# Homework Help: Local Extrema

1. Oct 28, 2013

### Qube

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. Oct 28, 2013

### eumyang

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.