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Jgeurts

  • 2 years 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
    • 2 years ago
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    what is the function?

  2. Abarnett
    • 2 years ago
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    ok

  3. Jgeurts
    • 2 years ago
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    Determine the intervals where the function is concave up and concave down.

  4. Jgeurts
    • 2 years 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
    • 2 years ago
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    for the first one the lowest point is at y=-4

  6. Jgeurts
    • 2 years 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
    • 2 years ago
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    Are you thinking of Leibniz's notation?

  8. agent0smith
    • 2 years 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
    • 2 years ago
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    f' is equal to 0 when x is 2

  10. Abarnett
    • 2 years ago
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    @Jgeurts

  11. Jgeurts
    • 2 years ago
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    oh i see

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

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

  14. Abarnett
    • 2 years ago
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    for f'... yes

  15. Jgeurts
    • 2 years ago
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    what about f(x)?

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

  17. Abarnett
    • 2 years ago
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    f(x)' will look like this

  18. Jgeurts
    • 2 years 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
    • 2 years ago
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    |dw:1367706115115:dw|

  20. Jgeurts
    • 2 years ago
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    what the.. how did you get that?

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

  22. Abarnett
    • 2 years ago
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    Graphing calculator

  23. Abarnett
    • 2 years ago
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    and when it is up it is the same numbers... just positive

  24. Abarnett
    • 2 years ago
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    @Jgeurts

  25. Jgeurts
    • 2 years ago
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    ahhh I see

  26. Jgeurts
    • 2 years ago
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    so what does that tell us?

  27. Jgeurts
    • 2 years ago
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    oh the slope! duh

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

  29. Jgeurts
    • 2 years ago
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    Thanks

  30. Abarnett
    • 2 years ago
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    np, tag me if you need me!

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

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

  33. Jgeurts
    • 2 years ago
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    Calc 1

  34. Abarnett
    • 2 years ago
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    cool

  35. agent0smith
    • 2 years 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
    • 2 years ago
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    Yea I needed to completely work the problem beginning to end

  37. agent0smith
    • 2 years 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
    • 2 years ago
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    @agent0smith ok, and that is really the only thing i need with the second derivative right?

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

  40. agent0smith
    • 2 years 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
    • 2 years ago
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    that is f''

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

  43. agent0smith
    • 2 years 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
    • 2 years ago
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    @agent0smith you are right my bad! :P

  45. agent0smith
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    looks good to me, what do you think @agent0smith

  49. Jgeurts
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    ha! you got me, no infinities now!

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

  53. agent0smith
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    @Jgeurts

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

  59. agent0smith
    • 2 years 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
    • 2 years ago
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    1!

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

  62. agent0smith
    • 2 years 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
    • 2 years ago
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    and x^2/x^3 would have a horizontal of y=0?

  64. Jgeurts
    • 2 years ago
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    no, and yes -1

  65. agent0smith
    • 2 years 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
    • 2 years ago
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    i see so its like the graph of X

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

  68. agent0smith
    • 2 years 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
    • 2 years ago
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    excellent

  70. agent0smith
    • 2 years 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
    • 2 years ago
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    ok :) thank you so much, i wish i could give out 2 best answers

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

  73. Jgeurts
    • 2 years ago
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    :) @abarnett

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