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How to find the interval of decrease on (x^3 - 1)/(x^3 + 1)?

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    f(x) = (x3 - 1)/(x3 + 1)

    2. Relevant equations
    (d/dx)[f(x)/g(x)] = [f`(x)⋅g(x) - f(x)⋅g`(x)]/[g(x)]2

    3. The attempt at a solution
    f`(x) = [3x2(x3 + 1) - 3x2(x3 - 1)]/[(x3 + 1)]2
    f`(x) = 3x2[(x3 + 1) - (x3 - 1)]/[(x3 + 1)]2
    f`(x) = 3x2(2)/[(x3 + 1)]2

    f`(x) = 6x2/(x3 + 1)2

    When I made a sign chart for this derivative, I couldn't find any x that could make the value of f`(x) negative. It's always positive, because both terms are squared. So I can't find any interval of decrease, and in turn, any local minimum or maximum. I'm asked to also write the intervals of concave up/down and any inflection points. I highly doubt my professor would assign a problem where I would just write N/A for six of these values. Am I doing something wrong? What x would make this derivative negative? Or perhaps, am I not deriving this correctly?
     
  2. jcsd
  3. Nov 2, 2015 #2

    andrewkirk

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    Your calculations look correct to me
     
  4. Nov 2, 2015 #3

    Mark44

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    Technically, f' is not always positive -- it's undefined for x = -1. Otherwise, f' > 0 for any ##x \ne -1##.
    As far as concavity, the 2nd derivative changes sign at a number of points, so writing N/A for these parts would be incorrect.
     
  5. Nov 2, 2015 #4
    You mean f``(x)? I looked into local minima/maxima a bit more just now, and it looks like it depends on the second derivative. Other than that, there is no interval of decrease, is there?
     
  6. Nov 2, 2015 #5

    Mark44

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    Yes, f'' determines the concavity. On any interval where f''(x) > 0, the graph of f is concave up, and on any interval where f''(x) < 0, the graph of f is concave down.
    No. Since f'(x) is never 0, which you showed, there are no local maxima or minima.
    No, there isn't.
     
  7. Nov 3, 2015 #6

    Ray Vickson

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    Actually, ##f'(0) = 0##, so ##f'(x)## can be zero in places. However, ##x=0## is a stationary inflection point, rather like in the graph of ##y = x^3##.
     
    Last edited by a moderator: Jan 28, 2016
  8. Nov 3, 2015 #7

    Mark44

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    Yes, you're correct about f'(0), Ray. I forgot that the numerator could be zero, and misspoke.
     
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