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Jgeurts

  • one year ago

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

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  1. Abarnett
    • one year ago
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    what is the function?

  2. Abarnett
    • one year ago
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    ok

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

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

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

  6. Jgeurts
    • one year ago
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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?

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

  8. agent0smith
    • one year ago
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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.

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

  10. Abarnett
    • one year ago
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    @Jgeurts

  11. Jgeurts
    • one year ago
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    oh i see

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

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

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

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

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

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

  18. Jgeurts
    • one year ago
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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?

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

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

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

  22. Abarnett
    • one year ago
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    Graphing calculator

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

  24. Abarnett
    • one year ago
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    @Jgeurts

  25. Jgeurts
    • one year ago
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    ahhh I see

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

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

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

  29. Jgeurts
    • one year ago
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    Thanks

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

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

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

  33. Jgeurts
    • one year ago
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    Calc 1

  34. Abarnett
    • one year ago
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    cool

  35. agent0smith
    • one year ago
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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.

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

  37. agent0smith
    • one year ago
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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.

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

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

  40. agent0smith
    • one year ago
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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.

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

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

  43. agent0smith
    • one year ago
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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

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

  45. agent0smith
    • one year ago
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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.

  46. Jgeurts
    • one year ago
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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

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

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

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

  50. agent0smith
    • one year ago
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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] ;)

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

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

  53. agent0smith
    • one year ago
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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.

  54. agent0smith
    • one year ago
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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.

  55. Jgeurts
    • one year ago
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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?

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

  57. Abarnett
    • one year ago
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    @Jgeurts

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

  59. agent0smith
    • one year ago
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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 ...?

  60. Jgeurts
    • one year ago
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    1!

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

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

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

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

  65. agent0smith
    • one year ago
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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.

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

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

  68. agent0smith
    • one year ago
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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.

  69. Jgeurts
    • one year ago
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    excellent

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

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

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

  73. Jgeurts
    • one year ago
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    :) @abarnett

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