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anonymous

  • 5 years ago

need help finding intervals of increse....max and min...intervals of concavity up and down...and point of inflection f '(x)=3e^x/(3+e^x)^2

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  1. amistre64
    • 5 years ago
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    so this is our y' right?

  2. amistre64
    • 5 years ago
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    e^x never equals zero, so there are no critical points

  3. anonymous
    • 5 years ago
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    ok but i get an interval of increase right from some number to infinity

  4. amistre64
    • 5 years ago
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    I would say yes, but to check for concavity we tend to use the 2nd derivative right?

  5. anonymous
    • 5 years ago
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    well i know for sure that my function is increasing so there has to be interval of increasing but i dont know the starting point

  6. amistre64
    • 5 years ago
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    e^x is never a negative number, so all of its positive right?

  7. anonymous
    • 5 years ago
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    figured it out it from -infinity to infinity for increase variables ok now for concavity...

  8. amistre64
    • 5 years ago
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    concave up is my guess.... but use 2nd derivative to be certain

  9. anonymous
    • 5 years ago
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    i meant increase interval is -infinity to infinity

  10. amistre64
    • 5 years ago
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    (3e^x) (3+e^x)^-2 -6 e^2x (3e^x) --------- + ----------- right? (3+e^x)^3 (3+e^x)^2

  11. amistre64
    • 5 years ago
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    3 e^2x --------- (3+e^x)^3

  12. amistre64
    • 5 years ago
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    its always concave up....

  13. anonymous
    • 5 years ago
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    my sources say otherwise and i am looking for the intervals from which it is concave up and down

  14. amistre64
    • 5 years ago
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    i mighta goofed on the 2nd integral.... better double check

  15. anonymous
    • 5 years ago
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    \[f"(x)=-(3e^{2x}-9e^x)/e ^{3x}+9e ^{2x}+27e^x+26\]

  16. amistre64
    • 5 years ago
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    3e^x( e^3x -2e^2x +9e^x ) ------------------------ (3+e^x)^4

  17. anonymous
    • 5 years ago
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    no it not to the fourth power...

  18. amistre64
    • 5 years ago
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    havent expanded the bottom to see if anything cancels, but thats just a left over from the quotient rule

  19. amistre64
    • 5 years ago
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    it either that, or my heads got all jumbled..... which is very likely :)

  20. anonymous
    • 5 years ago
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    oh no i think things cancel i am just saying the function i have is the correct function already checked it ....lol

  21. amistre64
    • 5 years ago
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    my top gets to be: -3e^x (e^x -1) i cant get it to match yours

  22. amistre64
    • 5 years ago
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    -3e^x (e^x -3)

  23. amistre64
    • 5 years ago
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    now it does lol

  24. anonymous
    • 5 years ago
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    yea there you go so x=ln 3

  25. amistre64
    • 5 years ago
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    yeah, when e^x = 3 x = ln(3) thats right... we get an inflection right?

  26. amistre64
    • 5 years ago
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    everything to the left of that would be (+) and everything to the right of it would be(-)

  27. anonymous
    • 5 years ago
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    yea...problem one complete now for two....consider the function f(x)=2x+4/3x+2 for this function there are two important intervals (-infinity, A) and (A, infinity) where the function is not defined at A. Find A

  28. amistre64
    • 5 years ago
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    leftside is concave up, and rightside of it os concave down

  29. amistre64
    • 5 years ago
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    A = -2/3 there is a vertical asymptote there and possess no real value.

  30. amistre64
    • 5 years ago
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    nothing calcels top to bottom, so it aint a hole; its just the place where the denominator goes to zero... -2/3

  31. anonymous
    • 5 years ago
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    ok yea that was correct....an i see you clarification

  32. anonymous
    • 5 years ago
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    problem 3 sin^2(x/5) defined at interval [-14.907963,3.226991]

  33. anonymous
    • 5 years ago
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    find where f(x) is concave down, global mini of the function, a local max which is not a global max, the function is increasing on a region....note: some answers must be given in interval notation

  34. amistre64
    • 5 years ago
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    the normal period for pi is about 6 and this stretches it past 6 and goes to 10pi for the period. 10pi is about 31.4

  35. amistre64
    • 5 years ago
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    the amplitude is exaggerated, but thats just the y value....

  36. amistre64
    • 5 years ago
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    sin normally has a max at pi/2 and a min at 3pi/2 so, this one will be at 5pi/2 and 15pi/2 if I am picturing it correctly

  37. amistre64
    • 5 years ago
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    but lets work it :)

  38. amistre64
    • 5 years ago
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    Dx (3 sin^2(x/5)) y' = (3/5) sin(x/5)cos(x/5)

  39. amistre64
    • 5 years ago
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    if sin or cos are zero, we get zeros...

  40. anonymous
    • 5 years ago
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    the fx is sin^2(x/5) sorry its problem 3....

  41. amistre64
    • 5 years ago
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    sin = 0 at 0 and 180, cos = 0 at 90 and 270

  42. amistre64
    • 5 years ago
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    replace (3/5) with (1/5) same setup tho

  43. amistre64
    • 5 years ago
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    but because the normal period is elongated to 10pi, we gotta find those numbers :)

  44. amistre64
    • 5 years ago
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    they are..... 0, 5pi, 5pi/2, 15pi/2

  45. amistre64
    • 5 years ago
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    we can go backwards to... -5pi, -5pi/2, -15pi/2, -10pi

  46. amistre64
    • 5 years ago
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    which of these numbers is in our interval?

  47. amistre64
    • 5 years ago
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    5pi is bigger than 15

  48. amistre64
    • 5 years ago
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    [-14.907963 , 3.226991] -15/2 = -7.5, so -5pi/2 is in our interval 0 is in our interval we got these 2 numbers (-5pi/2) and (0)

  49. anonymous
    • 5 years ago
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    are you using \[\sin ^2(x/5)\]

  50. amistre64
    • 5 years ago
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    yes, I am using the derivative of that and checking for zeros :)

  51. amistre64
    • 5 years ago
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    the 3 was superfluous and could be tossed on a whim...

  52. anonymous
    • 5 years ago
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    ok

  53. anonymous
    • 5 years ago
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    ok the global mini is at 0 but now we need a local max

  54. amistre64
    • 5 years ago
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    -5pi/2 would be the local max...

  55. anonymous
    • 5 years ago
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    nah it not...i checked

  56. amistre64
    • 5 years ago
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    this is what we look like I think

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  57. anonymous
    • 5 years ago
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    our local max was 3.22 from the interval

  58. anonymous
    • 5 years ago
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    i need the interval where it is concave down and region where it is increasing

  59. amistre64
    • 5 years ago
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    computer froze

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