's part of the concavity anyway, since maximums are concave down, minimums are concave up. If f'' is zero at a point where f' = 0, then it's an inflection point.

    • one year ago
  46. Jgeurts Group Title
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    @Abarnett @agent0smith So I have Global max (3,1/2), Global Min (0,-4), Decreasing on the interval [-2,0], Increasing on the interval [0,3], local max is f(3), local min is f(0), concave down at [-inf,-sqrt1/3]U[sqrt1/3, inf], concave up at [-sqrt1/3, sqrt1/3], inflection points occur at +-sqrt1/3

    • one year ago
  47. Jgeurts Group Title
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    by the way im supposed to be using the interval [-2,3] for everything i forgot to tell you, haha

    • one year ago
  48. Abarnett Group Title
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    looks good to me, what do you think @agent0smith

    • one year ago
  49. Jgeurts Group Title
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    \[f(x) = (x^2 - 4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(1-3x^2)/(x^2 +1)^3\]

    • one year ago
  50. agent0smith Group Title
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    lol that interval [-2,3] makes a big difference... as the original function has no maximum, it just has an asymptote. https://www.google.com/search?q=(x%5E2+-+4)%2F(x%5E2+%2B1)&aq=f&oq=(x%5E2+-+4)%2F(x%5E2+%2B1)&aqs=chrome.0.57j60l3j62l2.269j0&sourceid=chrome&ie=UTF-8 That's the original function... all your values look good. Except: concave down at [-inf,-sqrt1/3]U[sqrt1/3, inf], Something seems off here, with an interval [-2,3] ;)

    • one year ago
  51. Jgeurts Group Title
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    ha! you got me, no infinities now!

    • one year ago
  52. Jgeurts Group Title
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    so if there was no boundaries, the local and global maximum DNE right?

    • one year ago
  53. agent0smith Group Title
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    If you can learn to kinda read the graph, you can check your values. The max/mins should be easy to find (turning points), same with where the graph is increasing or decreasing (the slope f' is positive or negative). Points of inflection are more difficult to see... you kinda have to gauge where it changes from concave up to down which can be hard to see, but still possible to estimate where it is.

    • one year ago
  54. agent0smith Group Title
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    Correct, since f' will never equal 0 except at the minimum. \[\large f'(x) = 10x/(x^2 +1)^2\] - note that this can be only zero if x=0 (the denominator can never make the f' equal zero) - ie there's no other turning points/maximums, only a minimum.

    • one year ago
  55. Jgeurts Group Title
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    Ok, i feel so informed, haha. Hold up, last question! What about asymptotes? Horizontal/Vertical? Obviously theres one around y=1 but how do i prove it?

    • one year ago
  56. Abarnett Group Title
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    try this site it might help! never tried it http://calculator.tutorvista.com/math/601/asymptote-calculator.html

    • one year ago
  57. Abarnett Group Title
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    @Jgeurts

    • one year ago
  58. Jgeurts Group Title
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    I need to know how to do it from the ground up, no calculators

    • one year ago
  59. agent0smith Group Title
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    To find asymptotes... you just have to look at the original function, there's no calculus involved. \[\large f(x) = (x^2 - 4)/(x^2 +1)\] What happens when x gets really huge (positive or negative infinity? The -4 and +1 become tiny compared to the huge x^2, so it becomes \[\Large \frac{ x^2 }{ x^2 }\] which is equal to ...?

    • one year ago
  60. Jgeurts Group Title
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    1!

    • one year ago
  61. Jgeurts Group Title
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    so x^3/x^2 would have no horizontal, but have vertical?

    • one year ago
  62. agent0smith Group Title
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    Correct! :) that's for horizontal asymptotes. So y= 1 is a horizontal asymptote. To find vertical asymptotes, find values of x that cause the denominator to be zero \[\Large f(x) = \frac{ (x^2 - 4)}{(x^2 +1)}\]are there any x values that make the denominator zero? x^2 is always positive, and you're adding 1... What about instead for this function? \[\Large f(x) = \frac{ (x^2 - 4)}{(x +1)}\]

    • one year ago
  63. Jgeurts Group Title
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    and x^2/x^3 would have a horizontal of y=0?

    • one year ago
  64. Jgeurts Group Title
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    no, and yes -1

    • one year ago
  65. agent0smith Group Title
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    For x^3/x^2, that simplifies to just x - it's actually a slant asymptote (neither horizontal or vertical) - there is no horizontal. BUT x can't equal zero, so it has a vertical at x=0 (or just a 'hole' at x=0, not really an asymptote in this case), since the denominator is zero.

    • one year ago
  66. Jgeurts Group Title
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    i see so its like the graph of X

    • one year ago
  67. Jgeurts Group Title
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    and if its negative im guessing thats what flips it over

    • one year ago
  68. agent0smith Group Title
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    Yep. With a hole at x=0. So for vertical - look for values of x that give a denominator zero For horizontal - look for how the function behaves when x is approaching +infinity or -infinity.

    • one year ago
  69. Jgeurts Group Title
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    excellent

    • one year ago
  70. agent0smith Group Title
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    Slant asymptotes are a bit more confusing, sometimes you have to use polynomial long division to find them.

    • one year ago
  71. Jgeurts Group Title
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    ok :) thank you so much, i wish i could give out 2 best answers

    • one year ago
  72. Abarnett Group Title
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    @Jgeurts I gave him one for you!

    • one year ago
  73. Jgeurts Group Title
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    :) @abarnett

    • one year ago
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