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[apcalculus] Re: local minimum question
Posted:
Jan 31, 2005 9:58 AM


Hello everyone,
On Thu, 27 Jan 2005 LnMcmullin@aol.com wrote: ... I, and others, to quote Mark Howell's phrase, like to think of local minimum (and maximum) as the smallest (largest) value in the range "around here." Thus the endpoint of a closed interval could be, and in fact must be, a local maximum or minimum. ...
We have to be a little careful here. "Nice" continuous functions have either a local minimum or a local maximum at the endpoints of a closed interval. However, the function defined on [0,1] by
f(x) = xsin(1/x) for x in (0,1] while f(0)=0
is continuous on [0,1] but does not have either a local minimum or a local maximum at x=0. It is not too hard to see this, since for x>0, sin(1/x) takes on the values 1 and 1 on any interval of the form (0,d) for any d<1. As a result, xsin(1/x) takes on both positive and negative values no matter how close we get to zero.
Hope this helps,
Dick Maher
Richard J. Maher Mathematics and Statistics Loyola University Chicago 6525 N. Sheridan Rd. Chicago, Illinois 60626 17735083565 rjm@math.luc.edu
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