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Jgeurts

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

  • 11 months ago
  • 11 months ago

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  1. Abarnett
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    what is the function?

    • 11 months ago
  2. Abarnett
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    ok

    • 11 months ago
  3. Jgeurts
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    Determine the intervals where the function is concave up and concave down.

    • 11 months ago
  4. Jgeurts
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    \[f(x) = (x^2 - 4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(1-3x^2)/(x^2 +1)^3\]

    • 11 months ago
  5. Abarnett
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    for the first one the lowest point is at y=-4

    • 11 months ago
  6. Jgeurts
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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?

    • 11 months ago
  7. Abarnett
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    Are you thinking of Leibniz's notation?

    • 11 months ago
  8. agent0smith
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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.

    • 11 months ago
  9. Abarnett
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    f' is equal to 0 when x is 2

    • 11 months ago
  10. Abarnett
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    @Jgeurts

    • 11 months ago
  11. Jgeurts
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    oh i see

    • 11 months ago
  12. Jgeurts
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    i got negative concavity for both sides of zero in [-2,3]

    • 11 months ago
  13. Jgeurts
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    so does that mean it looks something like this? |dw:1367705711666:dw|

    • 11 months ago
  14. Abarnett
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    for f'... yes

    • 11 months ago
  15. Jgeurts
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    what about f(x)?

    • 11 months ago
  16. Abarnett
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    the graph you drew is f(x)... sorry my bad!

    • 11 months ago
  17. Abarnett
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    f(x)' will look like this

    • 11 months ago
  18. Jgeurts
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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?

    • 11 months ago
  19. Abarnett
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    |dw:1367706115115:dw|

    • 11 months ago
  20. Jgeurts
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    what the.. how did you get that?

    • 11 months ago
  21. Abarnett
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    when it is down x is approx. -0.638 and y is approx. -3.222

    • 11 months ago
  22. Abarnett
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    Graphing calculator

    • 11 months ago
  23. Abarnett
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    and when it is up it is the same numbers... just positive

    • 11 months ago
  24. Abarnett
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    @Jgeurts

    • 11 months ago
  25. Jgeurts
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    ahhh I see

    • 11 months ago
  26. Jgeurts
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    so what does that tell us?

    • 11 months ago
  27. Jgeurts
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    oh the slope! duh

    • 11 months ago
  28. Abarnett
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    Just type in the functions. https://www.desmos.com/calculator this is a good online graphing calculator

    • 11 months ago
  29. Jgeurts
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    Thanks

    • 11 months ago
  30. Abarnett
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    np, tag me if you need me!

    • 11 months ago
  31. Jgeurts
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    Thumps up! Im working on my practice final all afternoon so i might need you! @abarnett

    • 11 months ago
  32. Abarnett
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    ok I will be on as long as i can! what class are you taking?

    • 11 months ago
  33. Jgeurts
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    Calc 1

    • 11 months ago
  34. Abarnett
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    cool

    • 11 months ago
  35. agent0smith
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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.

    • 11 months ago
  36. Jgeurts
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    Yea I needed to completely work the problem beginning to end

    • 11 months ago
  37. agent0smith
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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.

    • 11 months ago
  38. Jgeurts
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    @agent0smith ok, and that is really the only thing i need with the second derivative right?

    • 11 months ago
  39. Abarnett
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    |dw:1367707501528:dw| very top y=10 and x=0

    • 11 months ago
  40. agent0smith
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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.

    • 11 months ago
  41. Abarnett
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    that is f''

    • 11 months ago
  42. Jgeurts
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    f'' just gives me inflection points at f'' = 0 and concavity.. cool! :)

    • 11 months ago
  43. agent0smith
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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

    • 11 months ago
  44. Abarnett
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    @agent0smith you are right my bad! :P

    • 11 months ago
  45. agent0smith
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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.

    • 11 months ago
  46. Jgeurts
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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

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

    • 11 months ago
  48. Abarnett
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    looks good to me, what do you think @agent0smith

    • 11 months ago
  49. Jgeurts
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    \[f(x) = (x^2 - 4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(1-3x^2)/(x^2 +1)^3\]

    • 11 months ago
  50. agent0smith
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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] ;)

    • 11 months ago
  51. Jgeurts
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    ha! you got me, no infinities now!

    • 11 months ago
  52. Jgeurts
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    so if there was no boundaries, the local and global maximum DNE right?

    • 11 months ago
  53. agent0smith
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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.

    • 11 months ago
  54. agent0smith
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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.

    • 11 months ago
  55. Jgeurts
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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?

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

    • 11 months ago
  57. Abarnett
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    @Jgeurts

    • 11 months ago
  58. Jgeurts
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    I need to know how to do it from the ground up, no calculators

    • 11 months ago
  59. agent0smith
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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 ...?

    • 11 months ago
  60. Jgeurts
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    1!

    • 11 months ago
  61. Jgeurts
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    so x^3/x^2 would have no horizontal, but have vertical?

    • 11 months ago
  62. agent0smith
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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? \[\Large f(x) = \frac{ (x^2 - 4)}{(x +1)}\]

    • 11 months ago
  63. Jgeurts
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    and x^2/x^3 would have a horizontal of y=0?

    • 11 months ago
  64. Jgeurts
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    no, and yes -1

    • 11 months ago
  65. agent0smith
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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.

    • 11 months ago
  66. Jgeurts
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    i see so its like the graph of X

    • 11 months ago
  67. Jgeurts
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    and if its negative im guessing thats what flips it over

    • 11 months ago
  68. agent0smith
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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.

    • 11 months ago
  69. Jgeurts
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    excellent

    • 11 months ago
  70. agent0smith
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    Slant asymptotes are a bit more confusing, sometimes you have to use polynomial long division to find them.

    • 11 months ago
  71. Jgeurts
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    ok :) thank you so much, i wish i could give out 2 best answers

    • 11 months ago
  72. Abarnett
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    @Jgeurts I gave him one for you!

    • 11 months ago
  73. Jgeurts
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    :) @abarnett

    • 11 months ago
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