1. Limited time only! Sign up for a free 30min personal 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!

Local Extrema

  1. Oct 28, 2013 #1

    Qube

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    http://i.minus.com/jZdpOtdOiChOn.jpg [Broken]

    2. Relevant equations

    Local extrema can be determined using the first derivative test.

    3. The attempt at a solution

    I ran the first derivative test to find the critical points, which were 0 and plus/minus 0.5. I plugged in the values into the original equation. x = 0 makes the function go to infinity, so x = 0 can be ruled out as any sort of local extrema. x = 0.5 makes the function = 2sqrt(e), while x = -0.5 makes the function = -2sqrt(e). Naively, I chose D, which pegs x = 0.5 as the local maximum, which makes sense, doesn't it?

    Unfortunately the formal definition of a local maxima is that the sign of the first derivative changes from positive to negative, and in the case of x = 0.5, the opposite happens; the sign actually flips from negative to positive around it, making it a local minima.

    I'm assuming this is the correct explanation.

    Who else would have fallen for this? Let's be honest :P.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 28, 2013 #2

    eumyang

    User Avatar
    Homework Helper

    I knew people who were fooled by this sort of problem. It is possible that a local maximum be "lower" than a local minimum. Look at the graphs of secant or cosecant, for example.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted