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Homework Help: Extrema and concavity of a function.

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x)={\frac {{x}^{2}-2}{{x}^{2}-1}}[/tex]
    List the critical points, extrema, points of inflection, and concavity.

    2. Relevant equations
    [tex]f'(x)={\frac {2x}{{x}^{2}-1}}-{\frac { \left( 2\,{x}^{2}-4 \right) x}{
    \left( {x}^{2}-1 \right) ^{2}}}[/tex]

    3. The attempt at a solution


    Solving for f'(x)=0 and checking for discontinuities, I get critical points of -1, 0, and 1.

    I can get that far. I'm unsure of how to make the intervals. Would my lowest interval be [tex]-\infty<x<-1[/tex] or [tex]-1<x<0[/tex]? I understand what's happening on the graph, it's just that discontinuities confuse me. :grumpy:


    Solving for f''(x)=0 and checking for discontinuities, I get points of inflection of -1 and 1 (and two non-real answers which are of no use to me.)

    Both of these points are discontinuities. How would I make an interval for this? Shouldn't there also be a point of inflection at (0,2)?
    Last edited: Nov 10, 2009
  2. jcsd
  3. Nov 11, 2009 #2
    The function seems to be perfectly well defined when x<-1.

    Are you sure? That function sure looks like it could change it's concavity.

    How do you make intervals? How about [tex] -\infty < x < -1[/tex], [tex] -1 < x < 1 [/tex] and [tex] 1 < x < \infty [/tex]? What's wrong with them? :)
  4. Nov 11, 2009 #3
    The extrema makes sense, thank you for that. I guess my brain was fried last night. :redface:

    The [tex]-1 < x < 1[/tex] interval for the concavity still don't make sense to me, though. If I used a test value of x=0 for said interval, wouldn't that give me the concavity for the parabola on the upper part of the graph? So shouldn't there be another point of concavity at (0,2)? But there obviously isn't, since f''(0)=2 (ie, it's not a point of inflection.) I guess that's what is confusing me.

    http://img42.imageshack.us/img42/5492/grapho.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  5. Nov 12, 2009 #4
    For a parabola f''(x) = constant for all x. However you are right in that there's something going on in the interval (-1,1). Perhaps you should check your calculations again.

    For motivation, check for example f''(0.5) and f''(0.7).
  6. Nov 12, 2009 #5

    Ok it just clicked in my head. There's a point of inflection whenever f''=0 or DNE. At (0,2) f''[tex]\neq[/tex]0, so there isn't a point of inflection there. Because the parabola is bowl-shaped I was thinking there had to be a point of concavity there, but there isn't.
    Last edited: Nov 12, 2009
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