1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    radou

    User Avatar
    Homework Helper

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

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [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

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That looks right to me.

    Damnit! Now the absolute value bars are showing up.... I swear, I'm going crazy or something
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Confusion of Concavity
  1. Concavity of entropy (Replies: 1)

  2. Concave Functions (Replies: 3)

Loading...