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Confusion of Concavity

  1. Oct 14, 2006 #1
    Given f'(x) = 1/(3+7x).
    Find the interval where the graph of f(x) concaves downward.

    First, I differentiate f'(x)= 1/(3+7x) to get f''(x) = -7/(3+7x)^2. So, for the graph to be concave downward, I put f''(x) = -7/(3+7x)^2<0, so 7/(3+7x)^2>0

    (3+7x)^2 is always more or equal 0. Then, I exclude -3/7. Therefore, my final answer is that the graph of f(x) concaves downward on interval (-infinity, -3/7) union (-3/7, infinity).

    But then again, if we integrate f'(x) = 1/(3+7x) to obtain f(x), we would obtain f(x) = [In (3+7x)]/7 + c.

    Therefore, the interval now would be (-3/7, infinity), since the domain of the graph f(x) = [In (3+7x)]/7 is (-3/7, infinity).

    But then again, we can integrate f'(x) = 1/(3+7x) to get f(x) = [In (3+7x)]/7, but this time, the term 3+7x is in the modulus form. By this new equation, we can conclude that the graph concave downward on (-infinity, -3/7) union (-3/7, infinity).

    Hm... Which solution is the correct one ?
  2. jcsd
  3. Oct 14, 2006 #2


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    Remember that f(x) = lnx is defined for x > 0.
  4. Oct 14, 2006 #3


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    [tex] \int \frac{1}{x}dx = \ln(|x|)[/tex]

    so it would be defined for x less than zero

    EDIT: My absolute value bars don't seem to be coming through.... :(
  5. Oct 14, 2006 #4
    Oh, yeah, then if I apply this formula I'll get the answer for the interval is (-infinity,-3/7) union (-3/7, infinity), right?
  6. Oct 14, 2006 #5


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    That looks right to me.

    Damnit! Now the absolute value bars are showing up.... I swear, I'm going crazy or something
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