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Homework Help: Extreme difficulty with a concavity problem

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    1. Determine the largest open interval where the function f(x)=2x-5/x+3 is increasing AND concave up (at the same time)

    2. Relevant equations



    3. The attempt at a solution

    I've run through the problem and arrived at a second derivative of f''(x)= -22/(x+3)^3. Setting this to zero I'm coming back with -3. However, I keep coming back with a situation in which this is concave down. Am I making some sort of error here?
     
  2. jcsd
  3. Mar 24, 2012 #2
    Pondering it, could it be as simple as this problem doesn't have a point of inflection; given that (-3+3)^2 returns as zero?
     
  4. Mar 24, 2012 #3

    ehild

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    You have to find the interval where the function is concave up. x=-3 is a point where the function is neither concave up nor concave down, as you said it is not even defined. What does concave up mean?

    ehild
     
  5. Mar 24, 2012 #4
    That there's a change of the sign upon reaching the point of inflection. In the case of concave up, from negative to positive.
     
  6. Mar 24, 2012 #5

    ehild

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    It is the definition of an inflection point, that concavity changes sign. What do you know about the second derivative if the function is concave up? What is the concavity of f(x)=x^2 at x=1, for example?

    ehild
     
  7. Mar 24, 2012 #6
    Perhaps I'm not following, but the second derivative of the function that you just posted is 2. There's no way to set that equal to 0.
     
  8. Mar 24, 2012 #7

    ehild

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  9. Mar 24, 2012 #8
    It's curved upward, so it's concave upward.
     
  10. Mar 24, 2012 #9

    ehild

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    When curved up, the second derivative is positive. Where is the second derivative of your function positive?

    ehild
     
  11. Mar 24, 2012 #10
    I'm not sure how this relates to my original question. In my original question I'm lost on whether or not the the function even has a point of inflection since my returned value when I set the second derivative to 0 is x=-3, which we've established isn't defined. Why we're going down this tangent of questioning is lost on me.
     
  12. Mar 24, 2012 #11

    ehild

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    The original question is in what open interval is the function concave up. Nothing is said about the inflection point. A function changes concavity at an inflection point, but the concavity depends on the sign of the second derivative. It is concave up if the second derivative is positive.

    ehild
     
  13. Mar 24, 2012 #12
    Correct. However, deriving to the second derivative and then setting to zero returns x = -3. That'd be the point of inflection, in which case I'd test the numbers around it to determine concavity. But as we've established, that's not defined in the function. Doesn't that mean that there's no points of inflection and thus no concavity?
     
  14. Mar 24, 2012 #13

    ehild

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    Why do you set the second derivative to zero?


    ehild
     
  15. Mar 24, 2012 #14
    To find the value to test for concavity.
     
  16. Mar 24, 2012 #15

    ehild

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    You find the point where the function is not defined. What about the concavity at x=-4? Does the function curved up or down there?

    ehild
     

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  17. Mar 24, 2012 #16
    At that point it doesn't, it's fairly constant right there.
     
  18. Mar 24, 2012 #17

    ehild

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    Well, it is curved a bit, see again. And what about -3.5?

    ehild
     
  19. Mar 24, 2012 #18

    Ray Vickson

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    As you wrote it, the function f(x) = 2x - 5/x + 3 has second derivative f''(x) = -10/x^3. Possibly you did NOT mean what you wrote; possibly you should have used brackets, to write f(x) = (2x-5)/(x+3). Is that what you meant? Certainly, that last f has the correct second derivative.

    RGV
     
  20. Mar 24, 2012 #19
    Sorry, Ray. Yes. I meant (2x-5)/(x+3).

    ehilde, if it's curving at all it's curving upward.
     
  21. Mar 24, 2012 #20

    ehild

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    So it is concave up at x=-3.5, isn't it? And it changes concavity only at x=-3.
    I can not help more. If you still do not get it, ask your teacher to explain concavity to you.


    ehild
     
  22. Mar 24, 2012 #21
    Ehild, you've not helped me one bit since we began speaking. Perhaps Ray would be more helpful at this point. I've made my question clear. I've performed the math as I've been taught. I've set the derivative equal to zero, which as we've said countless times now equals -3. I've selected random numbers both greater and lesser than -3. I've plugged those into the second derivative, as taught, and I've recorded the signs. Negative to the left of -3, positive to the right of -3. That, sir, is concave down if I've ever seen it. Again, all how I've been taught.

    If -3 in the denominator makes it equal zero, then is not in the domain of f? I thought I was fairly clear in what I was asking.
     
  23. Mar 24, 2012 #22

    Ray Vickson

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    In {x < -3}, f(x) = (2x-5)/(x+3) has f''(x) > 0, so f is certainly not concave in that region. For {x > -3} we have f''(x) < 0, so f is concave. In both regions, f is strictly increasing; just plot it and see.

    Personally, I don't like the terminology "concavity down or convex up", or whatever; I follow the practice in Operations Research and modern optimization, which just says "concave" and "convex". A concave function looks like (part of) a Macdonald's arch, while a convex function looks like (part of a) crater or a valley.

    RGV
     
  24. Mar 24, 2012 #23

    ehild

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    Hi Ray,

    Neither like I the terminology "concave up", but that was the question of the original post. Convex is called sometimes "concave up"
    http://www.mathwords.com/c/concave_up.htm

    ehild
     
  25. Mar 24, 2012 #24

    Ray Vickson

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    I know that was the terminology of the original problem, and that is why I explained what I meant by convex and concave (Macdonald arches, valleys, etc.) The OP can translate it into whatever terms he/she wants.

    RGV
     
  26. Mar 24, 2012 #25

    ehild

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    Solitare, I am sorry that I could not help.

    ehild
     
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